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-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0ac98582dd0b2497034e459e869a2a3bd28001d0d4c4b37a61a8ed5d05f228e3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 1: INTRODUCTION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.1: Page Number 8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "Eg=24.0;                     #Generated voltage in V\n",
-      "Ri=0.01;                   #Internal Resistance in \u03a9\n",
-      "P=100;                     #Power supplied in watts\n",
-      "\n",
-      "#Calculations\n",
-      "# (i)\n",
-      "I=P/Eg;                         #Load current in A\n",
-      "V_Ri=I*Ri;                      #Voltage drop in internal resistance\n",
-      "\n",
-      "# (ii)\n",
-      "V=Eg-(I*Ri);                     #Terminal Voltage\n",
-      "\n",
-      "#Results\n",
-      "print (\"The voltage drop in internal resistance is %.4f V\"%V_Ri);\n",
-      "print (\"The terminal voltage is %.2f V\"%V);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage drop in internal resistance is 0.0417 V\n",
-        "The terminal voltage is 23.96 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.2: Page number 10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Eg=500.0;                 #Generated voltage in V\n",
-      "Ri=1000.0;                #Internal Resistance in \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "# (i)\n",
-      "RL=10;                    #Load resistance of case 1 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "\n",
-      "print(\"The load current for RL=10\u03a9 is %.3f A\"%I);\n",
-      "\n",
-      "# (ii)\n",
-      "RL=50;                    #Load resistance of case 2 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "\n",
-      "print(\"The load current for RL=50\u03a9 is %.3f A\"%I);\n",
-      "\n",
-      "# (iii)\n",
-      "RL=100;                    #Load resistance of case 3 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "I=round(I,3);\n",
-      "\n",
-      "print(\"The load current for RL=100\u03a9 is %.3f A\"%I);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load current for RL=10\u03a9 is 0.495 A\n",
-        "The load current for RL=50\u03a9 is 0.476 A\n",
-        "The load current for RL=100\u03a9 is 0.455 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3: Page Number 11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "E=10.0;                #voltage of voltage source in V\n",
-      "Ri=10.0;               #Internal Resistance of the voltage source in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Isc=E/Ri;                #short circuit current in A\n",
-      "I=Isc;                   #Current value of current source in A\n",
-      "R=Ri;                    #Internal Resistence of the current source in \u03a9\n",
-      "\n",
-      "#Results\n",
-      "print(\"The current value of the current source= %d A\"%Isc);\n",
-      "print(\"The internal resistance of the current source =%d \u03a9 \"%R);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current value of the current source= 1 A\n",
-        "The internal resistance of the current source =10 \u03a9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.4: Page number 11-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I=6.0;                    # current value of current source in mA\n",
-      "Ri=2000.0;              #Internal Resistance of the current source in \u03a9\n",
-      "\n",
-      "#Calcultion\n",
-      "V=(I/1000)*Ri;                      #Voltage of voltage source in V\n",
-      "R=Ri;                         #Internal resistance of voltage source in \u03a9\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage of voltage source is %d V\"%V);\n",
-      "print(\"The internal resistance of the voltage source is %d \u03a9\"%R);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage of voltage source is 12 V\n",
-        "The internal resistance of the voltage source is 2000 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5: Page number 13\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "E=200.0;                    #Generated voltage in V\n",
-      "Ri=100.0;                   #Internal Resistance of generator in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "RL=100;                      #Load resistance for 1st case in \u03a9\n",
-      "I=E/(RL+Ri);                 #Load current in 1st case A\n",
-      "P=(I*I)*RL;                  #Power delivered to load of 2nd case in watts\n",
-      "Pt=(I*I)*(Ri+RL);            #Total power generated in watts\n",
-      "\n",
-      "print(\"Power delivered for RL=100\u03a9 is %d watts\"%P);\n",
-      "print(\"Total power generated for RL=100\u03a9 is %d watts\"%Pt);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "RL=300;                      #Load resistance for 2nd case in \u03a9\n",
-      "I=E/(RL+Ri);                 #Load current in 2nd case in A\n",
-      "P=(I*I)*RL;                  #Power delivered to load of 2nd case in watts\n",
-      "Pt=(I*I)*(Ri+RL);            #Total power generated in watts\n",
-      "\n",
-      "print(\"Power delivered for RL=300\u03a9 is %d watts\"%P);\n",
-      "print(\"Total power generated for RL=300\u03a9 is %d watts\"%Pt);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power delivered for RL=100\u03a9 is 100 watts\n",
-        "Total power generated for RL=100\u03a9 is 200 watts\n",
-        "Power delivered for RL=300\u03a9 is 75 watts\n",
-        "Total power generated for RL=300\u03a9 is 100 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6: Page number 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=12.0;                  #Output from amplifier in V\n",
-      "R_out_eq=15;           #Equivalent resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "RL=R_out_eq;              #Load resistance in \u03a9\n",
-      "Rt=RL+R_out_eq;           #Total resistance in \u03a9\n",
-      "I=V/Rt;                   #Circuit current in A\n",
-      "PL=pow(I,2)*RL;           #Power delivered to load in W\n",
-      "\n",
-      "#Results\n",
-      "print(\"Load resistance required is = %d \u03a9\"%RL);\n",
-      "print(\"Power delivered to load = %.1f W\"%PL);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load resistance required is = 15 \u03a9\n",
-        "Power delivered to load = 2.4 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7, Page number 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=50.0;                      #voltage from ac generator in V\n",
-      "R=100.0;                     #Resistance of internal impedance in \u03a9\n",
-      "XL=50.0;                     #inductive reactance of internal impedance in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Zi=100+(50j);               #Internal impedance in complex form (\u03a9)\n",
-      "ZL=conjugate(Zi);          #Load impedance (conjugate of internal impedance ) in \u03a9\n",
-      "Zt=Zi+ZL;                  #Total impedance in \u03a9\n",
-      "I=real(V/Zt);              #Circuit current in A\n",
-      "\n",
-      "Max_Power=pow(I,2)*R;      #Maximum power transferred to the load in watts\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print (\"Load impedance %d %dj \u03a9\"%(real(ZL),imag(ZL)));\n",
-      "print(\"Maximum power transferred to the load =%.2f W\"%Max_Power);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load impedance 100 -50j \u03a9\n",
-        "Maximum power transferred to the load =6.25 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.8: Page number 16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pR(R1,R2):\n",
-      "    return((R1*R2)/(R1+R2));\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "E=100.0;                              #Source voltage in V\n",
-      "R1=10.0;                              #Resistance of resistor 1 in \u03a9\n",
-      "R2=20.0;                              #Resistance of resistor 2 in \u03a9\n",
-      "R3=12.0;                              #Resistance of resistor 3 in \u03a9\n",
-      "R4=8.0;                               #Resistance of resistor 4 in \u03a9\n",
-      "RL=100.0;                             #Resistance of load in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Req=R1+pR(R3+R4,R2);                 #Equivalent resistance after removing RL ,in \u03a9\n",
-      "I=E/Req;                             #Total circuit current in A\n",
-      "I8=I*R2/(R2+R3+R4);\n",
-      "\n",
-      "#Thevenin's equivalent circuit's parameters\n",
-      "E0=I8*R4;                                         #Thevenin voltage V\n",
-      "R0=pR(pR(R1,R2)+R3,R4);                           #Thevenin resistance   \n",
-      "I_RL=E0/(R0+RL);                                  #Load current in A  \n",
-      "\n",
-      "#Result \n",
-      "print (\"Current through load = %.2f A.\"%I_RL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through load = 0.19 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.9: Page number 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pR(R1,R2):\n",
-      "    return((R1*R2)/(R1+R2));\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "V=20.0;                           #Voltage source in V\n",
-      "R1=1000.0;                        #resistance of resistor 1 in \u03a9\n",
-      "R2=1000.0;                        #resistance of resistor 2 in \u03a9\n",
-      "R3=1000.0;                        #resistance of resistor 3 in \u03a9\n",
-      "\n",
-      "#calculation\n",
-      "#parameter for Thevenin's equivalent circuit\n",
-      "E0=(V*R3)/(R1+R3);                         #thevenin voltage in V\n",
-      "R0=pR(R1,R3)+R2;                     #Thevenins resistance in \u03a9\n",
-      "\n",
-      "#result\n",
-      "print(\"The thevenin voltage = %d V\"%E0);\n",
-      "print(\"The thevenin resistance = %d \u03a9\"%R0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The thevenin voltage = 10 V\n",
-        "The thevenin resistance = 1500 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.10: Page number 18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=120.0;                          #Supply voltage in V\n",
-      "R1=40.0;                          #Resistor 1's resistance in \u03a9\n",
-      "R2=20.0;                          #Resistor 2's resistance in \u03a9\n",
-      "R3=60.0;                          #Resistor 3's resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, Thevenin's voltage and resistance are calculated\n",
-      "E0=(V*R2)/(R1+R2);                  #Thevenin voltage (voltage across the load resistance RL, after removing RL)in V\n",
-      "R0=(R1*R2)/(R1+R2) + R3;                 #Thevenin's resistance (Resistance between the terminals of load RL, with RL removed and source voltage shorted)in \u03a9                \n",
-      "RL=R0;                              #Value of load resistance to be connected for maximum power transfer in \u03a9\n",
-      "Pmax=pow(E0,2)/(4*RL);              #Maximum power transferred to load in watts\n",
-      "\n",
-      "#Results\n",
-      "print(\"The value of load resistance RL to which maximum power will be transferred = %.2f \u03a9.\"%RL);\n",
-      "print(\"The maximum power transferred to load =%.2f W.\"%Pmax);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of load resistance RL to which maximum power will be transferred = 73.33 \u03a9.\n",
-        "The maximum power transferred to load =5.45 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.11: Page number 18-19-20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=80.0;                           #Supply voltage in V\n",
-      "R1=100.0;                         #Resistor 1's resistance in \u03a9\n",
-      "R2=100.0;                         #Resistor 2's resistance in \u03a9\n",
-      "R3=30.0;                          #Resistor 3's resistance in \u03a9\n",
-      "R4=80.0;                          #Resistor 4's resistance in \u03a9\n",
-      "R5=20.0;                          #Resistor 5's resistance in \u03a9\n",
-      "R6=60.0;                          #Resistor 6's resistance in \u03a9\n",
-      "R7=20.0;                          #Resistor 7's resistance in \u03a9\n",
-      "R8=50.0;                          #Resistor 8's resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem,\n",
-      "E0=(V*R2)/(R1+R2);              #Thevenin's voltage for the circuit containing V, R1, R2 in V.\n",
-      "R0=(R1*R2)/(R1+R2);             #Thevenin's resistance for R1, R2 in \u03a9.\n",
-      "\n",
-      "#Using Thevenin's theorem again on E0, R0 and rest of the circuit resistors.\n",
-      "E0_1=(E0*R4)/(R0+R3+R4);                #Thevenin's voltage for the cicruit containing E0, R0, R3, R4 in V\n",
-      "R0_1=((R0+R3)*R4)/(R0+R3+R4);           #Thevenin's resistance of R0,R3,R4 (R0 and R3 in series and both in parallel with R4), in \u03a9 \n",
-      "\n",
-      "#Using Thevenin's theorem again on E0_1, R0_1, and rest of the circuit resistors.\n",
-      "E0_2=(E0_1*R6)/(R0_1+R5+R6);                       #Thevenin's voltage for the circuit containing E0_1, R0_1, R5, R6 in V\n",
-      "R0_2=((R0_1+R5)*R6)/(R0_1+R5+R6);                  #Thevenin's resistance of R0_1,R5,R6 (R0 and R3 in series and both in parallel with R4), in \u03a9\n",
-      "\n",
-      "\n",
-      "I_50=E0_2/(R0_2+R7+R8);                           #Current through the 50 \u03a9 resistor in A\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The current through the 50 \u03a9 resistor =%.1f A.\"%I_50);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the 50 \u03a9 resistor =0.1 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.12: Page number 22\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import floor\n",
-      "#Variable declaration\n",
-      "V=40.0;                       #Voltage supply in V\n",
-      "R1=4.0;                       #Resistor 1's resistance in \u03a9\n",
-      "R2=6.0;                       #Resistor 2's resistance in \u03a9\n",
-      "R3=5.0;                       #Resistor 3's resistance in \u03a9\n",
-      "R4=8.0;                       #Resistor 4's resistance in \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Using Norton's theorem,\n",
-      "#calculating Norton current by removing the load resistance R4 and short circuiting those two terminals of the circuit\n",
-      "R=R1 + (R2*R3)/(R2+R3);         #Load on source after removing R4 resistor, in  \u03a9\n",
-      "I=V/R;                          #Source current in A\n",
-      "\n",
-      "#Using current dividing rule ,calculating the short circuit current.\n",
-      "I_N=(I*R2)/(R2+R3);             #Norton's equivalent current or the short circuit current in A\n",
-      "\n",
-      "R_N=R3 + (R1*R2)/(R1+R2);       #Norton's equivalent resistance in \u03a9\n",
-      "\n",
-      "I_8=(I_N*R_N)/(R_N+R4);         #Current through the 8 \u03a9 resistance in A\n",
-      "\n",
-      " \n",
-      "\n",
-      "#Results\n",
-      "print(\"The current through the 8\u03a9 resistance =%.2f A.\"%I_8);\n",
-      "\n",
-      "#Note: The answer in the book is 1.55 A, but in the above code the approximate value is obtained, i.e not 1.55A but 1.56A\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the 8\u03a9 resistance =1.56 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.13 :Page number 23\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V1=30.0;                #Voltage source 1, V\n",
-      "V2=18.0;                #Voltage source 2, V\n",
-      "R1=20.0;                #1st resistor, \u03a9\n",
-      "R2=10.0;                #2nd resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Finding Thevenin's Equivalent circuit\n",
-      "I=(V1-V2)/(R1+R2);              #Current in the circuit, A\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to 1st loop of the circuit,\n",
-      "#V1-I*R1-E0=0, where E0 is the voltage across the points X-Y.\n",
-      "E0=V1-I*R1;                     #Thevenin's voltage source, V\n",
-      "\n",
-      "R0=R1*R2/(R1+R2);               #Thevenin's resistance, \u03a9\n",
-      "\n",
-      "#Finding Norton's equivalent circuit\n",
-      "IN=E0/R0;                   #Norton's equivalent current source, A\n",
-      "RN=R0;                      #Norton's equivanlent resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"IN=%.1fA and RN=%.2f \u03a9\"%(IN,RN));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IN=3.3A and RN=6.67 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10.ipynb
deleted file mode 100755
index c0b0e702..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10.ipynb
+++ /dev/null
@@ -1,1297 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:4bf16723750572755d7939b2e42621cb1f52fcda512b32f42476447ce96a6e93"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 10 : SINGLE STAGE TRANSISTOR AMPLIFIERS\n"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.2 : page number 243-244\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "f_min=2.0;                            #Minimum frequency of operation of amplifier, kHz\n",
-      "f_max=10.0;                           #Maximum frequency of operation of amplifier, kHz\n",
-      "RE=560.0;                             #Emitter resistor, \u2126\n",
-      "\n",
-      "#Calculations\n",
-      "#X_CE(Emitter capacitor's capacitive reactance)\n",
-      "#X_CE=1/(2*pi*f_min*CE)=RE/10\n",
-      "#From the above equation.\n",
-      "CE=1/(2*pi*f_min*1000*(RE/10));              #Emitter capacitor, F,\n",
-      "\n",
-      "CE=CE*10**6;                                 #Emitter capacitor, \ud835\udf07F\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print('The value of the emitter capacitor = %.2f \ud835\udf07F'%(CE));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of the emitter capacitor = 1.42 \ud835\udf07F\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.5: Page number 252-253\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                         #Collector supply voltage in V\n",
-      "VBE=0.7;                          #Base-emitter voltage, V\n",
-      "R1=10.0;                          #Resistor R1, k\u2126\n",
-      "R2=5.0;                           #Resistor R2, k\u2126\n",
-      "RC=1.0;                           #Collector resistor, k\u2126\n",
-      "RE=2.0;                           #Emitter resistor, k\u2126\n",
-      "RL=1.0;                           #Load resistor, k\u2126\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#For d.c load line, from the equation: VCE=VCC-IC*(RC+RE),\n",
-      "#VCE is maximum when IC=0 and IC is maximum when VCE=0.\n",
-      "VCE_max=VCC;                                        #Maximum collector-emitter voltage, V\n",
-      "IC_max=VCC/(RC+RE);                                 #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,20])\n",
-      "limit.set_ylim([0,6])\n",
-      "VCE_plot=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC_plot=[((VCC-i)/(RC+RE)) for i in (VCE_plot[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "#For operating point:\n",
-      "#Assuming VCC drops almost completely across R1 and R2,\n",
-      "V2=VCC*R2/(R1+R2);                                  #Voltage across resistor R2, V\n",
-      "IE=(V2-VBE)/RE;                                     #Emitter current, mA\n",
-      "IC=IE;                                              #Collector current, mA\n",
-      "VCE=VCC-IC*(RC+RE);                                 #Collector-emitter voltage , V\n",
-      "\n",
-      "print(\"The operating point: VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n",
-      "#(iii)\n",
-      "#For a.c load line\n",
-      "RAC=(RC*RL)/(RC+RL);                                #a.c load, k\u2126\n",
-      "VCE_ac_max=VCE+IC*RAC;                              #Maximum collector-emitter voltage, V\n",
-      "IC_ac_max=IC+VCE/RAC;                               #Maximum collector current, mA\n",
-      "print(\"Maximum v_CE=%.2fV and maximum i_C=%.2fmA\"%(VCE_ac_max,IC_ac_max));\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,10])\n",
-      "limit.set_ylim([0,20])\n",
-      "vCE_plot=[0,VCE_ac_max];          #Plot variable for V_CE\n",
-      "iC_plot=[IC_ac_max,0];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(vCE_plot,iC_plot);\n",
-      "xlabel(\"vCE(V)\");\n",
-      "ylabel(\"iC(mA)\");\n",
-      "title(\"a.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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ZKpWKr9l2mj9/Ps6ePYtjx47Bw8MDL7/8stQlWZ2qqirExMRgzZo1cHJyanJf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\nAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7fc7c93536d0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point: VCE=8.55V and IC=2.15mA.\n",
-        "Maximum v_CE=9.62V and maximum i_C=19.25mA\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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HHtPjjz+u//3vf7c8DJQCgBGoAVidXz4P4Nr2+Bu318+aNatx0zV0KBYAGIgjhWBVfCAM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\nAABsigUAAGyKBQAAbOr/AZgkaBTs6IVOAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7fc7c8684290>"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.6: Page number 253-254\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=10;                          #Collector resistor, k\u2126\n",
-      "RL=30;                          #Load resistor, k\u2126\n",
-      "VCC=20;                         #Collector supply voltage, V\n",
-      "IC=1;                           #Collector current, mA\n",
-      "VCE=10;                         #Collector-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#For d.c load line:\n",
-      "#From the equation: VCE=VCC-IC*(RC+RE),\n",
-      "#When VCE=0, IC is  maximum.\n",
-      "#Emitter resistor is neglected, assuming it as negligible\n",
-      "IC_max=VCC/RC;                      #Maximum collector current, mA\n",
-      "\n",
-      "#And, when IC=0, VCE is maximum\n",
-      "VCE_max=VCC;                        #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,20])\n",
-      "limit.set_ylim([0,5])\n",
-      "VCE_plot=[0,VCE_max];          #Plot variable for V_CE\n",
-      "IC_plot=[IC_max,0];            #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n",
-      "\n",
-      "#For a.c load line:\n",
-      "RAC=(RC*RL)/(RC+RL);                #a.c Load resistor, k\u2126\n",
-      "\n",
-      "VCE_ac_max=VCE+IC*RAC;                 #Maximum collector-emitter voltage, V\n",
-      "IC_ac_max=IC+ VCE/RAC;                #Maximum collector current, mA\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,25])\n",
-      "limit.set_ylim([0,5])\n",
-      "vCE_plot=[0,VCE_ac_max];          #Plot variable for V_CE\n",
-      "iC_plot=[IC_ac_max,0];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(vCE_plot,iC_plot);\n",
-      "xlabel(\"vCE(V)\");\n",
-      "ylabel(\"iC(mA)\");\n",
-      "title(\"a.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7fc7c8c8ec90>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7fc7c8a07890>"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.7 : Page number 254-255"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variabe declaration\n",
-      "VCE_Q=8.0;           #Q-point collector emitter voltage, V\n",
-      "IC_Q=1;               #Q-point collector current, mA\n",
-      "ic_positive_peak=1.5;        #Collector current at positive peak of signal, mA\n",
-      "ic_negative_peak=0.5;         #Collector current at negative peak of signal, mA\n",
-      "vce_positive_peak=7;          #Collector emitter voltage at positive peak of signal, V\n",
-      "vce_negative_peak=9;          #Collector emitter voltage at negative peak of signal, V\n",
-      "\n",
-      "#Plot\n",
-      "vce_plot=[vce_positive_peak,vce_negative_peak];    #Plot variable of vce\n",
-      "ic_plot=[ic_positive_peak,ic_negative_peak];    #Plot variable of ic\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,10])\n",
-      "limit.set_ylim([0,2])\n",
-      "p=plot(vce_plot,ic_plot);\n",
-      "xlabel(\"vCE(V)\");\n",
-      "ylabel(\"iC(mA)\");\n",
-      "title(\"a.c load line\");\n",
-      "plt.grid();\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7fc7c877c550>"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.8 : Page number 256\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                 #Collector supply voltage, V\n",
-      "RC=2.0;                   #Collector resistor, k\u2126\n",
-      "Rin=1.0;                  #Input resistance, k\u2126\n",
-      "beta=60.0;                #Base current amplification factor\n",
-      "RL=0.5;                   #Load resistor, k\u2126\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "RAC=(RC*RL)/(RC+RL);                #a.c load resistor, k\u2126\n",
-      "Av=beta*(RAC/Rin);                  #Voltage gain\n",
-      "\n",
-      "#Results\n",
-      "print(\"Voltage gain= %d.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain= 24.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.9 : Page number 256\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "V_in=1.0;                      #Input voltage , mV\n",
-      "RC=10.0;                      #Collector resistor, k\u2126\n",
-      "Rin=2.5;                      #Input resistance, k\u2126\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "RL=10.0;                      #Load resistor, k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "RAC=(RC*RL)/(RC+RL);                #Effective load, k\u2126\n",
-      "Av=beta*(RAC/Rin);                  #Voltage gain\n",
-      "\n",
-      "V_out=V_in*Av;                      #Output voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Output voltage= %dmV.\"%V_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage= 200mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10 : Page number 256-257\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "change_in_IB=10.0;                    #Change in base current, \ud835\udf07A\n",
-      "change_in_IC=1.0;                     #Change in collector current, mA\n",
-      "change_in_VBE=0.02;                   #Change in Base-emitter voltage, V\n",
-      "RC=5.0;                               #Collector resistor, k\u2126\n",
-      "RL=10.0;                              #Emitter resistor, k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "beta=(change_in_IC*1000)/change_in_IB;         #Base current amplification factor\n",
-      "\n",
-      "#(ii)\n",
-      "Rin=(change_in_VBE/change_in_IB)*1000;         #Input impedance, k\u2126\n",
-      "\n",
-      "#(iii)\n",
-      "RAC=round((RC*RL)/(RC+RL),1);                            #a.c load, k\u2126\n",
-      "\n",
-      "#(iv)\n",
-      "Av=beta*RAC/Rin;                                #Voltage gain\n",
-      "\n",
-      "#(v)\n",
-      "Ap=beta*Av;                                     #Power gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Beta= %d.\"%beta);\n",
-      "print(\"Input impedance=%d k\u2126.\"%Rin);\n",
-      "print(\"a.c load=%.1f k\u2126.\"%RAC);\n",
-      "print(\"Voltage gain= %d.\"%Av);\n",
-      "print(\"Power gain=%d.\"%Ap);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Beta= 100.\n",
-        "Input impedance=2 k\u2126.\n",
-        "a.c load=3.3 k\u2126.\n",
-        "Voltage gain= 165.\n",
-        "Power gain=16500.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.11 : Page number 257\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=50.0;                      #Base current amplification factor\n",
-      "RC=3.0;                         #Collector resistor,k\u2126\n",
-      "RL=6.0;                         #Load resistor, k\u2126\n",
-      "Rin=0.5;                        #Input impedance, k\u2126\n",
-      "Vin=1;                          #Input voltage,  mV\n",
-      "\n",
-      "#Calculation\n",
-      "RAC=(RC*RL)/(RC+RL);            #a.c load, k\u2126\n",
-      "Av=beta*RAC/Rin;                #Voltage gain\n",
-      "Vout=Vin*Av;                    #Output voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Output voltage=%dmV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=200mV\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.12 : Page number 257-258\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VT=6.0;                   #Collector potential, V\n",
-      "R1=1.0;                   #Resistor R1, k\u2126\n",
-      "R2=2.0;                   #Resistor R2, k\u2126\n",
-      "VB_found=4.0;             #Measured base voltage, V\n",
-      "\n",
-      "#Calculations\n",
-      "VB=(VT*R1)/(R1+R2);                     #Theoretical base voltage, V\n",
-      "\n",
-      "if(VB_found==VB):\n",
-      "    print(\"The circuit is operating properly.\");\n",
-      "else:\n",
-      "    print(\"The circuit is not operating properly.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The circuit is not operating properly.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.13 : Page number 258-259\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                   #Collector supply voltage, V\n",
-      "R1=40.0;                    #Resistor R1, k\u2126\n",
-      "R2=10.0;                    #Resistor R2, k\u2126\n",
-      "RC=6.0;                     #Collector resistor, k\u2126\n",
-      "RE=2.0;                     #Emitter resistor, k\u2126\n",
-      "beta=80;                    #Base current amplification factor\n",
-      "VBE=0.7;                    #Base emitter voltage, V\n",
-      "\n",
-      "#Calculations\n",
-      "V2=(VCC*R2)/(R1+R2);            #Voltage across resistor R2, V\n",
-      "VE=V2-VBE;                      #Emitter voltage, V\n",
-      "IE=VE/RE;                       #Emitter current,  mA\n",
-      "re=25/IE;                       #a.c emitter resistance, \u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"a.c emitter resistance= %.2f \u2126.\"%re);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "a.c emitter resistance= 38.46 \u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.14  : Page number 262-263\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage, V\n",
-      "VBE=0.7;                      #Base-emitter voltage, V\n",
-      "R1=150.0;                     #Resistor R1, k\u2126\n",
-      "R2=20.0                       #Resistor R2, k\u2126\n",
-      "RC=12.0;                      #Collector resistor, k\u2126\n",
-      "RE=2.2;                       #Emitter resistor, k\u2126\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V2=round(VCC*R2/(R1+R2),2);                  #Voltage across R2, V\n",
-      "VE=round(V2-VBE,2);                          #Voltage across emitter resistor, V\n",
-      "IE=round(VE/RE,2);                           #Emitter current, mA\n",
-      "re=round(25/IE,1);                           #a.c emitter resistance, \u2126\n",
-      "\n",
-      "\n",
-      "#(i)\n",
-      "#CE(emitter capacitor) connected in the circuit:\n",
-      "Av=(RC*1000)/re;                                   #Voltage gain for emitter capacitor connected.\n",
-      "\n",
-      "print(\"(i)Voltage gain= %d.\"%Av);\n",
-      "\n",
-      "#(ii)\n",
-      "#CE(emitter capacitor) removed from the circuit:\n",
-      "Av=(RC*1000)/(re+RE*1000);                          #Voltage gain for emitter capacitor removed.\n",
-      "\n",
-      "print(\"(ii)Voltage gain= %.2f.\"%Av);\n",
-      "\n",
-      "#Note: The answer in the text book has been approximated to 5.38 but it's actually coming 5.37.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Voltage gain= 360.\n",
-        "(ii)Voltage gain= 5.37.\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.15 : Page number 263\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=6.0;                       #Collector resistor, k\u2126\n",
-      "RL=12.0;                      #Load resistor, k\u2126\n",
-      "re=33.3;                      #a.c emitter resistance, \u2126\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "RAC=RC*RL/(RC+RL);              #a.c effective load, k\u2126\n",
-      "Av=RAC*1000/re;                 #Voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage gain= %d.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain= 120.\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.16 : Page number 263-264\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=9.0;                      #Collector supply voltage, V\n",
-      "VBE=0.7;                      #Base-emitter voltage, V\n",
-      "R1=240.0;                     #Resistor R1, k\u2126\n",
-      "R2=30.0                       #Resistor R2, k\u2126\n",
-      "RC=20.0;                      #Collector resistor, k\u2126\n",
-      "RE=3.0;                       #Emitter resistor, k\u2126\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "V2=round(VCC*R2/(R1+R2),1);                  #Voltage across R2, V\n",
-      "VE=round(V2-VBE,1);                          #Voltage across emitter resistor, V\n",
-      "IE=round(VE/RE,1);                           #Emitter current, mA\n",
-      "re=25/IE;                                    #a.c emitter resistance, \u2126\n",
-      "\n",
-      "#(ii)\n",
-      "Av=RC*1000/re;                                   #Voltage gain\n",
-      "\n",
-      "#(iii)\n",
-      "V_C_in=V2;                                  #d.c voltage across input capacitor, V\n",
-      "V_C_E=VE;                                   #d.c vooltage across emitter capacitor, V\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i) a.c emitter resistance=%d \u2126.\"%re);\n",
-      "print(\"(ii) Voltage gain =%d.\"%Av);\n",
-      "print(\"(iii) d.c voltage across input capacitor= %dV and emitter capacitor=%.1fV.\"%(V_C_in,V_C_E));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) a.c emitter resistance=250 \u2126.\n",
-        "(ii) Voltage gain =80.\n",
-        "(iii) d.c voltage across input capacitor= 1V and emitter capacitor=0.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.17 : Page number 264-265"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                      #Collector supply voltage, V\n",
-      "VBE=0.7;                      #Base-emitter voltage, V\n",
-      "R1=40.0;                     #Resistor R1, k\u2126\n",
-      "R2=10.0                       #Resistor R2, k\u2126\n",
-      "RC=2.0;                      #Collector resistor, k\u2126\n",
-      "RE=1.0;                       #Emitter resistor, k\u2126\n",
-      "RL=1.0;                       #Load resistor, k\u2126\n",
-      "beta=100;                     #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) D.C bias levels\n",
-      "V2=VCC*R2/(R1+R2);                  #Voltage across R2, V\n",
-      "VE=round(V2-VBE,1);                 #Voltage across emitter resistor, V\n",
-      "IE=round(VE/RE,1);                  #Emitter current, mA\n",
-      "IC=IE;                              #Collector current, mA\n",
-      "IB=IC/beta;                         #Base  current, mA\n",
-      "VC=VCC-IC*RC;                       #Collector voltage, V\n",
-      "print(\"(i) D.C bias levels: V2=%dV, VE=%.1fV, IE=%.1fmA, IC=%.1fmA, IB=%.3fmA and VC=%.1fV.\"%(V2,VE,IE,IC,IB,VC));\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "Cin_V=V2;                            #Voltage across Cin capacitor, V\n",
-      "CE_V=VE;                             #Voltage across CE capacitor, V \n",
-      "CC_V=VC;                             #Voltage across CC capacitor, V\n",
-      "print(\"(ii) D.c voltage across: Cin=%dV and CE=%.1fV and CC=%.1fV.\"%(Cin_V,CE_V,CC_V));\n",
-      "\n",
-      "#(iii)\n",
-      "re=round(25/IE,1);                           #a.c emitter resistance, \u2126\n",
-      "print(\"(iii) a.c emitter resistance=%.1f\u2126.\"%re);\n",
-      "\n",
-      "\n",
-      "#(iv)\n",
-      "RAC=round(RC*RL/(RC+RL),3);                #Total a.c collector resistance, k\u2126\n",
-      "Av=RAC/(re/1000);                        #Voltage gain\n",
-      "print(\"(iv) Voltage gain=%.1f.\"%Av);\n",
-      "\n",
-      "#(v)\n",
-      "print(\"(v) VC>VE. Therefore, the transistor is in active state.\" );\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) D.C bias levels: V2=3V, VE=2.3V, IE=2.3mA, IC=2.3mA, IB=0.023mA and VC=10.4V.\n",
-        "(ii) D.c voltage across: Cin=3V and CE=2.3V and CC=10.4V.\n",
-        "(iii) a.c emitter resistance=10.9\u2126.\n",
-        "(iv) Voltage gain=61.2.\n",
-        "(v) VC>VE. Therefore, the transistor is in active state.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.18 : page number 265\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=132.0;                   #Voltage gain\n",
-      "beta=200.0;                 #Base current amplification factor\n",
-      "P_in=60.0;                     #Input power, \ud835\udf07W\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "Ap=beta*Av;                          #Power gain\n",
-      "P_out=Ap*(P_in/10**6);                #Output power, W\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The power gain = %d and output power = %.3fW.\"%(Ap,P_out));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The power gain = 26400 and output power = 1.584W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.19 : page number 265-266\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IB=200.0;                       #Base current, microampere\n",
-      "IE=10.0;                        #Emitter current, mA\n",
-      "R1=27.0;                      #Resistor R1, kilo ohm\n",
-      "R2=13.0                       #Resistor R2, kilo ohm\n",
-      "RC=4.7;                       #Collector resistor, kilo ohm\n",
-      "RE=2.2;                       #Emitter resistor, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "IC=IE-(IB/1000);                              #Collector current, mA\n",
-      "beta=IC/(IB/1000);                            #Current gain\n",
-      "\n",
-      "print(\"(i) Current gain=%d\"%beta);\n",
-      "\n",
-      "#(ii)\n",
-      "#a.c emitter resistance is neglected, voltage gain=(collector resistor)/(emitter resistor)\n",
-      "Av=RC/RE;                           #Voltage gain\n",
-      "\n",
-      "print(\"(ii) Voltage gain=%.2f\"%Av);\n",
-      "\n",
-      "#(iii)\n",
-      "Ap=round(beta*Av,0);                         #Power gain\n",
-      "\n",
-      "#Results\n",
-      "print(\"(iii) Power gain=%d.\"%Ap);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Current gain=49\n",
-        "(ii) Voltage gain=2.14\n",
-        "(iii) Power gain=105.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.20 : Page number 266-267\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=30.0;                     #Collector supply voltage, V\n",
-      "VBE=0.7;                      #Base emitter voltage, V\n",
-      "R1=45.0;                      #Resistor R1, k\u2126\n",
-      "R2=15.0                       #Resistor R2, k\u2126\n",
-      "RC=10.0;                      #Collector resistor,k\u2126\n",
-      "RE=7.5;                       #Emitter resistor, k\u2126\n",
-      "beta=200.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "V2=round(VCC*R2/(R1+R2),1);                  #Voltage across R2, V (Voltage divider rule)\n",
-      "VE=V2;                                       #Voltage across emitter resistor(base-emitter voltage is neglected), V\n",
-      "IE=VE/RE;                                    #Emitter current, mA (OHM's LAW)\n",
-      "re=25/IE;                                    #a.c emitter resistance, ohm\n",
-      "Zin_base=(beta*re)/1000;                     #input impedance of transistor base,k\u2126\n",
-      "R1_R2=(R1*R2)/(R1+R2);                       #Parallel resistance between R1 and R2, k\u2126\n",
-      "Zin=((R1_R2)*Zin_base)/(R1_R2+Zin_base);     #Input impedance of the amplifier circuit, k\u2126\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The input impedance of the amplifier circuit= %.2f k\u2126.\"%Zin);       \n",
-      "\n",
-      "#Note: The input impedance of the amplifier circuit is approximated as 3.45 k\u2126 in the text book, but actually it's 3.46 k\u2126.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input impedance of the amplifier circuit= 3.46 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.21 : Page Number 268-269\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                           #Collector supply voltage, V.\n",
-      "RC=1.5;                             #Collector resistor, k\u2126.\n",
-      "R1=18.0;                            #Resistor R1, k\u2126.\n",
-      "R2=4.7;                             #Resistor R2, k\u2126.\n",
-      "RE1=300.0;                          #Emitter resistor 1, \u2126.\n",
-      "RE2=900.0;                          #Emitter resistor 2, \u2126.\n",
-      "VBE=0.7;                            #Base-emitter voltage, V.\n",
-      "beta=150.0;                         #Base current amplification factor.\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V2=round(VCC*R2/(R1+R2),1);                      #d.c voltage across R2, V. (Voltage divider rule)\n",
-      "VE=round(V2-VBE,1);                              #d.c voltage across RE, V.\n",
-      "IE=round((VE/(RE1+RE2))*1000,2);                 #d.c emitter current, mA.(OHM'S LAW)\n",
-      "re=round(25/IE,1);                               #a.c emitter resistance, \u2126.\n",
-      "Av=RC*1000/(re+RE1);                             #Voltage gain\n",
-      "Zin_base=(beta*(re+RE1))/1000;                   #Input impedance of transistor base, k\u2126.\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage gain of the swamped amplifier= %.2f.\"%Av);\n",
-      "print(\"Input impedance of transistor base of the swamped amplifier= %.2f k\u2126.\"%Zin_base);\n",
-      "\n",
-      "#Note:In the textbook Av is approximated to 4.66and Zin_base to 48.22 kilo ohm, but the actual answers come as 4.67 and 48.21 kilo ohm.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain of the swamped amplifier= 4.67.\n",
-        "Input impedance of transistor base of the swamped amplifier= 48.21 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.22 : Page number 269\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=1.5;                             #Collector resistor, k\u2126.\n",
-      "RE1=300.0;                          #Emitter resistor 1, \u2126.\n",
-      "re=21.5;                            #a.c emitter resistance, \u2126.\n",
-      "\n",
-      "#Calculations\n",
-      "Av=round(RC*1000/(re+RE1),2);                     #Voltage gain.\n",
-      "Av_1=round(RC*1000/(2*re+RE1),2);                 #Voltage gain when re doubles.\n",
-      "change_in_gain=round(Av-Av_1,2);                      #Change in voltage gain.\n",
-      "change_percentage=change_in_gain*100/Av;               #Change percentage\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "if(change_in_gain>0):\n",
-      "    print(\"The percentage change from the original value= %.2f%%(decrease)\"%change_percentage);\n",
-      "else:\n",
-      "    print(\"The percentage change from the original value= %.2f%%(increase)\"%change_percentage);\n",
-      "\n",
-      "\n",
-      "#Note: The percentage has been approximated in the text book as 6.22%, but the answer comes as 6.42%.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage change from the original value= 6.42%(decrease)\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.23 : Page number 269-270\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage, V\n",
-      "VBE=0.7;                      #Base emitter voltage, V\n",
-      "R1=10.0;                      #Resistor R1, kilo ohm\n",
-      "R2=2.2;                       #Resistor R2, kilo ohm\n",
-      "RC=4.0;                       #Collector resistor, kilo ohm\n",
-      "RE=1.1;                       #Emitter resistor, kilo ohm\n",
-      "beta=200.0;                   #Base current amplification factor\n",
-      "RE1=210.0;                    #Emitter resistor 1 of swamped amplifier, ohm.\n",
-      "RE2=900.0;                    #Emitter resistor 2 of swamped amplifier, ohm.\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V2=round(VCC*R2/(R1+R2),1);                      #d.c voltage across R2, V. (Voltage divider rule)\n",
-      "VE=round(V2-VBE,1);                              #d.c voltage across RE, V.\n",
-      "IE=(VE/RE);                                      #d.c emitter current, mA.(OHM'S LAW)\n",
-      "re=25/IE;                                        #a.c emitter resistance, ohm.\n",
-      "\n",
-      "\n",
-      "#(i) Zin_base:\n",
-      "Zin_base_standard=(beta*re)/1000;                      #input impedance of transistor base for standard amplifier , kilo ohm.\n",
-      "Zin_base_swamped=(beta*(re+RE1))/1000;                 #input impedance of transistor base for swamped amplifier, kilo ohm.\n",
-      "\n",
-      "\n",
-      "#(ii) Zin:\n",
-      "#input impedance for standard amplifier circuit\n",
-      "Zin_standard=(((R1*R2)/(R1+R2))*Zin_base_standard)/(Zin_base_standard +((R1*R2)/(R1+R2)));    #kilo ohm\n",
-      "\n",
-      "#input impedance for standard amplifier circuit\n",
-      "Zin_swamped=(((R1*R2)/(R1+R2))*Zin_base_swamped)/(Zin_base_swamped +((R1*R2)/(R1+R2)));       #kilo ohm\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i) input impedance of transistor base for standard amplifier= %d kilo ohm\"%Zin_base_standard);\n",
-      "print(\"    input impedance of transistor base for swamped amplifier= %d kilo ohm\"%Zin_base_swamped);\n",
-      "print(\"(ii) input impedance for standard amplifier= %.2f kilo ohm\"%Zin_standard);\n",
-      "print(\"     input impedance for swamped amplifier= %.2f kilo ohm\"%Zin_swamped);\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) input impedance of transistor base for standard amplifier= 5 kilo ohm\n",
-        "    input impedance of transistor base for swamped amplifier= 47 kilo ohm\n",
-        "(ii) input impedance for standard amplifier= 1.33 kilo ohm\n",
-        "     input impedance for swamped amplifier= 1.74 kilo ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.24 : Page number 270-271\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=4.0;                     #Collector resistor, kilo ohm\n",
-      "re=25.0;                    #a.c emitter resistance, ohm (calculated in example 10.23)\n",
-      "RE_1=210.0;                  #Emitter resistor 1 of swamped amplifier,ohm\n",
-      "\n",
-      "#Calculation\n",
-      "Av_standard=(RC*1000)/re;                       #Voltage gain of standard common emitter amplifier\n",
-      "Av_swamped=(RC*1000)/(re+RE_1);                 #Voltage gain of swamped amplifier\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage gain of standard amplifier=%d.\"%Av_standard);\n",
-      "print(\"The voltage gain of swamped amplifier=%d.\"%Av_swamped);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain of standard amplifier=160.\n",
-        "The voltage gain of swamped amplifier=17.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.26 : Page number 273-274\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_0=1000.0;                  #Open circuit voltage gain\n",
-      "R_in=2.0;                    #Input resistance, kilo ohm\n",
-      "R_out=1.0;                   #Output resistance, ohm\n",
-      "RL=4;                        #Load resistor across the output, ohm\n",
-      "I_2=0.5;                     #Output signal current, A.\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Since A_0*(I_1*R_in) = I_2*(R_out+RL)\n",
-      "I_1=I_2*(R_out+RL)/(A_0*(R_in*1000));           #Input current, A\n",
-      "V_1=I_1*(R_in*1000);                            #Input signal voltage, V\n",
-      "V_1=V_1*1000;                                   #Input signal voltage, mV\n",
-      "\n",
-      "print(\"The required input signal voltage =%.1fmV\"%V_1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required input signal voltage =2.5mV\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.27 : Page number 274\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_0=1000.0;                     #Open circuit voltage gain\n",
-      "R_in=7.0;                       #Input resistance, kilo ohm\n",
-      "R_out=15.0;                     #Output resistance, ohm\n",
-      "RL=35.0;                        #Load resistor across the output, ohm\n",
-      "R_s=3.0;                        #Internal resistance, kilo ohm\n",
-      "E_s=10.0;                       #Input signal voltage, mV.\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "I_1=E_s*(10**-3)/(R_s*1000+R_in*1000);                  #Input current, A\n",
-      "V_1=I_1*(R_in*1000);                                    #Voltage across input resistance, V\n",
-      "\n",
-      "#Since, A_v=V_2/V_1 = A_0*RL/(R_out+RL)\n",
-      "A_v=A_0*RL/(R_out+RL);                                 #Voltage gain\n",
-      "V_2=A_v*V_1;                                            #Outout voltage, V\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "P_2=V_2**2/RL;                              #Output power, W\n",
-      "P_1=V_1**2/(R_in*1000);                            #Input power, W\n",
-      "A_p=round(P_2/P_1,-6);                                #Power gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The magnitude of output voltage = %.1fV\"%V_2);\n",
-      "print(\"The power gain =%de-06.\"%(A_p/10**6));\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The magnitude of output voltage = 4.9V\n",
-        "The power gain =98e-06.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.28 : Page number 274-275\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v=80.0;                   #Voltage gain\n",
-      "V_2=1.0;                    #Output voltage, V\n",
-      "A_i=120.0;                  #Current gain\n",
-      "RL=2;                       #Load resistor, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "V_1=(V_2/A_v)*1000;                #Input signal voltage, mV\n",
-      "\n",
-      "#Since, A_i=A0*R_in/(R_out+RL) and A_v=A0*RL/(R_out+RL)\n",
-      "#So, A_v/A_i=RL/R_in\n",
-      "R_in=RL*A_i/A_v;                #Input resistance, kilo ohm\n",
-      "I_1=V_1/R_in;                   #Input current,  \u03bcA\n",
-      "A_p=A_i*A_v;                    #Power gain\n",
-      "\n",
-      "#Results\n",
-      "print(\"Necessary input signal voltage= %.1fmV\"%V_1);\n",
-      "print(\"Input signal current =%.2f  \u03bcA\"%I_1);\n",
-      "print(\"Power gain = %d.\"%A_p);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Necessary input signal voltage= 12.5mV\n",
-        "Input signal current =4.17  \u03bcA\n",
-        "Power gain = 9600.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_1.ipynb
deleted file mode 100755
index c0b0e702..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_1.ipynb
+++ /dev/null
@@ -1,1297 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:4bf16723750572755d7939b2e42621cb1f52fcda512b32f42476447ce96a6e93"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 10 : SINGLE STAGE TRANSISTOR AMPLIFIERS\n"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.2 : page number 243-244\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "f_min=2.0;                            #Minimum frequency of operation of amplifier, kHz\n",
-      "f_max=10.0;                           #Maximum frequency of operation of amplifier, kHz\n",
-      "RE=560.0;                             #Emitter resistor, \u2126\n",
-      "\n",
-      "#Calculations\n",
-      "#X_CE(Emitter capacitor's capacitive reactance)\n",
-      "#X_CE=1/(2*pi*f_min*CE)=RE/10\n",
-      "#From the above equation.\n",
-      "CE=1/(2*pi*f_min*1000*(RE/10));              #Emitter capacitor, F,\n",
-      "\n",
-      "CE=CE*10**6;                                 #Emitter capacitor, \ud835\udf07F\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print('The value of the emitter capacitor = %.2f \ud835\udf07F'%(CE));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of the emitter capacitor = 1.42 \ud835\udf07F\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.5: Page number 252-253\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                         #Collector supply voltage in V\n",
-      "VBE=0.7;                          #Base-emitter voltage, V\n",
-      "R1=10.0;                          #Resistor R1, k\u2126\n",
-      "R2=5.0;                           #Resistor R2, k\u2126\n",
-      "RC=1.0;                           #Collector resistor, k\u2126\n",
-      "RE=2.0;                           #Emitter resistor, k\u2126\n",
-      "RL=1.0;                           #Load resistor, k\u2126\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#For d.c load line, from the equation: VCE=VCC-IC*(RC+RE),\n",
-      "#VCE is maximum when IC=0 and IC is maximum when VCE=0.\n",
-      "VCE_max=VCC;                                        #Maximum collector-emitter voltage, V\n",
-      "IC_max=VCC/(RC+RE);                                 #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,20])\n",
-      "limit.set_ylim([0,6])\n",
-      "VCE_plot=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC_plot=[((VCC-i)/(RC+RE)) for i in (VCE_plot[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "#For operating point:\n",
-      "#Assuming VCC drops almost completely across R1 and R2,\n",
-      "V2=VCC*R2/(R1+R2);                                  #Voltage across resistor R2, V\n",
-      "IE=(V2-VBE)/RE;                                     #Emitter current, mA\n",
-      "IC=IE;                                              #Collector current, mA\n",
-      "VCE=VCC-IC*(RC+RE);                                 #Collector-emitter voltage , V\n",
-      "\n",
-      "print(\"The operating point: VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n",
-      "#(iii)\n",
-      "#For a.c load line\n",
-      "RAC=(RC*RL)/(RC+RL);                                #a.c load, k\u2126\n",
-      "VCE_ac_max=VCE+IC*RAC;                              #Maximum collector-emitter voltage, V\n",
-      "IC_ac_max=IC+VCE/RAC;                               #Maximum collector current, mA\n",
-      "print(\"Maximum v_CE=%.2fV and maximum i_C=%.2fmA\"%(VCE_ac_max,IC_ac_max));\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,10])\n",
-      "limit.set_ylim([0,20])\n",
-      "vCE_plot=[0,VCE_ac_max];          #Plot variable for V_CE\n",
-      "iC_plot=[IC_ac_max,0];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(vCE_plot,iC_plot);\n",
-      "xlabel(\"vCE(V)\");\n",
-      "ylabel(\"iC(mA)\");\n",
-      "title(\"a.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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ZKpWKr9l2mj9/Ps6ePYtjx47Bw8MDL7/8stQlWZ2qqirExMRgzZo1cHJyanJf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\nAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7fc7c93536d0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point: VCE=8.55V and IC=2.15mA.\n",
-        "Maximum v_CE=9.62V and maximum i_C=19.25mA\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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HHtPjjz+u//3vf7c8DJQCgBGoAVidXz4P4Nr2+Bu318+aNatx0zV0KBYAGIgjhWBVfCAM\n4APUAKzIp/sAsrKyPDtvb1RRUaFly5ZpxYoVjZsQsAD2DcAOvBZAXl6e5s2bpz179ighIUERERGq\nrKzUgQMHdPr0aU2aNEk/+clPdMcdd/h2KAoAJkINwCr8sgno7Nmz+ve//62SkhK1atVKvXr10n33\n3Xdbg3odigUAJsS+AZidTxeAY8eO6fjx4zd9/u++ffsUGRmpiIiIpk/qbSgWAJgUNQAz8+k+gJ/+\n9Kc6ceLETeefPHlSzz//fOOnAyyOfQMIJl4XgAMHDmjQoEE3nT9w4ED95z//8dtQgNnxN4UQDLwu\nAGfPnr3lZXw0JOyOGoDVeV0Aunfvro0bN950/qZNm3Tvvff6bSjASqgBWJXXncAFBQUaPny4Hnro\nIfXp00dut1u7du3S9u3btWHDBr8dCcROYFgVRwrBSD7dCRwbG6v8/HwNHDhQhYWFOnjwoAYNGqT8\n/Hy/HgYKWBU1ACvhT0EAfkININB8WgAPP/ywJKl169a6++67a53a8NMMeEUNwOwoACAAqAEEgl8+\nEAbA7aEGYEYUABBg1AD8hQIATI4agFlQAICBqAH4EgUAWAg1ACNRAIBJUAO4XRQAYFHUAAKNAgBM\niBpAU1AAQBCgBhAIFABgctQAGooCAIIMNQB/oQAAC6EG4A0FAAQxagC+RAEAFkUN4EYUAGAT1ABu\nFwUABAFqABIFANgSNYCmoACAIEMN2BcFANgcNYCGogCAIEYN2AsFAMCDGoA3FABgE9RA8KMAANSJ\nGsCNKADAhqiB4EQBAKgXNQCJAgBsjxoIHhQAgEahBuyLAgDgQQ1YGwUAoMmoAXuhAADUiRqwHgoA\ngE9QA8GPAgBQL2rAGigAAD5HDQQnwwogJiZGbdq00Te+8Q01b95cO3fuvD4UBQCYFjVgXpYpAIfD\nIZfLpby8vFov/gDMjRoIHoZuAuK3fMCaQkOlnJyrp6wsacoU6cwZo6dCYxlaAI8++qj69u2rnJwc\no8YAcBuoAWtrZtQDb9u2TR06dNDx48eVlpamnj17KiUlxXP57NmzPf9OTU1Vampq4IcEUK9rNfDR\nR1drgH0DgeNyueRyuZp8e1McBjpnzhy1bt1a06ZNk8ROYMCqTp+Wpk+/uhi88YaUlmb0RPZiiZ3A\n58+f19mzZyVJ586d00cffaTExEQjRgHgQ+wbsBZDFoCjR48qJSVFSUlJevDBBzV8+HANGTLEiFEA\n+AH7BqzBFJuAbsQmICB48L6BwLHEJiAA9kENmBcFACBgqAH/ogAAmBY1YC4UAABDUAO+RwEAsARq\nwHgUAADDUQO+QQEAsBxqwBgUAABToQaajgIAYGnUQOBQAABMixpoHAoAQNCgBvyLAgBgCdRA/SgA\nAEGJGvA9CgCA5VADdaMAAAQ9asA3KAAAlkYNXEcBALAVaqDpKAAAQcPuNUABALAtaqBxKAAAQcmO\nNUABAICogYagAAAEPbvUAAUAADegBupGAQCwlWCuAQoAALygBq6jAADYVrDVAAUAAA1k9xqgAABA\nwVEDFAAANIEda4ACAIAbWLUGKAAAuE12qQEKAAC8sFINUAAA4EPBXAMUAAA0kNlrgAIAAD8Jthqg\nAACgCcxYAxQAAARAMNQABQAAt8ksNUABAECAWbUGKAAA8CEja4ACAAADWakGKAAA8JNA1wAFAAAm\nYfYaoAAAIAACUQMUAACYkBlrgAIAgADzVw1YogA2b96snj17qkePHlqwYIERIwCAYcxSAwFfAKqq\nqvTss89q8+bN+uKLL7Ry5Urt378/0GNYhsvlMnoE0+C5uI7n4jqrPhehoVJOztVTVpY0ZYp05kxg\nZwj4ArBz5051795dMTExat68uTIyMrR27dpAj2EZVv3h9geei+t4Lq6z+nNhZA0EfAE4fPiwOnfu\n7Pk6Ojpahw8fDvQYAGAaRtVAwBcAh8MR6IcEAEu4sQa++sq/jxfwo4B27Nih2bNna/PmzZKk+fPn\nKyQkRL/4xS+uD8UiAQBN0piX9IAvAFeuXNF9992nTz75RB07dtQ3v/lNrVy5Ur169QrkGABge80C\n/oDNmun3v/+9vvOd76iqqkqTJ0/mxR8ADGDKN4IBAPzPdH8KgjeJXVVcXKzBgwcrPj5eCQkJ+t3v\nfmf0SIarqqpScnKy0tPTjR7FUKdOndKYMWPUq1cvxcXFaceOHUaPZJiFCxcqISFBiYmJGjdunC5e\nvGj0SAEzadIkOZ1OJSYmes4rKytTWlqaYmNjNWTIEJ06dcrrfZhqAeBNYtc1b95cCxcu1L59+7Rj\nxw4tWbLEts/FNYsWLVJcXJztDxJ4/vnnNWzYMO3fv1/5+fm23YR6+PBhLV68WLt27dKePXtUVVWl\nVatWGT1WwEycONFzMM012dnZSktLU0FBgR555BFlZ2d7vQ9TLQC8Sey6qKgoJSUlSZJat26tXr16\n6ciRIwZPZZxDhw5p06ZNysrKsvXfiTp9+rS2bt2qSZMmSbq6Ty00NNTgqYxz5coVnT9/3vPfTp06\nGT1SwKSkpCg8PLzWeevWrdOECRMkSRMmTND777/v9T5MtQDwJrG6FRUVKS8vTw8++KDRoxjmZz/7\nmV555RWFhJjqRzbgCgsLFRERoYkTJ+qBBx7QU089pfPnzxs9liE6deqkadOmqUuXLurYsaPCwsL0\n6KOPGj2WoY4ePSqn0ylJcjqdOnr0qNfrm+r/JrunfV0qKio0ZswYLVq0SK1btzZ6HENs2LBBkZGR\nSk5OtvVv/9LV33g///xzPfPMM/r8889111131Zv5waq8vFzr1q1TUVGRjhw5ooqKCr3zzjtGj2Ua\nDoej3tdUUy0AnTp1UnFxsefr4uJiRUdHGziRsS5fvqzRo0friSee0MiRI40exzDbt2/XunXr1K1b\nN2VmZurvf/+7xo8fb/RYhoiOjlZ0dLT69esnSRozZow+//xzg6cyxscff6xu3bqpXbt2atasmUaN\nGqXt27cbPZahnE6nSktLJUklJSWKjIz0en1TLQB9+/bVV199paKiIl26dEmrV6/WiBEjjB7LEG63\nW5MnT1ZcXJxeeOEFo8cx1Lx581RcXKzCwkKtWrVK3/72t7V8+XKjxzJEVFSUOnfurIKCAklXXwTj\n4+MNnsoYXbt21Y4dO3ThwgW53W59/PHHiouLM3osQ40YMUK5ubmSpNzc3Pp/cXSbzKZNm9yxsbHu\ne++91z1v3jyjxzHM1q1b3Q6Hw33//fe7k5KS3ElJSe4PPvjA6LEM53K53Onp6UaPYajdu3e7+/bt\n6+7du7f7+9//vvvUqVNGj2SYWbNmuXv27OlOSEhwjx8/3n3p0iWjRwqYjIwMd4cOHdzNmzd3R0dH\nu9988033yZMn3Y888oi7R48e7rS0NHd5ebnX++CNYABgU6baBAQACBwWAACwKRYAALApFgAAsCkW\nAACwKRYAALApFgAAsCkWAOD/Kygo0LBhwxQbG6s+ffpo7NixOnbsmFwul0JDQ5WcnOw5ffLJJ5Kk\nCxcuKDU1VdXV1brnnns879C95oUXXtDLL7+svXv3auLEiUZ8W8AtBfwjIQEzqqys1PDhw7Vw4UJ9\n73vfkyRt2bJFx48fl8Ph0MCBA7V+/fqbbvfmm29q9OjRCgkJUWZmplatWqWZM2dKkqqrq/WXv/xF\n27dvV+fOnXXo0CEVFxfX+ou3gJEoANjOiy++qKVLl3q+nj17thYvXqyHHnrI8+IvSYMGDVJ8fLzX\nv0D67rvv6rHHHpMkZWZmavXq1Z7LPv30U3Xt2tXzgp+enm6rDyyB+bEAwHbGjh2rNWvWeL5es2aN\n9u/frwceeOCWt9m6dWutTUCFhYW6dOmSvv76a3Xp0kWSlJCQoJCQEOXn50uSVq1apXHjxnnuo2/f\nvtq6daufviug8dgEBNtJSkrSsWPHVFJSomPHjqlt27Y3fbLSjVJSUm7aBHTkyBGFhYXVOu/aZqD4\n+HitXbtWc+fO9VwWERFh6091g/mwAMCWfvCDH+i9995TaWmpMjIy1KpVK23ZsqVR93HnnXeqsrKy\n1nkZGRkaMmSIBg0apN69eysiIsJzWWVlpe68806fzA/4AgsAbGns2LHKysrSyZMn9emnnyo0NFTz\n58/Xpk2bNGzYMElXt+G3a9fulvcRHh6uqqoqXbp0SS1atJAk3XPPPWrfvr1mzJhx0+c4FBQUKCEh\nwX/fFNBI7AOALcXFxamiokLR0dFyOp1q2bKlNmzYoMWLFys2Nlbx8fF67bXXFBERIYfDcdM+gL/+\n9a+SpCFDhty0XT8zM1P//e9/NWrUqFrn/+Mf/9Dw4cMD9j0C9eHzAIDbkJeXp4ULF9b7CWUXL15U\namqqtm3bZvsPtod58JMI3Ibk5GQNHjxY1dXVXq9XXFysBQsW8OIPU6EAAMCm+HUEAGyKBQAAbIoF\nAABsigUAAGyKBQAAbOr/AZgkaBTs6IVOAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7fc7c8684290>"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.6: Page number 253-254\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=10;                          #Collector resistor, k\u2126\n",
-      "RL=30;                          #Load resistor, k\u2126\n",
-      "VCC=20;                         #Collector supply voltage, V\n",
-      "IC=1;                           #Collector current, mA\n",
-      "VCE=10;                         #Collector-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#For d.c load line:\n",
-      "#From the equation: VCE=VCC-IC*(RC+RE),\n",
-      "#When VCE=0, IC is  maximum.\n",
-      "#Emitter resistor is neglected, assuming it as negligible\n",
-      "IC_max=VCC/RC;                      #Maximum collector current, mA\n",
-      "\n",
-      "#And, when IC=0, VCE is maximum\n",
-      "VCE_max=VCC;                        #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,20])\n",
-      "limit.set_ylim([0,5])\n",
-      "VCE_plot=[0,VCE_max];          #Plot variable for V_CE\n",
-      "IC_plot=[IC_max,0];            #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n",
-      "\n",
-      "#For a.c load line:\n",
-      "RAC=(RC*RL)/(RC+RL);                #a.c Load resistor, k\u2126\n",
-      "\n",
-      "VCE_ac_max=VCE+IC*RAC;                 #Maximum collector-emitter voltage, V\n",
-      "IC_ac_max=IC+ VCE/RAC;                #Maximum collector current, mA\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,25])\n",
-      "limit.set_ylim([0,5])\n",
-      "vCE_plot=[0,VCE_ac_max];          #Plot variable for V_CE\n",
-      "iC_plot=[IC_ac_max,0];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(vCE_plot,iC_plot);\n",
-      "xlabel(\"vCE(V)\");\n",
-      "ylabel(\"iC(mA)\");\n",
-      "title(\"a.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7fc7c8c8ec90>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7fc7c8a07890>"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.7 : Page number 254-255"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variabe declaration\n",
-      "VCE_Q=8.0;           #Q-point collector emitter voltage, V\n",
-      "IC_Q=1;               #Q-point collector current, mA\n",
-      "ic_positive_peak=1.5;        #Collector current at positive peak of signal, mA\n",
-      "ic_negative_peak=0.5;         #Collector current at negative peak of signal, mA\n",
-      "vce_positive_peak=7;          #Collector emitter voltage at positive peak of signal, V\n",
-      "vce_negative_peak=9;          #Collector emitter voltage at negative peak of signal, V\n",
-      "\n",
-      "#Plot\n",
-      "vce_plot=[vce_positive_peak,vce_negative_peak];    #Plot variable of vce\n",
-      "ic_plot=[ic_positive_peak,ic_negative_peak];    #Plot variable of ic\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,10])\n",
-      "limit.set_ylim([0,2])\n",
-      "p=plot(vce_plot,ic_plot);\n",
-      "xlabel(\"vCE(V)\");\n",
-      "ylabel(\"iC(mA)\");\n",
-      "title(\"a.c load line\");\n",
-      "plt.grid();\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7fc7c877c550>"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.8 : Page number 256\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                 #Collector supply voltage, V\n",
-      "RC=2.0;                   #Collector resistor, k\u2126\n",
-      "Rin=1.0;                  #Input resistance, k\u2126\n",
-      "beta=60.0;                #Base current amplification factor\n",
-      "RL=0.5;                   #Load resistor, k\u2126\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "RAC=(RC*RL)/(RC+RL);                #a.c load resistor, k\u2126\n",
-      "Av=beta*(RAC/Rin);                  #Voltage gain\n",
-      "\n",
-      "#Results\n",
-      "print(\"Voltage gain= %d.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain= 24.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.9 : Page number 256\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "V_in=1.0;                      #Input voltage , mV\n",
-      "RC=10.0;                      #Collector resistor, k\u2126\n",
-      "Rin=2.5;                      #Input resistance, k\u2126\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "RL=10.0;                      #Load resistor, k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "RAC=(RC*RL)/(RC+RL);                #Effective load, k\u2126\n",
-      "Av=beta*(RAC/Rin);                  #Voltage gain\n",
-      "\n",
-      "V_out=V_in*Av;                      #Output voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Output voltage= %dmV.\"%V_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage= 200mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.10 : Page number 256-257\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "change_in_IB=10.0;                    #Change in base current, \ud835\udf07A\n",
-      "change_in_IC=1.0;                     #Change in collector current, mA\n",
-      "change_in_VBE=0.02;                   #Change in Base-emitter voltage, V\n",
-      "RC=5.0;                               #Collector resistor, k\u2126\n",
-      "RL=10.0;                              #Emitter resistor, k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "beta=(change_in_IC*1000)/change_in_IB;         #Base current amplification factor\n",
-      "\n",
-      "#(ii)\n",
-      "Rin=(change_in_VBE/change_in_IB)*1000;         #Input impedance, k\u2126\n",
-      "\n",
-      "#(iii)\n",
-      "RAC=round((RC*RL)/(RC+RL),1);                            #a.c load, k\u2126\n",
-      "\n",
-      "#(iv)\n",
-      "Av=beta*RAC/Rin;                                #Voltage gain\n",
-      "\n",
-      "#(v)\n",
-      "Ap=beta*Av;                                     #Power gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Beta= %d.\"%beta);\n",
-      "print(\"Input impedance=%d k\u2126.\"%Rin);\n",
-      "print(\"a.c load=%.1f k\u2126.\"%RAC);\n",
-      "print(\"Voltage gain= %d.\"%Av);\n",
-      "print(\"Power gain=%d.\"%Ap);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Beta= 100.\n",
-        "Input impedance=2 k\u2126.\n",
-        "a.c load=3.3 k\u2126.\n",
-        "Voltage gain= 165.\n",
-        "Power gain=16500.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.11 : Page number 257\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=50.0;                      #Base current amplification factor\n",
-      "RC=3.0;                         #Collector resistor,k\u2126\n",
-      "RL=6.0;                         #Load resistor, k\u2126\n",
-      "Rin=0.5;                        #Input impedance, k\u2126\n",
-      "Vin=1;                          #Input voltage,  mV\n",
-      "\n",
-      "#Calculation\n",
-      "RAC=(RC*RL)/(RC+RL);            #a.c load, k\u2126\n",
-      "Av=beta*RAC/Rin;                #Voltage gain\n",
-      "Vout=Vin*Av;                    #Output voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Output voltage=%dmV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=200mV\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.12 : Page number 257-258\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VT=6.0;                   #Collector potential, V\n",
-      "R1=1.0;                   #Resistor R1, k\u2126\n",
-      "R2=2.0;                   #Resistor R2, k\u2126\n",
-      "VB_found=4.0;             #Measured base voltage, V\n",
-      "\n",
-      "#Calculations\n",
-      "VB=(VT*R1)/(R1+R2);                     #Theoretical base voltage, V\n",
-      "\n",
-      "if(VB_found==VB):\n",
-      "    print(\"The circuit is operating properly.\");\n",
-      "else:\n",
-      "    print(\"The circuit is not operating properly.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The circuit is not operating properly.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.13 : Page number 258-259\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                   #Collector supply voltage, V\n",
-      "R1=40.0;                    #Resistor R1, k\u2126\n",
-      "R2=10.0;                    #Resistor R2, k\u2126\n",
-      "RC=6.0;                     #Collector resistor, k\u2126\n",
-      "RE=2.0;                     #Emitter resistor, k\u2126\n",
-      "beta=80;                    #Base current amplification factor\n",
-      "VBE=0.7;                    #Base emitter voltage, V\n",
-      "\n",
-      "#Calculations\n",
-      "V2=(VCC*R2)/(R1+R2);            #Voltage across resistor R2, V\n",
-      "VE=V2-VBE;                      #Emitter voltage, V\n",
-      "IE=VE/RE;                       #Emitter current,  mA\n",
-      "re=25/IE;                       #a.c emitter resistance, \u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"a.c emitter resistance= %.2f \u2126.\"%re);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "a.c emitter resistance= 38.46 \u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.14  : Page number 262-263\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage, V\n",
-      "VBE=0.7;                      #Base-emitter voltage, V\n",
-      "R1=150.0;                     #Resistor R1, k\u2126\n",
-      "R2=20.0                       #Resistor R2, k\u2126\n",
-      "RC=12.0;                      #Collector resistor, k\u2126\n",
-      "RE=2.2;                       #Emitter resistor, k\u2126\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V2=round(VCC*R2/(R1+R2),2);                  #Voltage across R2, V\n",
-      "VE=round(V2-VBE,2);                          #Voltage across emitter resistor, V\n",
-      "IE=round(VE/RE,2);                           #Emitter current, mA\n",
-      "re=round(25/IE,1);                           #a.c emitter resistance, \u2126\n",
-      "\n",
-      "\n",
-      "#(i)\n",
-      "#CE(emitter capacitor) connected in the circuit:\n",
-      "Av=(RC*1000)/re;                                   #Voltage gain for emitter capacitor connected.\n",
-      "\n",
-      "print(\"(i)Voltage gain= %d.\"%Av);\n",
-      "\n",
-      "#(ii)\n",
-      "#CE(emitter capacitor) removed from the circuit:\n",
-      "Av=(RC*1000)/(re+RE*1000);                          #Voltage gain for emitter capacitor removed.\n",
-      "\n",
-      "print(\"(ii)Voltage gain= %.2f.\"%Av);\n",
-      "\n",
-      "#Note: The answer in the text book has been approximated to 5.38 but it's actually coming 5.37.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Voltage gain= 360.\n",
-        "(ii)Voltage gain= 5.37.\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.15 : Page number 263\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=6.0;                       #Collector resistor, k\u2126\n",
-      "RL=12.0;                      #Load resistor, k\u2126\n",
-      "re=33.3;                      #a.c emitter resistance, \u2126\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "RAC=RC*RL/(RC+RL);              #a.c effective load, k\u2126\n",
-      "Av=RAC*1000/re;                 #Voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage gain= %d.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain= 120.\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.16 : Page number 263-264\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=9.0;                      #Collector supply voltage, V\n",
-      "VBE=0.7;                      #Base-emitter voltage, V\n",
-      "R1=240.0;                     #Resistor R1, k\u2126\n",
-      "R2=30.0                       #Resistor R2, k\u2126\n",
-      "RC=20.0;                      #Collector resistor, k\u2126\n",
-      "RE=3.0;                       #Emitter resistor, k\u2126\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "V2=round(VCC*R2/(R1+R2),1);                  #Voltage across R2, V\n",
-      "VE=round(V2-VBE,1);                          #Voltage across emitter resistor, V\n",
-      "IE=round(VE/RE,1);                           #Emitter current, mA\n",
-      "re=25/IE;                                    #a.c emitter resistance, \u2126\n",
-      "\n",
-      "#(ii)\n",
-      "Av=RC*1000/re;                                   #Voltage gain\n",
-      "\n",
-      "#(iii)\n",
-      "V_C_in=V2;                                  #d.c voltage across input capacitor, V\n",
-      "V_C_E=VE;                                   #d.c vooltage across emitter capacitor, V\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i) a.c emitter resistance=%d \u2126.\"%re);\n",
-      "print(\"(ii) Voltage gain =%d.\"%Av);\n",
-      "print(\"(iii) d.c voltage across input capacitor= %dV and emitter capacitor=%.1fV.\"%(V_C_in,V_C_E));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) a.c emitter resistance=250 \u2126.\n",
-        "(ii) Voltage gain =80.\n",
-        "(iii) d.c voltage across input capacitor= 1V and emitter capacitor=0.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.17 : Page number 264-265"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                      #Collector supply voltage, V\n",
-      "VBE=0.7;                      #Base-emitter voltage, V\n",
-      "R1=40.0;                     #Resistor R1, k\u2126\n",
-      "R2=10.0                       #Resistor R2, k\u2126\n",
-      "RC=2.0;                      #Collector resistor, k\u2126\n",
-      "RE=1.0;                       #Emitter resistor, k\u2126\n",
-      "RL=1.0;                       #Load resistor, k\u2126\n",
-      "beta=100;                     #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) D.C bias levels\n",
-      "V2=VCC*R2/(R1+R2);                  #Voltage across R2, V\n",
-      "VE=round(V2-VBE,1);                 #Voltage across emitter resistor, V\n",
-      "IE=round(VE/RE,1);                  #Emitter current, mA\n",
-      "IC=IE;                              #Collector current, mA\n",
-      "IB=IC/beta;                         #Base  current, mA\n",
-      "VC=VCC-IC*RC;                       #Collector voltage, V\n",
-      "print(\"(i) D.C bias levels: V2=%dV, VE=%.1fV, IE=%.1fmA, IC=%.1fmA, IB=%.3fmA and VC=%.1fV.\"%(V2,VE,IE,IC,IB,VC));\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "Cin_V=V2;                            #Voltage across Cin capacitor, V\n",
-      "CE_V=VE;                             #Voltage across CE capacitor, V \n",
-      "CC_V=VC;                             #Voltage across CC capacitor, V\n",
-      "print(\"(ii) D.c voltage across: Cin=%dV and CE=%.1fV and CC=%.1fV.\"%(Cin_V,CE_V,CC_V));\n",
-      "\n",
-      "#(iii)\n",
-      "re=round(25/IE,1);                           #a.c emitter resistance, \u2126\n",
-      "print(\"(iii) a.c emitter resistance=%.1f\u2126.\"%re);\n",
-      "\n",
-      "\n",
-      "#(iv)\n",
-      "RAC=round(RC*RL/(RC+RL),3);                #Total a.c collector resistance, k\u2126\n",
-      "Av=RAC/(re/1000);                        #Voltage gain\n",
-      "print(\"(iv) Voltage gain=%.1f.\"%Av);\n",
-      "\n",
-      "#(v)\n",
-      "print(\"(v) VC>VE. Therefore, the transistor is in active state.\" );\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) D.C bias levels: V2=3V, VE=2.3V, IE=2.3mA, IC=2.3mA, IB=0.023mA and VC=10.4V.\n",
-        "(ii) D.c voltage across: Cin=3V and CE=2.3V and CC=10.4V.\n",
-        "(iii) a.c emitter resistance=10.9\u2126.\n",
-        "(iv) Voltage gain=61.2.\n",
-        "(v) VC>VE. Therefore, the transistor is in active state.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.18 : page number 265\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=132.0;                   #Voltage gain\n",
-      "beta=200.0;                 #Base current amplification factor\n",
-      "P_in=60.0;                     #Input power, \ud835\udf07W\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "Ap=beta*Av;                          #Power gain\n",
-      "P_out=Ap*(P_in/10**6);                #Output power, W\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The power gain = %d and output power = %.3fW.\"%(Ap,P_out));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The power gain = 26400 and output power = 1.584W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.19 : page number 265-266\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IB=200.0;                       #Base current, microampere\n",
-      "IE=10.0;                        #Emitter current, mA\n",
-      "R1=27.0;                      #Resistor R1, kilo ohm\n",
-      "R2=13.0                       #Resistor R2, kilo ohm\n",
-      "RC=4.7;                       #Collector resistor, kilo ohm\n",
-      "RE=2.2;                       #Emitter resistor, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "IC=IE-(IB/1000);                              #Collector current, mA\n",
-      "beta=IC/(IB/1000);                            #Current gain\n",
-      "\n",
-      "print(\"(i) Current gain=%d\"%beta);\n",
-      "\n",
-      "#(ii)\n",
-      "#a.c emitter resistance is neglected, voltage gain=(collector resistor)/(emitter resistor)\n",
-      "Av=RC/RE;                           #Voltage gain\n",
-      "\n",
-      "print(\"(ii) Voltage gain=%.2f\"%Av);\n",
-      "\n",
-      "#(iii)\n",
-      "Ap=round(beta*Av,0);                         #Power gain\n",
-      "\n",
-      "#Results\n",
-      "print(\"(iii) Power gain=%d.\"%Ap);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Current gain=49\n",
-        "(ii) Voltage gain=2.14\n",
-        "(iii) Power gain=105.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.20 : Page number 266-267\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=30.0;                     #Collector supply voltage, V\n",
-      "VBE=0.7;                      #Base emitter voltage, V\n",
-      "R1=45.0;                      #Resistor R1, k\u2126\n",
-      "R2=15.0                       #Resistor R2, k\u2126\n",
-      "RC=10.0;                      #Collector resistor,k\u2126\n",
-      "RE=7.5;                       #Emitter resistor, k\u2126\n",
-      "beta=200.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "V2=round(VCC*R2/(R1+R2),1);                  #Voltage across R2, V (Voltage divider rule)\n",
-      "VE=V2;                                       #Voltage across emitter resistor(base-emitter voltage is neglected), V\n",
-      "IE=VE/RE;                                    #Emitter current, mA (OHM's LAW)\n",
-      "re=25/IE;                                    #a.c emitter resistance, ohm\n",
-      "Zin_base=(beta*re)/1000;                     #input impedance of transistor base,k\u2126\n",
-      "R1_R2=(R1*R2)/(R1+R2);                       #Parallel resistance between R1 and R2, k\u2126\n",
-      "Zin=((R1_R2)*Zin_base)/(R1_R2+Zin_base);     #Input impedance of the amplifier circuit, k\u2126\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The input impedance of the amplifier circuit= %.2f k\u2126.\"%Zin);       \n",
-      "\n",
-      "#Note: The input impedance of the amplifier circuit is approximated as 3.45 k\u2126 in the text book, but actually it's 3.46 k\u2126.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input impedance of the amplifier circuit= 3.46 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.21 : Page Number 268-269\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                           #Collector supply voltage, V.\n",
-      "RC=1.5;                             #Collector resistor, k\u2126.\n",
-      "R1=18.0;                            #Resistor R1, k\u2126.\n",
-      "R2=4.7;                             #Resistor R2, k\u2126.\n",
-      "RE1=300.0;                          #Emitter resistor 1, \u2126.\n",
-      "RE2=900.0;                          #Emitter resistor 2, \u2126.\n",
-      "VBE=0.7;                            #Base-emitter voltage, V.\n",
-      "beta=150.0;                         #Base current amplification factor.\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V2=round(VCC*R2/(R1+R2),1);                      #d.c voltage across R2, V. (Voltage divider rule)\n",
-      "VE=round(V2-VBE,1);                              #d.c voltage across RE, V.\n",
-      "IE=round((VE/(RE1+RE2))*1000,2);                 #d.c emitter current, mA.(OHM'S LAW)\n",
-      "re=round(25/IE,1);                               #a.c emitter resistance, \u2126.\n",
-      "Av=RC*1000/(re+RE1);                             #Voltage gain\n",
-      "Zin_base=(beta*(re+RE1))/1000;                   #Input impedance of transistor base, k\u2126.\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage gain of the swamped amplifier= %.2f.\"%Av);\n",
-      "print(\"Input impedance of transistor base of the swamped amplifier= %.2f k\u2126.\"%Zin_base);\n",
-      "\n",
-      "#Note:In the textbook Av is approximated to 4.66and Zin_base to 48.22 kilo ohm, but the actual answers come as 4.67 and 48.21 kilo ohm.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain of the swamped amplifier= 4.67.\n",
-        "Input impedance of transistor base of the swamped amplifier= 48.21 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.22 : Page number 269\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=1.5;                             #Collector resistor, k\u2126.\n",
-      "RE1=300.0;                          #Emitter resistor 1, \u2126.\n",
-      "re=21.5;                            #a.c emitter resistance, \u2126.\n",
-      "\n",
-      "#Calculations\n",
-      "Av=round(RC*1000/(re+RE1),2);                     #Voltage gain.\n",
-      "Av_1=round(RC*1000/(2*re+RE1),2);                 #Voltage gain when re doubles.\n",
-      "change_in_gain=round(Av-Av_1,2);                      #Change in voltage gain.\n",
-      "change_percentage=change_in_gain*100/Av;               #Change percentage\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "if(change_in_gain>0):\n",
-      "    print(\"The percentage change from the original value= %.2f%%(decrease)\"%change_percentage);\n",
-      "else:\n",
-      "    print(\"The percentage change from the original value= %.2f%%(increase)\"%change_percentage);\n",
-      "\n",
-      "\n",
-      "#Note: The percentage has been approximated in the text book as 6.22%, but the answer comes as 6.42%.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage change from the original value= 6.42%(decrease)\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.23 : Page number 269-270\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage, V\n",
-      "VBE=0.7;                      #Base emitter voltage, V\n",
-      "R1=10.0;                      #Resistor R1, kilo ohm\n",
-      "R2=2.2;                       #Resistor R2, kilo ohm\n",
-      "RC=4.0;                       #Collector resistor, kilo ohm\n",
-      "RE=1.1;                       #Emitter resistor, kilo ohm\n",
-      "beta=200.0;                   #Base current amplification factor\n",
-      "RE1=210.0;                    #Emitter resistor 1 of swamped amplifier, ohm.\n",
-      "RE2=900.0;                    #Emitter resistor 2 of swamped amplifier, ohm.\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V2=round(VCC*R2/(R1+R2),1);                      #d.c voltage across R2, V. (Voltage divider rule)\n",
-      "VE=round(V2-VBE,1);                              #d.c voltage across RE, V.\n",
-      "IE=(VE/RE);                                      #d.c emitter current, mA.(OHM'S LAW)\n",
-      "re=25/IE;                                        #a.c emitter resistance, ohm.\n",
-      "\n",
-      "\n",
-      "#(i) Zin_base:\n",
-      "Zin_base_standard=(beta*re)/1000;                      #input impedance of transistor base for standard amplifier , kilo ohm.\n",
-      "Zin_base_swamped=(beta*(re+RE1))/1000;                 #input impedance of transistor base for swamped amplifier, kilo ohm.\n",
-      "\n",
-      "\n",
-      "#(ii) Zin:\n",
-      "#input impedance for standard amplifier circuit\n",
-      "Zin_standard=(((R1*R2)/(R1+R2))*Zin_base_standard)/(Zin_base_standard +((R1*R2)/(R1+R2)));    #kilo ohm\n",
-      "\n",
-      "#input impedance for standard amplifier circuit\n",
-      "Zin_swamped=(((R1*R2)/(R1+R2))*Zin_base_swamped)/(Zin_base_swamped +((R1*R2)/(R1+R2)));       #kilo ohm\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i) input impedance of transistor base for standard amplifier= %d kilo ohm\"%Zin_base_standard);\n",
-      "print(\"    input impedance of transistor base for swamped amplifier= %d kilo ohm\"%Zin_base_swamped);\n",
-      "print(\"(ii) input impedance for standard amplifier= %.2f kilo ohm\"%Zin_standard);\n",
-      "print(\"     input impedance for swamped amplifier= %.2f kilo ohm\"%Zin_swamped);\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) input impedance of transistor base for standard amplifier= 5 kilo ohm\n",
-        "    input impedance of transistor base for swamped amplifier= 47 kilo ohm\n",
-        "(ii) input impedance for standard amplifier= 1.33 kilo ohm\n",
-        "     input impedance for swamped amplifier= 1.74 kilo ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.24 : Page number 270-271\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=4.0;                     #Collector resistor, kilo ohm\n",
-      "re=25.0;                    #a.c emitter resistance, ohm (calculated in example 10.23)\n",
-      "RE_1=210.0;                  #Emitter resistor 1 of swamped amplifier,ohm\n",
-      "\n",
-      "#Calculation\n",
-      "Av_standard=(RC*1000)/re;                       #Voltage gain of standard common emitter amplifier\n",
-      "Av_swamped=(RC*1000)/(re+RE_1);                 #Voltage gain of swamped amplifier\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage gain of standard amplifier=%d.\"%Av_standard);\n",
-      "print(\"The voltage gain of swamped amplifier=%d.\"%Av_swamped);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain of standard amplifier=160.\n",
-        "The voltage gain of swamped amplifier=17.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.26 : Page number 273-274\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_0=1000.0;                  #Open circuit voltage gain\n",
-      "R_in=2.0;                    #Input resistance, kilo ohm\n",
-      "R_out=1.0;                   #Output resistance, ohm\n",
-      "RL=4;                        #Load resistor across the output, ohm\n",
-      "I_2=0.5;                     #Output signal current, A.\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Since A_0*(I_1*R_in) = I_2*(R_out+RL)\n",
-      "I_1=I_2*(R_out+RL)/(A_0*(R_in*1000));           #Input current, A\n",
-      "V_1=I_1*(R_in*1000);                            #Input signal voltage, V\n",
-      "V_1=V_1*1000;                                   #Input signal voltage, mV\n",
-      "\n",
-      "print(\"The required input signal voltage =%.1fmV\"%V_1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required input signal voltage =2.5mV\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.27 : Page number 274\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_0=1000.0;                     #Open circuit voltage gain\n",
-      "R_in=7.0;                       #Input resistance, kilo ohm\n",
-      "R_out=15.0;                     #Output resistance, ohm\n",
-      "RL=35.0;                        #Load resistor across the output, ohm\n",
-      "R_s=3.0;                        #Internal resistance, kilo ohm\n",
-      "E_s=10.0;                       #Input signal voltage, mV.\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "I_1=E_s*(10**-3)/(R_s*1000+R_in*1000);                  #Input current, A\n",
-      "V_1=I_1*(R_in*1000);                                    #Voltage across input resistance, V\n",
-      "\n",
-      "#Since, A_v=V_2/V_1 = A_0*RL/(R_out+RL)\n",
-      "A_v=A_0*RL/(R_out+RL);                                 #Voltage gain\n",
-      "V_2=A_v*V_1;                                            #Outout voltage, V\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "P_2=V_2**2/RL;                              #Output power, W\n",
-      "P_1=V_1**2/(R_in*1000);                            #Input power, W\n",
-      "A_p=round(P_2/P_1,-6);                                #Power gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The magnitude of output voltage = %.1fV\"%V_2);\n",
-      "print(\"The power gain =%de-06.\"%(A_p/10**6));\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The magnitude of output voltage = 4.9V\n",
-        "The power gain =98e-06.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 10.28 : Page number 274-275\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v=80.0;                   #Voltage gain\n",
-      "V_2=1.0;                    #Output voltage, V\n",
-      "A_i=120.0;                  #Current gain\n",
-      "RL=2;                       #Load resistor, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "V_1=(V_2/A_v)*1000;                #Input signal voltage, mV\n",
-      "\n",
-      "#Since, A_i=A0*R_in/(R_out+RL) and A_v=A0*RL/(R_out+RL)\n",
-      "#So, A_v/A_i=RL/R_in\n",
-      "R_in=RL*A_i/A_v;                #Input resistance, kilo ohm\n",
-      "I_1=V_1/R_in;                   #Input current,  \u03bcA\n",
-      "A_p=A_i*A_v;                    #Power gain\n",
-      "\n",
-      "#Results\n",
-      "print(\"Necessary input signal voltage= %.1fmV\"%V_1);\n",
-      "print(\"Input signal current =%.2f  \u03bcA\"%I_1);\n",
-      "print(\"Power gain = %d.\"%A_p);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Necessary input signal voltage= 12.5mV\n",
-        "Input signal current =4.17  \u03bcA\n",
-        "Power gain = 9600.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_2.ipynb
deleted file mode 100755
index 0aa8bc1a..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_2.ipynb
+++ /dev/null
@@ -1,1289 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 10 : SINGLE STAGE TRANSISTOR AMPLIFIERS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Using matplotlib backend: Qt4Agg\n"
-     ]
-    }
-   ],
-   "source": [
-    "%matplotlib "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.2 : page number 243-244"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The value of the emitter capacitor = 1.42 𝜇F\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "#Variable declaration\n",
-    "f_min=2.0;                            #Minimum frequency of operation of amplifier, kHz\n",
-    "f_max=10.0;                           #Maximum frequency of operation of amplifier, kHz\n",
-    "RE=560.0;                             #Emitter resistor, Ω\n",
-    "\n",
-    "#Calculations\n",
-    "#X_CE(Emitter capacitor's capacitive reactance)\n",
-    "#X_CE=1/(2*pi*f_min*CE)=RE/10\n",
-    "#From the above equation.\n",
-    "CE=1/(2*pi*f_min*1000*(RE/10));              #Emitter capacitor, F,\n",
-    "\n",
-    "CE=CE*10**6;                                 #Emitter capacitor, 𝜇F\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print('The value of the emitter capacitor = %.2f 𝜇F'%(CE));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.5: Page number 252-253"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The operating point: VCE=8.55V and IC=2.15mA.\n",
-      "Maximum v_CE=9.62V and maximum i_C=19.25mA\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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Og7rkLmiz7qlUGKwB7iXNVbw4W/Z0ROzS5UpLLc5hUPfcBW1WvkqFwXDgTNLo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\nRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fb5303cd278>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pylab as plt\n",
-    "\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=15.0;                         #Collector supply voltage in V\n",
-    "VBE=0.7;                          #Base-emitter voltage, V\n",
-    "R1=10.0;                          #Resistor R1, kΩ\n",
-    "R2=5.0;                           #Resistor R2, kΩ\n",
-    "RC=1.0;                           #Collector resistor, kΩ\n",
-    "RE=2.0;                           #Emitter resistor, kΩ\n",
-    "RL=1.0;                           #Load resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "#For d.c load line, from the equation: VCE=VCC-IC*(RC+RE),\n",
-    "#VCE is maximum when IC=0 and IC is maximum when VCE=0.\n",
-    "VCE_max=VCC;                                        #Maximum collector-emitter voltage, V\n",
-    "IC_max=VCC/(RC+RE);                                 #Maximum collector current, mA\n",
-    "\n",
-    "\n",
-    "#plot\n",
-    "VCE_plot=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-    "IC_plot=[((VCC-i)/(RC+RE)) for i in (VCE_plot[:])];      #Plot variable for I_C\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.xlim(0,20)\n",
-    "plt.ylim(0,6)\n",
-    "plt.plot(VCE_plot,IC_plot);\n",
-    "plt.xlabel(\"VCE(V)\");\n",
-    "plt.ylabel(\"IC(mA)\");\n",
-    "plt.title(\"d.c load line\");\n",
-    "\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "#For operating point:\n",
-    "#Assuming VCC drops almost completely across R1 and R2,\n",
-    "V2=VCC*R2/(R1+R2);                                  #Voltage across resistor R2, V\n",
-    "IE=(V2-VBE)/RE;                                     #Emitter current, mA\n",
-    "IC=IE;                                              #Collector current, mA\n",
-    "VCE=VCC-IC*(RC+RE);                                 #Collector-emitter voltage , V\n",
-    "\n",
-    "print(\"The operating point: VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n",
-    "\n",
-    "\n",
-    "#(iii)\n",
-    "#For a.c load line\n",
-    "RAC=(RC*RL)/(RC+RL);                                #a.c load, kΩ\n",
-    "VCE_ac_max=VCE+IC*RAC;                              #Maximum collector-emitter voltage, V\n",
-    "IC_ac_max=IC+VCE/RAC;                               #Maximum collector current, mA\n",
-    "print(\"Maximum v_CE=%.2fV and maximum i_C=%.2fmA\"%(VCE_ac_max,IC_ac_max));\n",
-    "\n",
-    "#plot\n",
-    "vCE_plot=[0,VCE_ac_max];          #Plot variable for V_CE\n",
-    "iC_plot=[IC_ac_max,0];      #Plot variable for I_C\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.xlim(0,10)\n",
-    "plt.ylim(0,20)\n",
-    "plt.plot(vCE_plot,iC_plot);\n",
-    "plt.xlabel(\"vCE(V)\");\n",
-    "plt.ylabel(\"iC(mA)\");\n",
-    "plt.title(\"a.c load line\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.6: Page number 253-254"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fb50cf60f28>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pylab as p\n",
-    "\n",
-    "#Variable declaration\n",
-    "RC=10;                          #Collector resistor, kΩ\n",
-    "RL=30;                          #Load resistor, kΩ\n",
-    "VCC=20;                         #Collector supply voltage, V\n",
-    "IC=1;                           #Collector current, mA\n",
-    "VCE=10;                         #Collector-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#For d.c load line:\n",
-    "#From the equation: VCE=VCC-IC*(RC+RE),\n",
-    "#When VCE=0, IC is  maximum.\n",
-    "#Emitter resistor is neglected, assuming it as negligible\n",
-    "IC_max=VCC/RC;                      #Maximum collector current, mA\n",
-    "\n",
-    "#And, when IC=0, VCE is maximum\n",
-    "VCE_max=VCC;                        #Maximum collector-emitter voltage, V\n",
-    "\n",
-    "#plot\n",
-    "p.subplot(211)\n",
-    "p.xlim(0,20)\n",
-    "p.ylim(0,5)\n",
-    "VCE_plot=[0,VCE_max];          #Plot variable for V_CE\n",
-    "IC_plot=[IC_max,0];            #Plot variable for I_C\n",
-    "\n",
-    "p.plot(VCE_plot,IC_plot);\n",
-    "p.xlabel(\"VCE(V)\");\n",
-    "p.ylabel(\"IC(mA)\");\n",
-    "p.title(\"d.c load line\");\n",
-    "\n",
-    "\n",
-    "#For a.c load line:\n",
-    "RAC=(RC*RL)/(RC+RL);                #a.c Load resistor, kΩ\n",
-    "\n",
-    "VCE_ac_max=VCE+IC*RAC;                 #Maximum collector-emitter voltage, V\n",
-    "IC_ac_max=IC+ VCE/RAC;                #Maximum collector current, mA\n",
-    "\n",
-    "#plot\n",
-    "p.subplot(212)\n",
-    "p.xlim([0,25])\n",
-    "p.ylim([0,5])\n",
-    "vCE_plot=[0,VCE_ac_max];          #Plot variable for V_CE\n",
-    "iC_plot=[IC_ac_max,0];      #Plot variable for I_C\n",
-    "\n",
-    "p.plot(vCE_plot,iC_plot);\n",
-    "p.xlabel(\"vCE(V)\");\n",
-    "p.ylabel(\"iC(mA)\");\n",
-    "p.title(\"a.c load line\");\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.7 : Page number 254-255"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fb5280aca20>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pylab as p\n",
-    "\n",
-    "#Variabe declaration\n",
-    "VCE_Q=8.0;           #Q-point collector emitter voltage, V\n",
-    "IC_Q=1;               #Q-point collector current, mA\n",
-    "ic_positive_peak=1.5;        #Collector current at positive peak of signal, mA\n",
-    "ic_negative_peak=0.5;         #Collector current at negative peak of signal, mA\n",
-    "vce_positive_peak=7;          #Collector emitter voltage at positive peak of signal, V\n",
-    "vce_negative_peak=9;          #Collector emitter voltage at negative peak of signal, V\n",
-    "\n",
-    "#Plot\n",
-    "vce_plot=[vce_positive_peak,vce_negative_peak];    #Plot variable of vce\n",
-    "ic_plot=[ic_positive_peak,ic_negative_peak];    #Plot variable of ic\n",
-    "\n",
-    "p.xlim(0,10)\n",
-    "p.ylim(0,2)\n",
-    "p.plot(vce_plot,ic_plot);\n",
-    "p.xlabel(\"vCE(V)\");\n",
-    "p.ylabel(\"iC(mA)\");\n",
-    "p.title(\"a.c load line\");\n",
-    "p.grid();\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.8 : Page number 256"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain= 24.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                 #Collector supply voltage, V\n",
-    "RC=2.0;                   #Collector resistor, kΩ\n",
-    "Rin=1.0;                  #Input resistance, kΩ\n",
-    "beta=60.0;                #Base current amplification factor\n",
-    "RL=0.5;                   #Load resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "RAC=(RC*RL)/(RC+RL);                #a.c load resistor, kΩ\n",
-    "Av=beta*(RAC/Rin);                  #Voltage gain\n",
-    "\n",
-    "#Results\n",
-    "print(\"Voltage gain= %d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.9 : Page number 256"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage= 200mV.\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "V_in=1.0;                      #Input voltage , mV\n",
-    "RC=10.0;                      #Collector resistor, kΩ\n",
-    "Rin=2.5;                      #Input resistance, kΩ\n",
-    "beta=100.0;                   #Base current amplification factor\n",
-    "RL=10.0;                      #Load resistor, kΩ\n",
-    "\n",
-    "#Calculations\n",
-    "RAC=(RC*RL)/(RC+RL);                #Effective load, kΩ\n",
-    "Av=beta*(RAC/Rin);                  #Voltage gain\n",
-    "\n",
-    "V_out=V_in*Av;                      #Output voltage, V\n",
-    "\n",
-    "#Results\n",
-    "print(\"Output voltage= %dmV.\"%V_out);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.10 : Page number 256-257"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Beta= 100.\n",
-      "Input impedance=2 kΩ.\n",
-      "a.c load=3.3 kΩ.\n",
-      "Voltage gain= 165.\n",
-      "Power gain=16500.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "change_in_IB=10.0;                    #Change in base current, 𝜇A\n",
-    "change_in_IC=1.0;                     #Change in collector current, mA\n",
-    "change_in_VBE=0.02;                   #Change in Base-emitter voltage, V\n",
-    "RC=5.0;                               #Collector resistor, kΩ\n",
-    "RL=10.0;                              #Emitter resistor, kΩ\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "beta=(change_in_IC*1000)/change_in_IB;         #Base current amplification factor\n",
-    "\n",
-    "#(ii)\n",
-    "Rin=(change_in_VBE/change_in_IB)*1000;         #Input impedance, kΩ\n",
-    "\n",
-    "#(iii)\n",
-    "RAC=round((RC*RL)/(RC+RL),1);                            #a.c load, kΩ\n",
-    "\n",
-    "#(iv)\n",
-    "Av=beta*RAC/Rin;                                #Voltage gain\n",
-    "\n",
-    "#(v)\n",
-    "Ap=beta*Av;                                     #Power gain\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"Beta= %d.\"%beta);\n",
-    "print(\"Input impedance=%d kΩ.\"%Rin);\n",
-    "print(\"a.c load=%.1f kΩ.\"%RAC);\n",
-    "print(\"Voltage gain= %d.\"%Av);\n",
-    "print(\"Power gain=%d.\"%Ap);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.11 : Page number 257"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=200mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "beta=50.0;                      #Base current amplification factor\n",
-    "RC=3.0;                         #Collector resistor,kΩ\n",
-    "RL=6.0;                         #Load resistor, kΩ\n",
-    "Rin=0.5;                        #Input impedance, kΩ\n",
-    "Vin=1;                          #Input voltage,  mV\n",
-    "\n",
-    "#Calculation\n",
-    "RAC=(RC*RL)/(RC+RL);            #a.c load, kΩ\n",
-    "Av=beta*RAC/Rin;                #Voltage gain\n",
-    "Vout=Vin*Av;                    #Output voltage, V\n",
-    "\n",
-    "#Results\n",
-    "print(\"Output voltage=%dmV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.12 : Page number 257-258"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The circuit is not operating properly.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VT=6.0;                   #Collector potential, V\n",
-    "R1=1.0;                   #Resistor R1, kΩ\n",
-    "R2=2.0;                   #Resistor R2, kΩ\n",
-    "VB_found=4.0;             #Measured base voltage, V\n",
-    "\n",
-    "#Calculations\n",
-    "VB=(VT*R1)/(R1+R2);                     #Theoretical base voltage, V\n",
-    "\n",
-    "if(VB_found==VB):\n",
-    "    print(\"The circuit is operating properly.\");\n",
-    "else:\n",
-    "    print(\"The circuit is not operating properly.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.13 : Page number 258-259"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a.c emitter resistance= 38.46 Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                   #Collector supply voltage, V\n",
-    "R1=40.0;                    #Resistor R1, kΩ\n",
-    "R2=10.0;                    #Resistor R2, kΩ\n",
-    "RC=6.0;                     #Collector resistor, kΩ\n",
-    "RE=2.0;                     #Emitter resistor, kΩ\n",
-    "beta=80;                    #Base current amplification factor\n",
-    "VBE=0.7;                    #Base emitter voltage, V\n",
-    "\n",
-    "#Calculations\n",
-    "V2=(VCC*R2)/(R1+R2);            #Voltage across resistor R2, V\n",
-    "VE=V2-VBE;                      #Emitter voltage, V\n",
-    "IE=VE/RE;                       #Emitter current,  mA\n",
-    "re=25/IE;                       #a.c emitter resistance, Ω\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"a.c emitter resistance= %.2f Ω.\"%re);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.14  : Page number 262-263"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)Voltage gain= 360.\n",
-      "(ii)Voltage gain= 5.37.\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "VCC=20.0;                     #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base-emitter voltage, V\n",
-    "R1=150.0;                     #Resistor R1, kΩ\n",
-    "R2=20.0                       #Resistor R2, kΩ\n",
-    "RC=12.0;                      #Collector resistor, kΩ\n",
-    "RE=2.2;                       #Emitter resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),2);                  #Voltage across R2, V\n",
-    "VE=round(V2-VBE,2);                          #Voltage across emitter resistor, V\n",
-    "IE=round(VE/RE,2);                           #Emitter current, mA\n",
-    "re=round(25/IE,1);                           #a.c emitter resistance, Ω\n",
-    "\n",
-    "\n",
-    "#(i)\n",
-    "#CE(emitter capacitor) connected in the circuit:\n",
-    "Av=(RC*1000)/re;                                   #Voltage gain for emitter capacitor connected.\n",
-    "\n",
-    "print(\"(i)Voltage gain= %d.\"%Av);\n",
-    "\n",
-    "#(ii)\n",
-    "#CE(emitter capacitor) removed from the circuit:\n",
-    "Av=(RC*1000)/(re+RE*1000);                          #Voltage gain for emitter capacitor removed.\n",
-    "\n",
-    "print(\"(ii)Voltage gain= %.2f.\"%Av);\n",
-    "\n",
-    "#Note: The answer in the text book has been approximated to 5.38 but it's actually coming 5.37.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.15 : Page number 263"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain= 120.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=6.0;                       #Collector resistor, kΩ\n",
-    "RL=12.0;                      #Load resistor, kΩ\n",
-    "re=33.3;                      #a.c emitter resistance, Ω\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "RAC=RC*RL/(RC+RL);              #a.c effective load, kΩ\n",
-    "Av=RAC*1000/re;                 #Voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"Voltage gain= %d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.16 : Page number 263-264"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) a.c emitter resistance=250 Ω.\n",
-      "(ii) Voltage gain =80.\n",
-      "(iii) d.c voltage across input capacitor= 1V and emitter capacitor=0.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=9.0;                      #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base-emitter voltage, V\n",
-    "R1=240.0;                     #Resistor R1, kΩ\n",
-    "R2=30.0                       #Resistor R2, kΩ\n",
-    "RC=20.0;                      #Collector resistor, kΩ\n",
-    "RE=3.0;                       #Emitter resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "V2=round(VCC*R2/(R1+R2),1);                  #Voltage across R2, V\n",
-    "VE=round(V2-VBE,1);                          #Voltage across emitter resistor, V\n",
-    "IE=round(VE/RE,1);                           #Emitter current, mA\n",
-    "re=25/IE;                                    #a.c emitter resistance, Ω\n",
-    "\n",
-    "#(ii)\n",
-    "Av=RC*1000/re;                                   #Voltage gain\n",
-    "\n",
-    "#(iii)\n",
-    "V_C_in=V2;                                  #d.c voltage across input capacitor, V\n",
-    "V_C_E=VE;                                   #d.c vooltage across emitter capacitor, V\n",
-    "\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"(i) a.c emitter resistance=%d Ω.\"%re);\n",
-    "print(\"(ii) Voltage gain =%d.\"%Av);\n",
-    "print(\"(iii) d.c voltage across input capacitor= %dV and emitter capacitor=%.1fV.\"%(V_C_in,V_C_E));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.17 : Page number 264-265"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) D.C bias levels: V2=3V, VE=2.3V, IE=2.3mA, IC=2.3mA, IB=0.023mA and VC=10.4V.\n",
-      "(ii) D.c voltage across: Cin=3V and CE=2.3V and CC=10.4V.\n",
-      "(iii) a.c emitter resistance=10.9Ω.\n",
-      "(iv) Voltage gain=61.2.\n",
-      "(v) VC>VE. Therefore, the transistor is in active state.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15.0;                      #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base-emitter voltage, V\n",
-    "R1=40.0;                     #Resistor R1, kΩ\n",
-    "R2=10.0                       #Resistor R2, kΩ\n",
-    "RC=2.0;                      #Collector resistor, kΩ\n",
-    "RE=1.0;                       #Emitter resistor, kΩ\n",
-    "RL=1.0;                       #Load resistor, kΩ\n",
-    "beta=100;                     #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) D.C bias levels\n",
-    "V2=VCC*R2/(R1+R2);                  #Voltage across R2, V\n",
-    "VE=round(V2-VBE,1);                 #Voltage across emitter resistor, V\n",
-    "IE=round(VE/RE,1);                  #Emitter current, mA\n",
-    "IC=IE;                              #Collector current, mA\n",
-    "IB=IC/beta;                         #Base  current, mA\n",
-    "VC=VCC-IC*RC;                       #Collector voltage, V\n",
-    "print(\"(i) D.C bias levels: V2=%dV, VE=%.1fV, IE=%.1fmA, IC=%.1fmA, IB=%.3fmA and VC=%.1fV.\"%(V2,VE,IE,IC,IB,VC));\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "Cin_V=V2;                            #Voltage across Cin capacitor, V\n",
-    "CE_V=VE;                             #Voltage across CE capacitor, V \n",
-    "CC_V=VC;                             #Voltage across CC capacitor, V\n",
-    "print(\"(ii) D.c voltage across: Cin=%dV and CE=%.1fV and CC=%.1fV.\"%(Cin_V,CE_V,CC_V));\n",
-    "\n",
-    "#(iii)\n",
-    "re=round(25/IE,1);                           #a.c emitter resistance, Ω\n",
-    "print(\"(iii) a.c emitter resistance=%.1fΩ.\"%re);\n",
-    "\n",
-    "\n",
-    "#(iv)\n",
-    "RAC=round(RC*RL/(RC+RL),3);                #Total a.c collector resistance, kΩ\n",
-    "Av=RAC/(re/1000);                        #Voltage gain\n",
-    "print(\"(iv) Voltage gain=%.1f.\"%Av);\n",
-    "\n",
-    "#(v)\n",
-    "print(\"(v) VC>VE. Therefore, the transistor is in active state.\" );\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.18 : page number 265"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The power gain = 26400 and output power = 1.584W.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=132.0;                   #Voltage gain\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "P_in=60.0;                     #Input power, 𝜇W\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "Ap=beta*Av;                          #Power gain\n",
-    "P_out=Ap*(P_in/10**6);                #Output power, W\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"The power gain = %d and output power = %.3fW.\"%(Ap,P_out));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.19 : page number 265-266"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Current gain=49\n",
-      "(ii) Voltage gain=2.14\n",
-      "(iii) Power gain=105.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IB=200.0;                       #Base current, microampere\n",
-    "IE=10.0;                        #Emitter current, mA\n",
-    "R1=27.0;                      #Resistor R1, kilo ohm\n",
-    "R2=13.0                       #Resistor R2, kilo ohm\n",
-    "RC=4.7;                       #Collector resistor, kilo ohm\n",
-    "RE=2.2;                       #Emitter resistor, kilo ohm\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "IC=IE-(IB/1000);                              #Collector current, mA\n",
-    "beta=IC/(IB/1000);                            #Current gain\n",
-    "\n",
-    "print(\"(i) Current gain=%d\"%beta);\n",
-    "\n",
-    "#(ii)\n",
-    "#a.c emitter resistance is neglected, voltage gain=(collector resistor)/(emitter resistor)\n",
-    "Av=RC/RE;                           #Voltage gain\n",
-    "\n",
-    "print(\"(ii) Voltage gain=%.2f\"%Av);\n",
-    "\n",
-    "#(iii)\n",
-    "Ap=round(beta*Av,0);                         #Power gain\n",
-    "\n",
-    "#Results\n",
-    "print(\"(iii) Power gain=%d.\"%Ap);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.20 : Page number 266-267"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance of the amplifier circuit= 3.46 kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=30.0;                     #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base emitter voltage, V\n",
-    "R1=45.0;                      #Resistor R1, kΩ\n",
-    "R2=15.0                       #Resistor R2, kΩ\n",
-    "RC=10.0;                      #Collector resistor,kΩ\n",
-    "RE=7.5;                       #Emitter resistor, kΩ\n",
-    "beta=200.0;                   #Base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),1);                  #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2;                                       #Voltage across emitter resistor(base-emitter voltage is neglected), V\n",
-    "IE=VE/RE;                                    #Emitter current, mA (OHM's LAW)\n",
-    "re=25/IE;                                    #a.c emitter resistance, ohm\n",
-    "Zin_base=(beta*re)/1000;                     #input impedance of transistor base,kΩ\n",
-    "R1_R2=(R1*R2)/(R1+R2);                       #Parallel resistance between R1 and R2, kΩ\n",
-    "Zin=((R1_R2)*Zin_base)/(R1_R2+Zin_base);     #Input impedance of the amplifier circuit, kΩ\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance of the amplifier circuit= %.2f kΩ.\"%Zin);       \n",
-    "\n",
-    "#Note: The input impedance of the amplifier circuit is approximated as 3.45 kΩ in the text book, but actually it's 3.46 kΩ.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.21 : Page Number 268-269"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the swamped amplifier= 4.67.\n",
-      "Input impedance of transistor base of the swamped amplifier= 48.21 kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                           #Collector supply voltage, V.\n",
-    "RC=1.5;                             #Collector resistor, kΩ.\n",
-    "R1=18.0;                            #Resistor R1, kΩ.\n",
-    "R2=4.7;                             #Resistor R2, kΩ.\n",
-    "RE1=300.0;                          #Emitter resistor 1, Ω.\n",
-    "RE2=900.0;                          #Emitter resistor 2, Ω.\n",
-    "VBE=0.7;                            #Base-emitter voltage, V.\n",
-    "beta=150.0;                         #Base current amplification factor.\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),1);                      #d.c voltage across R2, V. (Voltage divider rule)\n",
-    "VE=round(V2-VBE,1);                              #d.c voltage across RE, V.\n",
-    "IE=round((VE/(RE1+RE2))*1000,2);                 #d.c emitter current, mA.(OHM'S LAW)\n",
-    "re=round(25/IE,1);                               #a.c emitter resistance, Ω.\n",
-    "Av=RC*1000/(re+RE1);                             #Voltage gain\n",
-    "Zin_base=(beta*(re+RE1))/1000;                   #Input impedance of transistor base, kΩ.\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"The voltage gain of the swamped amplifier= %.2f.\"%Av);\n",
-    "print(\"Input impedance of transistor base of the swamped amplifier= %.2f kΩ.\"%Zin_base);\n",
-    "\n",
-    "#Note:In the textbook Av is approximated to 4.66and Zin_base to 48.22 kilo ohm, but the actual answers come as 4.67 and 48.21 kilo ohm.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.22 : Page number 269"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The percentage change from the original value= 6.42%(decrease)\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=1.5;                             #Collector resistor, kΩ.\n",
-    "RE1=300.0;                          #Emitter resistor 1, Ω.\n",
-    "re=21.5;                            #a.c emitter resistance, Ω.\n",
-    "\n",
-    "#Calculations\n",
-    "Av=round(RC*1000/(re+RE1),2);                     #Voltage gain.\n",
-    "Av_1=round(RC*1000/(2*re+RE1),2);                 #Voltage gain when re doubles.\n",
-    "change_in_gain=round(Av-Av_1,2);                      #Change in voltage gain.\n",
-    "change_percentage=change_in_gain*100/Av;               #Change percentage\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "if(change_in_gain>0):\n",
-    "    print(\"The percentage change from the original value= %.2f%%(decrease)\"%change_percentage);\n",
-    "else:\n",
-    "    print(\"The percentage change from the original value= %.2f%%(increase)\"%change_percentage);\n",
-    "\n",
-    "\n",
-    "#Note: The percentage has been approximated in the text book as 6.22%, but the answer comes as 6.42%.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.23 : Page number 269-270"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) input impedance of transistor base for standard amplifier= 5 kilo ohm\n",
-      "    input impedance of transistor base for swamped amplifier= 47 kilo ohm\n",
-      "(ii) input impedance for standard amplifier= 1.33 kilo ohm\n",
-      "     input impedance for swamped amplifier= 1.74 kilo ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                     #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base emitter voltage, V\n",
-    "R1=10.0;                      #Resistor R1, kilo ohm\n",
-    "R2=2.2;                       #Resistor R2, kilo ohm\n",
-    "RC=4.0;                       #Collector resistor, kilo ohm\n",
-    "RE=1.1;                       #Emitter resistor, kilo ohm\n",
-    "beta=200.0;                   #Base current amplification factor\n",
-    "RE1=210.0;                    #Emitter resistor 1 of swamped amplifier, ohm.\n",
-    "RE2=900.0;                    #Emitter resistor 2 of swamped amplifier, ohm.\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),1);                      #d.c voltage across R2, V. (Voltage divider rule)\n",
-    "VE=round(V2-VBE,1);                              #d.c voltage across RE, V.\n",
-    "IE=(VE/RE);                                      #d.c emitter current, mA.(OHM'S LAW)\n",
-    "re=25/IE;                                        #a.c emitter resistance, ohm.\n",
-    "\n",
-    "\n",
-    "#(i) Zin_base:\n",
-    "Zin_base_standard=(beta*re)/1000;                      #input impedance of transistor base for standard amplifier , kilo ohm.\n",
-    "Zin_base_swamped=(beta*(re+RE1))/1000;                 #input impedance of transistor base for swamped amplifier, kilo ohm.\n",
-    "\n",
-    "\n",
-    "#(ii) Zin:\n",
-    "#input impedance for standard amplifier circuit\n",
-    "Zin_standard=(((R1*R2)/(R1+R2))*Zin_base_standard)/(Zin_base_standard +((R1*R2)/(R1+R2)));    #kilo ohm\n",
-    "\n",
-    "#input impedance for standard amplifier circuit\n",
-    "Zin_swamped=(((R1*R2)/(R1+R2))*Zin_base_swamped)/(Zin_base_swamped +((R1*R2)/(R1+R2)));       #kilo ohm\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"(i) input impedance of transistor base for standard amplifier= %d kilo ohm\"%Zin_base_standard);\n",
-    "print(\"    input impedance of transistor base for swamped amplifier= %d kilo ohm\"%Zin_base_swamped);\n",
-    "print(\"(ii) input impedance for standard amplifier= %.2f kilo ohm\"%Zin_standard);\n",
-    "print(\"     input impedance for swamped amplifier= %.2f kilo ohm\"%Zin_swamped);\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.24 : Page number 270-271"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of standard amplifier=160.\n",
-      "The voltage gain of swamped amplifier=17.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=4.0;                     #Collector resistor, kilo ohm\n",
-    "re=25.0;                    #a.c emitter resistance, ohm (calculated in example 10.23)\n",
-    "RE_1=210.0;                  #Emitter resistor 1 of swamped amplifier,ohm\n",
-    "\n",
-    "#Calculation\n",
-    "Av_standard=(RC*1000)/re;                       #Voltage gain of standard common emitter amplifier\n",
-    "Av_swamped=(RC*1000)/(re+RE_1);                 #Voltage gain of swamped amplifier\n",
-    "\n",
-    "#Results\n",
-    "print(\"The voltage gain of standard amplifier=%d.\"%Av_standard);\n",
-    "print(\"The voltage gain of swamped amplifier=%d.\"%Av_swamped);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.26 : Page number 273-274"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The required input signal voltage =2.5mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_0=1000.0;                  #Open circuit voltage gain\n",
-    "R_in=2.0;                    #Input resistance, kilo ohm\n",
-    "R_out=1.0;                   #Output resistance, ohm\n",
-    "RL=4;                        #Load resistor across the output, ohm\n",
-    "I_2=0.5;                     #Output signal current, A.\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#Since A_0*(I_1*R_in) = I_2*(R_out+RL)\n",
-    "I_1=I_2*(R_out+RL)/(A_0*(R_in*1000));           #Input current, A\n",
-    "V_1=I_1*(R_in*1000);                            #Input signal voltage, V\n",
-    "V_1=V_1*1000;                                   #Input signal voltage, mV\n",
-    "\n",
-    "print(\"The required input signal voltage =%.1fmV\"%V_1);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.27 : Page number 274"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The magnitude of output voltage = 4.9V\n",
-      "The power gain =98e-06.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_0=1000.0;                     #Open circuit voltage gain\n",
-    "R_in=7.0;                       #Input resistance, kilo ohm\n",
-    "R_out=15.0;                     #Output resistance, ohm\n",
-    "RL=35.0;                        #Load resistor across the output, ohm\n",
-    "R_s=3.0;                        #Internal resistance, kilo ohm\n",
-    "E_s=10.0;                       #Input signal voltage, mV.\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "I_1=E_s*(10**-3)/(R_s*1000+R_in*1000);                  #Input current, A\n",
-    "V_1=I_1*(R_in*1000);                                    #Voltage across input resistance, V\n",
-    "\n",
-    "#Since, A_v=V_2/V_1 = A_0*RL/(R_out+RL)\n",
-    "A_v=A_0*RL/(R_out+RL);                                 #Voltage gain\n",
-    "V_2=A_v*V_1;                                            #Outout voltage, V\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "P_2=V_2**2/RL;                              #Output power, W\n",
-    "P_1=V_1**2/(R_in*1000);                            #Input power, W\n",
-    "A_p=round(P_2/P_1,-6);                                #Power gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The magnitude of output voltage = %.1fV\"%V_2);\n",
-    "print(\"The power gain =%de-06.\"%(A_p/10**6));\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.28 : Page number 274-275"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Necessary input signal voltage= 12.5mV\n",
-      "Input signal current =4.17  μA\n",
-      "Power gain = 9600.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_v=80.0;                   #Voltage gain\n",
-    "V_2=1.0;                    #Output voltage, V\n",
-    "A_i=120.0;                  #Current gain\n",
-    "RL=2;                       #Load resistor, kilo ohm\n",
-    "\n",
-    "#Calculation\n",
-    "V_1=(V_2/A_v)*1000;                #Input signal voltage, mV\n",
-    "\n",
-    "#Since, A_i=A0*R_in/(R_out+RL) and A_v=A0*RL/(R_out+RL)\n",
-    "#So, A_v/A_i=RL/R_in\n",
-    "R_in=RL*A_i/A_v;                #Input resistance, kilo ohm\n",
-    "I_1=V_1/R_in;                   #Input current,  μA\n",
-    "A_p=A_i*A_v;                    #Power gain\n",
-    "\n",
-    "#Results\n",
-    "print(\"Necessary input signal voltage= %.1fmV\"%V_1);\n",
-    "print(\"Input signal current =%.2f  μA\"%I_1);\n",
-    "print(\"Power gain = %d.\"%A_p);\n",
-    "\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_3.ipynb
deleted file mode 100755
index 0aa8bc1a..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_3.ipynb
+++ /dev/null
@@ -1,1289 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 10 : SINGLE STAGE TRANSISTOR AMPLIFIERS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Using matplotlib backend: Qt4Agg\n"
-     ]
-    }
-   ],
-   "source": [
-    "%matplotlib "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.2 : page number 243-244"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The value of the emitter capacitor = 1.42 𝜇F\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "#Variable declaration\n",
-    "f_min=2.0;                            #Minimum frequency of operation of amplifier, kHz\n",
-    "f_max=10.0;                           #Maximum frequency of operation of amplifier, kHz\n",
-    "RE=560.0;                             #Emitter resistor, Ω\n",
-    "\n",
-    "#Calculations\n",
-    "#X_CE(Emitter capacitor's capacitive reactance)\n",
-    "#X_CE=1/(2*pi*f_min*CE)=RE/10\n",
-    "#From the above equation.\n",
-    "CE=1/(2*pi*f_min*1000*(RE/10));              #Emitter capacitor, F,\n",
-    "\n",
-    "CE=CE*10**6;                                 #Emitter capacitor, 𝜇F\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print('The value of the emitter capacitor = %.2f 𝜇F'%(CE));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.5: Page number 252-253"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The operating point: VCE=8.55V and IC=2.15mA.\n",
-      "Maximum v_CE=9.62V and maximum i_C=19.25mA\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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Og7rkLmiz7qlUGKwB7iXNVbw4W/Z0ROzS5UpLLc5hUPfcBW1WvkqFwXDgTNLo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\nRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fb5303cd278>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pylab as plt\n",
-    "\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=15.0;                         #Collector supply voltage in V\n",
-    "VBE=0.7;                          #Base-emitter voltage, V\n",
-    "R1=10.0;                          #Resistor R1, kΩ\n",
-    "R2=5.0;                           #Resistor R2, kΩ\n",
-    "RC=1.0;                           #Collector resistor, kΩ\n",
-    "RE=2.0;                           #Emitter resistor, kΩ\n",
-    "RL=1.0;                           #Load resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "#For d.c load line, from the equation: VCE=VCC-IC*(RC+RE),\n",
-    "#VCE is maximum when IC=0 and IC is maximum when VCE=0.\n",
-    "VCE_max=VCC;                                        #Maximum collector-emitter voltage, V\n",
-    "IC_max=VCC/(RC+RE);                                 #Maximum collector current, mA\n",
-    "\n",
-    "\n",
-    "#plot\n",
-    "VCE_plot=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-    "IC_plot=[((VCC-i)/(RC+RE)) for i in (VCE_plot[:])];      #Plot variable for I_C\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.xlim(0,20)\n",
-    "plt.ylim(0,6)\n",
-    "plt.plot(VCE_plot,IC_plot);\n",
-    "plt.xlabel(\"VCE(V)\");\n",
-    "plt.ylabel(\"IC(mA)\");\n",
-    "plt.title(\"d.c load line\");\n",
-    "\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "#For operating point:\n",
-    "#Assuming VCC drops almost completely across R1 and R2,\n",
-    "V2=VCC*R2/(R1+R2);                                  #Voltage across resistor R2, V\n",
-    "IE=(V2-VBE)/RE;                                     #Emitter current, mA\n",
-    "IC=IE;                                              #Collector current, mA\n",
-    "VCE=VCC-IC*(RC+RE);                                 #Collector-emitter voltage , V\n",
-    "\n",
-    "print(\"The operating point: VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n",
-    "\n",
-    "\n",
-    "#(iii)\n",
-    "#For a.c load line\n",
-    "RAC=(RC*RL)/(RC+RL);                                #a.c load, kΩ\n",
-    "VCE_ac_max=VCE+IC*RAC;                              #Maximum collector-emitter voltage, V\n",
-    "IC_ac_max=IC+VCE/RAC;                               #Maximum collector current, mA\n",
-    "print(\"Maximum v_CE=%.2fV and maximum i_C=%.2fmA\"%(VCE_ac_max,IC_ac_max));\n",
-    "\n",
-    "#plot\n",
-    "vCE_plot=[0,VCE_ac_max];          #Plot variable for V_CE\n",
-    "iC_plot=[IC_ac_max,0];      #Plot variable for I_C\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.xlim(0,10)\n",
-    "plt.ylim(0,20)\n",
-    "plt.plot(vCE_plot,iC_plot);\n",
-    "plt.xlabel(\"vCE(V)\");\n",
-    "plt.ylabel(\"iC(mA)\");\n",
-    "plt.title(\"a.c load line\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.6: Page number 253-254"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fb50cf60f28>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pylab as p\n",
-    "\n",
-    "#Variable declaration\n",
-    "RC=10;                          #Collector resistor, kΩ\n",
-    "RL=30;                          #Load resistor, kΩ\n",
-    "VCC=20;                         #Collector supply voltage, V\n",
-    "IC=1;                           #Collector current, mA\n",
-    "VCE=10;                         #Collector-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#For d.c load line:\n",
-    "#From the equation: VCE=VCC-IC*(RC+RE),\n",
-    "#When VCE=0, IC is  maximum.\n",
-    "#Emitter resistor is neglected, assuming it as negligible\n",
-    "IC_max=VCC/RC;                      #Maximum collector current, mA\n",
-    "\n",
-    "#And, when IC=0, VCE is maximum\n",
-    "VCE_max=VCC;                        #Maximum collector-emitter voltage, V\n",
-    "\n",
-    "#plot\n",
-    "p.subplot(211)\n",
-    "p.xlim(0,20)\n",
-    "p.ylim(0,5)\n",
-    "VCE_plot=[0,VCE_max];          #Plot variable for V_CE\n",
-    "IC_plot=[IC_max,0];            #Plot variable for I_C\n",
-    "\n",
-    "p.plot(VCE_plot,IC_plot);\n",
-    "p.xlabel(\"VCE(V)\");\n",
-    "p.ylabel(\"IC(mA)\");\n",
-    "p.title(\"d.c load line\");\n",
-    "\n",
-    "\n",
-    "#For a.c load line:\n",
-    "RAC=(RC*RL)/(RC+RL);                #a.c Load resistor, kΩ\n",
-    "\n",
-    "VCE_ac_max=VCE+IC*RAC;                 #Maximum collector-emitter voltage, V\n",
-    "IC_ac_max=IC+ VCE/RAC;                #Maximum collector current, mA\n",
-    "\n",
-    "#plot\n",
-    "p.subplot(212)\n",
-    "p.xlim([0,25])\n",
-    "p.ylim([0,5])\n",
-    "vCE_plot=[0,VCE_ac_max];          #Plot variable for V_CE\n",
-    "iC_plot=[IC_ac_max,0];      #Plot variable for I_C\n",
-    "\n",
-    "p.plot(vCE_plot,iC_plot);\n",
-    "p.xlabel(\"vCE(V)\");\n",
-    "p.ylabel(\"iC(mA)\");\n",
-    "p.title(\"a.c load line\");\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.7 : Page number 254-255"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fb5280aca20>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pylab as p\n",
-    "\n",
-    "#Variabe declaration\n",
-    "VCE_Q=8.0;           #Q-point collector emitter voltage, V\n",
-    "IC_Q=1;               #Q-point collector current, mA\n",
-    "ic_positive_peak=1.5;        #Collector current at positive peak of signal, mA\n",
-    "ic_negative_peak=0.5;         #Collector current at negative peak of signal, mA\n",
-    "vce_positive_peak=7;          #Collector emitter voltage at positive peak of signal, V\n",
-    "vce_negative_peak=9;          #Collector emitter voltage at negative peak of signal, V\n",
-    "\n",
-    "#Plot\n",
-    "vce_plot=[vce_positive_peak,vce_negative_peak];    #Plot variable of vce\n",
-    "ic_plot=[ic_positive_peak,ic_negative_peak];    #Plot variable of ic\n",
-    "\n",
-    "p.xlim(0,10)\n",
-    "p.ylim(0,2)\n",
-    "p.plot(vce_plot,ic_plot);\n",
-    "p.xlabel(\"vCE(V)\");\n",
-    "p.ylabel(\"iC(mA)\");\n",
-    "p.title(\"a.c load line\");\n",
-    "p.grid();\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.8 : Page number 256"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain= 24.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                 #Collector supply voltage, V\n",
-    "RC=2.0;                   #Collector resistor, kΩ\n",
-    "Rin=1.0;                  #Input resistance, kΩ\n",
-    "beta=60.0;                #Base current amplification factor\n",
-    "RL=0.5;                   #Load resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "RAC=(RC*RL)/(RC+RL);                #a.c load resistor, kΩ\n",
-    "Av=beta*(RAC/Rin);                  #Voltage gain\n",
-    "\n",
-    "#Results\n",
-    "print(\"Voltage gain= %d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.9 : Page number 256"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage= 200mV.\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "V_in=1.0;                      #Input voltage , mV\n",
-    "RC=10.0;                      #Collector resistor, kΩ\n",
-    "Rin=2.5;                      #Input resistance, kΩ\n",
-    "beta=100.0;                   #Base current amplification factor\n",
-    "RL=10.0;                      #Load resistor, kΩ\n",
-    "\n",
-    "#Calculations\n",
-    "RAC=(RC*RL)/(RC+RL);                #Effective load, kΩ\n",
-    "Av=beta*(RAC/Rin);                  #Voltage gain\n",
-    "\n",
-    "V_out=V_in*Av;                      #Output voltage, V\n",
-    "\n",
-    "#Results\n",
-    "print(\"Output voltage= %dmV.\"%V_out);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.10 : Page number 256-257"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Beta= 100.\n",
-      "Input impedance=2 kΩ.\n",
-      "a.c load=3.3 kΩ.\n",
-      "Voltage gain= 165.\n",
-      "Power gain=16500.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "change_in_IB=10.0;                    #Change in base current, 𝜇A\n",
-    "change_in_IC=1.0;                     #Change in collector current, mA\n",
-    "change_in_VBE=0.02;                   #Change in Base-emitter voltage, V\n",
-    "RC=5.0;                               #Collector resistor, kΩ\n",
-    "RL=10.0;                              #Emitter resistor, kΩ\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "beta=(change_in_IC*1000)/change_in_IB;         #Base current amplification factor\n",
-    "\n",
-    "#(ii)\n",
-    "Rin=(change_in_VBE/change_in_IB)*1000;         #Input impedance, kΩ\n",
-    "\n",
-    "#(iii)\n",
-    "RAC=round((RC*RL)/(RC+RL),1);                            #a.c load, kΩ\n",
-    "\n",
-    "#(iv)\n",
-    "Av=beta*RAC/Rin;                                #Voltage gain\n",
-    "\n",
-    "#(v)\n",
-    "Ap=beta*Av;                                     #Power gain\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"Beta= %d.\"%beta);\n",
-    "print(\"Input impedance=%d kΩ.\"%Rin);\n",
-    "print(\"a.c load=%.1f kΩ.\"%RAC);\n",
-    "print(\"Voltage gain= %d.\"%Av);\n",
-    "print(\"Power gain=%d.\"%Ap);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.11 : Page number 257"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=200mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "beta=50.0;                      #Base current amplification factor\n",
-    "RC=3.0;                         #Collector resistor,kΩ\n",
-    "RL=6.0;                         #Load resistor, kΩ\n",
-    "Rin=0.5;                        #Input impedance, kΩ\n",
-    "Vin=1;                          #Input voltage,  mV\n",
-    "\n",
-    "#Calculation\n",
-    "RAC=(RC*RL)/(RC+RL);            #a.c load, kΩ\n",
-    "Av=beta*RAC/Rin;                #Voltage gain\n",
-    "Vout=Vin*Av;                    #Output voltage, V\n",
-    "\n",
-    "#Results\n",
-    "print(\"Output voltage=%dmV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.12 : Page number 257-258"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The circuit is not operating properly.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VT=6.0;                   #Collector potential, V\n",
-    "R1=1.0;                   #Resistor R1, kΩ\n",
-    "R2=2.0;                   #Resistor R2, kΩ\n",
-    "VB_found=4.0;             #Measured base voltage, V\n",
-    "\n",
-    "#Calculations\n",
-    "VB=(VT*R1)/(R1+R2);                     #Theoretical base voltage, V\n",
-    "\n",
-    "if(VB_found==VB):\n",
-    "    print(\"The circuit is operating properly.\");\n",
-    "else:\n",
-    "    print(\"The circuit is not operating properly.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.13 : Page number 258-259"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a.c emitter resistance= 38.46 Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                   #Collector supply voltage, V\n",
-    "R1=40.0;                    #Resistor R1, kΩ\n",
-    "R2=10.0;                    #Resistor R2, kΩ\n",
-    "RC=6.0;                     #Collector resistor, kΩ\n",
-    "RE=2.0;                     #Emitter resistor, kΩ\n",
-    "beta=80;                    #Base current amplification factor\n",
-    "VBE=0.7;                    #Base emitter voltage, V\n",
-    "\n",
-    "#Calculations\n",
-    "V2=(VCC*R2)/(R1+R2);            #Voltage across resistor R2, V\n",
-    "VE=V2-VBE;                      #Emitter voltage, V\n",
-    "IE=VE/RE;                       #Emitter current,  mA\n",
-    "re=25/IE;                       #a.c emitter resistance, Ω\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"a.c emitter resistance= %.2f Ω.\"%re);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.14  : Page number 262-263"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)Voltage gain= 360.\n",
-      "(ii)Voltage gain= 5.37.\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "VCC=20.0;                     #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base-emitter voltage, V\n",
-    "R1=150.0;                     #Resistor R1, kΩ\n",
-    "R2=20.0                       #Resistor R2, kΩ\n",
-    "RC=12.0;                      #Collector resistor, kΩ\n",
-    "RE=2.2;                       #Emitter resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),2);                  #Voltage across R2, V\n",
-    "VE=round(V2-VBE,2);                          #Voltage across emitter resistor, V\n",
-    "IE=round(VE/RE,2);                           #Emitter current, mA\n",
-    "re=round(25/IE,1);                           #a.c emitter resistance, Ω\n",
-    "\n",
-    "\n",
-    "#(i)\n",
-    "#CE(emitter capacitor) connected in the circuit:\n",
-    "Av=(RC*1000)/re;                                   #Voltage gain for emitter capacitor connected.\n",
-    "\n",
-    "print(\"(i)Voltage gain= %d.\"%Av);\n",
-    "\n",
-    "#(ii)\n",
-    "#CE(emitter capacitor) removed from the circuit:\n",
-    "Av=(RC*1000)/(re+RE*1000);                          #Voltage gain for emitter capacitor removed.\n",
-    "\n",
-    "print(\"(ii)Voltage gain= %.2f.\"%Av);\n",
-    "\n",
-    "#Note: The answer in the text book has been approximated to 5.38 but it's actually coming 5.37.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.15 : Page number 263"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain= 120.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=6.0;                       #Collector resistor, kΩ\n",
-    "RL=12.0;                      #Load resistor, kΩ\n",
-    "re=33.3;                      #a.c emitter resistance, Ω\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "RAC=RC*RL/(RC+RL);              #a.c effective load, kΩ\n",
-    "Av=RAC*1000/re;                 #Voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"Voltage gain= %d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.16 : Page number 263-264"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) a.c emitter resistance=250 Ω.\n",
-      "(ii) Voltage gain =80.\n",
-      "(iii) d.c voltage across input capacitor= 1V and emitter capacitor=0.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=9.0;                      #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base-emitter voltage, V\n",
-    "R1=240.0;                     #Resistor R1, kΩ\n",
-    "R2=30.0                       #Resistor R2, kΩ\n",
-    "RC=20.0;                      #Collector resistor, kΩ\n",
-    "RE=3.0;                       #Emitter resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "V2=round(VCC*R2/(R1+R2),1);                  #Voltage across R2, V\n",
-    "VE=round(V2-VBE,1);                          #Voltage across emitter resistor, V\n",
-    "IE=round(VE/RE,1);                           #Emitter current, mA\n",
-    "re=25/IE;                                    #a.c emitter resistance, Ω\n",
-    "\n",
-    "#(ii)\n",
-    "Av=RC*1000/re;                                   #Voltage gain\n",
-    "\n",
-    "#(iii)\n",
-    "V_C_in=V2;                                  #d.c voltage across input capacitor, V\n",
-    "V_C_E=VE;                                   #d.c vooltage across emitter capacitor, V\n",
-    "\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"(i) a.c emitter resistance=%d Ω.\"%re);\n",
-    "print(\"(ii) Voltage gain =%d.\"%Av);\n",
-    "print(\"(iii) d.c voltage across input capacitor= %dV and emitter capacitor=%.1fV.\"%(V_C_in,V_C_E));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.17 : Page number 264-265"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) D.C bias levels: V2=3V, VE=2.3V, IE=2.3mA, IC=2.3mA, IB=0.023mA and VC=10.4V.\n",
-      "(ii) D.c voltage across: Cin=3V and CE=2.3V and CC=10.4V.\n",
-      "(iii) a.c emitter resistance=10.9Ω.\n",
-      "(iv) Voltage gain=61.2.\n",
-      "(v) VC>VE. Therefore, the transistor is in active state.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15.0;                      #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base-emitter voltage, V\n",
-    "R1=40.0;                     #Resistor R1, kΩ\n",
-    "R2=10.0                       #Resistor R2, kΩ\n",
-    "RC=2.0;                      #Collector resistor, kΩ\n",
-    "RE=1.0;                       #Emitter resistor, kΩ\n",
-    "RL=1.0;                       #Load resistor, kΩ\n",
-    "beta=100;                     #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) D.C bias levels\n",
-    "V2=VCC*R2/(R1+R2);                  #Voltage across R2, V\n",
-    "VE=round(V2-VBE,1);                 #Voltage across emitter resistor, V\n",
-    "IE=round(VE/RE,1);                  #Emitter current, mA\n",
-    "IC=IE;                              #Collector current, mA\n",
-    "IB=IC/beta;                         #Base  current, mA\n",
-    "VC=VCC-IC*RC;                       #Collector voltage, V\n",
-    "print(\"(i) D.C bias levels: V2=%dV, VE=%.1fV, IE=%.1fmA, IC=%.1fmA, IB=%.3fmA and VC=%.1fV.\"%(V2,VE,IE,IC,IB,VC));\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "Cin_V=V2;                            #Voltage across Cin capacitor, V\n",
-    "CE_V=VE;                             #Voltage across CE capacitor, V \n",
-    "CC_V=VC;                             #Voltage across CC capacitor, V\n",
-    "print(\"(ii) D.c voltage across: Cin=%dV and CE=%.1fV and CC=%.1fV.\"%(Cin_V,CE_V,CC_V));\n",
-    "\n",
-    "#(iii)\n",
-    "re=round(25/IE,1);                           #a.c emitter resistance, Ω\n",
-    "print(\"(iii) a.c emitter resistance=%.1fΩ.\"%re);\n",
-    "\n",
-    "\n",
-    "#(iv)\n",
-    "RAC=round(RC*RL/(RC+RL),3);                #Total a.c collector resistance, kΩ\n",
-    "Av=RAC/(re/1000);                        #Voltage gain\n",
-    "print(\"(iv) Voltage gain=%.1f.\"%Av);\n",
-    "\n",
-    "#(v)\n",
-    "print(\"(v) VC>VE. Therefore, the transistor is in active state.\" );\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.18 : page number 265"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The power gain = 26400 and output power = 1.584W.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=132.0;                   #Voltage gain\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "P_in=60.0;                     #Input power, 𝜇W\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "Ap=beta*Av;                          #Power gain\n",
-    "P_out=Ap*(P_in/10**6);                #Output power, W\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"The power gain = %d and output power = %.3fW.\"%(Ap,P_out));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.19 : page number 265-266"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Current gain=49\n",
-      "(ii) Voltage gain=2.14\n",
-      "(iii) Power gain=105.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IB=200.0;                       #Base current, microampere\n",
-    "IE=10.0;                        #Emitter current, mA\n",
-    "R1=27.0;                      #Resistor R1, kilo ohm\n",
-    "R2=13.0                       #Resistor R2, kilo ohm\n",
-    "RC=4.7;                       #Collector resistor, kilo ohm\n",
-    "RE=2.2;                       #Emitter resistor, kilo ohm\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "IC=IE-(IB/1000);                              #Collector current, mA\n",
-    "beta=IC/(IB/1000);                            #Current gain\n",
-    "\n",
-    "print(\"(i) Current gain=%d\"%beta);\n",
-    "\n",
-    "#(ii)\n",
-    "#a.c emitter resistance is neglected, voltage gain=(collector resistor)/(emitter resistor)\n",
-    "Av=RC/RE;                           #Voltage gain\n",
-    "\n",
-    "print(\"(ii) Voltage gain=%.2f\"%Av);\n",
-    "\n",
-    "#(iii)\n",
-    "Ap=round(beta*Av,0);                         #Power gain\n",
-    "\n",
-    "#Results\n",
-    "print(\"(iii) Power gain=%d.\"%Ap);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.20 : Page number 266-267"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance of the amplifier circuit= 3.46 kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=30.0;                     #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base emitter voltage, V\n",
-    "R1=45.0;                      #Resistor R1, kΩ\n",
-    "R2=15.0                       #Resistor R2, kΩ\n",
-    "RC=10.0;                      #Collector resistor,kΩ\n",
-    "RE=7.5;                       #Emitter resistor, kΩ\n",
-    "beta=200.0;                   #Base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),1);                  #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2;                                       #Voltage across emitter resistor(base-emitter voltage is neglected), V\n",
-    "IE=VE/RE;                                    #Emitter current, mA (OHM's LAW)\n",
-    "re=25/IE;                                    #a.c emitter resistance, ohm\n",
-    "Zin_base=(beta*re)/1000;                     #input impedance of transistor base,kΩ\n",
-    "R1_R2=(R1*R2)/(R1+R2);                       #Parallel resistance between R1 and R2, kΩ\n",
-    "Zin=((R1_R2)*Zin_base)/(R1_R2+Zin_base);     #Input impedance of the amplifier circuit, kΩ\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance of the amplifier circuit= %.2f kΩ.\"%Zin);       \n",
-    "\n",
-    "#Note: The input impedance of the amplifier circuit is approximated as 3.45 kΩ in the text book, but actually it's 3.46 kΩ.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.21 : Page Number 268-269"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the swamped amplifier= 4.67.\n",
-      "Input impedance of transistor base of the swamped amplifier= 48.21 kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                           #Collector supply voltage, V.\n",
-    "RC=1.5;                             #Collector resistor, kΩ.\n",
-    "R1=18.0;                            #Resistor R1, kΩ.\n",
-    "R2=4.7;                             #Resistor R2, kΩ.\n",
-    "RE1=300.0;                          #Emitter resistor 1, Ω.\n",
-    "RE2=900.0;                          #Emitter resistor 2, Ω.\n",
-    "VBE=0.7;                            #Base-emitter voltage, V.\n",
-    "beta=150.0;                         #Base current amplification factor.\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),1);                      #d.c voltage across R2, V. (Voltage divider rule)\n",
-    "VE=round(V2-VBE,1);                              #d.c voltage across RE, V.\n",
-    "IE=round((VE/(RE1+RE2))*1000,2);                 #d.c emitter current, mA.(OHM'S LAW)\n",
-    "re=round(25/IE,1);                               #a.c emitter resistance, Ω.\n",
-    "Av=RC*1000/(re+RE1);                             #Voltage gain\n",
-    "Zin_base=(beta*(re+RE1))/1000;                   #Input impedance of transistor base, kΩ.\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"The voltage gain of the swamped amplifier= %.2f.\"%Av);\n",
-    "print(\"Input impedance of transistor base of the swamped amplifier= %.2f kΩ.\"%Zin_base);\n",
-    "\n",
-    "#Note:In the textbook Av is approximated to 4.66and Zin_base to 48.22 kilo ohm, but the actual answers come as 4.67 and 48.21 kilo ohm.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.22 : Page number 269"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The percentage change from the original value= 6.42%(decrease)\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=1.5;                             #Collector resistor, kΩ.\n",
-    "RE1=300.0;                          #Emitter resistor 1, Ω.\n",
-    "re=21.5;                            #a.c emitter resistance, Ω.\n",
-    "\n",
-    "#Calculations\n",
-    "Av=round(RC*1000/(re+RE1),2);                     #Voltage gain.\n",
-    "Av_1=round(RC*1000/(2*re+RE1),2);                 #Voltage gain when re doubles.\n",
-    "change_in_gain=round(Av-Av_1,2);                      #Change in voltage gain.\n",
-    "change_percentage=change_in_gain*100/Av;               #Change percentage\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "if(change_in_gain>0):\n",
-    "    print(\"The percentage change from the original value= %.2f%%(decrease)\"%change_percentage);\n",
-    "else:\n",
-    "    print(\"The percentage change from the original value= %.2f%%(increase)\"%change_percentage);\n",
-    "\n",
-    "\n",
-    "#Note: The percentage has been approximated in the text book as 6.22%, but the answer comes as 6.42%.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.23 : Page number 269-270"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) input impedance of transistor base for standard amplifier= 5 kilo ohm\n",
-      "    input impedance of transistor base for swamped amplifier= 47 kilo ohm\n",
-      "(ii) input impedance for standard amplifier= 1.33 kilo ohm\n",
-      "     input impedance for swamped amplifier= 1.74 kilo ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                     #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base emitter voltage, V\n",
-    "R1=10.0;                      #Resistor R1, kilo ohm\n",
-    "R2=2.2;                       #Resistor R2, kilo ohm\n",
-    "RC=4.0;                       #Collector resistor, kilo ohm\n",
-    "RE=1.1;                       #Emitter resistor, kilo ohm\n",
-    "beta=200.0;                   #Base current amplification factor\n",
-    "RE1=210.0;                    #Emitter resistor 1 of swamped amplifier, ohm.\n",
-    "RE2=900.0;                    #Emitter resistor 2 of swamped amplifier, ohm.\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),1);                      #d.c voltage across R2, V. (Voltage divider rule)\n",
-    "VE=round(V2-VBE,1);                              #d.c voltage across RE, V.\n",
-    "IE=(VE/RE);                                      #d.c emitter current, mA.(OHM'S LAW)\n",
-    "re=25/IE;                                        #a.c emitter resistance, ohm.\n",
-    "\n",
-    "\n",
-    "#(i) Zin_base:\n",
-    "Zin_base_standard=(beta*re)/1000;                      #input impedance of transistor base for standard amplifier , kilo ohm.\n",
-    "Zin_base_swamped=(beta*(re+RE1))/1000;                 #input impedance of transistor base for swamped amplifier, kilo ohm.\n",
-    "\n",
-    "\n",
-    "#(ii) Zin:\n",
-    "#input impedance for standard amplifier circuit\n",
-    "Zin_standard=(((R1*R2)/(R1+R2))*Zin_base_standard)/(Zin_base_standard +((R1*R2)/(R1+R2)));    #kilo ohm\n",
-    "\n",
-    "#input impedance for standard amplifier circuit\n",
-    "Zin_swamped=(((R1*R2)/(R1+R2))*Zin_base_swamped)/(Zin_base_swamped +((R1*R2)/(R1+R2)));       #kilo ohm\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"(i) input impedance of transistor base for standard amplifier= %d kilo ohm\"%Zin_base_standard);\n",
-    "print(\"    input impedance of transistor base for swamped amplifier= %d kilo ohm\"%Zin_base_swamped);\n",
-    "print(\"(ii) input impedance for standard amplifier= %.2f kilo ohm\"%Zin_standard);\n",
-    "print(\"     input impedance for swamped amplifier= %.2f kilo ohm\"%Zin_swamped);\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.24 : Page number 270-271"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of standard amplifier=160.\n",
-      "The voltage gain of swamped amplifier=17.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=4.0;                     #Collector resistor, kilo ohm\n",
-    "re=25.0;                    #a.c emitter resistance, ohm (calculated in example 10.23)\n",
-    "RE_1=210.0;                  #Emitter resistor 1 of swamped amplifier,ohm\n",
-    "\n",
-    "#Calculation\n",
-    "Av_standard=(RC*1000)/re;                       #Voltage gain of standard common emitter amplifier\n",
-    "Av_swamped=(RC*1000)/(re+RE_1);                 #Voltage gain of swamped amplifier\n",
-    "\n",
-    "#Results\n",
-    "print(\"The voltage gain of standard amplifier=%d.\"%Av_standard);\n",
-    "print(\"The voltage gain of swamped amplifier=%d.\"%Av_swamped);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.26 : Page number 273-274"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The required input signal voltage =2.5mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_0=1000.0;                  #Open circuit voltage gain\n",
-    "R_in=2.0;                    #Input resistance, kilo ohm\n",
-    "R_out=1.0;                   #Output resistance, ohm\n",
-    "RL=4;                        #Load resistor across the output, ohm\n",
-    "I_2=0.5;                     #Output signal current, A.\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#Since A_0*(I_1*R_in) = I_2*(R_out+RL)\n",
-    "I_1=I_2*(R_out+RL)/(A_0*(R_in*1000));           #Input current, A\n",
-    "V_1=I_1*(R_in*1000);                            #Input signal voltage, V\n",
-    "V_1=V_1*1000;                                   #Input signal voltage, mV\n",
-    "\n",
-    "print(\"The required input signal voltage =%.1fmV\"%V_1);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.27 : Page number 274"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The magnitude of output voltage = 4.9V\n",
-      "The power gain =98e-06.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_0=1000.0;                     #Open circuit voltage gain\n",
-    "R_in=7.0;                       #Input resistance, kilo ohm\n",
-    "R_out=15.0;                     #Output resistance, ohm\n",
-    "RL=35.0;                        #Load resistor across the output, ohm\n",
-    "R_s=3.0;                        #Internal resistance, kilo ohm\n",
-    "E_s=10.0;                       #Input signal voltage, mV.\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "I_1=E_s*(10**-3)/(R_s*1000+R_in*1000);                  #Input current, A\n",
-    "V_1=I_1*(R_in*1000);                                    #Voltage across input resistance, V\n",
-    "\n",
-    "#Since, A_v=V_2/V_1 = A_0*RL/(R_out+RL)\n",
-    "A_v=A_0*RL/(R_out+RL);                                 #Voltage gain\n",
-    "V_2=A_v*V_1;                                            #Outout voltage, V\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "P_2=V_2**2/RL;                              #Output power, W\n",
-    "P_1=V_1**2/(R_in*1000);                            #Input power, W\n",
-    "A_p=round(P_2/P_1,-6);                                #Power gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The magnitude of output voltage = %.1fV\"%V_2);\n",
-    "print(\"The power gain =%de-06.\"%(A_p/10**6));\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.28 : Page number 274-275"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Necessary input signal voltage= 12.5mV\n",
-      "Input signal current =4.17  μA\n",
-      "Power gain = 9600.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_v=80.0;                   #Voltage gain\n",
-    "V_2=1.0;                    #Output voltage, V\n",
-    "A_i=120.0;                  #Current gain\n",
-    "RL=2;                       #Load resistor, kilo ohm\n",
-    "\n",
-    "#Calculation\n",
-    "V_1=(V_2/A_v)*1000;                #Input signal voltage, mV\n",
-    "\n",
-    "#Since, A_i=A0*R_in/(R_out+RL) and A_v=A0*RL/(R_out+RL)\n",
-    "#So, A_v/A_i=RL/R_in\n",
-    "R_in=RL*A_i/A_v;                #Input resistance, kilo ohm\n",
-    "I_1=V_1/R_in;                   #Input current,  μA\n",
-    "A_p=A_i*A_v;                    #Power gain\n",
-    "\n",
-    "#Results\n",
-    "print(\"Necessary input signal voltage= %.1fmV\"%V_1);\n",
-    "print(\"Input signal current =%.2f  μA\"%I_1);\n",
-    "print(\"Power gain = %d.\"%A_p);\n",
-    "\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_4.ipynb
deleted file mode 100755
index c33a83d1..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_4.ipynb
+++ /dev/null
@@ -1,1298 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 10 : SINGLE STAGE TRANSISTOR AMPLIFIERS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Using matplotlib backend: Qt4Agg\n"
-     ]
-    }
-   ],
-   "source": [
-    "%matplotlib "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.2 : page number 243-244"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The value of the emitter capacitor = 1.42 𝜇F\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "#Variable declaration\n",
-    "f_min=2.0;                            #Minimum frequency of operation of amplifier, kHz\n",
-    "f_max=10.0;                           #Maximum frequency of operation of amplifier, kHz\n",
-    "RE=560.0;                             #Emitter resistor, Ω\n",
-    "\n",
-    "#Calculations\n",
-    "#X_CE(Emitter capacitor's capacitive reactance)\n",
-    "#X_CE=1/(2*pi*f_min*CE)=RE/10\n",
-    "#From the above equation.\n",
-    "CE=1/(2*pi*f_min*1000*(RE/10));              #Emitter capacitor, F,\n",
-    "\n",
-    "CE=CE*10**6;                                 #Emitter capacitor, 𝜇F\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print('The value of the emitter capacitor = %.2f 𝜇F'%(CE));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.5: Page number 252-253"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The operating point: VCE=8.55V and IC=2.15mA.\n",
-      "Maximum v_CE=9.62V and maximum i_C=19.25mA\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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Og7rkLmiz7qlUGKwB7iXNVbw4W/Z0ROzS5UpLLc5hUPfcBW1WvkqFwXDgTNLo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\nRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f895dcab908>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=15.0;                         #Collector supply voltage in V\n",
-    "VBE=0.7;                          #Base-emitter voltage, V\n",
-    "R1=10.0;                          #Resistor R1, kΩ\n",
-    "R2=5.0;                           #Resistor R2, kΩ\n",
-    "RC=1.0;                           #Collector resistor, kΩ\n",
-    "RE=2.0;                           #Emitter resistor, kΩ\n",
-    "RL=1.0;                           #Load resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "#For d.c load line, from the equation: VCE=VCC-IC*(RC+RE),\n",
-    "#VCE is maximum when IC=0 and IC is maximum when VCE=0.\n",
-    "VCE_max=VCC;                                        #Maximum collector-emitter voltage, V\n",
-    "IC_max=VCC/(RC+RE);                                 #Maximum collector current, mA\n",
-    "\n",
-    "\n",
-    "#plot\n",
-    "VCE_plot=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-    "IC_plot=[((VCC-i)/(RC+RE)) for i in (VCE_plot[:])];      #Plot variable for I_C\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.xlim(0,20)\n",
-    "plt.ylim(0,6)\n",
-    "plt.plot(VCE_plot,IC_plot);\n",
-    "plt.xlabel(\"VCE(V)\");\n",
-    "plt.ylabel(\"IC(mA)\");\n",
-    "plt.title(\"d.c load line\");\n",
-    "\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "#For operating point:\n",
-    "#Assuming VCC drops almost completely across R1 and R2,\n",
-    "V2=VCC*R2/(R1+R2);                                  #Voltage across resistor R2, V\n",
-    "IE=(V2-VBE)/RE;                                     #Emitter current, mA\n",
-    "IC=IE;                                              #Collector current, mA\n",
-    "VCE=VCC-IC*(RC+RE);                                 #Collector-emitter voltage , V\n",
-    "\n",
-    "print(\"The operating point: VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n",
-    "\n",
-    "\n",
-    "#(iii)\n",
-    "#For a.c load line\n",
-    "RAC=(RC*RL)/(RC+RL);                                #a.c load, kΩ\n",
-    "VCE_ac_max=VCE+IC*RAC;                              #Maximum collector-emitter voltage, V\n",
-    "IC_ac_max=IC+VCE/RAC;                               #Maximum collector current, mA\n",
-    "print(\"Maximum v_CE=%.2fV and maximum i_C=%.2fmA\"%(VCE_ac_max,IC_ac_max));\n",
-    "\n",
-    "#plot\n",
-    "vCE_plot=[0,VCE_ac_max];          #Plot variable for V_CE\n",
-    "iC_plot=[IC_ac_max,0];      #Plot variable for I_C\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.xlim(0,10)\n",
-    "plt.ylim(0,20)\n",
-    "plt.plot(vCE_plot,iC_plot);\n",
-    "plt.xlabel(\"vCE(V)\");\n",
-    "plt.ylabel(\"iC(mA)\");\n",
-    "plt.title(\"a.c load line\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.6: Page number 253-254"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f895dbf5390>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as p\n",
-    "\n",
-    "#Variable declaration\n",
-    "RC=10;                          #Collector resistor, kΩ\n",
-    "RL=30;                          #Load resistor, kΩ\n",
-    "VCC=20;                         #Collector supply voltage, V\n",
-    "IC=1;                           #Collector current, mA\n",
-    "VCE=10;                         #Collector-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#For d.c load line:\n",
-    "#From the equation: VCE=VCC-IC*(RC+RE),\n",
-    "#When VCE=0, IC is  maximum.\n",
-    "#Emitter resistor is neglected, assuming it as negligible\n",
-    "IC_max=VCC/RC;                      #Maximum collector current, mA\n",
-    "\n",
-    "#And, when IC=0, VCE is maximum\n",
-    "VCE_max=VCC;                        #Maximum collector-emitter voltage, V\n",
-    "\n",
-    "#plot\n",
-    "p.subplot(211)\n",
-    "p.xlim(0,20)\n",
-    "p.ylim(0,5)\n",
-    "VCE_plot=[0,VCE_max];          #Plot variable for V_CE\n",
-    "IC_plot=[IC_max,0];            #Plot variable for I_C\n",
-    "\n",
-    "p.plot(VCE_plot,IC_plot);\n",
-    "p.xlabel(\"VCE(V)\");\n",
-    "p.ylabel(\"IC(mA)\");\n",
-    "p.title(\"d.c load line\");\n",
-    "\n",
-    "\n",
-    "#For a.c load line:\n",
-    "RAC=(RC*RL)/(RC+RL);                #a.c Load resistor, kΩ\n",
-    "\n",
-    "VCE_ac_max=VCE+IC*RAC;                 #Maximum collector-emitter voltage, V\n",
-    "IC_ac_max=IC+ VCE/RAC;                #Maximum collector current, mA\n",
-    "\n",
-    "#plot\n",
-    "p.subplot(212)\n",
-    "p.xlim([0,25])\n",
-    "p.ylim([0,5])\n",
-    "vCE_plot=[0,VCE_ac_max];          #Plot variable for V_CE\n",
-    "iC_plot=[IC_ac_max,0];      #Plot variable for I_C\n",
-    "\n",
-    "p.plot(vCE_plot,iC_plot);\n",
-    "p.xlabel(\"vCE(V)\");\n",
-    "p.ylabel(\"iC(mA)\");\n",
-    "p.title(\"a.c load line\");\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.7 : Page number 254-255"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f898013a748>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as p\n",
-    "\n",
-    "#Variabe declaration\n",
-    "VCE_Q=8.0;           #Q-point collector emitter voltage, V\n",
-    "IC_Q=1;               #Q-point collector current, mA\n",
-    "ic_positive_peak=1.5;        #Collector current at positive peak of signal, mA\n",
-    "ic_negative_peak=0.5;         #Collector current at negative peak of signal, mA\n",
-    "vce_positive_peak=7;          #Collector emitter voltage at positive peak of signal, V\n",
-    "vce_negative_peak=9;          #Collector emitter voltage at negative peak of signal, V\n",
-    "\n",
-    "#Plot\n",
-    "vce_plot=[vce_positive_peak,vce_negative_peak];    #Plot variable of vce\n",
-    "ic_plot=[ic_positive_peak,ic_negative_peak];    #Plot variable of ic\n",
-    "\n",
-    "p.xlim(0,10)\n",
-    "p.ylim(0,2)\n",
-    "p.plot(vce_plot,ic_plot);\n",
-    "p.xlabel(\"vCE(V)\");\n",
-    "p.ylabel(\"iC(mA)\");\n",
-    "p.title(\"a.c load line\");\n",
-    "p.grid();\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.8 : Page number 256"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain= 24.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                 #Collector supply voltage, V\n",
-    "RC=2.0;                   #Collector resistor, kΩ\n",
-    "Rin=1.0;                  #Input resistance, kΩ\n",
-    "beta=60.0;                #Base current amplification factor\n",
-    "RL=0.5;                   #Load resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "RAC=(RC*RL)/(RC+RL);                #a.c load resistor, kΩ\n",
-    "Av=beta*(RAC/Rin);                  #Voltage gain\n",
-    "\n",
-    "#Results\n",
-    "print(\"Voltage gain= %d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.9 : Page number 256"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage= 200mV.\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "V_in=1.0;                      #Input voltage , mV\n",
-    "RC=10.0;                      #Collector resistor, kΩ\n",
-    "Rin=2.5;                      #Input resistance, kΩ\n",
-    "beta=100.0;                   #Base current amplification factor\n",
-    "RL=10.0;                      #Load resistor, kΩ\n",
-    "\n",
-    "#Calculations\n",
-    "RAC=(RC*RL)/(RC+RL);                #Effective load, kΩ\n",
-    "Av=beta*(RAC/Rin);                  #Voltage gain\n",
-    "\n",
-    "V_out=V_in*Av;                      #Output voltage, V\n",
-    "\n",
-    "#Results\n",
-    "print(\"Output voltage= %dmV.\"%V_out);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.10 : Page number 256-257"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Beta= 100.\n",
-      "Input impedance=2 kΩ.\n",
-      "a.c load=3.3 kΩ.\n",
-      "Voltage gain= 165.\n",
-      "Power gain=16500.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "change_in_IB=10.0;                    #Change in base current, 𝜇A\n",
-    "change_in_IC=1.0;                     #Change in collector current, mA\n",
-    "change_in_VBE=0.02;                   #Change in Base-emitter voltage, V\n",
-    "RC=5.0;                               #Collector resistor, kΩ\n",
-    "RL=10.0;                              #Emitter resistor, kΩ\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "beta=(change_in_IC*1000)/change_in_IB;         #Base current amplification factor\n",
-    "\n",
-    "#(ii)\n",
-    "Rin=(change_in_VBE/change_in_IB)*1000;         #Input impedance, kΩ\n",
-    "\n",
-    "#(iii)\n",
-    "RAC=round((RC*RL)/(RC+RL),1);                            #a.c load, kΩ\n",
-    "\n",
-    "#(iv)\n",
-    "Av=beta*RAC/Rin;                                #Voltage gain\n",
-    "\n",
-    "#(v)\n",
-    "Ap=beta*Av;                                     #Power gain\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"Beta= %d.\"%beta);\n",
-    "print(\"Input impedance=%d kΩ.\"%Rin);\n",
-    "print(\"a.c load=%.1f kΩ.\"%RAC);\n",
-    "print(\"Voltage gain= %d.\"%Av);\n",
-    "print(\"Power gain=%d.\"%Ap);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.11 : Page number 257"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=200mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "beta=50.0;                      #Base current amplification factor\n",
-    "RC=3.0;                         #Collector resistor,kΩ\n",
-    "RL=6.0;                         #Load resistor, kΩ\n",
-    "Rin=0.5;                        #Input impedance, kΩ\n",
-    "Vin=1;                          #Input voltage,  mV\n",
-    "\n",
-    "#Calculation\n",
-    "RAC=(RC*RL)/(RC+RL);            #a.c load, kΩ\n",
-    "Av=beta*RAC/Rin;                #Voltage gain\n",
-    "Vout=Vin*Av;                    #Output voltage, V\n",
-    "\n",
-    "#Results\n",
-    "print(\"Output voltage=%dmV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.12 : Page number 257-258"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The circuit is not operating properly.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VT=6.0;                   #Collector potential, V\n",
-    "R1=1.0;                   #Resistor R1, kΩ\n",
-    "R2=2.0;                   #Resistor R2, kΩ\n",
-    "VB_found=4.0;             #Measured base voltage, V\n",
-    "\n",
-    "#Calculations\n",
-    "VB=(VT*R1)/(R1+R2);                     #Theoretical base voltage, V\n",
-    "\n",
-    "if(VB_found==VB):\n",
-    "    print(\"The circuit is operating properly.\");\n",
-    "else:\n",
-    "    print(\"The circuit is not operating properly.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.13 : Page number 258-259"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a.c emitter resistance= 38.46 Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                   #Collector supply voltage, V\n",
-    "R1=40.0;                    #Resistor R1, kΩ\n",
-    "R2=10.0;                    #Resistor R2, kΩ\n",
-    "RC=6.0;                     #Collector resistor, kΩ\n",
-    "RE=2.0;                     #Emitter resistor, kΩ\n",
-    "beta=80;                    #Base current amplification factor\n",
-    "VBE=0.7;                    #Base emitter voltage, V\n",
-    "\n",
-    "#Calculations\n",
-    "V2=(VCC*R2)/(R1+R2);            #Voltage across resistor R2, V\n",
-    "VE=V2-VBE;                      #Emitter voltage, V\n",
-    "IE=VE/RE;                       #Emitter current,  mA\n",
-    "re=25/IE;                       #a.c emitter resistance, Ω\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"a.c emitter resistance= %.2f Ω.\"%re);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.14  : Page number 262-263"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)Voltage gain= 360.\n",
-      "(ii)Voltage gain= 5.37.\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "VCC=20.0;                     #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base-emitter voltage, V\n",
-    "R1=150.0;                     #Resistor R1, kΩ\n",
-    "R2=20.0                       #Resistor R2, kΩ\n",
-    "RC=12.0;                      #Collector resistor, kΩ\n",
-    "RE=2.2;                       #Emitter resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),2);                  #Voltage across R2, V\n",
-    "VE=round(V2-VBE,2);                          #Voltage across emitter resistor, V\n",
-    "IE=round(VE/RE,2);                           #Emitter current, mA\n",
-    "re=round(25/IE,1);                           #a.c emitter resistance, Ω\n",
-    "\n",
-    "\n",
-    "#(i)\n",
-    "#CE(emitter capacitor) connected in the circuit:\n",
-    "Av=(RC*1000)/re;                                   #Voltage gain for emitter capacitor connected.\n",
-    "\n",
-    "print(\"(i)Voltage gain= %d.\"%Av);\n",
-    "\n",
-    "#(ii)\n",
-    "#CE(emitter capacitor) removed from the circuit:\n",
-    "Av=(RC*1000)/(re+RE*1000);                          #Voltage gain for emitter capacitor removed.\n",
-    "\n",
-    "print(\"(ii)Voltage gain= %.2f.\"%Av);\n",
-    "\n",
-    "#Note: The answer in the text book has been approximated to 5.38 but it's actually coming 5.37.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.15 : Page number 263"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain= 120.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=6.0;                       #Collector resistor, kΩ\n",
-    "RL=12.0;                      #Load resistor, kΩ\n",
-    "re=33.3;                      #a.c emitter resistance, Ω\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "RAC=RC*RL/(RC+RL);              #a.c effective load, kΩ\n",
-    "Av=RAC*1000/re;                 #Voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"Voltage gain= %d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.16 : Page number 263-264"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) a.c emitter resistance=250 Ω.\n",
-      "(ii) Voltage gain =80.\n",
-      "(iii) d.c voltage across input capacitor= 1V and emitter capacitor=0.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=9.0;                      #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base-emitter voltage, V\n",
-    "R1=240.0;                     #Resistor R1, kΩ\n",
-    "R2=30.0                       #Resistor R2, kΩ\n",
-    "RC=20.0;                      #Collector resistor, kΩ\n",
-    "RE=3.0;                       #Emitter resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "V2=round(VCC*R2/(R1+R2),1);                  #Voltage across R2, V\n",
-    "VE=round(V2-VBE,1);                          #Voltage across emitter resistor, V\n",
-    "IE=round(VE/RE,1);                           #Emitter current, mA\n",
-    "re=25/IE;                                    #a.c emitter resistance, Ω\n",
-    "\n",
-    "#(ii)\n",
-    "Av=RC*1000/re;                                   #Voltage gain\n",
-    "\n",
-    "#(iii)\n",
-    "V_C_in=V2;                                  #d.c voltage across input capacitor, V\n",
-    "V_C_E=VE;                                   #d.c vooltage across emitter capacitor, V\n",
-    "\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"(i) a.c emitter resistance=%d Ω.\"%re);\n",
-    "print(\"(ii) Voltage gain =%d.\"%Av);\n",
-    "print(\"(iii) d.c voltage across input capacitor= %dV and emitter capacitor=%.1fV.\"%(V_C_in,V_C_E));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.17 : Page number 264-265"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) D.C bias levels: V2=3V, VE=2.3V, IE=2.3mA, IC=2.3mA, IB=0.023mA and VC=10.4V.\n",
-      "(ii) D.c voltage across: Cin=3V and CE=2.3V and CC=10.4V.\n",
-      "(iii) a.c emitter resistance=10.9Ω.\n",
-      "(iv) Voltage gain=61.2.\n",
-      "(v) VC>VE. Therefore, the transistor is in active state.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15.0;                      #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base-emitter voltage, V\n",
-    "R1=40.0;                     #Resistor R1, kΩ\n",
-    "R2=10.0                       #Resistor R2, kΩ\n",
-    "RC=2.0;                      #Collector resistor, kΩ\n",
-    "RE=1.0;                       #Emitter resistor, kΩ\n",
-    "RL=1.0;                       #Load resistor, kΩ\n",
-    "beta=100;                     #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) D.C bias levels\n",
-    "V2=VCC*R2/(R1+R2);                  #Voltage across R2, V\n",
-    "VE=round(V2-VBE,1);                 #Voltage across emitter resistor, V\n",
-    "IE=round(VE/RE,1);                  #Emitter current, mA\n",
-    "IC=IE;                              #Collector current, mA\n",
-    "IB=IC/beta;                         #Base  current, mA\n",
-    "VC=VCC-IC*RC;                       #Collector voltage, V\n",
-    "print(\"(i) D.C bias levels: V2=%dV, VE=%.1fV, IE=%.1fmA, IC=%.1fmA, IB=%.3fmA and VC=%.1fV.\"%(V2,VE,IE,IC,IB,VC));\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "Cin_V=V2;                            #Voltage across Cin capacitor, V\n",
-    "CE_V=VE;                             #Voltage across CE capacitor, V \n",
-    "CC_V=VC;                             #Voltage across CC capacitor, V\n",
-    "print(\"(ii) D.c voltage across: Cin=%dV and CE=%.1fV and CC=%.1fV.\"%(Cin_V,CE_V,CC_V));\n",
-    "\n",
-    "#(iii)\n",
-    "re=round(25/IE,1);                           #a.c emitter resistance, Ω\n",
-    "print(\"(iii) a.c emitter resistance=%.1fΩ.\"%re);\n",
-    "\n",
-    "\n",
-    "#(iv)\n",
-    "RAC=round(RC*RL/(RC+RL),3);                #Total a.c collector resistance, kΩ\n",
-    "Av=RAC/(re/1000);                        #Voltage gain\n",
-    "print(\"(iv) Voltage gain=%.1f.\"%Av);\n",
-    "\n",
-    "#(v)\n",
-    "print(\"(v) VC>VE. Therefore, the transistor is in active state.\" );\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.18 : page number 265"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The power gain = 26400 and output power = 1.584W.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=132.0;                   #Voltage gain\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "P_in=60.0;                     #Input power, 𝜇W\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "Ap=beta*Av;                          #Power gain\n",
-    "P_out=Ap*(P_in/10**6);                #Output power, W\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"The power gain = %d and output power = %.3fW.\"%(Ap,P_out));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.19 : page number 265-266"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Current gain=49\n",
-      "(ii) Voltage gain=2.14\n",
-      "(iii) Power gain=105.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IB=200.0;                       #Base current, microampere\n",
-    "IE=10.0;                        #Emitter current, mA\n",
-    "R1=27.0;                      #Resistor R1, kilo ohm\n",
-    "R2=13.0                       #Resistor R2, kilo ohm\n",
-    "RC=4.7;                       #Collector resistor, kilo ohm\n",
-    "RE=2.2;                       #Emitter resistor, kilo ohm\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "IC=IE-(IB/1000);                              #Collector current, mA\n",
-    "beta=IC/(IB/1000);                            #Current gain\n",
-    "\n",
-    "print(\"(i) Current gain=%d\"%beta);\n",
-    "\n",
-    "#(ii)\n",
-    "#a.c emitter resistance is neglected, voltage gain=(collector resistor)/(emitter resistor)\n",
-    "Av=RC/RE;                           #Voltage gain\n",
-    "\n",
-    "print(\"(ii) Voltage gain=%.2f\"%Av);\n",
-    "\n",
-    "#(iii)\n",
-    "Ap=round(beta*Av,0);                         #Power gain\n",
-    "\n",
-    "#Results\n",
-    "print(\"(iii) Power gain=%d.\"%Ap);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.20 : Page number 266-267"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance of the amplifier circuit= 3.46 kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=30.0;                     #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base emitter voltage, V\n",
-    "R1=45.0;                      #Resistor R1, kΩ\n",
-    "R2=15.0                       #Resistor R2, kΩ\n",
-    "RC=10.0;                      #Collector resistor,kΩ\n",
-    "RE=7.5;                       #Emitter resistor, kΩ\n",
-    "beta=200.0;                   #Base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),1);                  #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2;                                       #Voltage across emitter resistor(base-emitter voltage is neglected), V\n",
-    "IE=VE/RE;                                    #Emitter current, mA (OHM's LAW)\n",
-    "re=25/IE;                                    #a.c emitter resistance, ohm\n",
-    "Zin_base=(beta*re)/1000;                     #input impedance of transistor base,kΩ\n",
-    "R1_R2=(R1*R2)/(R1+R2);                       #Parallel resistance between R1 and R2, kΩ\n",
-    "Zin=((R1_R2)*Zin_base)/(R1_R2+Zin_base);     #Input impedance of the amplifier circuit, kΩ\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance of the amplifier circuit= %.2f kΩ.\"%Zin);       \n",
-    "\n",
-    "#Note: The input impedance of the amplifier circuit is approximated as 3.45 kΩ in the text book, but actually it's 3.46 kΩ.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.21 : Page Number 268-269"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the swamped amplifier= 4.67.\n",
-      "Input impedance of transistor base of the swamped amplifier= 48.21 kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                           #Collector supply voltage, V.\n",
-    "RC=1.5;                             #Collector resistor, kΩ.\n",
-    "R1=18.0;                            #Resistor R1, kΩ.\n",
-    "R2=4.7;                             #Resistor R2, kΩ.\n",
-    "RE1=300.0;                          #Emitter resistor 1, Ω.\n",
-    "RE2=900.0;                          #Emitter resistor 2, Ω.\n",
-    "VBE=0.7;                            #Base-emitter voltage, V.\n",
-    "beta=150.0;                         #Base current amplification factor.\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),1);                      #d.c voltage across R2, V. (Voltage divider rule)\n",
-    "VE=round(V2-VBE,1);                              #d.c voltage across RE, V.\n",
-    "IE=round((VE/(RE1+RE2))*1000,2);                 #d.c emitter current, mA.(OHM'S LAW)\n",
-    "re=round(25/IE,1);                               #a.c emitter resistance, Ω.\n",
-    "Av=RC*1000/(re+RE1);                             #Voltage gain\n",
-    "Zin_base=(beta*(re+RE1))/1000;                   #Input impedance of transistor base, kΩ.\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"The voltage gain of the swamped amplifier= %.2f.\"%Av);\n",
-    "print(\"Input impedance of transistor base of the swamped amplifier= %.2f kΩ.\"%Zin_base);\n",
-    "\n",
-    "#Note:In the textbook Av is approximated to 4.66and Zin_base to 48.22 kilo ohm, but the actual answers come as 4.67 and 48.21 kilo ohm.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.22 : Page number 269"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The percentage change from the original value= 6.42%(decrease)\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=1.5;                             #Collector resistor, kΩ.\n",
-    "RE1=300.0;                          #Emitter resistor 1, Ω.\n",
-    "re=21.5;                            #a.c emitter resistance, Ω.\n",
-    "\n",
-    "#Calculations\n",
-    "Av=round(RC*1000/(re+RE1),2);                     #Voltage gain.\n",
-    "Av_1=round(RC*1000/(2*re+RE1),2);                 #Voltage gain when re doubles.\n",
-    "change_in_gain=round(Av-Av_1,2);                      #Change in voltage gain.\n",
-    "change_percentage=change_in_gain*100/Av;               #Change percentage\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "if(change_in_gain>0):\n",
-    "    print(\"The percentage change from the original value= %.2f%%(decrease)\"%change_percentage);\n",
-    "else:\n",
-    "    print(\"The percentage change from the original value= %.2f%%(increase)\"%change_percentage);\n",
-    "\n",
-    "\n",
-    "#Note: The percentage has been approximated in the text book as 6.22%, but the answer comes as 6.42%.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.23 : Page number 269-270"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) input impedance of transistor base for standard amplifier= 5 kilo ohm\n",
-      "    input impedance of transistor base for swamped amplifier= 47 kilo ohm\n",
-      "(ii) input impedance for standard amplifier= 1.33 kilo ohm\n",
-      "     input impedance for swamped amplifier= 1.74 kilo ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                     #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base emitter voltage, V\n",
-    "R1=10.0;                      #Resistor R1, kilo ohm\n",
-    "R2=2.2;                       #Resistor R2, kilo ohm\n",
-    "RC=4.0;                       #Collector resistor, kilo ohm\n",
-    "RE=1.1;                       #Emitter resistor, kilo ohm\n",
-    "beta=200.0;                   #Base current amplification factor\n",
-    "RE1=210.0;                    #Emitter resistor 1 of swamped amplifier, ohm.\n",
-    "RE2=900.0;                    #Emitter resistor 2 of swamped amplifier, ohm.\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),1);                      #d.c voltage across R2, V. (Voltage divider rule)\n",
-    "VE=round(V2-VBE,1);                              #d.c voltage across RE, V.\n",
-    "IE=(VE/RE);                                      #d.c emitter current, mA.(OHM'S LAW)\n",
-    "re=25/IE;                                        #a.c emitter resistance, ohm.\n",
-    "\n",
-    "\n",
-    "#(i) Zin_base:\n",
-    "Zin_base_standard=(beta*re)/1000;                      #input impedance of transistor base for standard amplifier , kilo ohm.\n",
-    "Zin_base_swamped=(beta*(re+RE1))/1000;                 #input impedance of transistor base for swamped amplifier, kilo ohm.\n",
-    "\n",
-    "\n",
-    "#(ii) Zin:\n",
-    "#input impedance for standard amplifier circuit\n",
-    "Zin_standard=(((R1*R2)/(R1+R2))*Zin_base_standard)/(Zin_base_standard +((R1*R2)/(R1+R2)));    #kilo ohm\n",
-    "\n",
-    "#input impedance for standard amplifier circuit\n",
-    "Zin_swamped=(((R1*R2)/(R1+R2))*Zin_base_swamped)/(Zin_base_swamped +((R1*R2)/(R1+R2)));       #kilo ohm\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"(i) input impedance of transistor base for standard amplifier= %d kilo ohm\"%Zin_base_standard);\n",
-    "print(\"    input impedance of transistor base for swamped amplifier= %d kilo ohm\"%Zin_base_swamped);\n",
-    "print(\"(ii) input impedance for standard amplifier= %.2f kilo ohm\"%Zin_standard);\n",
-    "print(\"     input impedance for swamped amplifier= %.2f kilo ohm\"%Zin_swamped);\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.24 : Page number 270-271"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of standard amplifier=160.\n",
-      "The voltage gain of swamped amplifier=17.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=4.0;                     #Collector resistor, kilo ohm\n",
-    "re=25.0;                    #a.c emitter resistance, ohm (calculated in example 10.23)\n",
-    "RE_1=210.0;                  #Emitter resistor 1 of swamped amplifier,ohm\n",
-    "\n",
-    "#Calculation\n",
-    "Av_standard=(RC*1000)/re;                       #Voltage gain of standard common emitter amplifier\n",
-    "Av_swamped=(RC*1000)/(re+RE_1);                 #Voltage gain of swamped amplifier\n",
-    "\n",
-    "#Results\n",
-    "print(\"The voltage gain of standard amplifier=%d.\"%Av_standard);\n",
-    "print(\"The voltage gain of swamped amplifier=%d.\"%Av_swamped);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.26 : Page number 273-274"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The required input signal voltage =2.5mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_0=1000.0;                  #Open circuit voltage gain\n",
-    "R_in=2.0;                    #Input resistance, kilo ohm\n",
-    "R_out=1.0;                   #Output resistance, ohm\n",
-    "RL=4;                        #Load resistor across the output, ohm\n",
-    "I_2=0.5;                     #Output signal current, A.\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#Since A_0*(I_1*R_in) = I_2*(R_out+RL)\n",
-    "I_1=I_2*(R_out+RL)/(A_0*(R_in*1000));           #Input current, A\n",
-    "V_1=I_1*(R_in*1000);                            #Input signal voltage, V\n",
-    "V_1=V_1*1000;                                   #Input signal voltage, mV\n",
-    "\n",
-    "print(\"The required input signal voltage =%.1fmV\"%V_1);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.27 : Page number 274"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The magnitude of output voltage = 4.9V\n",
-      "The power gain =98e-06.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_0=1000.0;                     #Open circuit voltage gain\n",
-    "R_in=7.0;                       #Input resistance, kilo ohm\n",
-    "R_out=15.0;                     #Output resistance, ohm\n",
-    "RL=35.0;                        #Load resistor across the output, ohm\n",
-    "R_s=3.0;                        #Internal resistance, kilo ohm\n",
-    "E_s=10.0;                       #Input signal voltage, mV.\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "I_1=E_s*(10**-3)/(R_s*1000+R_in*1000);                  #Input current, A\n",
-    "V_1=I_1*(R_in*1000);                                    #Voltage across input resistance, V\n",
-    "\n",
-    "#Since, A_v=V_2/V_1 = A_0*RL/(R_out+RL)\n",
-    "A_v=A_0*RL/(R_out+RL);                                 #Voltage gain\n",
-    "V_2=A_v*V_1;                                            #Outout voltage, V\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "P_2=V_2**2/RL;                              #Output power, W\n",
-    "P_1=V_1**2/(R_in*1000);                            #Input power, W\n",
-    "A_p=round(P_2/P_1,-6);                                #Power gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The magnitude of output voltage = %.1fV\"%V_2);\n",
-    "print(\"The power gain =%de-06.\"%(A_p/10**6));\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.28 : Page number 274-275"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Necessary input signal voltage= 12.5mV\n",
-      "Input signal current =4.17  μA\n",
-      "Power gain = 9600.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_v=80.0;                   #Voltage gain\n",
-    "V_2=1.0;                    #Output voltage, V\n",
-    "A_i=120.0;                  #Current gain\n",
-    "RL=2;                       #Load resistor, kilo ohm\n",
-    "\n",
-    "#Calculation\n",
-    "V_1=(V_2/A_v)*1000;                #Input signal voltage, mV\n",
-    "\n",
-    "#Since, A_i=A0*R_in/(R_out+RL) and A_v=A0*RL/(R_out+RL)\n",
-    "#So, A_v/A_i=RL/R_in\n",
-    "R_in=RL*A_i/A_v;                #Input resistance, kilo ohm\n",
-    "I_1=V_1/R_in;                   #Input current,  μA\n",
-    "A_p=A_i*A_v;                    #Power gain\n",
-    "\n",
-    "#Results\n",
-    "print(\"Necessary input signal voltage= %.1fmV\"%V_1);\n",
-    "print(\"Input signal current =%.2f  μA\"%I_1);\n",
-    "print(\"Power gain = %d.\"%A_p);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_5.ipynb
deleted file mode 100755
index c33a83d1..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_5.ipynb
+++ /dev/null
@@ -1,1298 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 10 : SINGLE STAGE TRANSISTOR AMPLIFIERS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Using matplotlib backend: Qt4Agg\n"
-     ]
-    }
-   ],
-   "source": [
-    "%matplotlib "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.2 : page number 243-244"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The value of the emitter capacitor = 1.42 𝜇F\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "#Variable declaration\n",
-    "f_min=2.0;                            #Minimum frequency of operation of amplifier, kHz\n",
-    "f_max=10.0;                           #Maximum frequency of operation of amplifier, kHz\n",
-    "RE=560.0;                             #Emitter resistor, Ω\n",
-    "\n",
-    "#Calculations\n",
-    "#X_CE(Emitter capacitor's capacitive reactance)\n",
-    "#X_CE=1/(2*pi*f_min*CE)=RE/10\n",
-    "#From the above equation.\n",
-    "CE=1/(2*pi*f_min*1000*(RE/10));              #Emitter capacitor, F,\n",
-    "\n",
-    "CE=CE*10**6;                                 #Emitter capacitor, 𝜇F\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print('The value of the emitter capacitor = %.2f 𝜇F'%(CE));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.5: Page number 252-253"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The operating point: VCE=8.55V and IC=2.15mA.\n",
-      "Maximum v_CE=9.62V and maximum i_C=19.25mA\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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Og7rkLmiz7qlUGKwB7iXNVbw4W/Z0ROzS5UpLLc5hUPfcBW1WvkqFwXDgTNLo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\nRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f895dcab908>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=15.0;                         #Collector supply voltage in V\n",
-    "VBE=0.7;                          #Base-emitter voltage, V\n",
-    "R1=10.0;                          #Resistor R1, kΩ\n",
-    "R2=5.0;                           #Resistor R2, kΩ\n",
-    "RC=1.0;                           #Collector resistor, kΩ\n",
-    "RE=2.0;                           #Emitter resistor, kΩ\n",
-    "RL=1.0;                           #Load resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "#For d.c load line, from the equation: VCE=VCC-IC*(RC+RE),\n",
-    "#VCE is maximum when IC=0 and IC is maximum when VCE=0.\n",
-    "VCE_max=VCC;                                        #Maximum collector-emitter voltage, V\n",
-    "IC_max=VCC/(RC+RE);                                 #Maximum collector current, mA\n",
-    "\n",
-    "\n",
-    "#plot\n",
-    "VCE_plot=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-    "IC_plot=[((VCC-i)/(RC+RE)) for i in (VCE_plot[:])];      #Plot variable for I_C\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.xlim(0,20)\n",
-    "plt.ylim(0,6)\n",
-    "plt.plot(VCE_plot,IC_plot);\n",
-    "plt.xlabel(\"VCE(V)\");\n",
-    "plt.ylabel(\"IC(mA)\");\n",
-    "plt.title(\"d.c load line\");\n",
-    "\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "#For operating point:\n",
-    "#Assuming VCC drops almost completely across R1 and R2,\n",
-    "V2=VCC*R2/(R1+R2);                                  #Voltage across resistor R2, V\n",
-    "IE=(V2-VBE)/RE;                                     #Emitter current, mA\n",
-    "IC=IE;                                              #Collector current, mA\n",
-    "VCE=VCC-IC*(RC+RE);                                 #Collector-emitter voltage , V\n",
-    "\n",
-    "print(\"The operating point: VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n",
-    "\n",
-    "\n",
-    "#(iii)\n",
-    "#For a.c load line\n",
-    "RAC=(RC*RL)/(RC+RL);                                #a.c load, kΩ\n",
-    "VCE_ac_max=VCE+IC*RAC;                              #Maximum collector-emitter voltage, V\n",
-    "IC_ac_max=IC+VCE/RAC;                               #Maximum collector current, mA\n",
-    "print(\"Maximum v_CE=%.2fV and maximum i_C=%.2fmA\"%(VCE_ac_max,IC_ac_max));\n",
-    "\n",
-    "#plot\n",
-    "vCE_plot=[0,VCE_ac_max];          #Plot variable for V_CE\n",
-    "iC_plot=[IC_ac_max,0];      #Plot variable for I_C\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.xlim(0,10)\n",
-    "plt.ylim(0,20)\n",
-    "plt.plot(vCE_plot,iC_plot);\n",
-    "plt.xlabel(\"vCE(V)\");\n",
-    "plt.ylabel(\"iC(mA)\");\n",
-    "plt.title(\"a.c load line\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.6: Page number 253-254"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f895dbf5390>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as p\n",
-    "\n",
-    "#Variable declaration\n",
-    "RC=10;                          #Collector resistor, kΩ\n",
-    "RL=30;                          #Load resistor, kΩ\n",
-    "VCC=20;                         #Collector supply voltage, V\n",
-    "IC=1;                           #Collector current, mA\n",
-    "VCE=10;                         #Collector-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#For d.c load line:\n",
-    "#From the equation: VCE=VCC-IC*(RC+RE),\n",
-    "#When VCE=0, IC is  maximum.\n",
-    "#Emitter resistor is neglected, assuming it as negligible\n",
-    "IC_max=VCC/RC;                      #Maximum collector current, mA\n",
-    "\n",
-    "#And, when IC=0, VCE is maximum\n",
-    "VCE_max=VCC;                        #Maximum collector-emitter voltage, V\n",
-    "\n",
-    "#plot\n",
-    "p.subplot(211)\n",
-    "p.xlim(0,20)\n",
-    "p.ylim(0,5)\n",
-    "VCE_plot=[0,VCE_max];          #Plot variable for V_CE\n",
-    "IC_plot=[IC_max,0];            #Plot variable for I_C\n",
-    "\n",
-    "p.plot(VCE_plot,IC_plot);\n",
-    "p.xlabel(\"VCE(V)\");\n",
-    "p.ylabel(\"IC(mA)\");\n",
-    "p.title(\"d.c load line\");\n",
-    "\n",
-    "\n",
-    "#For a.c load line:\n",
-    "RAC=(RC*RL)/(RC+RL);                #a.c Load resistor, kΩ\n",
-    "\n",
-    "VCE_ac_max=VCE+IC*RAC;                 #Maximum collector-emitter voltage, V\n",
-    "IC_ac_max=IC+ VCE/RAC;                #Maximum collector current, mA\n",
-    "\n",
-    "#plot\n",
-    "p.subplot(212)\n",
-    "p.xlim([0,25])\n",
-    "p.ylim([0,5])\n",
-    "vCE_plot=[0,VCE_ac_max];          #Plot variable for V_CE\n",
-    "iC_plot=[IC_ac_max,0];      #Plot variable for I_C\n",
-    "\n",
-    "p.plot(vCE_plot,iC_plot);\n",
-    "p.xlabel(\"vCE(V)\");\n",
-    "p.ylabel(\"iC(mA)\");\n",
-    "p.title(\"a.c load line\");\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.7 : Page number 254-255"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f898013a748>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as p\n",
-    "\n",
-    "#Variabe declaration\n",
-    "VCE_Q=8.0;           #Q-point collector emitter voltage, V\n",
-    "IC_Q=1;               #Q-point collector current, mA\n",
-    "ic_positive_peak=1.5;        #Collector current at positive peak of signal, mA\n",
-    "ic_negative_peak=0.5;         #Collector current at negative peak of signal, mA\n",
-    "vce_positive_peak=7;          #Collector emitter voltage at positive peak of signal, V\n",
-    "vce_negative_peak=9;          #Collector emitter voltage at negative peak of signal, V\n",
-    "\n",
-    "#Plot\n",
-    "vce_plot=[vce_positive_peak,vce_negative_peak];    #Plot variable of vce\n",
-    "ic_plot=[ic_positive_peak,ic_negative_peak];    #Plot variable of ic\n",
-    "\n",
-    "p.xlim(0,10)\n",
-    "p.ylim(0,2)\n",
-    "p.plot(vce_plot,ic_plot);\n",
-    "p.xlabel(\"vCE(V)\");\n",
-    "p.ylabel(\"iC(mA)\");\n",
-    "p.title(\"a.c load line\");\n",
-    "p.grid();\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.8 : Page number 256"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain= 24.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                 #Collector supply voltage, V\n",
-    "RC=2.0;                   #Collector resistor, kΩ\n",
-    "Rin=1.0;                  #Input resistance, kΩ\n",
-    "beta=60.0;                #Base current amplification factor\n",
-    "RL=0.5;                   #Load resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "RAC=(RC*RL)/(RC+RL);                #a.c load resistor, kΩ\n",
-    "Av=beta*(RAC/Rin);                  #Voltage gain\n",
-    "\n",
-    "#Results\n",
-    "print(\"Voltage gain= %d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.9 : Page number 256"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage= 200mV.\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "V_in=1.0;                      #Input voltage , mV\n",
-    "RC=10.0;                      #Collector resistor, kΩ\n",
-    "Rin=2.5;                      #Input resistance, kΩ\n",
-    "beta=100.0;                   #Base current amplification factor\n",
-    "RL=10.0;                      #Load resistor, kΩ\n",
-    "\n",
-    "#Calculations\n",
-    "RAC=(RC*RL)/(RC+RL);                #Effective load, kΩ\n",
-    "Av=beta*(RAC/Rin);                  #Voltage gain\n",
-    "\n",
-    "V_out=V_in*Av;                      #Output voltage, V\n",
-    "\n",
-    "#Results\n",
-    "print(\"Output voltage= %dmV.\"%V_out);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.10 : Page number 256-257"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Beta= 100.\n",
-      "Input impedance=2 kΩ.\n",
-      "a.c load=3.3 kΩ.\n",
-      "Voltage gain= 165.\n",
-      "Power gain=16500.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "change_in_IB=10.0;                    #Change in base current, 𝜇A\n",
-    "change_in_IC=1.0;                     #Change in collector current, mA\n",
-    "change_in_VBE=0.02;                   #Change in Base-emitter voltage, V\n",
-    "RC=5.0;                               #Collector resistor, kΩ\n",
-    "RL=10.0;                              #Emitter resistor, kΩ\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "beta=(change_in_IC*1000)/change_in_IB;         #Base current amplification factor\n",
-    "\n",
-    "#(ii)\n",
-    "Rin=(change_in_VBE/change_in_IB)*1000;         #Input impedance, kΩ\n",
-    "\n",
-    "#(iii)\n",
-    "RAC=round((RC*RL)/(RC+RL),1);                            #a.c load, kΩ\n",
-    "\n",
-    "#(iv)\n",
-    "Av=beta*RAC/Rin;                                #Voltage gain\n",
-    "\n",
-    "#(v)\n",
-    "Ap=beta*Av;                                     #Power gain\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"Beta= %d.\"%beta);\n",
-    "print(\"Input impedance=%d kΩ.\"%Rin);\n",
-    "print(\"a.c load=%.1f kΩ.\"%RAC);\n",
-    "print(\"Voltage gain= %d.\"%Av);\n",
-    "print(\"Power gain=%d.\"%Ap);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.11 : Page number 257"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=200mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "beta=50.0;                      #Base current amplification factor\n",
-    "RC=3.0;                         #Collector resistor,kΩ\n",
-    "RL=6.0;                         #Load resistor, kΩ\n",
-    "Rin=0.5;                        #Input impedance, kΩ\n",
-    "Vin=1;                          #Input voltage,  mV\n",
-    "\n",
-    "#Calculation\n",
-    "RAC=(RC*RL)/(RC+RL);            #a.c load, kΩ\n",
-    "Av=beta*RAC/Rin;                #Voltage gain\n",
-    "Vout=Vin*Av;                    #Output voltage, V\n",
-    "\n",
-    "#Results\n",
-    "print(\"Output voltage=%dmV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.12 : Page number 257-258"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The circuit is not operating properly.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VT=6.0;                   #Collector potential, V\n",
-    "R1=1.0;                   #Resistor R1, kΩ\n",
-    "R2=2.0;                   #Resistor R2, kΩ\n",
-    "VB_found=4.0;             #Measured base voltage, V\n",
-    "\n",
-    "#Calculations\n",
-    "VB=(VT*R1)/(R1+R2);                     #Theoretical base voltage, V\n",
-    "\n",
-    "if(VB_found==VB):\n",
-    "    print(\"The circuit is operating properly.\");\n",
-    "else:\n",
-    "    print(\"The circuit is not operating properly.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.13 : Page number 258-259"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a.c emitter resistance= 38.46 Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                   #Collector supply voltage, V\n",
-    "R1=40.0;                    #Resistor R1, kΩ\n",
-    "R2=10.0;                    #Resistor R2, kΩ\n",
-    "RC=6.0;                     #Collector resistor, kΩ\n",
-    "RE=2.0;                     #Emitter resistor, kΩ\n",
-    "beta=80;                    #Base current amplification factor\n",
-    "VBE=0.7;                    #Base emitter voltage, V\n",
-    "\n",
-    "#Calculations\n",
-    "V2=(VCC*R2)/(R1+R2);            #Voltage across resistor R2, V\n",
-    "VE=V2-VBE;                      #Emitter voltage, V\n",
-    "IE=VE/RE;                       #Emitter current,  mA\n",
-    "re=25/IE;                       #a.c emitter resistance, Ω\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"a.c emitter resistance= %.2f Ω.\"%re);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.14  : Page number 262-263"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)Voltage gain= 360.\n",
-      "(ii)Voltage gain= 5.37.\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "VCC=20.0;                     #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base-emitter voltage, V\n",
-    "R1=150.0;                     #Resistor R1, kΩ\n",
-    "R2=20.0                       #Resistor R2, kΩ\n",
-    "RC=12.0;                      #Collector resistor, kΩ\n",
-    "RE=2.2;                       #Emitter resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),2);                  #Voltage across R2, V\n",
-    "VE=round(V2-VBE,2);                          #Voltage across emitter resistor, V\n",
-    "IE=round(VE/RE,2);                           #Emitter current, mA\n",
-    "re=round(25/IE,1);                           #a.c emitter resistance, Ω\n",
-    "\n",
-    "\n",
-    "#(i)\n",
-    "#CE(emitter capacitor) connected in the circuit:\n",
-    "Av=(RC*1000)/re;                                   #Voltage gain for emitter capacitor connected.\n",
-    "\n",
-    "print(\"(i)Voltage gain= %d.\"%Av);\n",
-    "\n",
-    "#(ii)\n",
-    "#CE(emitter capacitor) removed from the circuit:\n",
-    "Av=(RC*1000)/(re+RE*1000);                          #Voltage gain for emitter capacitor removed.\n",
-    "\n",
-    "print(\"(ii)Voltage gain= %.2f.\"%Av);\n",
-    "\n",
-    "#Note: The answer in the text book has been approximated to 5.38 but it's actually coming 5.37.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.15 : Page number 263"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain= 120.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=6.0;                       #Collector resistor, kΩ\n",
-    "RL=12.0;                      #Load resistor, kΩ\n",
-    "re=33.3;                      #a.c emitter resistance, Ω\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "RAC=RC*RL/(RC+RL);              #a.c effective load, kΩ\n",
-    "Av=RAC*1000/re;                 #Voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"Voltage gain= %d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.16 : Page number 263-264"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) a.c emitter resistance=250 Ω.\n",
-      "(ii) Voltage gain =80.\n",
-      "(iii) d.c voltage across input capacitor= 1V and emitter capacitor=0.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=9.0;                      #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base-emitter voltage, V\n",
-    "R1=240.0;                     #Resistor R1, kΩ\n",
-    "R2=30.0                       #Resistor R2, kΩ\n",
-    "RC=20.0;                      #Collector resistor, kΩ\n",
-    "RE=3.0;                       #Emitter resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "V2=round(VCC*R2/(R1+R2),1);                  #Voltage across R2, V\n",
-    "VE=round(V2-VBE,1);                          #Voltage across emitter resistor, V\n",
-    "IE=round(VE/RE,1);                           #Emitter current, mA\n",
-    "re=25/IE;                                    #a.c emitter resistance, Ω\n",
-    "\n",
-    "#(ii)\n",
-    "Av=RC*1000/re;                                   #Voltage gain\n",
-    "\n",
-    "#(iii)\n",
-    "V_C_in=V2;                                  #d.c voltage across input capacitor, V\n",
-    "V_C_E=VE;                                   #d.c vooltage across emitter capacitor, V\n",
-    "\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"(i) a.c emitter resistance=%d Ω.\"%re);\n",
-    "print(\"(ii) Voltage gain =%d.\"%Av);\n",
-    "print(\"(iii) d.c voltage across input capacitor= %dV and emitter capacitor=%.1fV.\"%(V_C_in,V_C_E));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.17 : Page number 264-265"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) D.C bias levels: V2=3V, VE=2.3V, IE=2.3mA, IC=2.3mA, IB=0.023mA and VC=10.4V.\n",
-      "(ii) D.c voltage across: Cin=3V and CE=2.3V and CC=10.4V.\n",
-      "(iii) a.c emitter resistance=10.9Ω.\n",
-      "(iv) Voltage gain=61.2.\n",
-      "(v) VC>VE. Therefore, the transistor is in active state.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15.0;                      #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base-emitter voltage, V\n",
-    "R1=40.0;                     #Resistor R1, kΩ\n",
-    "R2=10.0                       #Resistor R2, kΩ\n",
-    "RC=2.0;                      #Collector resistor, kΩ\n",
-    "RE=1.0;                       #Emitter resistor, kΩ\n",
-    "RL=1.0;                       #Load resistor, kΩ\n",
-    "beta=100;                     #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) D.C bias levels\n",
-    "V2=VCC*R2/(R1+R2);                  #Voltage across R2, V\n",
-    "VE=round(V2-VBE,1);                 #Voltage across emitter resistor, V\n",
-    "IE=round(VE/RE,1);                  #Emitter current, mA\n",
-    "IC=IE;                              #Collector current, mA\n",
-    "IB=IC/beta;                         #Base  current, mA\n",
-    "VC=VCC-IC*RC;                       #Collector voltage, V\n",
-    "print(\"(i) D.C bias levels: V2=%dV, VE=%.1fV, IE=%.1fmA, IC=%.1fmA, IB=%.3fmA and VC=%.1fV.\"%(V2,VE,IE,IC,IB,VC));\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "Cin_V=V2;                            #Voltage across Cin capacitor, V\n",
-    "CE_V=VE;                             #Voltage across CE capacitor, V \n",
-    "CC_V=VC;                             #Voltage across CC capacitor, V\n",
-    "print(\"(ii) D.c voltage across: Cin=%dV and CE=%.1fV and CC=%.1fV.\"%(Cin_V,CE_V,CC_V));\n",
-    "\n",
-    "#(iii)\n",
-    "re=round(25/IE,1);                           #a.c emitter resistance, Ω\n",
-    "print(\"(iii) a.c emitter resistance=%.1fΩ.\"%re);\n",
-    "\n",
-    "\n",
-    "#(iv)\n",
-    "RAC=round(RC*RL/(RC+RL),3);                #Total a.c collector resistance, kΩ\n",
-    "Av=RAC/(re/1000);                        #Voltage gain\n",
-    "print(\"(iv) Voltage gain=%.1f.\"%Av);\n",
-    "\n",
-    "#(v)\n",
-    "print(\"(v) VC>VE. Therefore, the transistor is in active state.\" );\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.18 : page number 265"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The power gain = 26400 and output power = 1.584W.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=132.0;                   #Voltage gain\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "P_in=60.0;                     #Input power, 𝜇W\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "Ap=beta*Av;                          #Power gain\n",
-    "P_out=Ap*(P_in/10**6);                #Output power, W\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"The power gain = %d and output power = %.3fW.\"%(Ap,P_out));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.19 : page number 265-266"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Current gain=49\n",
-      "(ii) Voltage gain=2.14\n",
-      "(iii) Power gain=105.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IB=200.0;                       #Base current, microampere\n",
-    "IE=10.0;                        #Emitter current, mA\n",
-    "R1=27.0;                      #Resistor R1, kilo ohm\n",
-    "R2=13.0                       #Resistor R2, kilo ohm\n",
-    "RC=4.7;                       #Collector resistor, kilo ohm\n",
-    "RE=2.2;                       #Emitter resistor, kilo ohm\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "IC=IE-(IB/1000);                              #Collector current, mA\n",
-    "beta=IC/(IB/1000);                            #Current gain\n",
-    "\n",
-    "print(\"(i) Current gain=%d\"%beta);\n",
-    "\n",
-    "#(ii)\n",
-    "#a.c emitter resistance is neglected, voltage gain=(collector resistor)/(emitter resistor)\n",
-    "Av=RC/RE;                           #Voltage gain\n",
-    "\n",
-    "print(\"(ii) Voltage gain=%.2f\"%Av);\n",
-    "\n",
-    "#(iii)\n",
-    "Ap=round(beta*Av,0);                         #Power gain\n",
-    "\n",
-    "#Results\n",
-    "print(\"(iii) Power gain=%d.\"%Ap);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.20 : Page number 266-267"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance of the amplifier circuit= 3.46 kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=30.0;                     #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base emitter voltage, V\n",
-    "R1=45.0;                      #Resistor R1, kΩ\n",
-    "R2=15.0                       #Resistor R2, kΩ\n",
-    "RC=10.0;                      #Collector resistor,kΩ\n",
-    "RE=7.5;                       #Emitter resistor, kΩ\n",
-    "beta=200.0;                   #Base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),1);                  #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2;                                       #Voltage across emitter resistor(base-emitter voltage is neglected), V\n",
-    "IE=VE/RE;                                    #Emitter current, mA (OHM's LAW)\n",
-    "re=25/IE;                                    #a.c emitter resistance, ohm\n",
-    "Zin_base=(beta*re)/1000;                     #input impedance of transistor base,kΩ\n",
-    "R1_R2=(R1*R2)/(R1+R2);                       #Parallel resistance between R1 and R2, kΩ\n",
-    "Zin=((R1_R2)*Zin_base)/(R1_R2+Zin_base);     #Input impedance of the amplifier circuit, kΩ\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance of the amplifier circuit= %.2f kΩ.\"%Zin);       \n",
-    "\n",
-    "#Note: The input impedance of the amplifier circuit is approximated as 3.45 kΩ in the text book, but actually it's 3.46 kΩ.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.21 : Page Number 268-269"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the swamped amplifier= 4.67.\n",
-      "Input impedance of transistor base of the swamped amplifier= 48.21 kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                           #Collector supply voltage, V.\n",
-    "RC=1.5;                             #Collector resistor, kΩ.\n",
-    "R1=18.0;                            #Resistor R1, kΩ.\n",
-    "R2=4.7;                             #Resistor R2, kΩ.\n",
-    "RE1=300.0;                          #Emitter resistor 1, Ω.\n",
-    "RE2=900.0;                          #Emitter resistor 2, Ω.\n",
-    "VBE=0.7;                            #Base-emitter voltage, V.\n",
-    "beta=150.0;                         #Base current amplification factor.\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),1);                      #d.c voltage across R2, V. (Voltage divider rule)\n",
-    "VE=round(V2-VBE,1);                              #d.c voltage across RE, V.\n",
-    "IE=round((VE/(RE1+RE2))*1000,2);                 #d.c emitter current, mA.(OHM'S LAW)\n",
-    "re=round(25/IE,1);                               #a.c emitter resistance, Ω.\n",
-    "Av=RC*1000/(re+RE1);                             #Voltage gain\n",
-    "Zin_base=(beta*(re+RE1))/1000;                   #Input impedance of transistor base, kΩ.\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"The voltage gain of the swamped amplifier= %.2f.\"%Av);\n",
-    "print(\"Input impedance of transistor base of the swamped amplifier= %.2f kΩ.\"%Zin_base);\n",
-    "\n",
-    "#Note:In the textbook Av is approximated to 4.66and Zin_base to 48.22 kilo ohm, but the actual answers come as 4.67 and 48.21 kilo ohm.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.22 : Page number 269"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The percentage change from the original value= 6.42%(decrease)\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=1.5;                             #Collector resistor, kΩ.\n",
-    "RE1=300.0;                          #Emitter resistor 1, Ω.\n",
-    "re=21.5;                            #a.c emitter resistance, Ω.\n",
-    "\n",
-    "#Calculations\n",
-    "Av=round(RC*1000/(re+RE1),2);                     #Voltage gain.\n",
-    "Av_1=round(RC*1000/(2*re+RE1),2);                 #Voltage gain when re doubles.\n",
-    "change_in_gain=round(Av-Av_1,2);                      #Change in voltage gain.\n",
-    "change_percentage=change_in_gain*100/Av;               #Change percentage\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "if(change_in_gain>0):\n",
-    "    print(\"The percentage change from the original value= %.2f%%(decrease)\"%change_percentage);\n",
-    "else:\n",
-    "    print(\"The percentage change from the original value= %.2f%%(increase)\"%change_percentage);\n",
-    "\n",
-    "\n",
-    "#Note: The percentage has been approximated in the text book as 6.22%, but the answer comes as 6.42%.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.23 : Page number 269-270"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) input impedance of transistor base for standard amplifier= 5 kilo ohm\n",
-      "    input impedance of transistor base for swamped amplifier= 47 kilo ohm\n",
-      "(ii) input impedance for standard amplifier= 1.33 kilo ohm\n",
-      "     input impedance for swamped amplifier= 1.74 kilo ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                     #Collector supply voltage, V\n",
-    "VBE=0.7;                      #Base emitter voltage, V\n",
-    "R1=10.0;                      #Resistor R1, kilo ohm\n",
-    "R2=2.2;                       #Resistor R2, kilo ohm\n",
-    "RC=4.0;                       #Collector resistor, kilo ohm\n",
-    "RE=1.1;                       #Emitter resistor, kilo ohm\n",
-    "beta=200.0;                   #Base current amplification factor\n",
-    "RE1=210.0;                    #Emitter resistor 1 of swamped amplifier, ohm.\n",
-    "RE2=900.0;                    #Emitter resistor 2 of swamped amplifier, ohm.\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "V2=round(VCC*R2/(R1+R2),1);                      #d.c voltage across R2, V. (Voltage divider rule)\n",
-    "VE=round(V2-VBE,1);                              #d.c voltage across RE, V.\n",
-    "IE=(VE/RE);                                      #d.c emitter current, mA.(OHM'S LAW)\n",
-    "re=25/IE;                                        #a.c emitter resistance, ohm.\n",
-    "\n",
-    "\n",
-    "#(i) Zin_base:\n",
-    "Zin_base_standard=(beta*re)/1000;                      #input impedance of transistor base for standard amplifier , kilo ohm.\n",
-    "Zin_base_swamped=(beta*(re+RE1))/1000;                 #input impedance of transistor base for swamped amplifier, kilo ohm.\n",
-    "\n",
-    "\n",
-    "#(ii) Zin:\n",
-    "#input impedance for standard amplifier circuit\n",
-    "Zin_standard=(((R1*R2)/(R1+R2))*Zin_base_standard)/(Zin_base_standard +((R1*R2)/(R1+R2)));    #kilo ohm\n",
-    "\n",
-    "#input impedance for standard amplifier circuit\n",
-    "Zin_swamped=(((R1*R2)/(R1+R2))*Zin_base_swamped)/(Zin_base_swamped +((R1*R2)/(R1+R2)));       #kilo ohm\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"(i) input impedance of transistor base for standard amplifier= %d kilo ohm\"%Zin_base_standard);\n",
-    "print(\"    input impedance of transistor base for swamped amplifier= %d kilo ohm\"%Zin_base_swamped);\n",
-    "print(\"(ii) input impedance for standard amplifier= %.2f kilo ohm\"%Zin_standard);\n",
-    "print(\"     input impedance for swamped amplifier= %.2f kilo ohm\"%Zin_swamped);\n",
-    "\n",
-    "\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.24 : Page number 270-271"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of standard amplifier=160.\n",
-      "The voltage gain of swamped amplifier=17.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=4.0;                     #Collector resistor, kilo ohm\n",
-    "re=25.0;                    #a.c emitter resistance, ohm (calculated in example 10.23)\n",
-    "RE_1=210.0;                  #Emitter resistor 1 of swamped amplifier,ohm\n",
-    "\n",
-    "#Calculation\n",
-    "Av_standard=(RC*1000)/re;                       #Voltage gain of standard common emitter amplifier\n",
-    "Av_swamped=(RC*1000)/(re+RE_1);                 #Voltage gain of swamped amplifier\n",
-    "\n",
-    "#Results\n",
-    "print(\"The voltage gain of standard amplifier=%d.\"%Av_standard);\n",
-    "print(\"The voltage gain of swamped amplifier=%d.\"%Av_swamped);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.26 : Page number 273-274"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The required input signal voltage =2.5mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_0=1000.0;                  #Open circuit voltage gain\n",
-    "R_in=2.0;                    #Input resistance, kilo ohm\n",
-    "R_out=1.0;                   #Output resistance, ohm\n",
-    "RL=4;                        #Load resistor across the output, ohm\n",
-    "I_2=0.5;                     #Output signal current, A.\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "#Since A_0*(I_1*R_in) = I_2*(R_out+RL)\n",
-    "I_1=I_2*(R_out+RL)/(A_0*(R_in*1000));           #Input current, A\n",
-    "V_1=I_1*(R_in*1000);                            #Input signal voltage, V\n",
-    "V_1=V_1*1000;                                   #Input signal voltage, mV\n",
-    "\n",
-    "print(\"The required input signal voltage =%.1fmV\"%V_1);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.27 : Page number 274"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The magnitude of output voltage = 4.9V\n",
-      "The power gain =98e-06.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_0=1000.0;                     #Open circuit voltage gain\n",
-    "R_in=7.0;                       #Input resistance, kilo ohm\n",
-    "R_out=15.0;                     #Output resistance, ohm\n",
-    "RL=35.0;                        #Load resistor across the output, ohm\n",
-    "R_s=3.0;                        #Internal resistance, kilo ohm\n",
-    "E_s=10.0;                       #Input signal voltage, mV.\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "I_1=E_s*(10**-3)/(R_s*1000+R_in*1000);                  #Input current, A\n",
-    "V_1=I_1*(R_in*1000);                                    #Voltage across input resistance, V\n",
-    "\n",
-    "#Since, A_v=V_2/V_1 = A_0*RL/(R_out+RL)\n",
-    "A_v=A_0*RL/(R_out+RL);                                 #Voltage gain\n",
-    "V_2=A_v*V_1;                                            #Outout voltage, V\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "P_2=V_2**2/RL;                              #Output power, W\n",
-    "P_1=V_1**2/(R_in*1000);                            #Input power, W\n",
-    "A_p=round(P_2/P_1,-6);                                #Power gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The magnitude of output voltage = %.1fV\"%V_2);\n",
-    "print(\"The power gain =%de-06.\"%(A_p/10**6));\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 10.28 : Page number 274-275"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Necessary input signal voltage= 12.5mV\n",
-      "Input signal current =4.17  μA\n",
-      "Power gain = 9600.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_v=80.0;                   #Voltage gain\n",
-    "V_2=1.0;                    #Output voltage, V\n",
-    "A_i=120.0;                  #Current gain\n",
-    "RL=2;                       #Load resistor, kilo ohm\n",
-    "\n",
-    "#Calculation\n",
-    "V_1=(V_2/A_v)*1000;                #Input signal voltage, mV\n",
-    "\n",
-    "#Since, A_i=A0*R_in/(R_out+RL) and A_v=A0*RL/(R_out+RL)\n",
-    "#So, A_v/A_i=RL/R_in\n",
-    "R_in=RL*A_i/A_v;                #Input resistance, kilo ohm\n",
-    "I_1=V_1/R_in;                   #Input current,  μA\n",
-    "A_p=A_i*A_v;                    #Power gain\n",
-    "\n",
-    "#Results\n",
-    "print(\"Necessary input signal voltage= %.1fmV\"%V_1);\n",
-    "print(\"Input signal current =%.2f  μA\"%I_1);\n",
-    "print(\"Power gain = %d.\"%A_p);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11.ipynb
deleted file mode 100755
index 6e0fb200..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11.ipynb
+++ /dev/null
@@ -1,1025 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8e425c9c2dfbfee43b3a89e44b0fd7936ba869da73ac3c372e9b23848f1cded1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 11 : MULTISTAGE TRANSISTOR AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 11.1 : Page number 285"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "#Variable declaration\n",
-      "#(i)\n",
-      "A_v=30;                 #Voltage gain\n",
-      "\n",
-      "#(ii)\n",
-      "A_p=100;                #Power gain\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_v_dB=20*log10(A_v);          #Voltage gain, dB\n",
-      "A_p_dB=10*log10(A_p);          #Power gain, dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i) Voltage gain in dB=%.2fdB\"%A_v_dB);\n",
-      "print(\"(ii) Power gain in dB=%ddB\"%A_p_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Voltage gain in dB=29.54dB\n",
-        "(ii) Power gain in dB=20dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.2 : Page number 285-286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#(i)\n",
-      "A_p_dB=40.0;                #Power gain in dB\n",
-      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
-      "A_p=10**A_p_b;              #Power gain in number\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) Power gain in number=%d\"%A_p);\n",
-      "\n",
-      "#(ii)\n",
-      "A_p_dB=43.0;                #Power gain in dB\n",
-      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
-      "A_p=round(10**A_p_b,-4);              #Power gain in number\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) Power gain in number=%d\"%A_p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Power gain in number=10000\n",
-        "(ii) Power gain in number=20000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.3 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "Av_1=100.0;                     #Voltage gain of stage 1\n",
-      "Av_2=200.0;                     #Voltage gain of stage 2\n",
-      "Av_3=400.0;                     #Voltage gain of stage 3\n",
-      "\n",
-      "#Calculations\n",
-      "Av_1_dB=20*log10(Av_1);                     #Voltage gain of stage 1, dB\n",
-      "Av_2_dB=20*log10(Av_2);                     #Voltage gain of stage 2, dB\n",
-      "Av_3_dB=20*log10(Av_3);                     #Voltage gain of stage 3, dB\n",
-      "\n",
-      "Av_T=Av_1_dB+Av_2_dB+Av_3_dB;            #Total voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total voltage gain=%ddB\"%Av_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total voltage gain=138dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.4 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_p_absolute=30.0;                          #Absolute gain of each stage\n",
-      "number_of_stages=5.0;                       #number of stages\n",
-      "negative_feedback=10.0;                     #negative feedback, dB\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_p_dB=round(10*log10(A_p_absolute),2);              #Power gain of one stage. dB\n",
-      "A_p_T=number_of_stages * A_p_dB;            #Total power gain, dB\n",
-      "\n",
-      "#(ii)\n",
-      "A_p_resultant=A_p_T-negative_feedback;      #Resultant power gain with negative feedback, dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total power gain = %.2fdB.\"%A_p_T);\n",
-      "print(\"The resultant power gain with negative feedback = %.2fdB.\"%A_p_resultant);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power gain = 73.85dB.\n",
-        "The resultant power gain with negative feedback = 63.85dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.5 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "P_out_2kHz=1.5;                 #Output power at 2 kHz, W\n",
-      "P_out_20kHz=0.3;                #Output power at 20 kHz, W\n",
-      "P_in=10.0;                      #Input power, mW\n",
-      "\n",
-      "#Calculations\n",
-      "A_p_dB_2kHz=10*log10(P_out_2kHz*1000/P_in);         #dB power gain at 2 kHz\n",
-      "A_p_dB_20kHz=10*log10(P_out_20kHz*1000/P_in);       #dB power gain at 20 kHz\n",
-      "Fall_in_gain=A_p_dB_2kHz-A_p_dB_20kHz;              #Fall in gain from 2kHz to 20kHz\n",
-      "\n",
-      "#Results\n",
-      "print(\"The fall in gain from 2kHz to 20kHz=%.2fdB\"%Fall_in_gain);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fall in gain from 2kHz to 20kHz=6.99dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.6 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_v=15.0;                   #Voltage gain, dB\n",
-      "V_1=0.8;                    #Input signal voltage, V\n",
-      "\n",
-      "#Calculations\n",
-      "#Since, Av(in decibel)=20*log10(V_2/V_1),\n",
-      "V_2=V_1*(10**(A_v/20));                 #Output voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The output voltage= %.1fV.\"%V_2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage= 4.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.7 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_0_dB=70.0;                   #Open circuit voltage gain, dB\n",
-      "A_v_dB=67.0;                   #Voltage gain, dB\n",
-      "R_out=1.5;                     #Output resistance, kilo ohm\n",
-      "\n",
-      "#Calculations\n",
-      "#Since, A_0_dB-A_v_dB=20*log10(A_0/A_v)\n",
-      "ratio_A0_Av=round(10**((A_0_dB-A_v_dB)/20),2);                  #Ratio of open-circuit voltage gain to normal voltage gain\n",
-      "\n",
-      "#Since, A_v/A_0 = RL/(R_out+RL)\n",
-      "RL=R_out/(ratio_A0_Av-1);                               #Load resistor, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The load resistance=%.2f kilo ohm.\"%RL);\n",
-      "\n",
-      "#Note: The value of load resistor is calculated to be 3.6585 kilo ohm and approximated to 3.66. But, in the text it has been approximated to 3.65.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load resistance=3.66 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.8 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=1.0;                 #Load resistance,  kilo ohm\n",
-      "A_v=40.0;               #Voltage gain, dB\n",
-      "V_in=10.0;              #Input signal voltage, mV\n",
-      "\n",
-      "#Calcultaions\n",
-      "#(i)\n",
-      "#Since, A_v=20*log10(V_out/V_in)\n",
-      "V_out=V_in*(10**(A_v/20))/1000;              #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "P_L=(V_out**2/RL);                            #The load power, mW\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i)The output voltage is %dV.\"%V_out);\n",
-      "print(\"(ii)The load poweris %dmW.\"%P_L);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The output voltage is 1V.\n",
-        "(ii)The load poweris 1mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.9 : Page number 287-288\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "P_2=40.0;                     #Output power, W\n",
-      "R=10.0;                        #Resistance of speaker, ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_p_dB=25.0;                                   #Power gain, dB\n",
-      "#Since, A_p_dB=10*log10(P_2/P_1)\n",
-      "P_1=(P_2/10**(A_p_dB/10))*1000;                #Input power, mW\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "A_v_dB=40.0;                                    #Voltage gain, dB\n",
-      "\n",
-      "#Since, P=(V**2)/R,\n",
-      "V_2=(P_2*R)**0.5;                               #Output voltage, V\n",
-      "\n",
-      "#Since, A_v_dB=20*log10(V_2/V_1)\n",
-      "V_1=(V_2/10**(A_v_dB/20))*1000;                #Input voltage, mV\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print(\"(i)The input power=%.1fmW.\"%P_1);\n",
-      "print(\"(ii)The input voltage=%dmV.\"%V_1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The input power=126.5mW.\n",
-        "(ii)The input voltage=200mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.10 : Page  number 288\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v_max=2000.0;                 #Maximum voltage gain\n",
-      "f_max=2.0;                      #Frequency at which maximum voltage gain occurs,kHz\n",
-      "A_v=1414.0;                     #Voltage gain at 50 Hz and 10kHz\n",
-      "f1=50;                          #Lower frequency at which gain is 1414, Hz\n",
-      "f2=10;                          #Upper frequency at which gain is 1414, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, bandwidth is the range of frequency  over which gain is greater than or equal to 70.7% of maximum gain\n",
-      "if((A_v/A_v_max)*100 ==70.7):                               \n",
-      "    print(\"(i)The bandwidth is from %dHz to %dkHz.\"%(f1,f2));\n",
-      "    print(\"(ii)The lower cut-off frequency=%dHz.\"%f1);\n",
-      "    print(\"(iii)The upper cut-off frequency=%dkHz.\"%f2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The bandwidth is from 50Hz to 10kHz.\n",
-        "(ii)The lower cut-off frequency=50Hz.\n",
-        "(iii)The upper cut-off frequency=10kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11 : Page number 291\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v=60.0;               #Voltage gain of single stage amplifier\n",
-      "R_C=500.0;              #Collector load, ohm\n",
-      "R_in=1.0;               #Input impedance, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, there is no loading , second stage gain remains at A_v\n",
-      "#But, due to loading effect of input impedance of second stage, gain of first stage decreases\n",
-      "A_v_2=A_v;                                                    #Voltage gain of second stage\n",
-      "R_AC=round((R_C*R_in*1000)/(R_C+R_in*1000),0);              #Effective load of first stage, ohm\n",
-      "A_v_1=A_v*R_AC/R_C;                                         #Gain of first stage\n",
-      "A_v_T=A_v_1*A_v_2;                                            #Total gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total gain=%d.\"%A_v_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total gain=2397.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.12 : Page number 291-292\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rin=1.0;                        #Input resistance, kilo ohm\n",
-      "beta=100.0;                     #base current amplification factor\n",
-      "RC=2.0;                         #Collector load, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_AC=(RC*Rin)/(RC+Rin);               #Effective load on first stage, kilo ohm\n",
-      "A_v_1=round(beta*(R_AC/Rin),0);                #Voltage gain of first stage\n",
-      "\n",
-      "#(ii)\n",
-      "A_v_2=round(beta*RC/Rin,0);                      #Voltage gain of second stage\n",
-      "\n",
-      "#(iii)\n",
-      "A_v_T=A_v_1*A_v_2;                      #Total voltage gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i)The voltage gain of first stage =%d.\"%A_v_1);\n",
-      "print(\"(ii)The voltage gain of second stage =%d.\"%A_v_2);\n",
-      "print(\"(iii)The total voltage gain =%d.\"%A_v_T);\n",
-      "\n",
-      "#Note: The approximation inthe text for A_v_1=66.66 is taken as 66 but here it has been taken 67 and therefore the total voltage is 13400 instead of 13200.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage gain of first stage =67.\n",
-        "(ii)The voltage gain of second stage =200.\n",
-        "(iii)The total voltage gain =13400.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.13 : Page number 292\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=10.0;                    #Collector load of single stage amplifier, kilo ohm\n",
-      "Rin=1.0;                    #Input resistance, kilo ohm\n",
-      "beta=100.0;                 #base current amplification factor\n",
-      "RL=100.0;                   #Load resistor, ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "R_AC=round((RC*1000)*RL/(RC*1000+RL),-1);                 #Effective collector load,\n",
-      "A_v=beta*R_AC/(Rin*1000);                              #Voltage gain\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage gain=%d.\"%A_v);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain=10.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.14 : Page number 292-293\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                       #Collector suppply voltage, V\n",
-      "R1=10.0;                        #resistor R1, kilo ohm\n",
-      "R2=2.2;                         #resistor R2, kilo ohm\n",
-      "R3=10.0;                        #resistor R3, kilo ohm\n",
-      "R4=2.2;                         #resistor R4, kilo ohm\n",
-      "RC_1=3.6;                         #Collector resistor of first stage, kilo ohm\n",
-      "RC_2=4.0;                         #Collector resistor of second stage, kilo ohm\n",
-      "RE_1=900.0;                     #Emitter resistor of first stage, ohm\n",
-      "RE_2=1.0;                       #Emitter resistor of second stage, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Biasing potential for the second stage is the voltage across R4 resistor,\n",
-      "#so, by voltage divider rule:\n",
-      "VB=VCC*R4/(R3+R4);                #Biasing potential for second stage,(Voltage across R4), V\n",
-      "\n",
-      "print(\"The biasing voltage for the second stage=%.1fV.\"%VB);\n",
-      "\n",
-      "#If coupling capacitor C_c is replaced by a wire, RC_1 and R3 become parallel\n",
-      "Req=round((RC_1*R3)/(RC_1+R3),2);        #Equivalent resistance of R3 parallel with RC_1, kilo ohm\n",
-      "VB=VCC*R4/(Req+R4);             #Biasing voltage if coupling capacitor is replaced by a wire, V\n",
-      "\n",
-      "print(\"The biasing voltage after replacing coupling capacitor by wire=%.2fV.\"%VB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The biasing voltage for the second stage=3.6V.\n",
-        "The biasing voltage after replacing coupling capacitor by wire=9.07V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.15 : Page number 293-294\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return (r1*r2)/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage, V\n",
-      "R1=22.0;                            #Resistor R1, kilo ohm\n",
-      "R2=3.3;                            #Resistor R2, kilo ohm\n",
-      "R3=5.0;                            #Resistor R3, kilo ohm\n",
-      "R4=1.0;                            #Resistor R4, kilo ohm\n",
-      "R5=15.0;                            #Resistor R5, kilo ohm\n",
-      "R6=2.5;                            #Resistor R6, kilo ohm\n",
-      "R7=5.0;                            #Resistor R7, kilo ohm\n",
-      "R8=1.0;                            #Resistor R8, kilo ohm\n",
-      "beta=200;                           #Base current amplification factor\n",
-      "RL=10.0;                            #Load resistor, kilo ohm\n",
-      "V_BE=0.7;                           #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#for 2nd stage\n",
-      "V_R6=round(VCC*R6/(R5+R6),2);                #Voltage across R6, V (voltage divider rule)\n",
-      "V_R8=round(V_R6-V_BE,2);                     #Voltage across R8, V\n",
-      "IE_2=round(V_R8/R8,2);                         #Emitter current through R8, mA (OHM's LAW)\n",
-      "re_2nd_stage=round(25/IE_2,1);                 #a.c emitter resistance for 2nd stage, ohm\n",
-      "\n",
-      "#For 1st stage\n",
-      "V_R2=round(VCC*R2/(R1+R2),2);                     #Voltage across R2, V (voltage divider rule)\n",
-      "V_R4=round(V_R2-V_BE,2);                          #Voltage across R4, V\n",
-      "IE_1=round(V_R4/R4,2);                            #Emitter current through R4, mA (OHM's LAW)\n",
-      "re_1st_stage=round(25/IE_1,1);                    #a.c emitter resistance for 1st stage, ohm\n",
-      "\n",
-      "#(i)\n",
-      "Zin_base_2nd_stage=round((beta*re_2nd_stage)/1000,2);               #input resistance of transistor base of 2nd stage, kilo ohm\n",
-      "Zin=round(pr(pr(R5,R6),Zin_base_2nd_stage),2);                      #Input impedance of the 2nd stage, kilo ohm\n",
-      "R_AC_1st_stage=round(pr(R3,Zin),2);                                          #Effective collector load for 1st stage, kilo ohm\n",
-      "A_v_1=round(R_AC_1st_stage*1000/re_1st_stage,0);                                  #voltage gain of 1st stage\n",
-      "\n",
-      "#(ii)\n",
-      "R_AC_2nd_stage=round(pr(R7,RL),2);                          #Effective collector load for 2nd stage, kilo ohm\n",
-      "A_v_2=round(R_AC_2nd_stage*1000/re_2nd_stage,1);                  #voltage gain of 2nd stage\n",
-      "\n",
-      "#(iii)\n",
-      "A_v_overall=A_v_1*A_v_2;                            #overall voltage gain\n",
-      "\n",
-      "\n",
-      "#results\n",
-      "print(\"(i)The voltage gain of 1st stage=%.0f.\"%A_v_1);\n",
-      "print(\"(i)The voltage gain of 2nd stage=%.1f.\"%A_v_2);\n",
-      "print(\"(i)The overall voltage gain =%d.\"%A_v_overall);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage gain of 1st stage=53.\n",
-        "(i)The voltage gain of 2nd stage=191.4.\n",
-        "(i)The overall voltage gain =10144.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.16 : Page number 297\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Primary_impedance=1000.0;                       #Primary impedance, ohm\n",
-      "Load_impedance=10.0;                            #Load impedance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#since,for maximum power transfer primary impedance should be equal to output impedance\n",
-      "#and, impedance of secondary should be equal to load impedance\n",
-      "#therfore, primary_impedance/load_impedance=square of(primary to secondary turn ratio)\n",
-      "n=(Primary_impedance/Load_impedance)**0.5;           #Primary to secondary turn ratio\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print('The primary to secondary turn ratio for maximum power transfer=%d.'%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary to secondary turn ratio for maximum power transfer=10.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.17 : Page number 297\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=16.0;                    #Load resistor, ohm\n",
-      "R_p=10.0;                   #Output impedance of primary, kilo ohm\n",
-      "Vp=10.0;                    #Terminal voltage of the source, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, for maximum power transfer, the impedance of the primary should be equal to output impedance of the source\n",
-      "n=(R_p*1000/RL)**0.5;                        #Primary to secondary turns ratio\n",
-      "\n",
-      "#Since, power in a transformer remains constant,\n",
-      "#ratio of primary to secondary voltageis equal to primary to secondary turns ratio\n",
-      "Vs=Vp/n;                                #Voltage across the external load, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The primary to secondary turns ratio=%d.\"%n);\n",
-      "print(\"The voltage across the external load=%.1fV.\"%Vs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary to secondary turns ratio=25.\n",
-        "The voltage across the external load=0.4V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.18 : Page number 297-298\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rp=300.0;                   #D.C resistance of primary, ohm\n",
-      "RL=3.0;                     #Load resistance, ohm\n",
-      "R_out=3.0;                  #Ouput resistance of the transistor, kilo ohm               \n",
-      "\n",
-      "#Calculation\n",
-      "#when no signal is applied, only Rp is seen to be the load.\n",
-      "#But, when a.c signal is applied, RL in secondary reflects as RL*(squre of turns ratio).\n",
-      "#Therefore, load is seen to be Rp in series with the reflected RL in primary.\n",
-      "#i.e, R_out=Rp+(n**2 * RL), where n is the turns ratio\n",
-      "n=((R_out*1000-Rp)/RL)**0.5;                     #turns ratio\n",
-      "\n",
-      "#Result\n",
-      "print(\"Turns ratio for maximum power transfer=%d.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Turns ratio for maximum power transfer=30.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.19 : Page number 298"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "f=200.0;                        #Frequency, Hz\n",
-      "Z_out=10.0;                     #Output impedance of the transistor, kilo ohm\n",
-      "Z_in=2.5;                       #Input impedance of the next stage, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#For perfect impedance matching,\n",
-      "#Z_out should be equal to primary impedance\n",
-      "#Z_out=2*pi*f*(primary inductance)\n",
-      "Lp=(Z_out*1000)/(2*pi*f);                   #Primary inductance, H\n",
-      "\n",
-      "#for the secondary side,\n",
-      "#Z_in should be equal to impedance of secondary\n",
-      "Ls=(Z_in*1000)/(2*pi*f);                    #Secondary inductance, H\n",
-      "\n",
-      "\n",
-      "#result\n",
-      "print(\"The primary inductance=%.0fH.\"%Lp);\n",
-      "print(\"The secondary inductance=%.0fH.\"%Ls);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary inductance=8H.\n",
-        "The secondary inductance=2H.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.20 : Page number 299\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Lp=8.0;                     #Primary inductance, H\n",
-      "Ls=2.0;                     #Secondary inductance, H\n",
-      "K=10**-5;                   #Inductance to turns ratio, constant\n",
-      "\n",
-      "#Calculations\n",
-      "Np=(Lp/K)**0.5;                  #Primary turns\n",
-      "Ns=(Ls/K)**0.5;                  #Secondary turns\n",
-      "\n",
-      "#Result\n",
-      "print(\"The primary turns=%.0f.\"%Np);\n",
-      "print(\"The secondary turns=%.0f.\"%Ns);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary turns=894.\n",
-        "The secondary turns=447.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.21 : Page number 300-301\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                   #Collector supply voltage, V\n",
-      "R1=100.0;                   #Resistor R1, kilo ohm\n",
-      "R2=22.0;                    #Resistor R2, kilo ohm\n",
-      "R3=22.0;                    #Resistor R3, kilo ohm\n",
-      "R4=4.7;                     #Resistor R4, kilo ohm\n",
-      "R5=10.0;                    #Resistor R5, kilo ohm\n",
-      "R6=10.0;                    #Resistor R6, kilo ohm\n",
-      "beta=125;                   #Base current amplification factor\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) D.C voltages\n",
-      "#For 1st stage:\n",
-      "V_B1=VCC*R2/(R1+R2);                    #Voltage at the base of 1st transistor, V (Voltage across R2, using voltage divider rule)\n",
-      "V_E1=V_B1-V_BE;                         #Emitter voltage of the 1st transistor, V\n",
-      "I_E1=round(V_E1/R4,2);                           #Emitter current of 1st transistor, mA (OHM's LAW)\n",
-      "I_C1=I_E1;                              #Collector current of 1st transistor, mA(approximately equals to emitter current)\n",
-      "V_C1=VCC-I_C1*R3;                       #Collector voltage of 1st transistor, V\n",
-      "\n",
-      "#For 2nd stage:\n",
-      "V_B2=V_C1;                              #Voltage at the base of 2nd transistor, V (equals to collector voltage of 1st transistor)\n",
-      "V_E2=V_C1-V_BE;                         #Emitter voltage of the 2nd transistor, V\n",
-      "I_E2=V_E2/R6;                           #Emitter current of 2nd transistor, mA (OHM's LAW)\n",
-      "I_C2=I_E2;                              #Collector current 2nd transistor, mA(approximately equals to emitter current)\n",
-      "V_C2=VCC-I_C2*R5;                       #Collector voltage of 2nd transistor, V\n",
-      "\n",
-      "print(\"(i) D.C voltages\");\n",
-      "print(\"First stage: VB1=%.2fV , VE1=%.2fV and VC1=%.2fV\"%(V_B1,V_E1,V_C1));\n",
-      "print(\"First stage: VB2=%.2fV , VE2=%.2fV and VC2=%.2fV\"%(V_B2,V_E2,V_C2));\n",
-      "\n",
-      "#(ii)Voltage gain\n",
-      "#First stage\n",
-      "re_1=25/I_E1;                                   #a.c emitter resistance of 1st transistor, ohm\n",
-      "re_2=25/I_E2;                                   #a.c emitter resistance of 2nd transistor, ohm\n",
-      "Zin_2nd_stage=beta*re_2/1000;                   #Input impedance of 2nd stage, kilo ohm\n",
-      "R_AC=R3*Zin_2nd_stage/(R3+Zin_2nd_stage);       #Total a.c collector load, kilo ohm\n",
-      "A_v1=round(R_AC*1000/re_1,0);                                 #Voltage gain of first stage\n",
-      "\n",
-      "print(\"The voltage gain of first stage=%d.\"%A_v1);\n",
-      "\n",
-      "#Second stage\n",
-      "R_AC=R5;                                        #Total a.c collector load for 2nd stage, kilo ohm(Due to no loading effect, equal to R5)\n",
-      "A_v2=round(R5*1000/re_2,0);                                   #Voltage gain of 2nd stage\n",
-      "\n",
-      "print(\"The voltage gain of second stage=%d.\"%A_v2);\n",
-      "\n",
-      "A_vT=A_v1*A_v2;                                 #Overall voltage gain\n",
-      "\n",
-      "print(\"Overall voltage gain=%d.\"%A_vT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) D.C voltages\n",
-        "First stage: VB1=2.16V , VE1=1.46V and VC1=5.18V\n",
-        "First stage: VB2=5.18V , VE2=4.48V and VC2=7.52V\n",
-        "The voltage gain of first stage=66.\n",
-        "The voltage gain of second stage=179.\n",
-        "Overall voltage gain=11814.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_1.ipynb
deleted file mode 100755
index 6e0fb200..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_1.ipynb
+++ /dev/null
@@ -1,1025 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8e425c9c2dfbfee43b3a89e44b0fd7936ba869da73ac3c372e9b23848f1cded1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 11 : MULTISTAGE TRANSISTOR AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 11.1 : Page number 285"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "#Variable declaration\n",
-      "#(i)\n",
-      "A_v=30;                 #Voltage gain\n",
-      "\n",
-      "#(ii)\n",
-      "A_p=100;                #Power gain\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_v_dB=20*log10(A_v);          #Voltage gain, dB\n",
-      "A_p_dB=10*log10(A_p);          #Power gain, dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i) Voltage gain in dB=%.2fdB\"%A_v_dB);\n",
-      "print(\"(ii) Power gain in dB=%ddB\"%A_p_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Voltage gain in dB=29.54dB\n",
-        "(ii) Power gain in dB=20dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.2 : Page number 285-286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#(i)\n",
-      "A_p_dB=40.0;                #Power gain in dB\n",
-      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
-      "A_p=10**A_p_b;              #Power gain in number\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) Power gain in number=%d\"%A_p);\n",
-      "\n",
-      "#(ii)\n",
-      "A_p_dB=43.0;                #Power gain in dB\n",
-      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
-      "A_p=round(10**A_p_b,-4);              #Power gain in number\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) Power gain in number=%d\"%A_p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Power gain in number=10000\n",
-        "(ii) Power gain in number=20000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.3 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "Av_1=100.0;                     #Voltage gain of stage 1\n",
-      "Av_2=200.0;                     #Voltage gain of stage 2\n",
-      "Av_3=400.0;                     #Voltage gain of stage 3\n",
-      "\n",
-      "#Calculations\n",
-      "Av_1_dB=20*log10(Av_1);                     #Voltage gain of stage 1, dB\n",
-      "Av_2_dB=20*log10(Av_2);                     #Voltage gain of stage 2, dB\n",
-      "Av_3_dB=20*log10(Av_3);                     #Voltage gain of stage 3, dB\n",
-      "\n",
-      "Av_T=Av_1_dB+Av_2_dB+Av_3_dB;            #Total voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total voltage gain=%ddB\"%Av_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total voltage gain=138dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.4 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_p_absolute=30.0;                          #Absolute gain of each stage\n",
-      "number_of_stages=5.0;                       #number of stages\n",
-      "negative_feedback=10.0;                     #negative feedback, dB\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_p_dB=round(10*log10(A_p_absolute),2);              #Power gain of one stage. dB\n",
-      "A_p_T=number_of_stages * A_p_dB;            #Total power gain, dB\n",
-      "\n",
-      "#(ii)\n",
-      "A_p_resultant=A_p_T-negative_feedback;      #Resultant power gain with negative feedback, dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total power gain = %.2fdB.\"%A_p_T);\n",
-      "print(\"The resultant power gain with negative feedback = %.2fdB.\"%A_p_resultant);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power gain = 73.85dB.\n",
-        "The resultant power gain with negative feedback = 63.85dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.5 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "P_out_2kHz=1.5;                 #Output power at 2 kHz, W\n",
-      "P_out_20kHz=0.3;                #Output power at 20 kHz, W\n",
-      "P_in=10.0;                      #Input power, mW\n",
-      "\n",
-      "#Calculations\n",
-      "A_p_dB_2kHz=10*log10(P_out_2kHz*1000/P_in);         #dB power gain at 2 kHz\n",
-      "A_p_dB_20kHz=10*log10(P_out_20kHz*1000/P_in);       #dB power gain at 20 kHz\n",
-      "Fall_in_gain=A_p_dB_2kHz-A_p_dB_20kHz;              #Fall in gain from 2kHz to 20kHz\n",
-      "\n",
-      "#Results\n",
-      "print(\"The fall in gain from 2kHz to 20kHz=%.2fdB\"%Fall_in_gain);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fall in gain from 2kHz to 20kHz=6.99dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.6 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_v=15.0;                   #Voltage gain, dB\n",
-      "V_1=0.8;                    #Input signal voltage, V\n",
-      "\n",
-      "#Calculations\n",
-      "#Since, Av(in decibel)=20*log10(V_2/V_1),\n",
-      "V_2=V_1*(10**(A_v/20));                 #Output voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The output voltage= %.1fV.\"%V_2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage= 4.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.7 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_0_dB=70.0;                   #Open circuit voltage gain, dB\n",
-      "A_v_dB=67.0;                   #Voltage gain, dB\n",
-      "R_out=1.5;                     #Output resistance, kilo ohm\n",
-      "\n",
-      "#Calculations\n",
-      "#Since, A_0_dB-A_v_dB=20*log10(A_0/A_v)\n",
-      "ratio_A0_Av=round(10**((A_0_dB-A_v_dB)/20),2);                  #Ratio of open-circuit voltage gain to normal voltage gain\n",
-      "\n",
-      "#Since, A_v/A_0 = RL/(R_out+RL)\n",
-      "RL=R_out/(ratio_A0_Av-1);                               #Load resistor, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The load resistance=%.2f kilo ohm.\"%RL);\n",
-      "\n",
-      "#Note: The value of load resistor is calculated to be 3.6585 kilo ohm and approximated to 3.66. But, in the text it has been approximated to 3.65.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load resistance=3.66 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.8 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=1.0;                 #Load resistance,  kilo ohm\n",
-      "A_v=40.0;               #Voltage gain, dB\n",
-      "V_in=10.0;              #Input signal voltage, mV\n",
-      "\n",
-      "#Calcultaions\n",
-      "#(i)\n",
-      "#Since, A_v=20*log10(V_out/V_in)\n",
-      "V_out=V_in*(10**(A_v/20))/1000;              #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "P_L=(V_out**2/RL);                            #The load power, mW\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i)The output voltage is %dV.\"%V_out);\n",
-      "print(\"(ii)The load poweris %dmW.\"%P_L);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The output voltage is 1V.\n",
-        "(ii)The load poweris 1mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.9 : Page number 287-288\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "P_2=40.0;                     #Output power, W\n",
-      "R=10.0;                        #Resistance of speaker, ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_p_dB=25.0;                                   #Power gain, dB\n",
-      "#Since, A_p_dB=10*log10(P_2/P_1)\n",
-      "P_1=(P_2/10**(A_p_dB/10))*1000;                #Input power, mW\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "A_v_dB=40.0;                                    #Voltage gain, dB\n",
-      "\n",
-      "#Since, P=(V**2)/R,\n",
-      "V_2=(P_2*R)**0.5;                               #Output voltage, V\n",
-      "\n",
-      "#Since, A_v_dB=20*log10(V_2/V_1)\n",
-      "V_1=(V_2/10**(A_v_dB/20))*1000;                #Input voltage, mV\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print(\"(i)The input power=%.1fmW.\"%P_1);\n",
-      "print(\"(ii)The input voltage=%dmV.\"%V_1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The input power=126.5mW.\n",
-        "(ii)The input voltage=200mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.10 : Page  number 288\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v_max=2000.0;                 #Maximum voltage gain\n",
-      "f_max=2.0;                      #Frequency at which maximum voltage gain occurs,kHz\n",
-      "A_v=1414.0;                     #Voltage gain at 50 Hz and 10kHz\n",
-      "f1=50;                          #Lower frequency at which gain is 1414, Hz\n",
-      "f2=10;                          #Upper frequency at which gain is 1414, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, bandwidth is the range of frequency  over which gain is greater than or equal to 70.7% of maximum gain\n",
-      "if((A_v/A_v_max)*100 ==70.7):                               \n",
-      "    print(\"(i)The bandwidth is from %dHz to %dkHz.\"%(f1,f2));\n",
-      "    print(\"(ii)The lower cut-off frequency=%dHz.\"%f1);\n",
-      "    print(\"(iii)The upper cut-off frequency=%dkHz.\"%f2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The bandwidth is from 50Hz to 10kHz.\n",
-        "(ii)The lower cut-off frequency=50Hz.\n",
-        "(iii)The upper cut-off frequency=10kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11 : Page number 291\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v=60.0;               #Voltage gain of single stage amplifier\n",
-      "R_C=500.0;              #Collector load, ohm\n",
-      "R_in=1.0;               #Input impedance, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, there is no loading , second stage gain remains at A_v\n",
-      "#But, due to loading effect of input impedance of second stage, gain of first stage decreases\n",
-      "A_v_2=A_v;                                                    #Voltage gain of second stage\n",
-      "R_AC=round((R_C*R_in*1000)/(R_C+R_in*1000),0);              #Effective load of first stage, ohm\n",
-      "A_v_1=A_v*R_AC/R_C;                                         #Gain of first stage\n",
-      "A_v_T=A_v_1*A_v_2;                                            #Total gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total gain=%d.\"%A_v_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total gain=2397.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.12 : Page number 291-292\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rin=1.0;                        #Input resistance, kilo ohm\n",
-      "beta=100.0;                     #base current amplification factor\n",
-      "RC=2.0;                         #Collector load, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_AC=(RC*Rin)/(RC+Rin);               #Effective load on first stage, kilo ohm\n",
-      "A_v_1=round(beta*(R_AC/Rin),0);                #Voltage gain of first stage\n",
-      "\n",
-      "#(ii)\n",
-      "A_v_2=round(beta*RC/Rin,0);                      #Voltage gain of second stage\n",
-      "\n",
-      "#(iii)\n",
-      "A_v_T=A_v_1*A_v_2;                      #Total voltage gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i)The voltage gain of first stage =%d.\"%A_v_1);\n",
-      "print(\"(ii)The voltage gain of second stage =%d.\"%A_v_2);\n",
-      "print(\"(iii)The total voltage gain =%d.\"%A_v_T);\n",
-      "\n",
-      "#Note: The approximation inthe text for A_v_1=66.66 is taken as 66 but here it has been taken 67 and therefore the total voltage is 13400 instead of 13200.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage gain of first stage =67.\n",
-        "(ii)The voltage gain of second stage =200.\n",
-        "(iii)The total voltage gain =13400.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.13 : Page number 292\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=10.0;                    #Collector load of single stage amplifier, kilo ohm\n",
-      "Rin=1.0;                    #Input resistance, kilo ohm\n",
-      "beta=100.0;                 #base current amplification factor\n",
-      "RL=100.0;                   #Load resistor, ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "R_AC=round((RC*1000)*RL/(RC*1000+RL),-1);                 #Effective collector load,\n",
-      "A_v=beta*R_AC/(Rin*1000);                              #Voltage gain\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage gain=%d.\"%A_v);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain=10.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.14 : Page number 292-293\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                       #Collector suppply voltage, V\n",
-      "R1=10.0;                        #resistor R1, kilo ohm\n",
-      "R2=2.2;                         #resistor R2, kilo ohm\n",
-      "R3=10.0;                        #resistor R3, kilo ohm\n",
-      "R4=2.2;                         #resistor R4, kilo ohm\n",
-      "RC_1=3.6;                         #Collector resistor of first stage, kilo ohm\n",
-      "RC_2=4.0;                         #Collector resistor of second stage, kilo ohm\n",
-      "RE_1=900.0;                     #Emitter resistor of first stage, ohm\n",
-      "RE_2=1.0;                       #Emitter resistor of second stage, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Biasing potential for the second stage is the voltage across R4 resistor,\n",
-      "#so, by voltage divider rule:\n",
-      "VB=VCC*R4/(R3+R4);                #Biasing potential for second stage,(Voltage across R4), V\n",
-      "\n",
-      "print(\"The biasing voltage for the second stage=%.1fV.\"%VB);\n",
-      "\n",
-      "#If coupling capacitor C_c is replaced by a wire, RC_1 and R3 become parallel\n",
-      "Req=round((RC_1*R3)/(RC_1+R3),2);        #Equivalent resistance of R3 parallel with RC_1, kilo ohm\n",
-      "VB=VCC*R4/(Req+R4);             #Biasing voltage if coupling capacitor is replaced by a wire, V\n",
-      "\n",
-      "print(\"The biasing voltage after replacing coupling capacitor by wire=%.2fV.\"%VB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The biasing voltage for the second stage=3.6V.\n",
-        "The biasing voltage after replacing coupling capacitor by wire=9.07V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.15 : Page number 293-294\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return (r1*r2)/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage, V\n",
-      "R1=22.0;                            #Resistor R1, kilo ohm\n",
-      "R2=3.3;                            #Resistor R2, kilo ohm\n",
-      "R3=5.0;                            #Resistor R3, kilo ohm\n",
-      "R4=1.0;                            #Resistor R4, kilo ohm\n",
-      "R5=15.0;                            #Resistor R5, kilo ohm\n",
-      "R6=2.5;                            #Resistor R6, kilo ohm\n",
-      "R7=5.0;                            #Resistor R7, kilo ohm\n",
-      "R8=1.0;                            #Resistor R8, kilo ohm\n",
-      "beta=200;                           #Base current amplification factor\n",
-      "RL=10.0;                            #Load resistor, kilo ohm\n",
-      "V_BE=0.7;                           #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#for 2nd stage\n",
-      "V_R6=round(VCC*R6/(R5+R6),2);                #Voltage across R6, V (voltage divider rule)\n",
-      "V_R8=round(V_R6-V_BE,2);                     #Voltage across R8, V\n",
-      "IE_2=round(V_R8/R8,2);                         #Emitter current through R8, mA (OHM's LAW)\n",
-      "re_2nd_stage=round(25/IE_2,1);                 #a.c emitter resistance for 2nd stage, ohm\n",
-      "\n",
-      "#For 1st stage\n",
-      "V_R2=round(VCC*R2/(R1+R2),2);                     #Voltage across R2, V (voltage divider rule)\n",
-      "V_R4=round(V_R2-V_BE,2);                          #Voltage across R4, V\n",
-      "IE_1=round(V_R4/R4,2);                            #Emitter current through R4, mA (OHM's LAW)\n",
-      "re_1st_stage=round(25/IE_1,1);                    #a.c emitter resistance for 1st stage, ohm\n",
-      "\n",
-      "#(i)\n",
-      "Zin_base_2nd_stage=round((beta*re_2nd_stage)/1000,2);               #input resistance of transistor base of 2nd stage, kilo ohm\n",
-      "Zin=round(pr(pr(R5,R6),Zin_base_2nd_stage),2);                      #Input impedance of the 2nd stage, kilo ohm\n",
-      "R_AC_1st_stage=round(pr(R3,Zin),2);                                          #Effective collector load for 1st stage, kilo ohm\n",
-      "A_v_1=round(R_AC_1st_stage*1000/re_1st_stage,0);                                  #voltage gain of 1st stage\n",
-      "\n",
-      "#(ii)\n",
-      "R_AC_2nd_stage=round(pr(R7,RL),2);                          #Effective collector load for 2nd stage, kilo ohm\n",
-      "A_v_2=round(R_AC_2nd_stage*1000/re_2nd_stage,1);                  #voltage gain of 2nd stage\n",
-      "\n",
-      "#(iii)\n",
-      "A_v_overall=A_v_1*A_v_2;                            #overall voltage gain\n",
-      "\n",
-      "\n",
-      "#results\n",
-      "print(\"(i)The voltage gain of 1st stage=%.0f.\"%A_v_1);\n",
-      "print(\"(i)The voltage gain of 2nd stage=%.1f.\"%A_v_2);\n",
-      "print(\"(i)The overall voltage gain =%d.\"%A_v_overall);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage gain of 1st stage=53.\n",
-        "(i)The voltage gain of 2nd stage=191.4.\n",
-        "(i)The overall voltage gain =10144.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.16 : Page number 297\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Primary_impedance=1000.0;                       #Primary impedance, ohm\n",
-      "Load_impedance=10.0;                            #Load impedance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#since,for maximum power transfer primary impedance should be equal to output impedance\n",
-      "#and, impedance of secondary should be equal to load impedance\n",
-      "#therfore, primary_impedance/load_impedance=square of(primary to secondary turn ratio)\n",
-      "n=(Primary_impedance/Load_impedance)**0.5;           #Primary to secondary turn ratio\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print('The primary to secondary turn ratio for maximum power transfer=%d.'%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary to secondary turn ratio for maximum power transfer=10.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.17 : Page number 297\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=16.0;                    #Load resistor, ohm\n",
-      "R_p=10.0;                   #Output impedance of primary, kilo ohm\n",
-      "Vp=10.0;                    #Terminal voltage of the source, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, for maximum power transfer, the impedance of the primary should be equal to output impedance of the source\n",
-      "n=(R_p*1000/RL)**0.5;                        #Primary to secondary turns ratio\n",
-      "\n",
-      "#Since, power in a transformer remains constant,\n",
-      "#ratio of primary to secondary voltageis equal to primary to secondary turns ratio\n",
-      "Vs=Vp/n;                                #Voltage across the external load, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The primary to secondary turns ratio=%d.\"%n);\n",
-      "print(\"The voltage across the external load=%.1fV.\"%Vs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary to secondary turns ratio=25.\n",
-        "The voltage across the external load=0.4V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.18 : Page number 297-298\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rp=300.0;                   #D.C resistance of primary, ohm\n",
-      "RL=3.0;                     #Load resistance, ohm\n",
-      "R_out=3.0;                  #Ouput resistance of the transistor, kilo ohm               \n",
-      "\n",
-      "#Calculation\n",
-      "#when no signal is applied, only Rp is seen to be the load.\n",
-      "#But, when a.c signal is applied, RL in secondary reflects as RL*(squre of turns ratio).\n",
-      "#Therefore, load is seen to be Rp in series with the reflected RL in primary.\n",
-      "#i.e, R_out=Rp+(n**2 * RL), where n is the turns ratio\n",
-      "n=((R_out*1000-Rp)/RL)**0.5;                     #turns ratio\n",
-      "\n",
-      "#Result\n",
-      "print(\"Turns ratio for maximum power transfer=%d.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Turns ratio for maximum power transfer=30.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.19 : Page number 298"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "f=200.0;                        #Frequency, Hz\n",
-      "Z_out=10.0;                     #Output impedance of the transistor, kilo ohm\n",
-      "Z_in=2.5;                       #Input impedance of the next stage, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#For perfect impedance matching,\n",
-      "#Z_out should be equal to primary impedance\n",
-      "#Z_out=2*pi*f*(primary inductance)\n",
-      "Lp=(Z_out*1000)/(2*pi*f);                   #Primary inductance, H\n",
-      "\n",
-      "#for the secondary side,\n",
-      "#Z_in should be equal to impedance of secondary\n",
-      "Ls=(Z_in*1000)/(2*pi*f);                    #Secondary inductance, H\n",
-      "\n",
-      "\n",
-      "#result\n",
-      "print(\"The primary inductance=%.0fH.\"%Lp);\n",
-      "print(\"The secondary inductance=%.0fH.\"%Ls);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary inductance=8H.\n",
-        "The secondary inductance=2H.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.20 : Page number 299\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Lp=8.0;                     #Primary inductance, H\n",
-      "Ls=2.0;                     #Secondary inductance, H\n",
-      "K=10**-5;                   #Inductance to turns ratio, constant\n",
-      "\n",
-      "#Calculations\n",
-      "Np=(Lp/K)**0.5;                  #Primary turns\n",
-      "Ns=(Ls/K)**0.5;                  #Secondary turns\n",
-      "\n",
-      "#Result\n",
-      "print(\"The primary turns=%.0f.\"%Np);\n",
-      "print(\"The secondary turns=%.0f.\"%Ns);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary turns=894.\n",
-        "The secondary turns=447.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.21 : Page number 300-301\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                   #Collector supply voltage, V\n",
-      "R1=100.0;                   #Resistor R1, kilo ohm\n",
-      "R2=22.0;                    #Resistor R2, kilo ohm\n",
-      "R3=22.0;                    #Resistor R3, kilo ohm\n",
-      "R4=4.7;                     #Resistor R4, kilo ohm\n",
-      "R5=10.0;                    #Resistor R5, kilo ohm\n",
-      "R6=10.0;                    #Resistor R6, kilo ohm\n",
-      "beta=125;                   #Base current amplification factor\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) D.C voltages\n",
-      "#For 1st stage:\n",
-      "V_B1=VCC*R2/(R1+R2);                    #Voltage at the base of 1st transistor, V (Voltage across R2, using voltage divider rule)\n",
-      "V_E1=V_B1-V_BE;                         #Emitter voltage of the 1st transistor, V\n",
-      "I_E1=round(V_E1/R4,2);                           #Emitter current of 1st transistor, mA (OHM's LAW)\n",
-      "I_C1=I_E1;                              #Collector current of 1st transistor, mA(approximately equals to emitter current)\n",
-      "V_C1=VCC-I_C1*R3;                       #Collector voltage of 1st transistor, V\n",
-      "\n",
-      "#For 2nd stage:\n",
-      "V_B2=V_C1;                              #Voltage at the base of 2nd transistor, V (equals to collector voltage of 1st transistor)\n",
-      "V_E2=V_C1-V_BE;                         #Emitter voltage of the 2nd transistor, V\n",
-      "I_E2=V_E2/R6;                           #Emitter current of 2nd transistor, mA (OHM's LAW)\n",
-      "I_C2=I_E2;                              #Collector current 2nd transistor, mA(approximately equals to emitter current)\n",
-      "V_C2=VCC-I_C2*R5;                       #Collector voltage of 2nd transistor, V\n",
-      "\n",
-      "print(\"(i) D.C voltages\");\n",
-      "print(\"First stage: VB1=%.2fV , VE1=%.2fV and VC1=%.2fV\"%(V_B1,V_E1,V_C1));\n",
-      "print(\"First stage: VB2=%.2fV , VE2=%.2fV and VC2=%.2fV\"%(V_B2,V_E2,V_C2));\n",
-      "\n",
-      "#(ii)Voltage gain\n",
-      "#First stage\n",
-      "re_1=25/I_E1;                                   #a.c emitter resistance of 1st transistor, ohm\n",
-      "re_2=25/I_E2;                                   #a.c emitter resistance of 2nd transistor, ohm\n",
-      "Zin_2nd_stage=beta*re_2/1000;                   #Input impedance of 2nd stage, kilo ohm\n",
-      "R_AC=R3*Zin_2nd_stage/(R3+Zin_2nd_stage);       #Total a.c collector load, kilo ohm\n",
-      "A_v1=round(R_AC*1000/re_1,0);                                 #Voltage gain of first stage\n",
-      "\n",
-      "print(\"The voltage gain of first stage=%d.\"%A_v1);\n",
-      "\n",
-      "#Second stage\n",
-      "R_AC=R5;                                        #Total a.c collector load for 2nd stage, kilo ohm(Due to no loading effect, equal to R5)\n",
-      "A_v2=round(R5*1000/re_2,0);                                   #Voltage gain of 2nd stage\n",
-      "\n",
-      "print(\"The voltage gain of second stage=%d.\"%A_v2);\n",
-      "\n",
-      "A_vT=A_v1*A_v2;                                 #Overall voltage gain\n",
-      "\n",
-      "print(\"Overall voltage gain=%d.\"%A_vT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) D.C voltages\n",
-        "First stage: VB1=2.16V , VE1=1.46V and VC1=5.18V\n",
-        "First stage: VB2=5.18V , VE2=4.48V and VC2=7.52V\n",
-        "The voltage gain of first stage=66.\n",
-        "The voltage gain of second stage=179.\n",
-        "Overall voltage gain=11814.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_2.ipynb
deleted file mode 100755
index 6e0fb200..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_2.ipynb
+++ /dev/null
@@ -1,1025 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8e425c9c2dfbfee43b3a89e44b0fd7936ba869da73ac3c372e9b23848f1cded1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 11 : MULTISTAGE TRANSISTOR AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 11.1 : Page number 285"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "#Variable declaration\n",
-      "#(i)\n",
-      "A_v=30;                 #Voltage gain\n",
-      "\n",
-      "#(ii)\n",
-      "A_p=100;                #Power gain\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_v_dB=20*log10(A_v);          #Voltage gain, dB\n",
-      "A_p_dB=10*log10(A_p);          #Power gain, dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i) Voltage gain in dB=%.2fdB\"%A_v_dB);\n",
-      "print(\"(ii) Power gain in dB=%ddB\"%A_p_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Voltage gain in dB=29.54dB\n",
-        "(ii) Power gain in dB=20dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.2 : Page number 285-286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#(i)\n",
-      "A_p_dB=40.0;                #Power gain in dB\n",
-      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
-      "A_p=10**A_p_b;              #Power gain in number\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) Power gain in number=%d\"%A_p);\n",
-      "\n",
-      "#(ii)\n",
-      "A_p_dB=43.0;                #Power gain in dB\n",
-      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
-      "A_p=round(10**A_p_b,-4);              #Power gain in number\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) Power gain in number=%d\"%A_p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Power gain in number=10000\n",
-        "(ii) Power gain in number=20000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.3 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "Av_1=100.0;                     #Voltage gain of stage 1\n",
-      "Av_2=200.0;                     #Voltage gain of stage 2\n",
-      "Av_3=400.0;                     #Voltage gain of stage 3\n",
-      "\n",
-      "#Calculations\n",
-      "Av_1_dB=20*log10(Av_1);                     #Voltage gain of stage 1, dB\n",
-      "Av_2_dB=20*log10(Av_2);                     #Voltage gain of stage 2, dB\n",
-      "Av_3_dB=20*log10(Av_3);                     #Voltage gain of stage 3, dB\n",
-      "\n",
-      "Av_T=Av_1_dB+Av_2_dB+Av_3_dB;            #Total voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total voltage gain=%ddB\"%Av_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total voltage gain=138dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.4 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_p_absolute=30.0;                          #Absolute gain of each stage\n",
-      "number_of_stages=5.0;                       #number of stages\n",
-      "negative_feedback=10.0;                     #negative feedback, dB\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_p_dB=round(10*log10(A_p_absolute),2);              #Power gain of one stage. dB\n",
-      "A_p_T=number_of_stages * A_p_dB;            #Total power gain, dB\n",
-      "\n",
-      "#(ii)\n",
-      "A_p_resultant=A_p_T-negative_feedback;      #Resultant power gain with negative feedback, dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total power gain = %.2fdB.\"%A_p_T);\n",
-      "print(\"The resultant power gain with negative feedback = %.2fdB.\"%A_p_resultant);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power gain = 73.85dB.\n",
-        "The resultant power gain with negative feedback = 63.85dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.5 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "P_out_2kHz=1.5;                 #Output power at 2 kHz, W\n",
-      "P_out_20kHz=0.3;                #Output power at 20 kHz, W\n",
-      "P_in=10.0;                      #Input power, mW\n",
-      "\n",
-      "#Calculations\n",
-      "A_p_dB_2kHz=10*log10(P_out_2kHz*1000/P_in);         #dB power gain at 2 kHz\n",
-      "A_p_dB_20kHz=10*log10(P_out_20kHz*1000/P_in);       #dB power gain at 20 kHz\n",
-      "Fall_in_gain=A_p_dB_2kHz-A_p_dB_20kHz;              #Fall in gain from 2kHz to 20kHz\n",
-      "\n",
-      "#Results\n",
-      "print(\"The fall in gain from 2kHz to 20kHz=%.2fdB\"%Fall_in_gain);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fall in gain from 2kHz to 20kHz=6.99dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.6 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_v=15.0;                   #Voltage gain, dB\n",
-      "V_1=0.8;                    #Input signal voltage, V\n",
-      "\n",
-      "#Calculations\n",
-      "#Since, Av(in decibel)=20*log10(V_2/V_1),\n",
-      "V_2=V_1*(10**(A_v/20));                 #Output voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The output voltage= %.1fV.\"%V_2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage= 4.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.7 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_0_dB=70.0;                   #Open circuit voltage gain, dB\n",
-      "A_v_dB=67.0;                   #Voltage gain, dB\n",
-      "R_out=1.5;                     #Output resistance, kilo ohm\n",
-      "\n",
-      "#Calculations\n",
-      "#Since, A_0_dB-A_v_dB=20*log10(A_0/A_v)\n",
-      "ratio_A0_Av=round(10**((A_0_dB-A_v_dB)/20),2);                  #Ratio of open-circuit voltage gain to normal voltage gain\n",
-      "\n",
-      "#Since, A_v/A_0 = RL/(R_out+RL)\n",
-      "RL=R_out/(ratio_A0_Av-1);                               #Load resistor, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The load resistance=%.2f kilo ohm.\"%RL);\n",
-      "\n",
-      "#Note: The value of load resistor is calculated to be 3.6585 kilo ohm and approximated to 3.66. But, in the text it has been approximated to 3.65.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load resistance=3.66 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.8 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=1.0;                 #Load resistance,  kilo ohm\n",
-      "A_v=40.0;               #Voltage gain, dB\n",
-      "V_in=10.0;              #Input signal voltage, mV\n",
-      "\n",
-      "#Calcultaions\n",
-      "#(i)\n",
-      "#Since, A_v=20*log10(V_out/V_in)\n",
-      "V_out=V_in*(10**(A_v/20))/1000;              #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "P_L=(V_out**2/RL);                            #The load power, mW\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i)The output voltage is %dV.\"%V_out);\n",
-      "print(\"(ii)The load poweris %dmW.\"%P_L);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The output voltage is 1V.\n",
-        "(ii)The load poweris 1mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.9 : Page number 287-288\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "P_2=40.0;                     #Output power, W\n",
-      "R=10.0;                        #Resistance of speaker, ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_p_dB=25.0;                                   #Power gain, dB\n",
-      "#Since, A_p_dB=10*log10(P_2/P_1)\n",
-      "P_1=(P_2/10**(A_p_dB/10))*1000;                #Input power, mW\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "A_v_dB=40.0;                                    #Voltage gain, dB\n",
-      "\n",
-      "#Since, P=(V**2)/R,\n",
-      "V_2=(P_2*R)**0.5;                               #Output voltage, V\n",
-      "\n",
-      "#Since, A_v_dB=20*log10(V_2/V_1)\n",
-      "V_1=(V_2/10**(A_v_dB/20))*1000;                #Input voltage, mV\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print(\"(i)The input power=%.1fmW.\"%P_1);\n",
-      "print(\"(ii)The input voltage=%dmV.\"%V_1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The input power=126.5mW.\n",
-        "(ii)The input voltage=200mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.10 : Page  number 288\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v_max=2000.0;                 #Maximum voltage gain\n",
-      "f_max=2.0;                      #Frequency at which maximum voltage gain occurs,kHz\n",
-      "A_v=1414.0;                     #Voltage gain at 50 Hz and 10kHz\n",
-      "f1=50;                          #Lower frequency at which gain is 1414, Hz\n",
-      "f2=10;                          #Upper frequency at which gain is 1414, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, bandwidth is the range of frequency  over which gain is greater than or equal to 70.7% of maximum gain\n",
-      "if((A_v/A_v_max)*100 ==70.7):                               \n",
-      "    print(\"(i)The bandwidth is from %dHz to %dkHz.\"%(f1,f2));\n",
-      "    print(\"(ii)The lower cut-off frequency=%dHz.\"%f1);\n",
-      "    print(\"(iii)The upper cut-off frequency=%dkHz.\"%f2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The bandwidth is from 50Hz to 10kHz.\n",
-        "(ii)The lower cut-off frequency=50Hz.\n",
-        "(iii)The upper cut-off frequency=10kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11 : Page number 291\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v=60.0;               #Voltage gain of single stage amplifier\n",
-      "R_C=500.0;              #Collector load, ohm\n",
-      "R_in=1.0;               #Input impedance, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, there is no loading , second stage gain remains at A_v\n",
-      "#But, due to loading effect of input impedance of second stage, gain of first stage decreases\n",
-      "A_v_2=A_v;                                                    #Voltage gain of second stage\n",
-      "R_AC=round((R_C*R_in*1000)/(R_C+R_in*1000),0);              #Effective load of first stage, ohm\n",
-      "A_v_1=A_v*R_AC/R_C;                                         #Gain of first stage\n",
-      "A_v_T=A_v_1*A_v_2;                                            #Total gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total gain=%d.\"%A_v_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total gain=2397.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.12 : Page number 291-292\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rin=1.0;                        #Input resistance, kilo ohm\n",
-      "beta=100.0;                     #base current amplification factor\n",
-      "RC=2.0;                         #Collector load, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_AC=(RC*Rin)/(RC+Rin);               #Effective load on first stage, kilo ohm\n",
-      "A_v_1=round(beta*(R_AC/Rin),0);                #Voltage gain of first stage\n",
-      "\n",
-      "#(ii)\n",
-      "A_v_2=round(beta*RC/Rin,0);                      #Voltage gain of second stage\n",
-      "\n",
-      "#(iii)\n",
-      "A_v_T=A_v_1*A_v_2;                      #Total voltage gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i)The voltage gain of first stage =%d.\"%A_v_1);\n",
-      "print(\"(ii)The voltage gain of second stage =%d.\"%A_v_2);\n",
-      "print(\"(iii)The total voltage gain =%d.\"%A_v_T);\n",
-      "\n",
-      "#Note: The approximation inthe text for A_v_1=66.66 is taken as 66 but here it has been taken 67 and therefore the total voltage is 13400 instead of 13200.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage gain of first stage =67.\n",
-        "(ii)The voltage gain of second stage =200.\n",
-        "(iii)The total voltage gain =13400.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.13 : Page number 292\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=10.0;                    #Collector load of single stage amplifier, kilo ohm\n",
-      "Rin=1.0;                    #Input resistance, kilo ohm\n",
-      "beta=100.0;                 #base current amplification factor\n",
-      "RL=100.0;                   #Load resistor, ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "R_AC=round((RC*1000)*RL/(RC*1000+RL),-1);                 #Effective collector load,\n",
-      "A_v=beta*R_AC/(Rin*1000);                              #Voltage gain\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage gain=%d.\"%A_v);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain=10.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.14 : Page number 292-293\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                       #Collector suppply voltage, V\n",
-      "R1=10.0;                        #resistor R1, kilo ohm\n",
-      "R2=2.2;                         #resistor R2, kilo ohm\n",
-      "R3=10.0;                        #resistor R3, kilo ohm\n",
-      "R4=2.2;                         #resistor R4, kilo ohm\n",
-      "RC_1=3.6;                         #Collector resistor of first stage, kilo ohm\n",
-      "RC_2=4.0;                         #Collector resistor of second stage, kilo ohm\n",
-      "RE_1=900.0;                     #Emitter resistor of first stage, ohm\n",
-      "RE_2=1.0;                       #Emitter resistor of second stage, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Biasing potential for the second stage is the voltage across R4 resistor,\n",
-      "#so, by voltage divider rule:\n",
-      "VB=VCC*R4/(R3+R4);                #Biasing potential for second stage,(Voltage across R4), V\n",
-      "\n",
-      "print(\"The biasing voltage for the second stage=%.1fV.\"%VB);\n",
-      "\n",
-      "#If coupling capacitor C_c is replaced by a wire, RC_1 and R3 become parallel\n",
-      "Req=round((RC_1*R3)/(RC_1+R3),2);        #Equivalent resistance of R3 parallel with RC_1, kilo ohm\n",
-      "VB=VCC*R4/(Req+R4);             #Biasing voltage if coupling capacitor is replaced by a wire, V\n",
-      "\n",
-      "print(\"The biasing voltage after replacing coupling capacitor by wire=%.2fV.\"%VB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The biasing voltage for the second stage=3.6V.\n",
-        "The biasing voltage after replacing coupling capacitor by wire=9.07V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.15 : Page number 293-294\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return (r1*r2)/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage, V\n",
-      "R1=22.0;                            #Resistor R1, kilo ohm\n",
-      "R2=3.3;                            #Resistor R2, kilo ohm\n",
-      "R3=5.0;                            #Resistor R3, kilo ohm\n",
-      "R4=1.0;                            #Resistor R4, kilo ohm\n",
-      "R5=15.0;                            #Resistor R5, kilo ohm\n",
-      "R6=2.5;                            #Resistor R6, kilo ohm\n",
-      "R7=5.0;                            #Resistor R7, kilo ohm\n",
-      "R8=1.0;                            #Resistor R8, kilo ohm\n",
-      "beta=200;                           #Base current amplification factor\n",
-      "RL=10.0;                            #Load resistor, kilo ohm\n",
-      "V_BE=0.7;                           #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#for 2nd stage\n",
-      "V_R6=round(VCC*R6/(R5+R6),2);                #Voltage across R6, V (voltage divider rule)\n",
-      "V_R8=round(V_R6-V_BE,2);                     #Voltage across R8, V\n",
-      "IE_2=round(V_R8/R8,2);                         #Emitter current through R8, mA (OHM's LAW)\n",
-      "re_2nd_stage=round(25/IE_2,1);                 #a.c emitter resistance for 2nd stage, ohm\n",
-      "\n",
-      "#For 1st stage\n",
-      "V_R2=round(VCC*R2/(R1+R2),2);                     #Voltage across R2, V (voltage divider rule)\n",
-      "V_R4=round(V_R2-V_BE,2);                          #Voltage across R4, V\n",
-      "IE_1=round(V_R4/R4,2);                            #Emitter current through R4, mA (OHM's LAW)\n",
-      "re_1st_stage=round(25/IE_1,1);                    #a.c emitter resistance for 1st stage, ohm\n",
-      "\n",
-      "#(i)\n",
-      "Zin_base_2nd_stage=round((beta*re_2nd_stage)/1000,2);               #input resistance of transistor base of 2nd stage, kilo ohm\n",
-      "Zin=round(pr(pr(R5,R6),Zin_base_2nd_stage),2);                      #Input impedance of the 2nd stage, kilo ohm\n",
-      "R_AC_1st_stage=round(pr(R3,Zin),2);                                          #Effective collector load for 1st stage, kilo ohm\n",
-      "A_v_1=round(R_AC_1st_stage*1000/re_1st_stage,0);                                  #voltage gain of 1st stage\n",
-      "\n",
-      "#(ii)\n",
-      "R_AC_2nd_stage=round(pr(R7,RL),2);                          #Effective collector load for 2nd stage, kilo ohm\n",
-      "A_v_2=round(R_AC_2nd_stage*1000/re_2nd_stage,1);                  #voltage gain of 2nd stage\n",
-      "\n",
-      "#(iii)\n",
-      "A_v_overall=A_v_1*A_v_2;                            #overall voltage gain\n",
-      "\n",
-      "\n",
-      "#results\n",
-      "print(\"(i)The voltage gain of 1st stage=%.0f.\"%A_v_1);\n",
-      "print(\"(i)The voltage gain of 2nd stage=%.1f.\"%A_v_2);\n",
-      "print(\"(i)The overall voltage gain =%d.\"%A_v_overall);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage gain of 1st stage=53.\n",
-        "(i)The voltage gain of 2nd stage=191.4.\n",
-        "(i)The overall voltage gain =10144.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.16 : Page number 297\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Primary_impedance=1000.0;                       #Primary impedance, ohm\n",
-      "Load_impedance=10.0;                            #Load impedance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#since,for maximum power transfer primary impedance should be equal to output impedance\n",
-      "#and, impedance of secondary should be equal to load impedance\n",
-      "#therfore, primary_impedance/load_impedance=square of(primary to secondary turn ratio)\n",
-      "n=(Primary_impedance/Load_impedance)**0.5;           #Primary to secondary turn ratio\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print('The primary to secondary turn ratio for maximum power transfer=%d.'%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary to secondary turn ratio for maximum power transfer=10.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.17 : Page number 297\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=16.0;                    #Load resistor, ohm\n",
-      "R_p=10.0;                   #Output impedance of primary, kilo ohm\n",
-      "Vp=10.0;                    #Terminal voltage of the source, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, for maximum power transfer, the impedance of the primary should be equal to output impedance of the source\n",
-      "n=(R_p*1000/RL)**0.5;                        #Primary to secondary turns ratio\n",
-      "\n",
-      "#Since, power in a transformer remains constant,\n",
-      "#ratio of primary to secondary voltageis equal to primary to secondary turns ratio\n",
-      "Vs=Vp/n;                                #Voltage across the external load, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The primary to secondary turns ratio=%d.\"%n);\n",
-      "print(\"The voltage across the external load=%.1fV.\"%Vs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary to secondary turns ratio=25.\n",
-        "The voltage across the external load=0.4V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.18 : Page number 297-298\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rp=300.0;                   #D.C resistance of primary, ohm\n",
-      "RL=3.0;                     #Load resistance, ohm\n",
-      "R_out=3.0;                  #Ouput resistance of the transistor, kilo ohm               \n",
-      "\n",
-      "#Calculation\n",
-      "#when no signal is applied, only Rp is seen to be the load.\n",
-      "#But, when a.c signal is applied, RL in secondary reflects as RL*(squre of turns ratio).\n",
-      "#Therefore, load is seen to be Rp in series with the reflected RL in primary.\n",
-      "#i.e, R_out=Rp+(n**2 * RL), where n is the turns ratio\n",
-      "n=((R_out*1000-Rp)/RL)**0.5;                     #turns ratio\n",
-      "\n",
-      "#Result\n",
-      "print(\"Turns ratio for maximum power transfer=%d.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Turns ratio for maximum power transfer=30.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.19 : Page number 298"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "f=200.0;                        #Frequency, Hz\n",
-      "Z_out=10.0;                     #Output impedance of the transistor, kilo ohm\n",
-      "Z_in=2.5;                       #Input impedance of the next stage, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#For perfect impedance matching,\n",
-      "#Z_out should be equal to primary impedance\n",
-      "#Z_out=2*pi*f*(primary inductance)\n",
-      "Lp=(Z_out*1000)/(2*pi*f);                   #Primary inductance, H\n",
-      "\n",
-      "#for the secondary side,\n",
-      "#Z_in should be equal to impedance of secondary\n",
-      "Ls=(Z_in*1000)/(2*pi*f);                    #Secondary inductance, H\n",
-      "\n",
-      "\n",
-      "#result\n",
-      "print(\"The primary inductance=%.0fH.\"%Lp);\n",
-      "print(\"The secondary inductance=%.0fH.\"%Ls);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary inductance=8H.\n",
-        "The secondary inductance=2H.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.20 : Page number 299\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Lp=8.0;                     #Primary inductance, H\n",
-      "Ls=2.0;                     #Secondary inductance, H\n",
-      "K=10**-5;                   #Inductance to turns ratio, constant\n",
-      "\n",
-      "#Calculations\n",
-      "Np=(Lp/K)**0.5;                  #Primary turns\n",
-      "Ns=(Ls/K)**0.5;                  #Secondary turns\n",
-      "\n",
-      "#Result\n",
-      "print(\"The primary turns=%.0f.\"%Np);\n",
-      "print(\"The secondary turns=%.0f.\"%Ns);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary turns=894.\n",
-        "The secondary turns=447.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.21 : Page number 300-301\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                   #Collector supply voltage, V\n",
-      "R1=100.0;                   #Resistor R1, kilo ohm\n",
-      "R2=22.0;                    #Resistor R2, kilo ohm\n",
-      "R3=22.0;                    #Resistor R3, kilo ohm\n",
-      "R4=4.7;                     #Resistor R4, kilo ohm\n",
-      "R5=10.0;                    #Resistor R5, kilo ohm\n",
-      "R6=10.0;                    #Resistor R6, kilo ohm\n",
-      "beta=125;                   #Base current amplification factor\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) D.C voltages\n",
-      "#For 1st stage:\n",
-      "V_B1=VCC*R2/(R1+R2);                    #Voltage at the base of 1st transistor, V (Voltage across R2, using voltage divider rule)\n",
-      "V_E1=V_B1-V_BE;                         #Emitter voltage of the 1st transistor, V\n",
-      "I_E1=round(V_E1/R4,2);                           #Emitter current of 1st transistor, mA (OHM's LAW)\n",
-      "I_C1=I_E1;                              #Collector current of 1st transistor, mA(approximately equals to emitter current)\n",
-      "V_C1=VCC-I_C1*R3;                       #Collector voltage of 1st transistor, V\n",
-      "\n",
-      "#For 2nd stage:\n",
-      "V_B2=V_C1;                              #Voltage at the base of 2nd transistor, V (equals to collector voltage of 1st transistor)\n",
-      "V_E2=V_C1-V_BE;                         #Emitter voltage of the 2nd transistor, V\n",
-      "I_E2=V_E2/R6;                           #Emitter current of 2nd transistor, mA (OHM's LAW)\n",
-      "I_C2=I_E2;                              #Collector current 2nd transistor, mA(approximately equals to emitter current)\n",
-      "V_C2=VCC-I_C2*R5;                       #Collector voltage of 2nd transistor, V\n",
-      "\n",
-      "print(\"(i) D.C voltages\");\n",
-      "print(\"First stage: VB1=%.2fV , VE1=%.2fV and VC1=%.2fV\"%(V_B1,V_E1,V_C1));\n",
-      "print(\"First stage: VB2=%.2fV , VE2=%.2fV and VC2=%.2fV\"%(V_B2,V_E2,V_C2));\n",
-      "\n",
-      "#(ii)Voltage gain\n",
-      "#First stage\n",
-      "re_1=25/I_E1;                                   #a.c emitter resistance of 1st transistor, ohm\n",
-      "re_2=25/I_E2;                                   #a.c emitter resistance of 2nd transistor, ohm\n",
-      "Zin_2nd_stage=beta*re_2/1000;                   #Input impedance of 2nd stage, kilo ohm\n",
-      "R_AC=R3*Zin_2nd_stage/(R3+Zin_2nd_stage);       #Total a.c collector load, kilo ohm\n",
-      "A_v1=round(R_AC*1000/re_1,0);                                 #Voltage gain of first stage\n",
-      "\n",
-      "print(\"The voltage gain of first stage=%d.\"%A_v1);\n",
-      "\n",
-      "#Second stage\n",
-      "R_AC=R5;                                        #Total a.c collector load for 2nd stage, kilo ohm(Due to no loading effect, equal to R5)\n",
-      "A_v2=round(R5*1000/re_2,0);                                   #Voltage gain of 2nd stage\n",
-      "\n",
-      "print(\"The voltage gain of second stage=%d.\"%A_v2);\n",
-      "\n",
-      "A_vT=A_v1*A_v2;                                 #Overall voltage gain\n",
-      "\n",
-      "print(\"Overall voltage gain=%d.\"%A_vT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) D.C voltages\n",
-        "First stage: VB1=2.16V , VE1=1.46V and VC1=5.18V\n",
-        "First stage: VB2=5.18V , VE2=4.48V and VC2=7.52V\n",
-        "The voltage gain of first stage=66.\n",
-        "The voltage gain of second stage=179.\n",
-        "Overall voltage gain=11814.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_3.ipynb
deleted file mode 100755
index 6e0fb200..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_3.ipynb
+++ /dev/null
@@ -1,1025 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8e425c9c2dfbfee43b3a89e44b0fd7936ba869da73ac3c372e9b23848f1cded1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 11 : MULTISTAGE TRANSISTOR AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 11.1 : Page number 285"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "#Variable declaration\n",
-      "#(i)\n",
-      "A_v=30;                 #Voltage gain\n",
-      "\n",
-      "#(ii)\n",
-      "A_p=100;                #Power gain\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_v_dB=20*log10(A_v);          #Voltage gain, dB\n",
-      "A_p_dB=10*log10(A_p);          #Power gain, dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i) Voltage gain in dB=%.2fdB\"%A_v_dB);\n",
-      "print(\"(ii) Power gain in dB=%ddB\"%A_p_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Voltage gain in dB=29.54dB\n",
-        "(ii) Power gain in dB=20dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.2 : Page number 285-286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#(i)\n",
-      "A_p_dB=40.0;                #Power gain in dB\n",
-      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
-      "A_p=10**A_p_b;              #Power gain in number\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) Power gain in number=%d\"%A_p);\n",
-      "\n",
-      "#(ii)\n",
-      "A_p_dB=43.0;                #Power gain in dB\n",
-      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
-      "A_p=round(10**A_p_b,-4);              #Power gain in number\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) Power gain in number=%d\"%A_p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Power gain in number=10000\n",
-        "(ii) Power gain in number=20000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.3 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "Av_1=100.0;                     #Voltage gain of stage 1\n",
-      "Av_2=200.0;                     #Voltage gain of stage 2\n",
-      "Av_3=400.0;                     #Voltage gain of stage 3\n",
-      "\n",
-      "#Calculations\n",
-      "Av_1_dB=20*log10(Av_1);                     #Voltage gain of stage 1, dB\n",
-      "Av_2_dB=20*log10(Av_2);                     #Voltage gain of stage 2, dB\n",
-      "Av_3_dB=20*log10(Av_3);                     #Voltage gain of stage 3, dB\n",
-      "\n",
-      "Av_T=Av_1_dB+Av_2_dB+Av_3_dB;            #Total voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total voltage gain=%ddB\"%Av_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total voltage gain=138dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.4 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_p_absolute=30.0;                          #Absolute gain of each stage\n",
-      "number_of_stages=5.0;                       #number of stages\n",
-      "negative_feedback=10.0;                     #negative feedback, dB\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_p_dB=round(10*log10(A_p_absolute),2);              #Power gain of one stage. dB\n",
-      "A_p_T=number_of_stages * A_p_dB;            #Total power gain, dB\n",
-      "\n",
-      "#(ii)\n",
-      "A_p_resultant=A_p_T-negative_feedback;      #Resultant power gain with negative feedback, dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total power gain = %.2fdB.\"%A_p_T);\n",
-      "print(\"The resultant power gain with negative feedback = %.2fdB.\"%A_p_resultant);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power gain = 73.85dB.\n",
-        "The resultant power gain with negative feedback = 63.85dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.5 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "P_out_2kHz=1.5;                 #Output power at 2 kHz, W\n",
-      "P_out_20kHz=0.3;                #Output power at 20 kHz, W\n",
-      "P_in=10.0;                      #Input power, mW\n",
-      "\n",
-      "#Calculations\n",
-      "A_p_dB_2kHz=10*log10(P_out_2kHz*1000/P_in);         #dB power gain at 2 kHz\n",
-      "A_p_dB_20kHz=10*log10(P_out_20kHz*1000/P_in);       #dB power gain at 20 kHz\n",
-      "Fall_in_gain=A_p_dB_2kHz-A_p_dB_20kHz;              #Fall in gain from 2kHz to 20kHz\n",
-      "\n",
-      "#Results\n",
-      "print(\"The fall in gain from 2kHz to 20kHz=%.2fdB\"%Fall_in_gain);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fall in gain from 2kHz to 20kHz=6.99dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.6 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_v=15.0;                   #Voltage gain, dB\n",
-      "V_1=0.8;                    #Input signal voltage, V\n",
-      "\n",
-      "#Calculations\n",
-      "#Since, Av(in decibel)=20*log10(V_2/V_1),\n",
-      "V_2=V_1*(10**(A_v/20));                 #Output voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The output voltage= %.1fV.\"%V_2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage= 4.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.7 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_0_dB=70.0;                   #Open circuit voltage gain, dB\n",
-      "A_v_dB=67.0;                   #Voltage gain, dB\n",
-      "R_out=1.5;                     #Output resistance, kilo ohm\n",
-      "\n",
-      "#Calculations\n",
-      "#Since, A_0_dB-A_v_dB=20*log10(A_0/A_v)\n",
-      "ratio_A0_Av=round(10**((A_0_dB-A_v_dB)/20),2);                  #Ratio of open-circuit voltage gain to normal voltage gain\n",
-      "\n",
-      "#Since, A_v/A_0 = RL/(R_out+RL)\n",
-      "RL=R_out/(ratio_A0_Av-1);                               #Load resistor, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The load resistance=%.2f kilo ohm.\"%RL);\n",
-      "\n",
-      "#Note: The value of load resistor is calculated to be 3.6585 kilo ohm and approximated to 3.66. But, in the text it has been approximated to 3.65.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load resistance=3.66 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.8 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=1.0;                 #Load resistance,  kilo ohm\n",
-      "A_v=40.0;               #Voltage gain, dB\n",
-      "V_in=10.0;              #Input signal voltage, mV\n",
-      "\n",
-      "#Calcultaions\n",
-      "#(i)\n",
-      "#Since, A_v=20*log10(V_out/V_in)\n",
-      "V_out=V_in*(10**(A_v/20))/1000;              #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "P_L=(V_out**2/RL);                            #The load power, mW\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i)The output voltage is %dV.\"%V_out);\n",
-      "print(\"(ii)The load poweris %dmW.\"%P_L);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The output voltage is 1V.\n",
-        "(ii)The load poweris 1mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.9 : Page number 287-288\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "P_2=40.0;                     #Output power, W\n",
-      "R=10.0;                        #Resistance of speaker, ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_p_dB=25.0;                                   #Power gain, dB\n",
-      "#Since, A_p_dB=10*log10(P_2/P_1)\n",
-      "P_1=(P_2/10**(A_p_dB/10))*1000;                #Input power, mW\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "A_v_dB=40.0;                                    #Voltage gain, dB\n",
-      "\n",
-      "#Since, P=(V**2)/R,\n",
-      "V_2=(P_2*R)**0.5;                               #Output voltage, V\n",
-      "\n",
-      "#Since, A_v_dB=20*log10(V_2/V_1)\n",
-      "V_1=(V_2/10**(A_v_dB/20))*1000;                #Input voltage, mV\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print(\"(i)The input power=%.1fmW.\"%P_1);\n",
-      "print(\"(ii)The input voltage=%dmV.\"%V_1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The input power=126.5mW.\n",
-        "(ii)The input voltage=200mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.10 : Page  number 288\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v_max=2000.0;                 #Maximum voltage gain\n",
-      "f_max=2.0;                      #Frequency at which maximum voltage gain occurs,kHz\n",
-      "A_v=1414.0;                     #Voltage gain at 50 Hz and 10kHz\n",
-      "f1=50;                          #Lower frequency at which gain is 1414, Hz\n",
-      "f2=10;                          #Upper frequency at which gain is 1414, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, bandwidth is the range of frequency  over which gain is greater than or equal to 70.7% of maximum gain\n",
-      "if((A_v/A_v_max)*100 ==70.7):                               \n",
-      "    print(\"(i)The bandwidth is from %dHz to %dkHz.\"%(f1,f2));\n",
-      "    print(\"(ii)The lower cut-off frequency=%dHz.\"%f1);\n",
-      "    print(\"(iii)The upper cut-off frequency=%dkHz.\"%f2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The bandwidth is from 50Hz to 10kHz.\n",
-        "(ii)The lower cut-off frequency=50Hz.\n",
-        "(iii)The upper cut-off frequency=10kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11 : Page number 291\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v=60.0;               #Voltage gain of single stage amplifier\n",
-      "R_C=500.0;              #Collector load, ohm\n",
-      "R_in=1.0;               #Input impedance, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, there is no loading , second stage gain remains at A_v\n",
-      "#But, due to loading effect of input impedance of second stage, gain of first stage decreases\n",
-      "A_v_2=A_v;                                                    #Voltage gain of second stage\n",
-      "R_AC=round((R_C*R_in*1000)/(R_C+R_in*1000),0);              #Effective load of first stage, ohm\n",
-      "A_v_1=A_v*R_AC/R_C;                                         #Gain of first stage\n",
-      "A_v_T=A_v_1*A_v_2;                                            #Total gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total gain=%d.\"%A_v_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total gain=2397.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.12 : Page number 291-292\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rin=1.0;                        #Input resistance, kilo ohm\n",
-      "beta=100.0;                     #base current amplification factor\n",
-      "RC=2.0;                         #Collector load, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_AC=(RC*Rin)/(RC+Rin);               #Effective load on first stage, kilo ohm\n",
-      "A_v_1=round(beta*(R_AC/Rin),0);                #Voltage gain of first stage\n",
-      "\n",
-      "#(ii)\n",
-      "A_v_2=round(beta*RC/Rin,0);                      #Voltage gain of second stage\n",
-      "\n",
-      "#(iii)\n",
-      "A_v_T=A_v_1*A_v_2;                      #Total voltage gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i)The voltage gain of first stage =%d.\"%A_v_1);\n",
-      "print(\"(ii)The voltage gain of second stage =%d.\"%A_v_2);\n",
-      "print(\"(iii)The total voltage gain =%d.\"%A_v_T);\n",
-      "\n",
-      "#Note: The approximation inthe text for A_v_1=66.66 is taken as 66 but here it has been taken 67 and therefore the total voltage is 13400 instead of 13200.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage gain of first stage =67.\n",
-        "(ii)The voltage gain of second stage =200.\n",
-        "(iii)The total voltage gain =13400.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.13 : Page number 292\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=10.0;                    #Collector load of single stage amplifier, kilo ohm\n",
-      "Rin=1.0;                    #Input resistance, kilo ohm\n",
-      "beta=100.0;                 #base current amplification factor\n",
-      "RL=100.0;                   #Load resistor, ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "R_AC=round((RC*1000)*RL/(RC*1000+RL),-1);                 #Effective collector load,\n",
-      "A_v=beta*R_AC/(Rin*1000);                              #Voltage gain\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage gain=%d.\"%A_v);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain=10.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.14 : Page number 292-293\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                       #Collector suppply voltage, V\n",
-      "R1=10.0;                        #resistor R1, kilo ohm\n",
-      "R2=2.2;                         #resistor R2, kilo ohm\n",
-      "R3=10.0;                        #resistor R3, kilo ohm\n",
-      "R4=2.2;                         #resistor R4, kilo ohm\n",
-      "RC_1=3.6;                         #Collector resistor of first stage, kilo ohm\n",
-      "RC_2=4.0;                         #Collector resistor of second stage, kilo ohm\n",
-      "RE_1=900.0;                     #Emitter resistor of first stage, ohm\n",
-      "RE_2=1.0;                       #Emitter resistor of second stage, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Biasing potential for the second stage is the voltage across R4 resistor,\n",
-      "#so, by voltage divider rule:\n",
-      "VB=VCC*R4/(R3+R4);                #Biasing potential for second stage,(Voltage across R4), V\n",
-      "\n",
-      "print(\"The biasing voltage for the second stage=%.1fV.\"%VB);\n",
-      "\n",
-      "#If coupling capacitor C_c is replaced by a wire, RC_1 and R3 become parallel\n",
-      "Req=round((RC_1*R3)/(RC_1+R3),2);        #Equivalent resistance of R3 parallel with RC_1, kilo ohm\n",
-      "VB=VCC*R4/(Req+R4);             #Biasing voltage if coupling capacitor is replaced by a wire, V\n",
-      "\n",
-      "print(\"The biasing voltage after replacing coupling capacitor by wire=%.2fV.\"%VB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The biasing voltage for the second stage=3.6V.\n",
-        "The biasing voltage after replacing coupling capacitor by wire=9.07V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.15 : Page number 293-294\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return (r1*r2)/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage, V\n",
-      "R1=22.0;                            #Resistor R1, kilo ohm\n",
-      "R2=3.3;                            #Resistor R2, kilo ohm\n",
-      "R3=5.0;                            #Resistor R3, kilo ohm\n",
-      "R4=1.0;                            #Resistor R4, kilo ohm\n",
-      "R5=15.0;                            #Resistor R5, kilo ohm\n",
-      "R6=2.5;                            #Resistor R6, kilo ohm\n",
-      "R7=5.0;                            #Resistor R7, kilo ohm\n",
-      "R8=1.0;                            #Resistor R8, kilo ohm\n",
-      "beta=200;                           #Base current amplification factor\n",
-      "RL=10.0;                            #Load resistor, kilo ohm\n",
-      "V_BE=0.7;                           #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#for 2nd stage\n",
-      "V_R6=round(VCC*R6/(R5+R6),2);                #Voltage across R6, V (voltage divider rule)\n",
-      "V_R8=round(V_R6-V_BE,2);                     #Voltage across R8, V\n",
-      "IE_2=round(V_R8/R8,2);                         #Emitter current through R8, mA (OHM's LAW)\n",
-      "re_2nd_stage=round(25/IE_2,1);                 #a.c emitter resistance for 2nd stage, ohm\n",
-      "\n",
-      "#For 1st stage\n",
-      "V_R2=round(VCC*R2/(R1+R2),2);                     #Voltage across R2, V (voltage divider rule)\n",
-      "V_R4=round(V_R2-V_BE,2);                          #Voltage across R4, V\n",
-      "IE_1=round(V_R4/R4,2);                            #Emitter current through R4, mA (OHM's LAW)\n",
-      "re_1st_stage=round(25/IE_1,1);                    #a.c emitter resistance for 1st stage, ohm\n",
-      "\n",
-      "#(i)\n",
-      "Zin_base_2nd_stage=round((beta*re_2nd_stage)/1000,2);               #input resistance of transistor base of 2nd stage, kilo ohm\n",
-      "Zin=round(pr(pr(R5,R6),Zin_base_2nd_stage),2);                      #Input impedance of the 2nd stage, kilo ohm\n",
-      "R_AC_1st_stage=round(pr(R3,Zin),2);                                          #Effective collector load for 1st stage, kilo ohm\n",
-      "A_v_1=round(R_AC_1st_stage*1000/re_1st_stage,0);                                  #voltage gain of 1st stage\n",
-      "\n",
-      "#(ii)\n",
-      "R_AC_2nd_stage=round(pr(R7,RL),2);                          #Effective collector load for 2nd stage, kilo ohm\n",
-      "A_v_2=round(R_AC_2nd_stage*1000/re_2nd_stage,1);                  #voltage gain of 2nd stage\n",
-      "\n",
-      "#(iii)\n",
-      "A_v_overall=A_v_1*A_v_2;                            #overall voltage gain\n",
-      "\n",
-      "\n",
-      "#results\n",
-      "print(\"(i)The voltage gain of 1st stage=%.0f.\"%A_v_1);\n",
-      "print(\"(i)The voltage gain of 2nd stage=%.1f.\"%A_v_2);\n",
-      "print(\"(i)The overall voltage gain =%d.\"%A_v_overall);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage gain of 1st stage=53.\n",
-        "(i)The voltage gain of 2nd stage=191.4.\n",
-        "(i)The overall voltage gain =10144.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.16 : Page number 297\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Primary_impedance=1000.0;                       #Primary impedance, ohm\n",
-      "Load_impedance=10.0;                            #Load impedance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#since,for maximum power transfer primary impedance should be equal to output impedance\n",
-      "#and, impedance of secondary should be equal to load impedance\n",
-      "#therfore, primary_impedance/load_impedance=square of(primary to secondary turn ratio)\n",
-      "n=(Primary_impedance/Load_impedance)**0.5;           #Primary to secondary turn ratio\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print('The primary to secondary turn ratio for maximum power transfer=%d.'%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary to secondary turn ratio for maximum power transfer=10.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.17 : Page number 297\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=16.0;                    #Load resistor, ohm\n",
-      "R_p=10.0;                   #Output impedance of primary, kilo ohm\n",
-      "Vp=10.0;                    #Terminal voltage of the source, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, for maximum power transfer, the impedance of the primary should be equal to output impedance of the source\n",
-      "n=(R_p*1000/RL)**0.5;                        #Primary to secondary turns ratio\n",
-      "\n",
-      "#Since, power in a transformer remains constant,\n",
-      "#ratio of primary to secondary voltageis equal to primary to secondary turns ratio\n",
-      "Vs=Vp/n;                                #Voltage across the external load, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The primary to secondary turns ratio=%d.\"%n);\n",
-      "print(\"The voltage across the external load=%.1fV.\"%Vs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary to secondary turns ratio=25.\n",
-        "The voltage across the external load=0.4V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.18 : Page number 297-298\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rp=300.0;                   #D.C resistance of primary, ohm\n",
-      "RL=3.0;                     #Load resistance, ohm\n",
-      "R_out=3.0;                  #Ouput resistance of the transistor, kilo ohm               \n",
-      "\n",
-      "#Calculation\n",
-      "#when no signal is applied, only Rp is seen to be the load.\n",
-      "#But, when a.c signal is applied, RL in secondary reflects as RL*(squre of turns ratio).\n",
-      "#Therefore, load is seen to be Rp in series with the reflected RL in primary.\n",
-      "#i.e, R_out=Rp+(n**2 * RL), where n is the turns ratio\n",
-      "n=((R_out*1000-Rp)/RL)**0.5;                     #turns ratio\n",
-      "\n",
-      "#Result\n",
-      "print(\"Turns ratio for maximum power transfer=%d.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Turns ratio for maximum power transfer=30.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.19 : Page number 298"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "f=200.0;                        #Frequency, Hz\n",
-      "Z_out=10.0;                     #Output impedance of the transistor, kilo ohm\n",
-      "Z_in=2.5;                       #Input impedance of the next stage, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#For perfect impedance matching,\n",
-      "#Z_out should be equal to primary impedance\n",
-      "#Z_out=2*pi*f*(primary inductance)\n",
-      "Lp=(Z_out*1000)/(2*pi*f);                   #Primary inductance, H\n",
-      "\n",
-      "#for the secondary side,\n",
-      "#Z_in should be equal to impedance of secondary\n",
-      "Ls=(Z_in*1000)/(2*pi*f);                    #Secondary inductance, H\n",
-      "\n",
-      "\n",
-      "#result\n",
-      "print(\"The primary inductance=%.0fH.\"%Lp);\n",
-      "print(\"The secondary inductance=%.0fH.\"%Ls);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary inductance=8H.\n",
-        "The secondary inductance=2H.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.20 : Page number 299\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Lp=8.0;                     #Primary inductance, H\n",
-      "Ls=2.0;                     #Secondary inductance, H\n",
-      "K=10**-5;                   #Inductance to turns ratio, constant\n",
-      "\n",
-      "#Calculations\n",
-      "Np=(Lp/K)**0.5;                  #Primary turns\n",
-      "Ns=(Ls/K)**0.5;                  #Secondary turns\n",
-      "\n",
-      "#Result\n",
-      "print(\"The primary turns=%.0f.\"%Np);\n",
-      "print(\"The secondary turns=%.0f.\"%Ns);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary turns=894.\n",
-        "The secondary turns=447.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.21 : Page number 300-301\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                   #Collector supply voltage, V\n",
-      "R1=100.0;                   #Resistor R1, kilo ohm\n",
-      "R2=22.0;                    #Resistor R2, kilo ohm\n",
-      "R3=22.0;                    #Resistor R3, kilo ohm\n",
-      "R4=4.7;                     #Resistor R4, kilo ohm\n",
-      "R5=10.0;                    #Resistor R5, kilo ohm\n",
-      "R6=10.0;                    #Resistor R6, kilo ohm\n",
-      "beta=125;                   #Base current amplification factor\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) D.C voltages\n",
-      "#For 1st stage:\n",
-      "V_B1=VCC*R2/(R1+R2);                    #Voltage at the base of 1st transistor, V (Voltage across R2, using voltage divider rule)\n",
-      "V_E1=V_B1-V_BE;                         #Emitter voltage of the 1st transistor, V\n",
-      "I_E1=round(V_E1/R4,2);                           #Emitter current of 1st transistor, mA (OHM's LAW)\n",
-      "I_C1=I_E1;                              #Collector current of 1st transistor, mA(approximately equals to emitter current)\n",
-      "V_C1=VCC-I_C1*R3;                       #Collector voltage of 1st transistor, V\n",
-      "\n",
-      "#For 2nd stage:\n",
-      "V_B2=V_C1;                              #Voltage at the base of 2nd transistor, V (equals to collector voltage of 1st transistor)\n",
-      "V_E2=V_C1-V_BE;                         #Emitter voltage of the 2nd transistor, V\n",
-      "I_E2=V_E2/R6;                           #Emitter current of 2nd transistor, mA (OHM's LAW)\n",
-      "I_C2=I_E2;                              #Collector current 2nd transistor, mA(approximately equals to emitter current)\n",
-      "V_C2=VCC-I_C2*R5;                       #Collector voltage of 2nd transistor, V\n",
-      "\n",
-      "print(\"(i) D.C voltages\");\n",
-      "print(\"First stage: VB1=%.2fV , VE1=%.2fV and VC1=%.2fV\"%(V_B1,V_E1,V_C1));\n",
-      "print(\"First stage: VB2=%.2fV , VE2=%.2fV and VC2=%.2fV\"%(V_B2,V_E2,V_C2));\n",
-      "\n",
-      "#(ii)Voltage gain\n",
-      "#First stage\n",
-      "re_1=25/I_E1;                                   #a.c emitter resistance of 1st transistor, ohm\n",
-      "re_2=25/I_E2;                                   #a.c emitter resistance of 2nd transistor, ohm\n",
-      "Zin_2nd_stage=beta*re_2/1000;                   #Input impedance of 2nd stage, kilo ohm\n",
-      "R_AC=R3*Zin_2nd_stage/(R3+Zin_2nd_stage);       #Total a.c collector load, kilo ohm\n",
-      "A_v1=round(R_AC*1000/re_1,0);                                 #Voltage gain of first stage\n",
-      "\n",
-      "print(\"The voltage gain of first stage=%d.\"%A_v1);\n",
-      "\n",
-      "#Second stage\n",
-      "R_AC=R5;                                        #Total a.c collector load for 2nd stage, kilo ohm(Due to no loading effect, equal to R5)\n",
-      "A_v2=round(R5*1000/re_2,0);                                   #Voltage gain of 2nd stage\n",
-      "\n",
-      "print(\"The voltage gain of second stage=%d.\"%A_v2);\n",
-      "\n",
-      "A_vT=A_v1*A_v2;                                 #Overall voltage gain\n",
-      "\n",
-      "print(\"Overall voltage gain=%d.\"%A_vT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) D.C voltages\n",
-        "First stage: VB1=2.16V , VE1=1.46V and VC1=5.18V\n",
-        "First stage: VB2=5.18V , VE2=4.48V and VC2=7.52V\n",
-        "The voltage gain of first stage=66.\n",
-        "The voltage gain of second stage=179.\n",
-        "Overall voltage gain=11814.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_4.ipynb
deleted file mode 100755
index 6e0fb200..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_4.ipynb
+++ /dev/null
@@ -1,1025 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8e425c9c2dfbfee43b3a89e44b0fd7936ba869da73ac3c372e9b23848f1cded1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 11 : MULTISTAGE TRANSISTOR AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 11.1 : Page number 285"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "#Variable declaration\n",
-      "#(i)\n",
-      "A_v=30;                 #Voltage gain\n",
-      "\n",
-      "#(ii)\n",
-      "A_p=100;                #Power gain\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_v_dB=20*log10(A_v);          #Voltage gain, dB\n",
-      "A_p_dB=10*log10(A_p);          #Power gain, dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i) Voltage gain in dB=%.2fdB\"%A_v_dB);\n",
-      "print(\"(ii) Power gain in dB=%ddB\"%A_p_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Voltage gain in dB=29.54dB\n",
-        "(ii) Power gain in dB=20dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.2 : Page number 285-286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#(i)\n",
-      "A_p_dB=40.0;                #Power gain in dB\n",
-      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
-      "A_p=10**A_p_b;              #Power gain in number\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) Power gain in number=%d\"%A_p);\n",
-      "\n",
-      "#(ii)\n",
-      "A_p_dB=43.0;                #Power gain in dB\n",
-      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
-      "A_p=round(10**A_p_b,-4);              #Power gain in number\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) Power gain in number=%d\"%A_p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Power gain in number=10000\n",
-        "(ii) Power gain in number=20000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.3 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "Av_1=100.0;                     #Voltage gain of stage 1\n",
-      "Av_2=200.0;                     #Voltage gain of stage 2\n",
-      "Av_3=400.0;                     #Voltage gain of stage 3\n",
-      "\n",
-      "#Calculations\n",
-      "Av_1_dB=20*log10(Av_1);                     #Voltage gain of stage 1, dB\n",
-      "Av_2_dB=20*log10(Av_2);                     #Voltage gain of stage 2, dB\n",
-      "Av_3_dB=20*log10(Av_3);                     #Voltage gain of stage 3, dB\n",
-      "\n",
-      "Av_T=Av_1_dB+Av_2_dB+Av_3_dB;            #Total voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total voltage gain=%ddB\"%Av_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total voltage gain=138dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.4 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_p_absolute=30.0;                          #Absolute gain of each stage\n",
-      "number_of_stages=5.0;                       #number of stages\n",
-      "negative_feedback=10.0;                     #negative feedback, dB\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_p_dB=round(10*log10(A_p_absolute),2);              #Power gain of one stage. dB\n",
-      "A_p_T=number_of_stages * A_p_dB;            #Total power gain, dB\n",
-      "\n",
-      "#(ii)\n",
-      "A_p_resultant=A_p_T-negative_feedback;      #Resultant power gain with negative feedback, dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total power gain = %.2fdB.\"%A_p_T);\n",
-      "print(\"The resultant power gain with negative feedback = %.2fdB.\"%A_p_resultant);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power gain = 73.85dB.\n",
-        "The resultant power gain with negative feedback = 63.85dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.5 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "P_out_2kHz=1.5;                 #Output power at 2 kHz, W\n",
-      "P_out_20kHz=0.3;                #Output power at 20 kHz, W\n",
-      "P_in=10.0;                      #Input power, mW\n",
-      "\n",
-      "#Calculations\n",
-      "A_p_dB_2kHz=10*log10(P_out_2kHz*1000/P_in);         #dB power gain at 2 kHz\n",
-      "A_p_dB_20kHz=10*log10(P_out_20kHz*1000/P_in);       #dB power gain at 20 kHz\n",
-      "Fall_in_gain=A_p_dB_2kHz-A_p_dB_20kHz;              #Fall in gain from 2kHz to 20kHz\n",
-      "\n",
-      "#Results\n",
-      "print(\"The fall in gain from 2kHz to 20kHz=%.2fdB\"%Fall_in_gain);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fall in gain from 2kHz to 20kHz=6.99dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.6 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_v=15.0;                   #Voltage gain, dB\n",
-      "V_1=0.8;                    #Input signal voltage, V\n",
-      "\n",
-      "#Calculations\n",
-      "#Since, Av(in decibel)=20*log10(V_2/V_1),\n",
-      "V_2=V_1*(10**(A_v/20));                 #Output voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The output voltage= %.1fV.\"%V_2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage= 4.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.7 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_0_dB=70.0;                   #Open circuit voltage gain, dB\n",
-      "A_v_dB=67.0;                   #Voltage gain, dB\n",
-      "R_out=1.5;                     #Output resistance, kilo ohm\n",
-      "\n",
-      "#Calculations\n",
-      "#Since, A_0_dB-A_v_dB=20*log10(A_0/A_v)\n",
-      "ratio_A0_Av=round(10**((A_0_dB-A_v_dB)/20),2);                  #Ratio of open-circuit voltage gain to normal voltage gain\n",
-      "\n",
-      "#Since, A_v/A_0 = RL/(R_out+RL)\n",
-      "RL=R_out/(ratio_A0_Av-1);                               #Load resistor, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The load resistance=%.2f kilo ohm.\"%RL);\n",
-      "\n",
-      "#Note: The value of load resistor is calculated to be 3.6585 kilo ohm and approximated to 3.66. But, in the text it has been approximated to 3.65.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load resistance=3.66 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.8 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=1.0;                 #Load resistance,  kilo ohm\n",
-      "A_v=40.0;               #Voltage gain, dB\n",
-      "V_in=10.0;              #Input signal voltage, mV\n",
-      "\n",
-      "#Calcultaions\n",
-      "#(i)\n",
-      "#Since, A_v=20*log10(V_out/V_in)\n",
-      "V_out=V_in*(10**(A_v/20))/1000;              #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "P_L=(V_out**2/RL);                            #The load power, mW\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i)The output voltage is %dV.\"%V_out);\n",
-      "print(\"(ii)The load poweris %dmW.\"%P_L);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The output voltage is 1V.\n",
-        "(ii)The load poweris 1mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.9 : Page number 287-288\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "P_2=40.0;                     #Output power, W\n",
-      "R=10.0;                        #Resistance of speaker, ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_p_dB=25.0;                                   #Power gain, dB\n",
-      "#Since, A_p_dB=10*log10(P_2/P_1)\n",
-      "P_1=(P_2/10**(A_p_dB/10))*1000;                #Input power, mW\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "A_v_dB=40.0;                                    #Voltage gain, dB\n",
-      "\n",
-      "#Since, P=(V**2)/R,\n",
-      "V_2=(P_2*R)**0.5;                               #Output voltage, V\n",
-      "\n",
-      "#Since, A_v_dB=20*log10(V_2/V_1)\n",
-      "V_1=(V_2/10**(A_v_dB/20))*1000;                #Input voltage, mV\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print(\"(i)The input power=%.1fmW.\"%P_1);\n",
-      "print(\"(ii)The input voltage=%dmV.\"%V_1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The input power=126.5mW.\n",
-        "(ii)The input voltage=200mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.10 : Page  number 288\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v_max=2000.0;                 #Maximum voltage gain\n",
-      "f_max=2.0;                      #Frequency at which maximum voltage gain occurs,kHz\n",
-      "A_v=1414.0;                     #Voltage gain at 50 Hz and 10kHz\n",
-      "f1=50;                          #Lower frequency at which gain is 1414, Hz\n",
-      "f2=10;                          #Upper frequency at which gain is 1414, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, bandwidth is the range of frequency  over which gain is greater than or equal to 70.7% of maximum gain\n",
-      "if((A_v/A_v_max)*100 ==70.7):                               \n",
-      "    print(\"(i)The bandwidth is from %dHz to %dkHz.\"%(f1,f2));\n",
-      "    print(\"(ii)The lower cut-off frequency=%dHz.\"%f1);\n",
-      "    print(\"(iii)The upper cut-off frequency=%dkHz.\"%f2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The bandwidth is from 50Hz to 10kHz.\n",
-        "(ii)The lower cut-off frequency=50Hz.\n",
-        "(iii)The upper cut-off frequency=10kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11 : Page number 291\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v=60.0;               #Voltage gain of single stage amplifier\n",
-      "R_C=500.0;              #Collector load, ohm\n",
-      "R_in=1.0;               #Input impedance, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, there is no loading , second stage gain remains at A_v\n",
-      "#But, due to loading effect of input impedance of second stage, gain of first stage decreases\n",
-      "A_v_2=A_v;                                                    #Voltage gain of second stage\n",
-      "R_AC=round((R_C*R_in*1000)/(R_C+R_in*1000),0);              #Effective load of first stage, ohm\n",
-      "A_v_1=A_v*R_AC/R_C;                                         #Gain of first stage\n",
-      "A_v_T=A_v_1*A_v_2;                                            #Total gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total gain=%d.\"%A_v_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total gain=2397.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.12 : Page number 291-292\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rin=1.0;                        #Input resistance, kilo ohm\n",
-      "beta=100.0;                     #base current amplification factor\n",
-      "RC=2.0;                         #Collector load, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_AC=(RC*Rin)/(RC+Rin);               #Effective load on first stage, kilo ohm\n",
-      "A_v_1=round(beta*(R_AC/Rin),0);                #Voltage gain of first stage\n",
-      "\n",
-      "#(ii)\n",
-      "A_v_2=round(beta*RC/Rin,0);                      #Voltage gain of second stage\n",
-      "\n",
-      "#(iii)\n",
-      "A_v_T=A_v_1*A_v_2;                      #Total voltage gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i)The voltage gain of first stage =%d.\"%A_v_1);\n",
-      "print(\"(ii)The voltage gain of second stage =%d.\"%A_v_2);\n",
-      "print(\"(iii)The total voltage gain =%d.\"%A_v_T);\n",
-      "\n",
-      "#Note: The approximation inthe text for A_v_1=66.66 is taken as 66 but here it has been taken 67 and therefore the total voltage is 13400 instead of 13200.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage gain of first stage =67.\n",
-        "(ii)The voltage gain of second stage =200.\n",
-        "(iii)The total voltage gain =13400.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.13 : Page number 292\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=10.0;                    #Collector load of single stage amplifier, kilo ohm\n",
-      "Rin=1.0;                    #Input resistance, kilo ohm\n",
-      "beta=100.0;                 #base current amplification factor\n",
-      "RL=100.0;                   #Load resistor, ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "R_AC=round((RC*1000)*RL/(RC*1000+RL),-1);                 #Effective collector load,\n",
-      "A_v=beta*R_AC/(Rin*1000);                              #Voltage gain\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage gain=%d.\"%A_v);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain=10.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.14 : Page number 292-293\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                       #Collector suppply voltage, V\n",
-      "R1=10.0;                        #resistor R1, kilo ohm\n",
-      "R2=2.2;                         #resistor R2, kilo ohm\n",
-      "R3=10.0;                        #resistor R3, kilo ohm\n",
-      "R4=2.2;                         #resistor R4, kilo ohm\n",
-      "RC_1=3.6;                         #Collector resistor of first stage, kilo ohm\n",
-      "RC_2=4.0;                         #Collector resistor of second stage, kilo ohm\n",
-      "RE_1=900.0;                     #Emitter resistor of first stage, ohm\n",
-      "RE_2=1.0;                       #Emitter resistor of second stage, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Biasing potential for the second stage is the voltage across R4 resistor,\n",
-      "#so, by voltage divider rule:\n",
-      "VB=VCC*R4/(R3+R4);                #Biasing potential for second stage,(Voltage across R4), V\n",
-      "\n",
-      "print(\"The biasing voltage for the second stage=%.1fV.\"%VB);\n",
-      "\n",
-      "#If coupling capacitor C_c is replaced by a wire, RC_1 and R3 become parallel\n",
-      "Req=round((RC_1*R3)/(RC_1+R3),2);        #Equivalent resistance of R3 parallel with RC_1, kilo ohm\n",
-      "VB=VCC*R4/(Req+R4);             #Biasing voltage if coupling capacitor is replaced by a wire, V\n",
-      "\n",
-      "print(\"The biasing voltage after replacing coupling capacitor by wire=%.2fV.\"%VB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The biasing voltage for the second stage=3.6V.\n",
-        "The biasing voltage after replacing coupling capacitor by wire=9.07V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.15 : Page number 293-294\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return (r1*r2)/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage, V\n",
-      "R1=22.0;                            #Resistor R1, kilo ohm\n",
-      "R2=3.3;                            #Resistor R2, kilo ohm\n",
-      "R3=5.0;                            #Resistor R3, kilo ohm\n",
-      "R4=1.0;                            #Resistor R4, kilo ohm\n",
-      "R5=15.0;                            #Resistor R5, kilo ohm\n",
-      "R6=2.5;                            #Resistor R6, kilo ohm\n",
-      "R7=5.0;                            #Resistor R7, kilo ohm\n",
-      "R8=1.0;                            #Resistor R8, kilo ohm\n",
-      "beta=200;                           #Base current amplification factor\n",
-      "RL=10.0;                            #Load resistor, kilo ohm\n",
-      "V_BE=0.7;                           #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#for 2nd stage\n",
-      "V_R6=round(VCC*R6/(R5+R6),2);                #Voltage across R6, V (voltage divider rule)\n",
-      "V_R8=round(V_R6-V_BE,2);                     #Voltage across R8, V\n",
-      "IE_2=round(V_R8/R8,2);                         #Emitter current through R8, mA (OHM's LAW)\n",
-      "re_2nd_stage=round(25/IE_2,1);                 #a.c emitter resistance for 2nd stage, ohm\n",
-      "\n",
-      "#For 1st stage\n",
-      "V_R2=round(VCC*R2/(R1+R2),2);                     #Voltage across R2, V (voltage divider rule)\n",
-      "V_R4=round(V_R2-V_BE,2);                          #Voltage across R4, V\n",
-      "IE_1=round(V_R4/R4,2);                            #Emitter current through R4, mA (OHM's LAW)\n",
-      "re_1st_stage=round(25/IE_1,1);                    #a.c emitter resistance for 1st stage, ohm\n",
-      "\n",
-      "#(i)\n",
-      "Zin_base_2nd_stage=round((beta*re_2nd_stage)/1000,2);               #input resistance of transistor base of 2nd stage, kilo ohm\n",
-      "Zin=round(pr(pr(R5,R6),Zin_base_2nd_stage),2);                      #Input impedance of the 2nd stage, kilo ohm\n",
-      "R_AC_1st_stage=round(pr(R3,Zin),2);                                          #Effective collector load for 1st stage, kilo ohm\n",
-      "A_v_1=round(R_AC_1st_stage*1000/re_1st_stage,0);                                  #voltage gain of 1st stage\n",
-      "\n",
-      "#(ii)\n",
-      "R_AC_2nd_stage=round(pr(R7,RL),2);                          #Effective collector load for 2nd stage, kilo ohm\n",
-      "A_v_2=round(R_AC_2nd_stage*1000/re_2nd_stage,1);                  #voltage gain of 2nd stage\n",
-      "\n",
-      "#(iii)\n",
-      "A_v_overall=A_v_1*A_v_2;                            #overall voltage gain\n",
-      "\n",
-      "\n",
-      "#results\n",
-      "print(\"(i)The voltage gain of 1st stage=%.0f.\"%A_v_1);\n",
-      "print(\"(i)The voltage gain of 2nd stage=%.1f.\"%A_v_2);\n",
-      "print(\"(i)The overall voltage gain =%d.\"%A_v_overall);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage gain of 1st stage=53.\n",
-        "(i)The voltage gain of 2nd stage=191.4.\n",
-        "(i)The overall voltage gain =10144.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.16 : Page number 297\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Primary_impedance=1000.0;                       #Primary impedance, ohm\n",
-      "Load_impedance=10.0;                            #Load impedance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#since,for maximum power transfer primary impedance should be equal to output impedance\n",
-      "#and, impedance of secondary should be equal to load impedance\n",
-      "#therfore, primary_impedance/load_impedance=square of(primary to secondary turn ratio)\n",
-      "n=(Primary_impedance/Load_impedance)**0.5;           #Primary to secondary turn ratio\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print('The primary to secondary turn ratio for maximum power transfer=%d.'%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary to secondary turn ratio for maximum power transfer=10.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.17 : Page number 297\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=16.0;                    #Load resistor, ohm\n",
-      "R_p=10.0;                   #Output impedance of primary, kilo ohm\n",
-      "Vp=10.0;                    #Terminal voltage of the source, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, for maximum power transfer, the impedance of the primary should be equal to output impedance of the source\n",
-      "n=(R_p*1000/RL)**0.5;                        #Primary to secondary turns ratio\n",
-      "\n",
-      "#Since, power in a transformer remains constant,\n",
-      "#ratio of primary to secondary voltageis equal to primary to secondary turns ratio\n",
-      "Vs=Vp/n;                                #Voltage across the external load, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The primary to secondary turns ratio=%d.\"%n);\n",
-      "print(\"The voltage across the external load=%.1fV.\"%Vs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary to secondary turns ratio=25.\n",
-        "The voltage across the external load=0.4V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.18 : Page number 297-298\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rp=300.0;                   #D.C resistance of primary, ohm\n",
-      "RL=3.0;                     #Load resistance, ohm\n",
-      "R_out=3.0;                  #Ouput resistance of the transistor, kilo ohm               \n",
-      "\n",
-      "#Calculation\n",
-      "#when no signal is applied, only Rp is seen to be the load.\n",
-      "#But, when a.c signal is applied, RL in secondary reflects as RL*(squre of turns ratio).\n",
-      "#Therefore, load is seen to be Rp in series with the reflected RL in primary.\n",
-      "#i.e, R_out=Rp+(n**2 * RL), where n is the turns ratio\n",
-      "n=((R_out*1000-Rp)/RL)**0.5;                     #turns ratio\n",
-      "\n",
-      "#Result\n",
-      "print(\"Turns ratio for maximum power transfer=%d.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Turns ratio for maximum power transfer=30.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.19 : Page number 298"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "f=200.0;                        #Frequency, Hz\n",
-      "Z_out=10.0;                     #Output impedance of the transistor, kilo ohm\n",
-      "Z_in=2.5;                       #Input impedance of the next stage, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#For perfect impedance matching,\n",
-      "#Z_out should be equal to primary impedance\n",
-      "#Z_out=2*pi*f*(primary inductance)\n",
-      "Lp=(Z_out*1000)/(2*pi*f);                   #Primary inductance, H\n",
-      "\n",
-      "#for the secondary side,\n",
-      "#Z_in should be equal to impedance of secondary\n",
-      "Ls=(Z_in*1000)/(2*pi*f);                    #Secondary inductance, H\n",
-      "\n",
-      "\n",
-      "#result\n",
-      "print(\"The primary inductance=%.0fH.\"%Lp);\n",
-      "print(\"The secondary inductance=%.0fH.\"%Ls);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary inductance=8H.\n",
-        "The secondary inductance=2H.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.20 : Page number 299\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Lp=8.0;                     #Primary inductance, H\n",
-      "Ls=2.0;                     #Secondary inductance, H\n",
-      "K=10**-5;                   #Inductance to turns ratio, constant\n",
-      "\n",
-      "#Calculations\n",
-      "Np=(Lp/K)**0.5;                  #Primary turns\n",
-      "Ns=(Ls/K)**0.5;                  #Secondary turns\n",
-      "\n",
-      "#Result\n",
-      "print(\"The primary turns=%.0f.\"%Np);\n",
-      "print(\"The secondary turns=%.0f.\"%Ns);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary turns=894.\n",
-        "The secondary turns=447.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.21 : Page number 300-301\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                   #Collector supply voltage, V\n",
-      "R1=100.0;                   #Resistor R1, kilo ohm\n",
-      "R2=22.0;                    #Resistor R2, kilo ohm\n",
-      "R3=22.0;                    #Resistor R3, kilo ohm\n",
-      "R4=4.7;                     #Resistor R4, kilo ohm\n",
-      "R5=10.0;                    #Resistor R5, kilo ohm\n",
-      "R6=10.0;                    #Resistor R6, kilo ohm\n",
-      "beta=125;                   #Base current amplification factor\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) D.C voltages\n",
-      "#For 1st stage:\n",
-      "V_B1=VCC*R2/(R1+R2);                    #Voltage at the base of 1st transistor, V (Voltage across R2, using voltage divider rule)\n",
-      "V_E1=V_B1-V_BE;                         #Emitter voltage of the 1st transistor, V\n",
-      "I_E1=round(V_E1/R4,2);                           #Emitter current of 1st transistor, mA (OHM's LAW)\n",
-      "I_C1=I_E1;                              #Collector current of 1st transistor, mA(approximately equals to emitter current)\n",
-      "V_C1=VCC-I_C1*R3;                       #Collector voltage of 1st transistor, V\n",
-      "\n",
-      "#For 2nd stage:\n",
-      "V_B2=V_C1;                              #Voltage at the base of 2nd transistor, V (equals to collector voltage of 1st transistor)\n",
-      "V_E2=V_C1-V_BE;                         #Emitter voltage of the 2nd transistor, V\n",
-      "I_E2=V_E2/R6;                           #Emitter current of 2nd transistor, mA (OHM's LAW)\n",
-      "I_C2=I_E2;                              #Collector current 2nd transistor, mA(approximately equals to emitter current)\n",
-      "V_C2=VCC-I_C2*R5;                       #Collector voltage of 2nd transistor, V\n",
-      "\n",
-      "print(\"(i) D.C voltages\");\n",
-      "print(\"First stage: VB1=%.2fV , VE1=%.2fV and VC1=%.2fV\"%(V_B1,V_E1,V_C1));\n",
-      "print(\"First stage: VB2=%.2fV , VE2=%.2fV and VC2=%.2fV\"%(V_B2,V_E2,V_C2));\n",
-      "\n",
-      "#(ii)Voltage gain\n",
-      "#First stage\n",
-      "re_1=25/I_E1;                                   #a.c emitter resistance of 1st transistor, ohm\n",
-      "re_2=25/I_E2;                                   #a.c emitter resistance of 2nd transistor, ohm\n",
-      "Zin_2nd_stage=beta*re_2/1000;                   #Input impedance of 2nd stage, kilo ohm\n",
-      "R_AC=R3*Zin_2nd_stage/(R3+Zin_2nd_stage);       #Total a.c collector load, kilo ohm\n",
-      "A_v1=round(R_AC*1000/re_1,0);                                 #Voltage gain of first stage\n",
-      "\n",
-      "print(\"The voltage gain of first stage=%d.\"%A_v1);\n",
-      "\n",
-      "#Second stage\n",
-      "R_AC=R5;                                        #Total a.c collector load for 2nd stage, kilo ohm(Due to no loading effect, equal to R5)\n",
-      "A_v2=round(R5*1000/re_2,0);                                   #Voltage gain of 2nd stage\n",
-      "\n",
-      "print(\"The voltage gain of second stage=%d.\"%A_v2);\n",
-      "\n",
-      "A_vT=A_v1*A_v2;                                 #Overall voltage gain\n",
-      "\n",
-      "print(\"Overall voltage gain=%d.\"%A_vT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) D.C voltages\n",
-        "First stage: VB1=2.16V , VE1=1.46V and VC1=5.18V\n",
-        "First stage: VB2=5.18V , VE2=4.48V and VC2=7.52V\n",
-        "The voltage gain of first stage=66.\n",
-        "The voltage gain of second stage=179.\n",
-        "Overall voltage gain=11814.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_5.ipynb
deleted file mode 100755
index 6e0fb200..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_5.ipynb
+++ /dev/null
@@ -1,1025 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:8e425c9c2dfbfee43b3a89e44b0fd7936ba869da73ac3c372e9b23848f1cded1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 11 : MULTISTAGE TRANSISTOR AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 11.1 : Page number 285"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "#Variable declaration\n",
-      "#(i)\n",
-      "A_v=30;                 #Voltage gain\n",
-      "\n",
-      "#(ii)\n",
-      "A_p=100;                #Power gain\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_v_dB=20*log10(A_v);          #Voltage gain, dB\n",
-      "A_p_dB=10*log10(A_p);          #Power gain, dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i) Voltage gain in dB=%.2fdB\"%A_v_dB);\n",
-      "print(\"(ii) Power gain in dB=%ddB\"%A_p_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Voltage gain in dB=29.54dB\n",
-        "(ii) Power gain in dB=20dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.2 : Page number 285-286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#(i)\n",
-      "A_p_dB=40.0;                #Power gain in dB\n",
-      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
-      "A_p=10**A_p_b;              #Power gain in number\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) Power gain in number=%d\"%A_p);\n",
-      "\n",
-      "#(ii)\n",
-      "A_p_dB=43.0;                #Power gain in dB\n",
-      "A_p_b=A_p_dB/10;            #Power gain in bel\n",
-      "A_p=round(10**A_p_b,-4);              #Power gain in number\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) Power gain in number=%d\"%A_p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Power gain in number=10000\n",
-        "(ii) Power gain in number=20000\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.3 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "Av_1=100.0;                     #Voltage gain of stage 1\n",
-      "Av_2=200.0;                     #Voltage gain of stage 2\n",
-      "Av_3=400.0;                     #Voltage gain of stage 3\n",
-      "\n",
-      "#Calculations\n",
-      "Av_1_dB=20*log10(Av_1);                     #Voltage gain of stage 1, dB\n",
-      "Av_2_dB=20*log10(Av_2);                     #Voltage gain of stage 2, dB\n",
-      "Av_3_dB=20*log10(Av_3);                     #Voltage gain of stage 3, dB\n",
-      "\n",
-      "Av_T=Av_1_dB+Av_2_dB+Av_3_dB;            #Total voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total voltage gain=%ddB\"%Av_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total voltage gain=138dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.4 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_p_absolute=30.0;                          #Absolute gain of each stage\n",
-      "number_of_stages=5.0;                       #number of stages\n",
-      "negative_feedback=10.0;                     #negative feedback, dB\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_p_dB=round(10*log10(A_p_absolute),2);              #Power gain of one stage. dB\n",
-      "A_p_T=number_of_stages * A_p_dB;            #Total power gain, dB\n",
-      "\n",
-      "#(ii)\n",
-      "A_p_resultant=A_p_T-negative_feedback;      #Resultant power gain with negative feedback, dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total power gain = %.2fdB.\"%A_p_T);\n",
-      "print(\"The resultant power gain with negative feedback = %.2fdB.\"%A_p_resultant);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power gain = 73.85dB.\n",
-        "The resultant power gain with negative feedback = 63.85dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.5 : Page number 286\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "P_out_2kHz=1.5;                 #Output power at 2 kHz, W\n",
-      "P_out_20kHz=0.3;                #Output power at 20 kHz, W\n",
-      "P_in=10.0;                      #Input power, mW\n",
-      "\n",
-      "#Calculations\n",
-      "A_p_dB_2kHz=10*log10(P_out_2kHz*1000/P_in);         #dB power gain at 2 kHz\n",
-      "A_p_dB_20kHz=10*log10(P_out_20kHz*1000/P_in);       #dB power gain at 20 kHz\n",
-      "Fall_in_gain=A_p_dB_2kHz-A_p_dB_20kHz;              #Fall in gain from 2kHz to 20kHz\n",
-      "\n",
-      "#Results\n",
-      "print(\"The fall in gain from 2kHz to 20kHz=%.2fdB\"%Fall_in_gain);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fall in gain from 2kHz to 20kHz=6.99dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.6 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_v=15.0;                   #Voltage gain, dB\n",
-      "V_1=0.8;                    #Input signal voltage, V\n",
-      "\n",
-      "#Calculations\n",
-      "#Since, Av(in decibel)=20*log10(V_2/V_1),\n",
-      "V_2=V_1*(10**(A_v/20));                 #Output voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The output voltage= %.1fV.\"%V_2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage= 4.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.7 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_0_dB=70.0;                   #Open circuit voltage gain, dB\n",
-      "A_v_dB=67.0;                   #Voltage gain, dB\n",
-      "R_out=1.5;                     #Output resistance, kilo ohm\n",
-      "\n",
-      "#Calculations\n",
-      "#Since, A_0_dB-A_v_dB=20*log10(A_0/A_v)\n",
-      "ratio_A0_Av=round(10**((A_0_dB-A_v_dB)/20),2);                  #Ratio of open-circuit voltage gain to normal voltage gain\n",
-      "\n",
-      "#Since, A_v/A_0 = RL/(R_out+RL)\n",
-      "RL=R_out/(ratio_A0_Av-1);                               #Load resistor, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The load resistance=%.2f kilo ohm.\"%RL);\n",
-      "\n",
-      "#Note: The value of load resistor is calculated to be 3.6585 kilo ohm and approximated to 3.66. But, in the text it has been approximated to 3.65.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load resistance=3.66 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.8 : Page number 287\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=1.0;                 #Load resistance,  kilo ohm\n",
-      "A_v=40.0;               #Voltage gain, dB\n",
-      "V_in=10.0;              #Input signal voltage, mV\n",
-      "\n",
-      "#Calcultaions\n",
-      "#(i)\n",
-      "#Since, A_v=20*log10(V_out/V_in)\n",
-      "V_out=V_in*(10**(A_v/20))/1000;              #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "P_L=(V_out**2/RL);                            #The load power, mW\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i)The output voltage is %dV.\"%V_out);\n",
-      "print(\"(ii)The load poweris %dmW.\"%P_L);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The output voltage is 1V.\n",
-        "(ii)The load poweris 1mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.9 : Page number 287-288\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "P_2=40.0;                     #Output power, W\n",
-      "R=10.0;                        #Resistance of speaker, ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "A_p_dB=25.0;                                   #Power gain, dB\n",
-      "#Since, A_p_dB=10*log10(P_2/P_1)\n",
-      "P_1=(P_2/10**(A_p_dB/10))*1000;                #Input power, mW\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "A_v_dB=40.0;                                    #Voltage gain, dB\n",
-      "\n",
-      "#Since, P=(V**2)/R,\n",
-      "V_2=(P_2*R)**0.5;                               #Output voltage, V\n",
-      "\n",
-      "#Since, A_v_dB=20*log10(V_2/V_1)\n",
-      "V_1=(V_2/10**(A_v_dB/20))*1000;                #Input voltage, mV\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print(\"(i)The input power=%.1fmW.\"%P_1);\n",
-      "print(\"(ii)The input voltage=%dmV.\"%V_1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The input power=126.5mW.\n",
-        "(ii)The input voltage=200mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.10 : Page  number 288\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v_max=2000.0;                 #Maximum voltage gain\n",
-      "f_max=2.0;                      #Frequency at which maximum voltage gain occurs,kHz\n",
-      "A_v=1414.0;                     #Voltage gain at 50 Hz and 10kHz\n",
-      "f1=50;                          #Lower frequency at which gain is 1414, Hz\n",
-      "f2=10;                          #Upper frequency at which gain is 1414, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, bandwidth is the range of frequency  over which gain is greater than or equal to 70.7% of maximum gain\n",
-      "if((A_v/A_v_max)*100 ==70.7):                               \n",
-      "    print(\"(i)The bandwidth is from %dHz to %dkHz.\"%(f1,f2));\n",
-      "    print(\"(ii)The lower cut-off frequency=%dHz.\"%f1);\n",
-      "    print(\"(iii)The upper cut-off frequency=%dkHz.\"%f2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The bandwidth is from 50Hz to 10kHz.\n",
-        "(ii)The lower cut-off frequency=50Hz.\n",
-        "(iii)The upper cut-off frequency=10kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.11 : Page number 291\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_v=60.0;               #Voltage gain of single stage amplifier\n",
-      "R_C=500.0;              #Collector load, ohm\n",
-      "R_in=1.0;               #Input impedance, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, there is no loading , second stage gain remains at A_v\n",
-      "#But, due to loading effect of input impedance of second stage, gain of first stage decreases\n",
-      "A_v_2=A_v;                                                    #Voltage gain of second stage\n",
-      "R_AC=round((R_C*R_in*1000)/(R_C+R_in*1000),0);              #Effective load of first stage, ohm\n",
-      "A_v_1=A_v*R_AC/R_C;                                         #Gain of first stage\n",
-      "A_v_T=A_v_1*A_v_2;                                            #Total gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total gain=%d.\"%A_v_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total gain=2397.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.12 : Page number 291-292\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rin=1.0;                        #Input resistance, kilo ohm\n",
-      "beta=100.0;                     #base current amplification factor\n",
-      "RC=2.0;                         #Collector load, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_AC=(RC*Rin)/(RC+Rin);               #Effective load on first stage, kilo ohm\n",
-      "A_v_1=round(beta*(R_AC/Rin),0);                #Voltage gain of first stage\n",
-      "\n",
-      "#(ii)\n",
-      "A_v_2=round(beta*RC/Rin,0);                      #Voltage gain of second stage\n",
-      "\n",
-      "#(iii)\n",
-      "A_v_T=A_v_1*A_v_2;                      #Total voltage gain\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"(i)The voltage gain of first stage =%d.\"%A_v_1);\n",
-      "print(\"(ii)The voltage gain of second stage =%d.\"%A_v_2);\n",
-      "print(\"(iii)The total voltage gain =%d.\"%A_v_T);\n",
-      "\n",
-      "#Note: The approximation inthe text for A_v_1=66.66 is taken as 66 but here it has been taken 67 and therefore the total voltage is 13400 instead of 13200.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage gain of first stage =67.\n",
-        "(ii)The voltage gain of second stage =200.\n",
-        "(iii)The total voltage gain =13400.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.13 : Page number 292\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=10.0;                    #Collector load of single stage amplifier, kilo ohm\n",
-      "Rin=1.0;                    #Input resistance, kilo ohm\n",
-      "beta=100.0;                 #base current amplification factor\n",
-      "RL=100.0;                   #Load resistor, ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "R_AC=round((RC*1000)*RL/(RC*1000+RL),-1);                 #Effective collector load,\n",
-      "A_v=beta*R_AC/(Rin*1000);                              #Voltage gain\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage gain=%d.\"%A_v);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain=10.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.14 : Page number 292-293\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                       #Collector suppply voltage, V\n",
-      "R1=10.0;                        #resistor R1, kilo ohm\n",
-      "R2=2.2;                         #resistor R2, kilo ohm\n",
-      "R3=10.0;                        #resistor R3, kilo ohm\n",
-      "R4=2.2;                         #resistor R4, kilo ohm\n",
-      "RC_1=3.6;                         #Collector resistor of first stage, kilo ohm\n",
-      "RC_2=4.0;                         #Collector resistor of second stage, kilo ohm\n",
-      "RE_1=900.0;                     #Emitter resistor of first stage, ohm\n",
-      "RE_2=1.0;                       #Emitter resistor of second stage, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Biasing potential for the second stage is the voltage across R4 resistor,\n",
-      "#so, by voltage divider rule:\n",
-      "VB=VCC*R4/(R3+R4);                #Biasing potential for second stage,(Voltage across R4), V\n",
-      "\n",
-      "print(\"The biasing voltage for the second stage=%.1fV.\"%VB);\n",
-      "\n",
-      "#If coupling capacitor C_c is replaced by a wire, RC_1 and R3 become parallel\n",
-      "Req=round((RC_1*R3)/(RC_1+R3),2);        #Equivalent resistance of R3 parallel with RC_1, kilo ohm\n",
-      "VB=VCC*R4/(Req+R4);             #Biasing voltage if coupling capacitor is replaced by a wire, V\n",
-      "\n",
-      "print(\"The biasing voltage after replacing coupling capacitor by wire=%.2fV.\"%VB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The biasing voltage for the second stage=3.6V.\n",
-        "The biasing voltage after replacing coupling capacitor by wire=9.07V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.15 : Page number 293-294\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return (r1*r2)/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage, V\n",
-      "R1=22.0;                            #Resistor R1, kilo ohm\n",
-      "R2=3.3;                            #Resistor R2, kilo ohm\n",
-      "R3=5.0;                            #Resistor R3, kilo ohm\n",
-      "R4=1.0;                            #Resistor R4, kilo ohm\n",
-      "R5=15.0;                            #Resistor R5, kilo ohm\n",
-      "R6=2.5;                            #Resistor R6, kilo ohm\n",
-      "R7=5.0;                            #Resistor R7, kilo ohm\n",
-      "R8=1.0;                            #Resistor R8, kilo ohm\n",
-      "beta=200;                           #Base current amplification factor\n",
-      "RL=10.0;                            #Load resistor, kilo ohm\n",
-      "V_BE=0.7;                           #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#for 2nd stage\n",
-      "V_R6=round(VCC*R6/(R5+R6),2);                #Voltage across R6, V (voltage divider rule)\n",
-      "V_R8=round(V_R6-V_BE,2);                     #Voltage across R8, V\n",
-      "IE_2=round(V_R8/R8,2);                         #Emitter current through R8, mA (OHM's LAW)\n",
-      "re_2nd_stage=round(25/IE_2,1);                 #a.c emitter resistance for 2nd stage, ohm\n",
-      "\n",
-      "#For 1st stage\n",
-      "V_R2=round(VCC*R2/(R1+R2),2);                     #Voltage across R2, V (voltage divider rule)\n",
-      "V_R4=round(V_R2-V_BE,2);                          #Voltage across R4, V\n",
-      "IE_1=round(V_R4/R4,2);                            #Emitter current through R4, mA (OHM's LAW)\n",
-      "re_1st_stage=round(25/IE_1,1);                    #a.c emitter resistance for 1st stage, ohm\n",
-      "\n",
-      "#(i)\n",
-      "Zin_base_2nd_stage=round((beta*re_2nd_stage)/1000,2);               #input resistance of transistor base of 2nd stage, kilo ohm\n",
-      "Zin=round(pr(pr(R5,R6),Zin_base_2nd_stage),2);                      #Input impedance of the 2nd stage, kilo ohm\n",
-      "R_AC_1st_stage=round(pr(R3,Zin),2);                                          #Effective collector load for 1st stage, kilo ohm\n",
-      "A_v_1=round(R_AC_1st_stage*1000/re_1st_stage,0);                                  #voltage gain of 1st stage\n",
-      "\n",
-      "#(ii)\n",
-      "R_AC_2nd_stage=round(pr(R7,RL),2);                          #Effective collector load for 2nd stage, kilo ohm\n",
-      "A_v_2=round(R_AC_2nd_stage*1000/re_2nd_stage,1);                  #voltage gain of 2nd stage\n",
-      "\n",
-      "#(iii)\n",
-      "A_v_overall=A_v_1*A_v_2;                            #overall voltage gain\n",
-      "\n",
-      "\n",
-      "#results\n",
-      "print(\"(i)The voltage gain of 1st stage=%.0f.\"%A_v_1);\n",
-      "print(\"(i)The voltage gain of 2nd stage=%.1f.\"%A_v_2);\n",
-      "print(\"(i)The overall voltage gain =%d.\"%A_v_overall);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage gain of 1st stage=53.\n",
-        "(i)The voltage gain of 2nd stage=191.4.\n",
-        "(i)The overall voltage gain =10144.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.16 : Page number 297\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Primary_impedance=1000.0;                       #Primary impedance, ohm\n",
-      "Load_impedance=10.0;                            #Load impedance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#since,for maximum power transfer primary impedance should be equal to output impedance\n",
-      "#and, impedance of secondary should be equal to load impedance\n",
-      "#therfore, primary_impedance/load_impedance=square of(primary to secondary turn ratio)\n",
-      "n=(Primary_impedance/Load_impedance)**0.5;           #Primary to secondary turn ratio\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print('The primary to secondary turn ratio for maximum power transfer=%d.'%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary to secondary turn ratio for maximum power transfer=10.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.17 : Page number 297\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=16.0;                    #Load resistor, ohm\n",
-      "R_p=10.0;                   #Output impedance of primary, kilo ohm\n",
-      "Vp=10.0;                    #Terminal voltage of the source, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, for maximum power transfer, the impedance of the primary should be equal to output impedance of the source\n",
-      "n=(R_p*1000/RL)**0.5;                        #Primary to secondary turns ratio\n",
-      "\n",
-      "#Since, power in a transformer remains constant,\n",
-      "#ratio of primary to secondary voltageis equal to primary to secondary turns ratio\n",
-      "Vs=Vp/n;                                #Voltage across the external load, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The primary to secondary turns ratio=%d.\"%n);\n",
-      "print(\"The voltage across the external load=%.1fV.\"%Vs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary to secondary turns ratio=25.\n",
-        "The voltage across the external load=0.4V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.18 : Page number 297-298\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rp=300.0;                   #D.C resistance of primary, ohm\n",
-      "RL=3.0;                     #Load resistance, ohm\n",
-      "R_out=3.0;                  #Ouput resistance of the transistor, kilo ohm               \n",
-      "\n",
-      "#Calculation\n",
-      "#when no signal is applied, only Rp is seen to be the load.\n",
-      "#But, when a.c signal is applied, RL in secondary reflects as RL*(squre of turns ratio).\n",
-      "#Therefore, load is seen to be Rp in series with the reflected RL in primary.\n",
-      "#i.e, R_out=Rp+(n**2 * RL), where n is the turns ratio\n",
-      "n=((R_out*1000-Rp)/RL)**0.5;                     #turns ratio\n",
-      "\n",
-      "#Result\n",
-      "print(\"Turns ratio for maximum power transfer=%d.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Turns ratio for maximum power transfer=30.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.19 : Page number 298"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "f=200.0;                        #Frequency, Hz\n",
-      "Z_out=10.0;                     #Output impedance of the transistor, kilo ohm\n",
-      "Z_in=2.5;                       #Input impedance of the next stage, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#For perfect impedance matching,\n",
-      "#Z_out should be equal to primary impedance\n",
-      "#Z_out=2*pi*f*(primary inductance)\n",
-      "Lp=(Z_out*1000)/(2*pi*f);                   #Primary inductance, H\n",
-      "\n",
-      "#for the secondary side,\n",
-      "#Z_in should be equal to impedance of secondary\n",
-      "Ls=(Z_in*1000)/(2*pi*f);                    #Secondary inductance, H\n",
-      "\n",
-      "\n",
-      "#result\n",
-      "print(\"The primary inductance=%.0fH.\"%Lp);\n",
-      "print(\"The secondary inductance=%.0fH.\"%Ls);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary inductance=8H.\n",
-        "The secondary inductance=2H.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.20 : Page number 299\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Lp=8.0;                     #Primary inductance, H\n",
-      "Ls=2.0;                     #Secondary inductance, H\n",
-      "K=10**-5;                   #Inductance to turns ratio, constant\n",
-      "\n",
-      "#Calculations\n",
-      "Np=(Lp/K)**0.5;                  #Primary turns\n",
-      "Ns=(Ls/K)**0.5;                  #Secondary turns\n",
-      "\n",
-      "#Result\n",
-      "print(\"The primary turns=%.0f.\"%Np);\n",
-      "print(\"The secondary turns=%.0f.\"%Ns);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The primary turns=894.\n",
-        "The secondary turns=447.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 11.21 : Page number 300-301\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                   #Collector supply voltage, V\n",
-      "R1=100.0;                   #Resistor R1, kilo ohm\n",
-      "R2=22.0;                    #Resistor R2, kilo ohm\n",
-      "R3=22.0;                    #Resistor R3, kilo ohm\n",
-      "R4=4.7;                     #Resistor R4, kilo ohm\n",
-      "R5=10.0;                    #Resistor R5, kilo ohm\n",
-      "R6=10.0;                    #Resistor R6, kilo ohm\n",
-      "beta=125;                   #Base current amplification factor\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) D.C voltages\n",
-      "#For 1st stage:\n",
-      "V_B1=VCC*R2/(R1+R2);                    #Voltage at the base of 1st transistor, V (Voltage across R2, using voltage divider rule)\n",
-      "V_E1=V_B1-V_BE;                         #Emitter voltage of the 1st transistor, V\n",
-      "I_E1=round(V_E1/R4,2);                           #Emitter current of 1st transistor, mA (OHM's LAW)\n",
-      "I_C1=I_E1;                              #Collector current of 1st transistor, mA(approximately equals to emitter current)\n",
-      "V_C1=VCC-I_C1*R3;                       #Collector voltage of 1st transistor, V\n",
-      "\n",
-      "#For 2nd stage:\n",
-      "V_B2=V_C1;                              #Voltage at the base of 2nd transistor, V (equals to collector voltage of 1st transistor)\n",
-      "V_E2=V_C1-V_BE;                         #Emitter voltage of the 2nd transistor, V\n",
-      "I_E2=V_E2/R6;                           #Emitter current of 2nd transistor, mA (OHM's LAW)\n",
-      "I_C2=I_E2;                              #Collector current 2nd transistor, mA(approximately equals to emitter current)\n",
-      "V_C2=VCC-I_C2*R5;                       #Collector voltage of 2nd transistor, V\n",
-      "\n",
-      "print(\"(i) D.C voltages\");\n",
-      "print(\"First stage: VB1=%.2fV , VE1=%.2fV and VC1=%.2fV\"%(V_B1,V_E1,V_C1));\n",
-      "print(\"First stage: VB2=%.2fV , VE2=%.2fV and VC2=%.2fV\"%(V_B2,V_E2,V_C2));\n",
-      "\n",
-      "#(ii)Voltage gain\n",
-      "#First stage\n",
-      "re_1=25/I_E1;                                   #a.c emitter resistance of 1st transistor, ohm\n",
-      "re_2=25/I_E2;                                   #a.c emitter resistance of 2nd transistor, ohm\n",
-      "Zin_2nd_stage=beta*re_2/1000;                   #Input impedance of 2nd stage, kilo ohm\n",
-      "R_AC=R3*Zin_2nd_stage/(R3+Zin_2nd_stage);       #Total a.c collector load, kilo ohm\n",
-      "A_v1=round(R_AC*1000/re_1,0);                                 #Voltage gain of first stage\n",
-      "\n",
-      "print(\"The voltage gain of first stage=%d.\"%A_v1);\n",
-      "\n",
-      "#Second stage\n",
-      "R_AC=R5;                                        #Total a.c collector load for 2nd stage, kilo ohm(Due to no loading effect, equal to R5)\n",
-      "A_v2=round(R5*1000/re_2,0);                                   #Voltage gain of 2nd stage\n",
-      "\n",
-      "print(\"The voltage gain of second stage=%d.\"%A_v2);\n",
-      "\n",
-      "A_vT=A_v1*A_v2;                                 #Overall voltage gain\n",
-      "\n",
-      "print(\"Overall voltage gain=%d.\"%A_vT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) D.C voltages\n",
-        "First stage: VB1=2.16V , VE1=1.46V and VC1=5.18V\n",
-        "First stage: VB2=5.18V , VE2=4.48V and VC2=7.52V\n",
-        "The voltage gain of first stage=66.\n",
-        "The voltage gain of second stage=179.\n",
-        "Overall voltage gain=11814.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12.ipynb
deleted file mode 100755
index 05e3d9d8..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12.ipynb
+++ /dev/null
@@ -1,968 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1e9209171152811793fc18d1ee8c80ddcef574d69421ec87eeaa8fb87a304f6d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 12: TRANSISTOR AUDIO POWER AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1 : Page number 308\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                               #Collector supply voltage, V\n",
-      "R1=10.0;                                #Resistor R1, kilo ohm\n",
-      "R2=2.2;                                 #Resistor R2, kilo ohm\n",
-      "RC=3.6;                                 #Collector resistor, kilo ohm\n",
-      "RE=1.1;                                 #Emitter resistor, kilo ohm\n",
-      "VBE=0.7;                                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "I1=VCC/(R1+R2);                     #Current through R1 and R2, mA (OHM's LAW)\n",
-      "V2=I1*R2;                           #Voltage across R2 resistor, V (OHM's LAW)\n",
-      "VE=V2-VBE;                          #Emitter voltage, V\n",
-      "IE=VE/RE;                           #Emitter current, mA (OHM's LAW)\n",
-      "IC=IE;                              #Collector current, mA (approximately equal to emitter current)\n",
-      "I_T=I1+IC;                          #Total current drawn from the supply, mA\n",
-      "P_dc=VCC*I_T;                       #Total power drawn from the supply, mW\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total power drawn from the supply=%.1fmW.\"%P_dc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power drawn from the supply=18.2mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2 : Page number 309\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_L=10.6;                       #Voltage across load, V.(from a.c voltmeter, therfore r.m.s value)\n",
-      "R_L=200.0;                      #Load resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, power =V**2/R,\n",
-      "P_O=(V_L**2/R_L)*1000;                     #A.C output power, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power = %.1fmW.\"%P_O);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power = 561.8mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.3 : Page number 309\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=100.0;                   #Load resistance, ohm\n",
-      "V_PP=18.0;                  #Peak-to-peak a.c voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, V(r.m.s)=(V(peak-to-peak)/2)/sqrt(2)\n",
-      "VL=V_PP/(2*(2**0.5));                      #r.m.s value, V\n",
-      "\n",
-      "#Since, power=(square of voltage)/resistance\n",
-      "P_O_max=(VL**2/RL)*1000;                   #Maximum possible a.c load power, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum possible a.c load power=%dmW.\"%P_O_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum possible a.c load power=405mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.4 : Page number 310\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_battery=12.0;                 #Battery voltage, V\n",
-      "P_out=2.0;                      #Output power, W\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Power=Current*Voltage\n",
-      "IC=(P_out/V_battery)*1000;                 #Maximum collector current , mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector current=%.1fmA.\"%IC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current=166.7mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.5 : Page number 310\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_battery=12.0;                     #Battery voltage, V\n",
-      "RL=4.0;                             #Collector load, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "IC_max=V_battery/RL;                #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector current=%dmA.\"%IC_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current=3mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.6 : Page number 310-311\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P=50.0;                     #Power supplied by power amplifier, W\n",
-      "R=8.0;                      #Resistance of speaker, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, Power=Voltage _square/Resistance,\n",
-      "V=(P*R)**0.5;               #a.c output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "I=V/R;                          #a.c output current, A (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The a.c output voltage=%dV.\"%V);\n",
-      "print(\"(ii) The a.c output current=%.1fA.\"%I);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The a.c output voltage=20V.\n",
-        "(ii) The a.c output current=2.5A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.7 : Page number 315\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                   #Collector supply voltage, V\n",
-      "ib_peak=10.0;               #Base current(peak), mA\n",
-      "RB=1.0;                     #Base resistance, kilo ohm\n",
-      "RC=20.0;                    #Collector resistance, ohm\n",
-      "beta=25.0;                  #Base current amplification factor\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "IB=round(VCC-VBE/RB,1);                         #Base current, mA (OHM's LAW)\n",
-      "IC=int(beta*IB);                            #Collector current, mA\n",
-      "VCE=VCC-(IC/1000)*RC;                  #Collector emitter voltage, V (KVL)\n",
-      "\n",
-      "#(i)\n",
-      "ic_peak=beta*ib_peak;                          #Collector current(peak), mA\n",
-      "P_o_ac=(ic_peak/1000)**2*RC/2;                 #Output power, W\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=VCC*IC/1000;                                    #Input power, W\n",
-      "\n",
-      "#(iii)\n",
-      "collector_efficiency=(P_o_ac/P_dc)*100;                   #Collector efficiency of the amplifier circuit,\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output power=%.3fW.\"%P_o_ac);\n",
-      "print(\"(ii) The input power=%.1fW.\"%P_dc);\n",
-      "print(\"(iii) The collector efficiency=%.1f%%.\"%collector_efficiency);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output power=0.625W.\n",
-        "(ii) The input power=9.6W.\n",
-        "(iii) The collector efficiency=6.5%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.8 : Page number 317\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_dc=10.0;                  #zero signal power dissipation, W\n",
-      "P_o=4.0;                    #a.c output power, W\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Collector_eff=(P_o/P_dc)*100;                 #collector efficiency\n",
-      "\n",
-      "#(ii)\n",
-      "#Zero signal power is the maximum power dissipation in a transistor, therefore,\n",
-      "Power_rating=P_dc;                  #Power rating of the transistor, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The collector efficiency=%d%%.\"%Collector_eff);\n",
-      "print(\"(i) The power rating of the transistor=%dW.\"%Power_rating);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The collector efficiency=40%.\n",
-        "(i) The power rating of the transistor=10W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.9 : Page number 317-318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=100.0;                   #Secondary load, ohm\n",
-      "n=10.0;                     #Transformer turn ratio\n",
-      "IC=100.0;                   #Zero signal collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "RL_reflected=n**2*RL;                        #Reflected load as seen by the primary of the transformer, ohm\n",
-      "P_o_ac_max=(IC/1000)**2*RL_reflected/2;             #Maximum a.c power output, W      \n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum a.c power output=%dW.\"%P_o_ac_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum a.c power output=50W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.10 : Page number 318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=5.0;                    #Collector supply voltage, V\n",
-      "IC=50.0;                    #Zero signal collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "P_o_max=VCC*IC/2;           #Maximum a.c output power, mW\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=VCC*IC;                #D.C input power, mW\n",
-      "#Since, maximum power is dissipated in the zero signal conditions\n",
-      "Power_rating=P_dc;              #Power rating of transistor, mW\n",
-      "\n",
-      "#(iii)\n",
-      "Max_collector_eff=(P_o_max/P_dc)*100;             #Maximum collector efficiency\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The maximum a.c output power=%dmW\"%P_o_max);\n",
-      "print(\"(ii) The power rating of the transistor=%dmW.\"%Power_rating);\n",
-      "print(\"(iii) The maximum collector efficiency =%d%%.\"%Max_collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The maximum a.c output power=125mW\n",
-        "(ii) The power rating of the transistor=250mW.\n",
-        "(iii) The maximum collector efficiency =50%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.11 : Page number 318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "ic_max=160.0;               #Maximum a.c collector current, mA\n",
-      "ic_min=10.0;                #Minimum a.c collector current, mA\n",
-      "vce_max=12.0;               #Maximum collector-emitter voltage, V\n",
-      "vce_min=2.0;                #Minimum collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "vce_pp=vce_max-vce_min;                                #peak to peak collector emitter voltage, V\n",
-      "ic_pp=ic_max-ic_min;                                   #peak to peak collector current, V\n",
-      "P_o=(vce_pp/(2*sqrt(2)))*(ic_pp/(2*sqrt(2)));          #a.c output power, mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power=%.1fmW.\"%P_o);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power=187.5mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12 : Page number 319-320\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                           #Battey voltage, V\n",
-      "IC_max_change=100.0;                #maximum collector current change, mA\n",
-      "RL=5.0;                             #Loudspeaker resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "VCE_max_change=VCC;                                         #Maximum collector-emitter voltage change\n",
-      "#(i) Loud speaker directly connected in the collector\n",
-      "Vmax_speaker=(IC_max_change/1000)*RL;                              #Maximum voltage across the loudspeaker, V\n",
-      "P_speaker_directly_coupled=Vmax_speaker*IC_max_change;                       #Power developed in the loudspeaker,mW\n",
-      "\n",
-      "#(ii) Loudspeaker transformer coupled\n",
-      "Z_out=(VCE_max_change/IC_max_change)*1000;                     #Output impedance of transistor, ohm\n",
-      "\n",
-      "#For max power transfer, primary impedance should be Z_out\n",
-      "RL_reflected=Z_out;                                         #Load resistance as seen by primary, ohm\n",
-      "n=sqrt(RL_reflected/RL);                                    #Turns ratio of transformer\n",
-      "Vp=VCC;                                                     #Transformer primary voltage, V\n",
-      "Vs=Vp/n;                                                    #Transformer secondary voltage, V\n",
-      "IL=Vs/RL;                                                   #Load current, A\n",
-      "P_speaker_transformer_coupled=IL**2*RL*1000;                #Power delivered to the speaker, mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The power transferred to the speaker when directly coupled=%dmW.\"%P_speaker_directly_coupled);\n",
-      "print(\"(ii) The power trasnferred to the speaker when transformer-coupled=%dmW.\"%P_speaker_transformer_coupled);\n",
-      "print(\"     The turns ratio=%.1f.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The power transferred to the speaker when directly coupled=50mW.\n",
-        "(ii) The power trasnferred to the speaker when transformer-coupled=1200mW.\n",
-        "     The turns ratio=4.9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.13 : Page number 320-321\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "beta=100.0;                                 #Base current amplification factor\n",
-      "RL=81.6;                                    #Load resistance, ohm\n",
-      "VCE_peak=30.0;                               #Peak value of collector voltage, V\n",
-      "IC_peak=35.0;                               #Peak value of collector current, mA\n",
-      "VCE_min=5.0;                                 #Minimum value of collector voltage, V\n",
-      "IC_min=1.0;                                 #Minimum value of collector current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC_zero_signal=(IC_peak-IC_min)/2 +1;                  #Zero signal collector current, mA\n",
-      "\n",
-      "#(ii)\n",
-      "IB_zero_signal=IC_zero_signal/beta;                 #Zero signal base current, mA\n",
-      "\n",
-      "#(iii)\n",
-      "VCE_zero_signal=(VCE_peak-VCE_min)/2 +5;                    #Zero signal collector-emitter voltage, V\n",
-      "VCC=VCE_zero_signal;                                    #Collector supply voltage,V (due to transformer coupling, aproximately equal to zero signal VCE)\n",
-      "P_dc=VCC*IC_zero_signal;                                #d.c input power, mW\n",
-      "VCE_ac=(VCE_peak-VCE_min)/(2*sqrt(2));                         #a.c output voltage, V\n",
-      "IC_ac=(IC_peak-IC_min)/(2*sqrt(2));                           #a.c output  current, mA\n",
-      "P_ac=VCE_ac*IC_ac;                                      #a.c output power, mW\n",
-      "\n",
-      "#(iv)\n",
-      "collector_eff=(P_ac/P_dc)*100;                          #Collector efficiency\n",
-      "\n",
-      "#(v)\n",
-      "#a.c resistance RL'=negative inverse of slope of the d.c load line\n",
-      "slope=(IC_peak-IC_min)/(VCE_min-VCE_peak);                                  #Slope of he d.c load line, kilo mho\n",
-      "RL_ac=-(1/slope)*1000;                                                     #a.c resistance, ohm\n",
-      "n=sqrt(RL_ac/RL);                                                           #Transformer turn ratio\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The approximate value of zero signal collector current=%dmA.\"%IC_zero_signal);\n",
-      "print(\"(ii)  The zero signal base current=%.2fmA.\"%IB_zero_signal);\n",
-      "print(\"(iii) The d.c input power= %dmW and a.c output power =%dmW.\"%(P_dc,P_ac));\n",
-      "print(\"(iv)  The collector efficiency=%.1f%%.\"%collector_eff);\n",
-      "print(\"(v)   The turn ratio of the transformer=%d.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The approximate value of zero signal collector current=18mA.\n",
-        "(ii)  The zero signal base current=0.18mA.\n",
-        "(iii) The d.c input power= 315mW and a.c output power =106mW.\n",
-        "(iv)  The collector efficiency=33.7%.\n",
-        "(v)   The turn ratio of the transformer=3.\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.14 : Page number 321-322\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=13.0;                        #Load resistance, ohm\n",
-      "RL_reflected=325.0;             #Load resistance, when referred to primary, ohm\n",
-      "VCC=20.0;                       #Supply voltage, V\n",
-      "IC=58.0;                      #Quiscent value of collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "n=sqrt(RL_reflected/RL);                    #Transformer turn ratio\n",
-      "\n",
-      "#(ii)\n",
-      "P_ac=(((IC/1000)**2)*RL_reflected/2)*1000;                   #A.C output power, mW\n",
-      "\n",
-      "#(iii)\n",
-      "P_dc=VCC*IC;                                    #d.c input power, mW\n",
-      "collector_eff=(P_ac/P_dc)*100;                        #Collector efficiency\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Transformer turn ratio=%d.\"%n);\n",
-      "print(\"(ii)  The a.c output power=%dmW.\"%P_ac);\n",
-      "print(\"(iii) The collector efficiency=%d%%.\"%collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Transformer turn ratio=5.\n",
-        "(ii)  The a.c output power=546mW.\n",
-        "(iii) The collector efficiency=47%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.15 : Page number 323\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_total=4.0;                #Total power dissipated by the power transistor, W\n",
-      "T_j_max=90.0;               #Maximum junction temperature, degree celsius\n",
-      "theta=10.0;                 #Thermal resistance, degree celsius per watt\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Total power dissipation=half of(max. junc. temp. - ambient temp.)\n",
-      "T_amb=T_j_max-(P_total*theta);                      #Ambient temperature, degree celsius\n",
-      "\n",
-      "#Result\n",
-      "print(\"The ambient temperature=%d degree celsius.\"%T_amb);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ambient temperature=50 degree celsius.\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.16 : Page number 323-324\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "theta=300.0;                #Thermal resistance, degree celsius per watt\n",
-      "T_j_max=90.0;               #Maximum junction temperature, degree celsius\n",
-      "T_amb=30.0;                 #Ambient temperature, degree celsius\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Without heat sink\n",
-      "P_total=((T_j_max-T_amb)/theta)*1000;              #Maximum permissible power dissipation without sink, mW\n",
-      "\n",
-      "print(\"(i)The maximum permissible power dissipation without heat sink=%dmW.\"%P_total);\n",
-      "\n",
-      "#(ii) With heat sink\n",
-      "theta=60.0;                                    #reduced thermal resistance, degree celsius per watt\n",
-      "P_total=((T_j_max-T_amb)/theta)*1000;          #Maximum permissible power dissipation with heat sink, mW\n",
-      "\n",
-      "print(\"(ii)The maximum permissible power dissipation with heat sink=%dmW.\"%P_total);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The maximum permissible power dissipation without heat sink=200mW.\n",
-        "(ii)The maximum permissible power dissipation with heat sink=1000mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.17 : Page number 324\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "theta=20.0;                         #Thermal resistance, degree celsius per watt\n",
-      "T_j_max=200.0;                      #Maximum junction temperature, degree celsius\n",
-      "T_amb=25.0;                         #Ambient temperature, degree celsius\n",
-      "VCE=4.0;                            #Collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "P_total=(T_j_max-T_amb)/theta;          #Maximum permissible power dissipation, W\n",
-      "\n",
-      "#since, the max. power dissipation=VCE_max*IC_max,therefore\n",
-      "IC_max=P_total/VCE;                                             #Maximum collector current, A\n",
-      "\n",
-      "print(\"The maximum collector current that the transistor can carry without destruction=%.2fA.\"%IC_max);\n",
-      "\n",
-      "#The ambient temperature rises\n",
-      "T_amb=75.0;                     #The risen ambibent temperature, degree celsius\n",
-      "P_total=(T_j_max-T_amb)/theta;          #Maximum permissible power dissipation, W\n",
-      "IC_max=P_total/VCE;                                             #Maximum collector current, A\n",
-      "\n",
-      "print(\"The maximum collector current for the risen ambient temperature=%.2fA.\"%IC_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current that the transistor can carry without destruction=2.19A.\n",
-        "The maximum collector current for the risen ambient temperature=1.56A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.18 : Page  number 328-329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                   #Supply voltage, V\n",
-      "RL=8.0;                     #Driving load, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC_sat=VCC/(2*RL);                  #Collector saturation current, A\n",
-      "P_o_max=round(VCC*IC_sat*0.25,2);            #Maximum load power, W\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=round(VCC*IC_sat/round(pi,2),2);                 #d.c input power, W\n",
-      "\n",
-      "#(iii)\n",
-      "Collector_eff=(P_o_max/P_dc)*100;       #Collector efficiency\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The maximum load power =%.2fW.\"%P_o_max);\n",
-      "print(\"(ii)  The d.c input power=%.2fW.\"%P_dc);\n",
-      "print(\"(iii) The collector efficiency=%.1f%%.\"%Collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The maximum load power =2.25W.\n",
-        "(ii)  The d.c input power=2.87W.\n",
-        "(iii) The collector efficiency=78.4%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.19 : Page number 329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_T=10.0;                   #Power rating of each transistor, W\n",
-      "max_eff=0.785;              #Maximum collector effciency\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, input power=max. a.c power + Power rating of transistor\n",
-      "#And, max. efficiency=max. a.c power/input d.c power\n",
-      "P_2T=2*P_T;                                   #Total power dissipation by two transistors\n",
-      "P_o_max=(max_eff*P_2T)/(1-max_eff);           #Maximum output a.c power, W\n",
-      "\n",
-      "#result\n",
-      "print(\"The maximum output power that can be obtained=%.2fW.\"%P_o_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum output power that can be obtained=73.02W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.20 : Page number 329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "eff=60.0/100;                   #Efficiency of the amplifier\n",
-      "P_T=2.5;                        #Power dissipated by each transistor, W\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, input power=max. a.c power + Power rating of transistor\n",
-      "#And, max. efficiency=max. a.c power/input d.c power\n",
-      "P_2T=2*P_T;                        #Total power dissipated by both transistors, W\n",
-      "P_ac=(eff*P_2T)/(1-eff);           #Output a.c power, W\n",
-      "P_dc=P_ac+P_2T;                    #Input d.c power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power= %.1fW.\"%P_ac);\n",
-      "print(\"The d.c input power= %.1fW.\"%P_dc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power= 7.5W.\n",
-        "The d.c input power= 12.5W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.21 : Page number 329-330\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;               #Supply voltage, V\n",
-      "RL=10.0;                #Load resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "IC_sat=(VCC/(2*RL))*1000;          #Saturated collector current, mA\n",
-      "VCE_off=VCC/2;                     #Collector-emitter voltage in off state, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"1st end point of a.c load line, IC(sat)=%dmA.\"%IC_sat);\n",
-      "print(\"2nd end point of a.c load line, VCE(off)=%dV.\"%VCE_off);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1st end point of a.c load line, IC(sat)=500mA.\n",
-        "2nd end point of a.c load line, VCE(off)=5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_1.ipynb
deleted file mode 100755
index 05e3d9d8..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_1.ipynb
+++ /dev/null
@@ -1,968 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1e9209171152811793fc18d1ee8c80ddcef574d69421ec87eeaa8fb87a304f6d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 12: TRANSISTOR AUDIO POWER AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1 : Page number 308\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                               #Collector supply voltage, V\n",
-      "R1=10.0;                                #Resistor R1, kilo ohm\n",
-      "R2=2.2;                                 #Resistor R2, kilo ohm\n",
-      "RC=3.6;                                 #Collector resistor, kilo ohm\n",
-      "RE=1.1;                                 #Emitter resistor, kilo ohm\n",
-      "VBE=0.7;                                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "I1=VCC/(R1+R2);                     #Current through R1 and R2, mA (OHM's LAW)\n",
-      "V2=I1*R2;                           #Voltage across R2 resistor, V (OHM's LAW)\n",
-      "VE=V2-VBE;                          #Emitter voltage, V\n",
-      "IE=VE/RE;                           #Emitter current, mA (OHM's LAW)\n",
-      "IC=IE;                              #Collector current, mA (approximately equal to emitter current)\n",
-      "I_T=I1+IC;                          #Total current drawn from the supply, mA\n",
-      "P_dc=VCC*I_T;                       #Total power drawn from the supply, mW\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total power drawn from the supply=%.1fmW.\"%P_dc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power drawn from the supply=18.2mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2 : Page number 309\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_L=10.6;                       #Voltage across load, V.(from a.c voltmeter, therfore r.m.s value)\n",
-      "R_L=200.0;                      #Load resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, power =V**2/R,\n",
-      "P_O=(V_L**2/R_L)*1000;                     #A.C output power, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power = %.1fmW.\"%P_O);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power = 561.8mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.3 : Page number 309\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=100.0;                   #Load resistance, ohm\n",
-      "V_PP=18.0;                  #Peak-to-peak a.c voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, V(r.m.s)=(V(peak-to-peak)/2)/sqrt(2)\n",
-      "VL=V_PP/(2*(2**0.5));                      #r.m.s value, V\n",
-      "\n",
-      "#Since, power=(square of voltage)/resistance\n",
-      "P_O_max=(VL**2/RL)*1000;                   #Maximum possible a.c load power, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum possible a.c load power=%dmW.\"%P_O_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum possible a.c load power=405mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.4 : Page number 310\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_battery=12.0;                 #Battery voltage, V\n",
-      "P_out=2.0;                      #Output power, W\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Power=Current*Voltage\n",
-      "IC=(P_out/V_battery)*1000;                 #Maximum collector current , mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector current=%.1fmA.\"%IC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current=166.7mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.5 : Page number 310\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_battery=12.0;                     #Battery voltage, V\n",
-      "RL=4.0;                             #Collector load, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "IC_max=V_battery/RL;                #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector current=%dmA.\"%IC_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current=3mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.6 : Page number 310-311\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P=50.0;                     #Power supplied by power amplifier, W\n",
-      "R=8.0;                      #Resistance of speaker, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, Power=Voltage _square/Resistance,\n",
-      "V=(P*R)**0.5;               #a.c output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "I=V/R;                          #a.c output current, A (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The a.c output voltage=%dV.\"%V);\n",
-      "print(\"(ii) The a.c output current=%.1fA.\"%I);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The a.c output voltage=20V.\n",
-        "(ii) The a.c output current=2.5A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.7 : Page number 315\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                   #Collector supply voltage, V\n",
-      "ib_peak=10.0;               #Base current(peak), mA\n",
-      "RB=1.0;                     #Base resistance, kilo ohm\n",
-      "RC=20.0;                    #Collector resistance, ohm\n",
-      "beta=25.0;                  #Base current amplification factor\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "IB=round(VCC-VBE/RB,1);                         #Base current, mA (OHM's LAW)\n",
-      "IC=int(beta*IB);                            #Collector current, mA\n",
-      "VCE=VCC-(IC/1000)*RC;                  #Collector emitter voltage, V (KVL)\n",
-      "\n",
-      "#(i)\n",
-      "ic_peak=beta*ib_peak;                          #Collector current(peak), mA\n",
-      "P_o_ac=(ic_peak/1000)**2*RC/2;                 #Output power, W\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=VCC*IC/1000;                                    #Input power, W\n",
-      "\n",
-      "#(iii)\n",
-      "collector_efficiency=(P_o_ac/P_dc)*100;                   #Collector efficiency of the amplifier circuit,\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output power=%.3fW.\"%P_o_ac);\n",
-      "print(\"(ii) The input power=%.1fW.\"%P_dc);\n",
-      "print(\"(iii) The collector efficiency=%.1f%%.\"%collector_efficiency);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output power=0.625W.\n",
-        "(ii) The input power=9.6W.\n",
-        "(iii) The collector efficiency=6.5%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.8 : Page number 317\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_dc=10.0;                  #zero signal power dissipation, W\n",
-      "P_o=4.0;                    #a.c output power, W\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Collector_eff=(P_o/P_dc)*100;                 #collector efficiency\n",
-      "\n",
-      "#(ii)\n",
-      "#Zero signal power is the maximum power dissipation in a transistor, therefore,\n",
-      "Power_rating=P_dc;                  #Power rating of the transistor, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The collector efficiency=%d%%.\"%Collector_eff);\n",
-      "print(\"(i) The power rating of the transistor=%dW.\"%Power_rating);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The collector efficiency=40%.\n",
-        "(i) The power rating of the transistor=10W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.9 : Page number 317-318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=100.0;                   #Secondary load, ohm\n",
-      "n=10.0;                     #Transformer turn ratio\n",
-      "IC=100.0;                   #Zero signal collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "RL_reflected=n**2*RL;                        #Reflected load as seen by the primary of the transformer, ohm\n",
-      "P_o_ac_max=(IC/1000)**2*RL_reflected/2;             #Maximum a.c power output, W      \n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum a.c power output=%dW.\"%P_o_ac_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum a.c power output=50W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.10 : Page number 318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=5.0;                    #Collector supply voltage, V\n",
-      "IC=50.0;                    #Zero signal collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "P_o_max=VCC*IC/2;           #Maximum a.c output power, mW\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=VCC*IC;                #D.C input power, mW\n",
-      "#Since, maximum power is dissipated in the zero signal conditions\n",
-      "Power_rating=P_dc;              #Power rating of transistor, mW\n",
-      "\n",
-      "#(iii)\n",
-      "Max_collector_eff=(P_o_max/P_dc)*100;             #Maximum collector efficiency\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The maximum a.c output power=%dmW\"%P_o_max);\n",
-      "print(\"(ii) The power rating of the transistor=%dmW.\"%Power_rating);\n",
-      "print(\"(iii) The maximum collector efficiency =%d%%.\"%Max_collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The maximum a.c output power=125mW\n",
-        "(ii) The power rating of the transistor=250mW.\n",
-        "(iii) The maximum collector efficiency =50%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.11 : Page number 318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "ic_max=160.0;               #Maximum a.c collector current, mA\n",
-      "ic_min=10.0;                #Minimum a.c collector current, mA\n",
-      "vce_max=12.0;               #Maximum collector-emitter voltage, V\n",
-      "vce_min=2.0;                #Minimum collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "vce_pp=vce_max-vce_min;                                #peak to peak collector emitter voltage, V\n",
-      "ic_pp=ic_max-ic_min;                                   #peak to peak collector current, V\n",
-      "P_o=(vce_pp/(2*sqrt(2)))*(ic_pp/(2*sqrt(2)));          #a.c output power, mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power=%.1fmW.\"%P_o);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power=187.5mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12 : Page number 319-320\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                           #Battey voltage, V\n",
-      "IC_max_change=100.0;                #maximum collector current change, mA\n",
-      "RL=5.0;                             #Loudspeaker resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "VCE_max_change=VCC;                                         #Maximum collector-emitter voltage change\n",
-      "#(i) Loud speaker directly connected in the collector\n",
-      "Vmax_speaker=(IC_max_change/1000)*RL;                              #Maximum voltage across the loudspeaker, V\n",
-      "P_speaker_directly_coupled=Vmax_speaker*IC_max_change;                       #Power developed in the loudspeaker,mW\n",
-      "\n",
-      "#(ii) Loudspeaker transformer coupled\n",
-      "Z_out=(VCE_max_change/IC_max_change)*1000;                     #Output impedance of transistor, ohm\n",
-      "\n",
-      "#For max power transfer, primary impedance should be Z_out\n",
-      "RL_reflected=Z_out;                                         #Load resistance as seen by primary, ohm\n",
-      "n=sqrt(RL_reflected/RL);                                    #Turns ratio of transformer\n",
-      "Vp=VCC;                                                     #Transformer primary voltage, V\n",
-      "Vs=Vp/n;                                                    #Transformer secondary voltage, V\n",
-      "IL=Vs/RL;                                                   #Load current, A\n",
-      "P_speaker_transformer_coupled=IL**2*RL*1000;                #Power delivered to the speaker, mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The power transferred to the speaker when directly coupled=%dmW.\"%P_speaker_directly_coupled);\n",
-      "print(\"(ii) The power trasnferred to the speaker when transformer-coupled=%dmW.\"%P_speaker_transformer_coupled);\n",
-      "print(\"     The turns ratio=%.1f.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The power transferred to the speaker when directly coupled=50mW.\n",
-        "(ii) The power trasnferred to the speaker when transformer-coupled=1200mW.\n",
-        "     The turns ratio=4.9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.13 : Page number 320-321\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "beta=100.0;                                 #Base current amplification factor\n",
-      "RL=81.6;                                    #Load resistance, ohm\n",
-      "VCE_peak=30.0;                               #Peak value of collector voltage, V\n",
-      "IC_peak=35.0;                               #Peak value of collector current, mA\n",
-      "VCE_min=5.0;                                 #Minimum value of collector voltage, V\n",
-      "IC_min=1.0;                                 #Minimum value of collector current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC_zero_signal=(IC_peak-IC_min)/2 +1;                  #Zero signal collector current, mA\n",
-      "\n",
-      "#(ii)\n",
-      "IB_zero_signal=IC_zero_signal/beta;                 #Zero signal base current, mA\n",
-      "\n",
-      "#(iii)\n",
-      "VCE_zero_signal=(VCE_peak-VCE_min)/2 +5;                    #Zero signal collector-emitter voltage, V\n",
-      "VCC=VCE_zero_signal;                                    #Collector supply voltage,V (due to transformer coupling, aproximately equal to zero signal VCE)\n",
-      "P_dc=VCC*IC_zero_signal;                                #d.c input power, mW\n",
-      "VCE_ac=(VCE_peak-VCE_min)/(2*sqrt(2));                         #a.c output voltage, V\n",
-      "IC_ac=(IC_peak-IC_min)/(2*sqrt(2));                           #a.c output  current, mA\n",
-      "P_ac=VCE_ac*IC_ac;                                      #a.c output power, mW\n",
-      "\n",
-      "#(iv)\n",
-      "collector_eff=(P_ac/P_dc)*100;                          #Collector efficiency\n",
-      "\n",
-      "#(v)\n",
-      "#a.c resistance RL'=negative inverse of slope of the d.c load line\n",
-      "slope=(IC_peak-IC_min)/(VCE_min-VCE_peak);                                  #Slope of he d.c load line, kilo mho\n",
-      "RL_ac=-(1/slope)*1000;                                                     #a.c resistance, ohm\n",
-      "n=sqrt(RL_ac/RL);                                                           #Transformer turn ratio\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The approximate value of zero signal collector current=%dmA.\"%IC_zero_signal);\n",
-      "print(\"(ii)  The zero signal base current=%.2fmA.\"%IB_zero_signal);\n",
-      "print(\"(iii) The d.c input power= %dmW and a.c output power =%dmW.\"%(P_dc,P_ac));\n",
-      "print(\"(iv)  The collector efficiency=%.1f%%.\"%collector_eff);\n",
-      "print(\"(v)   The turn ratio of the transformer=%d.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The approximate value of zero signal collector current=18mA.\n",
-        "(ii)  The zero signal base current=0.18mA.\n",
-        "(iii) The d.c input power= 315mW and a.c output power =106mW.\n",
-        "(iv)  The collector efficiency=33.7%.\n",
-        "(v)   The turn ratio of the transformer=3.\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.14 : Page number 321-322\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=13.0;                        #Load resistance, ohm\n",
-      "RL_reflected=325.0;             #Load resistance, when referred to primary, ohm\n",
-      "VCC=20.0;                       #Supply voltage, V\n",
-      "IC=58.0;                      #Quiscent value of collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "n=sqrt(RL_reflected/RL);                    #Transformer turn ratio\n",
-      "\n",
-      "#(ii)\n",
-      "P_ac=(((IC/1000)**2)*RL_reflected/2)*1000;                   #A.C output power, mW\n",
-      "\n",
-      "#(iii)\n",
-      "P_dc=VCC*IC;                                    #d.c input power, mW\n",
-      "collector_eff=(P_ac/P_dc)*100;                        #Collector efficiency\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Transformer turn ratio=%d.\"%n);\n",
-      "print(\"(ii)  The a.c output power=%dmW.\"%P_ac);\n",
-      "print(\"(iii) The collector efficiency=%d%%.\"%collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Transformer turn ratio=5.\n",
-        "(ii)  The a.c output power=546mW.\n",
-        "(iii) The collector efficiency=47%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.15 : Page number 323\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_total=4.0;                #Total power dissipated by the power transistor, W\n",
-      "T_j_max=90.0;               #Maximum junction temperature, degree celsius\n",
-      "theta=10.0;                 #Thermal resistance, degree celsius per watt\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Total power dissipation=half of(max. junc. temp. - ambient temp.)\n",
-      "T_amb=T_j_max-(P_total*theta);                      #Ambient temperature, degree celsius\n",
-      "\n",
-      "#Result\n",
-      "print(\"The ambient temperature=%d degree celsius.\"%T_amb);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ambient temperature=50 degree celsius.\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.16 : Page number 323-324\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "theta=300.0;                #Thermal resistance, degree celsius per watt\n",
-      "T_j_max=90.0;               #Maximum junction temperature, degree celsius\n",
-      "T_amb=30.0;                 #Ambient temperature, degree celsius\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Without heat sink\n",
-      "P_total=((T_j_max-T_amb)/theta)*1000;              #Maximum permissible power dissipation without sink, mW\n",
-      "\n",
-      "print(\"(i)The maximum permissible power dissipation without heat sink=%dmW.\"%P_total);\n",
-      "\n",
-      "#(ii) With heat sink\n",
-      "theta=60.0;                                    #reduced thermal resistance, degree celsius per watt\n",
-      "P_total=((T_j_max-T_amb)/theta)*1000;          #Maximum permissible power dissipation with heat sink, mW\n",
-      "\n",
-      "print(\"(ii)The maximum permissible power dissipation with heat sink=%dmW.\"%P_total);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The maximum permissible power dissipation without heat sink=200mW.\n",
-        "(ii)The maximum permissible power dissipation with heat sink=1000mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.17 : Page number 324\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "theta=20.0;                         #Thermal resistance, degree celsius per watt\n",
-      "T_j_max=200.0;                      #Maximum junction temperature, degree celsius\n",
-      "T_amb=25.0;                         #Ambient temperature, degree celsius\n",
-      "VCE=4.0;                            #Collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "P_total=(T_j_max-T_amb)/theta;          #Maximum permissible power dissipation, W\n",
-      "\n",
-      "#since, the max. power dissipation=VCE_max*IC_max,therefore\n",
-      "IC_max=P_total/VCE;                                             #Maximum collector current, A\n",
-      "\n",
-      "print(\"The maximum collector current that the transistor can carry without destruction=%.2fA.\"%IC_max);\n",
-      "\n",
-      "#The ambient temperature rises\n",
-      "T_amb=75.0;                     #The risen ambibent temperature, degree celsius\n",
-      "P_total=(T_j_max-T_amb)/theta;          #Maximum permissible power dissipation, W\n",
-      "IC_max=P_total/VCE;                                             #Maximum collector current, A\n",
-      "\n",
-      "print(\"The maximum collector current for the risen ambient temperature=%.2fA.\"%IC_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current that the transistor can carry without destruction=2.19A.\n",
-        "The maximum collector current for the risen ambient temperature=1.56A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.18 : Page  number 328-329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                   #Supply voltage, V\n",
-      "RL=8.0;                     #Driving load, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC_sat=VCC/(2*RL);                  #Collector saturation current, A\n",
-      "P_o_max=round(VCC*IC_sat*0.25,2);            #Maximum load power, W\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=round(VCC*IC_sat/round(pi,2),2);                 #d.c input power, W\n",
-      "\n",
-      "#(iii)\n",
-      "Collector_eff=(P_o_max/P_dc)*100;       #Collector efficiency\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The maximum load power =%.2fW.\"%P_o_max);\n",
-      "print(\"(ii)  The d.c input power=%.2fW.\"%P_dc);\n",
-      "print(\"(iii) The collector efficiency=%.1f%%.\"%Collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The maximum load power =2.25W.\n",
-        "(ii)  The d.c input power=2.87W.\n",
-        "(iii) The collector efficiency=78.4%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.19 : Page number 329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_T=10.0;                   #Power rating of each transistor, W\n",
-      "max_eff=0.785;              #Maximum collector effciency\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, input power=max. a.c power + Power rating of transistor\n",
-      "#And, max. efficiency=max. a.c power/input d.c power\n",
-      "P_2T=2*P_T;                                   #Total power dissipation by two transistors\n",
-      "P_o_max=(max_eff*P_2T)/(1-max_eff);           #Maximum output a.c power, W\n",
-      "\n",
-      "#result\n",
-      "print(\"The maximum output power that can be obtained=%.2fW.\"%P_o_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum output power that can be obtained=73.02W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.20 : Page number 329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "eff=60.0/100;                   #Efficiency of the amplifier\n",
-      "P_T=2.5;                        #Power dissipated by each transistor, W\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, input power=max. a.c power + Power rating of transistor\n",
-      "#And, max. efficiency=max. a.c power/input d.c power\n",
-      "P_2T=2*P_T;                        #Total power dissipated by both transistors, W\n",
-      "P_ac=(eff*P_2T)/(1-eff);           #Output a.c power, W\n",
-      "P_dc=P_ac+P_2T;                    #Input d.c power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power= %.1fW.\"%P_ac);\n",
-      "print(\"The d.c input power= %.1fW.\"%P_dc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power= 7.5W.\n",
-        "The d.c input power= 12.5W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.21 : Page number 329-330\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;               #Supply voltage, V\n",
-      "RL=10.0;                #Load resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "IC_sat=(VCC/(2*RL))*1000;          #Saturated collector current, mA\n",
-      "VCE_off=VCC/2;                     #Collector-emitter voltage in off state, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"1st end point of a.c load line, IC(sat)=%dmA.\"%IC_sat);\n",
-      "print(\"2nd end point of a.c load line, VCE(off)=%dV.\"%VCE_off);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1st end point of a.c load line, IC(sat)=500mA.\n",
-        "2nd end point of a.c load line, VCE(off)=5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_2.ipynb
deleted file mode 100755
index 05e3d9d8..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_2.ipynb
+++ /dev/null
@@ -1,968 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1e9209171152811793fc18d1ee8c80ddcef574d69421ec87eeaa8fb87a304f6d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 12: TRANSISTOR AUDIO POWER AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1 : Page number 308\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                               #Collector supply voltage, V\n",
-      "R1=10.0;                                #Resistor R1, kilo ohm\n",
-      "R2=2.2;                                 #Resistor R2, kilo ohm\n",
-      "RC=3.6;                                 #Collector resistor, kilo ohm\n",
-      "RE=1.1;                                 #Emitter resistor, kilo ohm\n",
-      "VBE=0.7;                                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "I1=VCC/(R1+R2);                     #Current through R1 and R2, mA (OHM's LAW)\n",
-      "V2=I1*R2;                           #Voltage across R2 resistor, V (OHM's LAW)\n",
-      "VE=V2-VBE;                          #Emitter voltage, V\n",
-      "IE=VE/RE;                           #Emitter current, mA (OHM's LAW)\n",
-      "IC=IE;                              #Collector current, mA (approximately equal to emitter current)\n",
-      "I_T=I1+IC;                          #Total current drawn from the supply, mA\n",
-      "P_dc=VCC*I_T;                       #Total power drawn from the supply, mW\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total power drawn from the supply=%.1fmW.\"%P_dc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power drawn from the supply=18.2mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2 : Page number 309\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_L=10.6;                       #Voltage across load, V.(from a.c voltmeter, therfore r.m.s value)\n",
-      "R_L=200.0;                      #Load resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, power =V**2/R,\n",
-      "P_O=(V_L**2/R_L)*1000;                     #A.C output power, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power = %.1fmW.\"%P_O);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power = 561.8mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.3 : Page number 309\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=100.0;                   #Load resistance, ohm\n",
-      "V_PP=18.0;                  #Peak-to-peak a.c voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, V(r.m.s)=(V(peak-to-peak)/2)/sqrt(2)\n",
-      "VL=V_PP/(2*(2**0.5));                      #r.m.s value, V\n",
-      "\n",
-      "#Since, power=(square of voltage)/resistance\n",
-      "P_O_max=(VL**2/RL)*1000;                   #Maximum possible a.c load power, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum possible a.c load power=%dmW.\"%P_O_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum possible a.c load power=405mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.4 : Page number 310\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_battery=12.0;                 #Battery voltage, V\n",
-      "P_out=2.0;                      #Output power, W\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Power=Current*Voltage\n",
-      "IC=(P_out/V_battery)*1000;                 #Maximum collector current , mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector current=%.1fmA.\"%IC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current=166.7mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.5 : Page number 310\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_battery=12.0;                     #Battery voltage, V\n",
-      "RL=4.0;                             #Collector load, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "IC_max=V_battery/RL;                #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector current=%dmA.\"%IC_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current=3mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.6 : Page number 310-311\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P=50.0;                     #Power supplied by power amplifier, W\n",
-      "R=8.0;                      #Resistance of speaker, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, Power=Voltage _square/Resistance,\n",
-      "V=(P*R)**0.5;               #a.c output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "I=V/R;                          #a.c output current, A (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The a.c output voltage=%dV.\"%V);\n",
-      "print(\"(ii) The a.c output current=%.1fA.\"%I);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The a.c output voltage=20V.\n",
-        "(ii) The a.c output current=2.5A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.7 : Page number 315\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                   #Collector supply voltage, V\n",
-      "ib_peak=10.0;               #Base current(peak), mA\n",
-      "RB=1.0;                     #Base resistance, kilo ohm\n",
-      "RC=20.0;                    #Collector resistance, ohm\n",
-      "beta=25.0;                  #Base current amplification factor\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "IB=round(VCC-VBE/RB,1);                         #Base current, mA (OHM's LAW)\n",
-      "IC=int(beta*IB);                            #Collector current, mA\n",
-      "VCE=VCC-(IC/1000)*RC;                  #Collector emitter voltage, V (KVL)\n",
-      "\n",
-      "#(i)\n",
-      "ic_peak=beta*ib_peak;                          #Collector current(peak), mA\n",
-      "P_o_ac=(ic_peak/1000)**2*RC/2;                 #Output power, W\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=VCC*IC/1000;                                    #Input power, W\n",
-      "\n",
-      "#(iii)\n",
-      "collector_efficiency=(P_o_ac/P_dc)*100;                   #Collector efficiency of the amplifier circuit,\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output power=%.3fW.\"%P_o_ac);\n",
-      "print(\"(ii) The input power=%.1fW.\"%P_dc);\n",
-      "print(\"(iii) The collector efficiency=%.1f%%.\"%collector_efficiency);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output power=0.625W.\n",
-        "(ii) The input power=9.6W.\n",
-        "(iii) The collector efficiency=6.5%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.8 : Page number 317\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_dc=10.0;                  #zero signal power dissipation, W\n",
-      "P_o=4.0;                    #a.c output power, W\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Collector_eff=(P_o/P_dc)*100;                 #collector efficiency\n",
-      "\n",
-      "#(ii)\n",
-      "#Zero signal power is the maximum power dissipation in a transistor, therefore,\n",
-      "Power_rating=P_dc;                  #Power rating of the transistor, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The collector efficiency=%d%%.\"%Collector_eff);\n",
-      "print(\"(i) The power rating of the transistor=%dW.\"%Power_rating);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The collector efficiency=40%.\n",
-        "(i) The power rating of the transistor=10W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.9 : Page number 317-318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=100.0;                   #Secondary load, ohm\n",
-      "n=10.0;                     #Transformer turn ratio\n",
-      "IC=100.0;                   #Zero signal collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "RL_reflected=n**2*RL;                        #Reflected load as seen by the primary of the transformer, ohm\n",
-      "P_o_ac_max=(IC/1000)**2*RL_reflected/2;             #Maximum a.c power output, W      \n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum a.c power output=%dW.\"%P_o_ac_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum a.c power output=50W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.10 : Page number 318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=5.0;                    #Collector supply voltage, V\n",
-      "IC=50.0;                    #Zero signal collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "P_o_max=VCC*IC/2;           #Maximum a.c output power, mW\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=VCC*IC;                #D.C input power, mW\n",
-      "#Since, maximum power is dissipated in the zero signal conditions\n",
-      "Power_rating=P_dc;              #Power rating of transistor, mW\n",
-      "\n",
-      "#(iii)\n",
-      "Max_collector_eff=(P_o_max/P_dc)*100;             #Maximum collector efficiency\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The maximum a.c output power=%dmW\"%P_o_max);\n",
-      "print(\"(ii) The power rating of the transistor=%dmW.\"%Power_rating);\n",
-      "print(\"(iii) The maximum collector efficiency =%d%%.\"%Max_collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The maximum a.c output power=125mW\n",
-        "(ii) The power rating of the transistor=250mW.\n",
-        "(iii) The maximum collector efficiency =50%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.11 : Page number 318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "ic_max=160.0;               #Maximum a.c collector current, mA\n",
-      "ic_min=10.0;                #Minimum a.c collector current, mA\n",
-      "vce_max=12.0;               #Maximum collector-emitter voltage, V\n",
-      "vce_min=2.0;                #Minimum collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "vce_pp=vce_max-vce_min;                                #peak to peak collector emitter voltage, V\n",
-      "ic_pp=ic_max-ic_min;                                   #peak to peak collector current, V\n",
-      "P_o=(vce_pp/(2*sqrt(2)))*(ic_pp/(2*sqrt(2)));          #a.c output power, mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power=%.1fmW.\"%P_o);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power=187.5mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12 : Page number 319-320\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                           #Battey voltage, V\n",
-      "IC_max_change=100.0;                #maximum collector current change, mA\n",
-      "RL=5.0;                             #Loudspeaker resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "VCE_max_change=VCC;                                         #Maximum collector-emitter voltage change\n",
-      "#(i) Loud speaker directly connected in the collector\n",
-      "Vmax_speaker=(IC_max_change/1000)*RL;                              #Maximum voltage across the loudspeaker, V\n",
-      "P_speaker_directly_coupled=Vmax_speaker*IC_max_change;                       #Power developed in the loudspeaker,mW\n",
-      "\n",
-      "#(ii) Loudspeaker transformer coupled\n",
-      "Z_out=(VCE_max_change/IC_max_change)*1000;                     #Output impedance of transistor, ohm\n",
-      "\n",
-      "#For max power transfer, primary impedance should be Z_out\n",
-      "RL_reflected=Z_out;                                         #Load resistance as seen by primary, ohm\n",
-      "n=sqrt(RL_reflected/RL);                                    #Turns ratio of transformer\n",
-      "Vp=VCC;                                                     #Transformer primary voltage, V\n",
-      "Vs=Vp/n;                                                    #Transformer secondary voltage, V\n",
-      "IL=Vs/RL;                                                   #Load current, A\n",
-      "P_speaker_transformer_coupled=IL**2*RL*1000;                #Power delivered to the speaker, mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The power transferred to the speaker when directly coupled=%dmW.\"%P_speaker_directly_coupled);\n",
-      "print(\"(ii) The power trasnferred to the speaker when transformer-coupled=%dmW.\"%P_speaker_transformer_coupled);\n",
-      "print(\"     The turns ratio=%.1f.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The power transferred to the speaker when directly coupled=50mW.\n",
-        "(ii) The power trasnferred to the speaker when transformer-coupled=1200mW.\n",
-        "     The turns ratio=4.9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.13 : Page number 320-321\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "beta=100.0;                                 #Base current amplification factor\n",
-      "RL=81.6;                                    #Load resistance, ohm\n",
-      "VCE_peak=30.0;                               #Peak value of collector voltage, V\n",
-      "IC_peak=35.0;                               #Peak value of collector current, mA\n",
-      "VCE_min=5.0;                                 #Minimum value of collector voltage, V\n",
-      "IC_min=1.0;                                 #Minimum value of collector current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC_zero_signal=(IC_peak-IC_min)/2 +1;                  #Zero signal collector current, mA\n",
-      "\n",
-      "#(ii)\n",
-      "IB_zero_signal=IC_zero_signal/beta;                 #Zero signal base current, mA\n",
-      "\n",
-      "#(iii)\n",
-      "VCE_zero_signal=(VCE_peak-VCE_min)/2 +5;                    #Zero signal collector-emitter voltage, V\n",
-      "VCC=VCE_zero_signal;                                    #Collector supply voltage,V (due to transformer coupling, aproximately equal to zero signal VCE)\n",
-      "P_dc=VCC*IC_zero_signal;                                #d.c input power, mW\n",
-      "VCE_ac=(VCE_peak-VCE_min)/(2*sqrt(2));                         #a.c output voltage, V\n",
-      "IC_ac=(IC_peak-IC_min)/(2*sqrt(2));                           #a.c output  current, mA\n",
-      "P_ac=VCE_ac*IC_ac;                                      #a.c output power, mW\n",
-      "\n",
-      "#(iv)\n",
-      "collector_eff=(P_ac/P_dc)*100;                          #Collector efficiency\n",
-      "\n",
-      "#(v)\n",
-      "#a.c resistance RL'=negative inverse of slope of the d.c load line\n",
-      "slope=(IC_peak-IC_min)/(VCE_min-VCE_peak);                                  #Slope of he d.c load line, kilo mho\n",
-      "RL_ac=-(1/slope)*1000;                                                     #a.c resistance, ohm\n",
-      "n=sqrt(RL_ac/RL);                                                           #Transformer turn ratio\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The approximate value of zero signal collector current=%dmA.\"%IC_zero_signal);\n",
-      "print(\"(ii)  The zero signal base current=%.2fmA.\"%IB_zero_signal);\n",
-      "print(\"(iii) The d.c input power= %dmW and a.c output power =%dmW.\"%(P_dc,P_ac));\n",
-      "print(\"(iv)  The collector efficiency=%.1f%%.\"%collector_eff);\n",
-      "print(\"(v)   The turn ratio of the transformer=%d.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The approximate value of zero signal collector current=18mA.\n",
-        "(ii)  The zero signal base current=0.18mA.\n",
-        "(iii) The d.c input power= 315mW and a.c output power =106mW.\n",
-        "(iv)  The collector efficiency=33.7%.\n",
-        "(v)   The turn ratio of the transformer=3.\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.14 : Page number 321-322\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=13.0;                        #Load resistance, ohm\n",
-      "RL_reflected=325.0;             #Load resistance, when referred to primary, ohm\n",
-      "VCC=20.0;                       #Supply voltage, V\n",
-      "IC=58.0;                      #Quiscent value of collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "n=sqrt(RL_reflected/RL);                    #Transformer turn ratio\n",
-      "\n",
-      "#(ii)\n",
-      "P_ac=(((IC/1000)**2)*RL_reflected/2)*1000;                   #A.C output power, mW\n",
-      "\n",
-      "#(iii)\n",
-      "P_dc=VCC*IC;                                    #d.c input power, mW\n",
-      "collector_eff=(P_ac/P_dc)*100;                        #Collector efficiency\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Transformer turn ratio=%d.\"%n);\n",
-      "print(\"(ii)  The a.c output power=%dmW.\"%P_ac);\n",
-      "print(\"(iii) The collector efficiency=%d%%.\"%collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Transformer turn ratio=5.\n",
-        "(ii)  The a.c output power=546mW.\n",
-        "(iii) The collector efficiency=47%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.15 : Page number 323\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_total=4.0;                #Total power dissipated by the power transistor, W\n",
-      "T_j_max=90.0;               #Maximum junction temperature, degree celsius\n",
-      "theta=10.0;                 #Thermal resistance, degree celsius per watt\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Total power dissipation=half of(max. junc. temp. - ambient temp.)\n",
-      "T_amb=T_j_max-(P_total*theta);                      #Ambient temperature, degree celsius\n",
-      "\n",
-      "#Result\n",
-      "print(\"The ambient temperature=%d degree celsius.\"%T_amb);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ambient temperature=50 degree celsius.\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.16 : Page number 323-324\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "theta=300.0;                #Thermal resistance, degree celsius per watt\n",
-      "T_j_max=90.0;               #Maximum junction temperature, degree celsius\n",
-      "T_amb=30.0;                 #Ambient temperature, degree celsius\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Without heat sink\n",
-      "P_total=((T_j_max-T_amb)/theta)*1000;              #Maximum permissible power dissipation without sink, mW\n",
-      "\n",
-      "print(\"(i)The maximum permissible power dissipation without heat sink=%dmW.\"%P_total);\n",
-      "\n",
-      "#(ii) With heat sink\n",
-      "theta=60.0;                                    #reduced thermal resistance, degree celsius per watt\n",
-      "P_total=((T_j_max-T_amb)/theta)*1000;          #Maximum permissible power dissipation with heat sink, mW\n",
-      "\n",
-      "print(\"(ii)The maximum permissible power dissipation with heat sink=%dmW.\"%P_total);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The maximum permissible power dissipation without heat sink=200mW.\n",
-        "(ii)The maximum permissible power dissipation with heat sink=1000mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.17 : Page number 324\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "theta=20.0;                         #Thermal resistance, degree celsius per watt\n",
-      "T_j_max=200.0;                      #Maximum junction temperature, degree celsius\n",
-      "T_amb=25.0;                         #Ambient temperature, degree celsius\n",
-      "VCE=4.0;                            #Collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "P_total=(T_j_max-T_amb)/theta;          #Maximum permissible power dissipation, W\n",
-      "\n",
-      "#since, the max. power dissipation=VCE_max*IC_max,therefore\n",
-      "IC_max=P_total/VCE;                                             #Maximum collector current, A\n",
-      "\n",
-      "print(\"The maximum collector current that the transistor can carry without destruction=%.2fA.\"%IC_max);\n",
-      "\n",
-      "#The ambient temperature rises\n",
-      "T_amb=75.0;                     #The risen ambibent temperature, degree celsius\n",
-      "P_total=(T_j_max-T_amb)/theta;          #Maximum permissible power dissipation, W\n",
-      "IC_max=P_total/VCE;                                             #Maximum collector current, A\n",
-      "\n",
-      "print(\"The maximum collector current for the risen ambient temperature=%.2fA.\"%IC_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current that the transistor can carry without destruction=2.19A.\n",
-        "The maximum collector current for the risen ambient temperature=1.56A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.18 : Page  number 328-329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                   #Supply voltage, V\n",
-      "RL=8.0;                     #Driving load, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC_sat=VCC/(2*RL);                  #Collector saturation current, A\n",
-      "P_o_max=round(VCC*IC_sat*0.25,2);            #Maximum load power, W\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=round(VCC*IC_sat/round(pi,2),2);                 #d.c input power, W\n",
-      "\n",
-      "#(iii)\n",
-      "Collector_eff=(P_o_max/P_dc)*100;       #Collector efficiency\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The maximum load power =%.2fW.\"%P_o_max);\n",
-      "print(\"(ii)  The d.c input power=%.2fW.\"%P_dc);\n",
-      "print(\"(iii) The collector efficiency=%.1f%%.\"%Collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The maximum load power =2.25W.\n",
-        "(ii)  The d.c input power=2.87W.\n",
-        "(iii) The collector efficiency=78.4%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.19 : Page number 329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_T=10.0;                   #Power rating of each transistor, W\n",
-      "max_eff=0.785;              #Maximum collector effciency\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, input power=max. a.c power + Power rating of transistor\n",
-      "#And, max. efficiency=max. a.c power/input d.c power\n",
-      "P_2T=2*P_T;                                   #Total power dissipation by two transistors\n",
-      "P_o_max=(max_eff*P_2T)/(1-max_eff);           #Maximum output a.c power, W\n",
-      "\n",
-      "#result\n",
-      "print(\"The maximum output power that can be obtained=%.2fW.\"%P_o_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum output power that can be obtained=73.02W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.20 : Page number 329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "eff=60.0/100;                   #Efficiency of the amplifier\n",
-      "P_T=2.5;                        #Power dissipated by each transistor, W\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, input power=max. a.c power + Power rating of transistor\n",
-      "#And, max. efficiency=max. a.c power/input d.c power\n",
-      "P_2T=2*P_T;                        #Total power dissipated by both transistors, W\n",
-      "P_ac=(eff*P_2T)/(1-eff);           #Output a.c power, W\n",
-      "P_dc=P_ac+P_2T;                    #Input d.c power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power= %.1fW.\"%P_ac);\n",
-      "print(\"The d.c input power= %.1fW.\"%P_dc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power= 7.5W.\n",
-        "The d.c input power= 12.5W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.21 : Page number 329-330\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;               #Supply voltage, V\n",
-      "RL=10.0;                #Load resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "IC_sat=(VCC/(2*RL))*1000;          #Saturated collector current, mA\n",
-      "VCE_off=VCC/2;                     #Collector-emitter voltage in off state, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"1st end point of a.c load line, IC(sat)=%dmA.\"%IC_sat);\n",
-      "print(\"2nd end point of a.c load line, VCE(off)=%dV.\"%VCE_off);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1st end point of a.c load line, IC(sat)=500mA.\n",
-        "2nd end point of a.c load line, VCE(off)=5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_3.ipynb
deleted file mode 100755
index 05e3d9d8..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_3.ipynb
+++ /dev/null
@@ -1,968 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1e9209171152811793fc18d1ee8c80ddcef574d69421ec87eeaa8fb87a304f6d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 12: TRANSISTOR AUDIO POWER AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1 : Page number 308\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                               #Collector supply voltage, V\n",
-      "R1=10.0;                                #Resistor R1, kilo ohm\n",
-      "R2=2.2;                                 #Resistor R2, kilo ohm\n",
-      "RC=3.6;                                 #Collector resistor, kilo ohm\n",
-      "RE=1.1;                                 #Emitter resistor, kilo ohm\n",
-      "VBE=0.7;                                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "I1=VCC/(R1+R2);                     #Current through R1 and R2, mA (OHM's LAW)\n",
-      "V2=I1*R2;                           #Voltage across R2 resistor, V (OHM's LAW)\n",
-      "VE=V2-VBE;                          #Emitter voltage, V\n",
-      "IE=VE/RE;                           #Emitter current, mA (OHM's LAW)\n",
-      "IC=IE;                              #Collector current, mA (approximately equal to emitter current)\n",
-      "I_T=I1+IC;                          #Total current drawn from the supply, mA\n",
-      "P_dc=VCC*I_T;                       #Total power drawn from the supply, mW\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total power drawn from the supply=%.1fmW.\"%P_dc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power drawn from the supply=18.2mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2 : Page number 309\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_L=10.6;                       #Voltage across load, V.(from a.c voltmeter, therfore r.m.s value)\n",
-      "R_L=200.0;                      #Load resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, power =V**2/R,\n",
-      "P_O=(V_L**2/R_L)*1000;                     #A.C output power, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power = %.1fmW.\"%P_O);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power = 561.8mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.3 : Page number 309\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=100.0;                   #Load resistance, ohm\n",
-      "V_PP=18.0;                  #Peak-to-peak a.c voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, V(r.m.s)=(V(peak-to-peak)/2)/sqrt(2)\n",
-      "VL=V_PP/(2*(2**0.5));                      #r.m.s value, V\n",
-      "\n",
-      "#Since, power=(square of voltage)/resistance\n",
-      "P_O_max=(VL**2/RL)*1000;                   #Maximum possible a.c load power, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum possible a.c load power=%dmW.\"%P_O_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum possible a.c load power=405mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.4 : Page number 310\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_battery=12.0;                 #Battery voltage, V\n",
-      "P_out=2.0;                      #Output power, W\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Power=Current*Voltage\n",
-      "IC=(P_out/V_battery)*1000;                 #Maximum collector current , mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector current=%.1fmA.\"%IC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current=166.7mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.5 : Page number 310\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_battery=12.0;                     #Battery voltage, V\n",
-      "RL=4.0;                             #Collector load, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "IC_max=V_battery/RL;                #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector current=%dmA.\"%IC_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current=3mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.6 : Page number 310-311\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P=50.0;                     #Power supplied by power amplifier, W\n",
-      "R=8.0;                      #Resistance of speaker, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, Power=Voltage _square/Resistance,\n",
-      "V=(P*R)**0.5;               #a.c output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "I=V/R;                          #a.c output current, A (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The a.c output voltage=%dV.\"%V);\n",
-      "print(\"(ii) The a.c output current=%.1fA.\"%I);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The a.c output voltage=20V.\n",
-        "(ii) The a.c output current=2.5A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.7 : Page number 315\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                   #Collector supply voltage, V\n",
-      "ib_peak=10.0;               #Base current(peak), mA\n",
-      "RB=1.0;                     #Base resistance, kilo ohm\n",
-      "RC=20.0;                    #Collector resistance, ohm\n",
-      "beta=25.0;                  #Base current amplification factor\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "IB=round(VCC-VBE/RB,1);                         #Base current, mA (OHM's LAW)\n",
-      "IC=int(beta*IB);                            #Collector current, mA\n",
-      "VCE=VCC-(IC/1000)*RC;                  #Collector emitter voltage, V (KVL)\n",
-      "\n",
-      "#(i)\n",
-      "ic_peak=beta*ib_peak;                          #Collector current(peak), mA\n",
-      "P_o_ac=(ic_peak/1000)**2*RC/2;                 #Output power, W\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=VCC*IC/1000;                                    #Input power, W\n",
-      "\n",
-      "#(iii)\n",
-      "collector_efficiency=(P_o_ac/P_dc)*100;                   #Collector efficiency of the amplifier circuit,\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output power=%.3fW.\"%P_o_ac);\n",
-      "print(\"(ii) The input power=%.1fW.\"%P_dc);\n",
-      "print(\"(iii) The collector efficiency=%.1f%%.\"%collector_efficiency);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output power=0.625W.\n",
-        "(ii) The input power=9.6W.\n",
-        "(iii) The collector efficiency=6.5%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.8 : Page number 317\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_dc=10.0;                  #zero signal power dissipation, W\n",
-      "P_o=4.0;                    #a.c output power, W\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Collector_eff=(P_o/P_dc)*100;                 #collector efficiency\n",
-      "\n",
-      "#(ii)\n",
-      "#Zero signal power is the maximum power dissipation in a transistor, therefore,\n",
-      "Power_rating=P_dc;                  #Power rating of the transistor, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The collector efficiency=%d%%.\"%Collector_eff);\n",
-      "print(\"(i) The power rating of the transistor=%dW.\"%Power_rating);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The collector efficiency=40%.\n",
-        "(i) The power rating of the transistor=10W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.9 : Page number 317-318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=100.0;                   #Secondary load, ohm\n",
-      "n=10.0;                     #Transformer turn ratio\n",
-      "IC=100.0;                   #Zero signal collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "RL_reflected=n**2*RL;                        #Reflected load as seen by the primary of the transformer, ohm\n",
-      "P_o_ac_max=(IC/1000)**2*RL_reflected/2;             #Maximum a.c power output, W      \n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum a.c power output=%dW.\"%P_o_ac_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum a.c power output=50W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.10 : Page number 318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=5.0;                    #Collector supply voltage, V\n",
-      "IC=50.0;                    #Zero signal collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "P_o_max=VCC*IC/2;           #Maximum a.c output power, mW\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=VCC*IC;                #D.C input power, mW\n",
-      "#Since, maximum power is dissipated in the zero signal conditions\n",
-      "Power_rating=P_dc;              #Power rating of transistor, mW\n",
-      "\n",
-      "#(iii)\n",
-      "Max_collector_eff=(P_o_max/P_dc)*100;             #Maximum collector efficiency\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The maximum a.c output power=%dmW\"%P_o_max);\n",
-      "print(\"(ii) The power rating of the transistor=%dmW.\"%Power_rating);\n",
-      "print(\"(iii) The maximum collector efficiency =%d%%.\"%Max_collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The maximum a.c output power=125mW\n",
-        "(ii) The power rating of the transistor=250mW.\n",
-        "(iii) The maximum collector efficiency =50%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.11 : Page number 318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "ic_max=160.0;               #Maximum a.c collector current, mA\n",
-      "ic_min=10.0;                #Minimum a.c collector current, mA\n",
-      "vce_max=12.0;               #Maximum collector-emitter voltage, V\n",
-      "vce_min=2.0;                #Minimum collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "vce_pp=vce_max-vce_min;                                #peak to peak collector emitter voltage, V\n",
-      "ic_pp=ic_max-ic_min;                                   #peak to peak collector current, V\n",
-      "P_o=(vce_pp/(2*sqrt(2)))*(ic_pp/(2*sqrt(2)));          #a.c output power, mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power=%.1fmW.\"%P_o);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power=187.5mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12 : Page number 319-320\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                           #Battey voltage, V\n",
-      "IC_max_change=100.0;                #maximum collector current change, mA\n",
-      "RL=5.0;                             #Loudspeaker resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "VCE_max_change=VCC;                                         #Maximum collector-emitter voltage change\n",
-      "#(i) Loud speaker directly connected in the collector\n",
-      "Vmax_speaker=(IC_max_change/1000)*RL;                              #Maximum voltage across the loudspeaker, V\n",
-      "P_speaker_directly_coupled=Vmax_speaker*IC_max_change;                       #Power developed in the loudspeaker,mW\n",
-      "\n",
-      "#(ii) Loudspeaker transformer coupled\n",
-      "Z_out=(VCE_max_change/IC_max_change)*1000;                     #Output impedance of transistor, ohm\n",
-      "\n",
-      "#For max power transfer, primary impedance should be Z_out\n",
-      "RL_reflected=Z_out;                                         #Load resistance as seen by primary, ohm\n",
-      "n=sqrt(RL_reflected/RL);                                    #Turns ratio of transformer\n",
-      "Vp=VCC;                                                     #Transformer primary voltage, V\n",
-      "Vs=Vp/n;                                                    #Transformer secondary voltage, V\n",
-      "IL=Vs/RL;                                                   #Load current, A\n",
-      "P_speaker_transformer_coupled=IL**2*RL*1000;                #Power delivered to the speaker, mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The power transferred to the speaker when directly coupled=%dmW.\"%P_speaker_directly_coupled);\n",
-      "print(\"(ii) The power trasnferred to the speaker when transformer-coupled=%dmW.\"%P_speaker_transformer_coupled);\n",
-      "print(\"     The turns ratio=%.1f.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The power transferred to the speaker when directly coupled=50mW.\n",
-        "(ii) The power trasnferred to the speaker when transformer-coupled=1200mW.\n",
-        "     The turns ratio=4.9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.13 : Page number 320-321\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "beta=100.0;                                 #Base current amplification factor\n",
-      "RL=81.6;                                    #Load resistance, ohm\n",
-      "VCE_peak=30.0;                               #Peak value of collector voltage, V\n",
-      "IC_peak=35.0;                               #Peak value of collector current, mA\n",
-      "VCE_min=5.0;                                 #Minimum value of collector voltage, V\n",
-      "IC_min=1.0;                                 #Minimum value of collector current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC_zero_signal=(IC_peak-IC_min)/2 +1;                  #Zero signal collector current, mA\n",
-      "\n",
-      "#(ii)\n",
-      "IB_zero_signal=IC_zero_signal/beta;                 #Zero signal base current, mA\n",
-      "\n",
-      "#(iii)\n",
-      "VCE_zero_signal=(VCE_peak-VCE_min)/2 +5;                    #Zero signal collector-emitter voltage, V\n",
-      "VCC=VCE_zero_signal;                                    #Collector supply voltage,V (due to transformer coupling, aproximately equal to zero signal VCE)\n",
-      "P_dc=VCC*IC_zero_signal;                                #d.c input power, mW\n",
-      "VCE_ac=(VCE_peak-VCE_min)/(2*sqrt(2));                         #a.c output voltage, V\n",
-      "IC_ac=(IC_peak-IC_min)/(2*sqrt(2));                           #a.c output  current, mA\n",
-      "P_ac=VCE_ac*IC_ac;                                      #a.c output power, mW\n",
-      "\n",
-      "#(iv)\n",
-      "collector_eff=(P_ac/P_dc)*100;                          #Collector efficiency\n",
-      "\n",
-      "#(v)\n",
-      "#a.c resistance RL'=negative inverse of slope of the d.c load line\n",
-      "slope=(IC_peak-IC_min)/(VCE_min-VCE_peak);                                  #Slope of he d.c load line, kilo mho\n",
-      "RL_ac=-(1/slope)*1000;                                                     #a.c resistance, ohm\n",
-      "n=sqrt(RL_ac/RL);                                                           #Transformer turn ratio\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The approximate value of zero signal collector current=%dmA.\"%IC_zero_signal);\n",
-      "print(\"(ii)  The zero signal base current=%.2fmA.\"%IB_zero_signal);\n",
-      "print(\"(iii) The d.c input power= %dmW and a.c output power =%dmW.\"%(P_dc,P_ac));\n",
-      "print(\"(iv)  The collector efficiency=%.1f%%.\"%collector_eff);\n",
-      "print(\"(v)   The turn ratio of the transformer=%d.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The approximate value of zero signal collector current=18mA.\n",
-        "(ii)  The zero signal base current=0.18mA.\n",
-        "(iii) The d.c input power= 315mW and a.c output power =106mW.\n",
-        "(iv)  The collector efficiency=33.7%.\n",
-        "(v)   The turn ratio of the transformer=3.\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.14 : Page number 321-322\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=13.0;                        #Load resistance, ohm\n",
-      "RL_reflected=325.0;             #Load resistance, when referred to primary, ohm\n",
-      "VCC=20.0;                       #Supply voltage, V\n",
-      "IC=58.0;                      #Quiscent value of collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "n=sqrt(RL_reflected/RL);                    #Transformer turn ratio\n",
-      "\n",
-      "#(ii)\n",
-      "P_ac=(((IC/1000)**2)*RL_reflected/2)*1000;                   #A.C output power, mW\n",
-      "\n",
-      "#(iii)\n",
-      "P_dc=VCC*IC;                                    #d.c input power, mW\n",
-      "collector_eff=(P_ac/P_dc)*100;                        #Collector efficiency\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Transformer turn ratio=%d.\"%n);\n",
-      "print(\"(ii)  The a.c output power=%dmW.\"%P_ac);\n",
-      "print(\"(iii) The collector efficiency=%d%%.\"%collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Transformer turn ratio=5.\n",
-        "(ii)  The a.c output power=546mW.\n",
-        "(iii) The collector efficiency=47%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.15 : Page number 323\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_total=4.0;                #Total power dissipated by the power transistor, W\n",
-      "T_j_max=90.0;               #Maximum junction temperature, degree celsius\n",
-      "theta=10.0;                 #Thermal resistance, degree celsius per watt\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Total power dissipation=half of(max. junc. temp. - ambient temp.)\n",
-      "T_amb=T_j_max-(P_total*theta);                      #Ambient temperature, degree celsius\n",
-      "\n",
-      "#Result\n",
-      "print(\"The ambient temperature=%d degree celsius.\"%T_amb);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ambient temperature=50 degree celsius.\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.16 : Page number 323-324\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "theta=300.0;                #Thermal resistance, degree celsius per watt\n",
-      "T_j_max=90.0;               #Maximum junction temperature, degree celsius\n",
-      "T_amb=30.0;                 #Ambient temperature, degree celsius\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Without heat sink\n",
-      "P_total=((T_j_max-T_amb)/theta)*1000;              #Maximum permissible power dissipation without sink, mW\n",
-      "\n",
-      "print(\"(i)The maximum permissible power dissipation without heat sink=%dmW.\"%P_total);\n",
-      "\n",
-      "#(ii) With heat sink\n",
-      "theta=60.0;                                    #reduced thermal resistance, degree celsius per watt\n",
-      "P_total=((T_j_max-T_amb)/theta)*1000;          #Maximum permissible power dissipation with heat sink, mW\n",
-      "\n",
-      "print(\"(ii)The maximum permissible power dissipation with heat sink=%dmW.\"%P_total);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The maximum permissible power dissipation without heat sink=200mW.\n",
-        "(ii)The maximum permissible power dissipation with heat sink=1000mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.17 : Page number 324\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "theta=20.0;                         #Thermal resistance, degree celsius per watt\n",
-      "T_j_max=200.0;                      #Maximum junction temperature, degree celsius\n",
-      "T_amb=25.0;                         #Ambient temperature, degree celsius\n",
-      "VCE=4.0;                            #Collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "P_total=(T_j_max-T_amb)/theta;          #Maximum permissible power dissipation, W\n",
-      "\n",
-      "#since, the max. power dissipation=VCE_max*IC_max,therefore\n",
-      "IC_max=P_total/VCE;                                             #Maximum collector current, A\n",
-      "\n",
-      "print(\"The maximum collector current that the transistor can carry without destruction=%.2fA.\"%IC_max);\n",
-      "\n",
-      "#The ambient temperature rises\n",
-      "T_amb=75.0;                     #The risen ambibent temperature, degree celsius\n",
-      "P_total=(T_j_max-T_amb)/theta;          #Maximum permissible power dissipation, W\n",
-      "IC_max=P_total/VCE;                                             #Maximum collector current, A\n",
-      "\n",
-      "print(\"The maximum collector current for the risen ambient temperature=%.2fA.\"%IC_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current that the transistor can carry without destruction=2.19A.\n",
-        "The maximum collector current for the risen ambient temperature=1.56A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.18 : Page  number 328-329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                   #Supply voltage, V\n",
-      "RL=8.0;                     #Driving load, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC_sat=VCC/(2*RL);                  #Collector saturation current, A\n",
-      "P_o_max=round(VCC*IC_sat*0.25,2);            #Maximum load power, W\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=round(VCC*IC_sat/round(pi,2),2);                 #d.c input power, W\n",
-      "\n",
-      "#(iii)\n",
-      "Collector_eff=(P_o_max/P_dc)*100;       #Collector efficiency\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The maximum load power =%.2fW.\"%P_o_max);\n",
-      "print(\"(ii)  The d.c input power=%.2fW.\"%P_dc);\n",
-      "print(\"(iii) The collector efficiency=%.1f%%.\"%Collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The maximum load power =2.25W.\n",
-        "(ii)  The d.c input power=2.87W.\n",
-        "(iii) The collector efficiency=78.4%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.19 : Page number 329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_T=10.0;                   #Power rating of each transistor, W\n",
-      "max_eff=0.785;              #Maximum collector effciency\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, input power=max. a.c power + Power rating of transistor\n",
-      "#And, max. efficiency=max. a.c power/input d.c power\n",
-      "P_2T=2*P_T;                                   #Total power dissipation by two transistors\n",
-      "P_o_max=(max_eff*P_2T)/(1-max_eff);           #Maximum output a.c power, W\n",
-      "\n",
-      "#result\n",
-      "print(\"The maximum output power that can be obtained=%.2fW.\"%P_o_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum output power that can be obtained=73.02W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.20 : Page number 329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "eff=60.0/100;                   #Efficiency of the amplifier\n",
-      "P_T=2.5;                        #Power dissipated by each transistor, W\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, input power=max. a.c power + Power rating of transistor\n",
-      "#And, max. efficiency=max. a.c power/input d.c power\n",
-      "P_2T=2*P_T;                        #Total power dissipated by both transistors, W\n",
-      "P_ac=(eff*P_2T)/(1-eff);           #Output a.c power, W\n",
-      "P_dc=P_ac+P_2T;                    #Input d.c power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power= %.1fW.\"%P_ac);\n",
-      "print(\"The d.c input power= %.1fW.\"%P_dc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power= 7.5W.\n",
-        "The d.c input power= 12.5W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.21 : Page number 329-330\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;               #Supply voltage, V\n",
-      "RL=10.0;                #Load resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "IC_sat=(VCC/(2*RL))*1000;          #Saturated collector current, mA\n",
-      "VCE_off=VCC/2;                     #Collector-emitter voltage in off state, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"1st end point of a.c load line, IC(sat)=%dmA.\"%IC_sat);\n",
-      "print(\"2nd end point of a.c load line, VCE(off)=%dV.\"%VCE_off);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1st end point of a.c load line, IC(sat)=500mA.\n",
-        "2nd end point of a.c load line, VCE(off)=5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_4.ipynb
deleted file mode 100755
index 05e3d9d8..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_4.ipynb
+++ /dev/null
@@ -1,968 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1e9209171152811793fc18d1ee8c80ddcef574d69421ec87eeaa8fb87a304f6d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 12: TRANSISTOR AUDIO POWER AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1 : Page number 308\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                               #Collector supply voltage, V\n",
-      "R1=10.0;                                #Resistor R1, kilo ohm\n",
-      "R2=2.2;                                 #Resistor R2, kilo ohm\n",
-      "RC=3.6;                                 #Collector resistor, kilo ohm\n",
-      "RE=1.1;                                 #Emitter resistor, kilo ohm\n",
-      "VBE=0.7;                                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "I1=VCC/(R1+R2);                     #Current through R1 and R2, mA (OHM's LAW)\n",
-      "V2=I1*R2;                           #Voltage across R2 resistor, V (OHM's LAW)\n",
-      "VE=V2-VBE;                          #Emitter voltage, V\n",
-      "IE=VE/RE;                           #Emitter current, mA (OHM's LAW)\n",
-      "IC=IE;                              #Collector current, mA (approximately equal to emitter current)\n",
-      "I_T=I1+IC;                          #Total current drawn from the supply, mA\n",
-      "P_dc=VCC*I_T;                       #Total power drawn from the supply, mW\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total power drawn from the supply=%.1fmW.\"%P_dc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power drawn from the supply=18.2mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2 : Page number 309\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_L=10.6;                       #Voltage across load, V.(from a.c voltmeter, therfore r.m.s value)\n",
-      "R_L=200.0;                      #Load resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, power =V**2/R,\n",
-      "P_O=(V_L**2/R_L)*1000;                     #A.C output power, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power = %.1fmW.\"%P_O);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power = 561.8mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.3 : Page number 309\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=100.0;                   #Load resistance, ohm\n",
-      "V_PP=18.0;                  #Peak-to-peak a.c voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, V(r.m.s)=(V(peak-to-peak)/2)/sqrt(2)\n",
-      "VL=V_PP/(2*(2**0.5));                      #r.m.s value, V\n",
-      "\n",
-      "#Since, power=(square of voltage)/resistance\n",
-      "P_O_max=(VL**2/RL)*1000;                   #Maximum possible a.c load power, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum possible a.c load power=%dmW.\"%P_O_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum possible a.c load power=405mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.4 : Page number 310\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_battery=12.0;                 #Battery voltage, V\n",
-      "P_out=2.0;                      #Output power, W\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Power=Current*Voltage\n",
-      "IC=(P_out/V_battery)*1000;                 #Maximum collector current , mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector current=%.1fmA.\"%IC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current=166.7mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.5 : Page number 310\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_battery=12.0;                     #Battery voltage, V\n",
-      "RL=4.0;                             #Collector load, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "IC_max=V_battery/RL;                #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector current=%dmA.\"%IC_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current=3mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.6 : Page number 310-311\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P=50.0;                     #Power supplied by power amplifier, W\n",
-      "R=8.0;                      #Resistance of speaker, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, Power=Voltage _square/Resistance,\n",
-      "V=(P*R)**0.5;               #a.c output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "I=V/R;                          #a.c output current, A (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The a.c output voltage=%dV.\"%V);\n",
-      "print(\"(ii) The a.c output current=%.1fA.\"%I);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The a.c output voltage=20V.\n",
-        "(ii) The a.c output current=2.5A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.7 : Page number 315\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                   #Collector supply voltage, V\n",
-      "ib_peak=10.0;               #Base current(peak), mA\n",
-      "RB=1.0;                     #Base resistance, kilo ohm\n",
-      "RC=20.0;                    #Collector resistance, ohm\n",
-      "beta=25.0;                  #Base current amplification factor\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "IB=round(VCC-VBE/RB,1);                         #Base current, mA (OHM's LAW)\n",
-      "IC=int(beta*IB);                            #Collector current, mA\n",
-      "VCE=VCC-(IC/1000)*RC;                  #Collector emitter voltage, V (KVL)\n",
-      "\n",
-      "#(i)\n",
-      "ic_peak=beta*ib_peak;                          #Collector current(peak), mA\n",
-      "P_o_ac=(ic_peak/1000)**2*RC/2;                 #Output power, W\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=VCC*IC/1000;                                    #Input power, W\n",
-      "\n",
-      "#(iii)\n",
-      "collector_efficiency=(P_o_ac/P_dc)*100;                   #Collector efficiency of the amplifier circuit,\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output power=%.3fW.\"%P_o_ac);\n",
-      "print(\"(ii) The input power=%.1fW.\"%P_dc);\n",
-      "print(\"(iii) The collector efficiency=%.1f%%.\"%collector_efficiency);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output power=0.625W.\n",
-        "(ii) The input power=9.6W.\n",
-        "(iii) The collector efficiency=6.5%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.8 : Page number 317\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_dc=10.0;                  #zero signal power dissipation, W\n",
-      "P_o=4.0;                    #a.c output power, W\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Collector_eff=(P_o/P_dc)*100;                 #collector efficiency\n",
-      "\n",
-      "#(ii)\n",
-      "#Zero signal power is the maximum power dissipation in a transistor, therefore,\n",
-      "Power_rating=P_dc;                  #Power rating of the transistor, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The collector efficiency=%d%%.\"%Collector_eff);\n",
-      "print(\"(i) The power rating of the transistor=%dW.\"%Power_rating);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The collector efficiency=40%.\n",
-        "(i) The power rating of the transistor=10W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.9 : Page number 317-318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=100.0;                   #Secondary load, ohm\n",
-      "n=10.0;                     #Transformer turn ratio\n",
-      "IC=100.0;                   #Zero signal collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "RL_reflected=n**2*RL;                        #Reflected load as seen by the primary of the transformer, ohm\n",
-      "P_o_ac_max=(IC/1000)**2*RL_reflected/2;             #Maximum a.c power output, W      \n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum a.c power output=%dW.\"%P_o_ac_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum a.c power output=50W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.10 : Page number 318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=5.0;                    #Collector supply voltage, V\n",
-      "IC=50.0;                    #Zero signal collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "P_o_max=VCC*IC/2;           #Maximum a.c output power, mW\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=VCC*IC;                #D.C input power, mW\n",
-      "#Since, maximum power is dissipated in the zero signal conditions\n",
-      "Power_rating=P_dc;              #Power rating of transistor, mW\n",
-      "\n",
-      "#(iii)\n",
-      "Max_collector_eff=(P_o_max/P_dc)*100;             #Maximum collector efficiency\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The maximum a.c output power=%dmW\"%P_o_max);\n",
-      "print(\"(ii) The power rating of the transistor=%dmW.\"%Power_rating);\n",
-      "print(\"(iii) The maximum collector efficiency =%d%%.\"%Max_collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The maximum a.c output power=125mW\n",
-        "(ii) The power rating of the transistor=250mW.\n",
-        "(iii) The maximum collector efficiency =50%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.11 : Page number 318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "ic_max=160.0;               #Maximum a.c collector current, mA\n",
-      "ic_min=10.0;                #Minimum a.c collector current, mA\n",
-      "vce_max=12.0;               #Maximum collector-emitter voltage, V\n",
-      "vce_min=2.0;                #Minimum collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "vce_pp=vce_max-vce_min;                                #peak to peak collector emitter voltage, V\n",
-      "ic_pp=ic_max-ic_min;                                   #peak to peak collector current, V\n",
-      "P_o=(vce_pp/(2*sqrt(2)))*(ic_pp/(2*sqrt(2)));          #a.c output power, mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power=%.1fmW.\"%P_o);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power=187.5mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12 : Page number 319-320\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                           #Battey voltage, V\n",
-      "IC_max_change=100.0;                #maximum collector current change, mA\n",
-      "RL=5.0;                             #Loudspeaker resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "VCE_max_change=VCC;                                         #Maximum collector-emitter voltage change\n",
-      "#(i) Loud speaker directly connected in the collector\n",
-      "Vmax_speaker=(IC_max_change/1000)*RL;                              #Maximum voltage across the loudspeaker, V\n",
-      "P_speaker_directly_coupled=Vmax_speaker*IC_max_change;                       #Power developed in the loudspeaker,mW\n",
-      "\n",
-      "#(ii) Loudspeaker transformer coupled\n",
-      "Z_out=(VCE_max_change/IC_max_change)*1000;                     #Output impedance of transistor, ohm\n",
-      "\n",
-      "#For max power transfer, primary impedance should be Z_out\n",
-      "RL_reflected=Z_out;                                         #Load resistance as seen by primary, ohm\n",
-      "n=sqrt(RL_reflected/RL);                                    #Turns ratio of transformer\n",
-      "Vp=VCC;                                                     #Transformer primary voltage, V\n",
-      "Vs=Vp/n;                                                    #Transformer secondary voltage, V\n",
-      "IL=Vs/RL;                                                   #Load current, A\n",
-      "P_speaker_transformer_coupled=IL**2*RL*1000;                #Power delivered to the speaker, mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The power transferred to the speaker when directly coupled=%dmW.\"%P_speaker_directly_coupled);\n",
-      "print(\"(ii) The power trasnferred to the speaker when transformer-coupled=%dmW.\"%P_speaker_transformer_coupled);\n",
-      "print(\"     The turns ratio=%.1f.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The power transferred to the speaker when directly coupled=50mW.\n",
-        "(ii) The power trasnferred to the speaker when transformer-coupled=1200mW.\n",
-        "     The turns ratio=4.9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.13 : Page number 320-321\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "beta=100.0;                                 #Base current amplification factor\n",
-      "RL=81.6;                                    #Load resistance, ohm\n",
-      "VCE_peak=30.0;                               #Peak value of collector voltage, V\n",
-      "IC_peak=35.0;                               #Peak value of collector current, mA\n",
-      "VCE_min=5.0;                                 #Minimum value of collector voltage, V\n",
-      "IC_min=1.0;                                 #Minimum value of collector current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC_zero_signal=(IC_peak-IC_min)/2 +1;                  #Zero signal collector current, mA\n",
-      "\n",
-      "#(ii)\n",
-      "IB_zero_signal=IC_zero_signal/beta;                 #Zero signal base current, mA\n",
-      "\n",
-      "#(iii)\n",
-      "VCE_zero_signal=(VCE_peak-VCE_min)/2 +5;                    #Zero signal collector-emitter voltage, V\n",
-      "VCC=VCE_zero_signal;                                    #Collector supply voltage,V (due to transformer coupling, aproximately equal to zero signal VCE)\n",
-      "P_dc=VCC*IC_zero_signal;                                #d.c input power, mW\n",
-      "VCE_ac=(VCE_peak-VCE_min)/(2*sqrt(2));                         #a.c output voltage, V\n",
-      "IC_ac=(IC_peak-IC_min)/(2*sqrt(2));                           #a.c output  current, mA\n",
-      "P_ac=VCE_ac*IC_ac;                                      #a.c output power, mW\n",
-      "\n",
-      "#(iv)\n",
-      "collector_eff=(P_ac/P_dc)*100;                          #Collector efficiency\n",
-      "\n",
-      "#(v)\n",
-      "#a.c resistance RL'=negative inverse of slope of the d.c load line\n",
-      "slope=(IC_peak-IC_min)/(VCE_min-VCE_peak);                                  #Slope of he d.c load line, kilo mho\n",
-      "RL_ac=-(1/slope)*1000;                                                     #a.c resistance, ohm\n",
-      "n=sqrt(RL_ac/RL);                                                           #Transformer turn ratio\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The approximate value of zero signal collector current=%dmA.\"%IC_zero_signal);\n",
-      "print(\"(ii)  The zero signal base current=%.2fmA.\"%IB_zero_signal);\n",
-      "print(\"(iii) The d.c input power= %dmW and a.c output power =%dmW.\"%(P_dc,P_ac));\n",
-      "print(\"(iv)  The collector efficiency=%.1f%%.\"%collector_eff);\n",
-      "print(\"(v)   The turn ratio of the transformer=%d.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The approximate value of zero signal collector current=18mA.\n",
-        "(ii)  The zero signal base current=0.18mA.\n",
-        "(iii) The d.c input power= 315mW and a.c output power =106mW.\n",
-        "(iv)  The collector efficiency=33.7%.\n",
-        "(v)   The turn ratio of the transformer=3.\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.14 : Page number 321-322\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=13.0;                        #Load resistance, ohm\n",
-      "RL_reflected=325.0;             #Load resistance, when referred to primary, ohm\n",
-      "VCC=20.0;                       #Supply voltage, V\n",
-      "IC=58.0;                      #Quiscent value of collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "n=sqrt(RL_reflected/RL);                    #Transformer turn ratio\n",
-      "\n",
-      "#(ii)\n",
-      "P_ac=(((IC/1000)**2)*RL_reflected/2)*1000;                   #A.C output power, mW\n",
-      "\n",
-      "#(iii)\n",
-      "P_dc=VCC*IC;                                    #d.c input power, mW\n",
-      "collector_eff=(P_ac/P_dc)*100;                        #Collector efficiency\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Transformer turn ratio=%d.\"%n);\n",
-      "print(\"(ii)  The a.c output power=%dmW.\"%P_ac);\n",
-      "print(\"(iii) The collector efficiency=%d%%.\"%collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Transformer turn ratio=5.\n",
-        "(ii)  The a.c output power=546mW.\n",
-        "(iii) The collector efficiency=47%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.15 : Page number 323\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_total=4.0;                #Total power dissipated by the power transistor, W\n",
-      "T_j_max=90.0;               #Maximum junction temperature, degree celsius\n",
-      "theta=10.0;                 #Thermal resistance, degree celsius per watt\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Total power dissipation=half of(max. junc. temp. - ambient temp.)\n",
-      "T_amb=T_j_max-(P_total*theta);                      #Ambient temperature, degree celsius\n",
-      "\n",
-      "#Result\n",
-      "print(\"The ambient temperature=%d degree celsius.\"%T_amb);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ambient temperature=50 degree celsius.\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.16 : Page number 323-324\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "theta=300.0;                #Thermal resistance, degree celsius per watt\n",
-      "T_j_max=90.0;               #Maximum junction temperature, degree celsius\n",
-      "T_amb=30.0;                 #Ambient temperature, degree celsius\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Without heat sink\n",
-      "P_total=((T_j_max-T_amb)/theta)*1000;              #Maximum permissible power dissipation without sink, mW\n",
-      "\n",
-      "print(\"(i)The maximum permissible power dissipation without heat sink=%dmW.\"%P_total);\n",
-      "\n",
-      "#(ii) With heat sink\n",
-      "theta=60.0;                                    #reduced thermal resistance, degree celsius per watt\n",
-      "P_total=((T_j_max-T_amb)/theta)*1000;          #Maximum permissible power dissipation with heat sink, mW\n",
-      "\n",
-      "print(\"(ii)The maximum permissible power dissipation with heat sink=%dmW.\"%P_total);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The maximum permissible power dissipation without heat sink=200mW.\n",
-        "(ii)The maximum permissible power dissipation with heat sink=1000mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.17 : Page number 324\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "theta=20.0;                         #Thermal resistance, degree celsius per watt\n",
-      "T_j_max=200.0;                      #Maximum junction temperature, degree celsius\n",
-      "T_amb=25.0;                         #Ambient temperature, degree celsius\n",
-      "VCE=4.0;                            #Collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "P_total=(T_j_max-T_amb)/theta;          #Maximum permissible power dissipation, W\n",
-      "\n",
-      "#since, the max. power dissipation=VCE_max*IC_max,therefore\n",
-      "IC_max=P_total/VCE;                                             #Maximum collector current, A\n",
-      "\n",
-      "print(\"The maximum collector current that the transistor can carry without destruction=%.2fA.\"%IC_max);\n",
-      "\n",
-      "#The ambient temperature rises\n",
-      "T_amb=75.0;                     #The risen ambibent temperature, degree celsius\n",
-      "P_total=(T_j_max-T_amb)/theta;          #Maximum permissible power dissipation, W\n",
-      "IC_max=P_total/VCE;                                             #Maximum collector current, A\n",
-      "\n",
-      "print(\"The maximum collector current for the risen ambient temperature=%.2fA.\"%IC_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current that the transistor can carry without destruction=2.19A.\n",
-        "The maximum collector current for the risen ambient temperature=1.56A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.18 : Page  number 328-329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                   #Supply voltage, V\n",
-      "RL=8.0;                     #Driving load, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC_sat=VCC/(2*RL);                  #Collector saturation current, A\n",
-      "P_o_max=round(VCC*IC_sat*0.25,2);            #Maximum load power, W\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=round(VCC*IC_sat/round(pi,2),2);                 #d.c input power, W\n",
-      "\n",
-      "#(iii)\n",
-      "Collector_eff=(P_o_max/P_dc)*100;       #Collector efficiency\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The maximum load power =%.2fW.\"%P_o_max);\n",
-      "print(\"(ii)  The d.c input power=%.2fW.\"%P_dc);\n",
-      "print(\"(iii) The collector efficiency=%.1f%%.\"%Collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The maximum load power =2.25W.\n",
-        "(ii)  The d.c input power=2.87W.\n",
-        "(iii) The collector efficiency=78.4%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.19 : Page number 329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_T=10.0;                   #Power rating of each transistor, W\n",
-      "max_eff=0.785;              #Maximum collector effciency\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, input power=max. a.c power + Power rating of transistor\n",
-      "#And, max. efficiency=max. a.c power/input d.c power\n",
-      "P_2T=2*P_T;                                   #Total power dissipation by two transistors\n",
-      "P_o_max=(max_eff*P_2T)/(1-max_eff);           #Maximum output a.c power, W\n",
-      "\n",
-      "#result\n",
-      "print(\"The maximum output power that can be obtained=%.2fW.\"%P_o_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum output power that can be obtained=73.02W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.20 : Page number 329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "eff=60.0/100;                   #Efficiency of the amplifier\n",
-      "P_T=2.5;                        #Power dissipated by each transistor, W\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, input power=max. a.c power + Power rating of transistor\n",
-      "#And, max. efficiency=max. a.c power/input d.c power\n",
-      "P_2T=2*P_T;                        #Total power dissipated by both transistors, W\n",
-      "P_ac=(eff*P_2T)/(1-eff);           #Output a.c power, W\n",
-      "P_dc=P_ac+P_2T;                    #Input d.c power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power= %.1fW.\"%P_ac);\n",
-      "print(\"The d.c input power= %.1fW.\"%P_dc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power= 7.5W.\n",
-        "The d.c input power= 12.5W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.21 : Page number 329-330\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;               #Supply voltage, V\n",
-      "RL=10.0;                #Load resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "IC_sat=(VCC/(2*RL))*1000;          #Saturated collector current, mA\n",
-      "VCE_off=VCC/2;                     #Collector-emitter voltage in off state, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"1st end point of a.c load line, IC(sat)=%dmA.\"%IC_sat);\n",
-      "print(\"2nd end point of a.c load line, VCE(off)=%dV.\"%VCE_off);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1st end point of a.c load line, IC(sat)=500mA.\n",
-        "2nd end point of a.c load line, VCE(off)=5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_5.ipynb
deleted file mode 100755
index 05e3d9d8..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_5.ipynb
+++ /dev/null
@@ -1,968 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1e9209171152811793fc18d1ee8c80ddcef574d69421ec87eeaa8fb87a304f6d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 12: TRANSISTOR AUDIO POWER AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.1 : Page number 308\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                               #Collector supply voltage, V\n",
-      "R1=10.0;                                #Resistor R1, kilo ohm\n",
-      "R2=2.2;                                 #Resistor R2, kilo ohm\n",
-      "RC=3.6;                                 #Collector resistor, kilo ohm\n",
-      "RE=1.1;                                 #Emitter resistor, kilo ohm\n",
-      "VBE=0.7;                                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "I1=VCC/(R1+R2);                     #Current through R1 and R2, mA (OHM's LAW)\n",
-      "V2=I1*R2;                           #Voltage across R2 resistor, V (OHM's LAW)\n",
-      "VE=V2-VBE;                          #Emitter voltage, V\n",
-      "IE=VE/RE;                           #Emitter current, mA (OHM's LAW)\n",
-      "IC=IE;                              #Collector current, mA (approximately equal to emitter current)\n",
-      "I_T=I1+IC;                          #Total current drawn from the supply, mA\n",
-      "P_dc=VCC*I_T;                       #Total power drawn from the supply, mW\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The total power drawn from the supply=%.1fmW.\"%P_dc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total power drawn from the supply=18.2mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.2 : Page number 309\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_L=10.6;                       #Voltage across load, V.(from a.c voltmeter, therfore r.m.s value)\n",
-      "R_L=200.0;                      #Load resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, power =V**2/R,\n",
-      "P_O=(V_L**2/R_L)*1000;                     #A.C output power, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power = %.1fmW.\"%P_O);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power = 561.8mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.3 : Page number 309\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=100.0;                   #Load resistance, ohm\n",
-      "V_PP=18.0;                  #Peak-to-peak a.c voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, V(r.m.s)=(V(peak-to-peak)/2)/sqrt(2)\n",
-      "VL=V_PP/(2*(2**0.5));                      #r.m.s value, V\n",
-      "\n",
-      "#Since, power=(square of voltage)/resistance\n",
-      "P_O_max=(VL**2/RL)*1000;                   #Maximum possible a.c load power, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum possible a.c load power=%dmW.\"%P_O_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum possible a.c load power=405mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.4 : Page number 310\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_battery=12.0;                 #Battery voltage, V\n",
-      "P_out=2.0;                      #Output power, W\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Power=Current*Voltage\n",
-      "IC=(P_out/V_battery)*1000;                 #Maximum collector current , mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector current=%.1fmA.\"%IC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current=166.7mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.5 : Page number 310\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_battery=12.0;                     #Battery voltage, V\n",
-      "RL=4.0;                             #Collector load, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "IC_max=V_battery/RL;                #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector current=%dmA.\"%IC_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current=3mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.6 : Page number 310-311\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P=50.0;                     #Power supplied by power amplifier, W\n",
-      "R=8.0;                      #Resistance of speaker, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, Power=Voltage _square/Resistance,\n",
-      "V=(P*R)**0.5;               #a.c output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "I=V/R;                          #a.c output current, A (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The a.c output voltage=%dV.\"%V);\n",
-      "print(\"(ii) The a.c output current=%.1fA.\"%I);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The a.c output voltage=20V.\n",
-        "(ii) The a.c output current=2.5A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.7 : Page number 315\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                   #Collector supply voltage, V\n",
-      "ib_peak=10.0;               #Base current(peak), mA\n",
-      "RB=1.0;                     #Base resistance, kilo ohm\n",
-      "RC=20.0;                    #Collector resistance, ohm\n",
-      "beta=25.0;                  #Base current amplification factor\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "IB=round(VCC-VBE/RB,1);                         #Base current, mA (OHM's LAW)\n",
-      "IC=int(beta*IB);                            #Collector current, mA\n",
-      "VCE=VCC-(IC/1000)*RC;                  #Collector emitter voltage, V (KVL)\n",
-      "\n",
-      "#(i)\n",
-      "ic_peak=beta*ib_peak;                          #Collector current(peak), mA\n",
-      "P_o_ac=(ic_peak/1000)**2*RC/2;                 #Output power, W\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=VCC*IC/1000;                                    #Input power, W\n",
-      "\n",
-      "#(iii)\n",
-      "collector_efficiency=(P_o_ac/P_dc)*100;                   #Collector efficiency of the amplifier circuit,\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output power=%.3fW.\"%P_o_ac);\n",
-      "print(\"(ii) The input power=%.1fW.\"%P_dc);\n",
-      "print(\"(iii) The collector efficiency=%.1f%%.\"%collector_efficiency);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output power=0.625W.\n",
-        "(ii) The input power=9.6W.\n",
-        "(iii) The collector efficiency=6.5%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.8 : Page number 317\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_dc=10.0;                  #zero signal power dissipation, W\n",
-      "P_o=4.0;                    #a.c output power, W\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Collector_eff=(P_o/P_dc)*100;                 #collector efficiency\n",
-      "\n",
-      "#(ii)\n",
-      "#Zero signal power is the maximum power dissipation in a transistor, therefore,\n",
-      "Power_rating=P_dc;                  #Power rating of the transistor, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The collector efficiency=%d%%.\"%Collector_eff);\n",
-      "print(\"(i) The power rating of the transistor=%dW.\"%Power_rating);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The collector efficiency=40%.\n",
-        "(i) The power rating of the transistor=10W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.9 : Page number 317-318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=100.0;                   #Secondary load, ohm\n",
-      "n=10.0;                     #Transformer turn ratio\n",
-      "IC=100.0;                   #Zero signal collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "RL_reflected=n**2*RL;                        #Reflected load as seen by the primary of the transformer, ohm\n",
-      "P_o_ac_max=(IC/1000)**2*RL_reflected/2;             #Maximum a.c power output, W      \n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum a.c power output=%dW.\"%P_o_ac_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum a.c power output=50W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.10 : Page number 318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=5.0;                    #Collector supply voltage, V\n",
-      "IC=50.0;                    #Zero signal collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "P_o_max=VCC*IC/2;           #Maximum a.c output power, mW\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=VCC*IC;                #D.C input power, mW\n",
-      "#Since, maximum power is dissipated in the zero signal conditions\n",
-      "Power_rating=P_dc;              #Power rating of transistor, mW\n",
-      "\n",
-      "#(iii)\n",
-      "Max_collector_eff=(P_o_max/P_dc)*100;             #Maximum collector efficiency\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The maximum a.c output power=%dmW\"%P_o_max);\n",
-      "print(\"(ii) The power rating of the transistor=%dmW.\"%Power_rating);\n",
-      "print(\"(iii) The maximum collector efficiency =%d%%.\"%Max_collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The maximum a.c output power=125mW\n",
-        "(ii) The power rating of the transistor=250mW.\n",
-        "(iii) The maximum collector efficiency =50%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.11 : Page number 318\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "ic_max=160.0;               #Maximum a.c collector current, mA\n",
-      "ic_min=10.0;                #Minimum a.c collector current, mA\n",
-      "vce_max=12.0;               #Maximum collector-emitter voltage, V\n",
-      "vce_min=2.0;                #Minimum collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "vce_pp=vce_max-vce_min;                                #peak to peak collector emitter voltage, V\n",
-      "ic_pp=ic_max-ic_min;                                   #peak to peak collector current, V\n",
-      "P_o=(vce_pp/(2*sqrt(2)))*(ic_pp/(2*sqrt(2)));          #a.c output power, mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power=%.1fmW.\"%P_o);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power=187.5mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.12 : Page number 319-320\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                           #Battey voltage, V\n",
-      "IC_max_change=100.0;                #maximum collector current change, mA\n",
-      "RL=5.0;                             #Loudspeaker resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "VCE_max_change=VCC;                                         #Maximum collector-emitter voltage change\n",
-      "#(i) Loud speaker directly connected in the collector\n",
-      "Vmax_speaker=(IC_max_change/1000)*RL;                              #Maximum voltage across the loudspeaker, V\n",
-      "P_speaker_directly_coupled=Vmax_speaker*IC_max_change;                       #Power developed in the loudspeaker,mW\n",
-      "\n",
-      "#(ii) Loudspeaker transformer coupled\n",
-      "Z_out=(VCE_max_change/IC_max_change)*1000;                     #Output impedance of transistor, ohm\n",
-      "\n",
-      "#For max power transfer, primary impedance should be Z_out\n",
-      "RL_reflected=Z_out;                                         #Load resistance as seen by primary, ohm\n",
-      "n=sqrt(RL_reflected/RL);                                    #Turns ratio of transformer\n",
-      "Vp=VCC;                                                     #Transformer primary voltage, V\n",
-      "Vs=Vp/n;                                                    #Transformer secondary voltage, V\n",
-      "IL=Vs/RL;                                                   #Load current, A\n",
-      "P_speaker_transformer_coupled=IL**2*RL*1000;                #Power delivered to the speaker, mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The power transferred to the speaker when directly coupled=%dmW.\"%P_speaker_directly_coupled);\n",
-      "print(\"(ii) The power trasnferred to the speaker when transformer-coupled=%dmW.\"%P_speaker_transformer_coupled);\n",
-      "print(\"     The turns ratio=%.1f.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The power transferred to the speaker when directly coupled=50mW.\n",
-        "(ii) The power trasnferred to the speaker when transformer-coupled=1200mW.\n",
-        "     The turns ratio=4.9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.13 : Page number 320-321\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "beta=100.0;                                 #Base current amplification factor\n",
-      "RL=81.6;                                    #Load resistance, ohm\n",
-      "VCE_peak=30.0;                               #Peak value of collector voltage, V\n",
-      "IC_peak=35.0;                               #Peak value of collector current, mA\n",
-      "VCE_min=5.0;                                 #Minimum value of collector voltage, V\n",
-      "IC_min=1.0;                                 #Minimum value of collector current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC_zero_signal=(IC_peak-IC_min)/2 +1;                  #Zero signal collector current, mA\n",
-      "\n",
-      "#(ii)\n",
-      "IB_zero_signal=IC_zero_signal/beta;                 #Zero signal base current, mA\n",
-      "\n",
-      "#(iii)\n",
-      "VCE_zero_signal=(VCE_peak-VCE_min)/2 +5;                    #Zero signal collector-emitter voltage, V\n",
-      "VCC=VCE_zero_signal;                                    #Collector supply voltage,V (due to transformer coupling, aproximately equal to zero signal VCE)\n",
-      "P_dc=VCC*IC_zero_signal;                                #d.c input power, mW\n",
-      "VCE_ac=(VCE_peak-VCE_min)/(2*sqrt(2));                         #a.c output voltage, V\n",
-      "IC_ac=(IC_peak-IC_min)/(2*sqrt(2));                           #a.c output  current, mA\n",
-      "P_ac=VCE_ac*IC_ac;                                      #a.c output power, mW\n",
-      "\n",
-      "#(iv)\n",
-      "collector_eff=(P_ac/P_dc)*100;                          #Collector efficiency\n",
-      "\n",
-      "#(v)\n",
-      "#a.c resistance RL'=negative inverse of slope of the d.c load line\n",
-      "slope=(IC_peak-IC_min)/(VCE_min-VCE_peak);                                  #Slope of he d.c load line, kilo mho\n",
-      "RL_ac=-(1/slope)*1000;                                                     #a.c resistance, ohm\n",
-      "n=sqrt(RL_ac/RL);                                                           #Transformer turn ratio\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The approximate value of zero signal collector current=%dmA.\"%IC_zero_signal);\n",
-      "print(\"(ii)  The zero signal base current=%.2fmA.\"%IB_zero_signal);\n",
-      "print(\"(iii) The d.c input power= %dmW and a.c output power =%dmW.\"%(P_dc,P_ac));\n",
-      "print(\"(iv)  The collector efficiency=%.1f%%.\"%collector_eff);\n",
-      "print(\"(v)   The turn ratio of the transformer=%d.\"%n);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The approximate value of zero signal collector current=18mA.\n",
-        "(ii)  The zero signal base current=0.18mA.\n",
-        "(iii) The d.c input power= 315mW and a.c output power =106mW.\n",
-        "(iv)  The collector efficiency=33.7%.\n",
-        "(v)   The turn ratio of the transformer=3.\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.14 : Page number 321-322\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=13.0;                        #Load resistance, ohm\n",
-      "RL_reflected=325.0;             #Load resistance, when referred to primary, ohm\n",
-      "VCC=20.0;                       #Supply voltage, V\n",
-      "IC=58.0;                      #Quiscent value of collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "n=sqrt(RL_reflected/RL);                    #Transformer turn ratio\n",
-      "\n",
-      "#(ii)\n",
-      "P_ac=(((IC/1000)**2)*RL_reflected/2)*1000;                   #A.C output power, mW\n",
-      "\n",
-      "#(iii)\n",
-      "P_dc=VCC*IC;                                    #d.c input power, mW\n",
-      "collector_eff=(P_ac/P_dc)*100;                        #Collector efficiency\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Transformer turn ratio=%d.\"%n);\n",
-      "print(\"(ii)  The a.c output power=%dmW.\"%P_ac);\n",
-      "print(\"(iii) The collector efficiency=%d%%.\"%collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Transformer turn ratio=5.\n",
-        "(ii)  The a.c output power=546mW.\n",
-        "(iii) The collector efficiency=47%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.15 : Page number 323\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_total=4.0;                #Total power dissipated by the power transistor, W\n",
-      "T_j_max=90.0;               #Maximum junction temperature, degree celsius\n",
-      "theta=10.0;                 #Thermal resistance, degree celsius per watt\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Total power dissipation=half of(max. junc. temp. - ambient temp.)\n",
-      "T_amb=T_j_max-(P_total*theta);                      #Ambient temperature, degree celsius\n",
-      "\n",
-      "#Result\n",
-      "print(\"The ambient temperature=%d degree celsius.\"%T_amb);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ambient temperature=50 degree celsius.\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.16 : Page number 323-324\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "theta=300.0;                #Thermal resistance, degree celsius per watt\n",
-      "T_j_max=90.0;               #Maximum junction temperature, degree celsius\n",
-      "T_amb=30.0;                 #Ambient temperature, degree celsius\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Without heat sink\n",
-      "P_total=((T_j_max-T_amb)/theta)*1000;              #Maximum permissible power dissipation without sink, mW\n",
-      "\n",
-      "print(\"(i)The maximum permissible power dissipation without heat sink=%dmW.\"%P_total);\n",
-      "\n",
-      "#(ii) With heat sink\n",
-      "theta=60.0;                                    #reduced thermal resistance, degree celsius per watt\n",
-      "P_total=((T_j_max-T_amb)/theta)*1000;          #Maximum permissible power dissipation with heat sink, mW\n",
-      "\n",
-      "print(\"(ii)The maximum permissible power dissipation with heat sink=%dmW.\"%P_total);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The maximum permissible power dissipation without heat sink=200mW.\n",
-        "(ii)The maximum permissible power dissipation with heat sink=1000mW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.17 : Page number 324\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "theta=20.0;                         #Thermal resistance, degree celsius per watt\n",
-      "T_j_max=200.0;                      #Maximum junction temperature, degree celsius\n",
-      "T_amb=25.0;                         #Ambient temperature, degree celsius\n",
-      "VCE=4.0;                            #Collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "P_total=(T_j_max-T_amb)/theta;          #Maximum permissible power dissipation, W\n",
-      "\n",
-      "#since, the max. power dissipation=VCE_max*IC_max,therefore\n",
-      "IC_max=P_total/VCE;                                             #Maximum collector current, A\n",
-      "\n",
-      "print(\"The maximum collector current that the transistor can carry without destruction=%.2fA.\"%IC_max);\n",
-      "\n",
-      "#The ambient temperature rises\n",
-      "T_amb=75.0;                     #The risen ambibent temperature, degree celsius\n",
-      "P_total=(T_j_max-T_amb)/theta;          #Maximum permissible power dissipation, W\n",
-      "IC_max=P_total/VCE;                                             #Maximum collector current, A\n",
-      "\n",
-      "print(\"The maximum collector current for the risen ambient temperature=%.2fA.\"%IC_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector current that the transistor can carry without destruction=2.19A.\n",
-        "The maximum collector current for the risen ambient temperature=1.56A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.18 : Page  number 328-329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                   #Supply voltage, V\n",
-      "RL=8.0;                     #Driving load, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC_sat=VCC/(2*RL);                  #Collector saturation current, A\n",
-      "P_o_max=round(VCC*IC_sat*0.25,2);            #Maximum load power, W\n",
-      "\n",
-      "#(ii)\n",
-      "P_dc=round(VCC*IC_sat/round(pi,2),2);                 #d.c input power, W\n",
-      "\n",
-      "#(iii)\n",
-      "Collector_eff=(P_o_max/P_dc)*100;       #Collector efficiency\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The maximum load power =%.2fW.\"%P_o_max);\n",
-      "print(\"(ii)  The d.c input power=%.2fW.\"%P_dc);\n",
-      "print(\"(iii) The collector efficiency=%.1f%%.\"%Collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The maximum load power =2.25W.\n",
-        "(ii)  The d.c input power=2.87W.\n",
-        "(iii) The collector efficiency=78.4%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.19 : Page number 329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_T=10.0;                   #Power rating of each transistor, W\n",
-      "max_eff=0.785;              #Maximum collector effciency\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, input power=max. a.c power + Power rating of transistor\n",
-      "#And, max. efficiency=max. a.c power/input d.c power\n",
-      "P_2T=2*P_T;                                   #Total power dissipation by two transistors\n",
-      "P_o_max=(max_eff*P_2T)/(1-max_eff);           #Maximum output a.c power, W\n",
-      "\n",
-      "#result\n",
-      "print(\"The maximum output power that can be obtained=%.2fW.\"%P_o_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum output power that can be obtained=73.02W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.20 : Page number 329\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "eff=60.0/100;                   #Efficiency of the amplifier\n",
-      "P_T=2.5;                        #Power dissipated by each transistor, W\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, input power=max. a.c power + Power rating of transistor\n",
-      "#And, max. efficiency=max. a.c power/input d.c power\n",
-      "P_2T=2*P_T;                        #Total power dissipated by both transistors, W\n",
-      "P_ac=(eff*P_2T)/(1-eff);           #Output a.c power, W\n",
-      "P_dc=P_ac+P_2T;                    #Input d.c power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"The a.c output power= %.1fW.\"%P_ac);\n",
-      "print(\"The d.c input power= %.1fW.\"%P_dc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c output power= 7.5W.\n",
-        "The d.c input power= 12.5W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 12.21 : Page number 329-330\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;               #Supply voltage, V\n",
-      "RL=10.0;                #Load resistance, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "IC_sat=(VCC/(2*RL))*1000;          #Saturated collector current, mA\n",
-      "VCE_off=VCC/2;                     #Collector-emitter voltage in off state, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"1st end point of a.c load line, IC(sat)=%dmA.\"%IC_sat);\n",
-      "print(\"2nd end point of a.c load line, VCE(off)=%dV.\"%VCE_off);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "1st end point of a.c load line, IC(sat)=500mA.\n",
-        "2nd end point of a.c load line, VCE(off)=5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13.ipynb
deleted file mode 100755
index 8c036515..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13.ipynb
+++ /dev/null
@@ -1,1147 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:950272de386b8a006089886b37042be2cbcff413c494ad9f74eae7a1268e49ba"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 13: AMPLIFIERS WITH NEGATIVE FEEDBACK"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.1 : Page number 338\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=3000.0;              #Voltage gain without feedback\n",
-      "m_v=0.01;               #Feedback fraction\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Avf=Av/(1+Av*m_v);                  #Voltage gain of the amplifier with negative feedback\n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage gain of the amplifier with negative feedback=%.0f.\"%Avf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain of the amplifier with negative feedback=97.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.2 : Page number  339\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=140.0;                   #Voltage gain\n",
-      "Avf=17.5;                   #Voltage gain with negative feedback\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Avf=Av/(1+Av*mv), so,\n",
-      "mv=(Av-Avf)/(Av*Avf);                 #Fraction of output fedback to the input\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The fraction of output fedback to the input=1/%.0f.\"%(1.0/mv));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fraction of output fedback to the input=1/20.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.3 : Page number 339\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=100.0;                   #Voltage gain\n",
-      "Avf=50.0;                   #Voltage gain with negative feedback\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "mv=(Av-Avf)/(Av*Avf);       #The fraction of output fedback to input\n",
-      "\n",
-      "#(ii) Overall gain is to be 75:\n",
-      "Avf=75.0;                           #The required overall gain\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Av=Avf/(1-Avf*mv);                  #The required value of amplifier gain\n",
-      "\n",
-      "#result\n",
-      "print(\"(i)  The fraction of output fedback to input=%.2f.\"%mv);\n",
-      "print(\"(ii) The required amplifier gain for overall gain to be 75=%d.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The fraction of output fedback to input=0.01.\n",
-        "(ii) The required amplifier gain for overall gain to be 75=300.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.4 : Page number 339-340\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vout=10.0;              #output voltage , V\n",
-      "Vin_f=0.5;              #Input votage for amplifier with feedback, V\n",
-      "Vin=0.25;                #Input votage for amplifier without feedback, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Av=Vout/Vin;                #Voltage gain without negative feedback\n",
-      "\n",
-      "#(ii)\n",
-      "Avf=Vout/Vin_f;             #Voltage gain with negative feedback\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "mv=(Av-Avf)/(Av*Avf);       #Feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The voltage gain without feedback=%d.\"%Av);\n",
-      "print(\"(ii) The feedback fraction = 1/%d.\"%(1/mv));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The voltage gain without feedback=40.\n",
-        "(ii) The feedback fraction = 1/40.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.5 : Page number 340\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=50.0;                #Gain without feedback\n",
-      "Avf=25.0;               #Gain with negative feedback\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "mv=(Av-Avf)/(Av*Avf);               #Feedback fraction\n",
-      "\n",
-      "#(i)\n",
-      "#percentage of reduction without feedback\n",
-      "Av_reduced=40.0;                                        #Reduced amplifier gain due to ageing\n",
-      "percentage_of_reduction=((Av-Av_reduced)/Av)*100;       #Percentage of reduction in stage gain\n",
-      "\n",
-      "print(\"(i)  The percentage of reduction in stage gain without feedback=%d%%.\"%percentage_of_reduction);\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Avf_reduced=round(Av_reduced/(1+mv*Av_reduced),1);               #Reduced net gain with negative feedback \n",
-      "percentage_of_reduction_f=((Avf-Avf_reduced)/Avf)*100;          #Percentage of reduction in net gain with feedback\n",
-      "\n",
-      "print(\"(ii) The percentage of reduction in net gain with feedback=%.1f%%\"%percentage_of_reduction_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The percentage of reduction in stage gain without feedback=20%.\n",
-        "(ii) The percentage of reduction in net gain with feedback=11.2%\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.6 : Page number 340\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=100.0;               #Gain\n",
-      "mv=0.1;                 #feedback fraction\n",
-      "Av_fall=6.0;            #fall in gain, dB\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Avf=round(Av/(1+Av*mv),2);           #Total system gain with feedback\n",
-      "\n",
-      "#Since, fall in gain=20*log10(Av/Av_1)\n",
-      "Av1=round(Av/10**(Av_fall/20),0);            #New absolute voltage gain without feedback\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Avf_new=round(Av1/(1+Av1*mv),2);             #New net system gain with feedback\n",
-      "\n",
-      "percentage_change=((Avf-Avf_new)/Avf)*100;          #Percentage change in system gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"The percentage change in system gain=%.2f%%\"%percentage_change);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage change in system gain=8.36%\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.7 : Page number 341\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=500.0;                       #Voltage gain without feedback\n",
-      "Avf=100.0;                      #Voltage gain with negative feedback\n",
-      "Av_fall_percentage=20.0;          #Gain fall percentage due to ageing\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "mv=(Av-Avf)/(Av*Avf);                                           #Feedback fraction\n",
-      "Av_reduced=((100-Av_fall_percentage)/100)*Av;                   #Reduced voltage gain\n",
-      "Avf_reduced=round(Av_reduced/(1+Av_reduced*mv),1);                       #Reduced total gain of the system\n",
-      "percentage_fall=((Avf-Avf_reduced)/Avf)*100;                    #Percentage of fall in total system gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"The feedback fraction=%.3f.\"%mv);\n",
-      "print(\"The percentage fall in system gain=%.1f%%.\"%percentage_fall);\n",
-      "\n",
-      "#Note: The percentage gain is calculated in the text as 4.7% due to approximation of Avf to 95.3 whose actual approximation will be (95.238)~95.2. So, the percentage fall calculated here is 4.8%\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The feedback fraction=0.008.\n",
-        "The percentage fall in system gain=4.8%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.8 : Page number 341\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "Av=100000.0;                        #Open loop voltage gain\n",
-      "f_dB=10.0;                          #Negative feedback, dB\n",
-      "\n",
-      "#Calculation\n",
-      "Av_dB=20*log10(Av);                        #dB voltage gain without feedback, dB\n",
-      "Avf_dB=Av_dB-f_dB;                         #dB voltage gain with feedback, dB\n",
-      "Avf=10**(Avf_dB/20);                       #Voltage gain with feedback\n",
-      "\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "mv=(Av-Avf)/(Av*Avf);                       #feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage gain with feedback=%d.\"%Avf);\n",
-      "print(\"The feedback fraction=%.2e.\"%mv);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain with feedback=31622.\n",
-        "The feedback fraction=2.16e-05.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.9 : Page number 341-342\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ao=1000.0;                        #Open circuit voltage gain\n",
-      "Rout=100.0;                       #Output resistance, ohm\n",
-      "RL=900.0;                         #Resistive load, ohm\n",
-      "mv=1/50;                          #feedback fraction\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Av=Ao*RL/(Rout+RL)\n",
-      "Av=Ao*RL/(Rout+RL);                     #Voltage gain without feedback\n",
-      "Avf=Av/(1+Av*mv);                       #Voltage gain with feedback\n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage gain with feedback=%.1f.\"%Avf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain with feedback=900.0.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.10 : Page number 342\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Avf=100.0;                    #Voltage gain with feedback\n",
-      "vary_f=1;                     #Vary percentage in voltage gain with feedback\n",
-      "vary_wf=20;                   #Vary percentage in voltage gain without feedback\n",
-      "\n",
-      "#Calculation\n",
-      "#Avf=Av/(1+Av*mv)\n",
-      "print(\"%d=Av/(1+Av*mv)        ------Eq. 1\"%Avf);         #Equation 1\n",
-      "\n",
-      "#considering variation in gains\n",
-      "Avf_vary=Avf*(1- vary_f/100.0);               #Gain with feedback, considering variation\n",
-      "print(\"%d=%.1f*Av/(1+%.1f*Av*mv)        ------Eq. 2\"%(Avf_vary,(1-vary_wf/100.0),(1-vary_wf/100.0)));     #Equation 2\n",
-      "\n",
-      "#Solving the above two equations\n",
-      "print(\"%d + %.1f*Av*mv=%.1fAv        ------Eq. 3 from Eq. 2\"%(Avf_vary,Avf_vary*(1-vary_wf/100.0),(1-vary_wf/100.0)));     #Equation 3\n",
-      "\n",
-      "#multiplying Eq. 1 with (Avf_vary*(1-vary_wf/100.0))/100=0.792\n",
-      "print(\"%.1f + %.1f*Av*mv=%.3fAv        ------Eq. 4 from Eq. 1\"%(Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf_vary*(1-vary_wf/100.0)/100.0));     #Equation 4\n",
-      "\n",
-      "print(\"Subtracting Eq.4 from Eq.3\" );\n",
-      "print(\"%.1f = %.3f*Av\"%(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0,(1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0));\n",
-      "Av=(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0)/((1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0);\n",
-      "print(\"Av=%.0f.\"%Av);\n",
-      "mv=(Av-Avf)/(Av*Avf);\n",
-      "print(\"mv=%.4f.\"%mv);\n",
-      "      "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "100=Av/(1+Av*mv)        ------Eq. 1\n",
-        "99=0.8*Av/(1+0.8*Av*mv)        ------Eq. 2\n",
-        "99 + 79.2*Av*mv=0.8Av        ------Eq. 3 from Eq. 2\n",
-        "79.2 + 79.2*Av*mv=0.792Av        ------Eq. 4 from Eq. 1\n",
-        "Subtracting Eq.4 from Eq.3\n",
-        "19.8 = 0.008*Av\n",
-        "Av=2475.\n",
-        "mv=0.0096.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.11 : Page number 345\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=10000.0;             #Volage gain without feedback\n",
-      "R1=2.0;                 #Resistor R1, kilo ohm\n",
-      "R2=18.0;                #Resistor R2, kilo ohm\n",
-      "Vin=1.0;                #input voltage, mV\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "mv=R1/(R1+R2);              #feedback fraction\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Avf=round(Av/(1+Av*mv),0);           #Voltage gain with feedback\n",
-      "\n",
-      "#(iii)\n",
-      "Vout=Avf*Vin;               #Output voltage, mV\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Feedback fraction=%.1f.\"%mv);\n",
-      "print(\"(ii)  Voltage gain with feedback=%d.\"%Avf);\n",
-      "print(\"(iii) Output voltage=%dmV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Feedback fraction=0.1.\n",
-        "(ii)  Voltage gain with feedback=10.\n",
-        "(iii) Output voltage=10mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.12 : Page number 345-346\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=10000.0;             #Volateg gain without feedback\n",
-      "Zin=10.0;               #Input impedance, kilo ohm\n",
-      "Zout=100.0;             #Output impedance, ohm\n",
-      "R1=10.0;                #Resistor R1, kilo ohm\n",
-      "R2=90.0;                #Resistor R2, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "mv=R1/(R1+R2);                  #Feedback fraction\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Avf=round(Av/(1+Av*mv),0);               #Voltage gain with feedback\n",
-      "\n",
-      "#(iii)\n",
-      "Zin_feedback=((1+Av*mv)*Zin)/1000;             #Increased input impedance due to negative feedback, mega ohm\n",
-      "\n",
-      "#(iv)\n",
-      "Zout_feedback=Zout/(1+Av*mv);             #Decreased output impedance due to negative feedback, ohm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Feedback fraction=%.1f.\"%mv);\n",
-      "print(\"(ii)  The voltage gain with feedback=%d.\"%Avf);\n",
-      "print(\"(iii) Increased input impedance due to negative feedback=%.0f mega ohm\"%Zin_feedback);\n",
-      "print(\"(iv)  Decreased output impedance due to negative feedback=%.1f ohm.\"%Zout_feedback);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Feedback fraction=0.1.\n",
-        "(ii)  The voltage gain with feedback=10.\n",
-        "(iii) Increased input impedance due to negative feedback=10 mega ohm\n",
-        "(iv)  Decreased output impedance due to negative feedback=0.1 ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.13 : Page number 346\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=150.0;                       #Voltage gain\n",
-      "D=5/100.0;                        #Distortion\n",
-      "mv=10/100.0;                      #Feedback fraction\n",
-      "\n",
-      "#Calculation\n",
-      "Dvf=round((D/(1+Av*mv))*100,3);                #Distortion with negative feedback\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Distortion with negative feedback=%.3f%%\"%Dvf);\n",
-      "\n",
-      "#Note: In the text, value of Dvf=0.3125% has been approximated to 0.313%. But, here the approximation is done to 0.312%\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Distortion with negative feedback=0.313%\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.14 : Page number 346\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=1000.0;                      #Voltage gain\n",
-      "f1=1.5;                         #Lower cut-off frequency, kHz\n",
-      "f2=501.5;                       #Upper cut-off frequency, kHz\n",
-      "mv=1/100.0;                       #Feedbcack fraction\n",
-      "\n",
-      "#Calculation\n",
-      "f1_f=(f1/(1+mv*Av))*1000;              #New lower cut-off frequency, Hz\n",
-      "f2_f=(f2*(1+mv*Av))/1000;              #New upper cut-off frequency, MHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The new lower cut-off frequency=%.1fHz\"%f1_f);\n",
-      "print(\"The new upper cut-off frequency=%.2fMHz\"%f2_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The new lower cut-off frequency=136.4Hz\n",
-        "The new upper cut-off frequency=5.52MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.15 : Page number 348\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ai=200.0;                       #Current gain without feedback\n",
-      "mi=0.012;                       #Current attenuation\n",
-      "\n",
-      "#Calculation\n",
-      "Aif=Ai/(1+Ai*mi);\n",
-      "\n",
-      "#Result\n",
-      "print(\"The effective current gain=%.2f.\"%Aif);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The effective current gain=58.82.\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.16 : Page number 349\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ai=240.0;                       #Current gain\n",
-      "Zin=15.0;                       #Input impedance without feedback, kilo ohm\n",
-      "mi=0.015;                       #Current feedback fraction\n",
-      "\n",
-      "#Calculations\n",
-      "Zin_f=Zin/(1+mi*Ai);                #Input impedance with feedback, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The input impedance when negative feedback is applied=%.2f kilo ohm\"%Zin_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input impedance when negative feedback is applied=3.26 kilo ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.17 : Page number 349\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ai=200.0;                   #Current gain without feedback\n",
-      "Zout=3.0;                   #Output impedance without feedback, kilo ohm\n",
-      "mi=0.01;                    #current feedback fraction\n",
-      "\n",
-      "#Calculation\n",
-      "Zout_f=Zout*(1+mi*Ai);              #Output impedance with negative feedback, kilo ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output impedance with negative feedback=%dkilo ohm.\"%Zout_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output impedance with negative feedback=9kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.18 : Page number 349\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ai=250.0;                   #Current gain without feedback\n",
-      "BW=400.0;                   #Bandwidth, kHz\n",
-      "mi=0.01;                    #current feedback fraction\n",
-      "\n",
-      "#Calculation\n",
-      "BW_f=BW*(1+mi*Ai);              #Bandwidth when negative feedback is applied, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"Bandwidth when negative feedback is applied=%dkHz.\"%BW_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bandwidth when negative feedback is applied=1400kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.19 : Page number 350-351\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=18.0;                   #Supply voltage, V\n",
-      "R1=16.0;                    #Resistor R1, kilo ohm\n",
-      "R2=22.0;                    #Resistor R2, kilo ohm\n",
-      "RE=910.0;                   #Emitter resistor, ohm\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculations\n",
-      "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-      "VE=V2-VBE;                         #Emitter voltage, V\n",
-      "IE=(VE/RE)*1000;                   #Emitter current, mA (OHM's LAW)\n",
-      "\n",
-      "#D.C load line\n",
-      "IC_sat=(VCC/RE)*1000;                      #Collector saturation current, mA\n",
-      "VCE_off=VCC;                               #Collector-emitter voltage in off state, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Value of VE=%.2fV and IE=%.2fmA\"%(VE,IE));\n",
-      "\n",
-      "#Plotting\n",
-      "VCE_plot=[0,VCE_off];            #Plotting variable for VCE\n",
-      "IC_plot=[IC_sat,0];              #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,20])\n",
-      "limit.set_ylim([0,25])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of VE=9.72V and IE=10.68mA\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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E8Vwu6NfPtIgaNoTYWHj6aTh92u7IxBdK2/XCW7ZsoVatWhw9epQuXbrQuHFj\n2rdvn/f5gQMHEhERAUBYWBixsbHEx8cDP/cQ9bhoj2fNmqX8Wfg4mPM5aRJERrp58UVYsCCeGTOg\nUiU3Lpfy6Y+P3W43CxcuBMj7fVlcfnGL6JNPPknFihV5+OGHAbWDrOZ2u/PeQHL5nJLPDRtg5EjT\nGpo9Gxo39s7rOCWfvhAw7aAff/yRU6dOAfDDDz+wfv16oqOj7QjFEfR/MGs5JZ9dusBnn5nTzNq3\nh4cfhu++s/51nJJPf2VLETh8+DDt27cnNjaW1q1bk5CQQNeuXe0IRUQuoUwZswfRnj2mADRuDK++\nCrm5dkcmVvGLdtCvqR1kLV1uW8vJ+fz0U7Pq2OMxq45bt778n+nkfFotYNpBIhKYbrjBHGTzwANw\nxx0waBBkZtodlVwOXQmISIl8/z089RQsWGA2qhsxAq64wu6onE3nCYiIz/3nP2ZukJZm7iLSeM8+\nagdJgS7cVyzWUD7zu/56WLsWnnkGhg+H3r3N1tVFpXzaS0VARC6bywUJCeYuotatzezgT3/y34Ns\n5GdqB4mI5Q4cgEcfNZvSPfMM3HWXKRTiXZoJiIhf+fBDMzAODYU5c3SQjbdpJiAFUs/VWspn0bVr\nB9u3wz33mIHx8OEXH2SjfNpLRUBEvKpUKbj/frNLaalS5iCbF1+Ec+fsjkxA7SAR8bHdu+Ghh+D4\ncbPquGNHuyMKHpoJiEhA8HjgjTfMQTZt2pjhsQ6yuXyaCUiB1HO1lvJ5+Vwu6NvXtIjKlnUTG2tW\nH+sgG99TERAR21SoYPYf2r4dUlKgSRNYtcpcKYhvqB0kIn7j/ffNQTZXX222oIiMtDuiwKJ2kIgE\ntFtugZ074fbboUMHGD3aOwfZyM9UBBxAPWxrKZ/W+nU+y5QxVwN79sCpU+YgmwULdJCNt6gIiIhf\nqlED5s6Fd94x/3vjjfDJJ3ZHFXw0ExARv5ebC3//O4wda1YeT5kCNWvaHZX/0UxARIJSSAgMGABf\nfWWuEKKi4NlnISfH7sgCn4qAA6iHbS3l01rFyWdoKEybZo643LgRmjWDdeu8F5sTqAiISMBp1AjW\nrIHp0+HBB6FXL/jvf+2OKjBpJiAiAe3MGZg50xSE++835x1XrGh3VPbQTEBEHKdsWTMw/uwz2L/f\nLDBbskSrjotKRcAB1MO2lvJpLavyefXVsGgRLF1qNqTr0MEsPJNLUxEQkaBy003w6afmbqJbb4Vh\nwyAry+7pzXwnAAAIcUlEQVSo/JdmAiIStE6cgAkTzNXBhAlmZlC6tN1ReY/OExARKcDu3WYriqws\nc5BNfLzdEXmHBsNSIPWwraV8WssX+YyONusKxo+H5GS46y749luvv2xAUBEQEUdwueDOO81BNpGR\nEBcHf/4z/PST3ZHZS+0gEXGktDRzvOWOHTBjBvTubQpFINNMQESkmDZuNAff165tDrJp0sTuiEpO\nMwEpkHrY1lI+rWV3Pjt3NusJevSAjh3hD3+AkydtDcmnVARExPHKlDFXA198AT/8YA6ymT/fGQfZ\nqB0kIvIrO3bAiBFw9izMmWMOtAkEageJiFigRQvYssWsLUhMhIEDISPD7qi8w5YisG7dOho3bkzD\nhg2ZNm2aHSE4it0912CjfFrLX/PpcsG995qDbMLDzVqDZ54JvoNsfF4Ezp8/z4MPPsi6dev44osv\nWLJkCV9++aWvw3CUndpFy1LKp7X8PZ+/PMjG7TbF4L337I7KOj4vAtu2baNBgwZERERQpkwZ+vfv\nz9tvv+3rMBzlpJNudfAB5dNagZLPCwfZzJhhhsg9esDevXZHdfl8XgQOHjxI3bp18x7XqVOHgwcP\n+joMEZESuf12+PxzaN/eDIwffxyys+2OquR8XgRcgb4kLwClpaXZHUJQUT6tFYj5LFsWxoyBXbvg\nwAFzS2lKit1RlYzPN1W9+uqrSU9Pz3ucnp5OnTp1Lvo6FQtrvfbaa3aHEFSUT2sFQz6bN7c7gpLx\n+TqBc+fOcf3117Nx40Zq165Nq1atWLJkCZGRkb4MQ0REsOFKoHTp0jz//PPceuutnD9/niFDhqgA\niIjYxC9XDIuIiG/43YphLSSzVkREBM2aNSMuLo5WrVrZHU5AGTx4MOHh4URHR+c9d/z4cbp06UKj\nRo3o2rVrwNze6A8KyufEiROpU6cOcXFxxMXFsW7dOhsjDCzp6el06tSJpk2bEhUVxXPPPQcU/z3q\nV0VAC8ms53K5cLvdpKSksG3bNrvDCSiDBg266JfS1KlT6dKlC6mpqXTu3JmpU6faFF3gKSifLpeL\n0aNHk5KSQkpKCt26dbMpusBTpkwZZs6cyZ49e/j444954YUX+PLLL4v9HvWrIqCFZN6hjl/JtG/f\nnipVquR7bvXq1SQnJwOQnJzMqlWr7AgtIBWUT9D7s6Rq1qxJbGwsABUrViQyMpKDBw8W+z3qV0VA\nC8ms53K5uOWWW2jZsiVz5861O5yAd/jwYcLDwwEIDw/n8OHDNkcU+ObMmUNMTAxDhgxRe62E0tLS\nSElJoXXr1sV+j/pVEdDaAOtt2bKFlJQU3nvvPV544QU2b95sd0hBw+Vy6T17mYYNG8a+ffvYuXMn\ntWrV4uGHH7Y7pICTnZ1NYmIis2fPJjQ0NN/nivIe9asiUNSFZFJ0tWrVAqB69erccccdmgtcpvDw\ncDIzMwHIyMigRo0aNkcU2GrUqJH3i2ro0KF6fxbT2bNnSUxMZMCAAfTu3Rso/nvUr4pAy5Yt+frr\nr0lLSyMnJ4dly5bRs2dPu8MKWD/++COnTp0C4IcffmD9+vX57syQ4uvZs2fe6tbXXnst7/94UjIZ\nv9ik/6233tL7sxg8Hg9DhgyhSZMmjBo1Ku/5Yr9HPX5m7dq1nkaNGnmuu+46z+TJk+0OJ6B98803\nnpiYGE9MTIynadOmymcx9e/f31OrVi1PmTJlPHXq1PEsWLDAc+zYMU/nzp09DRs29HTp0sVz4sQJ\nu8MMGL/O5/z58z0DBgzwREdHe5o1a+bp1auXJzMz0+4wA8bmzZs9LpfLExMT44mNjfXExsZ63nvv\nvWK/R7VYTETEwfyqHSQiIr6lIiAi4mAqAiIiDqYiICLiYCoCIiIOpiIgIuJgKgIiIg6mIiCOcvPN\nN7N+/fp8z82aNYvhw4eTmprKbbfdRqNGjWjRogV33XUXR44cwe12U7ly5bw97+Pi4ti4cSMAP/30\nE/Hx8eTm5nLttdeSmpqa72ePGjWKv/zlL3z++ecMGjTIZ/9OkaJSERBHSUpKYunSpfmeW7ZsGUlJ\nSSQkJPDAAw+QmprKjh07GD58OEePHsXlctGhQ4e8Pe9TUlLo3LkzAAsWLCAxMZGQkJCLfnZubi5v\nvvkmSUlJREVFceDAgXx7Y4n4AxUBcZTExETWrFnDuXPnALMF76FDh/j6669p27Ytt99+e97XduzY\nkaZNm15yv/vFixfTq1cvwBSYZcuW5X3uX//6F9dcc03e9ug9evS4qACJ2E1FQBylatWqtGrVirVr\n1wKwdOlS+vXrx549e2jevHmh37d58+Z87aB9+/aRk5PDN998Q7169QCIiooiJCSEXbt25f3su+++\nO+9ntGzZUlt5i99RERDH+WXbZtmyZfl+URemffv2+dpB9evXJysri7CwsAJ/9vnz53n77bfp27dv\n3ueqV6/OoUOHrP3HiFwmFQFxnJ49e7Jx40ZSUlL48ccfiYuLo2nTpuzYsaNYP6d8+fKcPn0633P9\n+/dn+fLlvP/++zRr1ozq1avnfe706dOUL1/ekn+DiFVUBMRxKlasSKdOnRg0aFDeVcDdd9/N1q1b\n89pEYHr6e/bsKfTnVKlShfPnz5OTk5P33LXXXku1atUYO3bsRVcYqampREVFWfyvEbk8KgLiSElJ\nSezevZukpCQAypUrx7vvvsucOXNo1KgRTZs25eWXX6Z69eq4XK6LZgIrV64EoGvXrhf1+ZOSkvjP\nf/5Dnz598j3/wQcfkJCQ4Jt/oEgR6TwBkcuQkpLCzJkzef311y/5dWfOnCE+Pp4tW7YQEqK/vcR/\n6N0ochni4uLo1KkTubm5l/y69PR0pk2bpgIgfkdXAiIiDqY/S0REHExFQETEwVQEREQcTEVARMTB\nVARERBzs/wEyuwnCdMXkIQAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f109441f290>"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.20 : Page number 352\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                   #Supply voltage, V\n",
-      "R1=10.0;                    #Resistor R1, kilo ohm\n",
-      "R2=10.0;                    #Resistor R2, kilo ohm\n",
-      "RE=5.0;                     #Emitter resistance, kilo ohm\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-      "VE=V2-VBE;                         #Emitter voltage, V\n",
-      "IE=(VE/RE);                   #Emitter current, mA (OHM's LAW)\n",
-      "re=25/IE;                          #a.c emitter resistance, ohm\n",
-      "Av=RE*1000/(re+RE*1000);                     #Voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage gain of the emitter follower circuit=%.3f.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain of the emitter follower circuit=0.994.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.21 : Page number 352-353\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RE=5.0;                 #Emitter resistance, kilo ohm\n",
-      "re=29.1;                #a.c emitter resistance, ohm\n",
-      "RL=5.0;                 #Load resistance, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "RE_ac=(RE*RL)/(RE+RL);               #New effective value of emitter resistance, kilo ohm\n",
-      "Av=RE_ac*1000/(re+RE_ac*1000);     #Voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage gain=%.3f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain=0.988\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.22 : Page number 354\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def pr(r1,r2):                     #Function for calculating parallel resistance\n",
-      "    return (r1*r2)/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=10.0;                   #Supply voltage, V\n",
-      "R1=10.0;                    #Resistor R1, kilo ohm\n",
-      "R2=10.0;                    #Resistor R2, kilo ohm\n",
-      "RE=4.3;                     #Emitter resistor, kilo ohm\n",
-      "RL=10.0;                    #Load resistance, kilo ohm\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=200.0;                 #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-      "VE=V2-VBE;                         #Emitter voltage, V\n",
-      "IE=(VE/RE);                        #Emitter current, mA (OHM's LAW)\n",
-      "re=25/IE;                          #a.c emitter resistance, ohm\n",
-      "RE_eff=pr(RE,RL);                  #Effective external emitter resistance, kilo ohm\n",
-      "Zin_base=beta*(re/1000+RE_eff);         #Input impedance of the base of the transistor, kilo ohm\n",
-      "Zin=pr(pr(R1,R2),Zin_base);        #Input impedance of emitter follower, kilo ohm\n",
-      "#Approximate value of input impedance taken as parallel resistance of R1 and R2 and ignoring Zin_base due to its relatively large value\n",
-      "Zin_approx=pr(R1,R2);              #Approximate input impedance, kilo ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"The input impedance of the emitter follower =%.2f kilo ohm\"%Zin);\n",
-      "print(\"The approximate value of the input impedance=%d kilo ohm\"%Zin_approx);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input impedance of the emitter follower =4.96 kilo ohm\n",
-        "The approximate value of the input impedance=5 kilo ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.23 : Page number 355\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def pr(r1,r2):                     #Function for calculating parallel resistance\n",
-      "    return (r1*r2)/(r1+r2);\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "re=20.0;                    #a.c emitter resistance, ohm\n",
-      "R1=3.0;                     #Resistor R1, kilo ohm\n",
-      "R2=4.7;                     #Resistor R2, kilo ohm\n",
-      "RS=600.0;                   #Source resistance, kilo ohm\n",
-      "beta=200.0;                 #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "Rin_ac=pr(pr(R1,R2)*1000,RS);            #Input a.c resistance, ohm\n",
-      "Zout=re + Rin_ac/beta;                   #Output impedance, ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output impedance=%.1f ohm\"%Zout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output impedance=22.3 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.24 : Page number  358\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                    #Supply voltage, V\n",
-      "R1=120.0;                    #Resistor R1, kilo ohm\n",
-      "R2=120.0;                    #Resistor R2, kilo ohm\n",
-      "RE=3.3;                      #Emitter resistor, kilo ohm\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "beta_1=70.0;                 #Base current amplification factor of 1st transistor\n",
-      "beta_2=70.0;                 #Base current amplification factor of 2nd transistor\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-      "IE_2=(V2-2*VBE)/RE;                #Emitter current, mA (OHM's LAW)\n",
-      "\n",
-      "#(ii)\n",
-      "Zin=(beta_1*beta_2*RE)/1000;               #Input impedance, mega ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  d.c value of current in RE=%.2fmA\"%IE_2);\n",
-      "print(\"(ii) Input impedance=%.2f mega ohm.\"%Zin);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  d.c value of current in RE=1.09mA\n",
-        "(ii) Input impedance=16.17 mega ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.25 : Page number 358-359\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                    #Supply voltage, V\n",
-      "R1=20.0;                     #Resistor R1, kilo ohm\n",
-      "R2=10.0;                     #Resistor R2, kilo ohm\n",
-      "RC=4.0;                      #Collector resistor, kilo ohm\n",
-      "RE=2.0;                      #Emitter resistor, kilo ohm\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "beta=100.0;                  #Base current amplification factor of 1st transistor\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) D.C Bias levels\n",
-      "VB1=VCC*R2/(R1+R2);                 #Base voltage of 1st transistor, V (Voltage divider rule)\n",
-      "VE1=VB1-VBE;                        #Emitter voltage of 1st transistor, V\n",
-      "VB2=VE1;                            #Base voltage of 2nd transistor, V\n",
-      "VE2=VB2-VBE;                        #Emitter voltage of 2nd transistor, V\n",
-      "IE2=VE2/RE;                         #Emitter current of 2nd transistor, mA (OHM' LAW)\n",
-      "IE1=IE2/beta;                       #Emitter current of 1st transistor, mA (IE~IC=beta*IB, here IB2=IE1)\n",
-      "\n",
-      "#(ii) A.C analysis\n",
-      "re1=25/IE1;                         #a.c emitter resistance of 1st transistor\n",
-      "re2=25/IE2;                         #a.c emitter resistance of 2nd transistor\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) D.C Bias levels: \\n VB1= %dV, VE1=%.1fV, VB2=%.1fV, VE2=%.1fV, IE2=%.1fmA and IE1=%.3fmA.\"%(VB1,VE1,VB2,VE2,IE2,IE1));\n",
-      "print(\"(ii) A.C Analysis:   \\n re1=%d ohm and re2=%.2f ohm \"%(re1,re2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) D.C Bias levels: \n",
-        " VB1= 4V, VE1=3.3V, VB2=3.3V, VE2=2.6V, IE2=1.3mA and IE1=0.013mA.\n",
-        "(ii) A.C Analysis:   \n",
-        " re1=1923 ohm and re2=19.23 ohm \n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_1.ipynb
deleted file mode 100755
index 8c036515..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_1.ipynb
+++ /dev/null
@@ -1,1147 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:950272de386b8a006089886b37042be2cbcff413c494ad9f74eae7a1268e49ba"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 13: AMPLIFIERS WITH NEGATIVE FEEDBACK"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.1 : Page number 338\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=3000.0;              #Voltage gain without feedback\n",
-      "m_v=0.01;               #Feedback fraction\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Avf=Av/(1+Av*m_v);                  #Voltage gain of the amplifier with negative feedback\n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage gain of the amplifier with negative feedback=%.0f.\"%Avf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain of the amplifier with negative feedback=97.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.2 : Page number  339\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=140.0;                   #Voltage gain\n",
-      "Avf=17.5;                   #Voltage gain with negative feedback\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Avf=Av/(1+Av*mv), so,\n",
-      "mv=(Av-Avf)/(Av*Avf);                 #Fraction of output fedback to the input\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The fraction of output fedback to the input=1/%.0f.\"%(1.0/mv));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fraction of output fedback to the input=1/20.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.3 : Page number 339\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=100.0;                   #Voltage gain\n",
-      "Avf=50.0;                   #Voltage gain with negative feedback\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "mv=(Av-Avf)/(Av*Avf);       #The fraction of output fedback to input\n",
-      "\n",
-      "#(ii) Overall gain is to be 75:\n",
-      "Avf=75.0;                           #The required overall gain\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Av=Avf/(1-Avf*mv);                  #The required value of amplifier gain\n",
-      "\n",
-      "#result\n",
-      "print(\"(i)  The fraction of output fedback to input=%.2f.\"%mv);\n",
-      "print(\"(ii) The required amplifier gain for overall gain to be 75=%d.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The fraction of output fedback to input=0.01.\n",
-        "(ii) The required amplifier gain for overall gain to be 75=300.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.4 : Page number 339-340\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vout=10.0;              #output voltage , V\n",
-      "Vin_f=0.5;              #Input votage for amplifier with feedback, V\n",
-      "Vin=0.25;                #Input votage for amplifier without feedback, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Av=Vout/Vin;                #Voltage gain without negative feedback\n",
-      "\n",
-      "#(ii)\n",
-      "Avf=Vout/Vin_f;             #Voltage gain with negative feedback\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "mv=(Av-Avf)/(Av*Avf);       #Feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The voltage gain without feedback=%d.\"%Av);\n",
-      "print(\"(ii) The feedback fraction = 1/%d.\"%(1/mv));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The voltage gain without feedback=40.\n",
-        "(ii) The feedback fraction = 1/40.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.5 : Page number 340\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=50.0;                #Gain without feedback\n",
-      "Avf=25.0;               #Gain with negative feedback\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "mv=(Av-Avf)/(Av*Avf);               #Feedback fraction\n",
-      "\n",
-      "#(i)\n",
-      "#percentage of reduction without feedback\n",
-      "Av_reduced=40.0;                                        #Reduced amplifier gain due to ageing\n",
-      "percentage_of_reduction=((Av-Av_reduced)/Av)*100;       #Percentage of reduction in stage gain\n",
-      "\n",
-      "print(\"(i)  The percentage of reduction in stage gain without feedback=%d%%.\"%percentage_of_reduction);\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Avf_reduced=round(Av_reduced/(1+mv*Av_reduced),1);               #Reduced net gain with negative feedback \n",
-      "percentage_of_reduction_f=((Avf-Avf_reduced)/Avf)*100;          #Percentage of reduction in net gain with feedback\n",
-      "\n",
-      "print(\"(ii) The percentage of reduction in net gain with feedback=%.1f%%\"%percentage_of_reduction_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The percentage of reduction in stage gain without feedback=20%.\n",
-        "(ii) The percentage of reduction in net gain with feedback=11.2%\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.6 : Page number 340\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=100.0;               #Gain\n",
-      "mv=0.1;                 #feedback fraction\n",
-      "Av_fall=6.0;            #fall in gain, dB\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Avf=round(Av/(1+Av*mv),2);           #Total system gain with feedback\n",
-      "\n",
-      "#Since, fall in gain=20*log10(Av/Av_1)\n",
-      "Av1=round(Av/10**(Av_fall/20),0);            #New absolute voltage gain without feedback\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Avf_new=round(Av1/(1+Av1*mv),2);             #New net system gain with feedback\n",
-      "\n",
-      "percentage_change=((Avf-Avf_new)/Avf)*100;          #Percentage change in system gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"The percentage change in system gain=%.2f%%\"%percentage_change);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage change in system gain=8.36%\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.7 : Page number 341\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=500.0;                       #Voltage gain without feedback\n",
-      "Avf=100.0;                      #Voltage gain with negative feedback\n",
-      "Av_fall_percentage=20.0;          #Gain fall percentage due to ageing\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "mv=(Av-Avf)/(Av*Avf);                                           #Feedback fraction\n",
-      "Av_reduced=((100-Av_fall_percentage)/100)*Av;                   #Reduced voltage gain\n",
-      "Avf_reduced=round(Av_reduced/(1+Av_reduced*mv),1);                       #Reduced total gain of the system\n",
-      "percentage_fall=((Avf-Avf_reduced)/Avf)*100;                    #Percentage of fall in total system gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"The feedback fraction=%.3f.\"%mv);\n",
-      "print(\"The percentage fall in system gain=%.1f%%.\"%percentage_fall);\n",
-      "\n",
-      "#Note: The percentage gain is calculated in the text as 4.7% due to approximation of Avf to 95.3 whose actual approximation will be (95.238)~95.2. So, the percentage fall calculated here is 4.8%\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The feedback fraction=0.008.\n",
-        "The percentage fall in system gain=4.8%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.8 : Page number 341\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "Av=100000.0;                        #Open loop voltage gain\n",
-      "f_dB=10.0;                          #Negative feedback, dB\n",
-      "\n",
-      "#Calculation\n",
-      "Av_dB=20*log10(Av);                        #dB voltage gain without feedback, dB\n",
-      "Avf_dB=Av_dB-f_dB;                         #dB voltage gain with feedback, dB\n",
-      "Avf=10**(Avf_dB/20);                       #Voltage gain with feedback\n",
-      "\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "mv=(Av-Avf)/(Av*Avf);                       #feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage gain with feedback=%d.\"%Avf);\n",
-      "print(\"The feedback fraction=%.2e.\"%mv);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain with feedback=31622.\n",
-        "The feedback fraction=2.16e-05.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.9 : Page number 341-342\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ao=1000.0;                        #Open circuit voltage gain\n",
-      "Rout=100.0;                       #Output resistance, ohm\n",
-      "RL=900.0;                         #Resistive load, ohm\n",
-      "mv=1/50;                          #feedback fraction\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Av=Ao*RL/(Rout+RL)\n",
-      "Av=Ao*RL/(Rout+RL);                     #Voltage gain without feedback\n",
-      "Avf=Av/(1+Av*mv);                       #Voltage gain with feedback\n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage gain with feedback=%.1f.\"%Avf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain with feedback=900.0.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.10 : Page number 342\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Avf=100.0;                    #Voltage gain with feedback\n",
-      "vary_f=1;                     #Vary percentage in voltage gain with feedback\n",
-      "vary_wf=20;                   #Vary percentage in voltage gain without feedback\n",
-      "\n",
-      "#Calculation\n",
-      "#Avf=Av/(1+Av*mv)\n",
-      "print(\"%d=Av/(1+Av*mv)        ------Eq. 1\"%Avf);         #Equation 1\n",
-      "\n",
-      "#considering variation in gains\n",
-      "Avf_vary=Avf*(1- vary_f/100.0);               #Gain with feedback, considering variation\n",
-      "print(\"%d=%.1f*Av/(1+%.1f*Av*mv)        ------Eq. 2\"%(Avf_vary,(1-vary_wf/100.0),(1-vary_wf/100.0)));     #Equation 2\n",
-      "\n",
-      "#Solving the above two equations\n",
-      "print(\"%d + %.1f*Av*mv=%.1fAv        ------Eq. 3 from Eq. 2\"%(Avf_vary,Avf_vary*(1-vary_wf/100.0),(1-vary_wf/100.0)));     #Equation 3\n",
-      "\n",
-      "#multiplying Eq. 1 with (Avf_vary*(1-vary_wf/100.0))/100=0.792\n",
-      "print(\"%.1f + %.1f*Av*mv=%.3fAv        ------Eq. 4 from Eq. 1\"%(Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf_vary*(1-vary_wf/100.0)/100.0));     #Equation 4\n",
-      "\n",
-      "print(\"Subtracting Eq.4 from Eq.3\" );\n",
-      "print(\"%.1f = %.3f*Av\"%(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0,(1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0));\n",
-      "Av=(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0)/((1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0);\n",
-      "print(\"Av=%.0f.\"%Av);\n",
-      "mv=(Av-Avf)/(Av*Avf);\n",
-      "print(\"mv=%.4f.\"%mv);\n",
-      "      "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "100=Av/(1+Av*mv)        ------Eq. 1\n",
-        "99=0.8*Av/(1+0.8*Av*mv)        ------Eq. 2\n",
-        "99 + 79.2*Av*mv=0.8Av        ------Eq. 3 from Eq. 2\n",
-        "79.2 + 79.2*Av*mv=0.792Av        ------Eq. 4 from Eq. 1\n",
-        "Subtracting Eq.4 from Eq.3\n",
-        "19.8 = 0.008*Av\n",
-        "Av=2475.\n",
-        "mv=0.0096.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.11 : Page number 345\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=10000.0;             #Volage gain without feedback\n",
-      "R1=2.0;                 #Resistor R1, kilo ohm\n",
-      "R2=18.0;                #Resistor R2, kilo ohm\n",
-      "Vin=1.0;                #input voltage, mV\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "mv=R1/(R1+R2);              #feedback fraction\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Avf=round(Av/(1+Av*mv),0);           #Voltage gain with feedback\n",
-      "\n",
-      "#(iii)\n",
-      "Vout=Avf*Vin;               #Output voltage, mV\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Feedback fraction=%.1f.\"%mv);\n",
-      "print(\"(ii)  Voltage gain with feedback=%d.\"%Avf);\n",
-      "print(\"(iii) Output voltage=%dmV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Feedback fraction=0.1.\n",
-        "(ii)  Voltage gain with feedback=10.\n",
-        "(iii) Output voltage=10mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.12 : Page number 345-346\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=10000.0;             #Volateg gain without feedback\n",
-      "Zin=10.0;               #Input impedance, kilo ohm\n",
-      "Zout=100.0;             #Output impedance, ohm\n",
-      "R1=10.0;                #Resistor R1, kilo ohm\n",
-      "R2=90.0;                #Resistor R2, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "mv=R1/(R1+R2);                  #Feedback fraction\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-      "Avf=round(Av/(1+Av*mv),0);               #Voltage gain with feedback\n",
-      "\n",
-      "#(iii)\n",
-      "Zin_feedback=((1+Av*mv)*Zin)/1000;             #Increased input impedance due to negative feedback, mega ohm\n",
-      "\n",
-      "#(iv)\n",
-      "Zout_feedback=Zout/(1+Av*mv);             #Decreased output impedance due to negative feedback, ohm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Feedback fraction=%.1f.\"%mv);\n",
-      "print(\"(ii)  The voltage gain with feedback=%d.\"%Avf);\n",
-      "print(\"(iii) Increased input impedance due to negative feedback=%.0f mega ohm\"%Zin_feedback);\n",
-      "print(\"(iv)  Decreased output impedance due to negative feedback=%.1f ohm.\"%Zout_feedback);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Feedback fraction=0.1.\n",
-        "(ii)  The voltage gain with feedback=10.\n",
-        "(iii) Increased input impedance due to negative feedback=10 mega ohm\n",
-        "(iv)  Decreased output impedance due to negative feedback=0.1 ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.13 : Page number 346\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=150.0;                       #Voltage gain\n",
-      "D=5/100.0;                        #Distortion\n",
-      "mv=10/100.0;                      #Feedback fraction\n",
-      "\n",
-      "#Calculation\n",
-      "Dvf=round((D/(1+Av*mv))*100,3);                #Distortion with negative feedback\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Distortion with negative feedback=%.3f%%\"%Dvf);\n",
-      "\n",
-      "#Note: In the text, value of Dvf=0.3125% has been approximated to 0.313%. But, here the approximation is done to 0.312%\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Distortion with negative feedback=0.313%\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.14 : Page number 346\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Av=1000.0;                      #Voltage gain\n",
-      "f1=1.5;                         #Lower cut-off frequency, kHz\n",
-      "f2=501.5;                       #Upper cut-off frequency, kHz\n",
-      "mv=1/100.0;                       #Feedbcack fraction\n",
-      "\n",
-      "#Calculation\n",
-      "f1_f=(f1/(1+mv*Av))*1000;              #New lower cut-off frequency, Hz\n",
-      "f2_f=(f2*(1+mv*Av))/1000;              #New upper cut-off frequency, MHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The new lower cut-off frequency=%.1fHz\"%f1_f);\n",
-      "print(\"The new upper cut-off frequency=%.2fMHz\"%f2_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The new lower cut-off frequency=136.4Hz\n",
-        "The new upper cut-off frequency=5.52MHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.15 : Page number 348\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ai=200.0;                       #Current gain without feedback\n",
-      "mi=0.012;                       #Current attenuation\n",
-      "\n",
-      "#Calculation\n",
-      "Aif=Ai/(1+Ai*mi);\n",
-      "\n",
-      "#Result\n",
-      "print(\"The effective current gain=%.2f.\"%Aif);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The effective current gain=58.82.\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.16 : Page number 349\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ai=240.0;                       #Current gain\n",
-      "Zin=15.0;                       #Input impedance without feedback, kilo ohm\n",
-      "mi=0.015;                       #Current feedback fraction\n",
-      "\n",
-      "#Calculations\n",
-      "Zin_f=Zin/(1+mi*Ai);                #Input impedance with feedback, kilo ohm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The input impedance when negative feedback is applied=%.2f kilo ohm\"%Zin_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input impedance when negative feedback is applied=3.26 kilo ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.17 : Page number 349\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ai=200.0;                   #Current gain without feedback\n",
-      "Zout=3.0;                   #Output impedance without feedback, kilo ohm\n",
-      "mi=0.01;                    #current feedback fraction\n",
-      "\n",
-      "#Calculation\n",
-      "Zout_f=Zout*(1+mi*Ai);              #Output impedance with negative feedback, kilo ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output impedance with negative feedback=%dkilo ohm.\"%Zout_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output impedance with negative feedback=9kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.18 : Page number 349\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ai=250.0;                   #Current gain without feedback\n",
-      "BW=400.0;                   #Bandwidth, kHz\n",
-      "mi=0.01;                    #current feedback fraction\n",
-      "\n",
-      "#Calculation\n",
-      "BW_f=BW*(1+mi*Ai);              #Bandwidth when negative feedback is applied, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"Bandwidth when negative feedback is applied=%dkHz.\"%BW_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Bandwidth when negative feedback is applied=1400kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.19 : Page number 350-351\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=18.0;                   #Supply voltage, V\n",
-      "R1=16.0;                    #Resistor R1, kilo ohm\n",
-      "R2=22.0;                    #Resistor R2, kilo ohm\n",
-      "RE=910.0;                   #Emitter resistor, ohm\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculations\n",
-      "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-      "VE=V2-VBE;                         #Emitter voltage, V\n",
-      "IE=(VE/RE)*1000;                   #Emitter current, mA (OHM's LAW)\n",
-      "\n",
-      "#D.C load line\n",
-      "IC_sat=(VCC/RE)*1000;                      #Collector saturation current, mA\n",
-      "VCE_off=VCC;                               #Collector-emitter voltage in off state, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Value of VE=%.2fV and IE=%.2fmA\"%(VE,IE));\n",
-      "\n",
-      "#Plotting\n",
-      "VCE_plot=[0,VCE_off];            #Plotting variable for VCE\n",
-      "IC_plot=[IC_sat,0];              #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,20])\n",
-      "limit.set_ylim([0,25])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of VE=9.72V and IE=10.68mA\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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E8Vwu6NfPtIgaNoTYWHj6aTh92u7IxBdK2/XCW7ZsoVatWhw9epQuXbrQuHFj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\nVARERBzs/wEyuwnCdMXkIQAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f109441f290>"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.20 : Page number 352\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                   #Supply voltage, V\n",
-      "R1=10.0;                    #Resistor R1, kilo ohm\n",
-      "R2=10.0;                    #Resistor R2, kilo ohm\n",
-      "RE=5.0;                     #Emitter resistance, kilo ohm\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-      "VE=V2-VBE;                         #Emitter voltage, V\n",
-      "IE=(VE/RE);                   #Emitter current, mA (OHM's LAW)\n",
-      "re=25/IE;                          #a.c emitter resistance, ohm\n",
-      "Av=RE*1000/(re+RE*1000);                     #Voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage gain of the emitter follower circuit=%.3f.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain of the emitter follower circuit=0.994.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.21 : Page number 352-353\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RE=5.0;                 #Emitter resistance, kilo ohm\n",
-      "re=29.1;                #a.c emitter resistance, ohm\n",
-      "RL=5.0;                 #Load resistance, kilo ohm\n",
-      "\n",
-      "#Calculation\n",
-      "RE_ac=(RE*RL)/(RE+RL);               #New effective value of emitter resistance, kilo ohm\n",
-      "Av=RE_ac*1000/(re+RE_ac*1000);     #Voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage gain=%.3f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain=0.988\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.22 : Page number 354\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def pr(r1,r2):                     #Function for calculating parallel resistance\n",
-      "    return (r1*r2)/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=10.0;                   #Supply voltage, V\n",
-      "R1=10.0;                    #Resistor R1, kilo ohm\n",
-      "R2=10.0;                    #Resistor R2, kilo ohm\n",
-      "RE=4.3;                     #Emitter resistor, kilo ohm\n",
-      "RL=10.0;                    #Load resistance, kilo ohm\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=200.0;                 #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-      "VE=V2-VBE;                         #Emitter voltage, V\n",
-      "IE=(VE/RE);                        #Emitter current, mA (OHM's LAW)\n",
-      "re=25/IE;                          #a.c emitter resistance, ohm\n",
-      "RE_eff=pr(RE,RL);                  #Effective external emitter resistance, kilo ohm\n",
-      "Zin_base=beta*(re/1000+RE_eff);         #Input impedance of the base of the transistor, kilo ohm\n",
-      "Zin=pr(pr(R1,R2),Zin_base);        #Input impedance of emitter follower, kilo ohm\n",
-      "#Approximate value of input impedance taken as parallel resistance of R1 and R2 and ignoring Zin_base due to its relatively large value\n",
-      "Zin_approx=pr(R1,R2);              #Approximate input impedance, kilo ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"The input impedance of the emitter follower =%.2f kilo ohm\"%Zin);\n",
-      "print(\"The approximate value of the input impedance=%d kilo ohm\"%Zin_approx);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input impedance of the emitter follower =4.96 kilo ohm\n",
-        "The approximate value of the input impedance=5 kilo ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.23 : Page number 355\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "def pr(r1,r2):                     #Function for calculating parallel resistance\n",
-      "    return (r1*r2)/(r1+r2);\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "re=20.0;                    #a.c emitter resistance, ohm\n",
-      "R1=3.0;                     #Resistor R1, kilo ohm\n",
-      "R2=4.7;                     #Resistor R2, kilo ohm\n",
-      "RS=600.0;                   #Source resistance, kilo ohm\n",
-      "beta=200.0;                 #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "Rin_ac=pr(pr(R1,R2)*1000,RS);            #Input a.c resistance, ohm\n",
-      "Zout=re + Rin_ac/beta;                   #Output impedance, ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output impedance=%.1f ohm\"%Zout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output impedance=22.3 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.24 : Page number  358\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                    #Supply voltage, V\n",
-      "R1=120.0;                    #Resistor R1, kilo ohm\n",
-      "R2=120.0;                    #Resistor R2, kilo ohm\n",
-      "RE=3.3;                      #Emitter resistor, kilo ohm\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "beta_1=70.0;                 #Base current amplification factor of 1st transistor\n",
-      "beta_2=70.0;                 #Base current amplification factor of 2nd transistor\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-      "IE_2=(V2-2*VBE)/RE;                #Emitter current, mA (OHM's LAW)\n",
-      "\n",
-      "#(ii)\n",
-      "Zin=(beta_1*beta_2*RE)/1000;               #Input impedance, mega ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  d.c value of current in RE=%.2fmA\"%IE_2);\n",
-      "print(\"(ii) Input impedance=%.2f mega ohm.\"%Zin);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  d.c value of current in RE=1.09mA\n",
-        "(ii) Input impedance=16.17 mega ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.25 : Page number 358-359\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                    #Supply voltage, V\n",
-      "R1=20.0;                     #Resistor R1, kilo ohm\n",
-      "R2=10.0;                     #Resistor R2, kilo ohm\n",
-      "RC=4.0;                      #Collector resistor, kilo ohm\n",
-      "RE=2.0;                      #Emitter resistor, kilo ohm\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "beta=100.0;                  #Base current amplification factor of 1st transistor\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) D.C Bias levels\n",
-      "VB1=VCC*R2/(R1+R2);                 #Base voltage of 1st transistor, V (Voltage divider rule)\n",
-      "VE1=VB1-VBE;                        #Emitter voltage of 1st transistor, V\n",
-      "VB2=VE1;                            #Base voltage of 2nd transistor, V\n",
-      "VE2=VB2-VBE;                        #Emitter voltage of 2nd transistor, V\n",
-      "IE2=VE2/RE;                         #Emitter current of 2nd transistor, mA (OHM' LAW)\n",
-      "IE1=IE2/beta;                       #Emitter current of 1st transistor, mA (IE~IC=beta*IB, here IB2=IE1)\n",
-      "\n",
-      "#(ii) A.C analysis\n",
-      "re1=25/IE1;                         #a.c emitter resistance of 1st transistor\n",
-      "re2=25/IE2;                         #a.c emitter resistance of 2nd transistor\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) D.C Bias levels: \\n VB1= %dV, VE1=%.1fV, VB2=%.1fV, VE2=%.1fV, IE2=%.1fmA and IE1=%.3fmA.\"%(VB1,VE1,VB2,VE2,IE2,IE1));\n",
-      "print(\"(ii) A.C Analysis:   \\n re1=%d ohm and re2=%.2f ohm \"%(re1,re2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) D.C Bias levels: \n",
-        " VB1= 4V, VE1=3.3V, VB2=3.3V, VE2=2.6V, IE2=1.3mA and IE1=0.013mA.\n",
-        "(ii) A.C Analysis:   \n",
-        " re1=1923 ohm and re2=19.23 ohm \n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_2.ipynb
deleted file mode 100755
index afa98f3a..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_2.ipynb
+++ /dev/null
@@ -1,1139 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 13: AMPLIFIERS WITH NEGATIVE FEEDBACK"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.1 : Page number 338"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the amplifier with negative feedback=97.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=3000.0;              #Voltage gain without feedback\n",
-    "m_v=0.01;               #Feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=Av/(1+Av*m_v);                  #Voltage gain of the amplifier with negative feedback\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain of the amplifier with negative feedback=%.0f.\"%Avf);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.2 : Page number  339"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The fraction of output fedback to the input=1/20.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=140.0;                   #Voltage gain\n",
-    "Avf=17.5;                   #Voltage gain with negative feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Avf=Av/(1+Av*mv), so,\n",
-    "mv=(Av-Avf)/(Av*Avf);                 #Fraction of output fedback to the input\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The fraction of output fedback to the input=1/%.0f.\"%(1.0/mv));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.3 : Page number 339"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The fraction of output fedback to input=0.01.\n",
-      "(ii) The required amplifier gain for overall gain to be 75=300.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=100.0;                   #Voltage gain\n",
-    "Avf=50.0;                   #Voltage gain with negative feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);       #The fraction of output fedback to input\n",
-    "\n",
-    "#(ii) Overall gain is to be 75:\n",
-    "Avf=75.0;                           #The required overall gain\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Av=Avf/(1-Avf*mv);                  #The required value of amplifier gain\n",
-    "\n",
-    "#result\n",
-    "print(\"(i)  The fraction of output fedback to input=%.2f.\"%mv);\n",
-    "print(\"(ii) The required amplifier gain for overall gain to be 75=%d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.4 : Page number 339-340"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The voltage gain without feedback=40.\n",
-      "(ii) The feedback fraction = 1/40.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vout=10.0;              #output voltage , V\n",
-    "Vin_f=0.5;              #Input votage for amplifier with feedback, V\n",
-    "Vin=0.25;                #Input votage for amplifier without feedback, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "Av=Vout/Vin;                #Voltage gain without negative feedback\n",
-    "\n",
-    "#(ii)\n",
-    "Avf=Vout/Vin_f;             #Voltage gain with negative feedback\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);       #Feedback fraction\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  The voltage gain without feedback=%d.\"%Av);\n",
-    "print(\"(ii) The feedback fraction = 1/%d.\"%(1/mv));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.5 : Page number 340"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The percentage of reduction in stage gain without feedback=20%.\n",
-      "(ii) The percentage of reduction in net gain with feedback=11.2%\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=50.0;                #Gain without feedback\n",
-    "Avf=25.0;               #Gain with negative feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);               #Feedback fraction\n",
-    "\n",
-    "#(i)\n",
-    "#percentage of reduction without feedback\n",
-    "Av_reduced=40.0;                                        #Reduced amplifier gain due to ageing\n",
-    "percentage_of_reduction=((Av-Av_reduced)/Av)*100;       #Percentage of reduction in stage gain\n",
-    "\n",
-    "print(\"(i)  The percentage of reduction in stage gain without feedback=%d%%.\"%percentage_of_reduction);\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf_reduced=round(Av_reduced/(1+mv*Av_reduced),1);               #Reduced net gain with negative feedback \n",
-    "percentage_of_reduction_f=((Avf-Avf_reduced)/Avf)*100;          #Percentage of reduction in net gain with feedback\n",
-    "\n",
-    "print(\"(ii) The percentage of reduction in net gain with feedback=%.1f%%\"%percentage_of_reduction_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.6 : Page number 340"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The percentage change in system gain=8.36%\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=100.0;               #Gain\n",
-    "mv=0.1;                 #feedback fraction\n",
-    "Av_fall=6.0;            #fall in gain, dB\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=round(Av/(1+Av*mv),2);           #Total system gain with feedback\n",
-    "\n",
-    "#Since, fall in gain=20*log10(Av/Av_1)\n",
-    "Av1=round(Av/10**(Av_fall/20),0);            #New absolute voltage gain without feedback\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf_new=round(Av1/(1+Av1*mv),2);             #New net system gain with feedback\n",
-    "\n",
-    "percentage_change=((Avf-Avf_new)/Avf)*100;          #Percentage change in system gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"The percentage change in system gain=%.2f%%\"%percentage_change);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.7 : Page number 341"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The feedback fraction=0.008.\n",
-      "The percentage fall in system gain=4.8%.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=500.0;                       #Voltage gain without feedback\n",
-    "Avf=100.0;                      #Voltage gain with negative feedback\n",
-    "Av_fall_percentage=20.0;          #Gain fall percentage due to ageing\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);                                           #Feedback fraction\n",
-    "Av_reduced=((100-Av_fall_percentage)/100)*Av;                   #Reduced voltage gain\n",
-    "Avf_reduced=round(Av_reduced/(1+Av_reduced*mv),1);                       #Reduced total gain of the system\n",
-    "percentage_fall=((Avf-Avf_reduced)/Avf)*100;                    #Percentage of fall in total system gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"The feedback fraction=%.3f.\"%mv);\n",
-    "print(\"The percentage fall in system gain=%.1f%%.\"%percentage_fall);\n",
-    "\n",
-    "#Note: The percentage gain is calculated in the text as 4.7% due to approximation of Avf to 95.3 whose actual approximation will be (95.238)~95.2. So, the percentage fall calculated here is 4.8%\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.8 : Page number 341"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain with feedback=31622.\n",
-      "The feedback fraction=2.16e-05.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "Av=100000.0;                        #Open loop voltage gain\n",
-    "f_dB=10.0;                          #Negative feedback, dB\n",
-    "\n",
-    "#Calculation\n",
-    "Av_dB=20*log10(Av);                        #dB voltage gain without feedback, dB\n",
-    "Avf_dB=Av_dB-f_dB;                         #dB voltage gain with feedback, dB\n",
-    "Avf=10**(Avf_dB/20);                       #Voltage gain with feedback\n",
-    "\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);                       #feedback fraction\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain with feedback=%d.\"%Avf);\n",
-    "print(\"The feedback fraction=%.2e.\"%mv);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.9 : Page number 341-342"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain with feedback=47.4.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ao=1000.0;                        #Open circuit voltage gain\n",
-    "Rout=100.0;                       #Output resistance, ohm\n",
-    "RL=900.0;                         #Resistive load, ohm\n",
-    "mv=1/50;                          #feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Av=Ao*RL/(Rout+RL)\n",
-    "Av=Ao*RL/(Rout+RL);                     #Voltage gain without feedback\n",
-    "Avf=Av/(1+Av*mv);                       #Voltage gain with feedback\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain with feedback=%.1f.\"%Avf);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.10 : Page number 342"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "100=Av/(1+Av*mv)        ------Eq. 1\n",
-      "99=0.8*Av/(1+0.8*Av*mv)        ------Eq. 2\n",
-      "99 + 79.2*Av*mv=0.8Av        ------Eq. 3 from Eq. 2\n",
-      "79.2 + 79.2*Av*mv=0.792Av        ------Eq. 4 from Eq. 1\n",
-      "Subtracting Eq.4 from Eq.3\n",
-      "19.8 = 0.008*Av\n",
-      "Av=2475.\n",
-      "mv=0.0096.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Avf=100.0;                    #Voltage gain with feedback\n",
-    "vary_f=1;                     #Vary percentage in voltage gain with feedback\n",
-    "vary_wf=20;                   #Vary percentage in voltage gain without feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#Avf=Av/(1+Av*mv)\n",
-    "print(\"%d=Av/(1+Av*mv)        ------Eq. 1\"%Avf);         #Equation 1\n",
-    "\n",
-    "#considering variation in gains\n",
-    "Avf_vary=Avf*(1- vary_f/100.0);               #Gain with feedback, considering variation\n",
-    "print(\"%d=%.1f*Av/(1+%.1f*Av*mv)        ------Eq. 2\"%(Avf_vary,(1-vary_wf/100.0),(1-vary_wf/100.0)));     #Equation 2\n",
-    "\n",
-    "#Solving the above two equations\n",
-    "print(\"%d + %.1f*Av*mv=%.1fAv        ------Eq. 3 from Eq. 2\"%(Avf_vary,Avf_vary*(1-vary_wf/100.0),(1-vary_wf/100.0)));     #Equation 3\n",
-    "\n",
-    "#multiplying Eq. 1 with (Avf_vary*(1-vary_wf/100.0))/100=0.792\n",
-    "print(\"%.1f + %.1f*Av*mv=%.3fAv        ------Eq. 4 from Eq. 1\"%(Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf_vary*(1-vary_wf/100.0)/100.0));     #Equation 4\n",
-    "\n",
-    "print(\"Subtracting Eq.4 from Eq.3\" );\n",
-    "print(\"%.1f = %.3f*Av\"%(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0,(1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0));\n",
-    "Av=(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0)/((1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0);\n",
-    "print(\"Av=%.0f.\"%Av);\n",
-    "mv=(Av-Avf)/(Av*Avf);\n",
-    "print(\"mv=%.4f.\"%mv);\n",
-    "      "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.11 : Page number 345"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)   Feedback fraction=0.1.\n",
-      "(ii)  Voltage gain with feedback=10.\n",
-      "(iii) Output voltage=10mV.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=10000.0;             #Volage gain without feedback\n",
-    "R1=2.0;                 #Resistor R1, kilo ohm\n",
-    "R2=18.0;                #Resistor R2, kilo ohm\n",
-    "Vin=1.0;                #input voltage, mV\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "mv=R1/(R1+R2);              #feedback fraction\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=round(Av/(1+Av*mv),0);           #Voltage gain with feedback\n",
-    "\n",
-    "#(iii)\n",
-    "Vout=Avf*Vin;               #Output voltage, mV\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)   Feedback fraction=%.1f.\"%mv);\n",
-    "print(\"(ii)  Voltage gain with feedback=%d.\"%Avf);\n",
-    "print(\"(iii) Output voltage=%dmV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.12 : Page number 345-346"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)   Feedback fraction=0.1.\n",
-      "(ii)  The voltage gain with feedback=10.\n",
-      "(iii) Increased input impedance due to negative feedback=10 mega ohm\n",
-      "(iv)  Decreased output impedance due to negative feedback=0.1 ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=10000.0;             #Volateg gain without feedback\n",
-    "Zin=10.0;               #Input impedance, kilo ohm\n",
-    "Zout=100.0;             #Output impedance, ohm\n",
-    "R1=10.0;                #Resistor R1, kilo ohm\n",
-    "R2=90.0;                #Resistor R2, kilo ohm\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "mv=R1/(R1+R2);                  #Feedback fraction\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=round(Av/(1+Av*mv),0);               #Voltage gain with feedback\n",
-    "\n",
-    "#(iii)\n",
-    "Zin_feedback=((1+Av*mv)*Zin)/1000;             #Increased input impedance due to negative feedback, mega ohm\n",
-    "\n",
-    "#(iv)\n",
-    "Zout_feedback=Zout/(1+Av*mv);             #Decreased output impedance due to negative feedback, ohm\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)   Feedback fraction=%.1f.\"%mv);\n",
-    "print(\"(ii)  The voltage gain with feedback=%d.\"%Avf);\n",
-    "print(\"(iii) Increased input impedance due to negative feedback=%.0f mega ohm\"%Zin_feedback);\n",
-    "print(\"(iv)  Decreased output impedance due to negative feedback=%.1f ohm.\"%Zout_feedback);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.13 : Page number 346"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Distortion with negative feedback=0.312%\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=150.0;                       #Voltage gain\n",
-    "D=5/100.0;                        #Distortion\n",
-    "mv=10/100.0;                      #Feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "Dvf=round((D/(1+Av*mv))*100,3);                #Distortion with negative feedback\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Distortion with negative feedback=%.3f%%\"%Dvf);\n",
-    "\n",
-    "#Note: In the text, value of Dvf=0.3125% has been approximated to 0.313%. But, here the approximation is done to 0.312%\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.14 : Page number 346"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The new lower cut-off frequency=136.4Hz\n",
-      "The new upper cut-off frequency=5.52MHz\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=1000.0;                      #Voltage gain\n",
-    "f1=1.5;                         #Lower cut-off frequency, kHz\n",
-    "f2=501.5;                       #Upper cut-off frequency, kHz\n",
-    "mv=1/100.0;                       #Feedbcack fraction\n",
-    "\n",
-    "#Calculation\n",
-    "f1_f=(f1/(1+mv*Av))*1000;              #New lower cut-off frequency, Hz\n",
-    "f2_f=(f2*(1+mv*Av))/1000;              #New upper cut-off frequency, MHz\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The new lower cut-off frequency=%.1fHz\"%f1_f);\n",
-    "print(\"The new upper cut-off frequency=%.2fMHz\"%f2_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.15 : Page number 348"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The effective current gain=58.82.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=200.0;                       #Current gain without feedback\n",
-    "mi=0.012;                       #Current attenuation\n",
-    "\n",
-    "#Calculation\n",
-    "Aif=Ai/(1+Ai*mi);\n",
-    "\n",
-    "#Result\n",
-    "print(\"The effective current gain=%.2f.\"%Aif);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.16 : Page number 349"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance when negative feedback is applied=3.26 kilo ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=240.0;                       #Current gain\n",
-    "Zin=15.0;                       #Input impedance without feedback, kilo ohm\n",
-    "mi=0.015;                       #Current feedback fraction\n",
-    "\n",
-    "#Calculations\n",
-    "Zin_f=Zin/(1+mi*Ai);                #Input impedance with feedback, kilo ohm\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance when negative feedback is applied=%.2f kilo ohm\"%Zin_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.17 : Page number 349"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output impedance with negative feedback=9kilo ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=200.0;                   #Current gain without feedback\n",
-    "Zout=3.0;                   #Output impedance without feedback, kilo ohm\n",
-    "mi=0.01;                    #current feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "Zout_f=Zout*(1+mi*Ai);              #Output impedance with negative feedback, kilo ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output impedance with negative feedback=%dkilo ohm.\"%Zout_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.18 : Page number 349"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Bandwidth when negative feedback is applied=1400kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=250.0;                   #Current gain without feedback\n",
-    "BW=400.0;                   #Bandwidth, kHz\n",
-    "mi=0.01;                    #current feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "BW_f=BW*(1+mi*Ai);              #Bandwidth when negative feedback is applied, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Bandwidth when negative feedback is applied=%dkHz.\"%BW_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.19 : Page number 350-351"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Value of VE=9.72V and IE=10.68mA\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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zB6mcykQiUvBmzoRBg6BNG7jySmjbNumIcktlIhGRShx4ILz0UrgMtVu30FP4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\nAxERQclARESA/w8INYcwb/NOegAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f9a7522f0b8>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pylab as p\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=18.0;                   #Supply voltage, V\n",
-    "R1=16.0;                    #Resistor R1, kilo ohm\n",
-    "R2=22.0;                    #Resistor R2, kilo ohm\n",
-    "RE=910.0;                   #Emitter resistor, ohm\n",
-    "VBE=0.7;                    #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculations\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2-VBE;                         #Emitter voltage, V\n",
-    "IE=(VE/RE)*1000;                   #Emitter current, mA (OHM's LAW)\n",
-    "\n",
-    "#D.C load line\n",
-    "IC_sat=(VCC/RE)*1000;                      #Collector saturation current, mA\n",
-    "VCE_off=VCC;                               #Collector-emitter voltage in off state, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Value of VE=%.2fV and IE=%.2fmA\"%(VE,IE));\n",
-    "\n",
-    "#Plotting\n",
-    "VCE_plot=[0,VCE_off];            #Plotting variable for VCE\n",
-    "IC_plot=[IC_sat,0];              #Plotting variable for IC\n",
-    "p.plot(VCE_plot,IC_plot);\n",
-    "p.xlim(0,20)\n",
-    "p.ylim(0,25)\n",
-    "p.xlabel('VCE(V)');\n",
-    "p.ylabel('IC(mA)');\n",
-    "p.title('d.c load line');\n",
-    "p.grid();\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.20 : Page number 352"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the emitter follower circuit=0.994.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                   #Supply voltage, V\n",
-    "R1=10.0;                    #Resistor R1, kilo ohm\n",
-    "R2=10.0;                    #Resistor R2, kilo ohm\n",
-    "RE=5.0;                     #Emitter resistance, kilo ohm\n",
-    "VBE=0.7;                    #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2-VBE;                         #Emitter voltage, V\n",
-    "IE=(VE/RE);                   #Emitter current, mA (OHM's LAW)\n",
-    "re=25/IE;                          #a.c emitter resistance, ohm\n",
-    "Av=RE*1000/(re+RE*1000);                     #Voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain of the emitter follower circuit=%.3f.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.21 : Page number 352-353"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain=0.988\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RE=5.0;                 #Emitter resistance, kilo ohm\n",
-    "re=29.1;                #a.c emitter resistance, ohm\n",
-    "RL=5.0;                 #Load resistance, kilo ohm\n",
-    "\n",
-    "#Calculation\n",
-    "RE_ac=(RE*RL)/(RE+RL);               #New effective value of emitter resistance, kilo ohm\n",
-    "Av=RE_ac*1000/(re+RE_ac*1000);     #Voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain=%.3f\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.22 : Page number 354"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance of the emitter follower =4.96 kilo ohm\n",
-      "The approximate value of the input impedance=5 kilo ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "def pr(r1,r2):                     #Function for calculating parallel resistance\n",
-    "    return (r1*r2)/(r1+r2);\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=10.0;                   #Supply voltage, V\n",
-    "R1=10.0;                    #Resistor R1, kilo ohm\n",
-    "R2=10.0;                    #Resistor R2, kilo ohm\n",
-    "RE=4.3;                     #Emitter resistor, kilo ohm\n",
-    "RL=10.0;                    #Load resistance, kilo ohm\n",
-    "VBE=0.7;                    #Base-emitter voltage, V\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2-VBE;                         #Emitter voltage, V\n",
-    "IE=(VE/RE);                        #Emitter current, mA (OHM's LAW)\n",
-    "re=25/IE;                          #a.c emitter resistance, ohm\n",
-    "RE_eff=pr(RE,RL);                  #Effective external emitter resistance, kilo ohm\n",
-    "Zin_base=beta*(re/1000+RE_eff);         #Input impedance of the base of the transistor, kilo ohm\n",
-    "Zin=pr(pr(R1,R2),Zin_base);        #Input impedance of emitter follower, kilo ohm\n",
-    "#Approximate value of input impedance taken as parallel resistance of R1 and R2 and ignoring Zin_base due to its relatively large value\n",
-    "Zin_approx=pr(R1,R2);              #Approximate input impedance, kilo ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance of the emitter follower =%.2f kilo ohm\"%Zin);\n",
-    "print(\"The approximate value of the input impedance=%d kilo ohm\"%Zin_approx);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.23 : Page number 355"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output impedance=22.3 ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "def pr(r1,r2):                     #Function for calculating parallel resistance\n",
-    "    return (r1*r2)/(r1+r2);\n",
-    "\n",
-    "\n",
-    "#Variable declaration\n",
-    "re=20.0;                    #a.c emitter resistance, ohm\n",
-    "R1=3.0;                     #Resistor R1, kilo ohm\n",
-    "R2=4.7;                     #Resistor R2, kilo ohm\n",
-    "RS=600.0;                   #Source resistance, kilo ohm\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "Rin_ac=pr(pr(R1,R2)*1000,RS);            #Input a.c resistance, ohm\n",
-    "Zout=re + Rin_ac/beta;                   #Output impedance, ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output impedance=%.1f ohm\"%Zout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.24 : Page number  358"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  d.c value of current in RE=1.09mA\n",
-      "(ii) Input impedance=16.17 mega ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                    #Supply voltage, V\n",
-    "R1=120.0;                    #Resistor R1, kilo ohm\n",
-    "R2=120.0;                    #Resistor R2, kilo ohm\n",
-    "RE=3.3;                      #Emitter resistor, kilo ohm\n",
-    "VBE=0.7;                     #Base-emitter voltage, V\n",
-    "beta_1=70.0;                 #Base current amplification factor of 1st transistor\n",
-    "beta_2=70.0;                 #Base current amplification factor of 2nd transistor\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "IE_2=(V2-2*VBE)/RE;                #Emitter current, mA (OHM's LAW)\n",
-    "\n",
-    "#(ii)\n",
-    "Zin=(beta_1*beta_2*RE)/1000;               #Input impedance, mega ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  d.c value of current in RE=%.2fmA\"%IE_2);\n",
-    "print(\"(ii) Input impedance=%.2f mega ohm.\"%Zin);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.25 : Page number 358-359"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 28,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) D.C Bias levels: \n",
-      " VB1= 4V, VE1=3.3V, VB2=3.3V, VE2=2.6V, IE2=1.3mA and IE1=0.013mA.\n",
-      "(ii) A.C Analysis:   \n",
-      " re1=1923 ohm and re2=19.23 ohm \n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12.0;                    #Supply voltage, V\n",
-    "R1=20.0;                     #Resistor R1, kilo ohm\n",
-    "R2=10.0;                     #Resistor R2, kilo ohm\n",
-    "RC=4.0;                      #Collector resistor, kilo ohm\n",
-    "RE=2.0;                      #Emitter resistor, kilo ohm\n",
-    "VBE=0.7;                     #Base-emitter voltage, V\n",
-    "beta=100.0;                  #Base current amplification factor of 1st transistor\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) D.C Bias levels\n",
-    "VB1=VCC*R2/(R1+R2);                 #Base voltage of 1st transistor, V (Voltage divider rule)\n",
-    "VE1=VB1-VBE;                        #Emitter voltage of 1st transistor, V\n",
-    "VB2=VE1;                            #Base voltage of 2nd transistor, V\n",
-    "VE2=VB2-VBE;                        #Emitter voltage of 2nd transistor, V\n",
-    "IE2=VE2/RE;                         #Emitter current of 2nd transistor, mA (OHM' LAW)\n",
-    "IE1=IE2/beta;                       #Emitter current of 1st transistor, mA (IE~IC=beta*IB, here IB2=IE1)\n",
-    "\n",
-    "#(ii) A.C analysis\n",
-    "re1=25/IE1;                         #a.c emitter resistance of 1st transistor\n",
-    "re2=25/IE2;                         #a.c emitter resistance of 2nd transistor\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) D.C Bias levels: \\n VB1= %dV, VE1=%.1fV, VB2=%.1fV, VE2=%.1fV, IE2=%.1fmA and IE1=%.3fmA.\"%(VB1,VE1,VB2,VE2,IE2,IE1));\n",
-    "print(\"(ii) A.C Analysis:   \\n re1=%d ohm and re2=%.2f ohm \"%(re1,re2));\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_3.ipynb
deleted file mode 100755
index afa98f3a..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_3.ipynb
+++ /dev/null
@@ -1,1139 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 13: AMPLIFIERS WITH NEGATIVE FEEDBACK"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.1 : Page number 338"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the amplifier with negative feedback=97.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=3000.0;              #Voltage gain without feedback\n",
-    "m_v=0.01;               #Feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=Av/(1+Av*m_v);                  #Voltage gain of the amplifier with negative feedback\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain of the amplifier with negative feedback=%.0f.\"%Avf);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.2 : Page number  339"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The fraction of output fedback to the input=1/20.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=140.0;                   #Voltage gain\n",
-    "Avf=17.5;                   #Voltage gain with negative feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Avf=Av/(1+Av*mv), so,\n",
-    "mv=(Av-Avf)/(Av*Avf);                 #Fraction of output fedback to the input\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The fraction of output fedback to the input=1/%.0f.\"%(1.0/mv));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.3 : Page number 339"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The fraction of output fedback to input=0.01.\n",
-      "(ii) The required amplifier gain for overall gain to be 75=300.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=100.0;                   #Voltage gain\n",
-    "Avf=50.0;                   #Voltage gain with negative feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);       #The fraction of output fedback to input\n",
-    "\n",
-    "#(ii) Overall gain is to be 75:\n",
-    "Avf=75.0;                           #The required overall gain\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Av=Avf/(1-Avf*mv);                  #The required value of amplifier gain\n",
-    "\n",
-    "#result\n",
-    "print(\"(i)  The fraction of output fedback to input=%.2f.\"%mv);\n",
-    "print(\"(ii) The required amplifier gain for overall gain to be 75=%d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.4 : Page number 339-340"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The voltage gain without feedback=40.\n",
-      "(ii) The feedback fraction = 1/40.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vout=10.0;              #output voltage , V\n",
-    "Vin_f=0.5;              #Input votage for amplifier with feedback, V\n",
-    "Vin=0.25;                #Input votage for amplifier without feedback, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "Av=Vout/Vin;                #Voltage gain without negative feedback\n",
-    "\n",
-    "#(ii)\n",
-    "Avf=Vout/Vin_f;             #Voltage gain with negative feedback\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);       #Feedback fraction\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  The voltage gain without feedback=%d.\"%Av);\n",
-    "print(\"(ii) The feedback fraction = 1/%d.\"%(1/mv));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.5 : Page number 340"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The percentage of reduction in stage gain without feedback=20%.\n",
-      "(ii) The percentage of reduction in net gain with feedback=11.2%\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=50.0;                #Gain without feedback\n",
-    "Avf=25.0;               #Gain with negative feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);               #Feedback fraction\n",
-    "\n",
-    "#(i)\n",
-    "#percentage of reduction without feedback\n",
-    "Av_reduced=40.0;                                        #Reduced amplifier gain due to ageing\n",
-    "percentage_of_reduction=((Av-Av_reduced)/Av)*100;       #Percentage of reduction in stage gain\n",
-    "\n",
-    "print(\"(i)  The percentage of reduction in stage gain without feedback=%d%%.\"%percentage_of_reduction);\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf_reduced=round(Av_reduced/(1+mv*Av_reduced),1);               #Reduced net gain with negative feedback \n",
-    "percentage_of_reduction_f=((Avf-Avf_reduced)/Avf)*100;          #Percentage of reduction in net gain with feedback\n",
-    "\n",
-    "print(\"(ii) The percentage of reduction in net gain with feedback=%.1f%%\"%percentage_of_reduction_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.6 : Page number 340"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The percentage change in system gain=8.36%\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=100.0;               #Gain\n",
-    "mv=0.1;                 #feedback fraction\n",
-    "Av_fall=6.0;            #fall in gain, dB\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=round(Av/(1+Av*mv),2);           #Total system gain with feedback\n",
-    "\n",
-    "#Since, fall in gain=20*log10(Av/Av_1)\n",
-    "Av1=round(Av/10**(Av_fall/20),0);            #New absolute voltage gain without feedback\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf_new=round(Av1/(1+Av1*mv),2);             #New net system gain with feedback\n",
-    "\n",
-    "percentage_change=((Avf-Avf_new)/Avf)*100;          #Percentage change in system gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"The percentage change in system gain=%.2f%%\"%percentage_change);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.7 : Page number 341"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The feedback fraction=0.008.\n",
-      "The percentage fall in system gain=4.8%.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=500.0;                       #Voltage gain without feedback\n",
-    "Avf=100.0;                      #Voltage gain with negative feedback\n",
-    "Av_fall_percentage=20.0;          #Gain fall percentage due to ageing\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);                                           #Feedback fraction\n",
-    "Av_reduced=((100-Av_fall_percentage)/100)*Av;                   #Reduced voltage gain\n",
-    "Avf_reduced=round(Av_reduced/(1+Av_reduced*mv),1);                       #Reduced total gain of the system\n",
-    "percentage_fall=((Avf-Avf_reduced)/Avf)*100;                    #Percentage of fall in total system gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"The feedback fraction=%.3f.\"%mv);\n",
-    "print(\"The percentage fall in system gain=%.1f%%.\"%percentage_fall);\n",
-    "\n",
-    "#Note: The percentage gain is calculated in the text as 4.7% due to approximation of Avf to 95.3 whose actual approximation will be (95.238)~95.2. So, the percentage fall calculated here is 4.8%\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.8 : Page number 341"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain with feedback=31622.\n",
-      "The feedback fraction=2.16e-05.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "Av=100000.0;                        #Open loop voltage gain\n",
-    "f_dB=10.0;                          #Negative feedback, dB\n",
-    "\n",
-    "#Calculation\n",
-    "Av_dB=20*log10(Av);                        #dB voltage gain without feedback, dB\n",
-    "Avf_dB=Av_dB-f_dB;                         #dB voltage gain with feedback, dB\n",
-    "Avf=10**(Avf_dB/20);                       #Voltage gain with feedback\n",
-    "\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);                       #feedback fraction\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain with feedback=%d.\"%Avf);\n",
-    "print(\"The feedback fraction=%.2e.\"%mv);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.9 : Page number 341-342"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain with feedback=47.4.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ao=1000.0;                        #Open circuit voltage gain\n",
-    "Rout=100.0;                       #Output resistance, ohm\n",
-    "RL=900.0;                         #Resistive load, ohm\n",
-    "mv=1/50;                          #feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Av=Ao*RL/(Rout+RL)\n",
-    "Av=Ao*RL/(Rout+RL);                     #Voltage gain without feedback\n",
-    "Avf=Av/(1+Av*mv);                       #Voltage gain with feedback\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain with feedback=%.1f.\"%Avf);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.10 : Page number 342"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "100=Av/(1+Av*mv)        ------Eq. 1\n",
-      "99=0.8*Av/(1+0.8*Av*mv)        ------Eq. 2\n",
-      "99 + 79.2*Av*mv=0.8Av        ------Eq. 3 from Eq. 2\n",
-      "79.2 + 79.2*Av*mv=0.792Av        ------Eq. 4 from Eq. 1\n",
-      "Subtracting Eq.4 from Eq.3\n",
-      "19.8 = 0.008*Av\n",
-      "Av=2475.\n",
-      "mv=0.0096.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Avf=100.0;                    #Voltage gain with feedback\n",
-    "vary_f=1;                     #Vary percentage in voltage gain with feedback\n",
-    "vary_wf=20;                   #Vary percentage in voltage gain without feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#Avf=Av/(1+Av*mv)\n",
-    "print(\"%d=Av/(1+Av*mv)        ------Eq. 1\"%Avf);         #Equation 1\n",
-    "\n",
-    "#considering variation in gains\n",
-    "Avf_vary=Avf*(1- vary_f/100.0);               #Gain with feedback, considering variation\n",
-    "print(\"%d=%.1f*Av/(1+%.1f*Av*mv)        ------Eq. 2\"%(Avf_vary,(1-vary_wf/100.0),(1-vary_wf/100.0)));     #Equation 2\n",
-    "\n",
-    "#Solving the above two equations\n",
-    "print(\"%d + %.1f*Av*mv=%.1fAv        ------Eq. 3 from Eq. 2\"%(Avf_vary,Avf_vary*(1-vary_wf/100.0),(1-vary_wf/100.0)));     #Equation 3\n",
-    "\n",
-    "#multiplying Eq. 1 with (Avf_vary*(1-vary_wf/100.0))/100=0.792\n",
-    "print(\"%.1f + %.1f*Av*mv=%.3fAv        ------Eq. 4 from Eq. 1\"%(Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf_vary*(1-vary_wf/100.0)/100.0));     #Equation 4\n",
-    "\n",
-    "print(\"Subtracting Eq.4 from Eq.3\" );\n",
-    "print(\"%.1f = %.3f*Av\"%(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0,(1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0));\n",
-    "Av=(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0)/((1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0);\n",
-    "print(\"Av=%.0f.\"%Av);\n",
-    "mv=(Av-Avf)/(Av*Avf);\n",
-    "print(\"mv=%.4f.\"%mv);\n",
-    "      "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.11 : Page number 345"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)   Feedback fraction=0.1.\n",
-      "(ii)  Voltage gain with feedback=10.\n",
-      "(iii) Output voltage=10mV.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=10000.0;             #Volage gain without feedback\n",
-    "R1=2.0;                 #Resistor R1, kilo ohm\n",
-    "R2=18.0;                #Resistor R2, kilo ohm\n",
-    "Vin=1.0;                #input voltage, mV\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "mv=R1/(R1+R2);              #feedback fraction\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=round(Av/(1+Av*mv),0);           #Voltage gain with feedback\n",
-    "\n",
-    "#(iii)\n",
-    "Vout=Avf*Vin;               #Output voltage, mV\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)   Feedback fraction=%.1f.\"%mv);\n",
-    "print(\"(ii)  Voltage gain with feedback=%d.\"%Avf);\n",
-    "print(\"(iii) Output voltage=%dmV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.12 : Page number 345-346"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)   Feedback fraction=0.1.\n",
-      "(ii)  The voltage gain with feedback=10.\n",
-      "(iii) Increased input impedance due to negative feedback=10 mega ohm\n",
-      "(iv)  Decreased output impedance due to negative feedback=0.1 ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=10000.0;             #Volateg gain without feedback\n",
-    "Zin=10.0;               #Input impedance, kilo ohm\n",
-    "Zout=100.0;             #Output impedance, ohm\n",
-    "R1=10.0;                #Resistor R1, kilo ohm\n",
-    "R2=90.0;                #Resistor R2, kilo ohm\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "mv=R1/(R1+R2);                  #Feedback fraction\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=round(Av/(1+Av*mv),0);               #Voltage gain with feedback\n",
-    "\n",
-    "#(iii)\n",
-    "Zin_feedback=((1+Av*mv)*Zin)/1000;             #Increased input impedance due to negative feedback, mega ohm\n",
-    "\n",
-    "#(iv)\n",
-    "Zout_feedback=Zout/(1+Av*mv);             #Decreased output impedance due to negative feedback, ohm\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)   Feedback fraction=%.1f.\"%mv);\n",
-    "print(\"(ii)  The voltage gain with feedback=%d.\"%Avf);\n",
-    "print(\"(iii) Increased input impedance due to negative feedback=%.0f mega ohm\"%Zin_feedback);\n",
-    "print(\"(iv)  Decreased output impedance due to negative feedback=%.1f ohm.\"%Zout_feedback);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.13 : Page number 346"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Distortion with negative feedback=0.312%\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=150.0;                       #Voltage gain\n",
-    "D=5/100.0;                        #Distortion\n",
-    "mv=10/100.0;                      #Feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "Dvf=round((D/(1+Av*mv))*100,3);                #Distortion with negative feedback\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Distortion with negative feedback=%.3f%%\"%Dvf);\n",
-    "\n",
-    "#Note: In the text, value of Dvf=0.3125% has been approximated to 0.313%. But, here the approximation is done to 0.312%\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.14 : Page number 346"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The new lower cut-off frequency=136.4Hz\n",
-      "The new upper cut-off frequency=5.52MHz\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=1000.0;                      #Voltage gain\n",
-    "f1=1.5;                         #Lower cut-off frequency, kHz\n",
-    "f2=501.5;                       #Upper cut-off frequency, kHz\n",
-    "mv=1/100.0;                       #Feedbcack fraction\n",
-    "\n",
-    "#Calculation\n",
-    "f1_f=(f1/(1+mv*Av))*1000;              #New lower cut-off frequency, Hz\n",
-    "f2_f=(f2*(1+mv*Av))/1000;              #New upper cut-off frequency, MHz\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The new lower cut-off frequency=%.1fHz\"%f1_f);\n",
-    "print(\"The new upper cut-off frequency=%.2fMHz\"%f2_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.15 : Page number 348"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The effective current gain=58.82.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=200.0;                       #Current gain without feedback\n",
-    "mi=0.012;                       #Current attenuation\n",
-    "\n",
-    "#Calculation\n",
-    "Aif=Ai/(1+Ai*mi);\n",
-    "\n",
-    "#Result\n",
-    "print(\"The effective current gain=%.2f.\"%Aif);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.16 : Page number 349"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance when negative feedback is applied=3.26 kilo ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=240.0;                       #Current gain\n",
-    "Zin=15.0;                       #Input impedance without feedback, kilo ohm\n",
-    "mi=0.015;                       #Current feedback fraction\n",
-    "\n",
-    "#Calculations\n",
-    "Zin_f=Zin/(1+mi*Ai);                #Input impedance with feedback, kilo ohm\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance when negative feedback is applied=%.2f kilo ohm\"%Zin_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.17 : Page number 349"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output impedance with negative feedback=9kilo ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=200.0;                   #Current gain without feedback\n",
-    "Zout=3.0;                   #Output impedance without feedback, kilo ohm\n",
-    "mi=0.01;                    #current feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "Zout_f=Zout*(1+mi*Ai);              #Output impedance with negative feedback, kilo ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output impedance with negative feedback=%dkilo ohm.\"%Zout_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.18 : Page number 349"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Bandwidth when negative feedback is applied=1400kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=250.0;                   #Current gain without feedback\n",
-    "BW=400.0;                   #Bandwidth, kHz\n",
-    "mi=0.01;                    #current feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "BW_f=BW*(1+mi*Ai);              #Bandwidth when negative feedback is applied, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Bandwidth when negative feedback is applied=%dkHz.\"%BW_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.19 : Page number 350-351"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Value of VE=9.72V and IE=10.68mA\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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N7HlgcfTco3IWmcRuXXXZLbeEm2+G+++HG26Arl3h2WfrJ7a0UY07XhrP5NWkgfxHM3sM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\nAxERQclARESA/w8INYcwb/NOegAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f9a7522f0b8>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pylab as p\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=18.0;                   #Supply voltage, V\n",
-    "R1=16.0;                    #Resistor R1, kilo ohm\n",
-    "R2=22.0;                    #Resistor R2, kilo ohm\n",
-    "RE=910.0;                   #Emitter resistor, ohm\n",
-    "VBE=0.7;                    #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculations\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2-VBE;                         #Emitter voltage, V\n",
-    "IE=(VE/RE)*1000;                   #Emitter current, mA (OHM's LAW)\n",
-    "\n",
-    "#D.C load line\n",
-    "IC_sat=(VCC/RE)*1000;                      #Collector saturation current, mA\n",
-    "VCE_off=VCC;                               #Collector-emitter voltage in off state, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Value of VE=%.2fV and IE=%.2fmA\"%(VE,IE));\n",
-    "\n",
-    "#Plotting\n",
-    "VCE_plot=[0,VCE_off];            #Plotting variable for VCE\n",
-    "IC_plot=[IC_sat,0];              #Plotting variable for IC\n",
-    "p.plot(VCE_plot,IC_plot);\n",
-    "p.xlim(0,20)\n",
-    "p.ylim(0,25)\n",
-    "p.xlabel('VCE(V)');\n",
-    "p.ylabel('IC(mA)');\n",
-    "p.title('d.c load line');\n",
-    "p.grid();\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.20 : Page number 352"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the emitter follower circuit=0.994.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                   #Supply voltage, V\n",
-    "R1=10.0;                    #Resistor R1, kilo ohm\n",
-    "R2=10.0;                    #Resistor R2, kilo ohm\n",
-    "RE=5.0;                     #Emitter resistance, kilo ohm\n",
-    "VBE=0.7;                    #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2-VBE;                         #Emitter voltage, V\n",
-    "IE=(VE/RE);                   #Emitter current, mA (OHM's LAW)\n",
-    "re=25/IE;                          #a.c emitter resistance, ohm\n",
-    "Av=RE*1000/(re+RE*1000);                     #Voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain of the emitter follower circuit=%.3f.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.21 : Page number 352-353"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain=0.988\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RE=5.0;                 #Emitter resistance, kilo ohm\n",
-    "re=29.1;                #a.c emitter resistance, ohm\n",
-    "RL=5.0;                 #Load resistance, kilo ohm\n",
-    "\n",
-    "#Calculation\n",
-    "RE_ac=(RE*RL)/(RE+RL);               #New effective value of emitter resistance, kilo ohm\n",
-    "Av=RE_ac*1000/(re+RE_ac*1000);     #Voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain=%.3f\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.22 : Page number 354"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance of the emitter follower =4.96 kilo ohm\n",
-      "The approximate value of the input impedance=5 kilo ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "def pr(r1,r2):                     #Function for calculating parallel resistance\n",
-    "    return (r1*r2)/(r1+r2);\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=10.0;                   #Supply voltage, V\n",
-    "R1=10.0;                    #Resistor R1, kilo ohm\n",
-    "R2=10.0;                    #Resistor R2, kilo ohm\n",
-    "RE=4.3;                     #Emitter resistor, kilo ohm\n",
-    "RL=10.0;                    #Load resistance, kilo ohm\n",
-    "VBE=0.7;                    #Base-emitter voltage, V\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2-VBE;                         #Emitter voltage, V\n",
-    "IE=(VE/RE);                        #Emitter current, mA (OHM's LAW)\n",
-    "re=25/IE;                          #a.c emitter resistance, ohm\n",
-    "RE_eff=pr(RE,RL);                  #Effective external emitter resistance, kilo ohm\n",
-    "Zin_base=beta*(re/1000+RE_eff);         #Input impedance of the base of the transistor, kilo ohm\n",
-    "Zin=pr(pr(R1,R2),Zin_base);        #Input impedance of emitter follower, kilo ohm\n",
-    "#Approximate value of input impedance taken as parallel resistance of R1 and R2 and ignoring Zin_base due to its relatively large value\n",
-    "Zin_approx=pr(R1,R2);              #Approximate input impedance, kilo ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance of the emitter follower =%.2f kilo ohm\"%Zin);\n",
-    "print(\"The approximate value of the input impedance=%d kilo ohm\"%Zin_approx);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.23 : Page number 355"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output impedance=22.3 ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "def pr(r1,r2):                     #Function for calculating parallel resistance\n",
-    "    return (r1*r2)/(r1+r2);\n",
-    "\n",
-    "\n",
-    "#Variable declaration\n",
-    "re=20.0;                    #a.c emitter resistance, ohm\n",
-    "R1=3.0;                     #Resistor R1, kilo ohm\n",
-    "R2=4.7;                     #Resistor R2, kilo ohm\n",
-    "RS=600.0;                   #Source resistance, kilo ohm\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "Rin_ac=pr(pr(R1,R2)*1000,RS);            #Input a.c resistance, ohm\n",
-    "Zout=re + Rin_ac/beta;                   #Output impedance, ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output impedance=%.1f ohm\"%Zout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.24 : Page number  358"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  d.c value of current in RE=1.09mA\n",
-      "(ii) Input impedance=16.17 mega ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                    #Supply voltage, V\n",
-    "R1=120.0;                    #Resistor R1, kilo ohm\n",
-    "R2=120.0;                    #Resistor R2, kilo ohm\n",
-    "RE=3.3;                      #Emitter resistor, kilo ohm\n",
-    "VBE=0.7;                     #Base-emitter voltage, V\n",
-    "beta_1=70.0;                 #Base current amplification factor of 1st transistor\n",
-    "beta_2=70.0;                 #Base current amplification factor of 2nd transistor\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "IE_2=(V2-2*VBE)/RE;                #Emitter current, mA (OHM's LAW)\n",
-    "\n",
-    "#(ii)\n",
-    "Zin=(beta_1*beta_2*RE)/1000;               #Input impedance, mega ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  d.c value of current in RE=%.2fmA\"%IE_2);\n",
-    "print(\"(ii) Input impedance=%.2f mega ohm.\"%Zin);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.25 : Page number 358-359"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 28,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) D.C Bias levels: \n",
-      " VB1= 4V, VE1=3.3V, VB2=3.3V, VE2=2.6V, IE2=1.3mA and IE1=0.013mA.\n",
-      "(ii) A.C Analysis:   \n",
-      " re1=1923 ohm and re2=19.23 ohm \n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12.0;                    #Supply voltage, V\n",
-    "R1=20.0;                     #Resistor R1, kilo ohm\n",
-    "R2=10.0;                     #Resistor R2, kilo ohm\n",
-    "RC=4.0;                      #Collector resistor, kilo ohm\n",
-    "RE=2.0;                      #Emitter resistor, kilo ohm\n",
-    "VBE=0.7;                     #Base-emitter voltage, V\n",
-    "beta=100.0;                  #Base current amplification factor of 1st transistor\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) D.C Bias levels\n",
-    "VB1=VCC*R2/(R1+R2);                 #Base voltage of 1st transistor, V (Voltage divider rule)\n",
-    "VE1=VB1-VBE;                        #Emitter voltage of 1st transistor, V\n",
-    "VB2=VE1;                            #Base voltage of 2nd transistor, V\n",
-    "VE2=VB2-VBE;                        #Emitter voltage of 2nd transistor, V\n",
-    "IE2=VE2/RE;                         #Emitter current of 2nd transistor, mA (OHM' LAW)\n",
-    "IE1=IE2/beta;                       #Emitter current of 1st transistor, mA (IE~IC=beta*IB, here IB2=IE1)\n",
-    "\n",
-    "#(ii) A.C analysis\n",
-    "re1=25/IE1;                         #a.c emitter resistance of 1st transistor\n",
-    "re2=25/IE2;                         #a.c emitter resistance of 2nd transistor\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) D.C Bias levels: \\n VB1= %dV, VE1=%.1fV, VB2=%.1fV, VE2=%.1fV, IE2=%.1fmA and IE1=%.3fmA.\"%(VB1,VE1,VB2,VE2,IE2,IE1));\n",
-    "print(\"(ii) A.C Analysis:   \\n re1=%d ohm and re2=%.2f ohm \"%(re1,re2));\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_4.ipynb
deleted file mode 100755
index 7d378f3d..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_4.ipynb
+++ /dev/null
@@ -1,1148 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 13: AMPLIFIERS WITH NEGATIVE FEEDBACK"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.1 : Page number 338"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the amplifier with negative feedback=97.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=3000.0;              #Voltage gain without feedback\n",
-    "m_v=0.01;               #Feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=Av/(1+Av*m_v);                  #Voltage gain of the amplifier with negative feedback\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain of the amplifier with negative feedback=%.0f.\"%Avf);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.2 : Page number  339"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The fraction of output fedback to the input=1/20.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=140.0;                   #Voltage gain\n",
-    "Avf=17.5;                   #Voltage gain with negative feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Avf=Av/(1+Av*mv), so,\n",
-    "mv=(Av-Avf)/(Av*Avf);                 #Fraction of output fedback to the input\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The fraction of output fedback to the input=1/%.0f.\"%(1.0/mv));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.3 : Page number 339"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The fraction of output fedback to input=0.01.\n",
-      "(ii) The required amplifier gain for overall gain to be 75=300.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=100.0;                   #Voltage gain\n",
-    "Avf=50.0;                   #Voltage gain with negative feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);       #The fraction of output fedback to input\n",
-    "\n",
-    "#(ii) Overall gain is to be 75:\n",
-    "Avf=75.0;                           #The required overall gain\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Av=Avf/(1-Avf*mv);                  #The required value of amplifier gain\n",
-    "\n",
-    "#result\n",
-    "print(\"(i)  The fraction of output fedback to input=%.2f.\"%mv);\n",
-    "print(\"(ii) The required amplifier gain for overall gain to be 75=%d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.4 : Page number 339-340"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The voltage gain without feedback=40.\n",
-      "(ii) The feedback fraction = 1/40.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vout=10.0;              #output voltage , V\n",
-    "Vin_f=0.5;              #Input votage for amplifier with feedback, V\n",
-    "Vin=0.25;                #Input votage for amplifier without feedback, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "Av=Vout/Vin;                #Voltage gain without negative feedback\n",
-    "\n",
-    "#(ii)\n",
-    "Avf=Vout/Vin_f;             #Voltage gain with negative feedback\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);       #Feedback fraction\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  The voltage gain without feedback=%d.\"%Av);\n",
-    "print(\"(ii) The feedback fraction = 1/%d.\"%(1/mv));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.5 : Page number 340"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The percentage of reduction in stage gain without feedback=20%.\n",
-      "(ii) The percentage of reduction in net gain with feedback=11.2%\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=50.0;                #Gain without feedback\n",
-    "Avf=25.0;               #Gain with negative feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);               #Feedback fraction\n",
-    "\n",
-    "#(i)\n",
-    "#percentage of reduction without feedback\n",
-    "Av_reduced=40.0;                                        #Reduced amplifier gain due to ageing\n",
-    "percentage_of_reduction=((Av-Av_reduced)/Av)*100;       #Percentage of reduction in stage gain\n",
-    "\n",
-    "print(\"(i)  The percentage of reduction in stage gain without feedback=%d%%.\"%percentage_of_reduction);\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf_reduced=round(Av_reduced/(1+mv*Av_reduced),1);               #Reduced net gain with negative feedback \n",
-    "percentage_of_reduction_f=((Avf-Avf_reduced)/Avf)*100;          #Percentage of reduction in net gain with feedback\n",
-    "\n",
-    "print(\"(ii) The percentage of reduction in net gain with feedback=%.1f%%\"%percentage_of_reduction_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.6 : Page number 340"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The percentage change in system gain=8.36%\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=100.0;               #Gain\n",
-    "mv=0.1;                 #feedback fraction\n",
-    "Av_fall=6.0;            #fall in gain, dB\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=round(Av/(1+Av*mv),2);           #Total system gain with feedback\n",
-    "\n",
-    "#Since, fall in gain=20*log10(Av/Av_1)\n",
-    "Av1=round(Av/10**(Av_fall/20),0);            #New absolute voltage gain without feedback\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf_new=round(Av1/(1+Av1*mv),2);             #New net system gain with feedback\n",
-    "\n",
-    "percentage_change=((Avf-Avf_new)/Avf)*100;          #Percentage change in system gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"The percentage change in system gain=%.2f%%\"%percentage_change);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.7 : Page number 341"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The feedback fraction=0.008.\n",
-      "The percentage fall in system gain=4.8%.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=500.0;                       #Voltage gain without feedback\n",
-    "Avf=100.0;                      #Voltage gain with negative feedback\n",
-    "Av_fall_percentage=20.0;          #Gain fall percentage due to ageing\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);                                           #Feedback fraction\n",
-    "Av_reduced=((100-Av_fall_percentage)/100)*Av;                   #Reduced voltage gain\n",
-    "Avf_reduced=round(Av_reduced/(1+Av_reduced*mv),1);                       #Reduced total gain of the system\n",
-    "percentage_fall=((Avf-Avf_reduced)/Avf)*100;                    #Percentage of fall in total system gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"The feedback fraction=%.3f.\"%mv);\n",
-    "print(\"The percentage fall in system gain=%.1f%%.\"%percentage_fall);\n",
-    "\n",
-    "#Note: The percentage gain is calculated in the text as 4.7% due to approximation of Avf to 95.3 whose actual approximation will be (95.238)~95.2. So, the percentage fall calculated here is 4.8%\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.8 : Page number 341"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain with feedback=31622.\n",
-      "The feedback fraction=2.16e-05.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "Av=100000.0;                        #Open loop voltage gain\n",
-    "f_dB=10.0;                          #Negative feedback, dB\n",
-    "\n",
-    "#Calculation\n",
-    "Av_dB=20*log10(Av);                        #dB voltage gain without feedback, dB\n",
-    "Avf_dB=Av_dB-f_dB;                         #dB voltage gain with feedback, dB\n",
-    "Avf=10**(Avf_dB/20);                       #Voltage gain with feedback\n",
-    "\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);                       #feedback fraction\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain with feedback=%d.\"%Avf);\n",
-    "print(\"The feedback fraction=%.2e.\"%mv);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.9 : Page number 341-342"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain with feedback=47.4.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ao=1000.0;                        #Open circuit voltage gain\n",
-    "Rout=100.0;                       #Output resistance, ohm\n",
-    "RL=900.0;                         #Resistive load, ohm\n",
-    "mv=1/50;                          #feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Av=Ao*RL/(Rout+RL)\n",
-    "Av=Ao*RL/(Rout+RL);                     #Voltage gain without feedback\n",
-    "Avf=Av/(1+Av*mv);                       #Voltage gain with feedback\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain with feedback=%.1f.\"%Avf);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.10 : Page number 342"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "100=Av/(1+Av*mv)        ------Eq. 1\n",
-      "99=0.8*Av/(1+0.8*Av*mv)        ------Eq. 2\n",
-      "99 + 79.2*Av*mv=0.8Av        ------Eq. 3 from Eq. 2\n",
-      "79.2 + 79.2*Av*mv=0.792Av        ------Eq. 4 from Eq. 1\n",
-      "Subtracting Eq.4 from Eq.3\n",
-      "19.8 = 0.008*Av\n",
-      "Av=2475.\n",
-      "mv=0.0096.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Avf=100.0;                    #Voltage gain with feedback\n",
-    "vary_f=1;                     #Vary percentage in voltage gain with feedback\n",
-    "vary_wf=20;                   #Vary percentage in voltage gain without feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#Avf=Av/(1+Av*mv)\n",
-    "print(\"%d=Av/(1+Av*mv)        ------Eq. 1\"%Avf);         #Equation 1\n",
-    "\n",
-    "#considering variation in gains\n",
-    "Avf_vary=Avf*(1- vary_f/100.0);               #Gain with feedback, considering variation\n",
-    "print(\"%d=%.1f*Av/(1+%.1f*Av*mv)        ------Eq. 2\"%(Avf_vary,(1-vary_wf/100.0),(1-vary_wf/100.0)));     #Equation 2\n",
-    "\n",
-    "#Solving the above two equations\n",
-    "print(\"%d + %.1f*Av*mv=%.1fAv        ------Eq. 3 from Eq. 2\"%(Avf_vary,Avf_vary*(1-vary_wf/100.0),(1-vary_wf/100.0)));     #Equation 3\n",
-    "\n",
-    "#multiplying Eq. 1 with (Avf_vary*(1-vary_wf/100.0))/100=0.792\n",
-    "print(\"%.1f + %.1f*Av*mv=%.3fAv        ------Eq. 4 from Eq. 1\"%(Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf_vary*(1-vary_wf/100.0)/100.0));     #Equation 4\n",
-    "\n",
-    "print(\"Subtracting Eq.4 from Eq.3\" );\n",
-    "print(\"%.1f = %.3f*Av\"%(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0,(1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0));\n",
-    "Av=(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0)/((1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0);\n",
-    "print(\"Av=%.0f.\"%Av);\n",
-    "mv=(Av-Avf)/(Av*Avf);\n",
-    "print(\"mv=%.4f.\"%mv);\n",
-    "      "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.11 : Page number 345"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)   Feedback fraction=0.1.\n",
-      "(ii)  Voltage gain with feedback=10.\n",
-      "(iii) Output voltage=10mV.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=10000.0;             #Volage gain without feedback\n",
-    "R1=2.0;                 #Resistor R1, kilo ohm\n",
-    "R2=18.0;                #Resistor R2, kilo ohm\n",
-    "Vin=1.0;                #input voltage, mV\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "mv=R1/(R1+R2);              #feedback fraction\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=round(Av/(1+Av*mv),0);           #Voltage gain with feedback\n",
-    "\n",
-    "#(iii)\n",
-    "Vout=Avf*Vin;               #Output voltage, mV\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)   Feedback fraction=%.1f.\"%mv);\n",
-    "print(\"(ii)  Voltage gain with feedback=%d.\"%Avf);\n",
-    "print(\"(iii) Output voltage=%dmV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.12 : Page number 345-346"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)   Feedback fraction=0.1.\n",
-      "(ii)  The voltage gain with feedback=10.\n",
-      "(iii) Increased input impedance due to negative feedback=10 mega ohm\n",
-      "(iv)  Decreased output impedance due to negative feedback=0.1 ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=10000.0;             #Volateg gain without feedback\n",
-    "Zin=10.0;               #Input impedance, kilo ohm\n",
-    "Zout=100.0;             #Output impedance, ohm\n",
-    "R1=10.0;                #Resistor R1, kilo ohm\n",
-    "R2=90.0;                #Resistor R2, kilo ohm\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "mv=R1/(R1+R2);                  #Feedback fraction\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=round(Av/(1+Av*mv),0);               #Voltage gain with feedback\n",
-    "\n",
-    "#(iii)\n",
-    "Zin_feedback=((1+Av*mv)*Zin)/1000;             #Increased input impedance due to negative feedback, mega ohm\n",
-    "\n",
-    "#(iv)\n",
-    "Zout_feedback=Zout/(1+Av*mv);             #Decreased output impedance due to negative feedback, ohm\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)   Feedback fraction=%.1f.\"%mv);\n",
-    "print(\"(ii)  The voltage gain with feedback=%d.\"%Avf);\n",
-    "print(\"(iii) Increased input impedance due to negative feedback=%.0f mega ohm\"%Zin_feedback);\n",
-    "print(\"(iv)  Decreased output impedance due to negative feedback=%.1f ohm.\"%Zout_feedback);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.13 : Page number 346"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Distortion with negative feedback=0.312%\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=150.0;                       #Voltage gain\n",
-    "D=5/100.0;                        #Distortion\n",
-    "mv=10/100.0;                      #Feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "Dvf=round((D/(1+Av*mv))*100,3);                #Distortion with negative feedback\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Distortion with negative feedback=%.3f%%\"%Dvf);\n",
-    "\n",
-    "#Note: In the text, value of Dvf=0.3125% has been approximated to 0.313%. But, here the approximation is done to 0.312%\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.14 : Page number 346"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The new lower cut-off frequency=136.4Hz\n",
-      "The new upper cut-off frequency=5.52MHz\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=1000.0;                      #Voltage gain\n",
-    "f1=1.5;                         #Lower cut-off frequency, kHz\n",
-    "f2=501.5;                       #Upper cut-off frequency, kHz\n",
-    "mv=1/100.0;                       #Feedbcack fraction\n",
-    "\n",
-    "#Calculation\n",
-    "f1_f=(f1/(1+mv*Av))*1000;              #New lower cut-off frequency, Hz\n",
-    "f2_f=(f2*(1+mv*Av))/1000;              #New upper cut-off frequency, MHz\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The new lower cut-off frequency=%.1fHz\"%f1_f);\n",
-    "print(\"The new upper cut-off frequency=%.2fMHz\"%f2_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.15 : Page number 348"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The effective current gain=58.82.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=200.0;                       #Current gain without feedback\n",
-    "mi=0.012;                       #Current attenuation\n",
-    "\n",
-    "#Calculation\n",
-    "Aif=Ai/(1+Ai*mi);\n",
-    "\n",
-    "#Result\n",
-    "print(\"The effective current gain=%.2f.\"%Aif);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.16 : Page number 349"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance when negative feedback is applied=3.26 kilo ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=240.0;                       #Current gain\n",
-    "Zin=15.0;                       #Input impedance without feedback, kilo ohm\n",
-    "mi=0.015;                       #Current feedback fraction\n",
-    "\n",
-    "#Calculations\n",
-    "Zin_f=Zin/(1+mi*Ai);                #Input impedance with feedback, kilo ohm\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance when negative feedback is applied=%.2f kilo ohm\"%Zin_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.17 : Page number 349"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output impedance with negative feedback=9kilo ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=200.0;                   #Current gain without feedback\n",
-    "Zout=3.0;                   #Output impedance without feedback, kilo ohm\n",
-    "mi=0.01;                    #current feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "Zout_f=Zout*(1+mi*Ai);              #Output impedance with negative feedback, kilo ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output impedance with negative feedback=%dkilo ohm.\"%Zout_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.18 : Page number 349"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Bandwidth when negative feedback is applied=1400kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=250.0;                   #Current gain without feedback\n",
-    "BW=400.0;                   #Bandwidth, kHz\n",
-    "mi=0.01;                    #current feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "BW_f=BW*(1+mi*Ai);              #Bandwidth when negative feedback is applied, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Bandwidth when negative feedback is applied=%dkHz.\"%BW_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.19 : Page number 350-351"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Value of VE=9.72V and IE=10.68mA\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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N7HlgcfTco3IWmcRuXXXZLbeEm2+G+++HG26Arl3h2WfrJ7a0UY07XhrP5NWkgfxHM3sM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\nAxERQclARESA/w8INYcwb/NOegAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f84ace55f98>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as p\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=18.0;                   #Supply voltage, V\n",
-    "R1=16.0;                    #Resistor R1, kilo ohm\n",
-    "R2=22.0;                    #Resistor R2, kilo ohm\n",
-    "RE=910.0;                   #Emitter resistor, ohm\n",
-    "VBE=0.7;                    #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculations\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2-VBE;                         #Emitter voltage, V\n",
-    "IE=(VE/RE)*1000;                   #Emitter current, mA (OHM's LAW)\n",
-    "\n",
-    "#D.C load line\n",
-    "IC_sat=(VCC/RE)*1000;                      #Collector saturation current, mA\n",
-    "VCE_off=VCC;                               #Collector-emitter voltage in off state, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Value of VE=%.2fV and IE=%.2fmA\"%(VE,IE));\n",
-    "\n",
-    "#Plotting\n",
-    "VCE_plot=[0,VCE_off];            #Plotting variable for VCE\n",
-    "IC_plot=[IC_sat,0];              #Plotting variable for IC\n",
-    "p.plot(VCE_plot,IC_plot);\n",
-    "p.xlim(0,20)\n",
-    "p.ylim(0,25)\n",
-    "p.xlabel('VCE(V)');\n",
-    "p.ylabel('IC(mA)');\n",
-    "p.title('d.c load line');\n",
-    "p.grid();\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.20 : Page number 352"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the emitter follower circuit=0.994.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                   #Supply voltage, V\n",
-    "R1=10.0;                    #Resistor R1, kilo ohm\n",
-    "R2=10.0;                    #Resistor R2, kilo ohm\n",
-    "RE=5.0;                     #Emitter resistance, kilo ohm\n",
-    "VBE=0.7;                    #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2-VBE;                         #Emitter voltage, V\n",
-    "IE=(VE/RE);                   #Emitter current, mA (OHM's LAW)\n",
-    "re=25/IE;                          #a.c emitter resistance, ohm\n",
-    "Av=RE*1000/(re+RE*1000);                     #Voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain of the emitter follower circuit=%.3f.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.21 : Page number 352-353"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain=0.988\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RE=5.0;                 #Emitter resistance, kilo ohm\n",
-    "re=29.1;                #a.c emitter resistance, ohm\n",
-    "RL=5.0;                 #Load resistance, kilo ohm\n",
-    "\n",
-    "#Calculation\n",
-    "RE_ac=(RE*RL)/(RE+RL);               #New effective value of emitter resistance, kilo ohm\n",
-    "Av=RE_ac*1000/(re+RE_ac*1000);     #Voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain=%.3f\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.22 : Page number 354"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance of the emitter follower =4.96 kilo ohm\n",
-      "The approximate value of the input impedance=5 kilo ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "def pr(r1,r2):                     #Function for calculating parallel resistance\n",
-    "    return (r1*r2)/(r1+r2);\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=10.0;                   #Supply voltage, V\n",
-    "R1=10.0;                    #Resistor R1, kilo ohm\n",
-    "R2=10.0;                    #Resistor R2, kilo ohm\n",
-    "RE=4.3;                     #Emitter resistor, kilo ohm\n",
-    "RL=10.0;                    #Load resistance, kilo ohm\n",
-    "VBE=0.7;                    #Base-emitter voltage, V\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2-VBE;                         #Emitter voltage, V\n",
-    "IE=(VE/RE);                        #Emitter current, mA (OHM's LAW)\n",
-    "re=25/IE;                          #a.c emitter resistance, ohm\n",
-    "RE_eff=pr(RE,RL);                  #Effective external emitter resistance, kilo ohm\n",
-    "Zin_base=beta*(re/1000+RE_eff);         #Input impedance of the base of the transistor, kilo ohm\n",
-    "Zin=pr(pr(R1,R2),Zin_base);        #Input impedance of emitter follower, kilo ohm\n",
-    "#Approximate value of input impedance taken as parallel resistance of R1 and R2 and ignoring Zin_base due to its relatively large value\n",
-    "Zin_approx=pr(R1,R2);              #Approximate input impedance, kilo ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance of the emitter follower =%.2f kilo ohm\"%Zin);\n",
-    "print(\"The approximate value of the input impedance=%d kilo ohm\"%Zin_approx);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.23 : Page number 355"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output impedance=22.3 ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "def pr(r1,r2):                     #Function for calculating parallel resistance\n",
-    "    return (r1*r2)/(r1+r2);\n",
-    "\n",
-    "\n",
-    "#Variable declaration\n",
-    "re=20.0;                    #a.c emitter resistance, ohm\n",
-    "R1=3.0;                     #Resistor R1, kilo ohm\n",
-    "R2=4.7;                     #Resistor R2, kilo ohm\n",
-    "RS=600.0;                   #Source resistance, kilo ohm\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "Rin_ac=pr(pr(R1,R2)*1000,RS);            #Input a.c resistance, ohm\n",
-    "Zout=re + Rin_ac/beta;                   #Output impedance, ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output impedance=%.1f ohm\"%Zout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.24 : Page number  358"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  d.c value of current in RE=1.09mA\n",
-      "(ii) Input impedance=16.17 mega ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                    #Supply voltage, V\n",
-    "R1=120.0;                    #Resistor R1, kilo ohm\n",
-    "R2=120.0;                    #Resistor R2, kilo ohm\n",
-    "RE=3.3;                      #Emitter resistor, kilo ohm\n",
-    "VBE=0.7;                     #Base-emitter voltage, V\n",
-    "beta_1=70.0;                 #Base current amplification factor of 1st transistor\n",
-    "beta_2=70.0;                 #Base current amplification factor of 2nd transistor\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "IE_2=(V2-2*VBE)/RE;                #Emitter current, mA (OHM's LAW)\n",
-    "\n",
-    "#(ii)\n",
-    "Zin=(beta_1*beta_2*RE)/1000;               #Input impedance, mega ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  d.c value of current in RE=%.2fmA\"%IE_2);\n",
-    "print(\"(ii) Input impedance=%.2f mega ohm.\"%Zin);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.25 : Page number 358-359"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) D.C Bias levels: \n",
-      " VB1= 4V, VE1=3.3V, VB2=3.3V, VE2=2.6V, IE2=1.3mA and IE1=0.013mA.\n",
-      "(ii) A.C Analysis:   \n",
-      " re1=1923 ohm and re2=19.23 ohm \n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12.0;                    #Supply voltage, V\n",
-    "R1=20.0;                     #Resistor R1, kilo ohm\n",
-    "R2=10.0;                     #Resistor R2, kilo ohm\n",
-    "RC=4.0;                      #Collector resistor, kilo ohm\n",
-    "RE=2.0;                      #Emitter resistor, kilo ohm\n",
-    "VBE=0.7;                     #Base-emitter voltage, V\n",
-    "beta=100.0;                  #Base current amplification factor of 1st transistor\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) D.C Bias levels\n",
-    "VB1=VCC*R2/(R1+R2);                 #Base voltage of 1st transistor, V (Voltage divider rule)\n",
-    "VE1=VB1-VBE;                        #Emitter voltage of 1st transistor, V\n",
-    "VB2=VE1;                            #Base voltage of 2nd transistor, V\n",
-    "VE2=VB2-VBE;                        #Emitter voltage of 2nd transistor, V\n",
-    "IE2=VE2/RE;                         #Emitter current of 2nd transistor, mA (OHM' LAW)\n",
-    "IE1=IE2/beta;                       #Emitter current of 1st transistor, mA (IE~IC=beta*IB, here IB2=IE1)\n",
-    "\n",
-    "#(ii) A.C analysis\n",
-    "re1=25/IE1;                         #a.c emitter resistance of 1st transistor\n",
-    "re2=25/IE2;                         #a.c emitter resistance of 2nd transistor\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) D.C Bias levels: \\n VB1= %dV, VE1=%.1fV, VB2=%.1fV, VE2=%.1fV, IE2=%.1fmA and IE1=%.3fmA.\"%(VB1,VE1,VB2,VE2,IE2,IE1));\n",
-    "print(\"(ii) A.C Analysis:   \\n re1=%d ohm and re2=%.2f ohm \"%(re1,re2));\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_5.ipynb
deleted file mode 100755
index 7d378f3d..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_5.ipynb
+++ /dev/null
@@ -1,1148 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 13: AMPLIFIERS WITH NEGATIVE FEEDBACK"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.1 : Page number 338"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the amplifier with negative feedback=97.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=3000.0;              #Voltage gain without feedback\n",
-    "m_v=0.01;               #Feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=Av/(1+Av*m_v);                  #Voltage gain of the amplifier with negative feedback\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain of the amplifier with negative feedback=%.0f.\"%Avf);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.2 : Page number  339"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The fraction of output fedback to the input=1/20.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=140.0;                   #Voltage gain\n",
-    "Avf=17.5;                   #Voltage gain with negative feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Avf=Av/(1+Av*mv), so,\n",
-    "mv=(Av-Avf)/(Av*Avf);                 #Fraction of output fedback to the input\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The fraction of output fedback to the input=1/%.0f.\"%(1.0/mv));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.3 : Page number 339"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The fraction of output fedback to input=0.01.\n",
-      "(ii) The required amplifier gain for overall gain to be 75=300.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=100.0;                   #Voltage gain\n",
-    "Avf=50.0;                   #Voltage gain with negative feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);       #The fraction of output fedback to input\n",
-    "\n",
-    "#(ii) Overall gain is to be 75:\n",
-    "Avf=75.0;                           #The required overall gain\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Av=Avf/(1-Avf*mv);                  #The required value of amplifier gain\n",
-    "\n",
-    "#result\n",
-    "print(\"(i)  The fraction of output fedback to input=%.2f.\"%mv);\n",
-    "print(\"(ii) The required amplifier gain for overall gain to be 75=%d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.4 : Page number 339-340"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The voltage gain without feedback=40.\n",
-      "(ii) The feedback fraction = 1/40.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vout=10.0;              #output voltage , V\n",
-    "Vin_f=0.5;              #Input votage for amplifier with feedback, V\n",
-    "Vin=0.25;                #Input votage for amplifier without feedback, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "Av=Vout/Vin;                #Voltage gain without negative feedback\n",
-    "\n",
-    "#(ii)\n",
-    "Avf=Vout/Vin_f;             #Voltage gain with negative feedback\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);       #Feedback fraction\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  The voltage gain without feedback=%d.\"%Av);\n",
-    "print(\"(ii) The feedback fraction = 1/%d.\"%(1/mv));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.5 : Page number 340"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The percentage of reduction in stage gain without feedback=20%.\n",
-      "(ii) The percentage of reduction in net gain with feedback=11.2%\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=50.0;                #Gain without feedback\n",
-    "Avf=25.0;               #Gain with negative feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);               #Feedback fraction\n",
-    "\n",
-    "#(i)\n",
-    "#percentage of reduction without feedback\n",
-    "Av_reduced=40.0;                                        #Reduced amplifier gain due to ageing\n",
-    "percentage_of_reduction=((Av-Av_reduced)/Av)*100;       #Percentage of reduction in stage gain\n",
-    "\n",
-    "print(\"(i)  The percentage of reduction in stage gain without feedback=%d%%.\"%percentage_of_reduction);\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf_reduced=round(Av_reduced/(1+mv*Av_reduced),1);               #Reduced net gain with negative feedback \n",
-    "percentage_of_reduction_f=((Avf-Avf_reduced)/Avf)*100;          #Percentage of reduction in net gain with feedback\n",
-    "\n",
-    "print(\"(ii) The percentage of reduction in net gain with feedback=%.1f%%\"%percentage_of_reduction_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.6 : Page number 340"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The percentage change in system gain=8.36%\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=100.0;               #Gain\n",
-    "mv=0.1;                 #feedback fraction\n",
-    "Av_fall=6.0;            #fall in gain, dB\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=round(Av/(1+Av*mv),2);           #Total system gain with feedback\n",
-    "\n",
-    "#Since, fall in gain=20*log10(Av/Av_1)\n",
-    "Av1=round(Av/10**(Av_fall/20),0);            #New absolute voltage gain without feedback\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf_new=round(Av1/(1+Av1*mv),2);             #New net system gain with feedback\n",
-    "\n",
-    "percentage_change=((Avf-Avf_new)/Avf)*100;          #Percentage change in system gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"The percentage change in system gain=%.2f%%\"%percentage_change);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.7 : Page number 341"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The feedback fraction=0.008.\n",
-      "The percentage fall in system gain=4.8%.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=500.0;                       #Voltage gain without feedback\n",
-    "Avf=100.0;                      #Voltage gain with negative feedback\n",
-    "Av_fall_percentage=20.0;          #Gain fall percentage due to ageing\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);                                           #Feedback fraction\n",
-    "Av_reduced=((100-Av_fall_percentage)/100)*Av;                   #Reduced voltage gain\n",
-    "Avf_reduced=round(Av_reduced/(1+Av_reduced*mv),1);                       #Reduced total gain of the system\n",
-    "percentage_fall=((Avf-Avf_reduced)/Avf)*100;                    #Percentage of fall in total system gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"The feedback fraction=%.3f.\"%mv);\n",
-    "print(\"The percentage fall in system gain=%.1f%%.\"%percentage_fall);\n",
-    "\n",
-    "#Note: The percentage gain is calculated in the text as 4.7% due to approximation of Avf to 95.3 whose actual approximation will be (95.238)~95.2. So, the percentage fall calculated here is 4.8%\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.8 : Page number 341"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain with feedback=31622.\n",
-      "The feedback fraction=2.16e-05.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "Av=100000.0;                        #Open loop voltage gain\n",
-    "f_dB=10.0;                          #Negative feedback, dB\n",
-    "\n",
-    "#Calculation\n",
-    "Av_dB=20*log10(Av);                        #dB voltage gain without feedback, dB\n",
-    "Avf_dB=Av_dB-f_dB;                         #dB voltage gain with feedback, dB\n",
-    "Avf=10**(Avf_dB/20);                       #Voltage gain with feedback\n",
-    "\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "mv=(Av-Avf)/(Av*Avf);                       #feedback fraction\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain with feedback=%d.\"%Avf);\n",
-    "print(\"The feedback fraction=%.2e.\"%mv);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.9 : Page number 341-342"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain with feedback=47.4.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ao=1000.0;                        #Open circuit voltage gain\n",
-    "Rout=100.0;                       #Output resistance, ohm\n",
-    "RL=900.0;                         #Resistive load, ohm\n",
-    "mv=1/50;                          #feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Av=Ao*RL/(Rout+RL)\n",
-    "Av=Ao*RL/(Rout+RL);                     #Voltage gain without feedback\n",
-    "Avf=Av/(1+Av*mv);                       #Voltage gain with feedback\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain with feedback=%.1f.\"%Avf);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.10 : Page number 342"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "100=Av/(1+Av*mv)        ------Eq. 1\n",
-      "99=0.8*Av/(1+0.8*Av*mv)        ------Eq. 2\n",
-      "99 + 79.2*Av*mv=0.8Av        ------Eq. 3 from Eq. 2\n",
-      "79.2 + 79.2*Av*mv=0.792Av        ------Eq. 4 from Eq. 1\n",
-      "Subtracting Eq.4 from Eq.3\n",
-      "19.8 = 0.008*Av\n",
-      "Av=2475.\n",
-      "mv=0.0096.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Avf=100.0;                    #Voltage gain with feedback\n",
-    "vary_f=1;                     #Vary percentage in voltage gain with feedback\n",
-    "vary_wf=20;                   #Vary percentage in voltage gain without feedback\n",
-    "\n",
-    "#Calculation\n",
-    "#Avf=Av/(1+Av*mv)\n",
-    "print(\"%d=Av/(1+Av*mv)        ------Eq. 1\"%Avf);         #Equation 1\n",
-    "\n",
-    "#considering variation in gains\n",
-    "Avf_vary=Avf*(1- vary_f/100.0);               #Gain with feedback, considering variation\n",
-    "print(\"%d=%.1f*Av/(1+%.1f*Av*mv)        ------Eq. 2\"%(Avf_vary,(1-vary_wf/100.0),(1-vary_wf/100.0)));     #Equation 2\n",
-    "\n",
-    "#Solving the above two equations\n",
-    "print(\"%d + %.1f*Av*mv=%.1fAv        ------Eq. 3 from Eq. 2\"%(Avf_vary,Avf_vary*(1-vary_wf/100.0),(1-vary_wf/100.0)));     #Equation 3\n",
-    "\n",
-    "#multiplying Eq. 1 with (Avf_vary*(1-vary_wf/100.0))/100=0.792\n",
-    "print(\"%.1f + %.1f*Av*mv=%.3fAv        ------Eq. 4 from Eq. 1\"%(Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf_vary*(1-vary_wf/100.0)/100.0));     #Equation 4\n",
-    "\n",
-    "print(\"Subtracting Eq.4 from Eq.3\" );\n",
-    "print(\"%.1f = %.3f*Av\"%(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0,(1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0));\n",
-    "Av=(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0)/((1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0);\n",
-    "print(\"Av=%.0f.\"%Av);\n",
-    "mv=(Av-Avf)/(Av*Avf);\n",
-    "print(\"mv=%.4f.\"%mv);\n",
-    "      "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.11 : Page number 345"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)   Feedback fraction=0.1.\n",
-      "(ii)  Voltage gain with feedback=10.\n",
-      "(iii) Output voltage=10mV.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=10000.0;             #Volage gain without feedback\n",
-    "R1=2.0;                 #Resistor R1, kilo ohm\n",
-    "R2=18.0;                #Resistor R2, kilo ohm\n",
-    "Vin=1.0;                #input voltage, mV\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "mv=R1/(R1+R2);              #feedback fraction\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=round(Av/(1+Av*mv),0);           #Voltage gain with feedback\n",
-    "\n",
-    "#(iii)\n",
-    "Vout=Avf*Vin;               #Output voltage, mV\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)   Feedback fraction=%.1f.\"%mv);\n",
-    "print(\"(ii)  Voltage gain with feedback=%d.\"%Avf);\n",
-    "print(\"(iii) Output voltage=%dmV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.12 : Page number 345-346"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)   Feedback fraction=0.1.\n",
-      "(ii)  The voltage gain with feedback=10.\n",
-      "(iii) Increased input impedance due to negative feedback=10 mega ohm\n",
-      "(iv)  Decreased output impedance due to negative feedback=0.1 ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=10000.0;             #Volateg gain without feedback\n",
-    "Zin=10.0;               #Input impedance, kilo ohm\n",
-    "Zout=100.0;             #Output impedance, ohm\n",
-    "R1=10.0;                #Resistor R1, kilo ohm\n",
-    "R2=90.0;                #Resistor R2, kilo ohm\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "mv=R1/(R1+R2);                  #Feedback fraction\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n",
-    "Avf=round(Av/(1+Av*mv),0);               #Voltage gain with feedback\n",
-    "\n",
-    "#(iii)\n",
-    "Zin_feedback=((1+Av*mv)*Zin)/1000;             #Increased input impedance due to negative feedback, mega ohm\n",
-    "\n",
-    "#(iv)\n",
-    "Zout_feedback=Zout/(1+Av*mv);             #Decreased output impedance due to negative feedback, ohm\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)   Feedback fraction=%.1f.\"%mv);\n",
-    "print(\"(ii)  The voltage gain with feedback=%d.\"%Avf);\n",
-    "print(\"(iii) Increased input impedance due to negative feedback=%.0f mega ohm\"%Zin_feedback);\n",
-    "print(\"(iv)  Decreased output impedance due to negative feedback=%.1f ohm.\"%Zout_feedback);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.13 : Page number 346"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Distortion with negative feedback=0.312%\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=150.0;                       #Voltage gain\n",
-    "D=5/100.0;                        #Distortion\n",
-    "mv=10/100.0;                      #Feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "Dvf=round((D/(1+Av*mv))*100,3);                #Distortion with negative feedback\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Distortion with negative feedback=%.3f%%\"%Dvf);\n",
-    "\n",
-    "#Note: In the text, value of Dvf=0.3125% has been approximated to 0.313%. But, here the approximation is done to 0.312%\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.14 : Page number 346"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The new lower cut-off frequency=136.4Hz\n",
-      "The new upper cut-off frequency=5.52MHz\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Av=1000.0;                      #Voltage gain\n",
-    "f1=1.5;                         #Lower cut-off frequency, kHz\n",
-    "f2=501.5;                       #Upper cut-off frequency, kHz\n",
-    "mv=1/100.0;                       #Feedbcack fraction\n",
-    "\n",
-    "#Calculation\n",
-    "f1_f=(f1/(1+mv*Av))*1000;              #New lower cut-off frequency, Hz\n",
-    "f2_f=(f2*(1+mv*Av))/1000;              #New upper cut-off frequency, MHz\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The new lower cut-off frequency=%.1fHz\"%f1_f);\n",
-    "print(\"The new upper cut-off frequency=%.2fMHz\"%f2_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.15 : Page number 348"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The effective current gain=58.82.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=200.0;                       #Current gain without feedback\n",
-    "mi=0.012;                       #Current attenuation\n",
-    "\n",
-    "#Calculation\n",
-    "Aif=Ai/(1+Ai*mi);\n",
-    "\n",
-    "#Result\n",
-    "print(\"The effective current gain=%.2f.\"%Aif);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.16 : Page number 349"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance when negative feedback is applied=3.26 kilo ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=240.0;                       #Current gain\n",
-    "Zin=15.0;                       #Input impedance without feedback, kilo ohm\n",
-    "mi=0.015;                       #Current feedback fraction\n",
-    "\n",
-    "#Calculations\n",
-    "Zin_f=Zin/(1+mi*Ai);                #Input impedance with feedback, kilo ohm\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance when negative feedback is applied=%.2f kilo ohm\"%Zin_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.17 : Page number 349"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output impedance with negative feedback=9kilo ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=200.0;                   #Current gain without feedback\n",
-    "Zout=3.0;                   #Output impedance without feedback, kilo ohm\n",
-    "mi=0.01;                    #current feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "Zout_f=Zout*(1+mi*Ai);              #Output impedance with negative feedback, kilo ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output impedance with negative feedback=%dkilo ohm.\"%Zout_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.18 : Page number 349"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Bandwidth when negative feedback is applied=1400kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Ai=250.0;                   #Current gain without feedback\n",
-    "BW=400.0;                   #Bandwidth, kHz\n",
-    "mi=0.01;                    #current feedback fraction\n",
-    "\n",
-    "#Calculation\n",
-    "BW_f=BW*(1+mi*Ai);              #Bandwidth when negative feedback is applied, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Bandwidth when negative feedback is applied=%dkHz.\"%BW_f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.19 : Page number 350-351"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Value of VE=9.72V and IE=10.68mA\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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N7HlgcfTco3IWmcRuXXXZLbeEm2+G+++HG26Arl3h2WfrJ7a0UY07XhrP5NWkgfxHM3sM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\nAxERQclARESA/w8INYcwb/NOegAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f84ace55f98>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as p\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=18.0;                   #Supply voltage, V\n",
-    "R1=16.0;                    #Resistor R1, kilo ohm\n",
-    "R2=22.0;                    #Resistor R2, kilo ohm\n",
-    "RE=910.0;                   #Emitter resistor, ohm\n",
-    "VBE=0.7;                    #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculations\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2-VBE;                         #Emitter voltage, V\n",
-    "IE=(VE/RE)*1000;                   #Emitter current, mA (OHM's LAW)\n",
-    "\n",
-    "#D.C load line\n",
-    "IC_sat=(VCC/RE)*1000;                      #Collector saturation current, mA\n",
-    "VCE_off=VCC;                               #Collector-emitter voltage in off state, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Value of VE=%.2fV and IE=%.2fmA\"%(VE,IE));\n",
-    "\n",
-    "#Plotting\n",
-    "VCE_plot=[0,VCE_off];            #Plotting variable for VCE\n",
-    "IC_plot=[IC_sat,0];              #Plotting variable for IC\n",
-    "p.plot(VCE_plot,IC_plot);\n",
-    "p.xlim(0,20)\n",
-    "p.ylim(0,25)\n",
-    "p.xlabel('VCE(V)');\n",
-    "p.ylabel('IC(mA)');\n",
-    "p.title('d.c load line');\n",
-    "p.grid();\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.20 : Page number 352"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the emitter follower circuit=0.994.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                   #Supply voltage, V\n",
-    "R1=10.0;                    #Resistor R1, kilo ohm\n",
-    "R2=10.0;                    #Resistor R2, kilo ohm\n",
-    "RE=5.0;                     #Emitter resistance, kilo ohm\n",
-    "VBE=0.7;                    #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2-VBE;                         #Emitter voltage, V\n",
-    "IE=(VE/RE);                   #Emitter current, mA (OHM's LAW)\n",
-    "re=25/IE;                          #a.c emitter resistance, ohm\n",
-    "Av=RE*1000/(re+RE*1000);                     #Voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain of the emitter follower circuit=%.3f.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.21 : Page number 352-353"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain=0.988\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RE=5.0;                 #Emitter resistance, kilo ohm\n",
-    "re=29.1;                #a.c emitter resistance, ohm\n",
-    "RL=5.0;                 #Load resistance, kilo ohm\n",
-    "\n",
-    "#Calculation\n",
-    "RE_ac=(RE*RL)/(RE+RL);               #New effective value of emitter resistance, kilo ohm\n",
-    "Av=RE_ac*1000/(re+RE_ac*1000);     #Voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage gain=%.3f\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.22 : Page number 354"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance of the emitter follower =4.96 kilo ohm\n",
-      "The approximate value of the input impedance=5 kilo ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "def pr(r1,r2):                     #Function for calculating parallel resistance\n",
-    "    return (r1*r2)/(r1+r2);\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=10.0;                   #Supply voltage, V\n",
-    "R1=10.0;                    #Resistor R1, kilo ohm\n",
-    "R2=10.0;                    #Resistor R2, kilo ohm\n",
-    "RE=4.3;                     #Emitter resistor, kilo ohm\n",
-    "RL=10.0;                    #Load resistance, kilo ohm\n",
-    "VBE=0.7;                    #Base-emitter voltage, V\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "VE=V2-VBE;                         #Emitter voltage, V\n",
-    "IE=(VE/RE);                        #Emitter current, mA (OHM's LAW)\n",
-    "re=25/IE;                          #a.c emitter resistance, ohm\n",
-    "RE_eff=pr(RE,RL);                  #Effective external emitter resistance, kilo ohm\n",
-    "Zin_base=beta*(re/1000+RE_eff);         #Input impedance of the base of the transistor, kilo ohm\n",
-    "Zin=pr(pr(R1,R2),Zin_base);        #Input impedance of emitter follower, kilo ohm\n",
-    "#Approximate value of input impedance taken as parallel resistance of R1 and R2 and ignoring Zin_base due to its relatively large value\n",
-    "Zin_approx=pr(R1,R2);              #Approximate input impedance, kilo ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance of the emitter follower =%.2f kilo ohm\"%Zin);\n",
-    "print(\"The approximate value of the input impedance=%d kilo ohm\"%Zin_approx);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.23 : Page number 355"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output impedance=22.3 ohm\n"
-     ]
-    }
-   ],
-   "source": [
-    "def pr(r1,r2):                     #Function for calculating parallel resistance\n",
-    "    return (r1*r2)/(r1+r2);\n",
-    "\n",
-    "\n",
-    "#Variable declaration\n",
-    "re=20.0;                    #a.c emitter resistance, ohm\n",
-    "R1=3.0;                     #Resistor R1, kilo ohm\n",
-    "R2=4.7;                     #Resistor R2, kilo ohm\n",
-    "RS=600.0;                   #Source resistance, kilo ohm\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "Rin_ac=pr(pr(R1,R2)*1000,RS);            #Input a.c resistance, ohm\n",
-    "Zout=re + Rin_ac/beta;                   #Output impedance, ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output impedance=%.1f ohm\"%Zout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.24 : Page number  358"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  d.c value of current in RE=1.09mA\n",
-      "(ii) Input impedance=16.17 mega ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10.0;                    #Supply voltage, V\n",
-    "R1=120.0;                    #Resistor R1, kilo ohm\n",
-    "R2=120.0;                    #Resistor R2, kilo ohm\n",
-    "RE=3.3;                      #Emitter resistor, kilo ohm\n",
-    "VBE=0.7;                     #Base-emitter voltage, V\n",
-    "beta_1=70.0;                 #Base current amplification factor of 1st transistor\n",
-    "beta_2=70.0;                 #Base current amplification factor of 2nd transistor\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "V2=VCC*R2/(R1+R2);                 #Voltage across R2, V (Voltage divider rule)\n",
-    "IE_2=(V2-2*VBE)/RE;                #Emitter current, mA (OHM's LAW)\n",
-    "\n",
-    "#(ii)\n",
-    "Zin=(beta_1*beta_2*RE)/1000;               #Input impedance, mega ohm\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  d.c value of current in RE=%.2fmA\"%IE_2);\n",
-    "print(\"(ii) Input impedance=%.2f mega ohm.\"%Zin);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 13.25 : Page number 358-359"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) D.C Bias levels: \n",
-      " VB1= 4V, VE1=3.3V, VB2=3.3V, VE2=2.6V, IE2=1.3mA and IE1=0.013mA.\n",
-      "(ii) A.C Analysis:   \n",
-      " re1=1923 ohm and re2=19.23 ohm \n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12.0;                    #Supply voltage, V\n",
-    "R1=20.0;                     #Resistor R1, kilo ohm\n",
-    "R2=10.0;                     #Resistor R2, kilo ohm\n",
-    "RC=4.0;                      #Collector resistor, kilo ohm\n",
-    "RE=2.0;                      #Emitter resistor, kilo ohm\n",
-    "VBE=0.7;                     #Base-emitter voltage, V\n",
-    "beta=100.0;                  #Base current amplification factor of 1st transistor\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) D.C Bias levels\n",
-    "VB1=VCC*R2/(R1+R2);                 #Base voltage of 1st transistor, V (Voltage divider rule)\n",
-    "VE1=VB1-VBE;                        #Emitter voltage of 1st transistor, V\n",
-    "VB2=VE1;                            #Base voltage of 2nd transistor, V\n",
-    "VE2=VB2-VBE;                        #Emitter voltage of 2nd transistor, V\n",
-    "IE2=VE2/RE;                         #Emitter current of 2nd transistor, mA (OHM' LAW)\n",
-    "IE1=IE2/beta;                       #Emitter current of 1st transistor, mA (IE~IC=beta*IB, here IB2=IE1)\n",
-    "\n",
-    "#(ii) A.C analysis\n",
-    "re1=25/IE1;                         #a.c emitter resistance of 1st transistor\n",
-    "re2=25/IE2;                         #a.c emitter resistance of 2nd transistor\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) D.C Bias levels: \\n VB1= %dV, VE1=%.1fV, VB2=%.1fV, VE2=%.1fV, IE2=%.1fmA and IE1=%.3fmA.\"%(VB1,VE1,VB2,VE2,IE2,IE1));\n",
-    "print(\"(ii) A.C Analysis:   \\n re1=%d ohm and re2=%.2f ohm \"%(re1,re2));\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14.ipynb
deleted file mode 100755
index 5e35882c..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14.ipynb
+++ /dev/null
@@ -1,502 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:ec0d27209d08b0b95750f66ce9ee21af5ef586e23d0bf0ea218aa20a4ce63e43"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 14: SINUSOIDAL OSCILLATORS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.1 : Page number 371-372\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "L1=58.6;                #Inductance, micro henry\n",
-      "C1=300.0;               #Capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "f=(1/(2*round(pi,2)*sqrt(L1*10**-6*C1*10**-12)))/1000;         #Frequency of oscillation, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"frequency of oscillation=%dkHz\"%f);\n",
-      "\n",
-      "\n",
-      "#Note : The frequency has been calculated in the text  as 1199kHz but here the answer gets approximated to 1200kHz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "frequency of oscillation=1200kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.2 : Page number 372\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "L1=1.0;                     #Inductance , mH\n",
-      "f=1.0;                      #frequency of oscillation, GHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, f=1/(2*pi*sqrt(L1*C1)),\n",
-      "C1=(1/(L1*10**-3*(f*10**12*2*pi)**2))*10**12;               #Capacitance, pF\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The Capacitance of the capacitor of the LC oscillator=%.2epF\"%C1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Capacitance of the capacitor of the LC oscillator=2.53e-11pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.3 : Page number 373-374\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "C1=0.001;               #Capacitor C1, microfarad\n",
-      "C2=0.01;                #Capacitor C2, microfarad\n",
-      "L=15.0;                 #Inductance, microhenry\n",
-      "\n",
-      "#Calculation\n",
-      "CT=C1*C2/(C1+C2);               #Total capacitance\n",
-      "\n",
-      "#(i) Operating frequency\n",
-      "f=(1/(2*pi*sqrt(CT*10**-6*L*10**-6)))/1000;             #Operating frequency, kHz\n",
-      "\n",
-      "#(ii) Feedback fraction\n",
-      "mv=C1/C2;                                   #Feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The operating frequency=%dkHz\"%f);\n",
-      "print(\"(ii) The feedback fraction=%.1f\"%mv);\n",
-      "\n",
-      "#Note : The operating frequency is calculated in the text as 1361kHz but here it has been approximated to 1362kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The operating frequency=1362kHz\n",
-        "(ii) The feedback fraction=0.1\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.4 : Page number 374: Page number\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "mv=0.25;                    #Feedback fraction\n",
-      "L=1.0;                      #Inductance, mH\n",
-      "f=1.0;                      #Operating frequeny, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, f=1/(2*pi*sqrt(L*C))\n",
-      "CT=round((1/(L*10**-3*(2*pi*f*10**6)**2))*10**12,1);              #Total capacitance, pF\n",
-      "\n",
-      "#Since, mv=C1/C2 and CT=C1*C2/(C1+C2) or CT=C2/(1+ (C2/C1)),\n",
-      "#From the above equations, substituting value of mv and calculaing value of C2,\n",
-      "C2=CT*(1+(1/mv));               #Capacitance of C2 capactior, pF\n",
-      "C1=mv*C2;                       #Capacitance of C1 capacitor, pF\n",
-      "\n",
-      "#Result\n",
-      "print(\"C1=%.1fpF and C2=%.1fpF\"%(C1,C2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C1=31.6pF and C2=126.5pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.5 : Page number 375-376\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable decalaration\n",
-      "L1=1000.0;                  #Inductance of L1 inductor, microhenry\n",
-      "L2=100.0;                   #Inductance of L2 inductor, microhenry\n",
-      "M=20.0;                     #Mutual inductance, microhenry\n",
-      "C=20.0;                     #Capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "LT=L1+L2+2*M;                   #Total inductance, microhenry\n",
-      "\n",
-      "#(i) Operating frequency\n",
-      "f=(1/(2*pi*sqrt(LT*10**-6*C*10**-12)))/1000;           #Operating frequency, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "mv=L2/L1;                   #feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The operating frequency=%dkHz.\"%f);\n",
-      "print(\"(ii) The feedback fraction=%.1f.\"%mv);\n",
-      "\n",
-      "#Note : The operating frequecy has been calculated in the text as 1052kHz but here it gets approximated to 1054kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The operating frequency=1054kHz.\n",
-        "(ii) The feedback fraction=0.1.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.6 : Page number 376\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=1.0;                  #Capacitance, pF\n",
-      "f=1.0;                  #Frequency, MHz\n",
-      "mv=0.2;                 #Feedback frequency\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "LT=(1/(C*10**-12*(2*pi*f*10**6)**2))*1000;               #Total inductance, mH\n",
-      "\n",
-      "#Since, mv=L2/L1 or L2=mv*L1 and L1+L2=LT or L1(1+mv)=LT,\n",
-      "L1=LT/(1+mv);                   #Inductance of L1 inductor, mH\n",
-      "L2=L1*mv;                       #inductance of L2 inductor, mH\n",
-      "\n",
-      "#Result\n",
-      "print(\"L1=%.1fmH and L2=%.2fmH.\"%(L1,L2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "L1=21.1mH and L2=4.22mH.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.7 : Page number 378\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "R1=1.0;                 #Resistor R1, mega ohm\n",
-      "R2=R1;                  #Resistor R2, mega ohm\n",
-      "R3=R1;                  #Resistor R3, mega ohm\n",
-      "C1=68.0;                #Capacitor C1, pF\n",
-      "C2=C1;                  #Capacitor C2, pF\n",
-      "C3=C1;                  #Capacitor C3, pF\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "R=R1*10**6;                   #Resistance of the resistors of phase shift circuit, ohm\n",
-      "C=C1*10**-12;                 #Capacitance of the capacitors of phase shift circuit, F\n",
-      "fo=1/(2*pi*R*C*sqrt(6));      #Frequency of oscillation, Hz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency of oscillation=%dHz\"%fo);\n",
-      "\n",
-      "#Note: The frequency of oscillation had been calculated in the text as 954Hz, but here it gets approximated to 955 HZ.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of oscillation=955Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.8 : Page number 378\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=5.0;                      #Capacitance of the capacitors of phase shift circuit, pF\n",
-      "fo=800.0;                    #Required frequency of oscillation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, fo=1/(2*pi*R*C*sqrt(6))\n",
-      "R=(1/(2*pi*C*10**-12*fo*10**3*sqrt(6)))/1000;                   #Resistance of the resistors of phase shift circuit, kilo ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"R=%.1f kilo ohm.\"%R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R=16.2 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.9 : Page number 380\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "#Resistance of R1 and R2 resistors of the R-C bridge circuit\n",
-      "R1=220.0;                           #kilo ohm \n",
-      "R2=220.0;                           #kilo ohm\n",
-      "\n",
-      "#Capacitance of C1 and C2 the capacitors of the R-C bridge circuit\n",
-      "C1=250.0;                           #pF\n",
-      "C2=250.0;                           #pF\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, R1=R2 and C1=C2, R1=R2 is taken as R and C1=C2 is taken as C\n",
-      "#And, f=1/(2*pi*sqrt(R1*R2*C1*C2))is transformed to f=1/(2*pi*R*C).\n",
-      "R=R1*10**3;                                   #kilo ohm\n",
-      "C=C1*10**-12;                                   #pF\n",
-      "f=1/(2*pi*R*C);           #Frequency of oscillation, Hz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency of oscillation=%dHz.\"%f);\n",
-      "\n",
-      "\n",
-      "#Note : The frequency of oscillation is calculated in the text as 2892Hz but here it gets approximated to 2893 Hz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of oscillation=2893Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.11 : Page number 384\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "#a.c equivalent values of the crystal:\n",
-      "L=1.0;                          #Inductance , H\n",
-      "C=0.01;                         #Capacitance , pF\n",
-      "R=1000.0;                       #Resistance , ohm\n",
-      "Cm=20.0;                        #Mounting capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "fs=(1/(2*round(pi,2)*sqrt(L*C*10**-12)))/1000;               #Series resonant frrequency, kHz\n",
-      "CT=(C*Cm/(C+Cm));                                            #Total capacitance, pF\n",
-      "fp=(1/(2*round(pi,2)*sqrt(L*CT*10**-12)))/1000;              #Prallel resonant frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"fs=%.0fkHz and fp=%.0fkHz.\"%(fs,fp));\n",
-      "\n",
-      "#Note: fs and fp are calculated in the text as 1589kHz and 1590kHz, but here it gets approximated to 1592kHz and 1593kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fs=1592kHz and fp=1593kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_1.ipynb
deleted file mode 100755
index 5e35882c..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_1.ipynb
+++ /dev/null
@@ -1,502 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:ec0d27209d08b0b95750f66ce9ee21af5ef586e23d0bf0ea218aa20a4ce63e43"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 14: SINUSOIDAL OSCILLATORS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.1 : Page number 371-372\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "L1=58.6;                #Inductance, micro henry\n",
-      "C1=300.0;               #Capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "f=(1/(2*round(pi,2)*sqrt(L1*10**-6*C1*10**-12)))/1000;         #Frequency of oscillation, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"frequency of oscillation=%dkHz\"%f);\n",
-      "\n",
-      "\n",
-      "#Note : The frequency has been calculated in the text  as 1199kHz but here the answer gets approximated to 1200kHz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "frequency of oscillation=1200kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.2 : Page number 372\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "L1=1.0;                     #Inductance , mH\n",
-      "f=1.0;                      #frequency of oscillation, GHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, f=1/(2*pi*sqrt(L1*C1)),\n",
-      "C1=(1/(L1*10**-3*(f*10**12*2*pi)**2))*10**12;               #Capacitance, pF\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The Capacitance of the capacitor of the LC oscillator=%.2epF\"%C1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Capacitance of the capacitor of the LC oscillator=2.53e-11pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.3 : Page number 373-374\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "C1=0.001;               #Capacitor C1, microfarad\n",
-      "C2=0.01;                #Capacitor C2, microfarad\n",
-      "L=15.0;                 #Inductance, microhenry\n",
-      "\n",
-      "#Calculation\n",
-      "CT=C1*C2/(C1+C2);               #Total capacitance\n",
-      "\n",
-      "#(i) Operating frequency\n",
-      "f=(1/(2*pi*sqrt(CT*10**-6*L*10**-6)))/1000;             #Operating frequency, kHz\n",
-      "\n",
-      "#(ii) Feedback fraction\n",
-      "mv=C1/C2;                                   #Feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The operating frequency=%dkHz\"%f);\n",
-      "print(\"(ii) The feedback fraction=%.1f\"%mv);\n",
-      "\n",
-      "#Note : The operating frequency is calculated in the text as 1361kHz but here it has been approximated to 1362kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The operating frequency=1362kHz\n",
-        "(ii) The feedback fraction=0.1\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.4 : Page number 374: Page number\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "mv=0.25;                    #Feedback fraction\n",
-      "L=1.0;                      #Inductance, mH\n",
-      "f=1.0;                      #Operating frequeny, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, f=1/(2*pi*sqrt(L*C))\n",
-      "CT=round((1/(L*10**-3*(2*pi*f*10**6)**2))*10**12,1);              #Total capacitance, pF\n",
-      "\n",
-      "#Since, mv=C1/C2 and CT=C1*C2/(C1+C2) or CT=C2/(1+ (C2/C1)),\n",
-      "#From the above equations, substituting value of mv and calculaing value of C2,\n",
-      "C2=CT*(1+(1/mv));               #Capacitance of C2 capactior, pF\n",
-      "C1=mv*C2;                       #Capacitance of C1 capacitor, pF\n",
-      "\n",
-      "#Result\n",
-      "print(\"C1=%.1fpF and C2=%.1fpF\"%(C1,C2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C1=31.6pF and C2=126.5pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.5 : Page number 375-376\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable decalaration\n",
-      "L1=1000.0;                  #Inductance of L1 inductor, microhenry\n",
-      "L2=100.0;                   #Inductance of L2 inductor, microhenry\n",
-      "M=20.0;                     #Mutual inductance, microhenry\n",
-      "C=20.0;                     #Capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "LT=L1+L2+2*M;                   #Total inductance, microhenry\n",
-      "\n",
-      "#(i) Operating frequency\n",
-      "f=(1/(2*pi*sqrt(LT*10**-6*C*10**-12)))/1000;           #Operating frequency, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "mv=L2/L1;                   #feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The operating frequency=%dkHz.\"%f);\n",
-      "print(\"(ii) The feedback fraction=%.1f.\"%mv);\n",
-      "\n",
-      "#Note : The operating frequecy has been calculated in the text as 1052kHz but here it gets approximated to 1054kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The operating frequency=1054kHz.\n",
-        "(ii) The feedback fraction=0.1.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.6 : Page number 376\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=1.0;                  #Capacitance, pF\n",
-      "f=1.0;                  #Frequency, MHz\n",
-      "mv=0.2;                 #Feedback frequency\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "LT=(1/(C*10**-12*(2*pi*f*10**6)**2))*1000;               #Total inductance, mH\n",
-      "\n",
-      "#Since, mv=L2/L1 or L2=mv*L1 and L1+L2=LT or L1(1+mv)=LT,\n",
-      "L1=LT/(1+mv);                   #Inductance of L1 inductor, mH\n",
-      "L2=L1*mv;                       #inductance of L2 inductor, mH\n",
-      "\n",
-      "#Result\n",
-      "print(\"L1=%.1fmH and L2=%.2fmH.\"%(L1,L2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "L1=21.1mH and L2=4.22mH.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.7 : Page number 378\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "R1=1.0;                 #Resistor R1, mega ohm\n",
-      "R2=R1;                  #Resistor R2, mega ohm\n",
-      "R3=R1;                  #Resistor R3, mega ohm\n",
-      "C1=68.0;                #Capacitor C1, pF\n",
-      "C2=C1;                  #Capacitor C2, pF\n",
-      "C3=C1;                  #Capacitor C3, pF\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "R=R1*10**6;                   #Resistance of the resistors of phase shift circuit, ohm\n",
-      "C=C1*10**-12;                 #Capacitance of the capacitors of phase shift circuit, F\n",
-      "fo=1/(2*pi*R*C*sqrt(6));      #Frequency of oscillation, Hz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency of oscillation=%dHz\"%fo);\n",
-      "\n",
-      "#Note: The frequency of oscillation had been calculated in the text as 954Hz, but here it gets approximated to 955 HZ.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of oscillation=955Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.8 : Page number 378\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=5.0;                      #Capacitance of the capacitors of phase shift circuit, pF\n",
-      "fo=800.0;                    #Required frequency of oscillation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, fo=1/(2*pi*R*C*sqrt(6))\n",
-      "R=(1/(2*pi*C*10**-12*fo*10**3*sqrt(6)))/1000;                   #Resistance of the resistors of phase shift circuit, kilo ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"R=%.1f kilo ohm.\"%R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R=16.2 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.9 : Page number 380\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "#Resistance of R1 and R2 resistors of the R-C bridge circuit\n",
-      "R1=220.0;                           #kilo ohm \n",
-      "R2=220.0;                           #kilo ohm\n",
-      "\n",
-      "#Capacitance of C1 and C2 the capacitors of the R-C bridge circuit\n",
-      "C1=250.0;                           #pF\n",
-      "C2=250.0;                           #pF\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, R1=R2 and C1=C2, R1=R2 is taken as R and C1=C2 is taken as C\n",
-      "#And, f=1/(2*pi*sqrt(R1*R2*C1*C2))is transformed to f=1/(2*pi*R*C).\n",
-      "R=R1*10**3;                                   #kilo ohm\n",
-      "C=C1*10**-12;                                   #pF\n",
-      "f=1/(2*pi*R*C);           #Frequency of oscillation, Hz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency of oscillation=%dHz.\"%f);\n",
-      "\n",
-      "\n",
-      "#Note : The frequency of oscillation is calculated in the text as 2892Hz but here it gets approximated to 2893 Hz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of oscillation=2893Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.11 : Page number 384\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "#a.c equivalent values of the crystal:\n",
-      "L=1.0;                          #Inductance , H\n",
-      "C=0.01;                         #Capacitance , pF\n",
-      "R=1000.0;                       #Resistance , ohm\n",
-      "Cm=20.0;                        #Mounting capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "fs=(1/(2*round(pi,2)*sqrt(L*C*10**-12)))/1000;               #Series resonant frrequency, kHz\n",
-      "CT=(C*Cm/(C+Cm));                                            #Total capacitance, pF\n",
-      "fp=(1/(2*round(pi,2)*sqrt(L*CT*10**-12)))/1000;              #Prallel resonant frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"fs=%.0fkHz and fp=%.0fkHz.\"%(fs,fp));\n",
-      "\n",
-      "#Note: fs and fp are calculated in the text as 1589kHz and 1590kHz, but here it gets approximated to 1592kHz and 1593kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fs=1592kHz and fp=1593kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_2.ipynb
deleted file mode 100755
index 5e35882c..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_2.ipynb
+++ /dev/null
@@ -1,502 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:ec0d27209d08b0b95750f66ce9ee21af5ef586e23d0bf0ea218aa20a4ce63e43"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 14: SINUSOIDAL OSCILLATORS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.1 : Page number 371-372\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "L1=58.6;                #Inductance, micro henry\n",
-      "C1=300.0;               #Capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "f=(1/(2*round(pi,2)*sqrt(L1*10**-6*C1*10**-12)))/1000;         #Frequency of oscillation, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"frequency of oscillation=%dkHz\"%f);\n",
-      "\n",
-      "\n",
-      "#Note : The frequency has been calculated in the text  as 1199kHz but here the answer gets approximated to 1200kHz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "frequency of oscillation=1200kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.2 : Page number 372\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "L1=1.0;                     #Inductance , mH\n",
-      "f=1.0;                      #frequency of oscillation, GHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, f=1/(2*pi*sqrt(L1*C1)),\n",
-      "C1=(1/(L1*10**-3*(f*10**12*2*pi)**2))*10**12;               #Capacitance, pF\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The Capacitance of the capacitor of the LC oscillator=%.2epF\"%C1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Capacitance of the capacitor of the LC oscillator=2.53e-11pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.3 : Page number 373-374\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "C1=0.001;               #Capacitor C1, microfarad\n",
-      "C2=0.01;                #Capacitor C2, microfarad\n",
-      "L=15.0;                 #Inductance, microhenry\n",
-      "\n",
-      "#Calculation\n",
-      "CT=C1*C2/(C1+C2);               #Total capacitance\n",
-      "\n",
-      "#(i) Operating frequency\n",
-      "f=(1/(2*pi*sqrt(CT*10**-6*L*10**-6)))/1000;             #Operating frequency, kHz\n",
-      "\n",
-      "#(ii) Feedback fraction\n",
-      "mv=C1/C2;                                   #Feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The operating frequency=%dkHz\"%f);\n",
-      "print(\"(ii) The feedback fraction=%.1f\"%mv);\n",
-      "\n",
-      "#Note : The operating frequency is calculated in the text as 1361kHz but here it has been approximated to 1362kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The operating frequency=1362kHz\n",
-        "(ii) The feedback fraction=0.1\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.4 : Page number 374: Page number\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "mv=0.25;                    #Feedback fraction\n",
-      "L=1.0;                      #Inductance, mH\n",
-      "f=1.0;                      #Operating frequeny, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, f=1/(2*pi*sqrt(L*C))\n",
-      "CT=round((1/(L*10**-3*(2*pi*f*10**6)**2))*10**12,1);              #Total capacitance, pF\n",
-      "\n",
-      "#Since, mv=C1/C2 and CT=C1*C2/(C1+C2) or CT=C2/(1+ (C2/C1)),\n",
-      "#From the above equations, substituting value of mv and calculaing value of C2,\n",
-      "C2=CT*(1+(1/mv));               #Capacitance of C2 capactior, pF\n",
-      "C1=mv*C2;                       #Capacitance of C1 capacitor, pF\n",
-      "\n",
-      "#Result\n",
-      "print(\"C1=%.1fpF and C2=%.1fpF\"%(C1,C2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C1=31.6pF and C2=126.5pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.5 : Page number 375-376\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable decalaration\n",
-      "L1=1000.0;                  #Inductance of L1 inductor, microhenry\n",
-      "L2=100.0;                   #Inductance of L2 inductor, microhenry\n",
-      "M=20.0;                     #Mutual inductance, microhenry\n",
-      "C=20.0;                     #Capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "LT=L1+L2+2*M;                   #Total inductance, microhenry\n",
-      "\n",
-      "#(i) Operating frequency\n",
-      "f=(1/(2*pi*sqrt(LT*10**-6*C*10**-12)))/1000;           #Operating frequency, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "mv=L2/L1;                   #feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The operating frequency=%dkHz.\"%f);\n",
-      "print(\"(ii) The feedback fraction=%.1f.\"%mv);\n",
-      "\n",
-      "#Note : The operating frequecy has been calculated in the text as 1052kHz but here it gets approximated to 1054kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The operating frequency=1054kHz.\n",
-        "(ii) The feedback fraction=0.1.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.6 : Page number 376\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=1.0;                  #Capacitance, pF\n",
-      "f=1.0;                  #Frequency, MHz\n",
-      "mv=0.2;                 #Feedback frequency\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "LT=(1/(C*10**-12*(2*pi*f*10**6)**2))*1000;               #Total inductance, mH\n",
-      "\n",
-      "#Since, mv=L2/L1 or L2=mv*L1 and L1+L2=LT or L1(1+mv)=LT,\n",
-      "L1=LT/(1+mv);                   #Inductance of L1 inductor, mH\n",
-      "L2=L1*mv;                       #inductance of L2 inductor, mH\n",
-      "\n",
-      "#Result\n",
-      "print(\"L1=%.1fmH and L2=%.2fmH.\"%(L1,L2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "L1=21.1mH and L2=4.22mH.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.7 : Page number 378\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "R1=1.0;                 #Resistor R1, mega ohm\n",
-      "R2=R1;                  #Resistor R2, mega ohm\n",
-      "R3=R1;                  #Resistor R3, mega ohm\n",
-      "C1=68.0;                #Capacitor C1, pF\n",
-      "C2=C1;                  #Capacitor C2, pF\n",
-      "C3=C1;                  #Capacitor C3, pF\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "R=R1*10**6;                   #Resistance of the resistors of phase shift circuit, ohm\n",
-      "C=C1*10**-12;                 #Capacitance of the capacitors of phase shift circuit, F\n",
-      "fo=1/(2*pi*R*C*sqrt(6));      #Frequency of oscillation, Hz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency of oscillation=%dHz\"%fo);\n",
-      "\n",
-      "#Note: The frequency of oscillation had been calculated in the text as 954Hz, but here it gets approximated to 955 HZ.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of oscillation=955Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.8 : Page number 378\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=5.0;                      #Capacitance of the capacitors of phase shift circuit, pF\n",
-      "fo=800.0;                    #Required frequency of oscillation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, fo=1/(2*pi*R*C*sqrt(6))\n",
-      "R=(1/(2*pi*C*10**-12*fo*10**3*sqrt(6)))/1000;                   #Resistance of the resistors of phase shift circuit, kilo ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"R=%.1f kilo ohm.\"%R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R=16.2 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.9 : Page number 380\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "#Resistance of R1 and R2 resistors of the R-C bridge circuit\n",
-      "R1=220.0;                           #kilo ohm \n",
-      "R2=220.0;                           #kilo ohm\n",
-      "\n",
-      "#Capacitance of C1 and C2 the capacitors of the R-C bridge circuit\n",
-      "C1=250.0;                           #pF\n",
-      "C2=250.0;                           #pF\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, R1=R2 and C1=C2, R1=R2 is taken as R and C1=C2 is taken as C\n",
-      "#And, f=1/(2*pi*sqrt(R1*R2*C1*C2))is transformed to f=1/(2*pi*R*C).\n",
-      "R=R1*10**3;                                   #kilo ohm\n",
-      "C=C1*10**-12;                                   #pF\n",
-      "f=1/(2*pi*R*C);           #Frequency of oscillation, Hz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency of oscillation=%dHz.\"%f);\n",
-      "\n",
-      "\n",
-      "#Note : The frequency of oscillation is calculated in the text as 2892Hz but here it gets approximated to 2893 Hz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of oscillation=2893Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.11 : Page number 384\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "#a.c equivalent values of the crystal:\n",
-      "L=1.0;                          #Inductance , H\n",
-      "C=0.01;                         #Capacitance , pF\n",
-      "R=1000.0;                       #Resistance , ohm\n",
-      "Cm=20.0;                        #Mounting capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "fs=(1/(2*round(pi,2)*sqrt(L*C*10**-12)))/1000;               #Series resonant frrequency, kHz\n",
-      "CT=(C*Cm/(C+Cm));                                            #Total capacitance, pF\n",
-      "fp=(1/(2*round(pi,2)*sqrt(L*CT*10**-12)))/1000;              #Prallel resonant frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"fs=%.0fkHz and fp=%.0fkHz.\"%(fs,fp));\n",
-      "\n",
-      "#Note: fs and fp are calculated in the text as 1589kHz and 1590kHz, but here it gets approximated to 1592kHz and 1593kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fs=1592kHz and fp=1593kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_3.ipynb
deleted file mode 100755
index 5e35882c..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_3.ipynb
+++ /dev/null
@@ -1,502 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:ec0d27209d08b0b95750f66ce9ee21af5ef586e23d0bf0ea218aa20a4ce63e43"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 14: SINUSOIDAL OSCILLATORS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.1 : Page number 371-372\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "L1=58.6;                #Inductance, micro henry\n",
-      "C1=300.0;               #Capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "f=(1/(2*round(pi,2)*sqrt(L1*10**-6*C1*10**-12)))/1000;         #Frequency of oscillation, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"frequency of oscillation=%dkHz\"%f);\n",
-      "\n",
-      "\n",
-      "#Note : The frequency has been calculated in the text  as 1199kHz but here the answer gets approximated to 1200kHz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "frequency of oscillation=1200kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.2 : Page number 372\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "L1=1.0;                     #Inductance , mH\n",
-      "f=1.0;                      #frequency of oscillation, GHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, f=1/(2*pi*sqrt(L1*C1)),\n",
-      "C1=(1/(L1*10**-3*(f*10**12*2*pi)**2))*10**12;               #Capacitance, pF\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The Capacitance of the capacitor of the LC oscillator=%.2epF\"%C1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Capacitance of the capacitor of the LC oscillator=2.53e-11pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.3 : Page number 373-374\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "C1=0.001;               #Capacitor C1, microfarad\n",
-      "C2=0.01;                #Capacitor C2, microfarad\n",
-      "L=15.0;                 #Inductance, microhenry\n",
-      "\n",
-      "#Calculation\n",
-      "CT=C1*C2/(C1+C2);               #Total capacitance\n",
-      "\n",
-      "#(i) Operating frequency\n",
-      "f=(1/(2*pi*sqrt(CT*10**-6*L*10**-6)))/1000;             #Operating frequency, kHz\n",
-      "\n",
-      "#(ii) Feedback fraction\n",
-      "mv=C1/C2;                                   #Feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The operating frequency=%dkHz\"%f);\n",
-      "print(\"(ii) The feedback fraction=%.1f\"%mv);\n",
-      "\n",
-      "#Note : The operating frequency is calculated in the text as 1361kHz but here it has been approximated to 1362kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The operating frequency=1362kHz\n",
-        "(ii) The feedback fraction=0.1\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.4 : Page number 374: Page number\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "mv=0.25;                    #Feedback fraction\n",
-      "L=1.0;                      #Inductance, mH\n",
-      "f=1.0;                      #Operating frequeny, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, f=1/(2*pi*sqrt(L*C))\n",
-      "CT=round((1/(L*10**-3*(2*pi*f*10**6)**2))*10**12,1);              #Total capacitance, pF\n",
-      "\n",
-      "#Since, mv=C1/C2 and CT=C1*C2/(C1+C2) or CT=C2/(1+ (C2/C1)),\n",
-      "#From the above equations, substituting value of mv and calculaing value of C2,\n",
-      "C2=CT*(1+(1/mv));               #Capacitance of C2 capactior, pF\n",
-      "C1=mv*C2;                       #Capacitance of C1 capacitor, pF\n",
-      "\n",
-      "#Result\n",
-      "print(\"C1=%.1fpF and C2=%.1fpF\"%(C1,C2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C1=31.6pF and C2=126.5pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.5 : Page number 375-376\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable decalaration\n",
-      "L1=1000.0;                  #Inductance of L1 inductor, microhenry\n",
-      "L2=100.0;                   #Inductance of L2 inductor, microhenry\n",
-      "M=20.0;                     #Mutual inductance, microhenry\n",
-      "C=20.0;                     #Capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "LT=L1+L2+2*M;                   #Total inductance, microhenry\n",
-      "\n",
-      "#(i) Operating frequency\n",
-      "f=(1/(2*pi*sqrt(LT*10**-6*C*10**-12)))/1000;           #Operating frequency, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "mv=L2/L1;                   #feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The operating frequency=%dkHz.\"%f);\n",
-      "print(\"(ii) The feedback fraction=%.1f.\"%mv);\n",
-      "\n",
-      "#Note : The operating frequecy has been calculated in the text as 1052kHz but here it gets approximated to 1054kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The operating frequency=1054kHz.\n",
-        "(ii) The feedback fraction=0.1.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.6 : Page number 376\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=1.0;                  #Capacitance, pF\n",
-      "f=1.0;                  #Frequency, MHz\n",
-      "mv=0.2;                 #Feedback frequency\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "LT=(1/(C*10**-12*(2*pi*f*10**6)**2))*1000;               #Total inductance, mH\n",
-      "\n",
-      "#Since, mv=L2/L1 or L2=mv*L1 and L1+L2=LT or L1(1+mv)=LT,\n",
-      "L1=LT/(1+mv);                   #Inductance of L1 inductor, mH\n",
-      "L2=L1*mv;                       #inductance of L2 inductor, mH\n",
-      "\n",
-      "#Result\n",
-      "print(\"L1=%.1fmH and L2=%.2fmH.\"%(L1,L2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "L1=21.1mH and L2=4.22mH.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.7 : Page number 378\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "R1=1.0;                 #Resistor R1, mega ohm\n",
-      "R2=R1;                  #Resistor R2, mega ohm\n",
-      "R3=R1;                  #Resistor R3, mega ohm\n",
-      "C1=68.0;                #Capacitor C1, pF\n",
-      "C2=C1;                  #Capacitor C2, pF\n",
-      "C3=C1;                  #Capacitor C3, pF\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "R=R1*10**6;                   #Resistance of the resistors of phase shift circuit, ohm\n",
-      "C=C1*10**-12;                 #Capacitance of the capacitors of phase shift circuit, F\n",
-      "fo=1/(2*pi*R*C*sqrt(6));      #Frequency of oscillation, Hz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency of oscillation=%dHz\"%fo);\n",
-      "\n",
-      "#Note: The frequency of oscillation had been calculated in the text as 954Hz, but here it gets approximated to 955 HZ.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of oscillation=955Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.8 : Page number 378\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=5.0;                      #Capacitance of the capacitors of phase shift circuit, pF\n",
-      "fo=800.0;                    #Required frequency of oscillation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, fo=1/(2*pi*R*C*sqrt(6))\n",
-      "R=(1/(2*pi*C*10**-12*fo*10**3*sqrt(6)))/1000;                   #Resistance of the resistors of phase shift circuit, kilo ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"R=%.1f kilo ohm.\"%R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R=16.2 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.9 : Page number 380\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "#Resistance of R1 and R2 resistors of the R-C bridge circuit\n",
-      "R1=220.0;                           #kilo ohm \n",
-      "R2=220.0;                           #kilo ohm\n",
-      "\n",
-      "#Capacitance of C1 and C2 the capacitors of the R-C bridge circuit\n",
-      "C1=250.0;                           #pF\n",
-      "C2=250.0;                           #pF\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, R1=R2 and C1=C2, R1=R2 is taken as R and C1=C2 is taken as C\n",
-      "#And, f=1/(2*pi*sqrt(R1*R2*C1*C2))is transformed to f=1/(2*pi*R*C).\n",
-      "R=R1*10**3;                                   #kilo ohm\n",
-      "C=C1*10**-12;                                   #pF\n",
-      "f=1/(2*pi*R*C);           #Frequency of oscillation, Hz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency of oscillation=%dHz.\"%f);\n",
-      "\n",
-      "\n",
-      "#Note : The frequency of oscillation is calculated in the text as 2892Hz but here it gets approximated to 2893 Hz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of oscillation=2893Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.11 : Page number 384\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "#a.c equivalent values of the crystal:\n",
-      "L=1.0;                          #Inductance , H\n",
-      "C=0.01;                         #Capacitance , pF\n",
-      "R=1000.0;                       #Resistance , ohm\n",
-      "Cm=20.0;                        #Mounting capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "fs=(1/(2*round(pi,2)*sqrt(L*C*10**-12)))/1000;               #Series resonant frrequency, kHz\n",
-      "CT=(C*Cm/(C+Cm));                                            #Total capacitance, pF\n",
-      "fp=(1/(2*round(pi,2)*sqrt(L*CT*10**-12)))/1000;              #Prallel resonant frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"fs=%.0fkHz and fp=%.0fkHz.\"%(fs,fp));\n",
-      "\n",
-      "#Note: fs and fp are calculated in the text as 1589kHz and 1590kHz, but here it gets approximated to 1592kHz and 1593kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fs=1592kHz and fp=1593kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_4.ipynb
deleted file mode 100755
index 5e35882c..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_4.ipynb
+++ /dev/null
@@ -1,502 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:ec0d27209d08b0b95750f66ce9ee21af5ef586e23d0bf0ea218aa20a4ce63e43"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 14: SINUSOIDAL OSCILLATORS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.1 : Page number 371-372\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "L1=58.6;                #Inductance, micro henry\n",
-      "C1=300.0;               #Capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "f=(1/(2*round(pi,2)*sqrt(L1*10**-6*C1*10**-12)))/1000;         #Frequency of oscillation, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"frequency of oscillation=%dkHz\"%f);\n",
-      "\n",
-      "\n",
-      "#Note : The frequency has been calculated in the text  as 1199kHz but here the answer gets approximated to 1200kHz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "frequency of oscillation=1200kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.2 : Page number 372\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "L1=1.0;                     #Inductance , mH\n",
-      "f=1.0;                      #frequency of oscillation, GHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, f=1/(2*pi*sqrt(L1*C1)),\n",
-      "C1=(1/(L1*10**-3*(f*10**12*2*pi)**2))*10**12;               #Capacitance, pF\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The Capacitance of the capacitor of the LC oscillator=%.2epF\"%C1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Capacitance of the capacitor of the LC oscillator=2.53e-11pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.3 : Page number 373-374\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "C1=0.001;               #Capacitor C1, microfarad\n",
-      "C2=0.01;                #Capacitor C2, microfarad\n",
-      "L=15.0;                 #Inductance, microhenry\n",
-      "\n",
-      "#Calculation\n",
-      "CT=C1*C2/(C1+C2);               #Total capacitance\n",
-      "\n",
-      "#(i) Operating frequency\n",
-      "f=(1/(2*pi*sqrt(CT*10**-6*L*10**-6)))/1000;             #Operating frequency, kHz\n",
-      "\n",
-      "#(ii) Feedback fraction\n",
-      "mv=C1/C2;                                   #Feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The operating frequency=%dkHz\"%f);\n",
-      "print(\"(ii) The feedback fraction=%.1f\"%mv);\n",
-      "\n",
-      "#Note : The operating frequency is calculated in the text as 1361kHz but here it has been approximated to 1362kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The operating frequency=1362kHz\n",
-        "(ii) The feedback fraction=0.1\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.4 : Page number 374: Page number\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "mv=0.25;                    #Feedback fraction\n",
-      "L=1.0;                      #Inductance, mH\n",
-      "f=1.0;                      #Operating frequeny, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, f=1/(2*pi*sqrt(L*C))\n",
-      "CT=round((1/(L*10**-3*(2*pi*f*10**6)**2))*10**12,1);              #Total capacitance, pF\n",
-      "\n",
-      "#Since, mv=C1/C2 and CT=C1*C2/(C1+C2) or CT=C2/(1+ (C2/C1)),\n",
-      "#From the above equations, substituting value of mv and calculaing value of C2,\n",
-      "C2=CT*(1+(1/mv));               #Capacitance of C2 capactior, pF\n",
-      "C1=mv*C2;                       #Capacitance of C1 capacitor, pF\n",
-      "\n",
-      "#Result\n",
-      "print(\"C1=%.1fpF and C2=%.1fpF\"%(C1,C2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C1=31.6pF and C2=126.5pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.5 : Page number 375-376\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable decalaration\n",
-      "L1=1000.0;                  #Inductance of L1 inductor, microhenry\n",
-      "L2=100.0;                   #Inductance of L2 inductor, microhenry\n",
-      "M=20.0;                     #Mutual inductance, microhenry\n",
-      "C=20.0;                     #Capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "LT=L1+L2+2*M;                   #Total inductance, microhenry\n",
-      "\n",
-      "#(i) Operating frequency\n",
-      "f=(1/(2*pi*sqrt(LT*10**-6*C*10**-12)))/1000;           #Operating frequency, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "mv=L2/L1;                   #feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The operating frequency=%dkHz.\"%f);\n",
-      "print(\"(ii) The feedback fraction=%.1f.\"%mv);\n",
-      "\n",
-      "#Note : The operating frequecy has been calculated in the text as 1052kHz but here it gets approximated to 1054kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The operating frequency=1054kHz.\n",
-        "(ii) The feedback fraction=0.1.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.6 : Page number 376\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=1.0;                  #Capacitance, pF\n",
-      "f=1.0;                  #Frequency, MHz\n",
-      "mv=0.2;                 #Feedback frequency\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "LT=(1/(C*10**-12*(2*pi*f*10**6)**2))*1000;               #Total inductance, mH\n",
-      "\n",
-      "#Since, mv=L2/L1 or L2=mv*L1 and L1+L2=LT or L1(1+mv)=LT,\n",
-      "L1=LT/(1+mv);                   #Inductance of L1 inductor, mH\n",
-      "L2=L1*mv;                       #inductance of L2 inductor, mH\n",
-      "\n",
-      "#Result\n",
-      "print(\"L1=%.1fmH and L2=%.2fmH.\"%(L1,L2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "L1=21.1mH and L2=4.22mH.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.7 : Page number 378\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "R1=1.0;                 #Resistor R1, mega ohm\n",
-      "R2=R1;                  #Resistor R2, mega ohm\n",
-      "R3=R1;                  #Resistor R3, mega ohm\n",
-      "C1=68.0;                #Capacitor C1, pF\n",
-      "C2=C1;                  #Capacitor C2, pF\n",
-      "C3=C1;                  #Capacitor C3, pF\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "R=R1*10**6;                   #Resistance of the resistors of phase shift circuit, ohm\n",
-      "C=C1*10**-12;                 #Capacitance of the capacitors of phase shift circuit, F\n",
-      "fo=1/(2*pi*R*C*sqrt(6));      #Frequency of oscillation, Hz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency of oscillation=%dHz\"%fo);\n",
-      "\n",
-      "#Note: The frequency of oscillation had been calculated in the text as 954Hz, but here it gets approximated to 955 HZ.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of oscillation=955Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.8 : Page number 378\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=5.0;                      #Capacitance of the capacitors of phase shift circuit, pF\n",
-      "fo=800.0;                    #Required frequency of oscillation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, fo=1/(2*pi*R*C*sqrt(6))\n",
-      "R=(1/(2*pi*C*10**-12*fo*10**3*sqrt(6)))/1000;                   #Resistance of the resistors of phase shift circuit, kilo ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"R=%.1f kilo ohm.\"%R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R=16.2 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.9 : Page number 380\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "#Resistance of R1 and R2 resistors of the R-C bridge circuit\n",
-      "R1=220.0;                           #kilo ohm \n",
-      "R2=220.0;                           #kilo ohm\n",
-      "\n",
-      "#Capacitance of C1 and C2 the capacitors of the R-C bridge circuit\n",
-      "C1=250.0;                           #pF\n",
-      "C2=250.0;                           #pF\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, R1=R2 and C1=C2, R1=R2 is taken as R and C1=C2 is taken as C\n",
-      "#And, f=1/(2*pi*sqrt(R1*R2*C1*C2))is transformed to f=1/(2*pi*R*C).\n",
-      "R=R1*10**3;                                   #kilo ohm\n",
-      "C=C1*10**-12;                                   #pF\n",
-      "f=1/(2*pi*R*C);           #Frequency of oscillation, Hz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency of oscillation=%dHz.\"%f);\n",
-      "\n",
-      "\n",
-      "#Note : The frequency of oscillation is calculated in the text as 2892Hz but here it gets approximated to 2893 Hz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of oscillation=2893Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.11 : Page number 384\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "#a.c equivalent values of the crystal:\n",
-      "L=1.0;                          #Inductance , H\n",
-      "C=0.01;                         #Capacitance , pF\n",
-      "R=1000.0;                       #Resistance , ohm\n",
-      "Cm=20.0;                        #Mounting capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "fs=(1/(2*round(pi,2)*sqrt(L*C*10**-12)))/1000;               #Series resonant frrequency, kHz\n",
-      "CT=(C*Cm/(C+Cm));                                            #Total capacitance, pF\n",
-      "fp=(1/(2*round(pi,2)*sqrt(L*CT*10**-12)))/1000;              #Prallel resonant frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"fs=%.0fkHz and fp=%.0fkHz.\"%(fs,fp));\n",
-      "\n",
-      "#Note: fs and fp are calculated in the text as 1589kHz and 1590kHz, but here it gets approximated to 1592kHz and 1593kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fs=1592kHz and fp=1593kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_5.ipynb
deleted file mode 100755
index 5e35882c..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_5.ipynb
+++ /dev/null
@@ -1,502 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:ec0d27209d08b0b95750f66ce9ee21af5ef586e23d0bf0ea218aa20a4ce63e43"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 14: SINUSOIDAL OSCILLATORS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.1 : Page number 371-372\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "L1=58.6;                #Inductance, micro henry\n",
-      "C1=300.0;               #Capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "f=(1/(2*round(pi,2)*sqrt(L1*10**-6*C1*10**-12)))/1000;         #Frequency of oscillation, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"frequency of oscillation=%dkHz\"%f);\n",
-      "\n",
-      "\n",
-      "#Note : The frequency has been calculated in the text  as 1199kHz but here the answer gets approximated to 1200kHz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "frequency of oscillation=1200kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.2 : Page number 372\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "L1=1.0;                     #Inductance , mH\n",
-      "f=1.0;                      #frequency of oscillation, GHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, f=1/(2*pi*sqrt(L1*C1)),\n",
-      "C1=(1/(L1*10**-3*(f*10**12*2*pi)**2))*10**12;               #Capacitance, pF\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The Capacitance of the capacitor of the LC oscillator=%.2epF\"%C1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Capacitance of the capacitor of the LC oscillator=2.53e-11pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.3 : Page number 373-374\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "C1=0.001;               #Capacitor C1, microfarad\n",
-      "C2=0.01;                #Capacitor C2, microfarad\n",
-      "L=15.0;                 #Inductance, microhenry\n",
-      "\n",
-      "#Calculation\n",
-      "CT=C1*C2/(C1+C2);               #Total capacitance\n",
-      "\n",
-      "#(i) Operating frequency\n",
-      "f=(1/(2*pi*sqrt(CT*10**-6*L*10**-6)))/1000;             #Operating frequency, kHz\n",
-      "\n",
-      "#(ii) Feedback fraction\n",
-      "mv=C1/C2;                                   #Feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The operating frequency=%dkHz\"%f);\n",
-      "print(\"(ii) The feedback fraction=%.1f\"%mv);\n",
-      "\n",
-      "#Note : The operating frequency is calculated in the text as 1361kHz but here it has been approximated to 1362kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The operating frequency=1362kHz\n",
-        "(ii) The feedback fraction=0.1\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.4 : Page number 374: Page number\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "mv=0.25;                    #Feedback fraction\n",
-      "L=1.0;                      #Inductance, mH\n",
-      "f=1.0;                      #Operating frequeny, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, f=1/(2*pi*sqrt(L*C))\n",
-      "CT=round((1/(L*10**-3*(2*pi*f*10**6)**2))*10**12,1);              #Total capacitance, pF\n",
-      "\n",
-      "#Since, mv=C1/C2 and CT=C1*C2/(C1+C2) or CT=C2/(1+ (C2/C1)),\n",
-      "#From the above equations, substituting value of mv and calculaing value of C2,\n",
-      "C2=CT*(1+(1/mv));               #Capacitance of C2 capactior, pF\n",
-      "C1=mv*C2;                       #Capacitance of C1 capacitor, pF\n",
-      "\n",
-      "#Result\n",
-      "print(\"C1=%.1fpF and C2=%.1fpF\"%(C1,C2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "C1=31.6pF and C2=126.5pF\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.5 : Page number 375-376\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable decalaration\n",
-      "L1=1000.0;                  #Inductance of L1 inductor, microhenry\n",
-      "L2=100.0;                   #Inductance of L2 inductor, microhenry\n",
-      "M=20.0;                     #Mutual inductance, microhenry\n",
-      "C=20.0;                     #Capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "LT=L1+L2+2*M;                   #Total inductance, microhenry\n",
-      "\n",
-      "#(i) Operating frequency\n",
-      "f=(1/(2*pi*sqrt(LT*10**-6*C*10**-12)))/1000;           #Operating frequency, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "mv=L2/L1;                   #feedback fraction\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The operating frequency=%dkHz.\"%f);\n",
-      "print(\"(ii) The feedback fraction=%.1f.\"%mv);\n",
-      "\n",
-      "#Note : The operating frequecy has been calculated in the text as 1052kHz but here it gets approximated to 1054kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The operating frequency=1054kHz.\n",
-        "(ii) The feedback fraction=0.1.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.6 : Page number 376\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=1.0;                  #Capacitance, pF\n",
-      "f=1.0;                  #Frequency, MHz\n",
-      "mv=0.2;                 #Feedback frequency\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "LT=(1/(C*10**-12*(2*pi*f*10**6)**2))*1000;               #Total inductance, mH\n",
-      "\n",
-      "#Since, mv=L2/L1 or L2=mv*L1 and L1+L2=LT or L1(1+mv)=LT,\n",
-      "L1=LT/(1+mv);                   #Inductance of L1 inductor, mH\n",
-      "L2=L1*mv;                       #inductance of L2 inductor, mH\n",
-      "\n",
-      "#Result\n",
-      "print(\"L1=%.1fmH and L2=%.2fmH.\"%(L1,L2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "L1=21.1mH and L2=4.22mH.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.7 : Page number 378\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "R1=1.0;                 #Resistor R1, mega ohm\n",
-      "R2=R1;                  #Resistor R2, mega ohm\n",
-      "R3=R1;                  #Resistor R3, mega ohm\n",
-      "C1=68.0;                #Capacitor C1, pF\n",
-      "C2=C1;                  #Capacitor C2, pF\n",
-      "C3=C1;                  #Capacitor C3, pF\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "R=R1*10**6;                   #Resistance of the resistors of phase shift circuit, ohm\n",
-      "C=C1*10**-12;                 #Capacitance of the capacitors of phase shift circuit, F\n",
-      "fo=1/(2*pi*R*C*sqrt(6));      #Frequency of oscillation, Hz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency of oscillation=%dHz\"%fo);\n",
-      "\n",
-      "#Note: The frequency of oscillation had been calculated in the text as 954Hz, but here it gets approximated to 955 HZ.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of oscillation=955Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.8 : Page number 378\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=5.0;                      #Capacitance of the capacitors of phase shift circuit, pF\n",
-      "fo=800.0;                    #Required frequency of oscillation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, fo=1/(2*pi*R*C*sqrt(6))\n",
-      "R=(1/(2*pi*C*10**-12*fo*10**3*sqrt(6)))/1000;                   #Resistance of the resistors of phase shift circuit, kilo ohm\n",
-      "\n",
-      "#Result\n",
-      "print(\"R=%.1f kilo ohm.\"%R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R=16.2 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.9 : Page number 380\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "#Resistance of R1 and R2 resistors of the R-C bridge circuit\n",
-      "R1=220.0;                           #kilo ohm \n",
-      "R2=220.0;                           #kilo ohm\n",
-      "\n",
-      "#Capacitance of C1 and C2 the capacitors of the R-C bridge circuit\n",
-      "C1=250.0;                           #pF\n",
-      "C2=250.0;                           #pF\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, R1=R2 and C1=C2, R1=R2 is taken as R and C1=C2 is taken as C\n",
-      "#And, f=1/(2*pi*sqrt(R1*R2*C1*C2))is transformed to f=1/(2*pi*R*C).\n",
-      "R=R1*10**3;                                   #kilo ohm\n",
-      "C=C1*10**-12;                                   #pF\n",
-      "f=1/(2*pi*R*C);           #Frequency of oscillation, Hz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency of oscillation=%dHz.\"%f);\n",
-      "\n",
-      "\n",
-      "#Note : The frequency of oscillation is calculated in the text as 2892Hz but here it gets approximated to 2893 Hz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of oscillation=2893Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.11 : Page number 384\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "#a.c equivalent values of the crystal:\n",
-      "L=1.0;                          #Inductance , H\n",
-      "C=0.01;                         #Capacitance , pF\n",
-      "R=1000.0;                       #Resistance , ohm\n",
-      "Cm=20.0;                        #Mounting capacitance, pF\n",
-      "\n",
-      "#Calculation\n",
-      "fs=(1/(2*round(pi,2)*sqrt(L*C*10**-12)))/1000;               #Series resonant frrequency, kHz\n",
-      "CT=(C*Cm/(C+Cm));                                            #Total capacitance, pF\n",
-      "fp=(1/(2*round(pi,2)*sqrt(L*CT*10**-12)))/1000;              #Prallel resonant frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"fs=%.0fkHz and fp=%.0fkHz.\"%(fs,fp));\n",
-      "\n",
-      "#Note: fs and fp are calculated in the text as 1589kHz and 1590kHz, but here it gets approximated to 1592kHz and 1593kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "fs=1592kHz and fp=1593kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15.ipynb
deleted file mode 100755
index e649cc91..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15.ipynb
+++ /dev/null
@@ -1,482 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:034eec32676d4e7abdedfb3bf68426d81a2d1483fc668bcbfdb5be18cec2e406"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 15: TRANSISTOR TUNED AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.1 : Page number 394"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=250.0*10**-12;                       #Capacitor of parallel resonant circuit, F\n",
-      "L=1.25*10**-3;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                #Resistor of the parallel resonant circuit, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(ii)  Impedance of the circuit at resonance\n",
-      "Zr=(L/(C*R))/1000;                          #Impedance of the circuit at resonance, kilo ohm\n",
-      "\n",
-      "#(iii) Quality factor of the circuit\n",
-      "Q=2*pi*(fr*10**3)*L/R;                          #Quality factor of the circuit\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The resonant frequency=%.1fkHz.\"%fr);\n",
-      "print(\"(ii) The impedance of the circuit at resonance=%d kilo ohm.\"%Zr);\n",
-      "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The resonant frequency=284.7kHz.\n",
-        "(ii) The impedance of the circuit at resonance=500 kilo ohm.\n",
-        "(iii) The quality factor of the circuit=223.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.2 : Page number 394-395\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=100.0*10**-12;                        #Capacitor of parallel resonant circuit, F\n",
-      "L=100.0*10**-6;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                 #Resistor of the parallel resonant circuit, ohm\n",
-      "V=10.0;                                 #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(ii)  Impedance of the circuit at resonance\n",
-      "Zr=(L/(C*R))/10**6;                          #Impedance of the circuit at resonance, mega ohm\n",
-      "\n",
-      "I=V/Zr;                                     #Line current at resonance, microampere\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The resonant frequency=%.2fkHz.\"%fr);\n",
-      "print(\"(ii) The impedance of the circuit at resonance=%.1f mega ohm.\"%Zr);\n",
-      "print(\"The line current at resonance=%d micro ampere.\"%I);\n",
-      "\n",
-      "#Note : The resonant frequency in the text has been calculated as 1592.28 kHz, but here it gets approximated to 1591.47 kHz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The resonant frequency=1591.47kHz.\n",
-        "(ii) The impedance of the circuit at resonance=0.1 mega ohm.\n",
-        "The line current at resonance=100 micro ampere.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.3 : Page number 395\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=250.0*10**-12;                              #Capacitor of parallel resonant circuit, F\n",
-      "Zr=500.0*10**3;                               #Dynamic impedance, ohm\n",
-      "R=10.0;                                       #Resistance of the coil, ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since,Zr=L/CR,\n",
-      "L=(Zr*C*R)*10**3;                           #Inductance of the coil, mH\n",
-      "\n",
-      "#(ii) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*10**-3*C))-(R/(L*10**-3))**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(iii) Quality factor of the circuit\n",
-      "Q=2*pi*(fr*10**3)*(L*10**-3)/R;                          #Quality factor of the circuit\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The inductance of the coil=%.2fmH.\"%L);\n",
-      "print(\"(ii) The resonant frequency=%.1fkHz.\"%fr);\n",
-      "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The inductance of the coil=1.25mH.\n",
-        "(ii) The resonant frequency=284.7kHz.\n",
-        "(iii) The quality factor of the circuit=223.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.4  : Page number 397\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Q=60.0;                     #Quality factor of the tuned amplifier\n",
-      "fr=1200.0;                  #Resonant frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "BW=fr/Q;                    #Bandwidth, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "f1=fr-(BW/2);                       #Lower cut-off frequency, kHz\n",
-      "f2=fr+(BW/2);                       #Upper cut-off frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The bandwidth=%dkHz\"%BW);\n",
-      "print(\"(ii) The lower and upper cut-off frequencies are=%dkHz and %dkHz.\"%(f1,f2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The bandwidth=20kHz\n",
-        "(ii) The lower and upper cut-off frequencies are=1190kHz and 1210kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.5 : Page number 397\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fr=2.0;                     #Resonant frequency, MHz\n",
-      "BW=50.0;                    #Bandwidth, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, bandwidth=resonant_frequency/quality_factor\n",
-      "Q=(fr*10**6)/(BW*10**3);                            #Quality factor\n",
-      "\n",
-      "#Result\n",
-      "print(\"The quality factor=%d\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The quality factor=40\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.7 : Page number 400\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=0.1*10**-6;                          #Capacitor of parallel resonant circuit, F\n",
-      "L=33.0*10**-3;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=25.0;                                #Resistor of the parallel resonant circuit, ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fr=(1/(2*pi*sqrt(L*C)))/1000;                  #Resonant frequency, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "XL=2*pi*(fr*10**3)*L;                           #Inductive reactance, ohm\n",
-      "Q=round(XL/R,0);                                #Quality factor\n",
-      "\n",
-      "#(iii)\n",
-      "BW=(fr*10**3)/Q;                                #Bandwidth\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The resonant frequency=%.2fkHz\"%fr);\n",
-      "print(\"(ii)  The quality factor= %d.\"%Q);\n",
-      "print(\"(iii) The bandwidth=%dHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The resonant frequency=2.77kHz\n",
-        "(ii)  The quality factor= 23.\n",
-        "(iii) The bandwidth=120Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.8 : Page number 401-402\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "BW_dt=200.0;                       #Bandwidth, kHz\n",
-      "fr=10.0;                        #Operating frequency, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, BW_dt=k*fr (i.e.,co-efficient_of_coupling * operating_frequency)\n",
-      "k=BW_dt/(fr*10**3);                     #co-efficient of coupling\n",
-      "\n",
-      "#Result\n",
-      "print(\"The co-efficient of coupling=%.2f.\"%k);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The co-efficient of coupling=0.02.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.9 : Page number 405\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=500.0*10**-12;                          #Capacitor of parallel resonant circuit, F\n",
-      "L=50.7*10**-6;                            #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                   #Resistor of the parallel resonant circuit, ohm\n",
-      "RL=1.0;                                   #Load resistance, mega ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fr=round((1/(2*pi*sqrt(L*C)))/1000);                  #Resonant frequency, Hz\n",
-      "\n",
-      "#(ii)\n",
-      "R_dc=R;                                             #d.c load, ohm\n",
-      "XL=2*pi*(fr*1000)*L;                                       #Inductive reactance, ohm\n",
-      "Q_coil=round(XL/R,1);                               #Quality factor\n",
-      "R_P=(Q_coil*XL)/1000   ;                            #Equivalent parallel resistance, kilo ohm\n",
-      "R_AC=(R_P*RL*10**3)/(R_P+RL*10**3);                 #A.C load,kilo ohm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The resonant frequency=%dkHz\"%fr);\n",
-      "print(\"(ii)  d.c load=%d ohm and a.c load=%d kilo ohm.\"%(R_dc,R_AC));\n",
-      "\n",
-      "#Note: In the text resonant frequency has been wrongly calculated to 106kHz but its actual value is approximately 1000kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The resonant frequency=1000kHz\n",
-        "(ii)  d.c load=10 ohm and a.c load=10 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.10 : Page number 406-407\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=50.0;                #Load resistance, ohm\n",
-      "n=5;                    #Turns ratio of the transformer\n",
-      "VCC=50.0;               #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_ac=n**2*RL;               #A.C load, ohm\n",
-      "\n",
-      "#(ii)\n",
-      "P_o_max=VCC**2/(2*R_ac);                #Maximum load power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The a.c load=%d ohm\"%R_ac);\n",
-      "print(\"(ii) Maximum load power=%dW\"%P_o_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The a.c load=1250 ohm\n",
-        "(ii) Maximum load power=1W\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.11 : Page number 407\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_D=4.0;                    #Maximum power dissipation, mW\n",
-      "P_o_max=1.0;                #Maximum load power, W\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "max_collector_eff=(P_o_max/(P_o_max+(P_D/1000)))*100;         #Maximum collector efficiency\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector efficiency=%.1f%%\"%max_collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector efficiency=99.6%\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_1.ipynb
deleted file mode 100755
index e649cc91..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_1.ipynb
+++ /dev/null
@@ -1,482 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:034eec32676d4e7abdedfb3bf68426d81a2d1483fc668bcbfdb5be18cec2e406"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 15: TRANSISTOR TUNED AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.1 : Page number 394"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=250.0*10**-12;                       #Capacitor of parallel resonant circuit, F\n",
-      "L=1.25*10**-3;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                #Resistor of the parallel resonant circuit, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(ii)  Impedance of the circuit at resonance\n",
-      "Zr=(L/(C*R))/1000;                          #Impedance of the circuit at resonance, kilo ohm\n",
-      "\n",
-      "#(iii) Quality factor of the circuit\n",
-      "Q=2*pi*(fr*10**3)*L/R;                          #Quality factor of the circuit\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The resonant frequency=%.1fkHz.\"%fr);\n",
-      "print(\"(ii) The impedance of the circuit at resonance=%d kilo ohm.\"%Zr);\n",
-      "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The resonant frequency=284.7kHz.\n",
-        "(ii) The impedance of the circuit at resonance=500 kilo ohm.\n",
-        "(iii) The quality factor of the circuit=223.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.2 : Page number 394-395\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=100.0*10**-12;                        #Capacitor of parallel resonant circuit, F\n",
-      "L=100.0*10**-6;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                 #Resistor of the parallel resonant circuit, ohm\n",
-      "V=10.0;                                 #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(ii)  Impedance of the circuit at resonance\n",
-      "Zr=(L/(C*R))/10**6;                          #Impedance of the circuit at resonance, mega ohm\n",
-      "\n",
-      "I=V/Zr;                                     #Line current at resonance, microampere\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The resonant frequency=%.2fkHz.\"%fr);\n",
-      "print(\"(ii) The impedance of the circuit at resonance=%.1f mega ohm.\"%Zr);\n",
-      "print(\"The line current at resonance=%d micro ampere.\"%I);\n",
-      "\n",
-      "#Note : The resonant frequency in the text has been calculated as 1592.28 kHz, but here it gets approximated to 1591.47 kHz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The resonant frequency=1591.47kHz.\n",
-        "(ii) The impedance of the circuit at resonance=0.1 mega ohm.\n",
-        "The line current at resonance=100 micro ampere.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.3 : Page number 395\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=250.0*10**-12;                              #Capacitor of parallel resonant circuit, F\n",
-      "Zr=500.0*10**3;                               #Dynamic impedance, ohm\n",
-      "R=10.0;                                       #Resistance of the coil, ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since,Zr=L/CR,\n",
-      "L=(Zr*C*R)*10**3;                           #Inductance of the coil, mH\n",
-      "\n",
-      "#(ii) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*10**-3*C))-(R/(L*10**-3))**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(iii) Quality factor of the circuit\n",
-      "Q=2*pi*(fr*10**3)*(L*10**-3)/R;                          #Quality factor of the circuit\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The inductance of the coil=%.2fmH.\"%L);\n",
-      "print(\"(ii) The resonant frequency=%.1fkHz.\"%fr);\n",
-      "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The inductance of the coil=1.25mH.\n",
-        "(ii) The resonant frequency=284.7kHz.\n",
-        "(iii) The quality factor of the circuit=223.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.4  : Page number 397\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Q=60.0;                     #Quality factor of the tuned amplifier\n",
-      "fr=1200.0;                  #Resonant frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "BW=fr/Q;                    #Bandwidth, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "f1=fr-(BW/2);                       #Lower cut-off frequency, kHz\n",
-      "f2=fr+(BW/2);                       #Upper cut-off frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The bandwidth=%dkHz\"%BW);\n",
-      "print(\"(ii) The lower and upper cut-off frequencies are=%dkHz and %dkHz.\"%(f1,f2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The bandwidth=20kHz\n",
-        "(ii) The lower and upper cut-off frequencies are=1190kHz and 1210kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.5 : Page number 397\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fr=2.0;                     #Resonant frequency, MHz\n",
-      "BW=50.0;                    #Bandwidth, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, bandwidth=resonant_frequency/quality_factor\n",
-      "Q=(fr*10**6)/(BW*10**3);                            #Quality factor\n",
-      "\n",
-      "#Result\n",
-      "print(\"The quality factor=%d\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The quality factor=40\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.7 : Page number 400\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=0.1*10**-6;                          #Capacitor of parallel resonant circuit, F\n",
-      "L=33.0*10**-3;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=25.0;                                #Resistor of the parallel resonant circuit, ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fr=(1/(2*pi*sqrt(L*C)))/1000;                  #Resonant frequency, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "XL=2*pi*(fr*10**3)*L;                           #Inductive reactance, ohm\n",
-      "Q=round(XL/R,0);                                #Quality factor\n",
-      "\n",
-      "#(iii)\n",
-      "BW=(fr*10**3)/Q;                                #Bandwidth\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The resonant frequency=%.2fkHz\"%fr);\n",
-      "print(\"(ii)  The quality factor= %d.\"%Q);\n",
-      "print(\"(iii) The bandwidth=%dHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The resonant frequency=2.77kHz\n",
-        "(ii)  The quality factor= 23.\n",
-        "(iii) The bandwidth=120Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.8 : Page number 401-402\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "BW_dt=200.0;                       #Bandwidth, kHz\n",
-      "fr=10.0;                        #Operating frequency, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, BW_dt=k*fr (i.e.,co-efficient_of_coupling * operating_frequency)\n",
-      "k=BW_dt/(fr*10**3);                     #co-efficient of coupling\n",
-      "\n",
-      "#Result\n",
-      "print(\"The co-efficient of coupling=%.2f.\"%k);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The co-efficient of coupling=0.02.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.9 : Page number 405\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=500.0*10**-12;                          #Capacitor of parallel resonant circuit, F\n",
-      "L=50.7*10**-6;                            #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                   #Resistor of the parallel resonant circuit, ohm\n",
-      "RL=1.0;                                   #Load resistance, mega ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fr=round((1/(2*pi*sqrt(L*C)))/1000);                  #Resonant frequency, Hz\n",
-      "\n",
-      "#(ii)\n",
-      "R_dc=R;                                             #d.c load, ohm\n",
-      "XL=2*pi*(fr*1000)*L;                                       #Inductive reactance, ohm\n",
-      "Q_coil=round(XL/R,1);                               #Quality factor\n",
-      "R_P=(Q_coil*XL)/1000   ;                            #Equivalent parallel resistance, kilo ohm\n",
-      "R_AC=(R_P*RL*10**3)/(R_P+RL*10**3);                 #A.C load,kilo ohm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The resonant frequency=%dkHz\"%fr);\n",
-      "print(\"(ii)  d.c load=%d ohm and a.c load=%d kilo ohm.\"%(R_dc,R_AC));\n",
-      "\n",
-      "#Note: In the text resonant frequency has been wrongly calculated to 106kHz but its actual value is approximately 1000kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The resonant frequency=1000kHz\n",
-        "(ii)  d.c load=10 ohm and a.c load=10 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.10 : Page number 406-407\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=50.0;                #Load resistance, ohm\n",
-      "n=5;                    #Turns ratio of the transformer\n",
-      "VCC=50.0;               #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_ac=n**2*RL;               #A.C load, ohm\n",
-      "\n",
-      "#(ii)\n",
-      "P_o_max=VCC**2/(2*R_ac);                #Maximum load power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The a.c load=%d ohm\"%R_ac);\n",
-      "print(\"(ii) Maximum load power=%dW\"%P_o_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The a.c load=1250 ohm\n",
-        "(ii) Maximum load power=1W\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.11 : Page number 407\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_D=4.0;                    #Maximum power dissipation, mW\n",
-      "P_o_max=1.0;                #Maximum load power, W\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "max_collector_eff=(P_o_max/(P_o_max+(P_D/1000)))*100;         #Maximum collector efficiency\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector efficiency=%.1f%%\"%max_collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector efficiency=99.6%\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_2.ipynb
deleted file mode 100755
index e649cc91..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_2.ipynb
+++ /dev/null
@@ -1,482 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:034eec32676d4e7abdedfb3bf68426d81a2d1483fc668bcbfdb5be18cec2e406"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 15: TRANSISTOR TUNED AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.1 : Page number 394"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=250.0*10**-12;                       #Capacitor of parallel resonant circuit, F\n",
-      "L=1.25*10**-3;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                #Resistor of the parallel resonant circuit, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(ii)  Impedance of the circuit at resonance\n",
-      "Zr=(L/(C*R))/1000;                          #Impedance of the circuit at resonance, kilo ohm\n",
-      "\n",
-      "#(iii) Quality factor of the circuit\n",
-      "Q=2*pi*(fr*10**3)*L/R;                          #Quality factor of the circuit\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The resonant frequency=%.1fkHz.\"%fr);\n",
-      "print(\"(ii) The impedance of the circuit at resonance=%d kilo ohm.\"%Zr);\n",
-      "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The resonant frequency=284.7kHz.\n",
-        "(ii) The impedance of the circuit at resonance=500 kilo ohm.\n",
-        "(iii) The quality factor of the circuit=223.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.2 : Page number 394-395\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=100.0*10**-12;                        #Capacitor of parallel resonant circuit, F\n",
-      "L=100.0*10**-6;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                 #Resistor of the parallel resonant circuit, ohm\n",
-      "V=10.0;                                 #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(ii)  Impedance of the circuit at resonance\n",
-      "Zr=(L/(C*R))/10**6;                          #Impedance of the circuit at resonance, mega ohm\n",
-      "\n",
-      "I=V/Zr;                                     #Line current at resonance, microampere\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The resonant frequency=%.2fkHz.\"%fr);\n",
-      "print(\"(ii) The impedance of the circuit at resonance=%.1f mega ohm.\"%Zr);\n",
-      "print(\"The line current at resonance=%d micro ampere.\"%I);\n",
-      "\n",
-      "#Note : The resonant frequency in the text has been calculated as 1592.28 kHz, but here it gets approximated to 1591.47 kHz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The resonant frequency=1591.47kHz.\n",
-        "(ii) The impedance of the circuit at resonance=0.1 mega ohm.\n",
-        "The line current at resonance=100 micro ampere.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.3 : Page number 395\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=250.0*10**-12;                              #Capacitor of parallel resonant circuit, F\n",
-      "Zr=500.0*10**3;                               #Dynamic impedance, ohm\n",
-      "R=10.0;                                       #Resistance of the coil, ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since,Zr=L/CR,\n",
-      "L=(Zr*C*R)*10**3;                           #Inductance of the coil, mH\n",
-      "\n",
-      "#(ii) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*10**-3*C))-(R/(L*10**-3))**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(iii) Quality factor of the circuit\n",
-      "Q=2*pi*(fr*10**3)*(L*10**-3)/R;                          #Quality factor of the circuit\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The inductance of the coil=%.2fmH.\"%L);\n",
-      "print(\"(ii) The resonant frequency=%.1fkHz.\"%fr);\n",
-      "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The inductance of the coil=1.25mH.\n",
-        "(ii) The resonant frequency=284.7kHz.\n",
-        "(iii) The quality factor of the circuit=223.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.4  : Page number 397\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Q=60.0;                     #Quality factor of the tuned amplifier\n",
-      "fr=1200.0;                  #Resonant frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "BW=fr/Q;                    #Bandwidth, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "f1=fr-(BW/2);                       #Lower cut-off frequency, kHz\n",
-      "f2=fr+(BW/2);                       #Upper cut-off frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The bandwidth=%dkHz\"%BW);\n",
-      "print(\"(ii) The lower and upper cut-off frequencies are=%dkHz and %dkHz.\"%(f1,f2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The bandwidth=20kHz\n",
-        "(ii) The lower and upper cut-off frequencies are=1190kHz and 1210kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.5 : Page number 397\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fr=2.0;                     #Resonant frequency, MHz\n",
-      "BW=50.0;                    #Bandwidth, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, bandwidth=resonant_frequency/quality_factor\n",
-      "Q=(fr*10**6)/(BW*10**3);                            #Quality factor\n",
-      "\n",
-      "#Result\n",
-      "print(\"The quality factor=%d\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The quality factor=40\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.7 : Page number 400\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=0.1*10**-6;                          #Capacitor of parallel resonant circuit, F\n",
-      "L=33.0*10**-3;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=25.0;                                #Resistor of the parallel resonant circuit, ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fr=(1/(2*pi*sqrt(L*C)))/1000;                  #Resonant frequency, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "XL=2*pi*(fr*10**3)*L;                           #Inductive reactance, ohm\n",
-      "Q=round(XL/R,0);                                #Quality factor\n",
-      "\n",
-      "#(iii)\n",
-      "BW=(fr*10**3)/Q;                                #Bandwidth\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The resonant frequency=%.2fkHz\"%fr);\n",
-      "print(\"(ii)  The quality factor= %d.\"%Q);\n",
-      "print(\"(iii) The bandwidth=%dHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The resonant frequency=2.77kHz\n",
-        "(ii)  The quality factor= 23.\n",
-        "(iii) The bandwidth=120Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.8 : Page number 401-402\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "BW_dt=200.0;                       #Bandwidth, kHz\n",
-      "fr=10.0;                        #Operating frequency, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, BW_dt=k*fr (i.e.,co-efficient_of_coupling * operating_frequency)\n",
-      "k=BW_dt/(fr*10**3);                     #co-efficient of coupling\n",
-      "\n",
-      "#Result\n",
-      "print(\"The co-efficient of coupling=%.2f.\"%k);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The co-efficient of coupling=0.02.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.9 : Page number 405\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=500.0*10**-12;                          #Capacitor of parallel resonant circuit, F\n",
-      "L=50.7*10**-6;                            #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                   #Resistor of the parallel resonant circuit, ohm\n",
-      "RL=1.0;                                   #Load resistance, mega ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fr=round((1/(2*pi*sqrt(L*C)))/1000);                  #Resonant frequency, Hz\n",
-      "\n",
-      "#(ii)\n",
-      "R_dc=R;                                             #d.c load, ohm\n",
-      "XL=2*pi*(fr*1000)*L;                                       #Inductive reactance, ohm\n",
-      "Q_coil=round(XL/R,1);                               #Quality factor\n",
-      "R_P=(Q_coil*XL)/1000   ;                            #Equivalent parallel resistance, kilo ohm\n",
-      "R_AC=(R_P*RL*10**3)/(R_P+RL*10**3);                 #A.C load,kilo ohm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The resonant frequency=%dkHz\"%fr);\n",
-      "print(\"(ii)  d.c load=%d ohm and a.c load=%d kilo ohm.\"%(R_dc,R_AC));\n",
-      "\n",
-      "#Note: In the text resonant frequency has been wrongly calculated to 106kHz but its actual value is approximately 1000kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The resonant frequency=1000kHz\n",
-        "(ii)  d.c load=10 ohm and a.c load=10 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.10 : Page number 406-407\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=50.0;                #Load resistance, ohm\n",
-      "n=5;                    #Turns ratio of the transformer\n",
-      "VCC=50.0;               #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_ac=n**2*RL;               #A.C load, ohm\n",
-      "\n",
-      "#(ii)\n",
-      "P_o_max=VCC**2/(2*R_ac);                #Maximum load power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The a.c load=%d ohm\"%R_ac);\n",
-      "print(\"(ii) Maximum load power=%dW\"%P_o_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The a.c load=1250 ohm\n",
-        "(ii) Maximum load power=1W\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.11 : Page number 407\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_D=4.0;                    #Maximum power dissipation, mW\n",
-      "P_o_max=1.0;                #Maximum load power, W\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "max_collector_eff=(P_o_max/(P_o_max+(P_D/1000)))*100;         #Maximum collector efficiency\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector efficiency=%.1f%%\"%max_collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector efficiency=99.6%\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_3.ipynb
deleted file mode 100755
index e649cc91..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_3.ipynb
+++ /dev/null
@@ -1,482 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:034eec32676d4e7abdedfb3bf68426d81a2d1483fc668bcbfdb5be18cec2e406"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 15: TRANSISTOR TUNED AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.1 : Page number 394"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=250.0*10**-12;                       #Capacitor of parallel resonant circuit, F\n",
-      "L=1.25*10**-3;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                #Resistor of the parallel resonant circuit, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(ii)  Impedance of the circuit at resonance\n",
-      "Zr=(L/(C*R))/1000;                          #Impedance of the circuit at resonance, kilo ohm\n",
-      "\n",
-      "#(iii) Quality factor of the circuit\n",
-      "Q=2*pi*(fr*10**3)*L/R;                          #Quality factor of the circuit\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The resonant frequency=%.1fkHz.\"%fr);\n",
-      "print(\"(ii) The impedance of the circuit at resonance=%d kilo ohm.\"%Zr);\n",
-      "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The resonant frequency=284.7kHz.\n",
-        "(ii) The impedance of the circuit at resonance=500 kilo ohm.\n",
-        "(iii) The quality factor of the circuit=223.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.2 : Page number 394-395\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=100.0*10**-12;                        #Capacitor of parallel resonant circuit, F\n",
-      "L=100.0*10**-6;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                 #Resistor of the parallel resonant circuit, ohm\n",
-      "V=10.0;                                 #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(ii)  Impedance of the circuit at resonance\n",
-      "Zr=(L/(C*R))/10**6;                          #Impedance of the circuit at resonance, mega ohm\n",
-      "\n",
-      "I=V/Zr;                                     #Line current at resonance, microampere\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The resonant frequency=%.2fkHz.\"%fr);\n",
-      "print(\"(ii) The impedance of the circuit at resonance=%.1f mega ohm.\"%Zr);\n",
-      "print(\"The line current at resonance=%d micro ampere.\"%I);\n",
-      "\n",
-      "#Note : The resonant frequency in the text has been calculated as 1592.28 kHz, but here it gets approximated to 1591.47 kHz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The resonant frequency=1591.47kHz.\n",
-        "(ii) The impedance of the circuit at resonance=0.1 mega ohm.\n",
-        "The line current at resonance=100 micro ampere.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.3 : Page number 395\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=250.0*10**-12;                              #Capacitor of parallel resonant circuit, F\n",
-      "Zr=500.0*10**3;                               #Dynamic impedance, ohm\n",
-      "R=10.0;                                       #Resistance of the coil, ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since,Zr=L/CR,\n",
-      "L=(Zr*C*R)*10**3;                           #Inductance of the coil, mH\n",
-      "\n",
-      "#(ii) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*10**-3*C))-(R/(L*10**-3))**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(iii) Quality factor of the circuit\n",
-      "Q=2*pi*(fr*10**3)*(L*10**-3)/R;                          #Quality factor of the circuit\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The inductance of the coil=%.2fmH.\"%L);\n",
-      "print(\"(ii) The resonant frequency=%.1fkHz.\"%fr);\n",
-      "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The inductance of the coil=1.25mH.\n",
-        "(ii) The resonant frequency=284.7kHz.\n",
-        "(iii) The quality factor of the circuit=223.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.4  : Page number 397\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Q=60.0;                     #Quality factor of the tuned amplifier\n",
-      "fr=1200.0;                  #Resonant frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "BW=fr/Q;                    #Bandwidth, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "f1=fr-(BW/2);                       #Lower cut-off frequency, kHz\n",
-      "f2=fr+(BW/2);                       #Upper cut-off frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The bandwidth=%dkHz\"%BW);\n",
-      "print(\"(ii) The lower and upper cut-off frequencies are=%dkHz and %dkHz.\"%(f1,f2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The bandwidth=20kHz\n",
-        "(ii) The lower and upper cut-off frequencies are=1190kHz and 1210kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.5 : Page number 397\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fr=2.0;                     #Resonant frequency, MHz\n",
-      "BW=50.0;                    #Bandwidth, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, bandwidth=resonant_frequency/quality_factor\n",
-      "Q=(fr*10**6)/(BW*10**3);                            #Quality factor\n",
-      "\n",
-      "#Result\n",
-      "print(\"The quality factor=%d\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The quality factor=40\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.7 : Page number 400\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=0.1*10**-6;                          #Capacitor of parallel resonant circuit, F\n",
-      "L=33.0*10**-3;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=25.0;                                #Resistor of the parallel resonant circuit, ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fr=(1/(2*pi*sqrt(L*C)))/1000;                  #Resonant frequency, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "XL=2*pi*(fr*10**3)*L;                           #Inductive reactance, ohm\n",
-      "Q=round(XL/R,0);                                #Quality factor\n",
-      "\n",
-      "#(iii)\n",
-      "BW=(fr*10**3)/Q;                                #Bandwidth\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The resonant frequency=%.2fkHz\"%fr);\n",
-      "print(\"(ii)  The quality factor= %d.\"%Q);\n",
-      "print(\"(iii) The bandwidth=%dHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The resonant frequency=2.77kHz\n",
-        "(ii)  The quality factor= 23.\n",
-        "(iii) The bandwidth=120Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.8 : Page number 401-402\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "BW_dt=200.0;                       #Bandwidth, kHz\n",
-      "fr=10.0;                        #Operating frequency, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, BW_dt=k*fr (i.e.,co-efficient_of_coupling * operating_frequency)\n",
-      "k=BW_dt/(fr*10**3);                     #co-efficient of coupling\n",
-      "\n",
-      "#Result\n",
-      "print(\"The co-efficient of coupling=%.2f.\"%k);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The co-efficient of coupling=0.02.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.9 : Page number 405\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=500.0*10**-12;                          #Capacitor of parallel resonant circuit, F\n",
-      "L=50.7*10**-6;                            #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                   #Resistor of the parallel resonant circuit, ohm\n",
-      "RL=1.0;                                   #Load resistance, mega ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fr=round((1/(2*pi*sqrt(L*C)))/1000);                  #Resonant frequency, Hz\n",
-      "\n",
-      "#(ii)\n",
-      "R_dc=R;                                             #d.c load, ohm\n",
-      "XL=2*pi*(fr*1000)*L;                                       #Inductive reactance, ohm\n",
-      "Q_coil=round(XL/R,1);                               #Quality factor\n",
-      "R_P=(Q_coil*XL)/1000   ;                            #Equivalent parallel resistance, kilo ohm\n",
-      "R_AC=(R_P*RL*10**3)/(R_P+RL*10**3);                 #A.C load,kilo ohm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The resonant frequency=%dkHz\"%fr);\n",
-      "print(\"(ii)  d.c load=%d ohm and a.c load=%d kilo ohm.\"%(R_dc,R_AC));\n",
-      "\n",
-      "#Note: In the text resonant frequency has been wrongly calculated to 106kHz but its actual value is approximately 1000kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The resonant frequency=1000kHz\n",
-        "(ii)  d.c load=10 ohm and a.c load=10 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.10 : Page number 406-407\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=50.0;                #Load resistance, ohm\n",
-      "n=5;                    #Turns ratio of the transformer\n",
-      "VCC=50.0;               #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_ac=n**2*RL;               #A.C load, ohm\n",
-      "\n",
-      "#(ii)\n",
-      "P_o_max=VCC**2/(2*R_ac);                #Maximum load power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The a.c load=%d ohm\"%R_ac);\n",
-      "print(\"(ii) Maximum load power=%dW\"%P_o_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The a.c load=1250 ohm\n",
-        "(ii) Maximum load power=1W\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.11 : Page number 407\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_D=4.0;                    #Maximum power dissipation, mW\n",
-      "P_o_max=1.0;                #Maximum load power, W\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "max_collector_eff=(P_o_max/(P_o_max+(P_D/1000)))*100;         #Maximum collector efficiency\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector efficiency=%.1f%%\"%max_collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector efficiency=99.6%\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_4.ipynb
deleted file mode 100755
index e649cc91..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_4.ipynb
+++ /dev/null
@@ -1,482 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:034eec32676d4e7abdedfb3bf68426d81a2d1483fc668bcbfdb5be18cec2e406"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 15: TRANSISTOR TUNED AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.1 : Page number 394"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=250.0*10**-12;                       #Capacitor of parallel resonant circuit, F\n",
-      "L=1.25*10**-3;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                #Resistor of the parallel resonant circuit, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(ii)  Impedance of the circuit at resonance\n",
-      "Zr=(L/(C*R))/1000;                          #Impedance of the circuit at resonance, kilo ohm\n",
-      "\n",
-      "#(iii) Quality factor of the circuit\n",
-      "Q=2*pi*(fr*10**3)*L/R;                          #Quality factor of the circuit\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The resonant frequency=%.1fkHz.\"%fr);\n",
-      "print(\"(ii) The impedance of the circuit at resonance=%d kilo ohm.\"%Zr);\n",
-      "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The resonant frequency=284.7kHz.\n",
-        "(ii) The impedance of the circuit at resonance=500 kilo ohm.\n",
-        "(iii) The quality factor of the circuit=223.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.2 : Page number 394-395\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=100.0*10**-12;                        #Capacitor of parallel resonant circuit, F\n",
-      "L=100.0*10**-6;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                 #Resistor of the parallel resonant circuit, ohm\n",
-      "V=10.0;                                 #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(ii)  Impedance of the circuit at resonance\n",
-      "Zr=(L/(C*R))/10**6;                          #Impedance of the circuit at resonance, mega ohm\n",
-      "\n",
-      "I=V/Zr;                                     #Line current at resonance, microampere\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The resonant frequency=%.2fkHz.\"%fr);\n",
-      "print(\"(ii) The impedance of the circuit at resonance=%.1f mega ohm.\"%Zr);\n",
-      "print(\"The line current at resonance=%d micro ampere.\"%I);\n",
-      "\n",
-      "#Note : The resonant frequency in the text has been calculated as 1592.28 kHz, but here it gets approximated to 1591.47 kHz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The resonant frequency=1591.47kHz.\n",
-        "(ii) The impedance of the circuit at resonance=0.1 mega ohm.\n",
-        "The line current at resonance=100 micro ampere.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.3 : Page number 395\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=250.0*10**-12;                              #Capacitor of parallel resonant circuit, F\n",
-      "Zr=500.0*10**3;                               #Dynamic impedance, ohm\n",
-      "R=10.0;                                       #Resistance of the coil, ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since,Zr=L/CR,\n",
-      "L=(Zr*C*R)*10**3;                           #Inductance of the coil, mH\n",
-      "\n",
-      "#(ii) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*10**-3*C))-(R/(L*10**-3))**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(iii) Quality factor of the circuit\n",
-      "Q=2*pi*(fr*10**3)*(L*10**-3)/R;                          #Quality factor of the circuit\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The inductance of the coil=%.2fmH.\"%L);\n",
-      "print(\"(ii) The resonant frequency=%.1fkHz.\"%fr);\n",
-      "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The inductance of the coil=1.25mH.\n",
-        "(ii) The resonant frequency=284.7kHz.\n",
-        "(iii) The quality factor of the circuit=223.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.4  : Page number 397\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Q=60.0;                     #Quality factor of the tuned amplifier\n",
-      "fr=1200.0;                  #Resonant frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "BW=fr/Q;                    #Bandwidth, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "f1=fr-(BW/2);                       #Lower cut-off frequency, kHz\n",
-      "f2=fr+(BW/2);                       #Upper cut-off frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The bandwidth=%dkHz\"%BW);\n",
-      "print(\"(ii) The lower and upper cut-off frequencies are=%dkHz and %dkHz.\"%(f1,f2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The bandwidth=20kHz\n",
-        "(ii) The lower and upper cut-off frequencies are=1190kHz and 1210kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.5 : Page number 397\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fr=2.0;                     #Resonant frequency, MHz\n",
-      "BW=50.0;                    #Bandwidth, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, bandwidth=resonant_frequency/quality_factor\n",
-      "Q=(fr*10**6)/(BW*10**3);                            #Quality factor\n",
-      "\n",
-      "#Result\n",
-      "print(\"The quality factor=%d\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The quality factor=40\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.7 : Page number 400\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=0.1*10**-6;                          #Capacitor of parallel resonant circuit, F\n",
-      "L=33.0*10**-3;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=25.0;                                #Resistor of the parallel resonant circuit, ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fr=(1/(2*pi*sqrt(L*C)))/1000;                  #Resonant frequency, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "XL=2*pi*(fr*10**3)*L;                           #Inductive reactance, ohm\n",
-      "Q=round(XL/R,0);                                #Quality factor\n",
-      "\n",
-      "#(iii)\n",
-      "BW=(fr*10**3)/Q;                                #Bandwidth\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The resonant frequency=%.2fkHz\"%fr);\n",
-      "print(\"(ii)  The quality factor= %d.\"%Q);\n",
-      "print(\"(iii) The bandwidth=%dHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The resonant frequency=2.77kHz\n",
-        "(ii)  The quality factor= 23.\n",
-        "(iii) The bandwidth=120Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.8 : Page number 401-402\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "BW_dt=200.0;                       #Bandwidth, kHz\n",
-      "fr=10.0;                        #Operating frequency, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, BW_dt=k*fr (i.e.,co-efficient_of_coupling * operating_frequency)\n",
-      "k=BW_dt/(fr*10**3);                     #co-efficient of coupling\n",
-      "\n",
-      "#Result\n",
-      "print(\"The co-efficient of coupling=%.2f.\"%k);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The co-efficient of coupling=0.02.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.9 : Page number 405\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=500.0*10**-12;                          #Capacitor of parallel resonant circuit, F\n",
-      "L=50.7*10**-6;                            #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                   #Resistor of the parallel resonant circuit, ohm\n",
-      "RL=1.0;                                   #Load resistance, mega ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fr=round((1/(2*pi*sqrt(L*C)))/1000);                  #Resonant frequency, Hz\n",
-      "\n",
-      "#(ii)\n",
-      "R_dc=R;                                             #d.c load, ohm\n",
-      "XL=2*pi*(fr*1000)*L;                                       #Inductive reactance, ohm\n",
-      "Q_coil=round(XL/R,1);                               #Quality factor\n",
-      "R_P=(Q_coil*XL)/1000   ;                            #Equivalent parallel resistance, kilo ohm\n",
-      "R_AC=(R_P*RL*10**3)/(R_P+RL*10**3);                 #A.C load,kilo ohm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The resonant frequency=%dkHz\"%fr);\n",
-      "print(\"(ii)  d.c load=%d ohm and a.c load=%d kilo ohm.\"%(R_dc,R_AC));\n",
-      "\n",
-      "#Note: In the text resonant frequency has been wrongly calculated to 106kHz but its actual value is approximately 1000kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The resonant frequency=1000kHz\n",
-        "(ii)  d.c load=10 ohm and a.c load=10 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.10 : Page number 406-407\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=50.0;                #Load resistance, ohm\n",
-      "n=5;                    #Turns ratio of the transformer\n",
-      "VCC=50.0;               #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_ac=n**2*RL;               #A.C load, ohm\n",
-      "\n",
-      "#(ii)\n",
-      "P_o_max=VCC**2/(2*R_ac);                #Maximum load power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The a.c load=%d ohm\"%R_ac);\n",
-      "print(\"(ii) Maximum load power=%dW\"%P_o_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The a.c load=1250 ohm\n",
-        "(ii) Maximum load power=1W\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.11 : Page number 407\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_D=4.0;                    #Maximum power dissipation, mW\n",
-      "P_o_max=1.0;                #Maximum load power, W\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "max_collector_eff=(P_o_max/(P_o_max+(P_D/1000)))*100;         #Maximum collector efficiency\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector efficiency=%.1f%%\"%max_collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector efficiency=99.6%\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_5.ipynb
deleted file mode 100755
index e649cc91..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_5.ipynb
+++ /dev/null
@@ -1,482 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:034eec32676d4e7abdedfb3bf68426d81a2d1483fc668bcbfdb5be18cec2e406"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 15: TRANSISTOR TUNED AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.1 : Page number 394"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=250.0*10**-12;                       #Capacitor of parallel resonant circuit, F\n",
-      "L=1.25*10**-3;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                #Resistor of the parallel resonant circuit, ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(ii)  Impedance of the circuit at resonance\n",
-      "Zr=(L/(C*R))/1000;                          #Impedance of the circuit at resonance, kilo ohm\n",
-      "\n",
-      "#(iii) Quality factor of the circuit\n",
-      "Q=2*pi*(fr*10**3)*L/R;                          #Quality factor of the circuit\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The resonant frequency=%.1fkHz.\"%fr);\n",
-      "print(\"(ii) The impedance of the circuit at resonance=%d kilo ohm.\"%Zr);\n",
-      "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The resonant frequency=284.7kHz.\n",
-        "(ii) The impedance of the circuit at resonance=500 kilo ohm.\n",
-        "(iii) The quality factor of the circuit=223.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.2 : Page number 394-395\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=100.0*10**-12;                        #Capacitor of parallel resonant circuit, F\n",
-      "L=100.0*10**-6;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                 #Resistor of the parallel resonant circuit, ohm\n",
-      "V=10.0;                                 #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(ii)  Impedance of the circuit at resonance\n",
-      "Zr=(L/(C*R))/10**6;                          #Impedance of the circuit at resonance, mega ohm\n",
-      "\n",
-      "I=V/Zr;                                     #Line current at resonance, microampere\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The resonant frequency=%.2fkHz.\"%fr);\n",
-      "print(\"(ii) The impedance of the circuit at resonance=%.1f mega ohm.\"%Zr);\n",
-      "print(\"The line current at resonance=%d micro ampere.\"%I);\n",
-      "\n",
-      "#Note : The resonant frequency in the text has been calculated as 1592.28 kHz, but here it gets approximated to 1591.47 kHz.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The resonant frequency=1591.47kHz.\n",
-        "(ii) The impedance of the circuit at resonance=0.1 mega ohm.\n",
-        "The line current at resonance=100 micro ampere.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.3 : Page number 395\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=250.0*10**-12;                              #Capacitor of parallel resonant circuit, F\n",
-      "Zr=500.0*10**3;                               #Dynamic impedance, ohm\n",
-      "R=10.0;                                       #Resistance of the coil, ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since,Zr=L/CR,\n",
-      "L=(Zr*C*R)*10**3;                           #Inductance of the coil, mH\n",
-      "\n",
-      "#(ii) Resonant frequency\n",
-      "fr=((1/(2*pi))*sqrt((1/(L*10**-3*C))-(R/(L*10**-3))**2))/1000;                         #Resonant frequecy, kHz\n",
-      "\n",
-      "#(iii) Quality factor of the circuit\n",
-      "Q=2*pi*(fr*10**3)*(L*10**-3)/R;                          #Quality factor of the circuit\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The inductance of the coil=%.2fmH.\"%L);\n",
-      "print(\"(ii) The resonant frequency=%.1fkHz.\"%fr);\n",
-      "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The inductance of the coil=1.25mH.\n",
-        "(ii) The resonant frequency=284.7kHz.\n",
-        "(iii) The quality factor of the circuit=223.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.4  : Page number 397\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Q=60.0;                     #Quality factor of the tuned amplifier\n",
-      "fr=1200.0;                  #Resonant frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "BW=fr/Q;                    #Bandwidth, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "f1=fr-(BW/2);                       #Lower cut-off frequency, kHz\n",
-      "f2=fr+(BW/2);                       #Upper cut-off frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The bandwidth=%dkHz\"%BW);\n",
-      "print(\"(ii) The lower and upper cut-off frequencies are=%dkHz and %dkHz.\"%(f1,f2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The bandwidth=20kHz\n",
-        "(ii) The lower and upper cut-off frequencies are=1190kHz and 1210kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.5 : Page number 397\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fr=2.0;                     #Resonant frequency, MHz\n",
-      "BW=50.0;                    #Bandwidth, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, bandwidth=resonant_frequency/quality_factor\n",
-      "Q=(fr*10**6)/(BW*10**3);                            #Quality factor\n",
-      "\n",
-      "#Result\n",
-      "print(\"The quality factor=%d\"%Q);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The quality factor=40\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.7 : Page number 400\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=0.1*10**-6;                          #Capacitor of parallel resonant circuit, F\n",
-      "L=33.0*10**-3;                         #Inductor of the parallel resonant circuit, H\n",
-      "R=25.0;                                #Resistor of the parallel resonant circuit, ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fr=(1/(2*pi*sqrt(L*C)))/1000;                  #Resonant frequency, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "XL=2*pi*(fr*10**3)*L;                           #Inductive reactance, ohm\n",
-      "Q=round(XL/R,0);                                #Quality factor\n",
-      "\n",
-      "#(iii)\n",
-      "BW=(fr*10**3)/Q;                                #Bandwidth\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The resonant frequency=%.2fkHz\"%fr);\n",
-      "print(\"(ii)  The quality factor= %d.\"%Q);\n",
-      "print(\"(iii) The bandwidth=%dHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The resonant frequency=2.77kHz\n",
-        "(ii)  The quality factor= 23.\n",
-        "(iii) The bandwidth=120Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.8 : Page number 401-402\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "BW_dt=200.0;                       #Bandwidth, kHz\n",
-      "fr=10.0;                        #Operating frequency, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, BW_dt=k*fr (i.e.,co-efficient_of_coupling * operating_frequency)\n",
-      "k=BW_dt/(fr*10**3);                     #co-efficient of coupling\n",
-      "\n",
-      "#Result\n",
-      "print(\"The co-efficient of coupling=%.2f.\"%k);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The co-efficient of coupling=0.02.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.9 : Page number 405\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "C=500.0*10**-12;                          #Capacitor of parallel resonant circuit, F\n",
-      "L=50.7*10**-6;                            #Inductor of the parallel resonant circuit, H\n",
-      "R=10.0;                                   #Resistor of the parallel resonant circuit, ohm\n",
-      "RL=1.0;                                   #Load resistance, mega ohm\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fr=round((1/(2*pi*sqrt(L*C)))/1000);                  #Resonant frequency, Hz\n",
-      "\n",
-      "#(ii)\n",
-      "R_dc=R;                                             #d.c load, ohm\n",
-      "XL=2*pi*(fr*1000)*L;                                       #Inductive reactance, ohm\n",
-      "Q_coil=round(XL/R,1);                               #Quality factor\n",
-      "R_P=(Q_coil*XL)/1000   ;                            #Equivalent parallel resistance, kilo ohm\n",
-      "R_AC=(R_P*RL*10**3)/(R_P+RL*10**3);                 #A.C load,kilo ohm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The resonant frequency=%dkHz\"%fr);\n",
-      "print(\"(ii)  d.c load=%d ohm and a.c load=%d kilo ohm.\"%(R_dc,R_AC));\n",
-      "\n",
-      "#Note: In the text resonant frequency has been wrongly calculated to 106kHz but its actual value is approximately 1000kHz\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The resonant frequency=1000kHz\n",
-        "(ii)  d.c load=10 ohm and a.c load=10 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.10 : Page number 406-407\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=50.0;                #Load resistance, ohm\n",
-      "n=5;                    #Turns ratio of the transformer\n",
-      "VCC=50.0;               #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_ac=n**2*RL;               #A.C load, ohm\n",
-      "\n",
-      "#(ii)\n",
-      "P_o_max=VCC**2/(2*R_ac);                #Maximum load power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The a.c load=%d ohm\"%R_ac);\n",
-      "print(\"(ii) Maximum load power=%dW\"%P_o_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The a.c load=1250 ohm\n",
-        "(ii) Maximum load power=1W\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 15.11 : Page number 407\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_D=4.0;                    #Maximum power dissipation, mW\n",
-      "P_o_max=1.0;                #Maximum load power, W\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "max_collector_eff=(P_o_max/(P_o_max+(P_D/1000)))*100;         #Maximum collector efficiency\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum collector efficiency=%.1f%%\"%max_collector_eff);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum collector efficiency=99.6%\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16.ipynb
deleted file mode 100755
index b2262b8f..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16.ipynb
+++ /dev/null
@@ -1,936 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d4ffda068787fb0974622fa8de40f7d54b5df2a00735e870e01cb9df45be78f9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 16 : MODULATION AND DEMODULATION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.2 : Page number 416-417\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variabledeclaration\n",
-      "V_pp_max=16.0;                  #Maximum peak-to-peak voltage of an AM wave, mV\n",
-      "V_pp_min=4.0;                   #Minimum peak-to-peak voltage of an AM wave, mV\n",
-      "\n",
-      "#Calculation\n",
-      "Vmax=V_pp_max/2;                     #Maximum voltage of AM wave, mV\n",
-      "Vmin=V_pp_min/2;                     #Minimum voltage of AM wave, mV\n",
-      "m=(Vmax-Vmin)/(Vmax+Vmin);           #Modulation factor.\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation factor=%.1f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation factor=0.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.3 : Page number 417\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Es=50.0;                #signalvoltage amplitude,  V\n",
-      "Ec=100.0;               #Carrier voltage amplitude, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "m=Es/Ec;                    #Modulation factor\n",
-      "\n",
-      "#Result\n",
-      "print(\"Modulation factor=%.1f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Modulation factor=0.5.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.4 : Page number 419\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=2500.0;                  #Carrier frequency, kHz\n",
-      "f1=50.0;                    #Lower frequency of the audio signal, Hz\n",
-      "f2=15000.0;                 #Upper frequency of the audio signal, Hz\n",
-      "\n",
-      "#Calculation\n",
-      "fl_usb=fc+(f1/1000);            #Lower frequency of upper sideband, kHz\n",
-      "fu_usb=fc+(f2/1000);            #Upper frequency of upper sideband, kHz\n",
-      "\n",
-      "fu_lsb=fc-(f1/1000);            #Lower frequency of upper sideband, kHz\n",
-      "fl_lsb=fc-(f2/1000);            #Upper frequency of upper sideband, kHz\n",
-      "\n",
-      "#Since, f1=50Hz is negligible with respect to f2=15000Hz,\n",
-      "BW=(fc+(f2/1000))-(fc-(f2/1000));                 #Bandwidth, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The upper sideband=%.2fkHz to %dkHz.\"%(fl_usb,fu_usb));\n",
-      "print(\"The lower sideband=%dkHz to %.2fkHz.\"%(fl_lsb,fu_lsb));\n",
-      "print(\"The bandwidth=%dkHz\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The upper sideband=2500.05kHz to 2515kHz.\n",
-        "The lower sideband=2485kHz to 2499.95kHz.\n",
-        "The bandwidth=30kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.5 : Page number 420\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "EC=5.0;             #Carrier amplitude, V\n",
-      "m=0.6;              #modulation factor\n",
-      "ws=6280.0;          #angular frequency of signal, radians/s\n",
-      "wc=211*10**4;       #angular frequency of carrier, radians/s\n",
-      "\n",
-      "#Calculation\n",
-      "fs=(ws/(2*pi))/1000;                   #Signal frequency, kHz\n",
-      "fc=(wc/(2*pi))/1000;                   #Carrier frequency, kHz\n",
-      "\n",
-      "#(i)\n",
-      "Max_amp=EC+m*EC;                        #Maximum amplitude of AM wave, V\n",
-      "Min_amp=EC-m*EC;                        #Minimum amplitude of AM wave, V\n",
-      "\n",
-      "#(ii)\n",
-      "frequency_components=[fc-fs,fc,fc+fs];          #frequency components, kHz\n",
-      "amplitudes=[m*EC/2,EC,m*EC/2];                  #Corresponding amplitudes, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The maximum and minimum amplitudes of AM wave=%dV and %dV.\"%(Max_amp,Min_amp));\n",
-      "print(\"(ii) The frequency components of the AM wave=%.0f,%.0f,%.0f.\"%(frequency_components[0],frequency_components[1],frequency_components[2]));\n",
-      "print(\"     The corresponding amplitudes are =%.1fV, %dV, %.1fV.\"%(amplitudes[0],amplitudes[1],amplitudes[2]));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The maximum and minimum amplitudes of AM wave=8V and 2V.\n",
-        "(ii) The frequency components of the AM wave=335,336,337.\n",
-        "     The corresponding amplitudes are =1.5V, 5V, 1.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.6 : Page number 420-421\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=1000.0;                  #Carrier frequency, kHz\n",
-      "fs=5.0;                     #Signal frequency, kHz\n",
-      "m=0.5;                      #Modulation factor\n",
-      "EC=100.0;                   #Amplitude of the carrier, V\n",
-      "\n",
-      "#Calculation\n",
-      "f_lsb=fc-fs;                    #Lower sideband frequency,kHz\n",
-      "f_usb=fc+fs;                    #Upper sideband frequency, kHz\n",
-      "Amplitude=m*EC/2;               #Amplitude of each sideband, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The lower and upper sideband frequencies are=%dkHz and %dkHz.\"%(f_lsb,f_usb));\n",
-      "print(\"The amplitude of each sideband =%dV\"%Amplitude);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lower and upper sideband frequencies are=995kHz and 1005kHz.\n",
-        "The amplitude of each sideband =25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.7 : Page number 421\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "EC=10.0;            #Carrier amplitude, V\n",
-      "ES=6.0;             #Signal amplitude, V\n",
-      "fc=10.0;            #Carrier frequency, MHz\n",
-      "fs=5/1000.0;        #Signal frequency. MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "m=ES/EC;            #Modulation factor\n",
-      "\n",
-      "#(ii)\n",
-      "f_lsb=fc-fs;                    #Lower sideband frequency,MHz\n",
-      "f_usb=fc+fs;                    #Upper sideband frequency, MHz\n",
-      "\n",
-      "#(iii)\n",
-      "Amplitude=m*EC/2;               #Amplitude of each sideband, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The modulation factor=%.1f.\"%m);\n",
-      "print(\"(ii)  The lower and upper sideband frequencies are=%.3fMHz and %.3fMHz.\"%(f_lsb,f_usb));\n",
-      "print(\"(iii) The amplitude of each sideband =%dV\"%Amplitude);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The modulation factor=0.6.\n",
-        "(ii)  The lower and upper sideband frequencies are=9.995MHz and 10.005MHz.\n",
-        "(iii) The amplitude of each sideband =3V\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.8 : Page number 423\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=500.0;               #Carrier power, W\n",
-      "m=1.0;                  #Modulation factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband power, W\n",
-      "\n",
-      "#(ii)\n",
-      "PT=Pc+Ps;                       #Power of AM wave, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The power in sidebands=%dW\"%Ps);\n",
-      "print(\"(ii) The power of AM wave=%dW\"%PT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The power in sidebands=250W\n",
-        "(ii) The power of AM wave=750W\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exmaple 16.9 : Page number 423\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=50.0;                    #Power of carrier, kW\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "m=80/100.0;                       #Modulation factor\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband Power, kW\n",
-      "print(\"(i) The sideband power for 80%% modulation=%dkW.\"%Ps);\n",
-      "\n",
-      "#(ii)\n",
-      "m=10/100.0;                       #Modulation factor\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband Power, kW\n",
-      "print(\"(ii) The sideband power for 10%% modulation=%.2fkW.\"%Ps);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The sideband power for 80% modulation=16kW.\n",
-        "(ii) The sideband power for 10% modulation=0.25kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.10 : Page number 423-424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=40.0;                       #Carrier power, kW\n",
-      "m=100/100.0;                     #Modulation index\n",
-      "amplifier_eff=72/100.0;          #Efficiency of modulated RF amplifier\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)Carrier power remains same after modulation\n",
-      "\n",
-      "#(ii)\n",
-      "Ps=(1/2.0)*(m**2)*Pc;                         #Sideband power\n",
-      "P_audio=Ps/amplifier_eff;                   #Required audio power, kW\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The carrier power=%dkW.\"%Pc);\n",
-      "print(\"(ii) The required audio power=%.1fkW.\"%P_audio);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The carrier power=40kW.\n",
-        "(ii) The required audio power=27.8kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.11 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs=1.0;                         #Signal frequency, kHz\n",
-      "fc=500.0;                       #Carrier frequency, kHz\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "sideband_f=[fc-fs,fc+fs];               #Sideband frequencies, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "BW=(fc+fs)-(fc-fs);                     #Bandwidth required, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The sideband frequencies=%dkHz and %dkHz.\"%(sideband_f[0],sideband_f[1]));\n",
-      "print(\"(ii) The bandwidth required=%dkHz\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The sideband frequencies=499kHz and 501kHz.\n",
-        "(ii) The bandwidth required=2kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.12 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "IC=8.0;                     #Antenna current due to carrier,A\n",
-      "m=40/100.0;                   #Modulation index\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n",
-      "IT=IC*sqrt(1+(m**2/2.0));                     #Total current, A\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total antenna current=%.2fA.\"%IT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total antenna current=8.31A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.13 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "IC=8.0;                             #Antenna current when only carrier is sent, A\n",
-      "IT=8.93;                            #Total antenna current, A\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n",
-      "m=sqrt((((IT/IC)**2)-1)*2)*100;                 #The %age of modulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The %%age of modulation=%.1f%%.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The %age of modulation=70.1%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.14 : Page number 425\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "Vc=100.0;                       #Carrier voltage, V\n",
-      "V_T=110.0;                      #The total voltage after modulation, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_voltage/Carrier_voltage)=(V_T/Vc)\u00b2=1+(m\u00b2/2).\n",
-      "m=sqrt((((V_T/Vc)**2)-1)*2);                 #The %age of modulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation index =%.3f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index =0.648.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.15 : Page number 425-426\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vc=5.0;                       #Carrier voltage, V\n",
-      "V_lsb=2.5;                      #Lower sideband component, V\n",
-      "V_usb=2.5;                      #Upper sideband component, V\n",
-      "R=2.0;                          #Resistor driven by AM wave, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, power=(r.m.s_voltage)\u00b2/resistance\n",
-      "#(i)\n",
-      "Pc=round((0.707*Vc)**2/R,2);             #Carrier power mW\n",
-      "\n",
-      "#(ii)\n",
-      "P_lower=round((0.707*V_lsb)**2/R,3);            #Power delivered by lower sideband, mW\n",
-      "\n",
-      "#(iii)\n",
-      "P_upper=round((0.707*V_usb)**2/R,3);            #Power delivered by upper sideband, mW\n",
-      "\n",
-      "P_T=round(Pc+P_lower+P_upper,3);                     #Total power delivered by the AM wave, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The carrier power=%.2fmW\"%Pc);\n",
-      "print(\"(ii) The power delivered by lower sideband=%.3fmW\"%P_lower);\n",
-      "print(\"(iii) The power delivered by upper sideband=%.3fmW\"%P_upper);\n",
-      "print(\"The total power delivered by the AM wave=%.3fmW\"%P_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The carrier power=6.25mW\n",
-        "(ii) The power delivered by lower sideband=1.562mW\n",
-        "(iii) The power delivered by upper sideband=1.562mW\n",
-        "The total power delivered by the AM wave=9.374mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.16 : Page number 428\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "wc=6e08;            #Carrier angular frequency, rad/s\n",
-      "ws=1250.0;            #Signal angular frequency, rad/s\n",
-      "mf=5;                   #Modulation index\n",
-      "Ec=12.0;                #Carrier amplitude, V\n",
-      "R=10.0;                 #Resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fc=wc/(2*pi);                       #Carrier frequency, Hz\n",
-      "\n",
-      "#(ii)\n",
-      "fs=ws/(2*pi);                       #Signal frequency, Hz\n",
-      "\n",
-      "#(iv)\n",
-      "delta_f=mf*fs;                      #Maximum frequency deviation, Hz\n",
-      "\n",
-      "#(v)\n",
-      "P=(Ec/sqrt(2))**2/R;                    #Power dissipated, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The carrier frequency=%.1fe06 Hz.\"%(fc/10**6));\n",
-      "print(\"(ii)  The signal frequency=%.0f Hz.\"%fs);\n",
-      "print(\"(iii) The modulation index=%d.\"%mf);\n",
-      "print(\"(iv)  The maximum frequency deviation=%.0fHz.\"%delta_f);\n",
-      "print(\"(v)   The power dissipated=%.1fW.\"%P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The carrier frequency=95.5e06 Hz.\n",
-        "(ii)  The signal frequency=199 Hz.\n",
-        "(iii) The modulation index=5.\n",
-        "(iv)  The maximum frequency deviation=995Hz.\n",
-        "(v)   The power dissipated=7.2W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.17 : Page number 428-429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "fc=25.0;                       #Carrier frequency, MHz\n",
-      "fs=400.0;                      #Signal frequency, Hz\n",
-      "Ec=4.0;                         #Carrier amplitude, V\n",
-      "delta_f=10.0;                   #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "wc=2*pi*fc*10**6;                 #Carrier angular frequency, rad/s\n",
-      "ws=2*pi*fs;                 #Signal angular frequency, rad/s\n",
-      "mf=delta_f*1000/fs;              #Modulation index\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"e=%dcos(%.2et + %dsin%dt)\"%(Ec,wc,mf,ws));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "e=4cos(1.57e+08t + 25sin2513t)\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.18 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "delta_f=50.0;                   #Maximum frequency deviation, kHz\n",
-      "fs=5.0;                         #Modulating frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "mf=delta_f/fs;          #Modulation index\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation index=%d\"%mf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index=10\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.19 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=1000.0;                  #Carrier frequency, kHz\n",
-      "fs=15.0;                    #Modulating frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "first_3_usb_f=[fc+fs,fc+2*fs,fc+3*fs];              #First three upper sideband frequncies, kHz\n",
-      "first_3_lsb_f=[fc-fs,fc-2*fs,fc-3*fs];              #First three lowerr sideband frequncies, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The first three upper sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_usb_f[0],first_3_usb_f[1],first_3_usb_f[2]));\n",
-      "print(\"The first three lower sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_lsb_f[0],first_3_lsb_f[1],first_3_lsb_f[2]));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The first three upper sideband frequencies=1015kHz ,1030kHz and 1045kHz.\n",
-        "The first three lower sideband frequencies=985kHz ,970kHz and 955kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.20 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs=15.0;                    #Modulating frequency, kHz\n",
-      "delta_f=75.0;               #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "BW=2*(delta_f+fs);          #Bandwidth, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The bandwidth of the FM signal=%dkHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The bandwidth of the FM signal=180kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.21 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "k=75.0;             #Frequency deviation constant, kHz/V\n",
-      "Es=2.0;             #Amplitude  of signal, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "delta_f=k*Es;       #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum frequency deviation=%dkHz.\"%delta_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum frequency deviation=150kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.22 : Page number 429-430\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs1=500.0;              #First audio frequency, Hz\n",
-      "fs2=200.0;              #Second audio frequency (decreased), Hz\n",
-      "Es=2.4;                 #AF voltage, V\n",
-      "delta_f1=4.8;           #Frequency deviation,kHz\n",
-      "\n",
-      "#Calculation\n",
-      "k=delta_f1/Es;          #Frequency deviation constant, kHz/V\n",
-      "Es=7.2;                 #AF voltage, V (increased)\n",
-      "delta_f2=k*Es;          #2nd frequency deviation, kHz\n",
-      "Es=10.0;                #AF voltage, V (increased)\n",
-      "delta_f3=k*Es;          #3rd frequency deviation, kHz\n",
-      "\n",
-      "mf1=delta_f1/(fs1/1000);        #Modulation index in 1st case\n",
-      "mf2=delta_f2/(fs1/1000);        #Modulation index in 2nd case\n",
-      "mf3=delta_f3/(fs2/1000);        #Modulation index in 3rd case\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency deviation in second case=%.1fkHz.\"%delta_f2);\n",
-      "print(\"The frequency deviation in third case=%dkHz.\"%delta_f3);\n",
-      "print(\"The modulation index in 1st case=%.1f\"%mf1);\n",
-      "print(\"The modulation index in 2nd case=%.1f\"%mf2);\n",
-      "print(\"The modulation index in 3rd case=%d\"%mf3);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency deviation in second case=14.4kHz.\n",
-        "The frequency deviation in third case=20kHz.\n",
-        "The modulation index in 1st case=9.6\n",
-        "The modulation index in 2nd case=28.8\n",
-        "The modulation index in 3rd case=100\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_1.ipynb
deleted file mode 100755
index b2262b8f..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_1.ipynb
+++ /dev/null
@@ -1,936 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d4ffda068787fb0974622fa8de40f7d54b5df2a00735e870e01cb9df45be78f9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 16 : MODULATION AND DEMODULATION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.2 : Page number 416-417\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variabledeclaration\n",
-      "V_pp_max=16.0;                  #Maximum peak-to-peak voltage of an AM wave, mV\n",
-      "V_pp_min=4.0;                   #Minimum peak-to-peak voltage of an AM wave, mV\n",
-      "\n",
-      "#Calculation\n",
-      "Vmax=V_pp_max/2;                     #Maximum voltage of AM wave, mV\n",
-      "Vmin=V_pp_min/2;                     #Minimum voltage of AM wave, mV\n",
-      "m=(Vmax-Vmin)/(Vmax+Vmin);           #Modulation factor.\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation factor=%.1f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation factor=0.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.3 : Page number 417\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Es=50.0;                #signalvoltage amplitude,  V\n",
-      "Ec=100.0;               #Carrier voltage amplitude, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "m=Es/Ec;                    #Modulation factor\n",
-      "\n",
-      "#Result\n",
-      "print(\"Modulation factor=%.1f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Modulation factor=0.5.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.4 : Page number 419\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=2500.0;                  #Carrier frequency, kHz\n",
-      "f1=50.0;                    #Lower frequency of the audio signal, Hz\n",
-      "f2=15000.0;                 #Upper frequency of the audio signal, Hz\n",
-      "\n",
-      "#Calculation\n",
-      "fl_usb=fc+(f1/1000);            #Lower frequency of upper sideband, kHz\n",
-      "fu_usb=fc+(f2/1000);            #Upper frequency of upper sideband, kHz\n",
-      "\n",
-      "fu_lsb=fc-(f1/1000);            #Lower frequency of upper sideband, kHz\n",
-      "fl_lsb=fc-(f2/1000);            #Upper frequency of upper sideband, kHz\n",
-      "\n",
-      "#Since, f1=50Hz is negligible with respect to f2=15000Hz,\n",
-      "BW=(fc+(f2/1000))-(fc-(f2/1000));                 #Bandwidth, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The upper sideband=%.2fkHz to %dkHz.\"%(fl_usb,fu_usb));\n",
-      "print(\"The lower sideband=%dkHz to %.2fkHz.\"%(fl_lsb,fu_lsb));\n",
-      "print(\"The bandwidth=%dkHz\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The upper sideband=2500.05kHz to 2515kHz.\n",
-        "The lower sideband=2485kHz to 2499.95kHz.\n",
-        "The bandwidth=30kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.5 : Page number 420\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "EC=5.0;             #Carrier amplitude, V\n",
-      "m=0.6;              #modulation factor\n",
-      "ws=6280.0;          #angular frequency of signal, radians/s\n",
-      "wc=211*10**4;       #angular frequency of carrier, radians/s\n",
-      "\n",
-      "#Calculation\n",
-      "fs=(ws/(2*pi))/1000;                   #Signal frequency, kHz\n",
-      "fc=(wc/(2*pi))/1000;                   #Carrier frequency, kHz\n",
-      "\n",
-      "#(i)\n",
-      "Max_amp=EC+m*EC;                        #Maximum amplitude of AM wave, V\n",
-      "Min_amp=EC-m*EC;                        #Minimum amplitude of AM wave, V\n",
-      "\n",
-      "#(ii)\n",
-      "frequency_components=[fc-fs,fc,fc+fs];          #frequency components, kHz\n",
-      "amplitudes=[m*EC/2,EC,m*EC/2];                  #Corresponding amplitudes, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The maximum and minimum amplitudes of AM wave=%dV and %dV.\"%(Max_amp,Min_amp));\n",
-      "print(\"(ii) The frequency components of the AM wave=%.0f,%.0f,%.0f.\"%(frequency_components[0],frequency_components[1],frequency_components[2]));\n",
-      "print(\"     The corresponding amplitudes are =%.1fV, %dV, %.1fV.\"%(amplitudes[0],amplitudes[1],amplitudes[2]));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The maximum and minimum amplitudes of AM wave=8V and 2V.\n",
-        "(ii) The frequency components of the AM wave=335,336,337.\n",
-        "     The corresponding amplitudes are =1.5V, 5V, 1.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.6 : Page number 420-421\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=1000.0;                  #Carrier frequency, kHz\n",
-      "fs=5.0;                     #Signal frequency, kHz\n",
-      "m=0.5;                      #Modulation factor\n",
-      "EC=100.0;                   #Amplitude of the carrier, V\n",
-      "\n",
-      "#Calculation\n",
-      "f_lsb=fc-fs;                    #Lower sideband frequency,kHz\n",
-      "f_usb=fc+fs;                    #Upper sideband frequency, kHz\n",
-      "Amplitude=m*EC/2;               #Amplitude of each sideband, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The lower and upper sideband frequencies are=%dkHz and %dkHz.\"%(f_lsb,f_usb));\n",
-      "print(\"The amplitude of each sideband =%dV\"%Amplitude);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lower and upper sideband frequencies are=995kHz and 1005kHz.\n",
-        "The amplitude of each sideband =25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.7 : Page number 421\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "EC=10.0;            #Carrier amplitude, V\n",
-      "ES=6.0;             #Signal amplitude, V\n",
-      "fc=10.0;            #Carrier frequency, MHz\n",
-      "fs=5/1000.0;        #Signal frequency. MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "m=ES/EC;            #Modulation factor\n",
-      "\n",
-      "#(ii)\n",
-      "f_lsb=fc-fs;                    #Lower sideband frequency,MHz\n",
-      "f_usb=fc+fs;                    #Upper sideband frequency, MHz\n",
-      "\n",
-      "#(iii)\n",
-      "Amplitude=m*EC/2;               #Amplitude of each sideband, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The modulation factor=%.1f.\"%m);\n",
-      "print(\"(ii)  The lower and upper sideband frequencies are=%.3fMHz and %.3fMHz.\"%(f_lsb,f_usb));\n",
-      "print(\"(iii) The amplitude of each sideband =%dV\"%Amplitude);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The modulation factor=0.6.\n",
-        "(ii)  The lower and upper sideband frequencies are=9.995MHz and 10.005MHz.\n",
-        "(iii) The amplitude of each sideband =3V\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.8 : Page number 423\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=500.0;               #Carrier power, W\n",
-      "m=1.0;                  #Modulation factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband power, W\n",
-      "\n",
-      "#(ii)\n",
-      "PT=Pc+Ps;                       #Power of AM wave, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The power in sidebands=%dW\"%Ps);\n",
-      "print(\"(ii) The power of AM wave=%dW\"%PT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The power in sidebands=250W\n",
-        "(ii) The power of AM wave=750W\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exmaple 16.9 : Page number 423\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=50.0;                    #Power of carrier, kW\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "m=80/100.0;                       #Modulation factor\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband Power, kW\n",
-      "print(\"(i) The sideband power for 80%% modulation=%dkW.\"%Ps);\n",
-      "\n",
-      "#(ii)\n",
-      "m=10/100.0;                       #Modulation factor\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband Power, kW\n",
-      "print(\"(ii) The sideband power for 10%% modulation=%.2fkW.\"%Ps);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The sideband power for 80% modulation=16kW.\n",
-        "(ii) The sideband power for 10% modulation=0.25kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.10 : Page number 423-424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=40.0;                       #Carrier power, kW\n",
-      "m=100/100.0;                     #Modulation index\n",
-      "amplifier_eff=72/100.0;          #Efficiency of modulated RF amplifier\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)Carrier power remains same after modulation\n",
-      "\n",
-      "#(ii)\n",
-      "Ps=(1/2.0)*(m**2)*Pc;                         #Sideband power\n",
-      "P_audio=Ps/amplifier_eff;                   #Required audio power, kW\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The carrier power=%dkW.\"%Pc);\n",
-      "print(\"(ii) The required audio power=%.1fkW.\"%P_audio);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The carrier power=40kW.\n",
-        "(ii) The required audio power=27.8kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.11 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs=1.0;                         #Signal frequency, kHz\n",
-      "fc=500.0;                       #Carrier frequency, kHz\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "sideband_f=[fc-fs,fc+fs];               #Sideband frequencies, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "BW=(fc+fs)-(fc-fs);                     #Bandwidth required, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The sideband frequencies=%dkHz and %dkHz.\"%(sideband_f[0],sideband_f[1]));\n",
-      "print(\"(ii) The bandwidth required=%dkHz\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The sideband frequencies=499kHz and 501kHz.\n",
-        "(ii) The bandwidth required=2kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.12 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "IC=8.0;                     #Antenna current due to carrier,A\n",
-      "m=40/100.0;                   #Modulation index\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n",
-      "IT=IC*sqrt(1+(m**2/2.0));                     #Total current, A\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total antenna current=%.2fA.\"%IT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total antenna current=8.31A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.13 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "IC=8.0;                             #Antenna current when only carrier is sent, A\n",
-      "IT=8.93;                            #Total antenna current, A\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n",
-      "m=sqrt((((IT/IC)**2)-1)*2)*100;                 #The %age of modulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The %%age of modulation=%.1f%%.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The %age of modulation=70.1%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.14 : Page number 425\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "Vc=100.0;                       #Carrier voltage, V\n",
-      "V_T=110.0;                      #The total voltage after modulation, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_voltage/Carrier_voltage)=(V_T/Vc)\u00b2=1+(m\u00b2/2).\n",
-      "m=sqrt((((V_T/Vc)**2)-1)*2);                 #The %age of modulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation index =%.3f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index =0.648.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.15 : Page number 425-426\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vc=5.0;                       #Carrier voltage, V\n",
-      "V_lsb=2.5;                      #Lower sideband component, V\n",
-      "V_usb=2.5;                      #Upper sideband component, V\n",
-      "R=2.0;                          #Resistor driven by AM wave, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, power=(r.m.s_voltage)\u00b2/resistance\n",
-      "#(i)\n",
-      "Pc=round((0.707*Vc)**2/R,2);             #Carrier power mW\n",
-      "\n",
-      "#(ii)\n",
-      "P_lower=round((0.707*V_lsb)**2/R,3);            #Power delivered by lower sideband, mW\n",
-      "\n",
-      "#(iii)\n",
-      "P_upper=round((0.707*V_usb)**2/R,3);            #Power delivered by upper sideband, mW\n",
-      "\n",
-      "P_T=round(Pc+P_lower+P_upper,3);                     #Total power delivered by the AM wave, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The carrier power=%.2fmW\"%Pc);\n",
-      "print(\"(ii) The power delivered by lower sideband=%.3fmW\"%P_lower);\n",
-      "print(\"(iii) The power delivered by upper sideband=%.3fmW\"%P_upper);\n",
-      "print(\"The total power delivered by the AM wave=%.3fmW\"%P_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The carrier power=6.25mW\n",
-        "(ii) The power delivered by lower sideband=1.562mW\n",
-        "(iii) The power delivered by upper sideband=1.562mW\n",
-        "The total power delivered by the AM wave=9.374mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.16 : Page number 428\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "wc=6e08;            #Carrier angular frequency, rad/s\n",
-      "ws=1250.0;            #Signal angular frequency, rad/s\n",
-      "mf=5;                   #Modulation index\n",
-      "Ec=12.0;                #Carrier amplitude, V\n",
-      "R=10.0;                 #Resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fc=wc/(2*pi);                       #Carrier frequency, Hz\n",
-      "\n",
-      "#(ii)\n",
-      "fs=ws/(2*pi);                       #Signal frequency, Hz\n",
-      "\n",
-      "#(iv)\n",
-      "delta_f=mf*fs;                      #Maximum frequency deviation, Hz\n",
-      "\n",
-      "#(v)\n",
-      "P=(Ec/sqrt(2))**2/R;                    #Power dissipated, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The carrier frequency=%.1fe06 Hz.\"%(fc/10**6));\n",
-      "print(\"(ii)  The signal frequency=%.0f Hz.\"%fs);\n",
-      "print(\"(iii) The modulation index=%d.\"%mf);\n",
-      "print(\"(iv)  The maximum frequency deviation=%.0fHz.\"%delta_f);\n",
-      "print(\"(v)   The power dissipated=%.1fW.\"%P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The carrier frequency=95.5e06 Hz.\n",
-        "(ii)  The signal frequency=199 Hz.\n",
-        "(iii) The modulation index=5.\n",
-        "(iv)  The maximum frequency deviation=995Hz.\n",
-        "(v)   The power dissipated=7.2W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.17 : Page number 428-429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "fc=25.0;                       #Carrier frequency, MHz\n",
-      "fs=400.0;                      #Signal frequency, Hz\n",
-      "Ec=4.0;                         #Carrier amplitude, V\n",
-      "delta_f=10.0;                   #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "wc=2*pi*fc*10**6;                 #Carrier angular frequency, rad/s\n",
-      "ws=2*pi*fs;                 #Signal angular frequency, rad/s\n",
-      "mf=delta_f*1000/fs;              #Modulation index\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"e=%dcos(%.2et + %dsin%dt)\"%(Ec,wc,mf,ws));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "e=4cos(1.57e+08t + 25sin2513t)\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.18 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "delta_f=50.0;                   #Maximum frequency deviation, kHz\n",
-      "fs=5.0;                         #Modulating frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "mf=delta_f/fs;          #Modulation index\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation index=%d\"%mf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index=10\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.19 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=1000.0;                  #Carrier frequency, kHz\n",
-      "fs=15.0;                    #Modulating frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "first_3_usb_f=[fc+fs,fc+2*fs,fc+3*fs];              #First three upper sideband frequncies, kHz\n",
-      "first_3_lsb_f=[fc-fs,fc-2*fs,fc-3*fs];              #First three lowerr sideband frequncies, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The first three upper sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_usb_f[0],first_3_usb_f[1],first_3_usb_f[2]));\n",
-      "print(\"The first three lower sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_lsb_f[0],first_3_lsb_f[1],first_3_lsb_f[2]));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The first three upper sideband frequencies=1015kHz ,1030kHz and 1045kHz.\n",
-        "The first three lower sideband frequencies=985kHz ,970kHz and 955kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.20 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs=15.0;                    #Modulating frequency, kHz\n",
-      "delta_f=75.0;               #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "BW=2*(delta_f+fs);          #Bandwidth, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The bandwidth of the FM signal=%dkHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The bandwidth of the FM signal=180kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.21 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "k=75.0;             #Frequency deviation constant, kHz/V\n",
-      "Es=2.0;             #Amplitude  of signal, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "delta_f=k*Es;       #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum frequency deviation=%dkHz.\"%delta_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum frequency deviation=150kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.22 : Page number 429-430\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs1=500.0;              #First audio frequency, Hz\n",
-      "fs2=200.0;              #Second audio frequency (decreased), Hz\n",
-      "Es=2.4;                 #AF voltage, V\n",
-      "delta_f1=4.8;           #Frequency deviation,kHz\n",
-      "\n",
-      "#Calculation\n",
-      "k=delta_f1/Es;          #Frequency deviation constant, kHz/V\n",
-      "Es=7.2;                 #AF voltage, V (increased)\n",
-      "delta_f2=k*Es;          #2nd frequency deviation, kHz\n",
-      "Es=10.0;                #AF voltage, V (increased)\n",
-      "delta_f3=k*Es;          #3rd frequency deviation, kHz\n",
-      "\n",
-      "mf1=delta_f1/(fs1/1000);        #Modulation index in 1st case\n",
-      "mf2=delta_f2/(fs1/1000);        #Modulation index in 2nd case\n",
-      "mf3=delta_f3/(fs2/1000);        #Modulation index in 3rd case\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency deviation in second case=%.1fkHz.\"%delta_f2);\n",
-      "print(\"The frequency deviation in third case=%dkHz.\"%delta_f3);\n",
-      "print(\"The modulation index in 1st case=%.1f\"%mf1);\n",
-      "print(\"The modulation index in 2nd case=%.1f\"%mf2);\n",
-      "print(\"The modulation index in 3rd case=%d\"%mf3);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency deviation in second case=14.4kHz.\n",
-        "The frequency deviation in third case=20kHz.\n",
-        "The modulation index in 1st case=9.6\n",
-        "The modulation index in 2nd case=28.8\n",
-        "The modulation index in 3rd case=100\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_2.ipynb
deleted file mode 100755
index b2262b8f..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_2.ipynb
+++ /dev/null
@@ -1,936 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d4ffda068787fb0974622fa8de40f7d54b5df2a00735e870e01cb9df45be78f9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 16 : MODULATION AND DEMODULATION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.2 : Page number 416-417\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variabledeclaration\n",
-      "V_pp_max=16.0;                  #Maximum peak-to-peak voltage of an AM wave, mV\n",
-      "V_pp_min=4.0;                   #Minimum peak-to-peak voltage of an AM wave, mV\n",
-      "\n",
-      "#Calculation\n",
-      "Vmax=V_pp_max/2;                     #Maximum voltage of AM wave, mV\n",
-      "Vmin=V_pp_min/2;                     #Minimum voltage of AM wave, mV\n",
-      "m=(Vmax-Vmin)/(Vmax+Vmin);           #Modulation factor.\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation factor=%.1f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation factor=0.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.3 : Page number 417\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Es=50.0;                #signalvoltage amplitude,  V\n",
-      "Ec=100.0;               #Carrier voltage amplitude, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "m=Es/Ec;                    #Modulation factor\n",
-      "\n",
-      "#Result\n",
-      "print(\"Modulation factor=%.1f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Modulation factor=0.5.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.4 : Page number 419\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=2500.0;                  #Carrier frequency, kHz\n",
-      "f1=50.0;                    #Lower frequency of the audio signal, Hz\n",
-      "f2=15000.0;                 #Upper frequency of the audio signal, Hz\n",
-      "\n",
-      "#Calculation\n",
-      "fl_usb=fc+(f1/1000);            #Lower frequency of upper sideband, kHz\n",
-      "fu_usb=fc+(f2/1000);            #Upper frequency of upper sideband, kHz\n",
-      "\n",
-      "fu_lsb=fc-(f1/1000);            #Lower frequency of upper sideband, kHz\n",
-      "fl_lsb=fc-(f2/1000);            #Upper frequency of upper sideband, kHz\n",
-      "\n",
-      "#Since, f1=50Hz is negligible with respect to f2=15000Hz,\n",
-      "BW=(fc+(f2/1000))-(fc-(f2/1000));                 #Bandwidth, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The upper sideband=%.2fkHz to %dkHz.\"%(fl_usb,fu_usb));\n",
-      "print(\"The lower sideband=%dkHz to %.2fkHz.\"%(fl_lsb,fu_lsb));\n",
-      "print(\"The bandwidth=%dkHz\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The upper sideband=2500.05kHz to 2515kHz.\n",
-        "The lower sideband=2485kHz to 2499.95kHz.\n",
-        "The bandwidth=30kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.5 : Page number 420\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "EC=5.0;             #Carrier amplitude, V\n",
-      "m=0.6;              #modulation factor\n",
-      "ws=6280.0;          #angular frequency of signal, radians/s\n",
-      "wc=211*10**4;       #angular frequency of carrier, radians/s\n",
-      "\n",
-      "#Calculation\n",
-      "fs=(ws/(2*pi))/1000;                   #Signal frequency, kHz\n",
-      "fc=(wc/(2*pi))/1000;                   #Carrier frequency, kHz\n",
-      "\n",
-      "#(i)\n",
-      "Max_amp=EC+m*EC;                        #Maximum amplitude of AM wave, V\n",
-      "Min_amp=EC-m*EC;                        #Minimum amplitude of AM wave, V\n",
-      "\n",
-      "#(ii)\n",
-      "frequency_components=[fc-fs,fc,fc+fs];          #frequency components, kHz\n",
-      "amplitudes=[m*EC/2,EC,m*EC/2];                  #Corresponding amplitudes, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The maximum and minimum amplitudes of AM wave=%dV and %dV.\"%(Max_amp,Min_amp));\n",
-      "print(\"(ii) The frequency components of the AM wave=%.0f,%.0f,%.0f.\"%(frequency_components[0],frequency_components[1],frequency_components[2]));\n",
-      "print(\"     The corresponding amplitudes are =%.1fV, %dV, %.1fV.\"%(amplitudes[0],amplitudes[1],amplitudes[2]));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The maximum and minimum amplitudes of AM wave=8V and 2V.\n",
-        "(ii) The frequency components of the AM wave=335,336,337.\n",
-        "     The corresponding amplitudes are =1.5V, 5V, 1.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.6 : Page number 420-421\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=1000.0;                  #Carrier frequency, kHz\n",
-      "fs=5.0;                     #Signal frequency, kHz\n",
-      "m=0.5;                      #Modulation factor\n",
-      "EC=100.0;                   #Amplitude of the carrier, V\n",
-      "\n",
-      "#Calculation\n",
-      "f_lsb=fc-fs;                    #Lower sideband frequency,kHz\n",
-      "f_usb=fc+fs;                    #Upper sideband frequency, kHz\n",
-      "Amplitude=m*EC/2;               #Amplitude of each sideband, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The lower and upper sideband frequencies are=%dkHz and %dkHz.\"%(f_lsb,f_usb));\n",
-      "print(\"The amplitude of each sideband =%dV\"%Amplitude);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lower and upper sideband frequencies are=995kHz and 1005kHz.\n",
-        "The amplitude of each sideband =25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.7 : Page number 421\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "EC=10.0;            #Carrier amplitude, V\n",
-      "ES=6.0;             #Signal amplitude, V\n",
-      "fc=10.0;            #Carrier frequency, MHz\n",
-      "fs=5/1000.0;        #Signal frequency. MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "m=ES/EC;            #Modulation factor\n",
-      "\n",
-      "#(ii)\n",
-      "f_lsb=fc-fs;                    #Lower sideband frequency,MHz\n",
-      "f_usb=fc+fs;                    #Upper sideband frequency, MHz\n",
-      "\n",
-      "#(iii)\n",
-      "Amplitude=m*EC/2;               #Amplitude of each sideband, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The modulation factor=%.1f.\"%m);\n",
-      "print(\"(ii)  The lower and upper sideband frequencies are=%.3fMHz and %.3fMHz.\"%(f_lsb,f_usb));\n",
-      "print(\"(iii) The amplitude of each sideband =%dV\"%Amplitude);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The modulation factor=0.6.\n",
-        "(ii)  The lower and upper sideband frequencies are=9.995MHz and 10.005MHz.\n",
-        "(iii) The amplitude of each sideband =3V\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.8 : Page number 423\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=500.0;               #Carrier power, W\n",
-      "m=1.0;                  #Modulation factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband power, W\n",
-      "\n",
-      "#(ii)\n",
-      "PT=Pc+Ps;                       #Power of AM wave, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The power in sidebands=%dW\"%Ps);\n",
-      "print(\"(ii) The power of AM wave=%dW\"%PT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The power in sidebands=250W\n",
-        "(ii) The power of AM wave=750W\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exmaple 16.9 : Page number 423\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=50.0;                    #Power of carrier, kW\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "m=80/100.0;                       #Modulation factor\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband Power, kW\n",
-      "print(\"(i) The sideband power for 80%% modulation=%dkW.\"%Ps);\n",
-      "\n",
-      "#(ii)\n",
-      "m=10/100.0;                       #Modulation factor\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband Power, kW\n",
-      "print(\"(ii) The sideband power for 10%% modulation=%.2fkW.\"%Ps);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The sideband power for 80% modulation=16kW.\n",
-        "(ii) The sideband power for 10% modulation=0.25kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.10 : Page number 423-424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=40.0;                       #Carrier power, kW\n",
-      "m=100/100.0;                     #Modulation index\n",
-      "amplifier_eff=72/100.0;          #Efficiency of modulated RF amplifier\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)Carrier power remains same after modulation\n",
-      "\n",
-      "#(ii)\n",
-      "Ps=(1/2.0)*(m**2)*Pc;                         #Sideband power\n",
-      "P_audio=Ps/amplifier_eff;                   #Required audio power, kW\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The carrier power=%dkW.\"%Pc);\n",
-      "print(\"(ii) The required audio power=%.1fkW.\"%P_audio);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The carrier power=40kW.\n",
-        "(ii) The required audio power=27.8kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.11 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs=1.0;                         #Signal frequency, kHz\n",
-      "fc=500.0;                       #Carrier frequency, kHz\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "sideband_f=[fc-fs,fc+fs];               #Sideband frequencies, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "BW=(fc+fs)-(fc-fs);                     #Bandwidth required, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The sideband frequencies=%dkHz and %dkHz.\"%(sideband_f[0],sideband_f[1]));\n",
-      "print(\"(ii) The bandwidth required=%dkHz\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The sideband frequencies=499kHz and 501kHz.\n",
-        "(ii) The bandwidth required=2kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.12 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "IC=8.0;                     #Antenna current due to carrier,A\n",
-      "m=40/100.0;                   #Modulation index\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n",
-      "IT=IC*sqrt(1+(m**2/2.0));                     #Total current, A\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total antenna current=%.2fA.\"%IT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total antenna current=8.31A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.13 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "IC=8.0;                             #Antenna current when only carrier is sent, A\n",
-      "IT=8.93;                            #Total antenna current, A\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n",
-      "m=sqrt((((IT/IC)**2)-1)*2)*100;                 #The %age of modulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The %%age of modulation=%.1f%%.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The %age of modulation=70.1%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.14 : Page number 425\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "Vc=100.0;                       #Carrier voltage, V\n",
-      "V_T=110.0;                      #The total voltage after modulation, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_voltage/Carrier_voltage)=(V_T/Vc)\u00b2=1+(m\u00b2/2).\n",
-      "m=sqrt((((V_T/Vc)**2)-1)*2);                 #The %age of modulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation index =%.3f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index =0.648.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.15 : Page number 425-426\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vc=5.0;                       #Carrier voltage, V\n",
-      "V_lsb=2.5;                      #Lower sideband component, V\n",
-      "V_usb=2.5;                      #Upper sideband component, V\n",
-      "R=2.0;                          #Resistor driven by AM wave, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, power=(r.m.s_voltage)\u00b2/resistance\n",
-      "#(i)\n",
-      "Pc=round((0.707*Vc)**2/R,2);             #Carrier power mW\n",
-      "\n",
-      "#(ii)\n",
-      "P_lower=round((0.707*V_lsb)**2/R,3);            #Power delivered by lower sideband, mW\n",
-      "\n",
-      "#(iii)\n",
-      "P_upper=round((0.707*V_usb)**2/R,3);            #Power delivered by upper sideband, mW\n",
-      "\n",
-      "P_T=round(Pc+P_lower+P_upper,3);                     #Total power delivered by the AM wave, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The carrier power=%.2fmW\"%Pc);\n",
-      "print(\"(ii) The power delivered by lower sideband=%.3fmW\"%P_lower);\n",
-      "print(\"(iii) The power delivered by upper sideband=%.3fmW\"%P_upper);\n",
-      "print(\"The total power delivered by the AM wave=%.3fmW\"%P_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The carrier power=6.25mW\n",
-        "(ii) The power delivered by lower sideband=1.562mW\n",
-        "(iii) The power delivered by upper sideband=1.562mW\n",
-        "The total power delivered by the AM wave=9.374mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.16 : Page number 428\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "wc=6e08;            #Carrier angular frequency, rad/s\n",
-      "ws=1250.0;            #Signal angular frequency, rad/s\n",
-      "mf=5;                   #Modulation index\n",
-      "Ec=12.0;                #Carrier amplitude, V\n",
-      "R=10.0;                 #Resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fc=wc/(2*pi);                       #Carrier frequency, Hz\n",
-      "\n",
-      "#(ii)\n",
-      "fs=ws/(2*pi);                       #Signal frequency, Hz\n",
-      "\n",
-      "#(iv)\n",
-      "delta_f=mf*fs;                      #Maximum frequency deviation, Hz\n",
-      "\n",
-      "#(v)\n",
-      "P=(Ec/sqrt(2))**2/R;                    #Power dissipated, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The carrier frequency=%.1fe06 Hz.\"%(fc/10**6));\n",
-      "print(\"(ii)  The signal frequency=%.0f Hz.\"%fs);\n",
-      "print(\"(iii) The modulation index=%d.\"%mf);\n",
-      "print(\"(iv)  The maximum frequency deviation=%.0fHz.\"%delta_f);\n",
-      "print(\"(v)   The power dissipated=%.1fW.\"%P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The carrier frequency=95.5e06 Hz.\n",
-        "(ii)  The signal frequency=199 Hz.\n",
-        "(iii) The modulation index=5.\n",
-        "(iv)  The maximum frequency deviation=995Hz.\n",
-        "(v)   The power dissipated=7.2W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.17 : Page number 428-429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "fc=25.0;                       #Carrier frequency, MHz\n",
-      "fs=400.0;                      #Signal frequency, Hz\n",
-      "Ec=4.0;                         #Carrier amplitude, V\n",
-      "delta_f=10.0;                   #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "wc=2*pi*fc*10**6;                 #Carrier angular frequency, rad/s\n",
-      "ws=2*pi*fs;                 #Signal angular frequency, rad/s\n",
-      "mf=delta_f*1000/fs;              #Modulation index\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"e=%dcos(%.2et + %dsin%dt)\"%(Ec,wc,mf,ws));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "e=4cos(1.57e+08t + 25sin2513t)\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.18 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "delta_f=50.0;                   #Maximum frequency deviation, kHz\n",
-      "fs=5.0;                         #Modulating frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "mf=delta_f/fs;          #Modulation index\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation index=%d\"%mf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index=10\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.19 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=1000.0;                  #Carrier frequency, kHz\n",
-      "fs=15.0;                    #Modulating frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "first_3_usb_f=[fc+fs,fc+2*fs,fc+3*fs];              #First three upper sideband frequncies, kHz\n",
-      "first_3_lsb_f=[fc-fs,fc-2*fs,fc-3*fs];              #First three lowerr sideband frequncies, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The first three upper sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_usb_f[0],first_3_usb_f[1],first_3_usb_f[2]));\n",
-      "print(\"The first three lower sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_lsb_f[0],first_3_lsb_f[1],first_3_lsb_f[2]));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The first three upper sideband frequencies=1015kHz ,1030kHz and 1045kHz.\n",
-        "The first three lower sideband frequencies=985kHz ,970kHz and 955kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.20 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs=15.0;                    #Modulating frequency, kHz\n",
-      "delta_f=75.0;               #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "BW=2*(delta_f+fs);          #Bandwidth, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The bandwidth of the FM signal=%dkHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The bandwidth of the FM signal=180kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.21 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "k=75.0;             #Frequency deviation constant, kHz/V\n",
-      "Es=2.0;             #Amplitude  of signal, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "delta_f=k*Es;       #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum frequency deviation=%dkHz.\"%delta_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum frequency deviation=150kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.22 : Page number 429-430\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs1=500.0;              #First audio frequency, Hz\n",
-      "fs2=200.0;              #Second audio frequency (decreased), Hz\n",
-      "Es=2.4;                 #AF voltage, V\n",
-      "delta_f1=4.8;           #Frequency deviation,kHz\n",
-      "\n",
-      "#Calculation\n",
-      "k=delta_f1/Es;          #Frequency deviation constant, kHz/V\n",
-      "Es=7.2;                 #AF voltage, V (increased)\n",
-      "delta_f2=k*Es;          #2nd frequency deviation, kHz\n",
-      "Es=10.0;                #AF voltage, V (increased)\n",
-      "delta_f3=k*Es;          #3rd frequency deviation, kHz\n",
-      "\n",
-      "mf1=delta_f1/(fs1/1000);        #Modulation index in 1st case\n",
-      "mf2=delta_f2/(fs1/1000);        #Modulation index in 2nd case\n",
-      "mf3=delta_f3/(fs2/1000);        #Modulation index in 3rd case\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency deviation in second case=%.1fkHz.\"%delta_f2);\n",
-      "print(\"The frequency deviation in third case=%dkHz.\"%delta_f3);\n",
-      "print(\"The modulation index in 1st case=%.1f\"%mf1);\n",
-      "print(\"The modulation index in 2nd case=%.1f\"%mf2);\n",
-      "print(\"The modulation index in 3rd case=%d\"%mf3);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency deviation in second case=14.4kHz.\n",
-        "The frequency deviation in third case=20kHz.\n",
-        "The modulation index in 1st case=9.6\n",
-        "The modulation index in 2nd case=28.8\n",
-        "The modulation index in 3rd case=100\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_3.ipynb
deleted file mode 100755
index b2262b8f..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_3.ipynb
+++ /dev/null
@@ -1,936 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d4ffda068787fb0974622fa8de40f7d54b5df2a00735e870e01cb9df45be78f9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 16 : MODULATION AND DEMODULATION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.2 : Page number 416-417\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variabledeclaration\n",
-      "V_pp_max=16.0;                  #Maximum peak-to-peak voltage of an AM wave, mV\n",
-      "V_pp_min=4.0;                   #Minimum peak-to-peak voltage of an AM wave, mV\n",
-      "\n",
-      "#Calculation\n",
-      "Vmax=V_pp_max/2;                     #Maximum voltage of AM wave, mV\n",
-      "Vmin=V_pp_min/2;                     #Minimum voltage of AM wave, mV\n",
-      "m=(Vmax-Vmin)/(Vmax+Vmin);           #Modulation factor.\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation factor=%.1f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation factor=0.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.3 : Page number 417\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Es=50.0;                #signalvoltage amplitude,  V\n",
-      "Ec=100.0;               #Carrier voltage amplitude, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "m=Es/Ec;                    #Modulation factor\n",
-      "\n",
-      "#Result\n",
-      "print(\"Modulation factor=%.1f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Modulation factor=0.5.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.4 : Page number 419\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=2500.0;                  #Carrier frequency, kHz\n",
-      "f1=50.0;                    #Lower frequency of the audio signal, Hz\n",
-      "f2=15000.0;                 #Upper frequency of the audio signal, Hz\n",
-      "\n",
-      "#Calculation\n",
-      "fl_usb=fc+(f1/1000);            #Lower frequency of upper sideband, kHz\n",
-      "fu_usb=fc+(f2/1000);            #Upper frequency of upper sideband, kHz\n",
-      "\n",
-      "fu_lsb=fc-(f1/1000);            #Lower frequency of upper sideband, kHz\n",
-      "fl_lsb=fc-(f2/1000);            #Upper frequency of upper sideband, kHz\n",
-      "\n",
-      "#Since, f1=50Hz is negligible with respect to f2=15000Hz,\n",
-      "BW=(fc+(f2/1000))-(fc-(f2/1000));                 #Bandwidth, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The upper sideband=%.2fkHz to %dkHz.\"%(fl_usb,fu_usb));\n",
-      "print(\"The lower sideband=%dkHz to %.2fkHz.\"%(fl_lsb,fu_lsb));\n",
-      "print(\"The bandwidth=%dkHz\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The upper sideband=2500.05kHz to 2515kHz.\n",
-        "The lower sideband=2485kHz to 2499.95kHz.\n",
-        "The bandwidth=30kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.5 : Page number 420\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "EC=5.0;             #Carrier amplitude, V\n",
-      "m=0.6;              #modulation factor\n",
-      "ws=6280.0;          #angular frequency of signal, radians/s\n",
-      "wc=211*10**4;       #angular frequency of carrier, radians/s\n",
-      "\n",
-      "#Calculation\n",
-      "fs=(ws/(2*pi))/1000;                   #Signal frequency, kHz\n",
-      "fc=(wc/(2*pi))/1000;                   #Carrier frequency, kHz\n",
-      "\n",
-      "#(i)\n",
-      "Max_amp=EC+m*EC;                        #Maximum amplitude of AM wave, V\n",
-      "Min_amp=EC-m*EC;                        #Minimum amplitude of AM wave, V\n",
-      "\n",
-      "#(ii)\n",
-      "frequency_components=[fc-fs,fc,fc+fs];          #frequency components, kHz\n",
-      "amplitudes=[m*EC/2,EC,m*EC/2];                  #Corresponding amplitudes, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The maximum and minimum amplitudes of AM wave=%dV and %dV.\"%(Max_amp,Min_amp));\n",
-      "print(\"(ii) The frequency components of the AM wave=%.0f,%.0f,%.0f.\"%(frequency_components[0],frequency_components[1],frequency_components[2]));\n",
-      "print(\"     The corresponding amplitudes are =%.1fV, %dV, %.1fV.\"%(amplitudes[0],amplitudes[1],amplitudes[2]));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The maximum and minimum amplitudes of AM wave=8V and 2V.\n",
-        "(ii) The frequency components of the AM wave=335,336,337.\n",
-        "     The corresponding amplitudes are =1.5V, 5V, 1.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.6 : Page number 420-421\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=1000.0;                  #Carrier frequency, kHz\n",
-      "fs=5.0;                     #Signal frequency, kHz\n",
-      "m=0.5;                      #Modulation factor\n",
-      "EC=100.0;                   #Amplitude of the carrier, V\n",
-      "\n",
-      "#Calculation\n",
-      "f_lsb=fc-fs;                    #Lower sideband frequency,kHz\n",
-      "f_usb=fc+fs;                    #Upper sideband frequency, kHz\n",
-      "Amplitude=m*EC/2;               #Amplitude of each sideband, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The lower and upper sideband frequencies are=%dkHz and %dkHz.\"%(f_lsb,f_usb));\n",
-      "print(\"The amplitude of each sideband =%dV\"%Amplitude);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lower and upper sideband frequencies are=995kHz and 1005kHz.\n",
-        "The amplitude of each sideband =25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.7 : Page number 421\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "EC=10.0;            #Carrier amplitude, V\n",
-      "ES=6.0;             #Signal amplitude, V\n",
-      "fc=10.0;            #Carrier frequency, MHz\n",
-      "fs=5/1000.0;        #Signal frequency. MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "m=ES/EC;            #Modulation factor\n",
-      "\n",
-      "#(ii)\n",
-      "f_lsb=fc-fs;                    #Lower sideband frequency,MHz\n",
-      "f_usb=fc+fs;                    #Upper sideband frequency, MHz\n",
-      "\n",
-      "#(iii)\n",
-      "Amplitude=m*EC/2;               #Amplitude of each sideband, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The modulation factor=%.1f.\"%m);\n",
-      "print(\"(ii)  The lower and upper sideband frequencies are=%.3fMHz and %.3fMHz.\"%(f_lsb,f_usb));\n",
-      "print(\"(iii) The amplitude of each sideband =%dV\"%Amplitude);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The modulation factor=0.6.\n",
-        "(ii)  The lower and upper sideband frequencies are=9.995MHz and 10.005MHz.\n",
-        "(iii) The amplitude of each sideband =3V\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.8 : Page number 423\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=500.0;               #Carrier power, W\n",
-      "m=1.0;                  #Modulation factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband power, W\n",
-      "\n",
-      "#(ii)\n",
-      "PT=Pc+Ps;                       #Power of AM wave, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The power in sidebands=%dW\"%Ps);\n",
-      "print(\"(ii) The power of AM wave=%dW\"%PT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The power in sidebands=250W\n",
-        "(ii) The power of AM wave=750W\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exmaple 16.9 : Page number 423\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=50.0;                    #Power of carrier, kW\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "m=80/100.0;                       #Modulation factor\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband Power, kW\n",
-      "print(\"(i) The sideband power for 80%% modulation=%dkW.\"%Ps);\n",
-      "\n",
-      "#(ii)\n",
-      "m=10/100.0;                       #Modulation factor\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband Power, kW\n",
-      "print(\"(ii) The sideband power for 10%% modulation=%.2fkW.\"%Ps);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The sideband power for 80% modulation=16kW.\n",
-        "(ii) The sideband power for 10% modulation=0.25kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.10 : Page number 423-424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=40.0;                       #Carrier power, kW\n",
-      "m=100/100.0;                     #Modulation index\n",
-      "amplifier_eff=72/100.0;          #Efficiency of modulated RF amplifier\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)Carrier power remains same after modulation\n",
-      "\n",
-      "#(ii)\n",
-      "Ps=(1/2.0)*(m**2)*Pc;                         #Sideband power\n",
-      "P_audio=Ps/amplifier_eff;                   #Required audio power, kW\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The carrier power=%dkW.\"%Pc);\n",
-      "print(\"(ii) The required audio power=%.1fkW.\"%P_audio);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The carrier power=40kW.\n",
-        "(ii) The required audio power=27.8kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.11 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs=1.0;                         #Signal frequency, kHz\n",
-      "fc=500.0;                       #Carrier frequency, kHz\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "sideband_f=[fc-fs,fc+fs];               #Sideband frequencies, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "BW=(fc+fs)-(fc-fs);                     #Bandwidth required, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The sideband frequencies=%dkHz and %dkHz.\"%(sideband_f[0],sideband_f[1]));\n",
-      "print(\"(ii) The bandwidth required=%dkHz\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The sideband frequencies=499kHz and 501kHz.\n",
-        "(ii) The bandwidth required=2kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.12 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "IC=8.0;                     #Antenna current due to carrier,A\n",
-      "m=40/100.0;                   #Modulation index\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n",
-      "IT=IC*sqrt(1+(m**2/2.0));                     #Total current, A\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total antenna current=%.2fA.\"%IT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total antenna current=8.31A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.13 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "IC=8.0;                             #Antenna current when only carrier is sent, A\n",
-      "IT=8.93;                            #Total antenna current, A\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n",
-      "m=sqrt((((IT/IC)**2)-1)*2)*100;                 #The %age of modulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The %%age of modulation=%.1f%%.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The %age of modulation=70.1%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.14 : Page number 425\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "Vc=100.0;                       #Carrier voltage, V\n",
-      "V_T=110.0;                      #The total voltage after modulation, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_voltage/Carrier_voltage)=(V_T/Vc)\u00b2=1+(m\u00b2/2).\n",
-      "m=sqrt((((V_T/Vc)**2)-1)*2);                 #The %age of modulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation index =%.3f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index =0.648.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.15 : Page number 425-426\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vc=5.0;                       #Carrier voltage, V\n",
-      "V_lsb=2.5;                      #Lower sideband component, V\n",
-      "V_usb=2.5;                      #Upper sideband component, V\n",
-      "R=2.0;                          #Resistor driven by AM wave, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, power=(r.m.s_voltage)\u00b2/resistance\n",
-      "#(i)\n",
-      "Pc=round((0.707*Vc)**2/R,2);             #Carrier power mW\n",
-      "\n",
-      "#(ii)\n",
-      "P_lower=round((0.707*V_lsb)**2/R,3);            #Power delivered by lower sideband, mW\n",
-      "\n",
-      "#(iii)\n",
-      "P_upper=round((0.707*V_usb)**2/R,3);            #Power delivered by upper sideband, mW\n",
-      "\n",
-      "P_T=round(Pc+P_lower+P_upper,3);                     #Total power delivered by the AM wave, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The carrier power=%.2fmW\"%Pc);\n",
-      "print(\"(ii) The power delivered by lower sideband=%.3fmW\"%P_lower);\n",
-      "print(\"(iii) The power delivered by upper sideband=%.3fmW\"%P_upper);\n",
-      "print(\"The total power delivered by the AM wave=%.3fmW\"%P_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The carrier power=6.25mW\n",
-        "(ii) The power delivered by lower sideband=1.562mW\n",
-        "(iii) The power delivered by upper sideband=1.562mW\n",
-        "The total power delivered by the AM wave=9.374mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.16 : Page number 428\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "wc=6e08;            #Carrier angular frequency, rad/s\n",
-      "ws=1250.0;            #Signal angular frequency, rad/s\n",
-      "mf=5;                   #Modulation index\n",
-      "Ec=12.0;                #Carrier amplitude, V\n",
-      "R=10.0;                 #Resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fc=wc/(2*pi);                       #Carrier frequency, Hz\n",
-      "\n",
-      "#(ii)\n",
-      "fs=ws/(2*pi);                       #Signal frequency, Hz\n",
-      "\n",
-      "#(iv)\n",
-      "delta_f=mf*fs;                      #Maximum frequency deviation, Hz\n",
-      "\n",
-      "#(v)\n",
-      "P=(Ec/sqrt(2))**2/R;                    #Power dissipated, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The carrier frequency=%.1fe06 Hz.\"%(fc/10**6));\n",
-      "print(\"(ii)  The signal frequency=%.0f Hz.\"%fs);\n",
-      "print(\"(iii) The modulation index=%d.\"%mf);\n",
-      "print(\"(iv)  The maximum frequency deviation=%.0fHz.\"%delta_f);\n",
-      "print(\"(v)   The power dissipated=%.1fW.\"%P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The carrier frequency=95.5e06 Hz.\n",
-        "(ii)  The signal frequency=199 Hz.\n",
-        "(iii) The modulation index=5.\n",
-        "(iv)  The maximum frequency deviation=995Hz.\n",
-        "(v)   The power dissipated=7.2W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.17 : Page number 428-429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "fc=25.0;                       #Carrier frequency, MHz\n",
-      "fs=400.0;                      #Signal frequency, Hz\n",
-      "Ec=4.0;                         #Carrier amplitude, V\n",
-      "delta_f=10.0;                   #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "wc=2*pi*fc*10**6;                 #Carrier angular frequency, rad/s\n",
-      "ws=2*pi*fs;                 #Signal angular frequency, rad/s\n",
-      "mf=delta_f*1000/fs;              #Modulation index\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"e=%dcos(%.2et + %dsin%dt)\"%(Ec,wc,mf,ws));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "e=4cos(1.57e+08t + 25sin2513t)\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.18 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "delta_f=50.0;                   #Maximum frequency deviation, kHz\n",
-      "fs=5.0;                         #Modulating frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "mf=delta_f/fs;          #Modulation index\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation index=%d\"%mf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index=10\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.19 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=1000.0;                  #Carrier frequency, kHz\n",
-      "fs=15.0;                    #Modulating frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "first_3_usb_f=[fc+fs,fc+2*fs,fc+3*fs];              #First three upper sideband frequncies, kHz\n",
-      "first_3_lsb_f=[fc-fs,fc-2*fs,fc-3*fs];              #First three lowerr sideband frequncies, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The first three upper sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_usb_f[0],first_3_usb_f[1],first_3_usb_f[2]));\n",
-      "print(\"The first three lower sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_lsb_f[0],first_3_lsb_f[1],first_3_lsb_f[2]));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The first three upper sideband frequencies=1015kHz ,1030kHz and 1045kHz.\n",
-        "The first three lower sideband frequencies=985kHz ,970kHz and 955kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.20 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs=15.0;                    #Modulating frequency, kHz\n",
-      "delta_f=75.0;               #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "BW=2*(delta_f+fs);          #Bandwidth, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The bandwidth of the FM signal=%dkHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The bandwidth of the FM signal=180kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.21 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "k=75.0;             #Frequency deviation constant, kHz/V\n",
-      "Es=2.0;             #Amplitude  of signal, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "delta_f=k*Es;       #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum frequency deviation=%dkHz.\"%delta_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum frequency deviation=150kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.22 : Page number 429-430\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs1=500.0;              #First audio frequency, Hz\n",
-      "fs2=200.0;              #Second audio frequency (decreased), Hz\n",
-      "Es=2.4;                 #AF voltage, V\n",
-      "delta_f1=4.8;           #Frequency deviation,kHz\n",
-      "\n",
-      "#Calculation\n",
-      "k=delta_f1/Es;          #Frequency deviation constant, kHz/V\n",
-      "Es=7.2;                 #AF voltage, V (increased)\n",
-      "delta_f2=k*Es;          #2nd frequency deviation, kHz\n",
-      "Es=10.0;                #AF voltage, V (increased)\n",
-      "delta_f3=k*Es;          #3rd frequency deviation, kHz\n",
-      "\n",
-      "mf1=delta_f1/(fs1/1000);        #Modulation index in 1st case\n",
-      "mf2=delta_f2/(fs1/1000);        #Modulation index in 2nd case\n",
-      "mf3=delta_f3/(fs2/1000);        #Modulation index in 3rd case\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency deviation in second case=%.1fkHz.\"%delta_f2);\n",
-      "print(\"The frequency deviation in third case=%dkHz.\"%delta_f3);\n",
-      "print(\"The modulation index in 1st case=%.1f\"%mf1);\n",
-      "print(\"The modulation index in 2nd case=%.1f\"%mf2);\n",
-      "print(\"The modulation index in 3rd case=%d\"%mf3);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency deviation in second case=14.4kHz.\n",
-        "The frequency deviation in third case=20kHz.\n",
-        "The modulation index in 1st case=9.6\n",
-        "The modulation index in 2nd case=28.8\n",
-        "The modulation index in 3rd case=100\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_4.ipynb
deleted file mode 100755
index b2262b8f..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_4.ipynb
+++ /dev/null
@@ -1,936 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d4ffda068787fb0974622fa8de40f7d54b5df2a00735e870e01cb9df45be78f9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 16 : MODULATION AND DEMODULATION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.2 : Page number 416-417\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variabledeclaration\n",
-      "V_pp_max=16.0;                  #Maximum peak-to-peak voltage of an AM wave, mV\n",
-      "V_pp_min=4.0;                   #Minimum peak-to-peak voltage of an AM wave, mV\n",
-      "\n",
-      "#Calculation\n",
-      "Vmax=V_pp_max/2;                     #Maximum voltage of AM wave, mV\n",
-      "Vmin=V_pp_min/2;                     #Minimum voltage of AM wave, mV\n",
-      "m=(Vmax-Vmin)/(Vmax+Vmin);           #Modulation factor.\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation factor=%.1f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation factor=0.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.3 : Page number 417\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Es=50.0;                #signalvoltage amplitude,  V\n",
-      "Ec=100.0;               #Carrier voltage amplitude, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "m=Es/Ec;                    #Modulation factor\n",
-      "\n",
-      "#Result\n",
-      "print(\"Modulation factor=%.1f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Modulation factor=0.5.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.4 : Page number 419\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=2500.0;                  #Carrier frequency, kHz\n",
-      "f1=50.0;                    #Lower frequency of the audio signal, Hz\n",
-      "f2=15000.0;                 #Upper frequency of the audio signal, Hz\n",
-      "\n",
-      "#Calculation\n",
-      "fl_usb=fc+(f1/1000);            #Lower frequency of upper sideband, kHz\n",
-      "fu_usb=fc+(f2/1000);            #Upper frequency of upper sideband, kHz\n",
-      "\n",
-      "fu_lsb=fc-(f1/1000);            #Lower frequency of upper sideband, kHz\n",
-      "fl_lsb=fc-(f2/1000);            #Upper frequency of upper sideband, kHz\n",
-      "\n",
-      "#Since, f1=50Hz is negligible with respect to f2=15000Hz,\n",
-      "BW=(fc+(f2/1000))-(fc-(f2/1000));                 #Bandwidth, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The upper sideband=%.2fkHz to %dkHz.\"%(fl_usb,fu_usb));\n",
-      "print(\"The lower sideband=%dkHz to %.2fkHz.\"%(fl_lsb,fu_lsb));\n",
-      "print(\"The bandwidth=%dkHz\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The upper sideband=2500.05kHz to 2515kHz.\n",
-        "The lower sideband=2485kHz to 2499.95kHz.\n",
-        "The bandwidth=30kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.5 : Page number 420\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "EC=5.0;             #Carrier amplitude, V\n",
-      "m=0.6;              #modulation factor\n",
-      "ws=6280.0;          #angular frequency of signal, radians/s\n",
-      "wc=211*10**4;       #angular frequency of carrier, radians/s\n",
-      "\n",
-      "#Calculation\n",
-      "fs=(ws/(2*pi))/1000;                   #Signal frequency, kHz\n",
-      "fc=(wc/(2*pi))/1000;                   #Carrier frequency, kHz\n",
-      "\n",
-      "#(i)\n",
-      "Max_amp=EC+m*EC;                        #Maximum amplitude of AM wave, V\n",
-      "Min_amp=EC-m*EC;                        #Minimum amplitude of AM wave, V\n",
-      "\n",
-      "#(ii)\n",
-      "frequency_components=[fc-fs,fc,fc+fs];          #frequency components, kHz\n",
-      "amplitudes=[m*EC/2,EC,m*EC/2];                  #Corresponding amplitudes, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The maximum and minimum amplitudes of AM wave=%dV and %dV.\"%(Max_amp,Min_amp));\n",
-      "print(\"(ii) The frequency components of the AM wave=%.0f,%.0f,%.0f.\"%(frequency_components[0],frequency_components[1],frequency_components[2]));\n",
-      "print(\"     The corresponding amplitudes are =%.1fV, %dV, %.1fV.\"%(amplitudes[0],amplitudes[1],amplitudes[2]));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The maximum and minimum amplitudes of AM wave=8V and 2V.\n",
-        "(ii) The frequency components of the AM wave=335,336,337.\n",
-        "     The corresponding amplitudes are =1.5V, 5V, 1.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.6 : Page number 420-421\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=1000.0;                  #Carrier frequency, kHz\n",
-      "fs=5.0;                     #Signal frequency, kHz\n",
-      "m=0.5;                      #Modulation factor\n",
-      "EC=100.0;                   #Amplitude of the carrier, V\n",
-      "\n",
-      "#Calculation\n",
-      "f_lsb=fc-fs;                    #Lower sideband frequency,kHz\n",
-      "f_usb=fc+fs;                    #Upper sideband frequency, kHz\n",
-      "Amplitude=m*EC/2;               #Amplitude of each sideband, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The lower and upper sideband frequencies are=%dkHz and %dkHz.\"%(f_lsb,f_usb));\n",
-      "print(\"The amplitude of each sideband =%dV\"%Amplitude);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lower and upper sideband frequencies are=995kHz and 1005kHz.\n",
-        "The amplitude of each sideband =25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.7 : Page number 421\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "EC=10.0;            #Carrier amplitude, V\n",
-      "ES=6.0;             #Signal amplitude, V\n",
-      "fc=10.0;            #Carrier frequency, MHz\n",
-      "fs=5/1000.0;        #Signal frequency. MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "m=ES/EC;            #Modulation factor\n",
-      "\n",
-      "#(ii)\n",
-      "f_lsb=fc-fs;                    #Lower sideband frequency,MHz\n",
-      "f_usb=fc+fs;                    #Upper sideband frequency, MHz\n",
-      "\n",
-      "#(iii)\n",
-      "Amplitude=m*EC/2;               #Amplitude of each sideband, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The modulation factor=%.1f.\"%m);\n",
-      "print(\"(ii)  The lower and upper sideband frequencies are=%.3fMHz and %.3fMHz.\"%(f_lsb,f_usb));\n",
-      "print(\"(iii) The amplitude of each sideband =%dV\"%Amplitude);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The modulation factor=0.6.\n",
-        "(ii)  The lower and upper sideband frequencies are=9.995MHz and 10.005MHz.\n",
-        "(iii) The amplitude of each sideband =3V\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.8 : Page number 423\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=500.0;               #Carrier power, W\n",
-      "m=1.0;                  #Modulation factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband power, W\n",
-      "\n",
-      "#(ii)\n",
-      "PT=Pc+Ps;                       #Power of AM wave, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The power in sidebands=%dW\"%Ps);\n",
-      "print(\"(ii) The power of AM wave=%dW\"%PT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The power in sidebands=250W\n",
-        "(ii) The power of AM wave=750W\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exmaple 16.9 : Page number 423\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=50.0;                    #Power of carrier, kW\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "m=80/100.0;                       #Modulation factor\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband Power, kW\n",
-      "print(\"(i) The sideband power for 80%% modulation=%dkW.\"%Ps);\n",
-      "\n",
-      "#(ii)\n",
-      "m=10/100.0;                       #Modulation factor\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband Power, kW\n",
-      "print(\"(ii) The sideband power for 10%% modulation=%.2fkW.\"%Ps);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The sideband power for 80% modulation=16kW.\n",
-        "(ii) The sideband power for 10% modulation=0.25kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.10 : Page number 423-424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=40.0;                       #Carrier power, kW\n",
-      "m=100/100.0;                     #Modulation index\n",
-      "amplifier_eff=72/100.0;          #Efficiency of modulated RF amplifier\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)Carrier power remains same after modulation\n",
-      "\n",
-      "#(ii)\n",
-      "Ps=(1/2.0)*(m**2)*Pc;                         #Sideband power\n",
-      "P_audio=Ps/amplifier_eff;                   #Required audio power, kW\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The carrier power=%dkW.\"%Pc);\n",
-      "print(\"(ii) The required audio power=%.1fkW.\"%P_audio);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The carrier power=40kW.\n",
-        "(ii) The required audio power=27.8kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.11 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs=1.0;                         #Signal frequency, kHz\n",
-      "fc=500.0;                       #Carrier frequency, kHz\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "sideband_f=[fc-fs,fc+fs];               #Sideband frequencies, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "BW=(fc+fs)-(fc-fs);                     #Bandwidth required, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The sideband frequencies=%dkHz and %dkHz.\"%(sideband_f[0],sideband_f[1]));\n",
-      "print(\"(ii) The bandwidth required=%dkHz\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The sideband frequencies=499kHz and 501kHz.\n",
-        "(ii) The bandwidth required=2kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.12 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "IC=8.0;                     #Antenna current due to carrier,A\n",
-      "m=40/100.0;                   #Modulation index\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n",
-      "IT=IC*sqrt(1+(m**2/2.0));                     #Total current, A\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total antenna current=%.2fA.\"%IT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total antenna current=8.31A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.13 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "IC=8.0;                             #Antenna current when only carrier is sent, A\n",
-      "IT=8.93;                            #Total antenna current, A\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n",
-      "m=sqrt((((IT/IC)**2)-1)*2)*100;                 #The %age of modulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The %%age of modulation=%.1f%%.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The %age of modulation=70.1%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.14 : Page number 425\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "Vc=100.0;                       #Carrier voltage, V\n",
-      "V_T=110.0;                      #The total voltage after modulation, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_voltage/Carrier_voltage)=(V_T/Vc)\u00b2=1+(m\u00b2/2).\n",
-      "m=sqrt((((V_T/Vc)**2)-1)*2);                 #The %age of modulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation index =%.3f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index =0.648.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.15 : Page number 425-426\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vc=5.0;                       #Carrier voltage, V\n",
-      "V_lsb=2.5;                      #Lower sideband component, V\n",
-      "V_usb=2.5;                      #Upper sideband component, V\n",
-      "R=2.0;                          #Resistor driven by AM wave, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, power=(r.m.s_voltage)\u00b2/resistance\n",
-      "#(i)\n",
-      "Pc=round((0.707*Vc)**2/R,2);             #Carrier power mW\n",
-      "\n",
-      "#(ii)\n",
-      "P_lower=round((0.707*V_lsb)**2/R,3);            #Power delivered by lower sideband, mW\n",
-      "\n",
-      "#(iii)\n",
-      "P_upper=round((0.707*V_usb)**2/R,3);            #Power delivered by upper sideband, mW\n",
-      "\n",
-      "P_T=round(Pc+P_lower+P_upper,3);                     #Total power delivered by the AM wave, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The carrier power=%.2fmW\"%Pc);\n",
-      "print(\"(ii) The power delivered by lower sideband=%.3fmW\"%P_lower);\n",
-      "print(\"(iii) The power delivered by upper sideband=%.3fmW\"%P_upper);\n",
-      "print(\"The total power delivered by the AM wave=%.3fmW\"%P_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The carrier power=6.25mW\n",
-        "(ii) The power delivered by lower sideband=1.562mW\n",
-        "(iii) The power delivered by upper sideband=1.562mW\n",
-        "The total power delivered by the AM wave=9.374mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.16 : Page number 428\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "wc=6e08;            #Carrier angular frequency, rad/s\n",
-      "ws=1250.0;            #Signal angular frequency, rad/s\n",
-      "mf=5;                   #Modulation index\n",
-      "Ec=12.0;                #Carrier amplitude, V\n",
-      "R=10.0;                 #Resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fc=wc/(2*pi);                       #Carrier frequency, Hz\n",
-      "\n",
-      "#(ii)\n",
-      "fs=ws/(2*pi);                       #Signal frequency, Hz\n",
-      "\n",
-      "#(iv)\n",
-      "delta_f=mf*fs;                      #Maximum frequency deviation, Hz\n",
-      "\n",
-      "#(v)\n",
-      "P=(Ec/sqrt(2))**2/R;                    #Power dissipated, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The carrier frequency=%.1fe06 Hz.\"%(fc/10**6));\n",
-      "print(\"(ii)  The signal frequency=%.0f Hz.\"%fs);\n",
-      "print(\"(iii) The modulation index=%d.\"%mf);\n",
-      "print(\"(iv)  The maximum frequency deviation=%.0fHz.\"%delta_f);\n",
-      "print(\"(v)   The power dissipated=%.1fW.\"%P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The carrier frequency=95.5e06 Hz.\n",
-        "(ii)  The signal frequency=199 Hz.\n",
-        "(iii) The modulation index=5.\n",
-        "(iv)  The maximum frequency deviation=995Hz.\n",
-        "(v)   The power dissipated=7.2W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.17 : Page number 428-429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "fc=25.0;                       #Carrier frequency, MHz\n",
-      "fs=400.0;                      #Signal frequency, Hz\n",
-      "Ec=4.0;                         #Carrier amplitude, V\n",
-      "delta_f=10.0;                   #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "wc=2*pi*fc*10**6;                 #Carrier angular frequency, rad/s\n",
-      "ws=2*pi*fs;                 #Signal angular frequency, rad/s\n",
-      "mf=delta_f*1000/fs;              #Modulation index\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"e=%dcos(%.2et + %dsin%dt)\"%(Ec,wc,mf,ws));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "e=4cos(1.57e+08t + 25sin2513t)\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.18 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "delta_f=50.0;                   #Maximum frequency deviation, kHz\n",
-      "fs=5.0;                         #Modulating frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "mf=delta_f/fs;          #Modulation index\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation index=%d\"%mf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index=10\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.19 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=1000.0;                  #Carrier frequency, kHz\n",
-      "fs=15.0;                    #Modulating frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "first_3_usb_f=[fc+fs,fc+2*fs,fc+3*fs];              #First three upper sideband frequncies, kHz\n",
-      "first_3_lsb_f=[fc-fs,fc-2*fs,fc-3*fs];              #First three lowerr sideband frequncies, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The first three upper sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_usb_f[0],first_3_usb_f[1],first_3_usb_f[2]));\n",
-      "print(\"The first three lower sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_lsb_f[0],first_3_lsb_f[1],first_3_lsb_f[2]));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The first three upper sideband frequencies=1015kHz ,1030kHz and 1045kHz.\n",
-        "The first three lower sideband frequencies=985kHz ,970kHz and 955kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.20 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs=15.0;                    #Modulating frequency, kHz\n",
-      "delta_f=75.0;               #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "BW=2*(delta_f+fs);          #Bandwidth, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The bandwidth of the FM signal=%dkHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The bandwidth of the FM signal=180kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.21 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "k=75.0;             #Frequency deviation constant, kHz/V\n",
-      "Es=2.0;             #Amplitude  of signal, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "delta_f=k*Es;       #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum frequency deviation=%dkHz.\"%delta_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum frequency deviation=150kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.22 : Page number 429-430\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs1=500.0;              #First audio frequency, Hz\n",
-      "fs2=200.0;              #Second audio frequency (decreased), Hz\n",
-      "Es=2.4;                 #AF voltage, V\n",
-      "delta_f1=4.8;           #Frequency deviation,kHz\n",
-      "\n",
-      "#Calculation\n",
-      "k=delta_f1/Es;          #Frequency deviation constant, kHz/V\n",
-      "Es=7.2;                 #AF voltage, V (increased)\n",
-      "delta_f2=k*Es;          #2nd frequency deviation, kHz\n",
-      "Es=10.0;                #AF voltage, V (increased)\n",
-      "delta_f3=k*Es;          #3rd frequency deviation, kHz\n",
-      "\n",
-      "mf1=delta_f1/(fs1/1000);        #Modulation index in 1st case\n",
-      "mf2=delta_f2/(fs1/1000);        #Modulation index in 2nd case\n",
-      "mf3=delta_f3/(fs2/1000);        #Modulation index in 3rd case\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency deviation in second case=%.1fkHz.\"%delta_f2);\n",
-      "print(\"The frequency deviation in third case=%dkHz.\"%delta_f3);\n",
-      "print(\"The modulation index in 1st case=%.1f\"%mf1);\n",
-      "print(\"The modulation index in 2nd case=%.1f\"%mf2);\n",
-      "print(\"The modulation index in 3rd case=%d\"%mf3);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency deviation in second case=14.4kHz.\n",
-        "The frequency deviation in third case=20kHz.\n",
-        "The modulation index in 1st case=9.6\n",
-        "The modulation index in 2nd case=28.8\n",
-        "The modulation index in 3rd case=100\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_5.ipynb
deleted file mode 100755
index b2262b8f..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_5.ipynb
+++ /dev/null
@@ -1,936 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d4ffda068787fb0974622fa8de40f7d54b5df2a00735e870e01cb9df45be78f9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 16 : MODULATION AND DEMODULATION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.2 : Page number 416-417\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variabledeclaration\n",
-      "V_pp_max=16.0;                  #Maximum peak-to-peak voltage of an AM wave, mV\n",
-      "V_pp_min=4.0;                   #Minimum peak-to-peak voltage of an AM wave, mV\n",
-      "\n",
-      "#Calculation\n",
-      "Vmax=V_pp_max/2;                     #Maximum voltage of AM wave, mV\n",
-      "Vmin=V_pp_min/2;                     #Minimum voltage of AM wave, mV\n",
-      "m=(Vmax-Vmin)/(Vmax+Vmin);           #Modulation factor.\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation factor=%.1f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation factor=0.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.3 : Page number 417\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Es=50.0;                #signalvoltage amplitude,  V\n",
-      "Ec=100.0;               #Carrier voltage amplitude, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "m=Es/Ec;                    #Modulation factor\n",
-      "\n",
-      "#Result\n",
-      "print(\"Modulation factor=%.1f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Modulation factor=0.5.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.4 : Page number 419\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=2500.0;                  #Carrier frequency, kHz\n",
-      "f1=50.0;                    #Lower frequency of the audio signal, Hz\n",
-      "f2=15000.0;                 #Upper frequency of the audio signal, Hz\n",
-      "\n",
-      "#Calculation\n",
-      "fl_usb=fc+(f1/1000);            #Lower frequency of upper sideband, kHz\n",
-      "fu_usb=fc+(f2/1000);            #Upper frequency of upper sideband, kHz\n",
-      "\n",
-      "fu_lsb=fc-(f1/1000);            #Lower frequency of upper sideband, kHz\n",
-      "fl_lsb=fc-(f2/1000);            #Upper frequency of upper sideband, kHz\n",
-      "\n",
-      "#Since, f1=50Hz is negligible with respect to f2=15000Hz,\n",
-      "BW=(fc+(f2/1000))-(fc-(f2/1000));                 #Bandwidth, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The upper sideband=%.2fkHz to %dkHz.\"%(fl_usb,fu_usb));\n",
-      "print(\"The lower sideband=%dkHz to %.2fkHz.\"%(fl_lsb,fu_lsb));\n",
-      "print(\"The bandwidth=%dkHz\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The upper sideband=2500.05kHz to 2515kHz.\n",
-        "The lower sideband=2485kHz to 2499.95kHz.\n",
-        "The bandwidth=30kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.5 : Page number 420\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "EC=5.0;             #Carrier amplitude, V\n",
-      "m=0.6;              #modulation factor\n",
-      "ws=6280.0;          #angular frequency of signal, radians/s\n",
-      "wc=211*10**4;       #angular frequency of carrier, radians/s\n",
-      "\n",
-      "#Calculation\n",
-      "fs=(ws/(2*pi))/1000;                   #Signal frequency, kHz\n",
-      "fc=(wc/(2*pi))/1000;                   #Carrier frequency, kHz\n",
-      "\n",
-      "#(i)\n",
-      "Max_amp=EC+m*EC;                        #Maximum amplitude of AM wave, V\n",
-      "Min_amp=EC-m*EC;                        #Minimum amplitude of AM wave, V\n",
-      "\n",
-      "#(ii)\n",
-      "frequency_components=[fc-fs,fc,fc+fs];          #frequency components, kHz\n",
-      "amplitudes=[m*EC/2,EC,m*EC/2];                  #Corresponding amplitudes, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The maximum and minimum amplitudes of AM wave=%dV and %dV.\"%(Max_amp,Min_amp));\n",
-      "print(\"(ii) The frequency components of the AM wave=%.0f,%.0f,%.0f.\"%(frequency_components[0],frequency_components[1],frequency_components[2]));\n",
-      "print(\"     The corresponding amplitudes are =%.1fV, %dV, %.1fV.\"%(amplitudes[0],amplitudes[1],amplitudes[2]));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The maximum and minimum amplitudes of AM wave=8V and 2V.\n",
-        "(ii) The frequency components of the AM wave=335,336,337.\n",
-        "     The corresponding amplitudes are =1.5V, 5V, 1.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.6 : Page number 420-421\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=1000.0;                  #Carrier frequency, kHz\n",
-      "fs=5.0;                     #Signal frequency, kHz\n",
-      "m=0.5;                      #Modulation factor\n",
-      "EC=100.0;                   #Amplitude of the carrier, V\n",
-      "\n",
-      "#Calculation\n",
-      "f_lsb=fc-fs;                    #Lower sideband frequency,kHz\n",
-      "f_usb=fc+fs;                    #Upper sideband frequency, kHz\n",
-      "Amplitude=m*EC/2;               #Amplitude of each sideband, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The lower and upper sideband frequencies are=%dkHz and %dkHz.\"%(f_lsb,f_usb));\n",
-      "print(\"The amplitude of each sideband =%dV\"%Amplitude);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The lower and upper sideband frequencies are=995kHz and 1005kHz.\n",
-        "The amplitude of each sideband =25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.7 : Page number 421\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "EC=10.0;            #Carrier amplitude, V\n",
-      "ES=6.0;             #Signal amplitude, V\n",
-      "fc=10.0;            #Carrier frequency, MHz\n",
-      "fs=5/1000.0;        #Signal frequency. MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "m=ES/EC;            #Modulation factor\n",
-      "\n",
-      "#(ii)\n",
-      "f_lsb=fc-fs;                    #Lower sideband frequency,MHz\n",
-      "f_usb=fc+fs;                    #Upper sideband frequency, MHz\n",
-      "\n",
-      "#(iii)\n",
-      "Amplitude=m*EC/2;               #Amplitude of each sideband, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The modulation factor=%.1f.\"%m);\n",
-      "print(\"(ii)  The lower and upper sideband frequencies are=%.3fMHz and %.3fMHz.\"%(f_lsb,f_usb));\n",
-      "print(\"(iii) The amplitude of each sideband =%dV\"%Amplitude);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The modulation factor=0.6.\n",
-        "(ii)  The lower and upper sideband frequencies are=9.995MHz and 10.005MHz.\n",
-        "(iii) The amplitude of each sideband =3V\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.8 : Page number 423\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=500.0;               #Carrier power, W\n",
-      "m=1.0;                  #Modulation factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband power, W\n",
-      "\n",
-      "#(ii)\n",
-      "PT=Pc+Ps;                       #Power of AM wave, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The power in sidebands=%dW\"%Ps);\n",
-      "print(\"(ii) The power of AM wave=%dW\"%PT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The power in sidebands=250W\n",
-        "(ii) The power of AM wave=750W\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Exmaple 16.9 : Page number 423\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=50.0;                    #Power of carrier, kW\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "m=80/100.0;                       #Modulation factor\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband Power, kW\n",
-      "print(\"(i) The sideband power for 80%% modulation=%dkW.\"%Ps);\n",
-      "\n",
-      "#(ii)\n",
-      "m=10/100.0;                       #Modulation factor\n",
-      "Ps=(1/2.0)*m**2*Pc;               #Sideband Power, kW\n",
-      "print(\"(ii) The sideband power for 10%% modulation=%.2fkW.\"%Ps);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The sideband power for 80% modulation=16kW.\n",
-        "(ii) The sideband power for 10% modulation=0.25kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.10 : Page number 423-424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Pc=40.0;                       #Carrier power, kW\n",
-      "m=100/100.0;                     #Modulation index\n",
-      "amplifier_eff=72/100.0;          #Efficiency of modulated RF amplifier\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)Carrier power remains same after modulation\n",
-      "\n",
-      "#(ii)\n",
-      "Ps=(1/2.0)*(m**2)*Pc;                         #Sideband power\n",
-      "P_audio=Ps/amplifier_eff;                   #Required audio power, kW\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The carrier power=%dkW.\"%Pc);\n",
-      "print(\"(ii) The required audio power=%.1fkW.\"%P_audio);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The carrier power=40kW.\n",
-        "(ii) The required audio power=27.8kW.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.11 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs=1.0;                         #Signal frequency, kHz\n",
-      "fc=500.0;                       #Carrier frequency, kHz\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "sideband_f=[fc-fs,fc+fs];               #Sideband frequencies, kHz\n",
-      "\n",
-      "#(ii)\n",
-      "BW=(fc+fs)-(fc-fs);                     #Bandwidth required, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The sideband frequencies=%dkHz and %dkHz.\"%(sideband_f[0],sideband_f[1]));\n",
-      "print(\"(ii) The bandwidth required=%dkHz\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The sideband frequencies=499kHz and 501kHz.\n",
-        "(ii) The bandwidth required=2kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.12 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "IC=8.0;                     #Antenna current due to carrier,A\n",
-      "m=40/100.0;                   #Modulation index\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n",
-      "IT=IC*sqrt(1+(m**2/2.0));                     #Total current, A\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total antenna current=%.2fA.\"%IT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total antenna current=8.31A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.13 : Page number 424\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "IC=8.0;                             #Antenna current when only carrier is sent, A\n",
-      "IT=8.93;                            #Total antenna current, A\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n",
-      "m=sqrt((((IT/IC)**2)-1)*2)*100;                 #The %age of modulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The %%age of modulation=%.1f%%.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The %age of modulation=70.1%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.14 : Page number 425\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "Vc=100.0;                       #Carrier voltage, V\n",
-      "V_T=110.0;                      #The total voltage after modulation, V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n",
-      "#that implies, (PT/Pc)=1+(m\u00b2/2),\n",
-      "#So, square_of(Total_voltage/Carrier_voltage)=(V_T/Vc)\u00b2=1+(m\u00b2/2).\n",
-      "m=sqrt((((V_T/Vc)**2)-1)*2);                 #The %age of modulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation index =%.3f.\"%m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index =0.648.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.15 : Page number 425-426\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vc=5.0;                       #Carrier voltage, V\n",
-      "V_lsb=2.5;                      #Lower sideband component, V\n",
-      "V_usb=2.5;                      #Upper sideband component, V\n",
-      "R=2.0;                          #Resistor driven by AM wave, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, power=(r.m.s_voltage)\u00b2/resistance\n",
-      "#(i)\n",
-      "Pc=round((0.707*Vc)**2/R,2);             #Carrier power mW\n",
-      "\n",
-      "#(ii)\n",
-      "P_lower=round((0.707*V_lsb)**2/R,3);            #Power delivered by lower sideband, mW\n",
-      "\n",
-      "#(iii)\n",
-      "P_upper=round((0.707*V_usb)**2/R,3);            #Power delivered by upper sideband, mW\n",
-      "\n",
-      "P_T=round(Pc+P_lower+P_upper,3);                     #Total power delivered by the AM wave, mW\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The carrier power=%.2fmW\"%Pc);\n",
-      "print(\"(ii) The power delivered by lower sideband=%.3fmW\"%P_lower);\n",
-      "print(\"(iii) The power delivered by upper sideband=%.3fmW\"%P_upper);\n",
-      "print(\"The total power delivered by the AM wave=%.3fmW\"%P_T);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The carrier power=6.25mW\n",
-        "(ii) The power delivered by lower sideband=1.562mW\n",
-        "(iii) The power delivered by upper sideband=1.562mW\n",
-        "The total power delivered by the AM wave=9.374mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.16 : Page number 428\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "wc=6e08;            #Carrier angular frequency, rad/s\n",
-      "ws=1250.0;            #Signal angular frequency, rad/s\n",
-      "mf=5;                   #Modulation index\n",
-      "Ec=12.0;                #Carrier amplitude, V\n",
-      "R=10.0;                 #Resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "fc=wc/(2*pi);                       #Carrier frequency, Hz\n",
-      "\n",
-      "#(ii)\n",
-      "fs=ws/(2*pi);                       #Signal frequency, Hz\n",
-      "\n",
-      "#(iv)\n",
-      "delta_f=mf*fs;                      #Maximum frequency deviation, Hz\n",
-      "\n",
-      "#(v)\n",
-      "P=(Ec/sqrt(2))**2/R;                    #Power dissipated, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The carrier frequency=%.1fe06 Hz.\"%(fc/10**6));\n",
-      "print(\"(ii)  The signal frequency=%.0f Hz.\"%fs);\n",
-      "print(\"(iii) The modulation index=%d.\"%mf);\n",
-      "print(\"(iv)  The maximum frequency deviation=%.0fHz.\"%delta_f);\n",
-      "print(\"(v)   The power dissipated=%.1fW.\"%P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   The carrier frequency=95.5e06 Hz.\n",
-        "(ii)  The signal frequency=199 Hz.\n",
-        "(iii) The modulation index=5.\n",
-        "(iv)  The maximum frequency deviation=995Hz.\n",
-        "(v)   The power dissipated=7.2W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.17 : Page number 428-429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "fc=25.0;                       #Carrier frequency, MHz\n",
-      "fs=400.0;                      #Signal frequency, Hz\n",
-      "Ec=4.0;                         #Carrier amplitude, V\n",
-      "delta_f=10.0;                   #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "wc=2*pi*fc*10**6;                 #Carrier angular frequency, rad/s\n",
-      "ws=2*pi*fs;                 #Signal angular frequency, rad/s\n",
-      "mf=delta_f*1000/fs;              #Modulation index\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"e=%dcos(%.2et + %dsin%dt)\"%(Ec,wc,mf,ws));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "e=4cos(1.57e+08t + 25sin2513t)\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.18 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "delta_f=50.0;                   #Maximum frequency deviation, kHz\n",
-      "fs=5.0;                         #Modulating frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "mf=delta_f/fs;          #Modulation index\n",
-      "\n",
-      "#Result\n",
-      "print(\"The modulation index=%d\"%mf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The modulation index=10\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.19 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fc=1000.0;                  #Carrier frequency, kHz\n",
-      "fs=15.0;                    #Modulating frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "first_3_usb_f=[fc+fs,fc+2*fs,fc+3*fs];              #First three upper sideband frequncies, kHz\n",
-      "first_3_lsb_f=[fc-fs,fc-2*fs,fc-3*fs];              #First three lowerr sideband frequncies, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The first three upper sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_usb_f[0],first_3_usb_f[1],first_3_usb_f[2]));\n",
-      "print(\"The first three lower sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_lsb_f[0],first_3_lsb_f[1],first_3_lsb_f[2]));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The first three upper sideband frequencies=1015kHz ,1030kHz and 1045kHz.\n",
-        "The first three lower sideband frequencies=985kHz ,970kHz and 955kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.20 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs=15.0;                    #Modulating frequency, kHz\n",
-      "delta_f=75.0;               #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "BW=2*(delta_f+fs);          #Bandwidth, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The bandwidth of the FM signal=%dkHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The bandwidth of the FM signal=180kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.21 : Page number 429\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "k=75.0;             #Frequency deviation constant, kHz/V\n",
-      "Es=2.0;             #Amplitude  of signal, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "delta_f=k*Es;       #Maximum frequency deviation, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum frequency deviation=%dkHz.\"%delta_f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum frequency deviation=150kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 16.22 : Page number 429-430\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "fs1=500.0;              #First audio frequency, Hz\n",
-      "fs2=200.0;              #Second audio frequency (decreased), Hz\n",
-      "Es=2.4;                 #AF voltage, V\n",
-      "delta_f1=4.8;           #Frequency deviation,kHz\n",
-      "\n",
-      "#Calculation\n",
-      "k=delta_f1/Es;          #Frequency deviation constant, kHz/V\n",
-      "Es=7.2;                 #AF voltage, V (increased)\n",
-      "delta_f2=k*Es;          #2nd frequency deviation, kHz\n",
-      "Es=10.0;                #AF voltage, V (increased)\n",
-      "delta_f3=k*Es;          #3rd frequency deviation, kHz\n",
-      "\n",
-      "mf1=delta_f1/(fs1/1000);        #Modulation index in 1st case\n",
-      "mf2=delta_f2/(fs1/1000);        #Modulation index in 2nd case\n",
-      "mf3=delta_f3/(fs2/1000);        #Modulation index in 3rd case\n",
-      "\n",
-      "#Result\n",
-      "print(\"The frequency deviation in second case=%.1fkHz.\"%delta_f2);\n",
-      "print(\"The frequency deviation in third case=%dkHz.\"%delta_f3);\n",
-      "print(\"The modulation index in 1st case=%.1f\"%mf1);\n",
-      "print(\"The modulation index in 2nd case=%.1f\"%mf2);\n",
-      "print(\"The modulation index in 3rd case=%d\"%mf3);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency deviation in second case=14.4kHz.\n",
-        "The frequency deviation in third case=20kHz.\n",
-        "The modulation index in 1st case=9.6\n",
-        "The modulation index in 2nd case=28.8\n",
-        "The modulation index in 3rd case=100\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17.ipynb
deleted file mode 100755
index 2c9a49a5..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17.ipynb
+++ /dev/null
@@ -1,877 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:33d9a7285654630dd24f2d6229210244e78961e0605c11041ed1b2f130cb19a1"
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- "nbformat": 3,
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- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 17 : REGULATED D.C POWER SUPPLY"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.1 : Page number 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL=400.0;                 #Output voltage with no-load, V\n",
-      "V_FL=300.0;                 #Output voltage with full-load, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "percentage_voltage_regulation=((V_NL-V_FL)/V_FL)*100;           #Percentage of voltage regulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The percentage of voltage regulation=%.2f%%.\"%percentage_voltage_regulation);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage of voltage regulation=33.33%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.2 : Page number 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_regulation=1.0;               #%age voltage regulation\n",
-      "V_NL=30.0;                      #Output voltage with no-load,V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, %age_of_voltage_regulation=((V_NL-V_FL)/V_FL)*100\n",
-      "V_FL=V_NL/(1+(V_regulation/100));                   #Output voltage with full-load, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The full-load voltage=%.1fV.\"%V_FL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The full-load voltage=29.7V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL_A=30.0;                #Output voltage of supply A with no-load, V\n",
-      "V_FL_A=25.0;                #Output voltage of supply A with full-load, V\n",
-      "V_NL_B=30.0;                #Output voltage of supply B with no-load, V\n",
-      "V_FL_B=29.0;                #Output voltage of supply B with full-load, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_regulation_A=((V_NL_A-V_FL_A)/V_FL_A)*100;            #%age of voltage regulation in power supply A\n",
-      "V_regulation_B=((V_NL_B-V_FL_B)/V_FL_B)*100;            #%age of voltage regulation in power supply B\n",
-      "\n",
-      "#Result\n",
-      "if(V_regulation_A<V_regulation_B):\n",
-      "    print(\"Power supply A is better than B.\");\n",
-      "else :\n",
-      "    print(\"Power supply B is better than A.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supply B is better than A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.4 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL=500.0;         #Output voltage with no-load, V\n",
-      "V_FL=300.0;         #Output voltage with full-load, V\n",
-      "I_FL=120.0;         #Output current with full-load, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Regulation=((V_NL-V_FL)/V_FL)*100;          #Voltage regulation percentage\n",
-      "\n",
-      "#(ii)\n",
-      "RL_min=V_FL/I_FL;                       #Minimum load resistance, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The voltage regulation=%.1f%%.\"%Regulation);\n",
-      "print(\"(ii)The minimum load resistance=%.1fk\u03a9.\"%RL_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The voltage regulation=66.7%.\n",
-        "(ii)The minimum load resistance=2.5k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.5 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VL_1=10.5;              #Initial output voltage with load, V\n",
-      "VL_2=10.0;              #Decreased output voltage with additional load, V\n",
-      "IL_1=1.0;               #Initial load current, A\n",
-      "IL_added=1.0;           #Added load current, A\n",
-      "\n",
-      "#Calculation\n",
-      "delta_VL=VL_1-VL_2;                 #Change in output voltage, V\n",
-      "delta_IL=IL_added;                  #Change in load current, A\n",
-      "\n",
-      "#(i)\n",
-      "Zo=delta_VL/delta_IL;               #Output impedance of power supply, \u03a9 (OHM's LAW)\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, Output_impedance=change_in_output_voltage/change_in_output_current\n",
-      "#Zo=(V_NL-VL_1)/delta_IL,\n",
-      "delta_IL=IL_1;                          #Change in load current, A\n",
-      "V_NL=VL_1+(delta_IL*Zo);                #Output voltage with no load, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output impedance=%.1f\u03a9.\"%Zo);\n",
-      "print(\"(ii) The output voltage with no-load=%dV.\"%V_NL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output impedance=0.5\u03a9.\n",
-        "(ii) The output voltage with no-load=11V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.6 : Page number 446\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Zo=0.01;                    #Output impedance, \u03a9\n",
-      "IL_max=1.0;                 #Maximum output current, A\n",
-      "IL_min=0.5                  #Minimum output current, A\n",
-      "f=10.0;                     #Frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Zo=delta_VL/delta_IL\n",
-      "delta_IL=IL_max-IL_min;                        #Maximum change in output current, A\n",
-      "delta_VL=(Zo*delta_IL)*1000;                   #Fluctuations in output voltage, mV\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage will have %dmV peak-to-peak fluctuation at a rate of %dkHz.\"%(delta_VL,f));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage will have 5mV peak-to-peak fluctuation at a rate of 10kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.7 : Page number 446\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "delta_Vout=10.0;                    #Change in output voltage, \u03bcV\n",
-      "delta_Vin=5.0;                      #Change in input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Line_regulation=delta_Vout/delta_Vin;           #Line regulation, \u03bcV/V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The line regulation of the voltage regulator=%d\u03bcV/V.\"%Line_regulation);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The line regulation of the voltage regulator=2\u03bcV/V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.8 : Page number 449-450\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=24.0;                   #Input voltage, V\n",
-      "Vz=12.0;                    #Zener voltage, V\n",
-      "Rs=160.0;                   #Series resistance, \u03a9\n",
-      "RL_max=float('inf');        #Maximum load resistance, \u03a9\n",
-      "RL_min=200.0;               #Minimum load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz;                       #Output voltage,(equal to zener regulated voltage), V\n",
-      "Is=((Vin-Vout)/Rs)*1000;       #Current through series resistance, mA\n",
-      "\n",
-      "#(ii)\n",
-      "IL_min=Vout/RL_max;                    #Minimum load current, A\n",
-      "IL_max=(Vout/RL_min)*1000;             #Maximum load current, mA\n",
-      "\n",
-      "#(iii)\n",
-      "IZ_min=Is-IL_max;                       #Minimum zener current, mA\n",
-      "IZ_max=Is-IL_min;                       #Maximum zener current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The current through the series resistance=%dmA\"%Is);\n",
-      "print(\"(ii) The minimum and maximum load currents are=%dA and %dmA\"%(IL_min,IL_max));\n",
-      "print(\"(iii) The minimum and maximum zener currents are=%dmA and %dmA\"%(IZ_min,IZ_max));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The current through the series resistance=75mA\n",
-        "(ii) The minimum and maximum load currents are=0A and 60mA\n",
-        "(iii) The minimum and maximum zener currents are=15mA and 75mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.9 : Page number 450\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VZ=15.0;                #Zener voltage, V\n",
-      "Vin_min=22.0            #Minimum input voltage, V\n",
-      "Vin_max=40.0            #Maximum input voltage, V\n",
-      "Vout=VZ;                #Regulated output voltage, V\n",
-      "IL_max=100.0;           #Maximum load current, mA\n",
-      "IL_min=20.0;            #Minimum load current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "RS_max=(Vin_min-Vout)/(IL_max/1000);               #Maximum value of series resistance, \u03a9 (OHM'S lAW)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum load resistance to hold the voltage constant=%d\u03a9.\"%RS_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum load resistance to hold the voltage constant=70\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.10 : Page number 450-451\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=3.3;                     #Zener voltage, V\n",
-      "Iz_min=3.0;                 #Minimum zener current, mA\n",
-      "Iz_max=100.0;               #Maximum zener current, mA\n",
-      "RL_max=2.0;                 #Maximum load resistance, k\u03a9\n",
-      "RL_min=500.0;               #Minimum load resistance, \u03a9\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Rs_min=(Vin-Vz)/(Iz_max/1000);             #Minimum series resistance required, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum series resistance required to limit the zener current=%.0f\u03a9.\"%Rs_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum series resistance required to limit the zener current=167\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.11 : Page number 451\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=3.3;                     #Zener voltage, V\n",
-      "Iz_min=3.0;                 #Minimum zener current, mA\n",
-      "Iz_max=100.0;               #Maximum zener current, mA\n",
-      "RL_max=2.0;                 #Maximum load resistance, k\u03a9\n",
-      "RL_min=500.0;               #Minimum load resistance, \u03a9\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IL_max=(Vz/RL_min)*1000;                   #Maximum load current, mA\n",
-      "Rs_max=((Vin-Vz)/(IL_max+Iz_min))*1000;    #Maximum series resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable value of series resistance=%d\u03a9.\"%Rs_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable value of series resistance=1739\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.12 : Page number 452\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10.0;                #Zener voltage, V\n",
-      "beta=100.0;             #Base current amplification factor\n",
-      "RL=1000.0;              #Load resistance, \u03a9\n",
-      "VBE=0.5;                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=Vz-VBE;                    #Output voltage, V\n",
-      "IL=(Vout/RL)*1000;                     #Load current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"The load current=%.1fmA\"%IL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage=9.5V.\n",
-        "The load current=9.5mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 : Page number 452\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IC=1.0;                     #Required current(collector current), A\n",
-      "Vout=6.0;                   #Constant output voltage, V\n",
-      "Vin=10.0;                   #Supply voltage, V\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "VBE=0.5;                    #Base-emitter voltage, V\n",
-      "Iz=10.0;                #Minimum zener current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IB=(IC/beta)*1000;                 #Base current, mA\n",
-      "\n",
-      "#Since, Vout=Vz-VBE;\n",
-      "Vz=Vout+VBE;                    #Zener breakdown voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_Rs=Vin-Vz;                    #Voltage across series resistance Rs, V\n",
-      "Rs=(V_Rs/(IB+Iz))*1000;                #Series resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The zener breakdown voltage=%.1fV\"%Vz);\n",
-      "print(\"(ii)The series resistance=%.0f\u03a9.\"%Rs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The zener breakdown voltage=6.5V\n",
-        "(ii)The series resistance=117\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.14: Page number 452-453\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=12.0;                    #Zener voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "RS=220.0;                   #Series resistance, \u03a9\n",
-      "RL=1.0;                     #Load resistance, k\u03a9\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz-VBE;                #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_RS=Vin-Vz;                       #Voltage across series resistance, RS, V\n",
-      "IR=(V_RS/RS)*1000;                 #Current through series resistance, mA\n",
-      "IL=Vout/RL;                        #Load current, mA\n",
-      "\n",
-      "#Since, IL is emitter current  and emitter current is approx. equal to collector current,\n",
-      "IC=IL;                              #Collector current, mA\n",
-      "IB=IC/beta;                         #Base current, mA\n",
-      "Iz=IR-IB;                           #Zener current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"(ii) The zener current=%dmA\"%Iz);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage=11.3V.\n",
-        "(ii) The zener current=36mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.15 : Page number 453-454\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "IL_min=0;                   #Minimum load current, A\n",
-      "IL_max=1.0;                 #Maximum load current, A\n",
-      "Vin_min=12.0;               #Minimum input voltage, V\n",
-      "Vin_max=18.0;               #Maximum input voltage, V\n",
-      "Iz_min=1.0;                 #Minimum zener current, mA\n",
-      "Vz=8.5;                     #Zener voltage, V\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IB_max=(IL_max/beta)*1000;                  #Maximum base current, mA\n",
-      "I_RS=Iz_min+IB_max;                         #Current through the series resistance, mA\n",
-      "RS=((Vin_min-Vz)/I_RS)*1000;                #Series resistance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "V_RS_max=Vin_max-Vz;                                      #Maximum voltage across series resistance, V\n",
-      "P_max_RS=ceil((V_RS_max**2/RS)*1000)/1000;                #Maximum power dissipation in series resistance RS, W\n",
-      "\n",
-      "#(iii)\n",
-      "I_RS_max=V_RS_max/floor(RS);                       #Maximum current through series resistance,mA\n",
-      "Iz_max=I_RS_max;                            #Maximum zener current, mA\n",
-      "P_z_max=Vz*Iz_max;                          #Maximum power dissipated in zener diode, W\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The series resistance=%d\u03a9.\"%RS);\n",
-      "print(\"(ii) The maximum power dissipated in series resistance=%.3fW.\"%P_max_RS);\n",
-      "print(\"(iii)The maximum power dissipated in zener diode=%.3fW.\"%P_z_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The series resistance=166\u03a9.\n",
-        "(ii) The maximum power dissipated in series resistance=0.542W.\n",
-        "(iii)The maximum power dissipated in zener diode=0.486W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.16 : Page number 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=2.0;                     #Resistor R1, k\u03a9\n",
-      "R2=1.0;                     #Resistor R2, k\u03a9\n",
-      "Vz=6.0;                     #Zener voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "m=R2/(R1+R2);                   #Feedback fraction\n",
-      "A_CL=1/m;                       #Closed-loop voltage gain\n",
-      "Vout=A_CL*(Vz+VBE);             #Regulated output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The regulated output voltage=%.1fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated output voltage=20.1V\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.17 : Page number 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=30.0;                     #Resistor R1, k\u03a9\n",
-      "R2=10.0;                     #Resistor R2, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "m=R2/(R1+R2);                   #Feedback fraction\n",
-      "A_CL=1/m;                       #Closed-loop voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The closed-loop voltage gain=%d.\"%A_CL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The closed-loop voltage gain=4.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.18 : Page number 457\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=22.0;               #Input voltage, V\n",
-      "Rs=130.0;               #Series resistance, \u03a9\n",
-      "Vz=8.3;                 #Zener voltage, V\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "RL=100.0;               #Load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz+VBE;                #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "IL=(Vout/RL)*1000;                 #Load current, mA (OHM's LAW)\n",
-      "IS=((Vin-Vout)/Rs)*1000;           #Current through series resistance, mA (OHM's LAW)\n",
-      "IC=IS-IL;                          #Collector current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The regulated output voltage=%dV\"%Vout);\n",
-      "print(\"(ii) Various currents for the shunt regulator are: IL=%dmA , IS=%dmA and IC=%dmA\"%(IL,IS,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The regulated output voltage=9V\n",
-        "(ii) Various currents for the shunt regulator are: IL=90mA , IS=100mA and IC=10mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.20 : Page number 463\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=240.0;               #Resistor R1 of the regulator, \u03a9\n",
-      "R2=2.4;                 #Variable resistance R2 of the regulator, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=1.25*(R2*1000/R1 + 1);             #Regulated output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The regulated output voltage=%.2fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated output voltage=13.75V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.21 : Page number 463\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vout_adj=8.0;           #Output voltage (adjusted), V\n",
-      "Vd=40.0;                #Input/output differential rating, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vin_max=Vout_adj+Vd;            #Maximum allowable input voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable input voltage=%dV.\"%Vin_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable input voltage=48V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_1.ipynb
deleted file mode 100755
index 2c9a49a5..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_1.ipynb
+++ /dev/null
@@ -1,877 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:33d9a7285654630dd24f2d6229210244e78961e0605c11041ed1b2f130cb19a1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 17 : REGULATED D.C POWER SUPPLY"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.1 : Page number 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL=400.0;                 #Output voltage with no-load, V\n",
-      "V_FL=300.0;                 #Output voltage with full-load, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "percentage_voltage_regulation=((V_NL-V_FL)/V_FL)*100;           #Percentage of voltage regulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The percentage of voltage regulation=%.2f%%.\"%percentage_voltage_regulation);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage of voltage regulation=33.33%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.2 : Page number 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_regulation=1.0;               #%age voltage regulation\n",
-      "V_NL=30.0;                      #Output voltage with no-load,V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, %age_of_voltage_regulation=((V_NL-V_FL)/V_FL)*100\n",
-      "V_FL=V_NL/(1+(V_regulation/100));                   #Output voltage with full-load, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The full-load voltage=%.1fV.\"%V_FL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The full-load voltage=29.7V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL_A=30.0;                #Output voltage of supply A with no-load, V\n",
-      "V_FL_A=25.0;                #Output voltage of supply A with full-load, V\n",
-      "V_NL_B=30.0;                #Output voltage of supply B with no-load, V\n",
-      "V_FL_B=29.0;                #Output voltage of supply B with full-load, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_regulation_A=((V_NL_A-V_FL_A)/V_FL_A)*100;            #%age of voltage regulation in power supply A\n",
-      "V_regulation_B=((V_NL_B-V_FL_B)/V_FL_B)*100;            #%age of voltage regulation in power supply B\n",
-      "\n",
-      "#Result\n",
-      "if(V_regulation_A<V_regulation_B):\n",
-      "    print(\"Power supply A is better than B.\");\n",
-      "else :\n",
-      "    print(\"Power supply B is better than A.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supply B is better than A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.4 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL=500.0;         #Output voltage with no-load, V\n",
-      "V_FL=300.0;         #Output voltage with full-load, V\n",
-      "I_FL=120.0;         #Output current with full-load, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Regulation=((V_NL-V_FL)/V_FL)*100;          #Voltage regulation percentage\n",
-      "\n",
-      "#(ii)\n",
-      "RL_min=V_FL/I_FL;                       #Minimum load resistance, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The voltage regulation=%.1f%%.\"%Regulation);\n",
-      "print(\"(ii)The minimum load resistance=%.1fk\u03a9.\"%RL_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The voltage regulation=66.7%.\n",
-        "(ii)The minimum load resistance=2.5k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.5 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VL_1=10.5;              #Initial output voltage with load, V\n",
-      "VL_2=10.0;              #Decreased output voltage with additional load, V\n",
-      "IL_1=1.0;               #Initial load current, A\n",
-      "IL_added=1.0;           #Added load current, A\n",
-      "\n",
-      "#Calculation\n",
-      "delta_VL=VL_1-VL_2;                 #Change in output voltage, V\n",
-      "delta_IL=IL_added;                  #Change in load current, A\n",
-      "\n",
-      "#(i)\n",
-      "Zo=delta_VL/delta_IL;               #Output impedance of power supply, \u03a9 (OHM's LAW)\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, Output_impedance=change_in_output_voltage/change_in_output_current\n",
-      "#Zo=(V_NL-VL_1)/delta_IL,\n",
-      "delta_IL=IL_1;                          #Change in load current, A\n",
-      "V_NL=VL_1+(delta_IL*Zo);                #Output voltage with no load, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output impedance=%.1f\u03a9.\"%Zo);\n",
-      "print(\"(ii) The output voltage with no-load=%dV.\"%V_NL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output impedance=0.5\u03a9.\n",
-        "(ii) The output voltage with no-load=11V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.6 : Page number 446\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Zo=0.01;                    #Output impedance, \u03a9\n",
-      "IL_max=1.0;                 #Maximum output current, A\n",
-      "IL_min=0.5                  #Minimum output current, A\n",
-      "f=10.0;                     #Frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Zo=delta_VL/delta_IL\n",
-      "delta_IL=IL_max-IL_min;                        #Maximum change in output current, A\n",
-      "delta_VL=(Zo*delta_IL)*1000;                   #Fluctuations in output voltage, mV\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage will have %dmV peak-to-peak fluctuation at a rate of %dkHz.\"%(delta_VL,f));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage will have 5mV peak-to-peak fluctuation at a rate of 10kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.7 : Page number 446\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "delta_Vout=10.0;                    #Change in output voltage, \u03bcV\n",
-      "delta_Vin=5.0;                      #Change in input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Line_regulation=delta_Vout/delta_Vin;           #Line regulation, \u03bcV/V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The line regulation of the voltage regulator=%d\u03bcV/V.\"%Line_regulation);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The line regulation of the voltage regulator=2\u03bcV/V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.8 : Page number 449-450\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=24.0;                   #Input voltage, V\n",
-      "Vz=12.0;                    #Zener voltage, V\n",
-      "Rs=160.0;                   #Series resistance, \u03a9\n",
-      "RL_max=float('inf');        #Maximum load resistance, \u03a9\n",
-      "RL_min=200.0;               #Minimum load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz;                       #Output voltage,(equal to zener regulated voltage), V\n",
-      "Is=((Vin-Vout)/Rs)*1000;       #Current through series resistance, mA\n",
-      "\n",
-      "#(ii)\n",
-      "IL_min=Vout/RL_max;                    #Minimum load current, A\n",
-      "IL_max=(Vout/RL_min)*1000;             #Maximum load current, mA\n",
-      "\n",
-      "#(iii)\n",
-      "IZ_min=Is-IL_max;                       #Minimum zener current, mA\n",
-      "IZ_max=Is-IL_min;                       #Maximum zener current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The current through the series resistance=%dmA\"%Is);\n",
-      "print(\"(ii) The minimum and maximum load currents are=%dA and %dmA\"%(IL_min,IL_max));\n",
-      "print(\"(iii) The minimum and maximum zener currents are=%dmA and %dmA\"%(IZ_min,IZ_max));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The current through the series resistance=75mA\n",
-        "(ii) The minimum and maximum load currents are=0A and 60mA\n",
-        "(iii) The minimum and maximum zener currents are=15mA and 75mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.9 : Page number 450\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VZ=15.0;                #Zener voltage, V\n",
-      "Vin_min=22.0            #Minimum input voltage, V\n",
-      "Vin_max=40.0            #Maximum input voltage, V\n",
-      "Vout=VZ;                #Regulated output voltage, V\n",
-      "IL_max=100.0;           #Maximum load current, mA\n",
-      "IL_min=20.0;            #Minimum load current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "RS_max=(Vin_min-Vout)/(IL_max/1000);               #Maximum value of series resistance, \u03a9 (OHM'S lAW)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum load resistance to hold the voltage constant=%d\u03a9.\"%RS_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum load resistance to hold the voltage constant=70\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.10 : Page number 450-451\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=3.3;                     #Zener voltage, V\n",
-      "Iz_min=3.0;                 #Minimum zener current, mA\n",
-      "Iz_max=100.0;               #Maximum zener current, mA\n",
-      "RL_max=2.0;                 #Maximum load resistance, k\u03a9\n",
-      "RL_min=500.0;               #Minimum load resistance, \u03a9\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Rs_min=(Vin-Vz)/(Iz_max/1000);             #Minimum series resistance required, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum series resistance required to limit the zener current=%.0f\u03a9.\"%Rs_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum series resistance required to limit the zener current=167\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.11 : Page number 451\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=3.3;                     #Zener voltage, V\n",
-      "Iz_min=3.0;                 #Minimum zener current, mA\n",
-      "Iz_max=100.0;               #Maximum zener current, mA\n",
-      "RL_max=2.0;                 #Maximum load resistance, k\u03a9\n",
-      "RL_min=500.0;               #Minimum load resistance, \u03a9\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IL_max=(Vz/RL_min)*1000;                   #Maximum load current, mA\n",
-      "Rs_max=((Vin-Vz)/(IL_max+Iz_min))*1000;    #Maximum series resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable value of series resistance=%d\u03a9.\"%Rs_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable value of series resistance=1739\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.12 : Page number 452\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10.0;                #Zener voltage, V\n",
-      "beta=100.0;             #Base current amplification factor\n",
-      "RL=1000.0;              #Load resistance, \u03a9\n",
-      "VBE=0.5;                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=Vz-VBE;                    #Output voltage, V\n",
-      "IL=(Vout/RL)*1000;                     #Load current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"The load current=%.1fmA\"%IL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage=9.5V.\n",
-        "The load current=9.5mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 : Page number 452\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IC=1.0;                     #Required current(collector current), A\n",
-      "Vout=6.0;                   #Constant output voltage, V\n",
-      "Vin=10.0;                   #Supply voltage, V\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "VBE=0.5;                    #Base-emitter voltage, V\n",
-      "Iz=10.0;                #Minimum zener current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IB=(IC/beta)*1000;                 #Base current, mA\n",
-      "\n",
-      "#Since, Vout=Vz-VBE;\n",
-      "Vz=Vout+VBE;                    #Zener breakdown voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_Rs=Vin-Vz;                    #Voltage across series resistance Rs, V\n",
-      "Rs=(V_Rs/(IB+Iz))*1000;                #Series resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The zener breakdown voltage=%.1fV\"%Vz);\n",
-      "print(\"(ii)The series resistance=%.0f\u03a9.\"%Rs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The zener breakdown voltage=6.5V\n",
-        "(ii)The series resistance=117\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.14: Page number 452-453\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=12.0;                    #Zener voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "RS=220.0;                   #Series resistance, \u03a9\n",
-      "RL=1.0;                     #Load resistance, k\u03a9\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz-VBE;                #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_RS=Vin-Vz;                       #Voltage across series resistance, RS, V\n",
-      "IR=(V_RS/RS)*1000;                 #Current through series resistance, mA\n",
-      "IL=Vout/RL;                        #Load current, mA\n",
-      "\n",
-      "#Since, IL is emitter current  and emitter current is approx. equal to collector current,\n",
-      "IC=IL;                              #Collector current, mA\n",
-      "IB=IC/beta;                         #Base current, mA\n",
-      "Iz=IR-IB;                           #Zener current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"(ii) The zener current=%dmA\"%Iz);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage=11.3V.\n",
-        "(ii) The zener current=36mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.15 : Page number 453-454\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "IL_min=0;                   #Minimum load current, A\n",
-      "IL_max=1.0;                 #Maximum load current, A\n",
-      "Vin_min=12.0;               #Minimum input voltage, V\n",
-      "Vin_max=18.0;               #Maximum input voltage, V\n",
-      "Iz_min=1.0;                 #Minimum zener current, mA\n",
-      "Vz=8.5;                     #Zener voltage, V\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IB_max=(IL_max/beta)*1000;                  #Maximum base current, mA\n",
-      "I_RS=Iz_min+IB_max;                         #Current through the series resistance, mA\n",
-      "RS=((Vin_min-Vz)/I_RS)*1000;                #Series resistance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "V_RS_max=Vin_max-Vz;                                      #Maximum voltage across series resistance, V\n",
-      "P_max_RS=ceil((V_RS_max**2/RS)*1000)/1000;                #Maximum power dissipation in series resistance RS, W\n",
-      "\n",
-      "#(iii)\n",
-      "I_RS_max=V_RS_max/floor(RS);                       #Maximum current through series resistance,mA\n",
-      "Iz_max=I_RS_max;                            #Maximum zener current, mA\n",
-      "P_z_max=Vz*Iz_max;                          #Maximum power dissipated in zener diode, W\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The series resistance=%d\u03a9.\"%RS);\n",
-      "print(\"(ii) The maximum power dissipated in series resistance=%.3fW.\"%P_max_RS);\n",
-      "print(\"(iii)The maximum power dissipated in zener diode=%.3fW.\"%P_z_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The series resistance=166\u03a9.\n",
-        "(ii) The maximum power dissipated in series resistance=0.542W.\n",
-        "(iii)The maximum power dissipated in zener diode=0.486W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.16 : Page number 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=2.0;                     #Resistor R1, k\u03a9\n",
-      "R2=1.0;                     #Resistor R2, k\u03a9\n",
-      "Vz=6.0;                     #Zener voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "m=R2/(R1+R2);                   #Feedback fraction\n",
-      "A_CL=1/m;                       #Closed-loop voltage gain\n",
-      "Vout=A_CL*(Vz+VBE);             #Regulated output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The regulated output voltage=%.1fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated output voltage=20.1V\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.17 : Page number 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=30.0;                     #Resistor R1, k\u03a9\n",
-      "R2=10.0;                     #Resistor R2, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "m=R2/(R1+R2);                   #Feedback fraction\n",
-      "A_CL=1/m;                       #Closed-loop voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The closed-loop voltage gain=%d.\"%A_CL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The closed-loop voltage gain=4.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.18 : Page number 457\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=22.0;               #Input voltage, V\n",
-      "Rs=130.0;               #Series resistance, \u03a9\n",
-      "Vz=8.3;                 #Zener voltage, V\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "RL=100.0;               #Load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz+VBE;                #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "IL=(Vout/RL)*1000;                 #Load current, mA (OHM's LAW)\n",
-      "IS=((Vin-Vout)/Rs)*1000;           #Current through series resistance, mA (OHM's LAW)\n",
-      "IC=IS-IL;                          #Collector current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The regulated output voltage=%dV\"%Vout);\n",
-      "print(\"(ii) Various currents for the shunt regulator are: IL=%dmA , IS=%dmA and IC=%dmA\"%(IL,IS,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The regulated output voltage=9V\n",
-        "(ii) Various currents for the shunt regulator are: IL=90mA , IS=100mA and IC=10mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.20 : Page number 463\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=240.0;               #Resistor R1 of the regulator, \u03a9\n",
-      "R2=2.4;                 #Variable resistance R2 of the regulator, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=1.25*(R2*1000/R1 + 1);             #Regulated output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The regulated output voltage=%.2fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated output voltage=13.75V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.21 : Page number 463\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vout_adj=8.0;           #Output voltage (adjusted), V\n",
-      "Vd=40.0;                #Input/output differential rating, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vin_max=Vout_adj+Vd;            #Maximum allowable input voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable input voltage=%dV.\"%Vin_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable input voltage=48V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_2.ipynb
deleted file mode 100755
index 2c9a49a5..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_2.ipynb
+++ /dev/null
@@ -1,877 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:33d9a7285654630dd24f2d6229210244e78961e0605c11041ed1b2f130cb19a1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 17 : REGULATED D.C POWER SUPPLY"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.1 : Page number 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL=400.0;                 #Output voltage with no-load, V\n",
-      "V_FL=300.0;                 #Output voltage with full-load, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "percentage_voltage_regulation=((V_NL-V_FL)/V_FL)*100;           #Percentage of voltage regulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The percentage of voltage regulation=%.2f%%.\"%percentage_voltage_regulation);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage of voltage regulation=33.33%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.2 : Page number 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_regulation=1.0;               #%age voltage regulation\n",
-      "V_NL=30.0;                      #Output voltage with no-load,V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, %age_of_voltage_regulation=((V_NL-V_FL)/V_FL)*100\n",
-      "V_FL=V_NL/(1+(V_regulation/100));                   #Output voltage with full-load, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The full-load voltage=%.1fV.\"%V_FL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The full-load voltage=29.7V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL_A=30.0;                #Output voltage of supply A with no-load, V\n",
-      "V_FL_A=25.0;                #Output voltage of supply A with full-load, V\n",
-      "V_NL_B=30.0;                #Output voltage of supply B with no-load, V\n",
-      "V_FL_B=29.0;                #Output voltage of supply B with full-load, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_regulation_A=((V_NL_A-V_FL_A)/V_FL_A)*100;            #%age of voltage regulation in power supply A\n",
-      "V_regulation_B=((V_NL_B-V_FL_B)/V_FL_B)*100;            #%age of voltage regulation in power supply B\n",
-      "\n",
-      "#Result\n",
-      "if(V_regulation_A<V_regulation_B):\n",
-      "    print(\"Power supply A is better than B.\");\n",
-      "else :\n",
-      "    print(\"Power supply B is better than A.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supply B is better than A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.4 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL=500.0;         #Output voltage with no-load, V\n",
-      "V_FL=300.0;         #Output voltage with full-load, V\n",
-      "I_FL=120.0;         #Output current with full-load, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Regulation=((V_NL-V_FL)/V_FL)*100;          #Voltage regulation percentage\n",
-      "\n",
-      "#(ii)\n",
-      "RL_min=V_FL/I_FL;                       #Minimum load resistance, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The voltage regulation=%.1f%%.\"%Regulation);\n",
-      "print(\"(ii)The minimum load resistance=%.1fk\u03a9.\"%RL_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The voltage regulation=66.7%.\n",
-        "(ii)The minimum load resistance=2.5k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.5 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VL_1=10.5;              #Initial output voltage with load, V\n",
-      "VL_2=10.0;              #Decreased output voltage with additional load, V\n",
-      "IL_1=1.0;               #Initial load current, A\n",
-      "IL_added=1.0;           #Added load current, A\n",
-      "\n",
-      "#Calculation\n",
-      "delta_VL=VL_1-VL_2;                 #Change in output voltage, V\n",
-      "delta_IL=IL_added;                  #Change in load current, A\n",
-      "\n",
-      "#(i)\n",
-      "Zo=delta_VL/delta_IL;               #Output impedance of power supply, \u03a9 (OHM's LAW)\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, Output_impedance=change_in_output_voltage/change_in_output_current\n",
-      "#Zo=(V_NL-VL_1)/delta_IL,\n",
-      "delta_IL=IL_1;                          #Change in load current, A\n",
-      "V_NL=VL_1+(delta_IL*Zo);                #Output voltage with no load, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output impedance=%.1f\u03a9.\"%Zo);\n",
-      "print(\"(ii) The output voltage with no-load=%dV.\"%V_NL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output impedance=0.5\u03a9.\n",
-        "(ii) The output voltage with no-load=11V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.6 : Page number 446\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Zo=0.01;                    #Output impedance, \u03a9\n",
-      "IL_max=1.0;                 #Maximum output current, A\n",
-      "IL_min=0.5                  #Minimum output current, A\n",
-      "f=10.0;                     #Frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Zo=delta_VL/delta_IL\n",
-      "delta_IL=IL_max-IL_min;                        #Maximum change in output current, A\n",
-      "delta_VL=(Zo*delta_IL)*1000;                   #Fluctuations in output voltage, mV\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage will have %dmV peak-to-peak fluctuation at a rate of %dkHz.\"%(delta_VL,f));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage will have 5mV peak-to-peak fluctuation at a rate of 10kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.7 : Page number 446\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "delta_Vout=10.0;                    #Change in output voltage, \u03bcV\n",
-      "delta_Vin=5.0;                      #Change in input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Line_regulation=delta_Vout/delta_Vin;           #Line regulation, \u03bcV/V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The line regulation of the voltage regulator=%d\u03bcV/V.\"%Line_regulation);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The line regulation of the voltage regulator=2\u03bcV/V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.8 : Page number 449-450\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=24.0;                   #Input voltage, V\n",
-      "Vz=12.0;                    #Zener voltage, V\n",
-      "Rs=160.0;                   #Series resistance, \u03a9\n",
-      "RL_max=float('inf');        #Maximum load resistance, \u03a9\n",
-      "RL_min=200.0;               #Minimum load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz;                       #Output voltage,(equal to zener regulated voltage), V\n",
-      "Is=((Vin-Vout)/Rs)*1000;       #Current through series resistance, mA\n",
-      "\n",
-      "#(ii)\n",
-      "IL_min=Vout/RL_max;                    #Minimum load current, A\n",
-      "IL_max=(Vout/RL_min)*1000;             #Maximum load current, mA\n",
-      "\n",
-      "#(iii)\n",
-      "IZ_min=Is-IL_max;                       #Minimum zener current, mA\n",
-      "IZ_max=Is-IL_min;                       #Maximum zener current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The current through the series resistance=%dmA\"%Is);\n",
-      "print(\"(ii) The minimum and maximum load currents are=%dA and %dmA\"%(IL_min,IL_max));\n",
-      "print(\"(iii) The minimum and maximum zener currents are=%dmA and %dmA\"%(IZ_min,IZ_max));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The current through the series resistance=75mA\n",
-        "(ii) The minimum and maximum load currents are=0A and 60mA\n",
-        "(iii) The minimum and maximum zener currents are=15mA and 75mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.9 : Page number 450\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VZ=15.0;                #Zener voltage, V\n",
-      "Vin_min=22.0            #Minimum input voltage, V\n",
-      "Vin_max=40.0            #Maximum input voltage, V\n",
-      "Vout=VZ;                #Regulated output voltage, V\n",
-      "IL_max=100.0;           #Maximum load current, mA\n",
-      "IL_min=20.0;            #Minimum load current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "RS_max=(Vin_min-Vout)/(IL_max/1000);               #Maximum value of series resistance, \u03a9 (OHM'S lAW)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum load resistance to hold the voltage constant=%d\u03a9.\"%RS_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum load resistance to hold the voltage constant=70\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.10 : Page number 450-451\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=3.3;                     #Zener voltage, V\n",
-      "Iz_min=3.0;                 #Minimum zener current, mA\n",
-      "Iz_max=100.0;               #Maximum zener current, mA\n",
-      "RL_max=2.0;                 #Maximum load resistance, k\u03a9\n",
-      "RL_min=500.0;               #Minimum load resistance, \u03a9\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Rs_min=(Vin-Vz)/(Iz_max/1000);             #Minimum series resistance required, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum series resistance required to limit the zener current=%.0f\u03a9.\"%Rs_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum series resistance required to limit the zener current=167\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.11 : Page number 451\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=3.3;                     #Zener voltage, V\n",
-      "Iz_min=3.0;                 #Minimum zener current, mA\n",
-      "Iz_max=100.0;               #Maximum zener current, mA\n",
-      "RL_max=2.0;                 #Maximum load resistance, k\u03a9\n",
-      "RL_min=500.0;               #Minimum load resistance, \u03a9\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IL_max=(Vz/RL_min)*1000;                   #Maximum load current, mA\n",
-      "Rs_max=((Vin-Vz)/(IL_max+Iz_min))*1000;    #Maximum series resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable value of series resistance=%d\u03a9.\"%Rs_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable value of series resistance=1739\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.12 : Page number 452\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10.0;                #Zener voltage, V\n",
-      "beta=100.0;             #Base current amplification factor\n",
-      "RL=1000.0;              #Load resistance, \u03a9\n",
-      "VBE=0.5;                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=Vz-VBE;                    #Output voltage, V\n",
-      "IL=(Vout/RL)*1000;                     #Load current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"The load current=%.1fmA\"%IL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage=9.5V.\n",
-        "The load current=9.5mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 : Page number 452\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IC=1.0;                     #Required current(collector current), A\n",
-      "Vout=6.0;                   #Constant output voltage, V\n",
-      "Vin=10.0;                   #Supply voltage, V\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "VBE=0.5;                    #Base-emitter voltage, V\n",
-      "Iz=10.0;                #Minimum zener current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IB=(IC/beta)*1000;                 #Base current, mA\n",
-      "\n",
-      "#Since, Vout=Vz-VBE;\n",
-      "Vz=Vout+VBE;                    #Zener breakdown voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_Rs=Vin-Vz;                    #Voltage across series resistance Rs, V\n",
-      "Rs=(V_Rs/(IB+Iz))*1000;                #Series resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The zener breakdown voltage=%.1fV\"%Vz);\n",
-      "print(\"(ii)The series resistance=%.0f\u03a9.\"%Rs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The zener breakdown voltage=6.5V\n",
-        "(ii)The series resistance=117\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.14: Page number 452-453\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=12.0;                    #Zener voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "RS=220.0;                   #Series resistance, \u03a9\n",
-      "RL=1.0;                     #Load resistance, k\u03a9\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz-VBE;                #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_RS=Vin-Vz;                       #Voltage across series resistance, RS, V\n",
-      "IR=(V_RS/RS)*1000;                 #Current through series resistance, mA\n",
-      "IL=Vout/RL;                        #Load current, mA\n",
-      "\n",
-      "#Since, IL is emitter current  and emitter current is approx. equal to collector current,\n",
-      "IC=IL;                              #Collector current, mA\n",
-      "IB=IC/beta;                         #Base current, mA\n",
-      "Iz=IR-IB;                           #Zener current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"(ii) The zener current=%dmA\"%Iz);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage=11.3V.\n",
-        "(ii) The zener current=36mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.15 : Page number 453-454\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "IL_min=0;                   #Minimum load current, A\n",
-      "IL_max=1.0;                 #Maximum load current, A\n",
-      "Vin_min=12.0;               #Minimum input voltage, V\n",
-      "Vin_max=18.0;               #Maximum input voltage, V\n",
-      "Iz_min=1.0;                 #Minimum zener current, mA\n",
-      "Vz=8.5;                     #Zener voltage, V\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IB_max=(IL_max/beta)*1000;                  #Maximum base current, mA\n",
-      "I_RS=Iz_min+IB_max;                         #Current through the series resistance, mA\n",
-      "RS=((Vin_min-Vz)/I_RS)*1000;                #Series resistance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "V_RS_max=Vin_max-Vz;                                      #Maximum voltage across series resistance, V\n",
-      "P_max_RS=ceil((V_RS_max**2/RS)*1000)/1000;                #Maximum power dissipation in series resistance RS, W\n",
-      "\n",
-      "#(iii)\n",
-      "I_RS_max=V_RS_max/floor(RS);                       #Maximum current through series resistance,mA\n",
-      "Iz_max=I_RS_max;                            #Maximum zener current, mA\n",
-      "P_z_max=Vz*Iz_max;                          #Maximum power dissipated in zener diode, W\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The series resistance=%d\u03a9.\"%RS);\n",
-      "print(\"(ii) The maximum power dissipated in series resistance=%.3fW.\"%P_max_RS);\n",
-      "print(\"(iii)The maximum power dissipated in zener diode=%.3fW.\"%P_z_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The series resistance=166\u03a9.\n",
-        "(ii) The maximum power dissipated in series resistance=0.542W.\n",
-        "(iii)The maximum power dissipated in zener diode=0.486W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.16 : Page number 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=2.0;                     #Resistor R1, k\u03a9\n",
-      "R2=1.0;                     #Resistor R2, k\u03a9\n",
-      "Vz=6.0;                     #Zener voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "m=R2/(R1+R2);                   #Feedback fraction\n",
-      "A_CL=1/m;                       #Closed-loop voltage gain\n",
-      "Vout=A_CL*(Vz+VBE);             #Regulated output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The regulated output voltage=%.1fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated output voltage=20.1V\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.17 : Page number 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=30.0;                     #Resistor R1, k\u03a9\n",
-      "R2=10.0;                     #Resistor R2, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "m=R2/(R1+R2);                   #Feedback fraction\n",
-      "A_CL=1/m;                       #Closed-loop voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The closed-loop voltage gain=%d.\"%A_CL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The closed-loop voltage gain=4.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.18 : Page number 457\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=22.0;               #Input voltage, V\n",
-      "Rs=130.0;               #Series resistance, \u03a9\n",
-      "Vz=8.3;                 #Zener voltage, V\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "RL=100.0;               #Load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz+VBE;                #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "IL=(Vout/RL)*1000;                 #Load current, mA (OHM's LAW)\n",
-      "IS=((Vin-Vout)/Rs)*1000;           #Current through series resistance, mA (OHM's LAW)\n",
-      "IC=IS-IL;                          #Collector current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The regulated output voltage=%dV\"%Vout);\n",
-      "print(\"(ii) Various currents for the shunt regulator are: IL=%dmA , IS=%dmA and IC=%dmA\"%(IL,IS,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The regulated output voltage=9V\n",
-        "(ii) Various currents for the shunt regulator are: IL=90mA , IS=100mA and IC=10mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.20 : Page number 463\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=240.0;               #Resistor R1 of the regulator, \u03a9\n",
-      "R2=2.4;                 #Variable resistance R2 of the regulator, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=1.25*(R2*1000/R1 + 1);             #Regulated output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The regulated output voltage=%.2fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated output voltage=13.75V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.21 : Page number 463\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vout_adj=8.0;           #Output voltage (adjusted), V\n",
-      "Vd=40.0;                #Input/output differential rating, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vin_max=Vout_adj+Vd;            #Maximum allowable input voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable input voltage=%dV.\"%Vin_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable input voltage=48V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_3.ipynb
deleted file mode 100755
index 2c9a49a5..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_3.ipynb
+++ /dev/null
@@ -1,877 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:33d9a7285654630dd24f2d6229210244e78961e0605c11041ed1b2f130cb19a1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 17 : REGULATED D.C POWER SUPPLY"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.1 : Page number 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL=400.0;                 #Output voltage with no-load, V\n",
-      "V_FL=300.0;                 #Output voltage with full-load, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "percentage_voltage_regulation=((V_NL-V_FL)/V_FL)*100;           #Percentage of voltage regulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The percentage of voltage regulation=%.2f%%.\"%percentage_voltage_regulation);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage of voltage regulation=33.33%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.2 : Page number 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_regulation=1.0;               #%age voltage regulation\n",
-      "V_NL=30.0;                      #Output voltage with no-load,V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, %age_of_voltage_regulation=((V_NL-V_FL)/V_FL)*100\n",
-      "V_FL=V_NL/(1+(V_regulation/100));                   #Output voltage with full-load, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The full-load voltage=%.1fV.\"%V_FL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The full-load voltage=29.7V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL_A=30.0;                #Output voltage of supply A with no-load, V\n",
-      "V_FL_A=25.0;                #Output voltage of supply A with full-load, V\n",
-      "V_NL_B=30.0;                #Output voltage of supply B with no-load, V\n",
-      "V_FL_B=29.0;                #Output voltage of supply B with full-load, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_regulation_A=((V_NL_A-V_FL_A)/V_FL_A)*100;            #%age of voltage regulation in power supply A\n",
-      "V_regulation_B=((V_NL_B-V_FL_B)/V_FL_B)*100;            #%age of voltage regulation in power supply B\n",
-      "\n",
-      "#Result\n",
-      "if(V_regulation_A<V_regulation_B):\n",
-      "    print(\"Power supply A is better than B.\");\n",
-      "else :\n",
-      "    print(\"Power supply B is better than A.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supply B is better than A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.4 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL=500.0;         #Output voltage with no-load, V\n",
-      "V_FL=300.0;         #Output voltage with full-load, V\n",
-      "I_FL=120.0;         #Output current with full-load, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Regulation=((V_NL-V_FL)/V_FL)*100;          #Voltage regulation percentage\n",
-      "\n",
-      "#(ii)\n",
-      "RL_min=V_FL/I_FL;                       #Minimum load resistance, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The voltage regulation=%.1f%%.\"%Regulation);\n",
-      "print(\"(ii)The minimum load resistance=%.1fk\u03a9.\"%RL_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The voltage regulation=66.7%.\n",
-        "(ii)The minimum load resistance=2.5k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.5 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VL_1=10.5;              #Initial output voltage with load, V\n",
-      "VL_2=10.0;              #Decreased output voltage with additional load, V\n",
-      "IL_1=1.0;               #Initial load current, A\n",
-      "IL_added=1.0;           #Added load current, A\n",
-      "\n",
-      "#Calculation\n",
-      "delta_VL=VL_1-VL_2;                 #Change in output voltage, V\n",
-      "delta_IL=IL_added;                  #Change in load current, A\n",
-      "\n",
-      "#(i)\n",
-      "Zo=delta_VL/delta_IL;               #Output impedance of power supply, \u03a9 (OHM's LAW)\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, Output_impedance=change_in_output_voltage/change_in_output_current\n",
-      "#Zo=(V_NL-VL_1)/delta_IL,\n",
-      "delta_IL=IL_1;                          #Change in load current, A\n",
-      "V_NL=VL_1+(delta_IL*Zo);                #Output voltage with no load, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output impedance=%.1f\u03a9.\"%Zo);\n",
-      "print(\"(ii) The output voltage with no-load=%dV.\"%V_NL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output impedance=0.5\u03a9.\n",
-        "(ii) The output voltage with no-load=11V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.6 : Page number 446\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Zo=0.01;                    #Output impedance, \u03a9\n",
-      "IL_max=1.0;                 #Maximum output current, A\n",
-      "IL_min=0.5                  #Minimum output current, A\n",
-      "f=10.0;                     #Frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Zo=delta_VL/delta_IL\n",
-      "delta_IL=IL_max-IL_min;                        #Maximum change in output current, A\n",
-      "delta_VL=(Zo*delta_IL)*1000;                   #Fluctuations in output voltage, mV\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage will have %dmV peak-to-peak fluctuation at a rate of %dkHz.\"%(delta_VL,f));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage will have 5mV peak-to-peak fluctuation at a rate of 10kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.7 : Page number 446\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "delta_Vout=10.0;                    #Change in output voltage, \u03bcV\n",
-      "delta_Vin=5.0;                      #Change in input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Line_regulation=delta_Vout/delta_Vin;           #Line regulation, \u03bcV/V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The line regulation of the voltage regulator=%d\u03bcV/V.\"%Line_regulation);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The line regulation of the voltage regulator=2\u03bcV/V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.8 : Page number 449-450\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=24.0;                   #Input voltage, V\n",
-      "Vz=12.0;                    #Zener voltage, V\n",
-      "Rs=160.0;                   #Series resistance, \u03a9\n",
-      "RL_max=float('inf');        #Maximum load resistance, \u03a9\n",
-      "RL_min=200.0;               #Minimum load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz;                       #Output voltage,(equal to zener regulated voltage), V\n",
-      "Is=((Vin-Vout)/Rs)*1000;       #Current through series resistance, mA\n",
-      "\n",
-      "#(ii)\n",
-      "IL_min=Vout/RL_max;                    #Minimum load current, A\n",
-      "IL_max=(Vout/RL_min)*1000;             #Maximum load current, mA\n",
-      "\n",
-      "#(iii)\n",
-      "IZ_min=Is-IL_max;                       #Minimum zener current, mA\n",
-      "IZ_max=Is-IL_min;                       #Maximum zener current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The current through the series resistance=%dmA\"%Is);\n",
-      "print(\"(ii) The minimum and maximum load currents are=%dA and %dmA\"%(IL_min,IL_max));\n",
-      "print(\"(iii) The minimum and maximum zener currents are=%dmA and %dmA\"%(IZ_min,IZ_max));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The current through the series resistance=75mA\n",
-        "(ii) The minimum and maximum load currents are=0A and 60mA\n",
-        "(iii) The minimum and maximum zener currents are=15mA and 75mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.9 : Page number 450\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VZ=15.0;                #Zener voltage, V\n",
-      "Vin_min=22.0            #Minimum input voltage, V\n",
-      "Vin_max=40.0            #Maximum input voltage, V\n",
-      "Vout=VZ;                #Regulated output voltage, V\n",
-      "IL_max=100.0;           #Maximum load current, mA\n",
-      "IL_min=20.0;            #Minimum load current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "RS_max=(Vin_min-Vout)/(IL_max/1000);               #Maximum value of series resistance, \u03a9 (OHM'S lAW)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum load resistance to hold the voltage constant=%d\u03a9.\"%RS_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum load resistance to hold the voltage constant=70\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.10 : Page number 450-451\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=3.3;                     #Zener voltage, V\n",
-      "Iz_min=3.0;                 #Minimum zener current, mA\n",
-      "Iz_max=100.0;               #Maximum zener current, mA\n",
-      "RL_max=2.0;                 #Maximum load resistance, k\u03a9\n",
-      "RL_min=500.0;               #Minimum load resistance, \u03a9\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Rs_min=(Vin-Vz)/(Iz_max/1000);             #Minimum series resistance required, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum series resistance required to limit the zener current=%.0f\u03a9.\"%Rs_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum series resistance required to limit the zener current=167\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.11 : Page number 451\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=3.3;                     #Zener voltage, V\n",
-      "Iz_min=3.0;                 #Minimum zener current, mA\n",
-      "Iz_max=100.0;               #Maximum zener current, mA\n",
-      "RL_max=2.0;                 #Maximum load resistance, k\u03a9\n",
-      "RL_min=500.0;               #Minimum load resistance, \u03a9\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IL_max=(Vz/RL_min)*1000;                   #Maximum load current, mA\n",
-      "Rs_max=((Vin-Vz)/(IL_max+Iz_min))*1000;    #Maximum series resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable value of series resistance=%d\u03a9.\"%Rs_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable value of series resistance=1739\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.12 : Page number 452\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10.0;                #Zener voltage, V\n",
-      "beta=100.0;             #Base current amplification factor\n",
-      "RL=1000.0;              #Load resistance, \u03a9\n",
-      "VBE=0.5;                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=Vz-VBE;                    #Output voltage, V\n",
-      "IL=(Vout/RL)*1000;                     #Load current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"The load current=%.1fmA\"%IL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage=9.5V.\n",
-        "The load current=9.5mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 : Page number 452\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IC=1.0;                     #Required current(collector current), A\n",
-      "Vout=6.0;                   #Constant output voltage, V\n",
-      "Vin=10.0;                   #Supply voltage, V\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "VBE=0.5;                    #Base-emitter voltage, V\n",
-      "Iz=10.0;                #Minimum zener current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IB=(IC/beta)*1000;                 #Base current, mA\n",
-      "\n",
-      "#Since, Vout=Vz-VBE;\n",
-      "Vz=Vout+VBE;                    #Zener breakdown voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_Rs=Vin-Vz;                    #Voltage across series resistance Rs, V\n",
-      "Rs=(V_Rs/(IB+Iz))*1000;                #Series resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The zener breakdown voltage=%.1fV\"%Vz);\n",
-      "print(\"(ii)The series resistance=%.0f\u03a9.\"%Rs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The zener breakdown voltage=6.5V\n",
-        "(ii)The series resistance=117\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.14: Page number 452-453\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=12.0;                    #Zener voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "RS=220.0;                   #Series resistance, \u03a9\n",
-      "RL=1.0;                     #Load resistance, k\u03a9\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz-VBE;                #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_RS=Vin-Vz;                       #Voltage across series resistance, RS, V\n",
-      "IR=(V_RS/RS)*1000;                 #Current through series resistance, mA\n",
-      "IL=Vout/RL;                        #Load current, mA\n",
-      "\n",
-      "#Since, IL is emitter current  and emitter current is approx. equal to collector current,\n",
-      "IC=IL;                              #Collector current, mA\n",
-      "IB=IC/beta;                         #Base current, mA\n",
-      "Iz=IR-IB;                           #Zener current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"(ii) The zener current=%dmA\"%Iz);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage=11.3V.\n",
-        "(ii) The zener current=36mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.15 : Page number 453-454\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "IL_min=0;                   #Minimum load current, A\n",
-      "IL_max=1.0;                 #Maximum load current, A\n",
-      "Vin_min=12.0;               #Minimum input voltage, V\n",
-      "Vin_max=18.0;               #Maximum input voltage, V\n",
-      "Iz_min=1.0;                 #Minimum zener current, mA\n",
-      "Vz=8.5;                     #Zener voltage, V\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IB_max=(IL_max/beta)*1000;                  #Maximum base current, mA\n",
-      "I_RS=Iz_min+IB_max;                         #Current through the series resistance, mA\n",
-      "RS=((Vin_min-Vz)/I_RS)*1000;                #Series resistance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "V_RS_max=Vin_max-Vz;                                      #Maximum voltage across series resistance, V\n",
-      "P_max_RS=ceil((V_RS_max**2/RS)*1000)/1000;                #Maximum power dissipation in series resistance RS, W\n",
-      "\n",
-      "#(iii)\n",
-      "I_RS_max=V_RS_max/floor(RS);                       #Maximum current through series resistance,mA\n",
-      "Iz_max=I_RS_max;                            #Maximum zener current, mA\n",
-      "P_z_max=Vz*Iz_max;                          #Maximum power dissipated in zener diode, W\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The series resistance=%d\u03a9.\"%RS);\n",
-      "print(\"(ii) The maximum power dissipated in series resistance=%.3fW.\"%P_max_RS);\n",
-      "print(\"(iii)The maximum power dissipated in zener diode=%.3fW.\"%P_z_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The series resistance=166\u03a9.\n",
-        "(ii) The maximum power dissipated in series resistance=0.542W.\n",
-        "(iii)The maximum power dissipated in zener diode=0.486W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.16 : Page number 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=2.0;                     #Resistor R1, k\u03a9\n",
-      "R2=1.0;                     #Resistor R2, k\u03a9\n",
-      "Vz=6.0;                     #Zener voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "m=R2/(R1+R2);                   #Feedback fraction\n",
-      "A_CL=1/m;                       #Closed-loop voltage gain\n",
-      "Vout=A_CL*(Vz+VBE);             #Regulated output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The regulated output voltage=%.1fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated output voltage=20.1V\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.17 : Page number 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=30.0;                     #Resistor R1, k\u03a9\n",
-      "R2=10.0;                     #Resistor R2, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "m=R2/(R1+R2);                   #Feedback fraction\n",
-      "A_CL=1/m;                       #Closed-loop voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The closed-loop voltage gain=%d.\"%A_CL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The closed-loop voltage gain=4.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.18 : Page number 457\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=22.0;               #Input voltage, V\n",
-      "Rs=130.0;               #Series resistance, \u03a9\n",
-      "Vz=8.3;                 #Zener voltage, V\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "RL=100.0;               #Load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz+VBE;                #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "IL=(Vout/RL)*1000;                 #Load current, mA (OHM's LAW)\n",
-      "IS=((Vin-Vout)/Rs)*1000;           #Current through series resistance, mA (OHM's LAW)\n",
-      "IC=IS-IL;                          #Collector current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The regulated output voltage=%dV\"%Vout);\n",
-      "print(\"(ii) Various currents for the shunt regulator are: IL=%dmA , IS=%dmA and IC=%dmA\"%(IL,IS,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The regulated output voltage=9V\n",
-        "(ii) Various currents for the shunt regulator are: IL=90mA , IS=100mA and IC=10mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.20 : Page number 463\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=240.0;               #Resistor R1 of the regulator, \u03a9\n",
-      "R2=2.4;                 #Variable resistance R2 of the regulator, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=1.25*(R2*1000/R1 + 1);             #Regulated output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The regulated output voltage=%.2fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated output voltage=13.75V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.21 : Page number 463\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vout_adj=8.0;           #Output voltage (adjusted), V\n",
-      "Vd=40.0;                #Input/output differential rating, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vin_max=Vout_adj+Vd;            #Maximum allowable input voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable input voltage=%dV.\"%Vin_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable input voltage=48V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_4.ipynb
deleted file mode 100755
index 2c9a49a5..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_4.ipynb
+++ /dev/null
@@ -1,877 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:33d9a7285654630dd24f2d6229210244e78961e0605c11041ed1b2f130cb19a1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 17 : REGULATED D.C POWER SUPPLY"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.1 : Page number 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL=400.0;                 #Output voltage with no-load, V\n",
-      "V_FL=300.0;                 #Output voltage with full-load, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "percentage_voltage_regulation=((V_NL-V_FL)/V_FL)*100;           #Percentage of voltage regulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The percentage of voltage regulation=%.2f%%.\"%percentage_voltage_regulation);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage of voltage regulation=33.33%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.2 : Page number 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_regulation=1.0;               #%age voltage regulation\n",
-      "V_NL=30.0;                      #Output voltage with no-load,V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, %age_of_voltage_regulation=((V_NL-V_FL)/V_FL)*100\n",
-      "V_FL=V_NL/(1+(V_regulation/100));                   #Output voltage with full-load, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The full-load voltage=%.1fV.\"%V_FL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The full-load voltage=29.7V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL_A=30.0;                #Output voltage of supply A with no-load, V\n",
-      "V_FL_A=25.0;                #Output voltage of supply A with full-load, V\n",
-      "V_NL_B=30.0;                #Output voltage of supply B with no-load, V\n",
-      "V_FL_B=29.0;                #Output voltage of supply B with full-load, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_regulation_A=((V_NL_A-V_FL_A)/V_FL_A)*100;            #%age of voltage regulation in power supply A\n",
-      "V_regulation_B=((V_NL_B-V_FL_B)/V_FL_B)*100;            #%age of voltage regulation in power supply B\n",
-      "\n",
-      "#Result\n",
-      "if(V_regulation_A<V_regulation_B):\n",
-      "    print(\"Power supply A is better than B.\");\n",
-      "else :\n",
-      "    print(\"Power supply B is better than A.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supply B is better than A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.4 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL=500.0;         #Output voltage with no-load, V\n",
-      "V_FL=300.0;         #Output voltage with full-load, V\n",
-      "I_FL=120.0;         #Output current with full-load, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Regulation=((V_NL-V_FL)/V_FL)*100;          #Voltage regulation percentage\n",
-      "\n",
-      "#(ii)\n",
-      "RL_min=V_FL/I_FL;                       #Minimum load resistance, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The voltage regulation=%.1f%%.\"%Regulation);\n",
-      "print(\"(ii)The minimum load resistance=%.1fk\u03a9.\"%RL_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The voltage regulation=66.7%.\n",
-        "(ii)The minimum load resistance=2.5k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.5 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VL_1=10.5;              #Initial output voltage with load, V\n",
-      "VL_2=10.0;              #Decreased output voltage with additional load, V\n",
-      "IL_1=1.0;               #Initial load current, A\n",
-      "IL_added=1.0;           #Added load current, A\n",
-      "\n",
-      "#Calculation\n",
-      "delta_VL=VL_1-VL_2;                 #Change in output voltage, V\n",
-      "delta_IL=IL_added;                  #Change in load current, A\n",
-      "\n",
-      "#(i)\n",
-      "Zo=delta_VL/delta_IL;               #Output impedance of power supply, \u03a9 (OHM's LAW)\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, Output_impedance=change_in_output_voltage/change_in_output_current\n",
-      "#Zo=(V_NL-VL_1)/delta_IL,\n",
-      "delta_IL=IL_1;                          #Change in load current, A\n",
-      "V_NL=VL_1+(delta_IL*Zo);                #Output voltage with no load, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output impedance=%.1f\u03a9.\"%Zo);\n",
-      "print(\"(ii) The output voltage with no-load=%dV.\"%V_NL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output impedance=0.5\u03a9.\n",
-        "(ii) The output voltage with no-load=11V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.6 : Page number 446\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Zo=0.01;                    #Output impedance, \u03a9\n",
-      "IL_max=1.0;                 #Maximum output current, A\n",
-      "IL_min=0.5                  #Minimum output current, A\n",
-      "f=10.0;                     #Frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Zo=delta_VL/delta_IL\n",
-      "delta_IL=IL_max-IL_min;                        #Maximum change in output current, A\n",
-      "delta_VL=(Zo*delta_IL)*1000;                   #Fluctuations in output voltage, mV\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage will have %dmV peak-to-peak fluctuation at a rate of %dkHz.\"%(delta_VL,f));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage will have 5mV peak-to-peak fluctuation at a rate of 10kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.7 : Page number 446\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "delta_Vout=10.0;                    #Change in output voltage, \u03bcV\n",
-      "delta_Vin=5.0;                      #Change in input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Line_regulation=delta_Vout/delta_Vin;           #Line regulation, \u03bcV/V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The line regulation of the voltage regulator=%d\u03bcV/V.\"%Line_regulation);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The line regulation of the voltage regulator=2\u03bcV/V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.8 : Page number 449-450\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=24.0;                   #Input voltage, V\n",
-      "Vz=12.0;                    #Zener voltage, V\n",
-      "Rs=160.0;                   #Series resistance, \u03a9\n",
-      "RL_max=float('inf');        #Maximum load resistance, \u03a9\n",
-      "RL_min=200.0;               #Minimum load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz;                       #Output voltage,(equal to zener regulated voltage), V\n",
-      "Is=((Vin-Vout)/Rs)*1000;       #Current through series resistance, mA\n",
-      "\n",
-      "#(ii)\n",
-      "IL_min=Vout/RL_max;                    #Minimum load current, A\n",
-      "IL_max=(Vout/RL_min)*1000;             #Maximum load current, mA\n",
-      "\n",
-      "#(iii)\n",
-      "IZ_min=Is-IL_max;                       #Minimum zener current, mA\n",
-      "IZ_max=Is-IL_min;                       #Maximum zener current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The current through the series resistance=%dmA\"%Is);\n",
-      "print(\"(ii) The minimum and maximum load currents are=%dA and %dmA\"%(IL_min,IL_max));\n",
-      "print(\"(iii) The minimum and maximum zener currents are=%dmA and %dmA\"%(IZ_min,IZ_max));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The current through the series resistance=75mA\n",
-        "(ii) The minimum and maximum load currents are=0A and 60mA\n",
-        "(iii) The minimum and maximum zener currents are=15mA and 75mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.9 : Page number 450\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VZ=15.0;                #Zener voltage, V\n",
-      "Vin_min=22.0            #Minimum input voltage, V\n",
-      "Vin_max=40.0            #Maximum input voltage, V\n",
-      "Vout=VZ;                #Regulated output voltage, V\n",
-      "IL_max=100.0;           #Maximum load current, mA\n",
-      "IL_min=20.0;            #Minimum load current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "RS_max=(Vin_min-Vout)/(IL_max/1000);               #Maximum value of series resistance, \u03a9 (OHM'S lAW)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum load resistance to hold the voltage constant=%d\u03a9.\"%RS_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum load resistance to hold the voltage constant=70\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.10 : Page number 450-451\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=3.3;                     #Zener voltage, V\n",
-      "Iz_min=3.0;                 #Minimum zener current, mA\n",
-      "Iz_max=100.0;               #Maximum zener current, mA\n",
-      "RL_max=2.0;                 #Maximum load resistance, k\u03a9\n",
-      "RL_min=500.0;               #Minimum load resistance, \u03a9\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Rs_min=(Vin-Vz)/(Iz_max/1000);             #Minimum series resistance required, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum series resistance required to limit the zener current=%.0f\u03a9.\"%Rs_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum series resistance required to limit the zener current=167\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.11 : Page number 451\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=3.3;                     #Zener voltage, V\n",
-      "Iz_min=3.0;                 #Minimum zener current, mA\n",
-      "Iz_max=100.0;               #Maximum zener current, mA\n",
-      "RL_max=2.0;                 #Maximum load resistance, k\u03a9\n",
-      "RL_min=500.0;               #Minimum load resistance, \u03a9\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IL_max=(Vz/RL_min)*1000;                   #Maximum load current, mA\n",
-      "Rs_max=((Vin-Vz)/(IL_max+Iz_min))*1000;    #Maximum series resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable value of series resistance=%d\u03a9.\"%Rs_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable value of series resistance=1739\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.12 : Page number 452\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10.0;                #Zener voltage, V\n",
-      "beta=100.0;             #Base current amplification factor\n",
-      "RL=1000.0;              #Load resistance, \u03a9\n",
-      "VBE=0.5;                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=Vz-VBE;                    #Output voltage, V\n",
-      "IL=(Vout/RL)*1000;                     #Load current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"The load current=%.1fmA\"%IL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage=9.5V.\n",
-        "The load current=9.5mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 : Page number 452\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IC=1.0;                     #Required current(collector current), A\n",
-      "Vout=6.0;                   #Constant output voltage, V\n",
-      "Vin=10.0;                   #Supply voltage, V\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "VBE=0.5;                    #Base-emitter voltage, V\n",
-      "Iz=10.0;                #Minimum zener current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IB=(IC/beta)*1000;                 #Base current, mA\n",
-      "\n",
-      "#Since, Vout=Vz-VBE;\n",
-      "Vz=Vout+VBE;                    #Zener breakdown voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_Rs=Vin-Vz;                    #Voltage across series resistance Rs, V\n",
-      "Rs=(V_Rs/(IB+Iz))*1000;                #Series resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The zener breakdown voltage=%.1fV\"%Vz);\n",
-      "print(\"(ii)The series resistance=%.0f\u03a9.\"%Rs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The zener breakdown voltage=6.5V\n",
-        "(ii)The series resistance=117\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.14: Page number 452-453\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=12.0;                    #Zener voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "RS=220.0;                   #Series resistance, \u03a9\n",
-      "RL=1.0;                     #Load resistance, k\u03a9\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz-VBE;                #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_RS=Vin-Vz;                       #Voltage across series resistance, RS, V\n",
-      "IR=(V_RS/RS)*1000;                 #Current through series resistance, mA\n",
-      "IL=Vout/RL;                        #Load current, mA\n",
-      "\n",
-      "#Since, IL is emitter current  and emitter current is approx. equal to collector current,\n",
-      "IC=IL;                              #Collector current, mA\n",
-      "IB=IC/beta;                         #Base current, mA\n",
-      "Iz=IR-IB;                           #Zener current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"(ii) The zener current=%dmA\"%Iz);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage=11.3V.\n",
-        "(ii) The zener current=36mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.15 : Page number 453-454\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "IL_min=0;                   #Minimum load current, A\n",
-      "IL_max=1.0;                 #Maximum load current, A\n",
-      "Vin_min=12.0;               #Minimum input voltage, V\n",
-      "Vin_max=18.0;               #Maximum input voltage, V\n",
-      "Iz_min=1.0;                 #Minimum zener current, mA\n",
-      "Vz=8.5;                     #Zener voltage, V\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IB_max=(IL_max/beta)*1000;                  #Maximum base current, mA\n",
-      "I_RS=Iz_min+IB_max;                         #Current through the series resistance, mA\n",
-      "RS=((Vin_min-Vz)/I_RS)*1000;                #Series resistance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "V_RS_max=Vin_max-Vz;                                      #Maximum voltage across series resistance, V\n",
-      "P_max_RS=ceil((V_RS_max**2/RS)*1000)/1000;                #Maximum power dissipation in series resistance RS, W\n",
-      "\n",
-      "#(iii)\n",
-      "I_RS_max=V_RS_max/floor(RS);                       #Maximum current through series resistance,mA\n",
-      "Iz_max=I_RS_max;                            #Maximum zener current, mA\n",
-      "P_z_max=Vz*Iz_max;                          #Maximum power dissipated in zener diode, W\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The series resistance=%d\u03a9.\"%RS);\n",
-      "print(\"(ii) The maximum power dissipated in series resistance=%.3fW.\"%P_max_RS);\n",
-      "print(\"(iii)The maximum power dissipated in zener diode=%.3fW.\"%P_z_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The series resistance=166\u03a9.\n",
-        "(ii) The maximum power dissipated in series resistance=0.542W.\n",
-        "(iii)The maximum power dissipated in zener diode=0.486W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.16 : Page number 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=2.0;                     #Resistor R1, k\u03a9\n",
-      "R2=1.0;                     #Resistor R2, k\u03a9\n",
-      "Vz=6.0;                     #Zener voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "m=R2/(R1+R2);                   #Feedback fraction\n",
-      "A_CL=1/m;                       #Closed-loop voltage gain\n",
-      "Vout=A_CL*(Vz+VBE);             #Regulated output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The regulated output voltage=%.1fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated output voltage=20.1V\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.17 : Page number 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=30.0;                     #Resistor R1, k\u03a9\n",
-      "R2=10.0;                     #Resistor R2, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "m=R2/(R1+R2);                   #Feedback fraction\n",
-      "A_CL=1/m;                       #Closed-loop voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The closed-loop voltage gain=%d.\"%A_CL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The closed-loop voltage gain=4.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.18 : Page number 457\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=22.0;               #Input voltage, V\n",
-      "Rs=130.0;               #Series resistance, \u03a9\n",
-      "Vz=8.3;                 #Zener voltage, V\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "RL=100.0;               #Load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz+VBE;                #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "IL=(Vout/RL)*1000;                 #Load current, mA (OHM's LAW)\n",
-      "IS=((Vin-Vout)/Rs)*1000;           #Current through series resistance, mA (OHM's LAW)\n",
-      "IC=IS-IL;                          #Collector current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The regulated output voltage=%dV\"%Vout);\n",
-      "print(\"(ii) Various currents for the shunt regulator are: IL=%dmA , IS=%dmA and IC=%dmA\"%(IL,IS,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The regulated output voltage=9V\n",
-        "(ii) Various currents for the shunt regulator are: IL=90mA , IS=100mA and IC=10mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.20 : Page number 463\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=240.0;               #Resistor R1 of the regulator, \u03a9\n",
-      "R2=2.4;                 #Variable resistance R2 of the regulator, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=1.25*(R2*1000/R1 + 1);             #Regulated output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The regulated output voltage=%.2fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated output voltage=13.75V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.21 : Page number 463\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vout_adj=8.0;           #Output voltage (adjusted), V\n",
-      "Vd=40.0;                #Input/output differential rating, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vin_max=Vout_adj+Vd;            #Maximum allowable input voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable input voltage=%dV.\"%Vin_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable input voltage=48V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_5.ipynb
deleted file mode 100755
index 2c9a49a5..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_5.ipynb
+++ /dev/null
@@ -1,877 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:33d9a7285654630dd24f2d6229210244e78961e0605c11041ed1b2f130cb19a1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 17 : REGULATED D.C POWER SUPPLY"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.1 : Page number 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL=400.0;                 #Output voltage with no-load, V\n",
-      "V_FL=300.0;                 #Output voltage with full-load, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "percentage_voltage_regulation=((V_NL-V_FL)/V_FL)*100;           #Percentage of voltage regulation\n",
-      "\n",
-      "#Result\n",
-      "print(\"The percentage of voltage regulation=%.2f%%.\"%percentage_voltage_regulation);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage of voltage regulation=33.33%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.2 : Page number 444\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_regulation=1.0;               #%age voltage regulation\n",
-      "V_NL=30.0;                      #Output voltage with no-load,V\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, %age_of_voltage_regulation=((V_NL-V_FL)/V_FL)*100\n",
-      "V_FL=V_NL/(1+(V_regulation/100));                   #Output voltage with full-load, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The full-load voltage=%.1fV.\"%V_FL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The full-load voltage=29.7V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL_A=30.0;                #Output voltage of supply A with no-load, V\n",
-      "V_FL_A=25.0;                #Output voltage of supply A with full-load, V\n",
-      "V_NL_B=30.0;                #Output voltage of supply B with no-load, V\n",
-      "V_FL_B=29.0;                #Output voltage of supply B with full-load, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_regulation_A=((V_NL_A-V_FL_A)/V_FL_A)*100;            #%age of voltage regulation in power supply A\n",
-      "V_regulation_B=((V_NL_B-V_FL_B)/V_FL_B)*100;            #%age of voltage regulation in power supply B\n",
-      "\n",
-      "#Result\n",
-      "if(V_regulation_A<V_regulation_B):\n",
-      "    print(\"Power supply A is better than B.\");\n",
-      "else :\n",
-      "    print(\"Power supply B is better than A.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supply B is better than A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.4 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_NL=500.0;         #Output voltage with no-load, V\n",
-      "V_FL=300.0;         #Output voltage with full-load, V\n",
-      "I_FL=120.0;         #Output current with full-load, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Regulation=((V_NL-V_FL)/V_FL)*100;          #Voltage regulation percentage\n",
-      "\n",
-      "#(ii)\n",
-      "RL_min=V_FL/I_FL;                       #Minimum load resistance, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The voltage regulation=%.1f%%.\"%Regulation);\n",
-      "print(\"(ii)The minimum load resistance=%.1fk\u03a9.\"%RL_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The voltage regulation=66.7%.\n",
-        "(ii)The minimum load resistance=2.5k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.5 : Page number 445\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VL_1=10.5;              #Initial output voltage with load, V\n",
-      "VL_2=10.0;              #Decreased output voltage with additional load, V\n",
-      "IL_1=1.0;               #Initial load current, A\n",
-      "IL_added=1.0;           #Added load current, A\n",
-      "\n",
-      "#Calculation\n",
-      "delta_VL=VL_1-VL_2;                 #Change in output voltage, V\n",
-      "delta_IL=IL_added;                  #Change in load current, A\n",
-      "\n",
-      "#(i)\n",
-      "Zo=delta_VL/delta_IL;               #Output impedance of power supply, \u03a9 (OHM's LAW)\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, Output_impedance=change_in_output_voltage/change_in_output_current\n",
-      "#Zo=(V_NL-VL_1)/delta_IL,\n",
-      "delta_IL=IL_1;                          #Change in load current, A\n",
-      "V_NL=VL_1+(delta_IL*Zo);                #Output voltage with no load, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output impedance=%.1f\u03a9.\"%Zo);\n",
-      "print(\"(ii) The output voltage with no-load=%dV.\"%V_NL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output impedance=0.5\u03a9.\n",
-        "(ii) The output voltage with no-load=11V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.6 : Page number 446\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Zo=0.01;                    #Output impedance, \u03a9\n",
-      "IL_max=1.0;                 #Maximum output current, A\n",
-      "IL_min=0.5                  #Minimum output current, A\n",
-      "f=10.0;                     #Frequency, kHz\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Zo=delta_VL/delta_IL\n",
-      "delta_IL=IL_max-IL_min;                        #Maximum change in output current, A\n",
-      "delta_VL=(Zo*delta_IL)*1000;                   #Fluctuations in output voltage, mV\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage will have %dmV peak-to-peak fluctuation at a rate of %dkHz.\"%(delta_VL,f));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage will have 5mV peak-to-peak fluctuation at a rate of 10kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.7 : Page number 446\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "delta_Vout=10.0;                    #Change in output voltage, \u03bcV\n",
-      "delta_Vin=5.0;                      #Change in input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Line_regulation=delta_Vout/delta_Vin;           #Line regulation, \u03bcV/V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The line regulation of the voltage regulator=%d\u03bcV/V.\"%Line_regulation);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The line regulation of the voltage regulator=2\u03bcV/V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.8 : Page number 449-450\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=24.0;                   #Input voltage, V\n",
-      "Vz=12.0;                    #Zener voltage, V\n",
-      "Rs=160.0;                   #Series resistance, \u03a9\n",
-      "RL_max=float('inf');        #Maximum load resistance, \u03a9\n",
-      "RL_min=200.0;               #Minimum load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz;                       #Output voltage,(equal to zener regulated voltage), V\n",
-      "Is=((Vin-Vout)/Rs)*1000;       #Current through series resistance, mA\n",
-      "\n",
-      "#(ii)\n",
-      "IL_min=Vout/RL_max;                    #Minimum load current, A\n",
-      "IL_max=(Vout/RL_min)*1000;             #Maximum load current, mA\n",
-      "\n",
-      "#(iii)\n",
-      "IZ_min=Is-IL_max;                       #Minimum zener current, mA\n",
-      "IZ_max=Is-IL_min;                       #Maximum zener current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The current through the series resistance=%dmA\"%Is);\n",
-      "print(\"(ii) The minimum and maximum load currents are=%dA and %dmA\"%(IL_min,IL_max));\n",
-      "print(\"(iii) The minimum and maximum zener currents are=%dmA and %dmA\"%(IZ_min,IZ_max));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The current through the series resistance=75mA\n",
-        "(ii) The minimum and maximum load currents are=0A and 60mA\n",
-        "(iii) The minimum and maximum zener currents are=15mA and 75mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.9 : Page number 450\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VZ=15.0;                #Zener voltage, V\n",
-      "Vin_min=22.0            #Minimum input voltage, V\n",
-      "Vin_max=40.0            #Maximum input voltage, V\n",
-      "Vout=VZ;                #Regulated output voltage, V\n",
-      "IL_max=100.0;           #Maximum load current, mA\n",
-      "IL_min=20.0;            #Minimum load current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "RS_max=(Vin_min-Vout)/(IL_max/1000);               #Maximum value of series resistance, \u03a9 (OHM'S lAW)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum load resistance to hold the voltage constant=%d\u03a9.\"%RS_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum load resistance to hold the voltage constant=70\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.10 : Page number 450-451\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=3.3;                     #Zener voltage, V\n",
-      "Iz_min=3.0;                 #Minimum zener current, mA\n",
-      "Iz_max=100.0;               #Maximum zener current, mA\n",
-      "RL_max=2.0;                 #Maximum load resistance, k\u03a9\n",
-      "RL_min=500.0;               #Minimum load resistance, \u03a9\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Rs_min=(Vin-Vz)/(Iz_max/1000);             #Minimum series resistance required, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum series resistance required to limit the zener current=%.0f\u03a9.\"%Rs_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum series resistance required to limit the zener current=167\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.11 : Page number 451\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=3.3;                     #Zener voltage, V\n",
-      "Iz_min=3.0;                 #Minimum zener current, mA\n",
-      "Iz_max=100.0;               #Maximum zener current, mA\n",
-      "RL_max=2.0;                 #Maximum load resistance, k\u03a9\n",
-      "RL_min=500.0;               #Minimum load resistance, \u03a9\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IL_max=(Vz/RL_min)*1000;                   #Maximum load current, mA\n",
-      "Rs_max=((Vin-Vz)/(IL_max+Iz_min))*1000;    #Maximum series resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable value of series resistance=%d\u03a9.\"%Rs_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable value of series resistance=1739\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.12 : Page number 452\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10.0;                #Zener voltage, V\n",
-      "beta=100.0;             #Base current amplification factor\n",
-      "RL=1000.0;              #Load resistance, \u03a9\n",
-      "VBE=0.5;                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=Vz-VBE;                    #Output voltage, V\n",
-      "IL=(Vout/RL)*1000;                     #Load current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"The load current=%.1fmA\"%IL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage=9.5V.\n",
-        "The load current=9.5mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.3 : Page number 452\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IC=1.0;                     #Required current(collector current), A\n",
-      "Vout=6.0;                   #Constant output voltage, V\n",
-      "Vin=10.0;                   #Supply voltage, V\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "VBE=0.5;                    #Base-emitter voltage, V\n",
-      "Iz=10.0;                #Minimum zener current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IB=(IC/beta)*1000;                 #Base current, mA\n",
-      "\n",
-      "#Since, Vout=Vz-VBE;\n",
-      "Vz=Vout+VBE;                    #Zener breakdown voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_Rs=Vin-Vz;                    #Voltage across series resistance Rs, V\n",
-      "Rs=(V_Rs/(IB+Iz))*1000;                #Series resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The zener breakdown voltage=%.1fV\"%Vz);\n",
-      "print(\"(ii)The series resistance=%.0f\u03a9.\"%Rs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The zener breakdown voltage=6.5V\n",
-        "(ii)The series resistance=117\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.14: Page number 452-453\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=12.0;                    #Zener voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "Vin=20.0;                   #Input voltage, V\n",
-      "RS=220.0;                   #Series resistance, \u03a9\n",
-      "RL=1.0;                     #Load resistance, k\u03a9\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz-VBE;                #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_RS=Vin-Vz;                       #Voltage across series resistance, RS, V\n",
-      "IR=(V_RS/RS)*1000;                 #Current through series resistance, mA\n",
-      "IL=Vout/RL;                        #Load current, mA\n",
-      "\n",
-      "#Since, IL is emitter current  and emitter current is approx. equal to collector current,\n",
-      "IC=IL;                              #Collector current, mA\n",
-      "IB=IC/beta;                         #Base current, mA\n",
-      "Iz=IR-IB;                           #Zener current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"(ii) The zener current=%dmA\"%Iz);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage=11.3V.\n",
-        "(ii) The zener current=36mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.15 : Page number 453-454\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "IL_min=0;                   #Minimum load current, A\n",
-      "IL_max=1.0;                 #Maximum load current, A\n",
-      "Vin_min=12.0;               #Minimum input voltage, V\n",
-      "Vin_max=18.0;               #Maximum input voltage, V\n",
-      "Iz_min=1.0;                 #Minimum zener current, mA\n",
-      "Vz=8.5;                     #Zener voltage, V\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IB_max=(IL_max/beta)*1000;                  #Maximum base current, mA\n",
-      "I_RS=Iz_min+IB_max;                         #Current through the series resistance, mA\n",
-      "RS=((Vin_min-Vz)/I_RS)*1000;                #Series resistance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "V_RS_max=Vin_max-Vz;                                      #Maximum voltage across series resistance, V\n",
-      "P_max_RS=ceil((V_RS_max**2/RS)*1000)/1000;                #Maximum power dissipation in series resistance RS, W\n",
-      "\n",
-      "#(iii)\n",
-      "I_RS_max=V_RS_max/floor(RS);                       #Maximum current through series resistance,mA\n",
-      "Iz_max=I_RS_max;                            #Maximum zener current, mA\n",
-      "P_z_max=Vz*Iz_max;                          #Maximum power dissipated in zener diode, W\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The series resistance=%d\u03a9.\"%RS);\n",
-      "print(\"(ii) The maximum power dissipated in series resistance=%.3fW.\"%P_max_RS);\n",
-      "print(\"(iii)The maximum power dissipated in zener diode=%.3fW.\"%P_z_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The series resistance=166\u03a9.\n",
-        "(ii) The maximum power dissipated in series resistance=0.542W.\n",
-        "(iii)The maximum power dissipated in zener diode=0.486W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.16 : Page number 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=2.0;                     #Resistor R1, k\u03a9\n",
-      "R2=1.0;                     #Resistor R2, k\u03a9\n",
-      "Vz=6.0;                     #Zener voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "m=R2/(R1+R2);                   #Feedback fraction\n",
-      "A_CL=1/m;                       #Closed-loop voltage gain\n",
-      "Vout=A_CL*(Vz+VBE);             #Regulated output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The regulated output voltage=%.1fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated output voltage=20.1V\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.17 : Page number 456\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=30.0;                     #Resistor R1, k\u03a9\n",
-      "R2=10.0;                     #Resistor R2, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "m=R2/(R1+R2);                   #Feedback fraction\n",
-      "A_CL=1/m;                       #Closed-loop voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The closed-loop voltage gain=%d.\"%A_CL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The closed-loop voltage gain=4.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.18 : Page number 457\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=22.0;               #Input voltage, V\n",
-      "Rs=130.0;               #Series resistance, \u03a9\n",
-      "Vz=8.3;                 #Zener voltage, V\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "RL=100.0;               #Load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout=Vz+VBE;                #Output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "IL=(Vout/RL)*1000;                 #Load current, mA (OHM's LAW)\n",
-      "IS=((Vin-Vout)/Rs)*1000;           #Current through series resistance, mA (OHM's LAW)\n",
-      "IC=IS-IL;                          #Collector current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The regulated output voltage=%dV\"%Vout);\n",
-      "print(\"(ii) Various currents for the shunt regulator are: IL=%dmA , IS=%dmA and IC=%dmA\"%(IL,IS,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The regulated output voltage=9V\n",
-        "(ii) Various currents for the shunt regulator are: IL=90mA , IS=100mA and IC=10mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.20 : Page number 463\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=240.0;               #Resistor R1 of the regulator, \u03a9\n",
-      "R2=2.4;                 #Variable resistance R2 of the regulator, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=1.25*(R2*1000/R1 + 1);             #Regulated output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The regulated output voltage=%.2fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated output voltage=13.75V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 17.21 : Page number 463\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vout_adj=8.0;           #Output voltage (adjusted), V\n",
-      "Vd=40.0;                #Input/output differential rating, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vin_max=Vout_adj+Vd;            #Maximum allowable input voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable input voltage=%dV.\"%Vin_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable input voltage=48V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18.ipynb
deleted file mode 100755
index 34ff63b3..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18.ipynb
+++ /dev/null
@@ -1,815 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:5aefee63b41b58f0caeac1aa6be18e130e4530aaaa6bc5e9bbc45b3687d3f8e9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 18 : SOLID-STATE SWITCHING CIRCUITS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.1 : Page number 472"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10;             #Supply voltage, V\n",
-      "RC=1.0;               #Collector resistor, k\u03a9\n",
-      "RB=47.0;              #Base resistor, k\u03a9\n",
-      "beta=100.0;           #Base current amplification factor\n",
-      "VBE=0.7;            #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "IC_sat=VCC/RC;              #Collector saturation current, mA\n",
-      "IB=IC_sat/beta;             #Base current, mA\n",
-      "V=IB*RB+VBE;                #Input voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Input voltage required to saturate the transistor switch=%.1fV.\"%V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input voltage required to saturate the transistor switch=5.4V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.2 : Page number 475"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10;             #Supply voltage, V\n",
-      "RC=1.0;               #Collector resistor, k\u03a9\n",
-      "ICBO=10.0;            #Collector leakage current, \u03bcA\n",
-      "V_knee=0.7;         #Knee voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC=ICBO;                    #Collector current, \u03bcA\n",
-      "VCE=VCC-(ICBO/1000)*RC;            #Collector-emitter voltage, V\n",
-      "\n",
-      "print(\"(i) The collector emitter voltage at cut-off=%.2fV.\"%VCE);\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, saturation current=IC_sat=(VCC-V_knee)/RC;         \n",
-      "VCE=V_knee;                    #Collector-emitter voltage, V\n",
-      "\n",
-      "print(\"(ii) The collector emitter voltage at saturation=%.1fV.\"%VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The collector emitter voltage at cut-off=9.99V.\n",
-        "(ii) The collector emitter voltage at saturation=0.7V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.3 : Page number 475-476"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10;             #Supply voltage, V\n",
-      "RC=1;               #Collector resistor, k\u03a9\n",
-      "VBB=2;              #Supply voltage to base, V\n",
-      "RB=2.7;             #Base resistor, k\u03a9\n",
-      "V_knee=0.7;         #Knee voltage, V\n",
-      "VBE=0.7;            #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IB=round((VBB-VBE)/RB,2);           #Base current, mA\n",
-      "Ic_sat=(VCC-V_knee)/RC;             #Collector saturation current, mA\n",
-      "beta_min=Ic_sat/IB;                 #Minimum value of base current amplification factor\n",
-      "print(\"(i) Minimum \u03b2=%.1f.\"%beta_min);\n",
-      "\n",
-      "#(ii)\n",
-      "VBB=1;                       #Supply voltage to base(changed), V\n",
-      "beta=50;                     #Base current amplification factor\n",
-      "IB=(VBB-VBE)/RB;             #Base current, mA\n",
-      "IC=beta*IB;                  #Collector current,mA\n",
-      "\n",
-      "if(IC<Ic_sat):\n",
-      "    print(\"(ii) The transistor will not be saturated.\");\n",
-      "else:\n",
-      "    print(\"(ii) The transistor will be saturated.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Minimum \u03b2=19.4.\n",
-        "(ii) The transistor will not be saturated.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.4 : Page number 480"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R2=10;                  #Resistor R2, k\u03a9\n",
-      "R3=10;                  #Resistor R3, k\u03a9\n",
-      "C1=0.01;                #Capacitor of 1st transistor, \u03bcF\n",
-      "C2=0.01;                #Capacitor of 2nd transistor, \u03bcF\n",
-      "\n",
-      "#Calculation\n",
-      "R=R2*1000;                   #Resistance, \u03a9\n",
-      "C=C1*10**-6;                 #Capacitance, F\n",
-      "T=round((1.4*R*C)*1000,2);   #Time period,m sec\n",
-      "f=1/(T*10**-3);              #Frequency, Hz\n",
-      "f=f/1000;                    #Frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"Time period of the square wave=%.2f m sec.\"%T);\n",
-      "print(\"Time frequency of the square wave=%d kHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time period of the square wave=0.14 m sec.\n",
-        "Time frequency of the square wave=7 kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.6 : Page number 485"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=10;               #Resistance in differentiating circuit, k\u03a9\n",
-      "C=2.2;              #Capacitance in differentiating circuit, \u03bcF\n",
-      "d_ei=10;            #Change in input voltage, V\n",
-      "dt=0.4;             #Time in which change occurs, s\n",
-      "\n",
-      "#Calculation\n",
-      "eo=R*1000*C*10**-6*d_ei/dt\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage=%.2fV.\"%eo);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage=0.55V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.7 : Page number 489"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin_peak=12;                #Peak value of input voltage, V\n",
-      "V_D=0.7;                    #Forward bias voltage of diode, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vout_peak=Vin_peak-V_D;         #Peak value of output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The peak output voltage=%.1fV.\"%Vout_peak);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The peak output voltage=11.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.8 : Page number 489"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin_peak=10;                #Peak value of input voltage, V\n",
-      "R=1;                        #Input resistor, k\u03a9\n",
-      "RL=4;                       #Load resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Vout_peak=(Vin_peak*RL)/(R+RL);         #Peak output voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The peak output voltage=%dV.\"%Vout_peak);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The peak output voltage=8V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.9 : Page number 490"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=-10;                #Input voltage, V\n",
-      "V_D=0.7;                #Forward bias voltage of the diode, V\n",
-      "R=1;                    #Resistance, k\u03a9\n",
-      "\n",
-      "\n",
-      "print(\"The diode will be forward biased for the negative half-cycle of input signal.\");\n",
-      "Vout=-V_D;              #Output voltage, V\n",
-      "V_R=Vin-(-V_D);         #Voltage across resistor R, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"The voltage across R=%.1fV.\"%V_R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The diode will be forward biased for the negative half-cycle of input signal.\n",
-        "The output voltage=-0.7V.\n",
-        "The voltage across R=-9.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.10 : Page number 490-491"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_F=0.7;                        #Forward bias voltage of diode, V\n",
-      "R=200.0;                          #Input resistor of the circuit,  \u03a9\n",
-      "RL=1.0;                           #Load resistor, k\u03a9\n",
-      "Vin_peak=10.0;                    #Peak input voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Positive half-cycle:\n",
-      "print(\"During the positive half cycle, the diode is foward biased and can be replaced by battery of %.1fV.\"%V_F);\n",
-      "print(\"Therefore, Vout=%.1fV.\"%V_F);\n",
-      "\n",
-      "#Negative half-cycle:\n",
-      "print(\"During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\");\n",
-      "Vout_peak=RL*(-Vin_peak)/(R/1000+RL);\n",
-      "print(\"Therefore, Vout_peak=%.2fV.\"%Vout_peak);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "During the positive half cycle, the diode is foward biased and can be replaced by battery of 0.7V.\n",
-        "Therefore, Vout=0.7V.\n",
-        "During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\n",
-        "Therefore, Vout_peak=-8.33V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.12 : Page number 491"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sin\n",
-      "from math import pi\n",
-      "V_biasing=10.0;                        #Biasing voltage, V\n",
-      "vin=[30*sin(t/10.0) for t in range(0,(int)(2*pi*10))]      #input voltage  waveform, V\n",
-      "p=plot(vin);\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vin(V)');\n",
-      "title('Input waveform');\n",
-      "show(p);\n",
-      "vout=[];                             #Output voltage waveform, V\n",
-      "for v in vin[:]:\n",
-      "    if(v-V_biasing)>0 :              #Diode is forward biased.\n",
-      "        vout.append(v-V_biasing);\n",
-      "    else:                            #Diode is reverse biased.\n",
-      "        vout.append(0);\n",
-      "        \n",
-      "p=plot(vout);\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vout(V)');\n",
-      "title('Output waveform');\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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OH4bixXWnETpt3w49eqjfhWJ5OnRBPxmTEMLOmjSBWrVA1pKK99+H554zT4Eo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\nTt+QhXRCCOEQWorEmDFj2L9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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a24cd1d0>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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VJTsFkTFJKwGTyYSkpCS0atVKyvuzBLQjKAhYt052CiJjkjodJISQ9t4sAe3g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\nBIiIDOz/AUzB4hO0oBxxAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a23a9e90>"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.13 : Page number 492"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "Vin=[];      #Input voltage waveform, V\n",
-      "t1=50;         #Assumed time interval, s\n",
-      "t2=100;         #Assumed time interval, s\n",
-      "V_biasing=10;  #Biasing voltage, V\n",
-      "for t in range(0,151):                  #time interval from 0s to 151s\n",
-      "    if(t<=t1):                      \n",
-      "        Vin.append(15);               #Value of input voltage for time 0 to t1 seconds \n",
-      "    elif(t<=t2 and t>t1):\n",
-      "        Vin.append(-30);             #Value of input voltage for time t1 to t2 seconds\n",
-      "    else :\n",
-      "        Vin.append(15);             #Value of input voltage after t2 seconds\n",
-      "\n",
-      "p=plot(Vin);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,160])\n",
-      "limit.set_ylim([-35,20])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vin(V)');\n",
-      "title('Input waveform');\n",
-      "show(p);\n",
-      "\n",
-      "vout=[];                 #Output voltage waveform, V\n",
-      "for v in Vin[:]:                  #Loop iterating input voltage \n",
-      "    if(v<=0):\n",
-      "        vout.append(0);               #Diode reverse biased\n",
-      "    else:\n",
-      "        vout.append(v-V_biasing);      #Diode forward biased\n",
-      "\n",
-      "p=plot(vout);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,160])\n",
-      "limit.set_ylim([-35,20])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vout(V)');\n",
-      "title('Output waveform');\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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xsVG0traKGTNmiIKCAmnmrKioaPfA0dVc6enpIjMz0365qVOnikOHDuk259Xe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\nQ+VgQAAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a25d35d0>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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eOXJEGRkZCg0NVYcOHXT8+HH17NlTubm5bpVTKns+jRgxQpJ05513ysPDQ6dP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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a2496cd0>"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.14 : Page number 492-493"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "Vin=[];      #Input voltage waveform, V\n",
-      "t1=50;         #Assumed time interval, s\n",
-      "t2=100;         #Assumed time interval, s\n",
-      "V_biasing=5;  #Biasing voltage, V\n",
-      "for t in range(0,151):                  #time interval from 0s to 151s\n",
-      "    if(t<=t1):                      \n",
-      "        Vin.append(10);               #Value of input voltage for time 0 to t1 seconds \n",
-      "    elif(t<=t2 and t>t1):\n",
-      "        Vin.append(-10);             #Value of input voltage for time t1 to t2 seconds\n",
-      "    else :\n",
-      "        Vin.append(0);\n",
-      "        \n",
-      "p=plot(Vin);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,101])\n",
-      "limit.set_ylim([-20,20])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vin(V)');\n",
-      "title('Input waveform');\n",
-      "show(p);\n",
-      "\n",
-      "vout=[];                 #Output voltage waveform, V\n",
-      "for v in Vin[:]:                  #Loop iterating input voltage \n",
-      "    if(v<=0):\n",
-      "        vout.append(v);               #Diode reverse biased\n",
-      "    else:\n",
-      "        vout.append(v-V_biasing);      #Diode forward biased\n",
-      "\n",
-      "p=plot(vout);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,101])\n",
-      "limit.set_ylim([-20,20])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vout(V)');\n",
-      "title('Output waveform');\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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bVUDwiibg4sWLmDRpEt566y14enq2WicIgsO30c6dOzFgwABoNBqI7Vxf4Qzt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\nBHydAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f4f12db9190>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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pKRwdHeHr64suXZgtRDdiQJC1aTUgDAYDNm7ciO3bt+Pq1avo378/rly5gvLycowaNQpz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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f4f12c80dd0>"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.15 : Page number 493"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "Vin=[];      #Input voltage waveform, V\n",
-      "t1=50;         #Assumed time interval, s\n",
-      "t2=100;         #Assumed time interval, s\n",
-      "V_D1=0.6;          #Forward Biasing voltage of the 1st diode, V\n",
-      "V_D2=0.6;          #Forward Biasing voltage of the 2nd diode, V\n",
-      "for t in range(0,151):                  #time interval from 0s to 151s\n",
-      "    if(t<=t1):                      \n",
-      "        Vin.append(10);               #Value of input voltage for time 0 to t1 seconds \n",
-      "    elif(t<=t2 and t>t1):\n",
-      "        Vin.append(-10);             #Value of input voltage for time t1 to t2 seconds\n",
-      "    else :\n",
-      "        Vin.append(0);\n",
-      "        \n",
-      "p=plot(Vin);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,110])\n",
-      "limit.set_ylim([-20,20])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vin(V)');\n",
-      "title('Input waveform');\n",
-      "show(p);\n",
-      "\n",
-      "vout=[];                 #Output voltage waveform, V\n",
-      "for v in Vin[:]:                  #Loop iterating input voltage \n",
-      "    if(v<=0):\n",
-      "        vout.append(-V_D1);               #Diode D1 forward biased, \n",
-      "    else:\n",
-      "        vout.append(V_D2);      #Diode D2 forward biased\n",
-      "\n",
-      "p=plot(vout);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,110])\n",
-      "limit.set_ylim([-1,1])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vout(V)');\n",
-      "title('Output waveform');\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a2417c90>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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agIAA4+vra4KDg83KlSur7N+zzz5runbtanr06GFSU1PdVjcnuAEALBrsUBIA\noG4QDAAAC4IBAGBBMAAALAgGAIAFwQAAsCAYgBo6d+6c3njjDXeXAdQ6ggGoocLCQr3++usutwUa\nCoIBqKE5c+bo6NGjiomJ0ZNPPlll2zvvvFN33HGHNm/erNLS0nqqEKgZznwGauj48eMaPXq0Dhw4\n4FL7Xbt2aeXKldqzZ48mTJigqVOnqmvXrnVcJfDTccQA1NBP/UwVGxurVatWae/evZKk8PBwbdiw\noS5KA64LwQDUgrlz5yomJkZ9+vTR5cuXFR0drZiYGM2fP9/Z5sKFC1q9erXuuusu7dixQ8uWLdOw\nYcPcVzRQCYaSgBo6c+aMbrnlFmVnZ1fb9sknn9S6des0evRoTZs2Tb179677AoEaIhiA63Dffffp\n888/1+23367nn3++0nbbtm1TfHy8mjRpUo/VATVDMAAALJhjAABYEAwAAAuCAQBgQTAAACwIBgCA\nBcEAALAgGAAAFgQDAMDi/wAfN3vR+U6XwgAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a23368d0>"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.16 : Page number 493-494"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "from math import sin\n",
-      "from math import pi\n",
-      "\n",
-      "VZ=20;                   #Assumed zener voltage, V\n",
-      "VF=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-      "Vin=[];      #Input voltage waveform, V\n",
-      "for t in range(0,(int)(2*pi*10)):                  #time interval from 0s to 151s\n",
-      "    Vin.append(30*sin(t/10.0));\n",
-      "        \n",
-      "p=plot(Vin);\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vin(V)');\n",
-      "title('Input waveform');\n",
-      "show(p);\n",
-      "\n",
-      "vout=[];                 #Output voltage waveform, V\n",
-      "for v in Vin[:]:                  #Loop iterating input voltage \n",
-      "    if(v<=-VF):\n",
-      "        vout.append(-VF);               #Zener diode forward biased, \n",
-      "    elif(v>=VZ):\n",
-      "        vout.append(VZ);        #Input voltage exceeds zener voltage\n",
-      "    else:\n",
-      "        vout.append(v);           #Zener diode reverse biased\n",
-      "\n",
-      "p=plot(vout);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,80])\n",
-      "limit.set_ylim([-1,40])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vout(V)');\n",
-      "title('Output waveform');\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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OH4bixXWnETpt3w49eqjfhWJ5OnRBPxmTEMLOmjSBWrVA1pKK99+H554zT4Eo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\nTt+QhXRCCOEQWorEmDFj2L9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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a2639290>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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lTcwJfP01MGwYC8DWdeliWvfp8GGlkxBpl02WwP79pvFksn2cFyBSlmIH53l5\neeGRRx5B+/bt0aFDBxw7dqzZz83MBNatkzEcWU1oKPDyy0qnINIuxeYEvL298d1336FHjx73fbyx\nca3r102HFhYXm5aMINt2547pfI+8PKBbN6XTENk+m5oTaEn/HDgAjB3LArAXHTsCY8aYLhFKRNan\nWAlIkoSwsDCMGDECH374YbOfx/kA+8N5ASLlKDYncPjwYbi7u+Pq1asIDw/H4MGDMX78+Abfk5CQ\nYP7aYDDAYDAgMxPYuNHKYUlWoaHA/PlKpyCyTUajEcZW7Eqr4jyB119/HU5OTli6dKn5vvuNaxUU\nAEOGmNadad/e2ilJLrzkJFHbsYk5gdu3b+PmzZsAgFu3biEtLQ0BAQEPfF5mJmAwsADsTfv2pt8r\nzx4msj5FhoOKioowbdo0AEBNTQ3mzp2LiIiIBz4vM9O01ADZn/p5gblzlU5CpC2qGA66n/vt0nh7\nA7t3A35+CoUi2Zw+DUycCOTm8iJBRK1hE8NBLXHuHFBZCfj6Kp2E5DBoEFBdDZw/r3QSIm2xmRKo\nHwrip0T7JElcVZRICTZXAmS/QkJYAkTWZhNzAkKYDh08etS0ZATZp/PnTWcPFxRwj4+opexyTuDU\nKcDRkQVg77y9TdcV+PFHpZMQaYdNlACHgrSD8wJE1sUSIFVhCRBZl+rnBLikgLZcvgwEBJiWBmln\nEx9RiNTF7uYEsrMBNzcWgFb06gW4ugL/+79KJyHSBtWXAIeCtIdDQkTWo/oS+OorloDW8HwBIutR\n9ZxAVZVAz56m48d79lQ6EVlLcTHQv7/pfzt0UDoNkW2xqzmB48dNbwYsAG1xdjadM/DvfyudhMj+\nqboEOB+gXZwXILIOVZcA5wO0iyVAZB2qnhNwchLIzwceeUTpNGRtN24AHh7ApUv8/RNZwq7mBIYM\n4RuAVj3yCBAeDmzerHQSIvum6hLgUJC2Pfss8OGHSqcgsm+KlUBqaioGDx6MgQMHYvXq1ff9HpaA\ntoWHm5aPyMpSOgmR/VJkTqC2thaDBg1CRkYGevfujZEjR2Lz5s3wvevakZIk4dYtAUdHa6cjNfnj\nH4HCQuD995VOQmQbbGJO4NixYxgwYAC8vLzQoUMHzJkzB19++eU938cCoIULgeRk4NYtpZMQ2SdF\nSiA/Px+enp7m2x4eHsjPz1ciCqmchwcwdiywdavSSYjsk4MSLyo189qBCQkJ5q8NBgMMBoM8gUjV\nnnsOSEzN3V0YAAAH/klEQVQEFixQOgmR+hiNRhiNxhY/X5E5gaNHjyIhIQGpqakAgDfffBPt2rXD\nK6+88v/BLBzXIvtVUwP07QukpZkOGyaixtnEnMCIESPwn//8B7m5uaiqqsKWLVswefJkJaK0Wmsa\n2FpsISPQeE4HB9NegFoOF7X1n6faMKeyFBkOcnBwwHvvvYeJEyeitrYWcXFxDY4MsiVGo1H1w1S2\nkBFoOmdcHDBypGl+oJmjibLZutWIq1cNyoZoBlvN6eMDBAYql6cxtvJ3ZClFSgAAJk2ahEmTJin1\n8mRjvL2BJUuAzz9XOonpUqe2MFFtqzmjotRZAvZKsRIgstRrrymdwCQhwfSf2jEnNYeqF5AjIiLL\nWfK2rto9AZV2ExGRXVH1AnJERCQvlgARkYaprgSas7qoEhYuXAidToeAgADzfdevX0d4eDh8fHwQ\nERGB0tJSBROa5OXlISQkBEOGDIG/vz82bNgAQH1ZKysrMXr0aAQFBcHf3998drjacgKmBQ/1ej2i\noqIAqDOjl5cXhg4dCr1ej1GjRgFQZ87S0lLMmDEDvr6+8PPzw7fffqu6nD/99BP0er35v65du2LD\nhg2qywkAa9euhb+/PwICAhATE4M7d+5YnlOoSE1Njejfv784f/68qKqqEoGBgeLUqVNKxxJCCHHw\n4EFx4sQJ4e/vb77v5ZdfFqtXrxZCCJGYmCheeeUVpeKZFRQUiKysLCGEEDdv3hQ+Pj7i1KlTqsx6\n69YtIYQQ1dXVYvTo0eLo0aOqzLlmzRoRExMjoqKihBDq/L17eXmJa9euNbhPjTnnz58vPvroIyGE\n6fdeWlqqypz1amtrhZubm7h48aLqcl66dEl4e3uLyspKIYQQs2bNEp9++qnFOVVVAkeOHBETJ040\n337zzTfFm2++qWCihs6fP9+gBAYNGiQKCwuFEKY330GDBikVrVFTpkwR6enpqs5669YtMWzYMPHt\nt9+qLmdeXp4IDQ0VmZmZIjIyUgihzt+7l5eXKC4ubnCf2nKWlpYKb2/ve+5XW8677du3T4wbN04I\nob6cly5dEp6enuL69euiurpaREZGirS0NItzqmo4yNZWFy0qKoJOpwMA6HQ6FBUVKZyoodzcXGRl\nZWH06NGqzFpXV4egoCDodDpERERg1KhRqsv50ksv4a233kK7dv//p6K2jIDpkOqwsDCMGDECH/53\nfQ215Tx//jxcXFywYMECDBs2DM8++yxu3bqlupx3S05ORnR0NAD1/Tx79+6NpUuXok+fPujVqxe6\ndeuG8PBwi3OqqgRs+dwASZJUlb+8vBzTp0/H+vXr0aVLlwaPqSVru3btkJ2djUuXLuHbb7/FyZMn\nGzyudM5//etfcHV1hV6vb/SQZaUz1jt8+DCysrKwd+9e/PWvf8WhQ4caPK6GnDU1NThx4gQWLVqE\nEydOoHPnzkhMTGzwPWrIWa+qqgq7du3CzJkz73lMDTlLSkqQkpKC3NxcXL58GeXl5fjss88afE9z\ncqqqBHr37o28vDzz7by8PHh4eCiYqGk6nQ6FhYUAgIKCAri6uiqcyKS6uhrTp0/HvHnzMHXqVADq\nzQoAXbt2RUhICPbt26eqnEeOHEFKSgq8vb0RHR2NzMxMzJs3T1UZ67m7uwMAXFxcMG3aNBw7dkx1\nOT08PODh4YGRI0cCAGbMmIETJ07Azc1NVTnr7d27F8OHD4eLiwsA9f0NZWRkwNvbGz179oSDgwOe\neuopfPPNNxb/PFVVAra2uujkyZORlJQEAEhKSjK/4SpJCIG4uDj4+flh8eLF5vvVlrW4uNh81EJF\nRQXS09Ph6+urqpyrVq1CXl4ezp8/j+TkZDz22GPYuHGjqjICwO3bt3Hz5k0AwK1bt5CWloaAgADV\n5XRzc4OnpyfOnDkDwPQmNmTIEERFRakqZ73Nmzebh4IA9f0N9e3bF0ePHkVFRQWEEMjIyICfn5/l\nP0/ZZy8stGfPHuHj4yP69+8vVq1apXQcszlz5gh3d3fRoUMH4eHhIT7++GNx7do1ERoaKgYOHCjC\nw8NFSUmJ0jHFoUOHhCRJIjAwUAQFBYmgoCCxd+9e1WX9/vvvhV6vF0OHDhX+/v7iT3/6kxBCqC5n\nPaPRaD46SG0Zz507JwIDA0VgYKAYMmSI+e9GbTmFECI7O1uMGDFCDB06VEybNk2UlpaqMmd5ebno\n2bOnuHHjhvk+NeZcuXKlGDx4sPD39xfz588XVVVVFudU7dpBREQkP1UNBxERkXWxBIiINIwlQESk\nYSwBIiINYwkQEWkYS4CISMNYAkR3KSsrw9/+9jelYxBZDUuA6C4lJSV4//33m/29RLaOJUB0l/j4\neJw9exZ6vR7Lli1r8nunTZuGKVOmYNeuXaipqbFSQqK2xTOGie5y4cIFREZGIicnp1nff+DAAXz8\n8cf45ptvMHPmTCxcuBD9+/eXOSVR2+GeANFdLP1MFBwcjKSkJHz33XcAgMGDB2PHjh1yRCOSBUuA\nqBHLly+HXq/HsGHDzBfA0ev15ushA6YVUDdt2oSnnnoK6enp2LBhA8LCwpQLTWQhDgcR3eXatWsY\nPnw4cnNzH/i9y5YtwxdffIHIyEjExcUhMDBQ/oBEbYwlQPQLc+fOxffff49JkybhL3/5S6Pft3fv\nXoSGhuKhhx6yYjqitsUSICLSMM4JEBFpGEuAiEjDWAJERBrGEiAi0jCWABGRhrEEiIg0jCVARKRh\nLAEiIg37P3SDpk3uSVOuAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a232f110>"
-       ]
-      }
-     ],
-     "prompt_number": 54
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.17 : Page number 494"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "from math import sin\n",
-      "from math import pi\n",
-      "\n",
-      "VZ1=20;                   #Assumed zener voltage, V\n",
-      "VF1=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-      "VZ2=20;                   #Assumed zener voltage, V\n",
-      "VF2=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-      "Vin=[];      #Input voltage waveform, V\n",
-      "for t in range(0,(int)(2*pi*10)):                  #time interval from 0s to 151s\n",
-      "    Vin.append(30*sin(t/10.0));\n",
-      "        \n",
-      "p=plot(Vin);\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vin(V)');\n",
-      "title('Input waveform');\n",
-      "show(p);\n",
-      "\n",
-      "vout=[];                 #Output voltage waveform, V\n",
-      "for v in Vin[:]:                  #Loop iterating input voltage \n",
-      "    if(v<=-(VZ1+VF2)):\n",
-      "        vout.append(-(VZ1+VF2));               #Zener diode forward biased, \n",
-      "    elif(v>=VZ2+VF1):\n",
-      "        vout.append(VZ2+VF1);        #Input voltage exceeds zener voltage\n",
-      "    else:\n",
-      "        vout.append(v);           #Zener diode reverse biased\n",
-      "\n",
-      "p=plot(vout);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,80])\n",
-      "limit.set_ylim([-40,40])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vout(V)');\n",
-      "title('Output waveform');\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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OH4bixXWnETpt3w49eqjfhWJ5OnRBPxmTEMLOmjSBWrVA1pKK99+H554zT4Eo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\nTt+QhXRCCOEQWorEmDFj2L9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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a2723990>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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loAbtzz+Vi/+lpHAmjUxCAAEBwNixwD/+ITtN/cXLchCZqH17YPp0HpeQ7Ztv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\nXb8AAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a2583910>"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_1.ipynb
deleted file mode 100755
index 34ff63b3..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_1.ipynb
+++ /dev/null
@@ -1,815 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:5aefee63b41b58f0caeac1aa6be18e130e4530aaaa6bc5e9bbc45b3687d3f8e9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 18 : SOLID-STATE SWITCHING CIRCUITS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.1 : Page number 472"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10;             #Supply voltage, V\n",
-      "RC=1.0;               #Collector resistor, k\u03a9\n",
-      "RB=47.0;              #Base resistor, k\u03a9\n",
-      "beta=100.0;           #Base current amplification factor\n",
-      "VBE=0.7;            #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "IC_sat=VCC/RC;              #Collector saturation current, mA\n",
-      "IB=IC_sat/beta;             #Base current, mA\n",
-      "V=IB*RB+VBE;                #Input voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Input voltage required to saturate the transistor switch=%.1fV.\"%V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input voltage required to saturate the transistor switch=5.4V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.2 : Page number 475"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10;             #Supply voltage, V\n",
-      "RC=1.0;               #Collector resistor, k\u03a9\n",
-      "ICBO=10.0;            #Collector leakage current, \u03bcA\n",
-      "V_knee=0.7;         #Knee voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IC=ICBO;                    #Collector current, \u03bcA\n",
-      "VCE=VCC-(ICBO/1000)*RC;            #Collector-emitter voltage, V\n",
-      "\n",
-      "print(\"(i) The collector emitter voltage at cut-off=%.2fV.\"%VCE);\n",
-      "\n",
-      "#(ii)\n",
-      "#Since, saturation current=IC_sat=(VCC-V_knee)/RC;         \n",
-      "VCE=V_knee;                    #Collector-emitter voltage, V\n",
-      "\n",
-      "print(\"(ii) The collector emitter voltage at saturation=%.1fV.\"%VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The collector emitter voltage at cut-off=9.99V.\n",
-        "(ii) The collector emitter voltage at saturation=0.7V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.3 : Page number 475-476"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10;             #Supply voltage, V\n",
-      "RC=1;               #Collector resistor, k\u03a9\n",
-      "VBB=2;              #Supply voltage to base, V\n",
-      "RB=2.7;             #Base resistor, k\u03a9\n",
-      "V_knee=0.7;         #Knee voltage, V\n",
-      "VBE=0.7;            #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IB=round((VBB-VBE)/RB,2);           #Base current, mA\n",
-      "Ic_sat=(VCC-V_knee)/RC;             #Collector saturation current, mA\n",
-      "beta_min=Ic_sat/IB;                 #Minimum value of base current amplification factor\n",
-      "print(\"(i) Minimum \u03b2=%.1f.\"%beta_min);\n",
-      "\n",
-      "#(ii)\n",
-      "VBB=1;                       #Supply voltage to base(changed), V\n",
-      "beta=50;                     #Base current amplification factor\n",
-      "IB=(VBB-VBE)/RB;             #Base current, mA\n",
-      "IC=beta*IB;                  #Collector current,mA\n",
-      "\n",
-      "if(IC<Ic_sat):\n",
-      "    print(\"(ii) The transistor will not be saturated.\");\n",
-      "else:\n",
-      "    print(\"(ii) The transistor will be saturated.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Minimum \u03b2=19.4.\n",
-        "(ii) The transistor will not be saturated.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.4 : Page number 480"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R2=10;                  #Resistor R2, k\u03a9\n",
-      "R3=10;                  #Resistor R3, k\u03a9\n",
-      "C1=0.01;                #Capacitor of 1st transistor, \u03bcF\n",
-      "C2=0.01;                #Capacitor of 2nd transistor, \u03bcF\n",
-      "\n",
-      "#Calculation\n",
-      "R=R2*1000;                   #Resistance, \u03a9\n",
-      "C=C1*10**-6;                 #Capacitance, F\n",
-      "T=round((1.4*R*C)*1000,2);   #Time period,m sec\n",
-      "f=1/(T*10**-3);              #Frequency, Hz\n",
-      "f=f/1000;                    #Frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"Time period of the square wave=%.2f m sec.\"%T);\n",
-      "print(\"Time frequency of the square wave=%d kHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time period of the square wave=0.14 m sec.\n",
-        "Time frequency of the square wave=7 kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.6 : Page number 485"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=10;               #Resistance in differentiating circuit, k\u03a9\n",
-      "C=2.2;              #Capacitance in differentiating circuit, \u03bcF\n",
-      "d_ei=10;            #Change in input voltage, V\n",
-      "dt=0.4;             #Time in which change occurs, s\n",
-      "\n",
-      "#Calculation\n",
-      "eo=R*1000*C*10**-6*d_ei/dt\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage=%.2fV.\"%eo);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage=0.55V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.7 : Page number 489"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin_peak=12;                #Peak value of input voltage, V\n",
-      "V_D=0.7;                    #Forward bias voltage of diode, V\n",
-      "\n",
-      "#Calculation\n",
-      "Vout_peak=Vin_peak-V_D;         #Peak value of output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The peak output voltage=%.1fV.\"%Vout_peak);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The peak output voltage=11.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.8 : Page number 489"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin_peak=10;                #Peak value of input voltage, V\n",
-      "R=1;                        #Input resistor, k\u03a9\n",
-      "RL=4;                       #Load resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Vout_peak=(Vin_peak*RL)/(R+RL);         #Peak output voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The peak output voltage=%dV.\"%Vout_peak);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The peak output voltage=8V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.9 : Page number 490"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vin=-10;                #Input voltage, V\n",
-      "V_D=0.7;                #Forward bias voltage of the diode, V\n",
-      "R=1;                    #Resistance, k\u03a9\n",
-      "\n",
-      "\n",
-      "print(\"The diode will be forward biased for the negative half-cycle of input signal.\");\n",
-      "Vout=-V_D;              #Output voltage, V\n",
-      "V_R=Vin-(-V_D);         #Voltage across resistor R, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage=%.1fV.\"%Vout);\n",
-      "print(\"The voltage across R=%.1fV.\"%V_R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The diode will be forward biased for the negative half-cycle of input signal.\n",
-        "The output voltage=-0.7V.\n",
-        "The voltage across R=-9.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.10 : Page number 490-491"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_F=0.7;                        #Forward bias voltage of diode, V\n",
-      "R=200.0;                          #Input resistor of the circuit,  \u03a9\n",
-      "RL=1.0;                           #Load resistor, k\u03a9\n",
-      "Vin_peak=10.0;                    #Peak input voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Positive half-cycle:\n",
-      "print(\"During the positive half cycle, the diode is foward biased and can be replaced by battery of %.1fV.\"%V_F);\n",
-      "print(\"Therefore, Vout=%.1fV.\"%V_F);\n",
-      "\n",
-      "#Negative half-cycle:\n",
-      "print(\"During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\");\n",
-      "Vout_peak=RL*(-Vin_peak)/(R/1000+RL);\n",
-      "print(\"Therefore, Vout_peak=%.2fV.\"%Vout_peak);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "During the positive half cycle, the diode is foward biased and can be replaced by battery of 0.7V.\n",
-        "Therefore, Vout=0.7V.\n",
-        "During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\n",
-        "Therefore, Vout_peak=-8.33V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.12 : Page number 491"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sin\n",
-      "from math import pi\n",
-      "V_biasing=10.0;                        #Biasing voltage, V\n",
-      "vin=[30*sin(t/10.0) for t in range(0,(int)(2*pi*10))]      #input voltage  waveform, V\n",
-      "p=plot(vin);\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vin(V)');\n",
-      "title('Input waveform');\n",
-      "show(p);\n",
-      "vout=[];                             #Output voltage waveform, V\n",
-      "for v in vin[:]:\n",
-      "    if(v-V_biasing)>0 :              #Diode is forward biased.\n",
-      "        vout.append(v-V_biasing);\n",
-      "    else:                            #Diode is reverse biased.\n",
-      "        vout.append(0);\n",
-      "        \n",
-      "p=plot(vout);\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vout(V)');\n",
-      "title('Output waveform');\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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OH4bixXWnETpt3w49eqjfhWJ5OnRBPxmTEMLOmjSBWrVA1pKK99+H554zT4Eo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\nTt+QhXRCCOEQWorEmDFj2L9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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a24cd1d0>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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VJTsFkTFJKwGTyYSkpCS0atVKyvuzBLQjKAhYt052CiJjkjodJISQ9t4sAe3g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\nBIiIDOz/AUzB4hO0oBxxAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a23a9e90>"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.13 : Page number 492"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "Vin=[];      #Input voltage waveform, V\n",
-      "t1=50;         #Assumed time interval, s\n",
-      "t2=100;         #Assumed time interval, s\n",
-      "V_biasing=10;  #Biasing voltage, V\n",
-      "for t in range(0,151):                  #time interval from 0s to 151s\n",
-      "    if(t<=t1):                      \n",
-      "        Vin.append(15);               #Value of input voltage for time 0 to t1 seconds \n",
-      "    elif(t<=t2 and t>t1):\n",
-      "        Vin.append(-30);             #Value of input voltage for time t1 to t2 seconds\n",
-      "    else :\n",
-      "        Vin.append(15);             #Value of input voltage after t2 seconds\n",
-      "\n",
-      "p=plot(Vin);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,160])\n",
-      "limit.set_ylim([-35,20])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vin(V)');\n",
-      "title('Input waveform');\n",
-      "show(p);\n",
-      "\n",
-      "vout=[];                 #Output voltage waveform, V\n",
-      "for v in Vin[:]:                  #Loop iterating input voltage \n",
-      "    if(v<=0):\n",
-      "        vout.append(0);               #Diode reverse biased\n",
-      "    else:\n",
-      "        vout.append(v-V_biasing);      #Diode forward biased\n",
-      "\n",
-      "p=plot(vout);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,160])\n",
-      "limit.set_ylim([-35,20])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vout(V)');\n",
-      "title('Output waveform');\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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xsVG0traKGTNmiIKCAmnmrKioaPfA0dVc6enpIjMz0365qVOnikOHDuk259Xe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\nQ+VgQAAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a25d35d0>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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eOXJEGRkZCg0NVYcOHXT8+HH17NlTubm5bpVTKns+jRgxQpJ05513ysPDQ6dP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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a2496cd0>"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.14 : Page number 492-493"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "Vin=[];      #Input voltage waveform, V\n",
-      "t1=50;         #Assumed time interval, s\n",
-      "t2=100;         #Assumed time interval, s\n",
-      "V_biasing=5;  #Biasing voltage, V\n",
-      "for t in range(0,151):                  #time interval from 0s to 151s\n",
-      "    if(t<=t1):                      \n",
-      "        Vin.append(10);               #Value of input voltage for time 0 to t1 seconds \n",
-      "    elif(t<=t2 and t>t1):\n",
-      "        Vin.append(-10);             #Value of input voltage for time t1 to t2 seconds\n",
-      "    else :\n",
-      "        Vin.append(0);\n",
-      "        \n",
-      "p=plot(Vin);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,101])\n",
-      "limit.set_ylim([-20,20])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vin(V)');\n",
-      "title('Input waveform');\n",
-      "show(p);\n",
-      "\n",
-      "vout=[];                 #Output voltage waveform, V\n",
-      "for v in Vin[:]:                  #Loop iterating input voltage \n",
-      "    if(v<=0):\n",
-      "        vout.append(v);               #Diode reverse biased\n",
-      "    else:\n",
-      "        vout.append(v-V_biasing);      #Diode forward biased\n",
-      "\n",
-      "p=plot(vout);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,101])\n",
-      "limit.set_ylim([-20,20])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vout(V)');\n",
-      "title('Output waveform');\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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bVUDwiibg4sWLmDRpEt566y14enq2WicIgsO30c6dOzFgwABoNBqI7Vxf4Qzt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\nBHydAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f4f12db9190>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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pKRwdHeHr64suXZgtRDdiQJC1aTUgDAYDNm7ciO3bt+Pq1avo378/rly5gvLycowaNQpz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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f4f12c80dd0>"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.15 : Page number 493"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "Vin=[];      #Input voltage waveform, V\n",
-      "t1=50;         #Assumed time interval, s\n",
-      "t2=100;         #Assumed time interval, s\n",
-      "V_D1=0.6;          #Forward Biasing voltage of the 1st diode, V\n",
-      "V_D2=0.6;          #Forward Biasing voltage of the 2nd diode, V\n",
-      "for t in range(0,151):                  #time interval from 0s to 151s\n",
-      "    if(t<=t1):                      \n",
-      "        Vin.append(10);               #Value of input voltage for time 0 to t1 seconds \n",
-      "    elif(t<=t2 and t>t1):\n",
-      "        Vin.append(-10);             #Value of input voltage for time t1 to t2 seconds\n",
-      "    else :\n",
-      "        Vin.append(0);\n",
-      "        \n",
-      "p=plot(Vin);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,110])\n",
-      "limit.set_ylim([-20,20])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vin(V)');\n",
-      "title('Input waveform');\n",
-      "show(p);\n",
-      "\n",
-      "vout=[];                 #Output voltage waveform, V\n",
-      "for v in Vin[:]:                  #Loop iterating input voltage \n",
-      "    if(v<=0):\n",
-      "        vout.append(-V_D1);               #Diode D1 forward biased, \n",
-      "    else:\n",
-      "        vout.append(V_D2);      #Diode D2 forward biased\n",
-      "\n",
-      "p=plot(vout);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,110])\n",
-      "limit.set_ylim([-1,1])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vout(V)');\n",
-      "title('Output waveform');\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a2417c90>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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agIAA4+vra4KDg83KlSur7N+zzz5runbtanr06GFSU1PdVjcnuAEALBrsUBIA\noG4QDAAAC4IBAGBBMAAALAgGAIAFwQAAsCAYgBo6d+6c3njjDXeXAdQ6ggGoocLCQr3++usutwUa\nCoIBqKE5c+bo6NGjiomJ0ZNPPlll2zvvvFN33HGHNm/erNLS0nqqEKgZznwGauj48eMaPXq0Dhw4\n4FL7Xbt2aeXKldqzZ48mTJigqVOnqmvXrnVcJfDTccQA1NBP/UwVGxurVatWae/evZKk8PBwbdiw\noS5KA64LwQDUgrlz5yomJkZ9+vTR5cuXFR0drZiYGM2fP9/Z5sKFC1q9erXuuusu7dixQ8uWLdOw\nYcPcVzRQCYaSgBo6c+aMbrnlFmVnZ1fb9sknn9S6des0evRoTZs2Tb179677AoEaIhiA63Dffffp\n888/1+23367nn3++0nbbtm1TfHy8mjRpUo/VATVDMAAALJhjAABYEAwAAAuCAQBgQTAAACwIBgCA\nBcEAALAgGAAAFgQDAMDi/wAfN3vR+U6XwgAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a23368d0>"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.16 : Page number 493-494"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "from math import sin\n",
-      "from math import pi\n",
-      "\n",
-      "VZ=20;                   #Assumed zener voltage, V\n",
-      "VF=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-      "Vin=[];      #Input voltage waveform, V\n",
-      "for t in range(0,(int)(2*pi*10)):                  #time interval from 0s to 151s\n",
-      "    Vin.append(30*sin(t/10.0));\n",
-      "        \n",
-      "p=plot(Vin);\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vin(V)');\n",
-      "title('Input waveform');\n",
-      "show(p);\n",
-      "\n",
-      "vout=[];                 #Output voltage waveform, V\n",
-      "for v in Vin[:]:                  #Loop iterating input voltage \n",
-      "    if(v<=-VF):\n",
-      "        vout.append(-VF);               #Zener diode forward biased, \n",
-      "    elif(v>=VZ):\n",
-      "        vout.append(VZ);        #Input voltage exceeds zener voltage\n",
-      "    else:\n",
-      "        vout.append(v);           #Zener diode reverse biased\n",
-      "\n",
-      "p=plot(vout);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,80])\n",
-      "limit.set_ylim([-1,40])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vout(V)');\n",
-      "title('Output waveform');\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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OH4bixXWnETpt3w49eqjfhWJ5OnRBPxmTEMLOmjSBWrVA1pKK99+H554zT4Eo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\nTt+QhXRCCOEQWorEmDFj2L9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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a2639290>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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lTcwJfP01MGwYC8DWdeliWvfp8GGlkxBpl02WwP79pvFksn2cFyBSlmIH53l5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\nLAEiIg37P3SDpk3uSVOuAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a232f110>"
-       ]
-      }
-     ],
-     "prompt_number": 54
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 18.17 : Page number 494"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "from math import sin\n",
-      "from math import pi\n",
-      "\n",
-      "VZ1=20;                   #Assumed zener voltage, V\n",
-      "VF1=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-      "VZ2=20;                   #Assumed zener voltage, V\n",
-      "VF2=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-      "Vin=[];      #Input voltage waveform, V\n",
-      "for t in range(0,(int)(2*pi*10)):                  #time interval from 0s to 151s\n",
-      "    Vin.append(30*sin(t/10.0));\n",
-      "        \n",
-      "p=plot(Vin);\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vin(V)');\n",
-      "title('Input waveform');\n",
-      "show(p);\n",
-      "\n",
-      "vout=[];                 #Output voltage waveform, V\n",
-      "for v in Vin[:]:                  #Loop iterating input voltage \n",
-      "    if(v<=-(VZ1+VF2)):\n",
-      "        vout.append(-(VZ1+VF2));               #Zener diode forward biased, \n",
-      "    elif(v>=VZ2+VF1):\n",
-      "        vout.append(VZ2+VF1);        #Input voltage exceeds zener voltage\n",
-      "    else:\n",
-      "        vout.append(v);           #Zener diode reverse biased\n",
-      "\n",
-      "p=plot(vout);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,80])\n",
-      "limit.set_ylim([-40,40])\n",
-      "xlabel('t-->');\n",
-      "ylabel('Vout(V)');\n",
-      "title('Output waveform');\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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OH4bixXWnETpt3w49eqjfhWJ5OnRBPxmTEMLOmjSBWrVA1pKK99+H554zT4Eo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\nTt+QhXRCCOEQWorEmDFj2L9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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a2723990>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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loAbtzz+Vi/+lpHAmjUxCAAEBwNixwD/+ITtN/cXLchCZqH17YPp0HpeQ7Ztv\nlGG/8eNlJ6FbcU+CGrzLlwEPD+WAKZc5rXtlZYCPD/DOO8Djj8tOU79xT4KoBpo3V06umzmTF/+T\nYdUqZbrrY4/JTkJV4Z4EEfjbrCyXLikLCm3bBvTuLTtN/cc9CaIasrdXrjg6c6aykh3VjXfeAQYO\nZEFYM+5JEP1FCCAoCBg+HJgyRXaa+i8nR9l7O3QI6NRJdpqGgSvTEdVSaqpyeepjxwArWta9Xnrm\nGcDZWbmYH9UNlgSRGTz/PODgoCx1SpZx8CDw5JNKGbdsKTtNw8GSIDKDs2cBLy/g+++VE+3IvCoq\ngAcfVMo4MlJ2moaFB66JzMDJSVkLe8YM2Unqp7VrleM/ERGyk1B1cE+CqAolJcpB1ffe4/x9c7p0\nCfD0VNYZ79tXdpqGh3sSRGZy113A0qXKJTu4gp35LFwIDBrEgrAl3JMguoOQEOCBB5ThJ6qdkyeV\nn2VqKhcUkoUHronM7NQp5XpOP/4IdO4sO43tEkI5k33AAODVV2Wnabg43ERkZp07KwewX3pJdhLb\ntnEjcOYMJwPYIu5JEP2N69eBXr2AJUuU4ScyzcWLypTi+Hjg4Ydlp2nYONxEZCG7dwMTJwK//qpc\nNZaq78UXgStXgE8/lZ2EWBJEFhQeDuh0wKJFspPYjv/+Fxg6VCnXdu1kpyGWBJEF5eYCPXsqexU9\ne8pOY/3KypSpri+9BIwbJzsNATZ04HrmzJno3r07evXqhSeffBJFRUXGx2JiYuDu7g5PT0/s2rVL\nRjyiKnXoAMTEABMmKB+AdGf//CfQqhXPrLZ1UkoiODgYv/76K37++Wd4eHggJiYGAJCWlob169cj\nLS0NCQkJmDx5MioqKmREJKrSxIlAmzbKOgik7rvvlEL95BNAo5GdhmpDSkkEBQXBzk556759+yIr\nKwsAsGXLFoSFhcHBwQE6nQ5du3ZFSkqKjIhEVdJogJUrlZL47TfZaazT8ePAmDHAunWAu7vsNFRb\n0s+TWL16NR776+I4OTk5cHV1NT7m6uqK7OxsWdGIqqTTAW++qQw7lZfLTmNdzp1TDlQvWqSsOEe2\nz95SGw4KCkJeXt5t9y9atAjDhg0DACxcuBB33XUXwsPDVbejUdlXjY6ONn6t1+uh1+trlZfIFJMm\nARs2AMuW8QSxG65fV9aIGDlSGZYj+QwGAwwGQ622IW120+eff46VK1di9+7daNKkCQAg9q8lqqKi\nogAAgwdVEd01AAALgUlEQVQPxvz589H3lquBcXYTWYOTJ5XZO/36AS1aKOdPtGihLFjUEP38M3D3\n3cBXXwF20scoqCo2MwU2ISEBL7/8MpKTk+Ho6Gi8Py0tDeHh4UhJSUF2djYGDRqEkydP3rY3wZIg\na3HsGHDiBFBcDFy+rPy3oc58atxY2YNo1kx2ElJjMyXh7u6OkpIStP1rEeF+/fphxYoVAJThqNWr\nV8Pe3h7Lli3Do48+etvrWRJERKazmZKoLZYEEZHpbOZkOiIisg0sCSIiUsWSICIiVSwJIiJSxZIg\nIiJVLAkiIlLFkiAiIlUsCSIiUsWSICIiVSwJIiJSxZIgIiJVLAkiIlLFkiAiIlUsCSIiUsWSICIi\nVSwJIiJSxZIgIiJVLAkiIlIlpSTmzp2LXr16wc/PD48++ihyc3ONj8XExMDd3R2enp7YtWuXjHhE\nRPQXKWtcX7p0CS1btgQAfPDBB0hLS8NHH32EtLQ0hIeH48cff0R2djYGDRqE9PR02NlV7jKucU1E\nZDqbWeP6RkEAQHFxsbEEtmzZgrCwMDg4OECn06Fr165ISUmREZGIiADYy3rj119/HWvWrEGrVq1g\nMBgAADk5OXjggQeMz3F1dUV2drakhEREZLE9iaCgIPj4+Nz2Z9u2bQCAhQsX4syZMxg7diw++OAD\n1e1oNBpLRSQior9hsT2JxMTEaj0vPDwcjz/+OKKjo+Hi4oLMzEzjY1lZWXBxcanyddHR0cav9Xo9\n9Hp9beISEdU7BoPBOFJTU1IOXJ84cQLu7u4AlAPXe/fuxYYNG4wHrlNSUowHrk+ePHnb3gQPXBMR\nma4mn51SjknMnj0bx48fh52dHXQ6HT7++GMAgJeXF0JDQ+Hl5QV7e3usWLGCw01ERBJJ2ZOoLe5J\nEBGZzmamwBIRkW1gSRARkSqWBBERqWJJEBGRKpYEERGpYkkQEZEqlgQREaliSRARkSqWBBERqWJJ\nEBGRKpYEERGpYkkQEZEqlgQREaliSRARkSqWBBERqWJJEBGRKpYEERGpYkkQEZEqlgQREamSWhJL\nly6FnZ0dLly4YLwvJiYG7u7u8PT0xK5duySmIyIiaSWRmZmJxMREdOrUyXhfWloa1q9fj7S0NCQk\nJGDy5MmoqKiQFbHWDAaD7AjVwpzmxZzmZQs5bSFjTUkriRkzZmDJkiWV7tuyZQvCwsLg4OAAnU6H\nrl27IiUlRVLC2rOV/3GY07yY07xsIactZKwpKSWxZcsWuLq6omfPnpXuz8nJgaurq/G2q6srsrOz\n6zoeERH9xd5SGw4KCkJeXt5t9y9cuBAxMTGVjjcIIVS3o9FoLJKPiIj+nkbc6RPaAo4ePYrAwEA0\na9YMAJCVlQUXFxccPHgQn332GQAgKioKADB48GDMnz8fffv2rRyaxUFEVCOmfuTXeUncqnPnzvjp\np5/Qtm1bpKWlITw8HCkpKcjOzsagQYNw8uRJlgIRkSQWG26qrpsLwMvLC6GhofDy8oK9vT1WrFjB\ngiAikkj6ngQREVkvmzvjOiEhAZ6ennB3d8fixYtlxzGaMGECtFotfHx8jPdduHABQUFB8PDwQHBw\nMAoLCyUmVGRmZiIgIAA9evSAt7c3li9fDsC6sl67dg19+/aFr68vvL29ER0dbXUZb1ZeXg4/Pz8M\nGzYMgHXm1Ol06NmzJ/z8/ODv7w/AOnMWFhZi1KhR6N69O7y8vHDw4EGry3n8+HH4+fkZ/7Rq1QrL\nly+3upwA8N5778Hb2xs+Pj4IDw/H9evXTc8pbEhZWZno0qWLOHXqlCgpKRG9evUSaWlpsmMJIYT4\n/vvvxaFDh4S3t7fxvpkzZ4rFixcLIYSIjY0Vr776qqx4Rrm5ueLw4cNCCCEuXbokPDw8RFpamtVl\nvXz5shBCiNLSUtG3b19x4MABq8t4w9KlS0V4eLgYNmyYEMI6/951Op04f/58pfusMee4cePEqlWr\nhBDK331hYaFV5ryhvLxcODs7izNnzlhdzqysLNG5c2dx7do1IYQQoaGh4vPPPzc5p02VxL59+8Sj\njz5qvB0TEyNiYmIkJqrs1KlTlUqiW7duIi8vTwihfDh369ZNVjRVTzzxhEhMTLTarJcvXxb33Xef\nOHjwoFVmzMzMFIGBgWLPnj1i6NChQgjr/HvX6XTi3Llzle6ztpyFhYWic+fOt91vbTlv9u2334qH\nH35YCGF9ObOysoSbm5u4cOGCKC0tFUOHDhW7du0yOadNDTdlZ2fDzc3NeNvaT7bLz8+HVqsFAGi1\nWuTn50tOVFlGRgYOHz6Mvn37Wl3WiooK+Pr6QqvVIjg4GP7+/laXEQCmT5+Ot99+G3Z2//9PyRpz\najQaDBo0CH369MHKlSsBWF/OU6dOwcnJCePHj8d9992Hf/zjH7h8+bLV5bxZfHw8wsLCAFjfz9PF\nxQUvv/wy7rnnHnTs2BGtW7dGUFCQyTltqiRseaaTRqOxqvzFxcUYOXIkli1bhpYtW1Z6zBqy2tnZ\n4ciRI8jKysLBgwdx9OjRSo9bQ8ZvvvkG7du3h5+fn+rcc2vICQD/+7//i8OHD2Pnzp348MMPsXfv\n3kqPW0POsrIyHDp0CJMnT8ahQ4fQvHlzxMbGVnqONeS8oaSkBNu2bcNTTz1122PWkLOgoABbt25F\nRkYGcnJyUFxcjLVr11Z6TnVy2lRJuLi4IDMz03g7MzOz0mU8rI1WqzWedZ6bm4v27dtLTqQoLS3F\nyJEjERERgeHDhwOw3qytWrVCQEAAvv32W6vLuG/fPmzduhWdO3dGWFgY9uzZg4iICKvLCQAdOnQA\nADg5OWHEiBFISUmxupyurq5wdXXF/fffDwAYNWoUDh06BGdnZ6vKecPOnTvRu3dvODk5AbC+f0NJ\nSUno3Lkz2rVrB3t7ezz55JPYv3+/yT9PmyqJPn364MSJE8jIyEBJSQnWr1+PkJAQ2bFUhYSEIC4u\nDgAQFxdn/ECWSQiBiRMnwsvLC9OmTTPeb01Zz507Z5xxcfXqVSQmJqJ79+5WlREAFi1ahMzMTJw6\ndQrx8fEYOHAg1qxZY3U5r1y5gkuXLgEALl++jF27dsHHx8fqcjo7O8PNzQ3p6ekAlA+5Hj16YNiw\nYVaV84Z169YZh5oA6/o3BACdOnXCgQMHcPXqVQghkJSUBC8vL9N/nhY/emJmO3bsEB4eHqJLly5i\n0aJFsuMYjRkzRnTo0EE4ODgIV1dXsXr1anH+/HkRGBgo3N3dRVBQkCgoKJAdU+zdu1doNBrRq1cv\n4evrK3x9fcXOnTutKusvv/wi/Pz8RM+ePYW3t7d46623hBDCqjLeymAwGGc3WVvOP/74Q/Tq1Uv0\n6tVL9OjRw/jvxtpyCiHEkSNHRJ8+fUTPnj3FiBEjRGFhoVXmLC4uFu3atRMXL1403meNOefNmyc8\nPT2Ft7e3GDdunCgpKTE5J0+mIyIiVTY13ERERHWLJUFERKpYEkREpIolQUREqlgSRESkiiVBRESq\nWBJEJigqKsJHH30kOwZRnWFJEJmgoKAAK1asqPZziWwdS4LIBFFRUfj999/h5+eHWbNm3fG5I0aM\nwBNPPIFt27ahrKysjhISmRfPuCYywenTpzF06FCkpqZW6/nJyclYvXo19u/fj6eeegoTJkxAly5d\nLJySyHy4J0FkAlN/pxowYADi4uLw008/AQA8PT2xadMmS0QjsgiWBFENzZkzB35+frjvvvuMiyT5\n+fkZ1+QGlKvYfvnll3jyySeRmJiI5cuXY9CgQfJCE5mIw01EJjh//jx69+6NjIyMv33urFmz8J//\n/AdDhw7FxIkT0atXL8sHJDIzlgSRicaOHYtffvkFQ4YMwZIlS1Sft3PnTgQGBuKuu+6qw3RE5sWS\nICIiVTwmQUREqlgSRESkiiVBRESqWBJERKSKJUFERKpYEkREpIolQUREqlgSRESk6v8AbU8ybWOZ\nXb8AAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f15a2583910>"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_2.ipynb
deleted file mode 100755
index 8a5bc9dc..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_2.ipynb
+++ /dev/null
@@ -1,799 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 18 : SOLID-STATE SWITCHING CIRCUITS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.1 : Page number 472"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Input voltage required to saturate the transistor switch=5.4V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10;             #Supply voltage, V\n",
-    "RC=1.0;               #Collector resistor, kΩ\n",
-    "RB=47.0;              #Base resistor, kΩ\n",
-    "beta=100.0;           #Base current amplification factor\n",
-    "VBE=0.7;            #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "IC_sat=VCC/RC;              #Collector saturation current, mA\n",
-    "IB=IC_sat/beta;             #Base current, mA\n",
-    "V=IB*RB+VBE;                #Input voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Input voltage required to saturate the transistor switch=%.1fV.\"%V);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.2 : Page number 475"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The collector emitter voltage at cut-off=9.99V.\n",
-      "(ii) The collector emitter voltage at saturation=0.7V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10;             #Supply voltage, V\n",
-    "RC=1.0;               #Collector resistor, kΩ\n",
-    "ICBO=10.0;            #Collector leakage current, μA\n",
-    "V_knee=0.7;         #Knee voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IC=ICBO;                    #Collector current, μA\n",
-    "VCE=VCC-(ICBO/1000)*RC;            #Collector-emitter voltage, V\n",
-    "\n",
-    "print(\"(i) The collector emitter voltage at cut-off=%.2fV.\"%VCE);\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, saturation current=IC_sat=(VCC-V_knee)/RC;         \n",
-    "VCE=V_knee;                    #Collector-emitter voltage, V\n",
-    "\n",
-    "print(\"(ii) The collector emitter voltage at saturation=%.1fV.\"%VCE);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.3 : Page number 475-476"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Minimum β=19.4.\n",
-      "(ii) The transistor will not be saturated.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10;             #Supply voltage, V\n",
-    "RC=1;               #Collector resistor, kΩ\n",
-    "VBB=2;              #Supply voltage to base, V\n",
-    "RB=2.7;             #Base resistor, kΩ\n",
-    "V_knee=0.7;         #Knee voltage, V\n",
-    "VBE=0.7;            #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IB=round((VBB-VBE)/RB,2);           #Base current, mA\n",
-    "Ic_sat=(VCC-V_knee)/RC;             #Collector saturation current, mA\n",
-    "beta_min=Ic_sat/IB;                 #Minimum value of base current amplification factor\n",
-    "print(\"(i) Minimum β=%.1f.\"%beta_min);\n",
-    "\n",
-    "#(ii)\n",
-    "VBB=1;                       #Supply voltage to base(changed), V\n",
-    "beta=50;                     #Base current amplification factor\n",
-    "IB=(VBB-VBE)/RB;             #Base current, mA\n",
-    "IC=beta*IB;                  #Collector current,mA\n",
-    "\n",
-    "if(IC<Ic_sat):\n",
-    "    print(\"(ii) The transistor will not be saturated.\");\n",
-    "else:\n",
-    "    print(\"(ii) The transistor will be saturated.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.4 : Page number 480"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Time period of the square wave=0.14 m sec.\n",
-      "Time frequency of the square wave=7 kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R2=10;                  #Resistor R2, kΩ\n",
-    "R3=10;                  #Resistor R3, kΩ\n",
-    "C1=0.01;                #Capacitor of 1st transistor, μF\n",
-    "C2=0.01;                #Capacitor of 2nd transistor, μF\n",
-    "\n",
-    "#Calculation\n",
-    "R=R2*1000;                   #Resistance, Ω\n",
-    "C=C1*10**-6;                 #Capacitance, F\n",
-    "T=round((1.4*R*C)*1000,2);   #Time period,m sec\n",
-    "f=1/(T*10**-3);              #Frequency, Hz\n",
-    "f=f/1000;                    #Frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Time period of the square wave=%.2f m sec.\"%T);\n",
-    "print(\"Time frequency of the square wave=%d kHz.\"%f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.6 : Page number 485"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output voltage=0.55V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=10;               #Resistance in differentiating circuit, kΩ\n",
-    "C=2.2;              #Capacitance in differentiating circuit, μF\n",
-    "d_ei=10;            #Change in input voltage, V\n",
-    "dt=0.4;             #Time in which change occurs, s\n",
-    "\n",
-    "#Calculation\n",
-    "eo=R*1000*C*10**-6*d_ei/dt\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output voltage=%.2fV.\"%eo);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.7 : Page number 489"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The peak output voltage=11.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vin_peak=12;                #Peak value of input voltage, V\n",
-    "V_D=0.7;                    #Forward bias voltage of diode, V\n",
-    "\n",
-    "#Calculation\n",
-    "Vout_peak=Vin_peak-V_D;         #Peak value of output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The peak output voltage=%.1fV.\"%Vout_peak);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.8 : Page number 489"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The peak output voltage=8V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vin_peak=10;                #Peak value of input voltage, V\n",
-    "R=1;                        #Input resistor, kΩ\n",
-    "RL=4;                       #Load resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "Vout_peak=(Vin_peak*RL)/(R+RL);         #Peak output voltage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The peak output voltage=%dV.\"%Vout_peak);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.9 : Page number 490"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The diode will be forward biased for the negative half-cycle of input signal.\n",
-      "The output voltage=-0.7V.\n",
-      "The voltage across R=-9.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vin=-10;                #Input voltage, V\n",
-    "V_D=0.7;                #Forward bias voltage of the diode, V\n",
-    "R=1;                    #Resistance, kΩ\n",
-    "\n",
-    "\n",
-    "print(\"The diode will be forward biased for the negative half-cycle of input signal.\");\n",
-    "Vout=-V_D;              #Output voltage, V\n",
-    "V_R=Vin-(-V_D);         #Voltage across resistor R, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output voltage=%.1fV.\"%Vout);\n",
-    "print(\"The voltage across R=%.1fV.\"%V_R);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.10 : Page number 490-491"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "During the positive half cycle, the diode is foward biased and can be replaced by battery of 0.7V.\n",
-      "Therefore, Vout=0.7V.\n",
-      "During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\n",
-      "Therefore, Vout_peak=-8.33V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_F=0.7;                        #Forward bias voltage of diode, V\n",
-    "R=200.0;                          #Input resistor of the circuit,  Ω\n",
-    "RL=1.0;                           #Load resistor, kΩ\n",
-    "Vin_peak=10.0;                    #Peak input voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#Positive half-cycle:\n",
-    "print(\"During the positive half cycle, the diode is foward biased and can be replaced by battery of %.1fV.\"%V_F);\n",
-    "print(\"Therefore, Vout=%.1fV.\"%V_F);\n",
-    "\n",
-    "#Negative half-cycle:\n",
-    "print(\"During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\");\n",
-    "Vout_peak=RL*(-Vin_peak)/(R/1000+RL);\n",
-    "print(\"Therefore, Vout_peak=%.2fV.\"%Vout_peak);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.12 : Page number 491"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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BuYK3cmUodc6aBU8+CY0bx47IufzJ2QhuSQ0k9ZT0APAMMA04ogYx5szixaF7bMeOcNll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\nQmCC\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fd5e441dfd0>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "from math import sin\n",
-    "from math import pi\n",
-    "\n",
-    "V_biasing=10.0;                        #Biasing voltage, V\n",
-    "vin=[30*sin(t/10.0) for t in range(0,(int)(2*pi*10))]      #input voltage  waveform, V\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.plot(vin);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "vout=[];                             #Output voltage waveform, V\n",
-    "for v in vin[:]:\n",
-    "    if(v-V_biasing)>0 :              #Diode is forward biased.\n",
-    "        vout.append(v-V_biasing);\n",
-    "    else:                            #Diode is reverse biased.\n",
-    "        vout.append(0);\n",
-    "\n",
-    "plt.subplot(212)        \n",
-    "plt.plot(vout);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.13 : Page number 492"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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ZldfuMY6rJX1X0h6S1ii86eaSPiPpBmBCq29uZmbN0+tVVZIOAI4AdgNGAW+Q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\n5CKPkAAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fd5e4324828>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "t1=50;         #Assumed time interval, s\n",
-    "t2=100;         #Assumed time interval, s\n",
-    "V_biasing=10;  #Biasing voltage, V\n",
-    "for t in range(0,151):                  #time interval from 0s to 151s\n",
-    "    if(t<=t1):                      \n",
-    "        Vin.append(15);               #Value of input voltage for time 0 to t1 seconds \n",
-    "    elif(t<=t2 and t>t1):\n",
-    "        Vin.append(-30);             #Value of input voltage for time t1 to t2 seconds\n",
-    "    else :\n",
-    "        Vin.append(15);             #Value of input voltage after t2 seconds\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.plot(Vin);\n",
-    "plt.xlim([0,160])\n",
-    "plt.ylim([-35,20])\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=0):\n",
-    "        vout.append(0);               #Diode reverse biased\n",
-    "    else:\n",
-    "        vout.append(v-V_biasing);      #Diode forward biased\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout);\n",
-    "plt.xlim(0,160)\n",
-    "plt.ylim(-35,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.14 : Page number 492-493"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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2yMNXAZA0RtLXJN0KjG1W4SSNzaOvZuRncZuVnofP2mDUa80iIvaWtD9wFLCb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\ng4WZmdXkYGFmZjU5WJiZWU3/Hz1pk1JFdhIyAAAAAElFTkSuQmCC\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fd5e42e1668>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "t1=50;         #Assumed time interval, s\n",
-    "t2=100;         #Assumed time interval, s\n",
-    "V_biasing=5;  #Biasing voltage, V\n",
-    "for t in range(0,151):                  #time interval from 0s to 151s\n",
-    "    if(t<=t1):                      \n",
-    "        Vin.append(10);               #Value of input voltage for time 0 to t1 seconds \n",
-    "    elif(t<=t2 and t>t1):\n",
-    "        Vin.append(-10);             #Value of input voltage for time t1 to t2 seconds\n",
-    "    else :\n",
-    "        Vin.append(0);\n",
-    "\n",
-    "plt.subplot(211)        \n",
-    "plt.plot(Vin);\n",
-    "plt.xlim(0,101)\n",
-    "plt.ylim(-20,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=0):\n",
-    "        vout.append(v);               #Diode reverse biased\n",
-    "    else:\n",
-    "        vout.append(v-V_biasing);      #Diode forward biased\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout);\n",
-    "plt.xlim(0,101)\n",
-    "plt.ylim(-20,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.15 : Page number 493"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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xH0DSNGAiMLvmnInAj7NY7pM0StKYiFjWoJjMcrkGYVVT7zDXX9WzbwC2BhbW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\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fd5e42367b8>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "t1=50;         #Assumed time interval, s\n",
-    "t2=100;         #Assumed time interval, s\n",
-    "V_D1=0.6;          #Forward Biasing voltage of the 1st diode, V\n",
-    "V_D2=0.6;          #Forward Biasing voltage of the 2nd diode, V\n",
-    "for t in range(0,151):                  #time interval from 0s to 151s\n",
-    "    if(t<=t1):                      \n",
-    "        Vin.append(10);               #Value of input voltage for time 0 to t1 seconds \n",
-    "    elif(t<=t2 and t>t1):\n",
-    "        Vin.append(-10);             #Value of input voltage for time t1 to t2 seconds\n",
-    "    else :\n",
-    "        Vin.append(0);\n",
-    "\n",
-    "plt.subplot(211);\n",
-    "plt.plot(Vin);\n",
-    "plt.xlim(0,110)\n",
-    "plt.ylim(-20,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=0):\n",
-    "        vout.append(-V_D1);               #Diode D1 forward biased, \n",
-    "    else:\n",
-    "        vout.append(V_D2);      #Diode D2 forward biased\n",
-    "\n",
-    "plt.subplot(212)        \n",
-    "plt.plot(vout);\n",
-    "plt.xlim(0,110)\n",
-    "plt.ylim(-1,1)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.16 : Page number 493-494"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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XvVPSHdHt44HfAi8Rupx2kSTCbMYxwFaSviO0mpB0lKQPJC2Nxpp2zbju59Hr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\nTQAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fd5e4244470>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "import matplotlib.pyplot as plt\n",
-    "from math import sin\n",
-    "from math import pi\n",
-    "\n",
-    "VZ=20;                   #Assumed zener voltage, V\n",
-    "VF=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "for t in range(0,(int)(2*pi*10)):                  #time interval from 0s to 151s\n",
-    "    Vin.append(30*sin(t/10.0));\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.plot(Vin);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=-VF):\n",
-    "        vout.append(-VF);               #Zener diode forward biased, \n",
-    "    elif(v>=VZ):\n",
-    "        vout.append(VZ);        #Input voltage exceeds zener voltage\n",
-    "    else:\n",
-    "        vout.append(v);           #Zener diode reverse biased\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout);\n",
-    "plt.xlim([0,80])\n",
-    "plt.ylim([-1,40])\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.17 : Page number 494"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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2Bx7Fu5x2lSR8NuM4oJWkz/BWE5J6SnpN0tJkrGm3gud9O3m9V4AvJDVMzp2d\njCl8LulWSVtKejT5f4+rgf74jubdID3wbswDqMHfnyJrBLQHbkz+D8vw3oy8xA+ApMZ4OaWS37VK\nxZ/XZPEe/kdcYq0L9zJusaStAORVd/+bcjxrlXyzvw8YYmajktM1Gf9PgKZ4X+rXzGwZ/iG5rgWe\nltz3RGABcISZbWRmVxfcpxOwI94M3xr4HbAYWK+M+CvyfCWmJM8NcCAwL/kX4CBgcnJdwJ/x5n5b\n/Hd2UHLbv/FxlPWTWBoAvYGhye2X47/vX+J9/zsBz+At6p/h3XW7AG/Lp6MPA04DtgDG4MmusGXW\nFy/SuYmZrU7OHQMcmjxPT/w9Pw/YHGiYPF+Vmdn7yb8f4t2YHcjP7/9C4F0zeyE5vh9PHnmJv0R3\n4EUzW5IcVyr+vCaL54Gdkm+NTfBf/tEpx1QRSi4lRuN99AD9gFGlH5AhtwMzzey6gnM1Gf/meHfK\nmjJuez+5vaLKWtl/tZn9D/8w/xjvWhgN7Jzcvq7411UpYAqeFMCTxF8Kjg9KbsfM5pnZ42a2ysw+\nwruVDkpuWwC8BBydPO5QYJmZPZ/8MbcBNjez7fFuhJeBFcBDQLdS8fcBHjaziUkiuBpohifjEteZ\n2SLzgf8S15vZkuRD/UngWTN71cy+whP43ut4D9ZJUvOkVUqSELsAM8jJ73/SVfNukojBfz6vk5P4\nCxwPDC84rlT8udzu3MxWS/od3gRvANyW9NlmlqRh+DfQzSQtwEuyXwGMlDQAX5jYJ70I105SR+Dn\nwIykW8WAC4DBwIgain8JsLmkBmUkjK2T26vKgNuTrpsG+IdtQzz+XwCn4B/WffCZLpUxFdhF0pbJ\nY48ELpW0Gf7t+QmA5PbrgAOADZLXX1rwPMPxP+Z7kn9L+pW3AxoD73v4NMZbQ7Pw359x+PtzaBL/\nn/Gfhf/HzSzpVisc0ytr1tTigusryjjeoNx3Yu22Ah6Q13JrBAw1s3GSXqDmfn+K7TRgaNKV8xZw\nEv4zzEX8yRhLZ+BXBacr9feby2QBYGaPAbumHUdFmdnP1nJT51oNpArM7Gn8D6MsNRX/VLyb5Ri8\nuwv4epykO94lAt5f3LzgcVuXep6y+l0FHGdmc5LnvAKfmbNU0n+A/5nZ2cltFXm+b240WyHpRXxa\n62tmtkrSVOBM4E0zK0kIf8a7kPYws08l9QKuL3iqkcDVkrbBWxj7JeffBf6XxPudWJJJAUPMrEty\nvAj4fqnyGrvRAAAW6klEQVS7bcu3E0St9q2b2dv4pIjS55eSg99/ADN7BZ9BV1pe4l+Od0sWnqvU\n+5/XbqhQx5jZZ8AfgesldZXUSD7z6l583OCe5K7TgR7yBUUt8Q/pQh8AO5TxEhdLapYMbJ+EjxNU\n5/kKPYGPgUxJjieXOgbYEB9c/zxJCOcUPkHSjzwFX4/zlpnNTs5/gLcerpW0odwOkg6kbCOAwyUd\nnLyHZ+PJZupa7h9ChUSyCJlhZlfh3VtXA5/iH3Dzgc5mtjK52xB8HcY7wGN886Ff4go8MSyVdGbB\n+SnAm/hahSvN7PFqPl+hKXg3zROljguTxaX4gqhP8LGG+8t4nmF4d9LQUudPBJrgC/KW4q2QlpQh\naT2dgC9a/RA4HDjSfDMxKLtVkatZPSEdmagNlcz+eAFYaGY9JbXAv1G2wf+I+5jZpymGGHJKUhu8\nj7nxWgbPQwgVkJWWRZllDCzjy+hDbsS+JyFUU+rJQlJrfKHOvwpO9wLuSq7fhe93EUJVpd98DiHn\nUk8WrKWMQQ6X0YcMMrP5ZtYwuqBCqJ5Up85KOhxYbGbTJXVax13L/Gao2IM7hBCqxCq5LXXaLYuO\nQE9Jb+GLkg6RNAT4oKLL0ItVnbEmLwMHDkw9hogz4sxznHmIMU9xVkWqycLMLjCz7cxsB7xkx0Qz\n+wU+tbB/crc8LKMPIYQ6Le2WxdpcARwmaTY+7/yKlOMJIYR6LTPlPsxsCt8UXctNGYCK6NSpU9oh\nVEjEWbMizpqThxghP3FWRSYW5VWVJMtz/CGEkAZJWM4GuEMIIeRAJIsQQgjlimQRQgihXKkmC0lN\nJT0r6WVJMyQNTM63SPb9nS1pbA3s/xtCCKEaUh/gltTczJZLagg8je9I9VPgIzO7UtK5QAszO6+M\nx8YAdwghVFIuB7jNd3ACaIpP5TWikGAIIWRK6slCUoNkX+cPgPFm9jxRSDCEEDIl9UV55tVA95a0\nEb6p+x5UYueuQYMGfX29U6dOdXpRTAghVMXkyZOZPHlytZ4j9TGLQpIuBpYDJwOdzGxxUkhwkpm1\nLeP+MWYRQgiVlLsxC0mbl8x0ktQMOAyYBYwmCgmGEEJmpNqykPQDfAC7QXK518wul7QpMALYFpiP\n78H9SRmPj5ZFCCFUUlVaFpnqhqqsSBYhhFB5ueuGCiGEkA+RLEIIIZQrkkUIIYRypT0bqrWkiZJe\nT2pDnZacj9pQIYSQIWnPhmoJtDSz6ZI2AF7ES32cRNSGCiGEosjdALeZfWBm05PrX+BrLFoTtaFC\nCCFTUi/3UULS94C9gGmUqg0lKZO1oV59FW69Fd57L+1I0tGgAWyzDeywg1923BHatgVV6vtKCCEP\nMpEski6o+4DTzewLSRWuDVXbvvoK7r8fbrwR3nkHfvUrOPjgtKNKx6pVsHAhzJ0LY8fCa6/B7rvD\n7bdDq1ZpRxdCqEmpJwtJjfBEMcTMSsp6LJa0VUFtqP+u7fG1WUhw4kTo3x923hnOPBN69oRGqb+D\n2bFyJVx+ObRv78n0pz9NO6IQAtSRQoKS7gaWmNmZBecGA0vNbHAWBrhXroSBA+HOO+GOO6Br16K/\nZK49+yyccAJ07Ah//ztstFHaEYUQCuVugFtSR+DnwCHJ1qovSeoGDAYOkzQbOBS4Iq0Y33oLDjgA\npk/3SySK8u27L7z8MjRpAvvs4+9bCCHfUm9ZVEexWxb/+Q/8+tdwwQVw2mk+oBsqZ/hwf+8uu8zH\nd2LwO4T0RSHBGrJyJZx/Ptx3n19++MMaf4l6ZfZs6N0bfvAD+Oc/YYMN0o4ohPotd91QWfT++3Do\nofD66/Dii5EoasKuu8K0adCsGXToAHPmpB1RCKGyIlkUePppTw6dO8Mjj8Bmm6UdUd3RvDn8619w\nxhmw//7w8MNpRxRCqIzohkrcdpt3Pd15J/ToUSNPGdZi6lTvlioZD4qxoBBqVy7HLCTdBhwBLDaz\ndsm5FsC9QBvgHXynvE/LeGy1k8WqVXDWWfDYYzB6tHeZhOJbtAiOPRZatoQhQ2D99dOOKIT6I69j\nFncApSekngdMMLNdgYnA+cV44aVLoXt3H4B99tlIFLWpVSuYNAk22cS7pRYuTDuiEMK6lJssJP1Y\n0o2SXpX0oaQFkh6V9NuaKB1uZk8BH5c6XfRCgnPnwn77Qbt2Pj6xySY1/QqhPE2bevffz37mP4vn\nn087ohDC2qwzWUgaA5wMjAW6AVsDuwMXAesBoyT1LEJcWxYWEgRqtJDgk0/6Qruzz4ZrroGGDWvy\n2UNlSHDOOV4e5PDDYeTItCMKIZSlvMpGvzCzJaXOfQG8lFyukbR5USL7trUOTFS2NtSQIT5GMXQo\nHHZYTYUXqqtXL2jTxv996y34wx9iAV8INaXotaEk3QgMM7Onq/Uq5QUhtQEeKhjgngV0KigkOMnM\n2pbxuAoPcJvBoEGeLB5+2KujhuxZtMhbGB06eGsjCjWGUPOKMcA9B7ha0juSrpS0d9XDWycllxKj\ngf7J9X7AqNIPqIyVK2HAABgzxqdtRqLIrlat4IknYMECOPJI+PzztCMKIUAFp84m3/z7JpdmwHBg\nuJlVey2upGFAJ2AzYDEwEHgQGAlsC8zHp85+UsZjy21ZfPaZT9Fcbz2vUxRTNPNh1Sr47W99ltqj\nj8b+GCHUpFpZZ5G0Lm4H2plZqkPD5SWL997zLo2f/MRLZUeXRr6YwRVXeD2pMWN8F74QQvUVbZ2F\npEaSjpQ0FBgDzAaOqUKMtWbWLN9P4bjjou87ryRfVf/HP/puhE8XdeQshLAu5Q1wHwYcD/QAngP+\nDYwys2W1E966ra1l8cwzcMwxcNVV8ItfpBBYqHFjx/rP8pZb4KgaX3UTQv1S491Qkibi4xP3mVnp\nhXOpKytZjB4NJ5/ss55io6K65cUXfdD7kku8rlQIoWqKkSw2NLN1zkeRtIGZfVGZF60ppZPFrbf6\n9qejR0dp8bpq3jz/EnDCCf6zjrUYIVReMcYsHpR0jaQDJX09j0jSDpJ+KalkZXdRSOom6Q1Jc5K9\nuMtkBn/6kw+GPvFEJIq6bMcdfezioYe8dbF6ddoRhVA/lDsbSlIPfJ/sjkALYBU+wP0o8K+kHEfN\nByY1wNd5HAosAp4H+prZGwX3sVWrjNNP9w+QMWO8immo+z7/HI4+GjbaCIYN86nRIYSKyWWJ8rWR\ntB8w0My6J8fnAWZmgwvuY717G0uWwIMP+gdHqD++/BL69/dV36NGRTHIECqqmFNnH6/IuRq2DfBu\nwfHC5Ny3mPmirUgU9U/Tpl7ja8894aCDfEvcEEJxrHP1gaT1gObA5smGRCWZaCPK+OBOQ9u2g7ji\nCr9ekUKCoW5p0ACuu87Hqzp29Cm2O++cdlQhZEttFBI8Hfg90AofNyjxGXCrmd1QrVdfV2DeDTXI\nzLolx2V2Q2W1Gy3Uvn/9Cy6+2Ae/Y5JDCGtXtDELSaea2fVVjqwKJDXEB9IPBd7HFwUeb2azCu4T\nySJ8y6hRvs5m6FDo0iXtaELIpqoki4oWwfhU0omlT5rZ3ZV5scows9WSfgeMw8dWbitMFCGUpVcv\n2Gwz+OlP4dprfRe+EEL1VbRlUdiqWA//tv+SmR1brMAqIloWYW1eew169IAzzvBLCOEbtTZ1VtIm\nwL9LxhPSEskirMuCBdCtGxxxhA+AN6jQ3L8Q6r6iTZ0twzJg+yo+NoRasd12vt/600/DiSfCV1+l\nHVGoCatXwzXXwLJMlDOtPyq6zuIhSaOTyyP4wPMDxQ0thOrbbDOYMME/WHr08M2wQn6tWAG9e/vW\nyKtWpR1N/VLRMYuDCg5XAfPNbGG1Xlg6FhgEtAV+ZGYvFdx2PjAgea3TzWzcWp4juqFChaxeDaee\n6uXrY+e9fProI+jZE9q0gTvu8EWZoWqK1g1lZlOAN4AN8fpQNdGgnwEcDUwpPCmpLdAHTyLdgZuk\nqC0aqqdhQ98Eq08f3znx9dfTjihUxttv+6LL/feHe+6JRJGGinZD9cHXOfTGP8ifTVoGVWZms81s\nLt+sCi/RCx88X2Vm7wBzgQ7Vea0QwMuZX3CBVyg++GCYODHtiEJFvPCCJ4nf/Q4GD46JCmmp6DqL\nC/Guov8CSNoCmADcV4SYtgGmFhy/R0ZKi4S64Re/gG239S13r7wS+vVLO6KwNg88AL/6la/O79Ur\n7Wjqt4omiwYliSLxERVolUgaD2xVeAow4EIze6jCUYZQwzp1gsmT4fDD4a23YNCg2EgpS8zgr3/1\nhZWPPQb77JN2RKGiyeKxZKOj4cnxcfh+FutkZodVIab3gG0Ljlsn58o0aNCgr69HIcFQGW3bwtSp\nPmg6dy7cdhs0a5Z2VGHVKp+M8PTTPiFhu+3Sjij/aqOQ4I3AMDN7WtIxwP7JTU+aWY1MnZU0CTjb\nzF5MjncHhgL74t1P44Gdy5r2FLOhQk1YsQIGDPAWxoMPwtZbpx1R/fXxxz4JoVEjuPfe2HqgWIox\nG2oOcLWkd4D9gCFmdmZNJApJR0l6N3nehyWNATCzmcAIYCbeejklMkIopmbNfLe9I46AffeFl19O\nO6L66Y03/P1v184rB0eiyJaKrrNoA/RNLs3w7qjhZjanuOGVG1fkkVCjRo6EU06Bm2+GY1OtfFa/\njBnjEw0GD4aTTko7mrqvVmpDSdobuB1oZ2YNK/XgGhbJIhTDiy961dq+feHyy32NRigOM7j6ah/I\nHjnS11KE4ivmfhaN8AVyffGKs5PxlsWoKsRZYyJZhGL58ENPFo0aeRfVZpulHVHd8/nn3opYsADu\nuy8GsmtTjY9ZSDpM0u34/tf/D3gE2NHM+qadKEIopi228C1a27WDH/0Ipk9PO6K6ZdYs6NABNt/c\niz1Gosi+8mZDTQSGAfeb2ce1FlUFRcsi1IZ77/XVw5dd5gvEYj1G9YwYAb/9rS+IjPGJdNTafhY1\nQdKVwJHAl8A84CQz+yy5LQoJhkyZPdundLZtC7fcEjN1qmL5ct+I6vHHPQHHQrv01OZ+FjVhHLCH\nme2F1386H75eZxGFBEOm7LorTJsGm2wC7dvDSy+V/5jwjddf926nL77w9y4SRf6klizMbIKZrUkO\np+ErtQF6EoUEQwY1awb/+Id3R3Xt6rvvrV6ddlTZtmqVV/vt1AnOPtsrxkarLJ+yUr9xAN+UD9kG\neLfgtigkGDKlb1+vhDp2LBx4IMybl3ZE2WPmRQDbtfMupyefhP79Y7wnzypaG6pKKlJIUNKFwEoz\nG17GU5QrakOFNLRp433v113nq44vuQT23BMaN4YmTfzf+lpK+7334NJLfYzi6quhe/dIEmkrem2o\nYpPUH5+Se4iZfZmcOw8wMxucHD8GDDSzZ8t4fAxwh9TNnAkXXug7ua1c6ZevvvJv1/VR8+Zw2mlw\n/PH1N2FmXd5mQ3UDrgEONLOPCs5HIcEQQiiiqiSLonZDleN6oAkwPpnsNM3MTjGzmZJKCgmuJAoJ\nhhBC6lLthqquaFmEEELl5W2dRQghhJyIZBFCCKFckSxCCCGUK5JFCCGEcqWWLCT9UdIrkl6W9Jik\nlgW3nS9prqRZkrqkFWMIIQSX5jqLDczsi+T6qcDuZvabgnUWP8LrRU0g1lmEEEKNydVsqJJEkVgf\nKCkqGIUEQwghY9JclIeky4ATgU+Ag5PT2wBTC+4WhQRDCCFlqRYSNLOLgIsknQucCgyq7GtEIcEQ\nQli33BcS/DoIaVvgETNrF4UEQwihuHI1ZiFpp4LDo4A3kuujgb6SmkjaHtgJeK624wshhPCNNMcs\nrpC0Cz6wPR/4NUAUEgwhhOzJRDdUVUU3VAghVF6uuqFCCCHkRySLEEII5YpkEUIIoVypJwtJZ0la\nI2nTgnNRGyqEEDIk1WQhqTVwGD4bquRcW6AP0BboDtykZN/VvKruYpjaEnHWrIiz5uQhRshPnFWR\ndsviWuCcUud6UcdqQ+XlFyjirFkRZ83JQ4yQnzirIs1FeT2Bd81sRqmbtgHeLTiO2lAhhJCytGpD\nXQRcgHdBhRBCyLhUFuVJ+j6+T8VyPIG0xlsQHYABAGZ2RXLfddaGqq2YQwihLqnsorxMrOCW9DbQ\n3sw+Ltj8aF+8+2k8a9n8KIQQQu1IdT+LAoa3MKI2VAghZFAmWhYhhBCyLe2ps1UmqZukNyTNSTZP\nygRJt0laLOnVgnMtJI2TNFvSWEkbpxxja0kTJb0uaYak0zIaZ1NJz0p6OYlzYBbjLCGpgaSXJI1O\njjMXp6R3JL2SvKfPZTjOjSWNTBbmvi5p36zFKWmX5H18Kfn3U0mnZTDOMyS9JulVSUOT7R8qHWMu\nk4WkBsANQFdgD+B4SbulG9XX7sDjKnQeMMHMdgUmAufXelTftgo408z2AH4M/DZ5/zIVp5l9CRxs\nZnsDewHdJXUgY3EWOB3vPi2RxTjXAJ3MbG8zK1m/lMU4rwMeNbO2wJ74fjeZitPM5iTvY3tgH2AZ\n8AAZilNSK3wX0vZm1g4feji+SjGaWe4uwH7AmILj84Bz046rIJ42wKsFx28AWyXXWwJvpB1jqXgf\nBDpnOU6gOfAC8KMsxonP6BsPdAJGZ/XnDrwNbFbqXKbiBDYC5pVxPlNxloqtC/Bk1uIEWuEVMlok\niWJ0Vf/Wc9my4LsL9xaS7YV7W5rZYgAz+wDYMuV4vibpe/i39mn4L0+m4ky6dl4GPgDGm9nzZDBO\nvqlGUDgImMU4DRgv6XlJJyfnshbn9sASSXckXTy3SGpO9uIsdBwwLLmemTjNbBFwDbAAX57wqZlN\nqEqMeU0WeZeJWQWSNgDuA043sy/4blypx2lma8y7oVoDHSTtQcbilHQ4sNjMppPM6luL1N9PoKN5\nt0kPvPvxADL2fuLfgNsDNyaxLsN7D7IWJwCSGgM9gZHJqczEKWkTvIRSG7yVsb6kn5cRU7kx5jVZ\nvAdsV3BcsqgvqxZL2gpAUkvgvynHg6RGeKIYYmajktOZi7OEmX0GTAa6kb04OwI9Jb0FDAcOkTQE\n+CBjcWJm7yf/foh3P3Yge+/nQrwU0AvJ8f148shanCW6Ay+a2ZLkOEtxdgbeMrOlZrYaH1P5SVVi\nzGuyeB7YSVIbSU2AvnhfXFaIb3/DHA30T673A0aVfkAKbgdmmtl1BecyFaekzUtmaUhqhpeHmUXG\n4jSzC8xsOzPbAf9dnGhmvwAeIkNxSmqetCaRtD7ezz6D7L2fi4F3Je2SnDoUeJ2MxVngePxLQoks\nxbkA2E/SepKEv5czqUqMaQ8MVWPgphswG69Ke17a8RTENQxYBHyZ/KBOwgeXJiTxjgM2STnGjsBq\nYDrwMvBS8n5umrE4f5DENh14FbgwOZ+pOEvFfBDfDHBnKk58LKDkZz6j5O8ma3EmMe2JfymcDvwH\n2DijcTYHPgQ2LDiXqTiBgfiXrFeBu4DGVYkxFuWFEEIoV167oUIIIdSiSBYhhBDKFckihBBCuSJZ\nhBBCKFckixBCCOWKZBFCCKFckSxCqKSkfPZv0o4jhNoUySKEymsBnFKROya1eULIvUgWIVTeX4Ad\nkoqog8u574OSHpR0pKSGtRFcCMUQK7hDqCRJbYCHzDeTqcj9DwR+ie/DMhK4w8zmFTHEEGpctCxC\nKDIze8LM+gE/TE69IenoNGMKobIapR1ACHkm6TLgcHw/gB8CLybXR5vZoOQ+6wFHAwPwgnin4rvq\nhZAb0Q0VQiVJ2hTfv2D7Ctx3MHAs8Ahwm5m9Uuz4QiiGSBYhVIGke4B2+F7w567jft3w/S2+qrXg\nQiiCSBYhhBDKFQPcIYQQyhXJIoQQQrkiWYQQQihXJIsQQgjlimQRQgihXJEsQgghlCuSRQghhHJF\nsgghhFCu/w+ip4Q9s46W/wAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fd5e4362cf8>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "import matplotlib.pyplot as plt\n",
-    "from math import sin\n",
-    "from math import pi\n",
-    "\n",
-    "VZ1=20;                   #Assumed zener voltage, V\n",
-    "VF1=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-    "VZ2=20;                   #Assumed zener voltage, V\n",
-    "VF2=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "for t in range(0,(int)(2*pi*10)):                  #time interval from 0s to 151s\n",
-    "    Vin.append(30*sin(t/10.0));\n",
-    "    \n",
-    "plt.subplot(211)\n",
-    "plt.plot(Vin);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=-(VZ1+VF2)):\n",
-    "        vout.append(-(VZ1+VF2));               #Zener diode forward biased, \n",
-    "    elif(v>=VZ2+VF1):\n",
-    "        vout.append(VZ2+VF1);        #Input voltage exceeds zener voltage\n",
-    "    else:\n",
-    "        vout.append(v);           #Zener diode reverse biased\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout); \n",
-    "plt.xlim([0,80])\n",
-    "plt.ylim([-40,40])\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_3.ipynb
deleted file mode 100755
index 9d2e921c..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_3.ipynb
+++ /dev/null
@@ -1,804 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 18 : SOLID-STATE SWITCHING CIRCUITS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.1 : Page number 472"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Input voltage required to saturate the transistor switch=5.4V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10;             #Supply voltage, V\n",
-    "RC=1.0;               #Collector resistor, kΩ\n",
-    "RB=47.0;              #Base resistor, kΩ\n",
-    "beta=100.0;           #Base current amplification factor\n",
-    "VBE=0.7;            #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "IC_sat=VCC/RC;              #Collector saturation current, mA\n",
-    "IB=IC_sat/beta;             #Base current, mA\n",
-    "V=IB*RB+VBE;                #Input voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Input voltage required to saturate the transistor switch=%.1fV.\"%V);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.2 : Page number 475"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The collector emitter voltage at cut-off=9.99V.\n",
-      "(ii) The collector emitter voltage at saturation=0.7V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10;             #Supply voltage, V\n",
-    "RC=1.0;               #Collector resistor, kΩ\n",
-    "ICBO=10.0;            #Collector leakage current, μA\n",
-    "V_knee=0.7;         #Knee voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IC=ICBO;                    #Collector current, μA\n",
-    "VCE=VCC-(ICBO/1000)*RC;            #Collector-emitter voltage, V\n",
-    "\n",
-    "print(\"(i) The collector emitter voltage at cut-off=%.2fV.\"%VCE);\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, saturation current=IC_sat=(VCC-V_knee)/RC;         \n",
-    "VCE=V_knee;                    #Collector-emitter voltage, V\n",
-    "\n",
-    "print(\"(ii) The collector emitter voltage at saturation=%.1fV.\"%VCE);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.3 : Page number 475-476"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Minimum β=19.4.\n",
-      "(ii) The transistor will not be saturated.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10;             #Supply voltage, V\n",
-    "RC=1;               #Collector resistor, kΩ\n",
-    "VBB=2;              #Supply voltage to base, V\n",
-    "RB=2.7;             #Base resistor, kΩ\n",
-    "V_knee=0.7;         #Knee voltage, V\n",
-    "VBE=0.7;            #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IB=round((VBB-VBE)/RB,2);           #Base current, mA\n",
-    "Ic_sat=(VCC-V_knee)/RC;             #Collector saturation current, mA\n",
-    "beta_min=Ic_sat/IB;                 #Minimum value of base current amplification factor\n",
-    "print(\"(i) Minimum β=%.1f.\"%beta_min);\n",
-    "\n",
-    "#(ii)\n",
-    "VBB=1;                       #Supply voltage to base(changed), V\n",
-    "beta=50;                     #Base current amplification factor\n",
-    "IB=(VBB-VBE)/RB;             #Base current, mA\n",
-    "IC=beta*IB;                  #Collector current,mA\n",
-    "\n",
-    "if(IC<Ic_sat):\n",
-    "    print(\"(ii) The transistor will not be saturated.\");\n",
-    "else:\n",
-    "    print(\"(ii) The transistor will be saturated.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.4 : Page number 480"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Time period of the square wave=0.14 m sec.\n",
-      "Time frequency of the square wave=7 kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R2=10;                  #Resistor R2, kΩ\n",
-    "R3=10;                  #Resistor R3, kΩ\n",
-    "C1=0.01;                #Capacitor of 1st transistor, μF\n",
-    "C2=0.01;                #Capacitor of 2nd transistor, μF\n",
-    "\n",
-    "#Calculation\n",
-    "R=R2*1000;                   #Resistance, Ω\n",
-    "C=C1*10**-6;                 #Capacitance, F\n",
-    "T=round((1.4*R*C)*1000,2);   #Time period,m sec\n",
-    "f=1/(T*10**-3);              #Frequency, Hz\n",
-    "f=f/1000;                    #Frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Time period of the square wave=%.2f m sec.\"%T);\n",
-    "print(\"Time frequency of the square wave=%d kHz.\"%f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.6 : Page number 485"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output voltage=0.55V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=10;               #Resistance in differentiating circuit, kΩ\n",
-    "C=2.2;              #Capacitance in differentiating circuit, μF\n",
-    "d_ei=10;            #Change in input voltage, V\n",
-    "dt=0.4;             #Time in which change occurs, s\n",
-    "\n",
-    "#Calculation\n",
-    "eo=R*1000*C*10**-6*d_ei/dt\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output voltage=%.2fV.\"%eo);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.7 : Page number 489"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The peak output voltage=11.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vin_peak=12;                #Peak value of input voltage, V\n",
-    "V_D=0.7;                    #Forward bias voltage of diode, V\n",
-    "\n",
-    "#Calculation\n",
-    "Vout_peak=Vin_peak-V_D;         #Peak value of output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The peak output voltage=%.1fV.\"%Vout_peak);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.8 : Page number 489"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The peak output voltage=8V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vin_peak=10;                #Peak value of input voltage, V\n",
-    "R=1;                        #Input resistor, kΩ\n",
-    "RL=4;                       #Load resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "Vout_peak=(Vin_peak*RL)/(R+RL);         #Peak output voltage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The peak output voltage=%dV.\"%Vout_peak);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.9 : Page number 490"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The diode will be forward biased for the negative half-cycle of input signal.\n",
-      "The output voltage=-0.7V.\n",
-      "The voltage across R=-9.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vin=-10;                #Input voltage, V\n",
-    "V_D=0.7;                #Forward bias voltage of the diode, V\n",
-    "R=1;                    #Resistance, kΩ\n",
-    "\n",
-    "\n",
-    "print(\"The diode will be forward biased for the negative half-cycle of input signal.\");\n",
-    "Vout=-V_D;              #Output voltage, V\n",
-    "V_R=Vin-(-V_D);         #Voltage across resistor R, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output voltage=%.1fV.\"%Vout);\n",
-    "print(\"The voltage across R=%.1fV.\"%V_R);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.10 : Page number 490-491"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "During the positive half cycle, the diode is foward biased and can be replaced by battery of 0.7V.\n",
-      "Therefore, Vout=0.7V.\n",
-      "During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\n",
-      "Therefore, Vout_peak=-8.33V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_F=0.7;                        #Forward bias voltage of diode, V\n",
-    "R=200.0;                          #Input resistor of the circuit,  Ω\n",
-    "RL=1.0;                           #Load resistor, kΩ\n",
-    "Vin_peak=10.0;                    #Peak input voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#Positive half-cycle:\n",
-    "print(\"During the positive half cycle, the diode is foward biased and can be replaced by battery of %.1fV.\"%V_F);\n",
-    "print(\"Therefore, Vout=%.1fV.\"%V_F);\n",
-    "\n",
-    "#Negative half-cycle:\n",
-    "print(\"During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\");\n",
-    "Vout_peak=RL*(-Vin_peak)/(R/1000+RL);\n",
-    "print(\"Therefore, Vout_peak=%.2fV.\"%Vout_peak);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.12 : Page number 491"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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Nkq5P7h8J7AA8TahyaidJhN6Mo4DNJH1LKDUhqZekdyUtSNqatss478fJ9SYB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\nQmCC\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f8a844e3828>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "from math import sin\n",
-    "from math import pi\n",
-    "\n",
-    "V_biasing=10.0;                        #Biasing voltage, V\n",
-    "vin=[30*sin(t/10.0) for t in range(0,(int)(2*pi*10))]      #input voltage  waveform, V\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.plot(vin);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "vout=[];                             #Output voltage waveform, V\n",
-    "for v in vin[:]:\n",
-    "    if(v-V_biasing)>0 :              #Diode is forward biased.\n",
-    "        vout.append(v-V_biasing);\n",
-    "    else:                            #Diode is reverse biased.\n",
-    "        vout.append(0);\n",
-    "\n",
-    "plt.subplot(212)        \n",
-    "plt.plot(vout);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.13 : Page number 492"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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ZldfuMY6rJX1X0h6S1ii86eaSPiPpBmBCq29uZmbN0+tVVZIOAI4AdgNGAW+Q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\n5CKPkAAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f8a84f72fd0>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "t1=50;         #Assumed time interval, s\n",
-    "t2=100;         #Assumed time interval, s\n",
-    "V_biasing=10;  #Biasing voltage, V\n",
-    "for t in range(0,151):                  #time interval from 0s to 151s\n",
-    "    if(t<=t1):                      \n",
-    "        Vin.append(15);               #Value of input voltage for time 0 to t1 seconds \n",
-    "    elif(t<=t2 and t>t1):\n",
-    "        Vin.append(-30);             #Value of input voltage for time t1 to t2 seconds\n",
-    "    else :\n",
-    "        Vin.append(15);             #Value of input voltage after t2 seconds\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.plot(Vin);\n",
-    "plt.xlim([0,160])\n",
-    "plt.ylim([-35,20])\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=0):\n",
-    "        vout.append(0);               #Diode reverse biased\n",
-    "    else:\n",
-    "        vout.append(v-V_biasing);      #Diode forward biased\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout);\n",
-    "plt.xlim(0,160)\n",
-    "plt.ylim(-35,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.14 : Page number 492-493"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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2yMNXAZA0RtLXJN0KjG1W4SSNzaOvZuRncZuVnofP2mDUa80iIvaWtD9wFLCb\npBHAElIH903AERHxQjMKJmkY8Gtgb1LKkQck3RARTzTjemYDxcNnbTCqmUgwIm4iBYZW2xmYGRFz\nACRdCYwDHCys1NwMZYNRX4fO/r4v6wbYxsDciuVn8jqzUnMzlJXZ8883dlytobMrA6sC6+UmqK5J\nHGtSkj/cEyZMWPa+o6ODjo6OwspiBq5ZWPlUJhK8//7GzlGrGeoo4P8BG5GyzHZ5jdSf0EzPAptW\nLI/K696jMliYlYH7LKxsKr9IH3ss3HrryXWfo1YH91nAWZK+GRFnN1LIfngA2ELSaOB54DDSBEGz\nUnPNwsps9uzGjuvrY1UXSvpy95URcXFjl60tIpZKOga4jdS3cn5ETGvW9cwGSlewiADVlX3HrPlm\nzWrsuD4lEpRUWatYmTScdXJEfKaxyw4MJxK0slpzTZg7F9Zaq+iSmC0XAautBm+9NfBPyssXiG9W\nLktam5Qnysyq6KpdOFhYmcyb1xUs6j+2T0Nnq1gEbN7gsWaDnofPWhnNng2bN/iXu081C0k3Al3t\nPSsA2wB+cKRZDzwiyspo9mwYMwYeeKD+Y/vawX1GxfslwJyIeKb+y/VNft7214GuX7fvR8Qtzbqe\n2UDziCgro1mzGq9Z9KkZKiLuJqXZWAMYAbzT2OXqcmZE7JRfDhTWVhwsrIy6ahaN6Gu6j0NJDz/6\nLHAo8GdJzR4J5UGH1rbcZ2Fl1PSaBfAD4F8i4oiI+DIpyd8PG7tknx0j6WFJ50nymBJrK+6zsDJq\nes0CGBYRlT/6C+o4tipJt0uaUvF6NP97AHAOMCYidgBeAM7sz7XMWs3NUFY2ixenJIKbbNLY8X3t\n4L4lP+joirz8OfqZtjwi9u3jrr8BbuxpoxMJWhm5GcrKpLOzk//9305WWQV+8pPGztHrDG5J/w1c\nHhF/lHQIsHve9IeIuK6xS/ahUNIGXQ9VkvQtUhPY56vs5xncVkoLFsCWW8LLLxddErPk9tvh1FPh\nzjtBGvgZ3DOAMyRtSJpXcUlEPNRoYetwuqQdgHeBp0jZb83axogR8Prr8M47MHx40aUx619/BfQ9\n6+xoUtbXCyStQmqOuiIiZjR+6V6v+76khWbtZNgwWG89eOkl2LgUT36xoa4/I6Gg7/Ms5kTEaRGx\nIylN+EGAM8Ca9cKd3FYm/Un1AX2fZ7GipAMkXQbcDEwHDmn8smaDn4fPWpk0tRlK0r6kmsT+pEl5\nVwL/HhE589MiAAAHmElEQVSLGr+k2dDgmoWVSX+boWp1cJ8EXA4cHxGvNH4Zs6HHw2etLF57LaUl\nHzmy8XP02gwVEXtFxHnNCBSSPiPpMUlLJe3UbdtJkmZKmibpEwN9bbNWcDOUlUVXf0V/ntzYr1nY\n/fQocDBwd+VKSduQ8k9tA+wHnCP54ZTWftwMZWXR3/4KKDBYRMT0iJjJ+xMGjgOujIglEfEUMJOU\ni8qsrbgZysqiv/0V0Pd0H620MXBfxfKzeZ1ZWxk5MuXiefXVoktiQ9306bDttv07R1ODhaTbgfUr\nV5GeuPeDiOgx31M9nBvKymqzzVKw2GyzoktiQ92SJZ189rOdVPy5rFuvuaFaQdJdpNFWk/PyiUBE\nxGl5+RZgfET8ucqxzg1lZlanRnJDFdnBXamy0JOAwyQNl7Q5sAVpjoeZmRWksGAh6SBJc4FdgN9K\nuhkgIqaSkhZOJaVBP9rVBzOzYhXeDNUfboYyM6tfOzdDmZlZiTlYmJlZTQ4WZmZWk4OFmZnVVORo\nqKqJBCWNlvSmpMn5dU5RZTQzs6TIdB9diQT/p8q2v0XETlXWm5lZAQoLFhExHaCHjLLOMmtmViJl\n7bPYLDdB3SVp96ILY2Y21JUxkeBzwKYR8Uruy7he0rYR8Ua1nZ1I0Mysd52dnXR2dvbrHIXP4O6e\nSLCe7Z7BbWZWv3aewb2s0JLWkzQsvx9DSiQ4q6iCmZlZCRMJAnsAUyRNJiUUPCoi/PgYM7MCFd4M\n1R9uhjIzq187N0OZmVmJOViYmVlNDhZmZlZTkR3cp0uaJulhSddKWrNi20mSZubtnyiqjGZmlhRZ\ns7gN2C4idgBmAicBSNoWOBTYBtgPOKeHlCBWob8TbgYL34fE92E534ukv/ehsGAREXdExLt58X5g\nVH5/IHBlRCyJiKdIgWTnAorYVvwLkfg+JL4Py/leJG0bLLr5KnBTfr8xMLdi27N5nZmZFaTw3FCS\nfgAsjogrmlkWMzNrXKGT8iQdCXwd2Csi3s7rTgQiIk7Ly7cA4yPiz1WO94w8M7MG1Dspr7BgIWks\n8Atgj4hYULF+W+Ay4COk5qfbgS09VdvMrDhFPinvbGA4cHse7HR/RBwdEVMlXQ1MBRYDRztQmJkV\nq61zQ5mZWWuUZTRU3SSNlfSEpBmSTii6PK0iaZSkOyU9LulRScfm9SMk3SZpuqRbJa1VdFlbQdKw\n/FTFSXl5qN6HtSRNzBNZH5f0kaF4LyR9S9JjkqZIukzS8KFwHySdL2mepCkV63r83I1MfG7LYJGf\nd/Fr4JPAdsDhkj5cbKlaZgnw7YjYDtgV+Eb+7CcCd0TE1sCd5EmOQ8BxpCbLLkP1PpwF3BQR2wD/\nCDzBELsXkjYCvgnsFBHbk5rZD2do3IcLSX8PK1X93I1OfG7LYEGapDczIuZExGLgSmBcwWVqiYh4\nISIezu/fAKaRJjSOAy7Ku10EHFRMCVtH0ihgf+C8itVD8T6sCXwsIi4EyBNaFzIE7wWwArCapBWB\nVUjztAb9fYiIe4FXuq3u6XM3NPG5XYNF94l7zzAEJ+5J2gzYgTQDfv2ImAcpoAAjiytZy/wS+C5p\n7k6XoXgfNgfmS7owN8mdK2lVhti9iIjnSCMsnyYFiYURcQdD7D5UGNnD525o4nO7BoshT9LqwDXA\ncbmG0X2kwqAeuSDpU8C8XMvqrQo9qO9DtiKwE/DfEbETsIjUBDHUfibWJn2bHg1sRKphfIEhdh96\n0a/P3a7B4llg04rlUXndkJCr2NcAl0TEDXn1PEnr5+0bAC8WVb4W2Q04UNIs4ApgL0mXAC8MsfsA\nqWY9NyL+mpevJQWPofYzsQ8wKyJejoilwHXARxl696FLT5/7WWCTiv369PezXYPFA8AWkkZLGg4c\nBkwquEytdAEwNSLOqlg3CTgyvz8CuKH7QYNJRHw/IjaNiDGk//87I+JLwI0MofsAkJsa5kraKq/a\nG3icIfYzQWp+2kXSyrnDdm/S4Iehch/Ee2vZPX3uScBheaTY5sAWwF9qnrxd51nkGeBnkQLe+RHx\ns4KL1BKSdgPuAR4lVSsD+D7pP/tq0jeGOcChEfFqUeVsJUl7AsdHxIGS1mEI3gdJ/0jq6P8AMAv4\nCqmzd0jdC0njSV8eFgMPAf8GrMEgvw+SLgc6gHWBecB44HpgIlU+t6STgK+R7tNxEXFbzWu0a7Aw\nM7PWaddmKDMzayEHCzMzq8nBwszManKwMDOzmhwszMysJgcLMzOrycHCrE45Hfj/LbocZq3kYGFW\nvxHA0X3ZMecrMmt7DhZm9TsVGJMzvJ5WY9/rJV0v6QBJK7SicGbN4BncZnWSNBq4MT9gpy/770FK\nrbALKf3ChRHxZBOLaDbgXLMwa7KIuCcijgD+Oa96QtLBRZbJrF4rFl0As3Ym6cfAp0gJHf8ZeDC/\nnxQRE/I+KwMHA18F1iI9+vP2Ispr1ig3Q5nVKWe2fTAiNu/DvqcBnwF+R8qO/Eizy2fWDA4WZg2Q\ndCmwPXBzRJzQy35jSc/aeKdlhTNrAgcLMzOryR3cZmZWk4OFmZnV5GBhZmY1OViYmVlNDhZmZlaT\ng4WZmdXkYGFmZjU5WJiZWU3/Hz1pk1JFdhIyAAAAAElFTkSuQmCC\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f8a8455d630>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "t1=50;         #Assumed time interval, s\n",
-    "t2=100;         #Assumed time interval, s\n",
-    "V_biasing=5;  #Biasing voltage, V\n",
-    "for t in range(0,151):                  #time interval from 0s to 151s\n",
-    "    if(t<=t1):                      \n",
-    "        Vin.append(10);               #Value of input voltage for time 0 to t1 seconds \n",
-    "    elif(t<=t2 and t>t1):\n",
-    "        Vin.append(-10);             #Value of input voltage for time t1 to t2 seconds\n",
-    "    else :\n",
-    "        Vin.append(0);\n",
-    "\n",
-    "plt.subplot(211)        \n",
-    "plt.plot(Vin);\n",
-    "plt.xlim(0,101)\n",
-    "plt.ylim(-20,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=0):\n",
-    "        vout.append(v);               #Diode reverse biased\n",
-    "    else:\n",
-    "        vout.append(v-V_biasing);      #Diode forward biased\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout);\n",
-    "plt.xlim(0,101)\n",
-    "plt.ylim(-20,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.15 : Page number 493"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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xH0DSNGAiMLvmnInAj7NY7pM0StKYiFjWoJjMcrkGYVVT7zDXX9WzbwC2BhbW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\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f8a845679e8>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "t1=50;         #Assumed time interval, s\n",
-    "t2=100;         #Assumed time interval, s\n",
-    "V_D1=0.6;          #Forward Biasing voltage of the 1st diode, V\n",
-    "V_D2=0.6;          #Forward Biasing voltage of the 2nd diode, V\n",
-    "for t in range(0,151):                  #time interval from 0s to 151s\n",
-    "    if(t<=t1):                      \n",
-    "        Vin.append(10);               #Value of input voltage for time 0 to t1 seconds \n",
-    "    elif(t<=t2 and t>t1):\n",
-    "        Vin.append(-10);             #Value of input voltage for time t1 to t2 seconds\n",
-    "    else :\n",
-    "        Vin.append(0);\n",
-    "\n",
-    "plt.subplot(211);\n",
-    "plt.plot(Vin);\n",
-    "plt.xlim(0,110)\n",
-    "plt.ylim(-20,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=0):\n",
-    "        vout.append(-V_D1);               #Diode D1 forward biased, \n",
-    "    else:\n",
-    "        vout.append(V_D2);      #Diode D2 forward biased\n",
-    "\n",
-    "plt.subplot(212)        \n",
-    "plt.plot(vout);\n",
-    "plt.xlim(0,110)\n",
-    "plt.ylim(-1,1)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.16 : Page number 493-494"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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XvVPSHdHt44HfAi8Rupx2kSTCbMYxwFaSviO0mpB0lKQPJC2Nxpp2zbju59Hr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\nTQAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f8a940a7780>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "from math import sin\n",
-    "from math import pi\n",
-    "\n",
-    "VZ=20;                   #Assumed zener voltage, V\n",
-    "VF=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "for t in range(0,(int)(2*pi*10)):                  #time interval from 0s to 151s\n",
-    "    Vin.append(30*sin(t/10.0));\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.plot(Vin);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=-VF):\n",
-    "        vout.append(-VF);               #Zener diode forward biased, \n",
-    "    elif(v>=VZ):\n",
-    "        vout.append(VZ);        #Input voltage exceeds zener voltage\n",
-    "    else:\n",
-    "        vout.append(v);           #Zener diode reverse biased\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout);\n",
-    "plt.xlim([0,80])\n",
-    "plt.ylim([-1,40])\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.17 : Page number 494"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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2Bx7Fu5x2lSR8NuM4oJWkz/BWE5J6SnpN0tJkrGm3gud9O3m9V4AvJDVMzp2d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\nsgghhFCu/w+ip4Q9s46W/wAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f8a73eeb3c8>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "from math import sin\n",
-    "from math import pi\n",
-    "\n",
-    "VZ1=20;                   #Assumed zener voltage, V\n",
-    "VF1=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-    "VZ2=20;                   #Assumed zener voltage, V\n",
-    "VF2=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "for t in range(0,(int)(2*pi*10)):                  #time interval from 0s to 151s\n",
-    "    Vin.append(30*sin(t/10.0));\n",
-    "    \n",
-    "plt.subplot(211)\n",
-    "plt.plot(Vin);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=-(VZ1+VF2)):\n",
-    "        vout.append(-(VZ1+VF2));               #Zener diode forward biased, \n",
-    "    elif(v>=VZ2+VF1):\n",
-    "        vout.append(VZ2+VF1);        #Input voltage exceeds zener voltage\n",
-    "    else:\n",
-    "        vout.append(v);           #Zener diode reverse biased\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout); \n",
-    "plt.xlim([0,80])\n",
-    "plt.ylim([-40,40])\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_4.ipynb
deleted file mode 100755
index 5cc7f4af..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_4.ipynb
+++ /dev/null
@@ -1,804 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 18 : SOLID-STATE SWITCHING CIRCUITS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.1 : Page number 472"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Input voltage required to saturate the transistor switch=5.4V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10;             #Supply voltage, V\n",
-    "RC=1.0;               #Collector resistor, kΩ\n",
-    "RB=47.0;              #Base resistor, kΩ\n",
-    "beta=100.0;           #Base current amplification factor\n",
-    "VBE=0.7;            #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "IC_sat=VCC/RC;              #Collector saturation current, mA\n",
-    "IB=IC_sat/beta;             #Base current, mA\n",
-    "V=IB*RB+VBE;                #Input voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Input voltage required to saturate the transistor switch=%.1fV.\"%V);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.2 : Page number 475"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The collector emitter voltage at cut-off=9.99V.\n",
-      "(ii) The collector emitter voltage at saturation=0.7V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10;             #Supply voltage, V\n",
-    "RC=1.0;               #Collector resistor, kΩ\n",
-    "ICBO=10.0;            #Collector leakage current, μA\n",
-    "V_knee=0.7;         #Knee voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IC=ICBO;                    #Collector current, μA\n",
-    "VCE=VCC-(ICBO/1000)*RC;            #Collector-emitter voltage, V\n",
-    "\n",
-    "print(\"(i) The collector emitter voltage at cut-off=%.2fV.\"%VCE);\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, saturation current=IC_sat=(VCC-V_knee)/RC;         \n",
-    "VCE=V_knee;                    #Collector-emitter voltage, V\n",
-    "\n",
-    "print(\"(ii) The collector emitter voltage at saturation=%.1fV.\"%VCE);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.3 : Page number 475-476"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Minimum β=19.4.\n",
-      "(ii) The transistor will not be saturated.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10;             #Supply voltage, V\n",
-    "RC=1;               #Collector resistor, kΩ\n",
-    "VBB=2;              #Supply voltage to base, V\n",
-    "RB=2.7;             #Base resistor, kΩ\n",
-    "V_knee=0.7;         #Knee voltage, V\n",
-    "VBE=0.7;            #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IB=round((VBB-VBE)/RB,2);           #Base current, mA\n",
-    "Ic_sat=(VCC-V_knee)/RC;             #Collector saturation current, mA\n",
-    "beta_min=Ic_sat/IB;                 #Minimum value of base current amplification factor\n",
-    "print(\"(i) Minimum β=%.1f.\"%beta_min);\n",
-    "\n",
-    "#(ii)\n",
-    "VBB=1;                       #Supply voltage to base(changed), V\n",
-    "beta=50;                     #Base current amplification factor\n",
-    "IB=(VBB-VBE)/RB;             #Base current, mA\n",
-    "IC=beta*IB;                  #Collector current,mA\n",
-    "\n",
-    "if(IC<Ic_sat):\n",
-    "    print(\"(ii) The transistor will not be saturated.\");\n",
-    "else:\n",
-    "    print(\"(ii) The transistor will be saturated.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.4 : Page number 480"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Time period of the square wave=0.14 m sec.\n",
-      "Time frequency of the square wave=7 kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R2=10;                  #Resistor R2, kΩ\n",
-    "R3=10;                  #Resistor R3, kΩ\n",
-    "C1=0.01;                #Capacitor of 1st transistor, μF\n",
-    "C2=0.01;                #Capacitor of 2nd transistor, μF\n",
-    "\n",
-    "#Calculation\n",
-    "R=R2*1000;                   #Resistance, Ω\n",
-    "C=C1*10**-6;                 #Capacitance, F\n",
-    "T=round((1.4*R*C)*1000,2);   #Time period,m sec\n",
-    "f=1/(T*10**-3);              #Frequency, Hz\n",
-    "f=f/1000;                    #Frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Time period of the square wave=%.2f m sec.\"%T);\n",
-    "print(\"Time frequency of the square wave=%d kHz.\"%f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.6 : Page number 485"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output voltage=0.55V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=10;               #Resistance in differentiating circuit, kΩ\n",
-    "C=2.2;              #Capacitance in differentiating circuit, μF\n",
-    "d_ei=10;            #Change in input voltage, V\n",
-    "dt=0.4;             #Time in which change occurs, s\n",
-    "\n",
-    "#Calculation\n",
-    "eo=R*1000*C*10**-6*d_ei/dt\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output voltage=%.2fV.\"%eo);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.7 : Page number 489"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The peak output voltage=11.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vin_peak=12;                #Peak value of input voltage, V\n",
-    "V_D=0.7;                    #Forward bias voltage of diode, V\n",
-    "\n",
-    "#Calculation\n",
-    "Vout_peak=Vin_peak-V_D;         #Peak value of output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The peak output voltage=%.1fV.\"%Vout_peak);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.8 : Page number 489"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The peak output voltage=8V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vin_peak=10;                #Peak value of input voltage, V\n",
-    "R=1;                        #Input resistor, kΩ\n",
-    "RL=4;                       #Load resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "Vout_peak=(Vin_peak*RL)/(R+RL);         #Peak output voltage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The peak output voltage=%dV.\"%Vout_peak);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.9 : Page number 490"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The diode will be forward biased for the negative half-cycle of input signal.\n",
-      "The output voltage=-0.7V.\n",
-      "The voltage across R=-9.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vin=-10;                #Input voltage, V\n",
-    "V_D=0.7;                #Forward bias voltage of the diode, V\n",
-    "R=1;                    #Resistance, kΩ\n",
-    "\n",
-    "\n",
-    "print(\"The diode will be forward biased for the negative half-cycle of input signal.\");\n",
-    "Vout=-V_D;              #Output voltage, V\n",
-    "V_R=Vin-(-V_D);         #Voltage across resistor R, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output voltage=%.1fV.\"%Vout);\n",
-    "print(\"The voltage across R=%.1fV.\"%V_R);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.10 : Page number 490-491"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "During the positive half cycle, the diode is foward biased and can be replaced by battery of 0.7V.\n",
-      "Therefore, Vout=0.7V.\n",
-      "During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\n",
-      "Therefore, Vout_peak=-8.33V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_F=0.7;                        #Forward bias voltage of diode, V\n",
-    "R=200.0;                          #Input resistor of the circuit,  Ω\n",
-    "RL=1.0;                           #Load resistor, kΩ\n",
-    "Vin_peak=10.0;                    #Peak input voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#Positive half-cycle:\n",
-    "print(\"During the positive half cycle, the diode is foward biased and can be replaced by battery of %.1fV.\"%V_F);\n",
-    "print(\"Therefore, Vout=%.1fV.\"%V_F);\n",
-    "\n",
-    "#Negative half-cycle:\n",
-    "print(\"During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\");\n",
-    "Vout_peak=RL*(-Vin_peak)/(R/1000+RL);\n",
-    "print(\"Therefore, Vout_peak=%.2fV.\"%Vout_peak);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.12 : Page number 491"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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Nkq5P7h8J7AA8TahyaidJhN6Mo4DNJH1LKDUhqZekdyUtSNqatss478fJ9SYB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\n0dVmhIEDB/54v6SkhJKSkix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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fa23cd37e80>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "from math import sin\n",
-    "from math import pi\n",
-    "\n",
-    "V_biasing=10.0;                        #Biasing voltage, V\n",
-    "vin=[30*sin(t/10.0) for t in range(0,(int)(2*pi*10))]      #input voltage  waveform, V\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.plot(vin);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "vout=[];                             #Output voltage waveform, V\n",
-    "for v in vin[:]:\n",
-    "    if(v-V_biasing)>0 :              #Diode is forward biased.\n",
-    "        vout.append(v-V_biasing);\n",
-    "    else:                            #Diode is reverse biased.\n",
-    "        vout.append(0);\n",
-    "\n",
-    "plt.subplot(212)        \n",
-    "plt.plot(vout);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.13 : Page number 492"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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ZldfuMY6rJX1X0h6S1ii86eaSPiPpBmBCq29uZmbN0+tVVZIOAI4AdgNGAW+Q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\n5CKPkAAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fa23dcfd668>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "t1=50;         #Assumed time interval, s\n",
-    "t2=100;         #Assumed time interval, s\n",
-    "V_biasing=10;  #Biasing voltage, V\n",
-    "for t in range(0,151):                  #time interval from 0s to 151s\n",
-    "    if(t<=t1):                      \n",
-    "        Vin.append(15);               #Value of input voltage for time 0 to t1 seconds \n",
-    "    elif(t<=t2 and t>t1):\n",
-    "        Vin.append(-30);             #Value of input voltage for time t1 to t2 seconds\n",
-    "    else :\n",
-    "        Vin.append(15);             #Value of input voltage after t2 seconds\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.plot(Vin);\n",
-    "plt.xlim([0,160])\n",
-    "plt.ylim([-35,20])\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=0):\n",
-    "        vout.append(0);               #Diode reverse biased\n",
-    "    else:\n",
-    "        vout.append(v-V_biasing);      #Diode forward biased\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout);\n",
-    "plt.xlim(0,160)\n",
-    "plt.ylim(-35,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.14 : Page number 492-493"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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2yMNXAZA0RtLXJN0KjG1W4SSNzaOvZuRncZuVnofP2mDUa80iIvaWtD9wFLCb\npBHAElIH903AERHxQjMKJmkY8Gtgb1LKkQck3RARTzTjemYDxcNnbTCqmUgwIm4iBYZW2xmYGRFz\nACRdCYwDHCys1NwMZYNRX4fO/r4v6wbYxsDciuVn8jqzUnMzlJXZ8883dlytobMrA6sC6+UmqK5J\nHGtSkj/cEyZMWPa+o6ODjo6OwspiBq5ZWPlUJhK8//7GzlGrGeoo4P8BG5GyzHZ5jdSf0EzPAptW\nLI/K696jMliYlYH7LKxsKr9IH3ss3HrryXWfo1YH91nAWZK+GRFnN1LIfngA2ELSaOB54DDSBEGz\nUnPNwsps9uzGjuvrY1UXSvpy95URcXFjl60tIpZKOga4jdS3cn5ETGvW9cwGSlewiADVlX3HrPlm\nzWrsuD4lEpRUWatYmTScdXJEfKaxyw4MJxK0slpzTZg7F9Zaq+iSmC0XAautBm+9NfBPyssXiG9W\nLktam5Qnysyq6KpdOFhYmcyb1xUs6j+2T0Nnq1gEbN7gsWaDnofPWhnNng2bN/iXu081C0k3Al3t\nPSsA2wB+cKRZDzwiyspo9mwYMwYeeKD+Y/vawX1GxfslwJyIeKb+y/VNft7214GuX7fvR8Qtzbqe\n2UDziCgro1mzGq9Z9KkZKiLuJqXZWAMYAbzT2OXqcmZE7JRfDhTWVhwsrIy6ahaN6Gu6j0NJDz/6\nLHAo8GdJzR4J5UGH1rbcZ2Fl1PSaBfAD4F8i4oiI+DIpyd8PG7tknx0j6WFJ50nymBJrK+6zsDJq\nes0CGBYRlT/6C+o4tipJt0uaUvF6NP97AHAOMCYidgBeAM7sz7XMWs3NUFY2ixenJIKbbNLY8X3t\n4L4lP+joirz8OfqZtjwi9u3jrr8BbuxpoxMJWhm5GcrKpLOzk//9305WWQV+8pPGztHrDG5J/w1c\nHhF/lHQIsHve9IeIuK6xS/ahUNIGXQ9VkvQtUhPY56vs5xncVkoLFsCWW8LLLxddErPk9tvh1FPh\nzjtBGvgZ3DOAMyRtSJpXcUlEPNRoYetwuqQdgHeBp0jZb83axogR8Prr8M47MHx40aUx619/BfQ9\n6+xoUtbXCyStQmqOuiIiZjR+6V6v+76khWbtZNgwWG89eOkl2LgUT36xoa4/I6Gg7/Ms5kTEaRGx\nIylN+EGAM8Ca9cKd3FYm/Un1AX2fZ7GipAMkXQbcDEwHDmn8smaDn4fPWpk0tRlK0r6kmsT+pEl5\nVwL/HhE589MiAAAHmElEQVSLGr+k2dDgmoWVSX+boWp1cJ8EXA4cHxGvNH4Zs6HHw2etLF57LaUl\nHzmy8XP02gwVEXtFxHnNCBSSPiPpMUlLJe3UbdtJkmZKmibpEwN9bbNWcDOUlUVXf0V/ntzYr1nY\n/fQocDBwd+VKSduQ8k9tA+wHnCP54ZTWftwMZWXR3/4KKDBYRMT0iJjJ+xMGjgOujIglEfEUMJOU\ni8qsrbgZysqiv/0V0Pd0H620MXBfxfKzeZ1ZWxk5MuXiefXVoktiQ9306bDttv07R1ODhaTbgfUr\nV5GeuPeDiOgx31M9nBvKymqzzVKw2GyzoktiQ92SJZ189rOdVPy5rFuvuaFaQdJdpNFWk/PyiUBE\nxGl5+RZgfET8ucqxzg1lZlanRnJDFdnBXamy0JOAwyQNl7Q5sAVpjoeZmRWksGAh6SBJc4FdgN9K\nuhkgIqaSkhZOJaVBP9rVBzOzYhXeDNUfboYyM6tfOzdDmZlZiTlYmJlZTQ4WZmZWk4OFmZnVVORo\nqKqJBCWNlvSmpMn5dU5RZTQzs6TIdB9diQT/p8q2v0XETlXWm5lZAQoLFhExHaCHjLLOMmtmViJl\n7bPYLDdB3SVp96ILY2Y21JUxkeBzwKYR8Uruy7he0rYR8Ua1nZ1I0Mysd52dnXR2dvbrHIXP4O6e\nSLCe7Z7BbWZWv3aewb2s0JLWkzQsvx9DSiQ4q6iCmZlZCRMJAnsAUyRNJiUUPCoi/PgYM7MCFd4M\n1R9uhjIzq187N0OZmVmJOViYmVlNDhZmZlZTkR3cp0uaJulhSddKWrNi20mSZubtnyiqjGZmlhRZ\ns7gN2C4idgBmAicBSNoWOBTYBtgPOKeHlCBWob8TbgYL34fE92E534ukv/ehsGAREXdExLt58X5g\nVH5/IHBlRCyJiKdIgWTnAorYVvwLkfg+JL4Py/leJG0bLLr5KnBTfr8xMLdi27N5nZmZFaTw3FCS\nfgAsjogrmlkWMzNrXKGT8iQdCXwd2Csi3s7rTgQiIk7Ly7cA4yPiz1WO94w8M7MG1Dspr7BgIWks\n8Atgj4hYULF+W+Ay4COk5qfbgS09VdvMrDhFPinvbGA4cHse7HR/RBwdEVMlXQ1MBRYDRztQmJkV\nq61zQ5mZWWuUZTRU3SSNlfSEpBmSTii6PK0iaZSkOyU9LulRScfm9SMk3SZpuqRbJa1VdFlbQdKw\n/FTFSXl5qN6HtSRNzBNZH5f0kaF4LyR9S9JjkqZIukzS8KFwHySdL2mepCkV63r83I1MfG7LYJGf\nd/Fr4JPAdsDhkj5cbKlaZgnw7YjYDtgV+Eb+7CcCd0TE1sCd5EmOQ8BxpCbLLkP1PpwF3BQR2wD/\nCDzBELsXkjYCvgnsFBHbk5rZD2do3IcLSX8PK1X93I1OfG7LYEGapDczIuZExGLgSmBcwWVqiYh4\nISIezu/fAKaRJjSOAy7Ku10EHFRMCVtH0ihgf+C8itVD8T6sCXwsIi4EyBNaFzIE7wWwArCapBWB\nVUjztAb9fYiIe4FXuq3u6XM3NPG5XYNF94l7zzAEJ+5J2gzYgTQDfv2ImAcpoAAjiytZy/wS+C5p\n7k6XoXgfNgfmS7owN8mdK2lVhti9iIjnSCMsnyYFiYURcQdD7D5UGNnD525o4nO7BoshT9LqwDXA\ncbmG0X2kwqAeuSDpU8C8XMvqrQo9qO9DtiKwE/DfEbETsIjUBDHUfibWJn2bHg1sRKphfIEhdh96\n0a/P3a7B4llg04rlUXndkJCr2NcAl0TEDXn1PEnr5+0bAC8WVb4W2Q04UNIs4ApgL0mXAC8MsfsA\nqWY9NyL+mpevJQWPofYzsQ8wKyJejoilwHXARxl696FLT5/7WWCTiv369PezXYPFA8AWkkZLGg4c\nBkwquEytdAEwNSLOqlg3CTgyvz8CuKH7QYNJRHw/IjaNiDGk//87I+JLwI0MofsAkJsa5kraKq/a\nG3icIfYzQWp+2kXSyrnDdm/S4Iehch/Ee2vZPX3uScBheaTY5sAWwF9qnrxd51nkGeBnkQLe+RHx\ns4KL1BKSdgPuAR4lVSsD+D7pP/tq0jeGOcChEfFqUeVsJUl7AsdHxIGS1mEI3gdJ/0jq6P8AMAv4\nCqmzd0jdC0njSV8eFgMPAf8GrMEgvw+SLgc6gHWBecB44HpgIlU+t6STgK+R7tNxEXFbzWu0a7Aw\nM7PWaddmKDMzayEHCzMzq8nBwszManKwMDOzmhwszMysJgcLMzOrycHCrE45Hfj/LbocZq3kYGFW\nvxHA0X3ZMecrMmt7DhZm9TsVGJMzvJ5WY9/rJV0v6QBJK7SicGbN4BncZnWSNBq4MT9gpy/770FK\nrbALKf3ChRHxZBOLaDbgXLMwa7KIuCcijgD+Oa96QtLBRZbJrF4rFl0As3Ym6cfAp0gJHf8ZeDC/\nnxQRE/I+KwMHA18F1iI9+vP2Ispr1ig3Q5nVKWe2fTAiNu/DvqcBnwF+R8qO/Eizy2fWDA4WZg2Q\ndCmwPXBzRJzQy35jSc/aeKdlhTNrAgcLMzOryR3cZmZWk4OFmZnV5GBhZmY1OViYmVlNDhZmZlaT\ng4WZmdXkYGFmZjU5WJiZWU3/Hz1pk1JFdhIyAAAAAElFTkSuQmCC\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fa23c928d30>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "t1=50;         #Assumed time interval, s\n",
-    "t2=100;         #Assumed time interval, s\n",
-    "V_biasing=5;  #Biasing voltage, V\n",
-    "for t in range(0,151):                  #time interval from 0s to 151s\n",
-    "    if(t<=t1):                      \n",
-    "        Vin.append(10);               #Value of input voltage for time 0 to t1 seconds \n",
-    "    elif(t<=t2 and t>t1):\n",
-    "        Vin.append(-10);             #Value of input voltage for time t1 to t2 seconds\n",
-    "    else :\n",
-    "        Vin.append(0);\n",
-    "\n",
-    "plt.subplot(211)        \n",
-    "plt.plot(Vin);\n",
-    "plt.xlim(0,101)\n",
-    "plt.ylim(-20,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=0):\n",
-    "        vout.append(v);               #Diode reverse biased\n",
-    "    else:\n",
-    "        vout.append(v-V_biasing);      #Diode forward biased\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout);\n",
-    "plt.xlim(0,101)\n",
-    "plt.ylim(-20,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.15 : Page number 493"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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xH0DSNGAiMLvmnInAj7NY7pM0StKYiFjWoJjMcrkGYVVT7zDXX9WzbwC2BhbW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\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fa23cd370b8>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "t1=50;         #Assumed time interval, s\n",
-    "t2=100;         #Assumed time interval, s\n",
-    "V_D1=0.6;          #Forward Biasing voltage of the 1st diode, V\n",
-    "V_D2=0.6;          #Forward Biasing voltage of the 2nd diode, V\n",
-    "for t in range(0,151):                  #time interval from 0s to 151s\n",
-    "    if(t<=t1):                      \n",
-    "        Vin.append(10);               #Value of input voltage for time 0 to t1 seconds \n",
-    "    elif(t<=t2 and t>t1):\n",
-    "        Vin.append(-10);             #Value of input voltage for time t1 to t2 seconds\n",
-    "    else :\n",
-    "        Vin.append(0);\n",
-    "\n",
-    "plt.subplot(211);\n",
-    "plt.plot(Vin);\n",
-    "plt.xlim(0,110)\n",
-    "plt.ylim(-20,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=0):\n",
-    "        vout.append(-V_D1);               #Diode D1 forward biased, \n",
-    "    else:\n",
-    "        vout.append(V_D2);      #Diode D2 forward biased\n",
-    "\n",
-    "plt.subplot(212)        \n",
-    "plt.plot(vout);\n",
-    "plt.xlim(0,110)\n",
-    "plt.ylim(-1,1)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.16 : Page number 493-494"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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XvVPSHdHt44HfAi8Rupx2kSTCbMYxwFaSviO0mpB0lKQPJC2Nxpp2zbju59Hr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\nTQAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fa23cac80f0>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "from math import sin\n",
-    "from math import pi\n",
-    "\n",
-    "VZ=20;                   #Assumed zener voltage, V\n",
-    "VF=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "for t in range(0,(int)(2*pi*10)):                  #time interval from 0s to 151s\n",
-    "    Vin.append(30*sin(t/10.0));\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.plot(Vin);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=-VF):\n",
-    "        vout.append(-VF);               #Zener diode forward biased, \n",
-    "    elif(v>=VZ):\n",
-    "        vout.append(VZ);        #Input voltage exceeds zener voltage\n",
-    "    else:\n",
-    "        vout.append(v);           #Zener diode reverse biased\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout);\n",
-    "plt.xlim([0,80])\n",
-    "plt.ylim([-1,40])\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.17 : Page number 494"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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2Bx7Fu5x2lSR8NuM4oJWkz/BWE5J6SnpN0tJkrGm3gud9O3m9V4AvJDVMzp2d\njCl8LulWSVtKejT5f4+rgf74jubdID3wbswDqMHfnyJrBLQHbkz+D8vw3oy8xA+ApMZ4OaWS37VK\nxZ/XZPEe/kdcYq0L9zJusaStAORVd/+bcjxrlXyzvw8YYmajktM1Gf9PgKZ4X+rXzGwZ/iG5rgWe\nltz3RGABcISZbWRmVxfcpxOwI94M3xr4HbAYWK+M+CvyfCWmJM8NcCAwL/kX4CBgcnJdwJ/x5n5b\n/Hd2UHLbv/FxlPWTWBoAvYGhye2X47/vX+J9/zsBz+At6p/h3XW7AG/Lp6MPA04DtgDG4MmusGXW\nFy/SuYmZrU7OHQMcmjxPT/w9Pw/YHGiYPF+Vmdn7yb8f4t2YHcjP7/9C4F0zeyE5vh9PHnmJv0R3\n4EUzW5IcVyr+vCaL54Gdkm+NTfBf/tEpx1QRSi4lRuN99AD9gFGlH5AhtwMzzey6gnM1Gf/meHfK\nmjJuez+5vaLKWtl/tZn9D/8w/xjvWhgN7Jzcvq7411UpYAqeFMCTxF8Kjg9KbsfM5pnZ42a2ysw+\nwruVDkpuWwC8BBydPO5QYJmZPZ/8MbcBNjez7fFuhJeBFcBDQLdS8fcBHjaziUkiuBpohifjEteZ\n2SLzgf8S15vZkuRD/UngWTN71cy+whP43ut4D9ZJUvOkVUqSELsAM8jJ73/SVfNukojBfz6vk5P4\nCxwPDC84rlT8udzu3MxWS/od3gRvANyW9NlmlqRh+DfQzSQtwEuyXwGMlDQAX5jYJ70I105SR+Dn\nwIykW8WAC4DBwIgain8JsLmkBmUkjK2T26vKgNuTrpsG+IdtQzz+XwCn4B/WffCZLpUxFdhF0pbJ\nY48ELpW0Gf7t+QmA5PbrgAOADZLXX1rwPMPxP+Z7kn9L+pW3AxoD73v4NMZbQ7Pw359x+PtzaBL/\nn/Gfhf/HzSzpVisc0ytr1tTigusryjjeoNx3Yu22Ah6Q13JrBAw1s3GSXqDmfn+K7TRgaNKV8xZw\nEv4zzEX8yRhLZ+BXBacr9feby2QBYGaPAbumHUdFmdnP1nJT51oNpArM7Gn8D6MsNRX/VLyb5Ri8\nuwv4epykO94lAt5f3LzgcVuXep6y+l0FHGdmc5LnvAKfmbNU0n+A/5nZ2cltFXm+b240WyHpRXxa\n62tmtkrSVOBM4E0zK0kIf8a7kPYws08l9QKuL3iqkcDVkrbBWxj7JeffBf6XxPudWJJJAUPMrEty\nvAj4fqnyGrvRAAAW6klEQVS7bcu3E0St9q2b2dv4pIjS55eSg99/ADN7BZ9BV1pe4l+Od0sWnqvU\n+5/XbqhQx5jZZ8AfgesldZXUSD7z6l583OCe5K7TgR7yBUUt8Q/pQh8AO5TxEhdLapYMbJ+EjxNU\n5/kKPYGPgUxJjieXOgbYEB9c/zxJCOcUPkHSjzwFX4/zlpnNTs5/gLcerpW0odwOkg6kbCOAwyUd\nnLyHZ+PJZupa7h9ChUSyCJlhZlfh3VtXA5/iH3Dzgc5mtjK52xB8HcY7wGN886Ff4go8MSyVdGbB\n+SnAm/hahSvN7PFqPl+hKXg3zROljguTxaX4gqhP8LGG+8t4nmF4d9LQUudPBJrgC/KW4q2QlpQh\naT2dgC9a/RA4HDjSfDMxKLtVkatZPSEdmagNlcz+eAFYaGY9JbXAv1G2wf+I+5jZpymGGHJKUhu8\nj7nxWgbPQwgVkJWWRZllDCzjy+hDbsS+JyFUU+rJQlJrfKHOvwpO9wLuSq7fhe93EUJVpd98DiHn\nUk8WrKWMQQ6X0YcMMrP5ZtYwuqBCqJ5Up85KOhxYbGbTJXVax13L/Gao2IM7hBCqxCq5LXXaLYuO\nQE9Jb+GLkg6RNAT4oKLL0ItVnbEmLwMHDkw9hogz4sxznHmIMU9xVkWqycLMLjCz7cxsB7xkx0Qz\n+wU+tbB/crc8LKMPIYQ6Le2WxdpcARwmaTY+7/yKlOMJIYR6LTPlPsxsCt8UXctNGYCK6NSpU9oh\nVEjEWbMizpqThxghP3FWRSYW5VWVJMtz/CGEkAZJWM4GuEMIIeRAJIsQQgjlimQRQgihXKkmC0lN\nJT0r6WVJMyQNTM63SPb9nS1pbA3s/xtCCKEaUh/gltTczJZLagg8je9I9VPgIzO7UtK5QAszO6+M\nx8YAdwghVFIuB7jNd3ACaIpP5TWikGAIIWRK6slCUoNkX+cPgPFm9jxRSDCEEDIl9UV55tVA95a0\nEb6p+x5UYueuQYMGfX29U6dOdXpRTAghVMXkyZOZPHlytZ4j9TGLQpIuBpYDJwOdzGxxUkhwkpm1\nLeP+MWYRQgiVlLsxC0mbl8x0ktQMOAyYBYwmCgmGEEJmpNqykPQDfAC7QXK518wul7QpMALYFpiP\n78H9SRmPj5ZFCCFUUlVaFpnqhqqsSBYhhFB5ueuGCiGEkA+RLEIIIZQrkkUIIYRypT0bqrWkiZJe\nT2pDnZacj9pQIYSQIWnPhmoJtDSz6ZI2AF7ES32cRNSGCiGEosjdALeZfWBm05PrX+BrLFoTtaFC\nCCFTUi/3UULS94C9gGmUqg0lKZO1oV59FW69Fd57L+1I0tGgAWyzDeywg1923BHatgVV6vtKCCEP\nMpEski6o+4DTzewLSRWuDVXbvvoK7r8fbrwR3nkHfvUrOPjgtKNKx6pVsHAhzJ0LY8fCa6/B7rvD\n7bdDq1ZpRxdCqEmpJwtJjfBEMcTMSsp6LJa0VUFtqP+u7fG1WUhw4kTo3x923hnOPBN69oRGqb+D\n2bFyJVx+ObRv78n0pz9NO6IQAtSRQoKS7gaWmNmZBecGA0vNbHAWBrhXroSBA+HOO+GOO6Br16K/\nZK49+yyccAJ07Ah//ztstFHaEYUQCuVugFtSR+DnwCHJ1qovSeoGDAYOkzQbOBS4Iq0Y33oLDjgA\npk/3SySK8u27L7z8MjRpAvvs4+9bCCHfUm9ZVEexWxb/+Q/8+tdwwQVw2mk+oBsqZ/hwf+8uu8zH\nd2LwO4T0RSHBGrJyJZx/Ptx3n19++MMaf4l6ZfZs6N0bfvAD+Oc/YYMN0o4ohPotd91QWfT++3Do\nofD66/Dii5EoasKuu8K0adCsGXToAHPmpB1RCKGyIlkUePppTw6dO8Mjj8Bmm6UdUd3RvDn8619w\nxhmw//7w8MNpRxRCqIzohkrcdpt3Pd15J/ToUSNPGdZi6lTvlioZD4qxoBBqVy7HLCTdBhwBLDaz\ndsm5FsC9QBvgHXynvE/LeGy1k8WqVXDWWfDYYzB6tHeZhOJbtAiOPRZatoQhQ2D99dOOKIT6I69j\nFncApSekngdMMLNdgYnA+cV44aVLoXt3H4B99tlIFLWpVSuYNAk22cS7pRYuTDuiEMK6lJssJP1Y\n0o2SXpX0oaQFkh6V9NuaKB1uZk8BH5c6XfRCgnPnwn77Qbt2Pj6xySY1/QqhPE2bevffz37mP4vn\nn087ohDC2qwzWUgaA5wMjAW6AVsDuwMXAesBoyT1LEJcWxYWEgRqtJDgk0/6Qruzz4ZrroGGDWvy\n2UNlSHDOOV4e5PDDYeTItCMKIZSlvMpGvzCzJaXOfQG8lFyukbR5USL7trUOTFS2NtSQIT5GMXQo\nHHZYTYUXqqtXL2jTxv996y34wx9iAV8INaXotaEk3QgMM7Onq/Uq5QUhtQEeKhjgngV0KigkOMnM\n2pbxuAoPcJvBoEGeLB5+2KujhuxZtMhbGB06eGsjCjWGUPOKMcA9B7ha0juSrpS0d9XDWycllxKj\ngf7J9X7AqNIPqIyVK2HAABgzxqdtRqLIrlat4IknYMECOPJI+PzztCMKIUAFp84m3/z7JpdmwHBg\nuJlVey2upGFAJ2AzYDEwEHgQGAlsC8zHp85+UsZjy21ZfPaZT9Fcbz2vUxRTNPNh1Sr47W99ltqj\nj8b+GCHUpFpZZ5G0Lm4H2plZqkPD5SWL997zLo2f/MRLZUeXRr6YwRVXeD2pMWN8F74QQvUVbZ2F\npEaSjpQ0FBgDzAaOqUKMtWbWLN9P4bjjou87ryRfVf/HP/puhE8XdeQshLAu5Q1wHwYcD/QAngP+\nDYwys2W1E966ra1l8cwzcMwxcNVV8ItfpBBYqHFjx/rP8pZb4KgaX3UTQv1S491Qkibi4xP3mVnp\nhXOpKytZjB4NJ5/ss55io6K65cUXfdD7kku8rlQIoWqKkSw2NLN1zkeRtIGZfVGZF60ppZPFrbf6\n9qejR0dp8bpq3jz/EnDCCf6zjrUYIVReMcYsHpR0jaQDJX09j0jSDpJ+KalkZXdRSOom6Q1Jc5K9\nuMtkBn/6kw+GPvFEJIq6bMcdfezioYe8dbF6ddoRhVA/lDsbSlIPfJ/sjkALYBU+wP0o8K+kHEfN\nByY1wNd5HAosAp4H+prZGwX3sVWrjNNP9w+QMWO8immo+z7/HI4+GjbaCIYN86nRIYSKyWWJ8rWR\ntB8w0My6J8fnAWZmgwvuY717G0uWwIMP+gdHqD++/BL69/dV36NGRTHIECqqmFNnH6/IuRq2DfBu\nwfHC5Ny3mPmirUgU9U/Tpl7ja8894aCDfEvcEEJxrHP1gaT1gObA5smGRCWZaCPK+OBOQ9u2g7ji\nCr9ekUKCoW5p0ACuu87Hqzp29Cm2O++cdlQhZEttFBI8Hfg90AofNyjxGXCrmd1QrVdfV2DeDTXI\nzLolx2V2Q2W1Gy3Uvn/9Cy6+2Ae/Y5JDCGtXtDELSaea2fVVjqwKJDXEB9IPBd7HFwUeb2azCu4T\nySJ8y6hRvs5m6FDo0iXtaELIpqoki4oWwfhU0omlT5rZ3ZV5scows9WSfgeMw8dWbitMFCGUpVcv\n2Gwz+OlP4dprfRe+EEL1VbRlUdiqWA//tv+SmR1brMAqIloWYW1eew169IAzzvBLCOEbtTZ1VtIm\nwL9LxhPSEskirMuCBdCtGxxxhA+AN6jQ3L8Q6r6iTZ0twzJg+yo+NoRasd12vt/600/DiSfCV1+l\nHVGoCatXwzXXwLJMlDOtPyq6zuIhSaOTyyP4wPMDxQ0thOrbbDOYMME/WHr08M2wQn6tWAG9e/vW\nyKtWpR1N/VLRMYuDCg5XAfPNbGG1Xlg6FhgEtAV+ZGYvFdx2PjAgea3TzWzcWp4juqFChaxeDaee\n6uXrY+e9fProI+jZE9q0gTvu8EWZoWqK1g1lZlOAN4AN8fpQNdGgnwEcDUwpPCmpLdAHTyLdgZuk\nqC0aqqdhQ98Eq08f3znx9dfTjihUxttv+6LL/feHe+6JRJGGinZD9cHXOfTGP8ifTVoGVWZms81s\nLt+sCi/RCx88X2Vm7wBzgQ7Vea0QwMuZX3CBVyg++GCYODHtiEJFvPCCJ4nf/Q4GD46JCmmp6DqL\nC/Guov8CSNoCmADcV4SYtgGmFhy/R0ZKi4S64Re/gG239S13r7wS+vVLO6KwNg88AL/6la/O79Ur\n7Wjqt4omiwYliSLxERVolUgaD2xVeAow4EIze6jCUYZQwzp1gsmT4fDD4a23YNCg2EgpS8zgr3/1\nhZWPPQb77JN2RKGiyeKxZKOj4cnxcfh+FutkZodVIab3gG0Ljlsn58o0aNCgr69HIcFQGW3bwtSp\nPmg6dy7cdhs0a5Z2VGHVKp+M8PTTPiFhu+3Sjij/aqOQ4I3AMDN7WtIxwP7JTU+aWY1MnZU0CTjb\nzF5MjncHhgL74t1P44Gdy5r2FLOhQk1YsQIGDPAWxoMPwtZbpx1R/fXxxz4JoVEjuPfe2HqgWIox\nG2oOcLWkd4D9gCFmdmZNJApJR0l6N3nehyWNATCzmcAIYCbeejklMkIopmbNfLe9I46AffeFl19O\nO6L66Y03/P1v184rB0eiyJaKrrNoA/RNLs3w7qjhZjanuOGVG1fkkVCjRo6EU06Bm2+GY1OtfFa/\njBnjEw0GD4aTTko7mrqvVmpDSdobuB1oZ2YNK/XgGhbJIhTDiy961dq+feHyy32NRigOM7j6ah/I\nHjnS11KE4ivmfhaN8AVyffGKs5PxlsWoKsRZYyJZhGL58ENPFo0aeRfVZpulHVHd8/nn3opYsADu\nuy8GsmtTjY9ZSDpM0u34/tf/D3gE2NHM+qadKEIopi228C1a27WDH/0Ipk9PO6K6ZdYs6NABNt/c\niz1Gosi+8mZDTQSGAfeb2ce1FlUFRcsi1IZ77/XVw5dd5gvEYj1G9YwYAb/9rS+IjPGJdNTafhY1\nQdKVwJHAl8A84CQz+yy5LQoJhkyZPdundLZtC7fcEjN1qmL5ct+I6vHHPQHHQrv01OZ+FjVhHLCH\nme2F1386H75eZxGFBEOm7LorTJsGm2wC7dvDSy+V/5jwjddf926nL77w9y4SRf6klizMbIKZrUkO\np+ErtQF6EoUEQwY1awb/+Id3R3Xt6rvvrV6ddlTZtmqVV/vt1AnOPtsrxkarLJ+yUr9xAN+UD9kG\neLfgtigkGDKlb1+vhDp2LBx4IMybl3ZE2WPmRQDbtfMupyefhP79Y7wnzypaG6pKKlJIUNKFwEoz\nG17GU5QrakOFNLRp433v113nq44vuQT23BMaN4YmTfzf+lpK+7334NJLfYzi6quhe/dIEmkrem2o\nYpPUH5+Se4iZfZmcOw8wMxucHD8GDDSzZ8t4fAxwh9TNnAkXXug7ua1c6ZevvvJv1/VR8+Zw2mlw\n/PH1N2FmXd5mQ3UDrgEONLOPCs5HIcEQQiiiqiSLonZDleN6oAkwPpnsNM3MTjGzmZJKCgmuJAoJ\nhhBC6lLthqquaFmEEELl5W2dRQghhJyIZBFCCKFckSxCCCGUK5JFCCGEcqWWLCT9UdIrkl6W9Jik\nlgW3nS9prqRZkrqkFWMIIQSX5jqLDczsi+T6qcDuZvabgnUWP8LrRU0g1lmEEEKNydVsqJJEkVgf\nKCkqGIUEQwghY9JclIeky4ATgU+Ag5PT2wBTC+4WhQRDCCFlqRYSNLOLgIsknQucCgyq7GtEIcEQ\nQli33BcS/DoIaVvgETNrF4UEQwihuHI1ZiFpp4LDo4A3kuujgb6SmkjaHtgJeK624wshhPCNNMcs\nrpC0Cz6wPR/4NUAUEgwhhOzJRDdUVUU3VAghVF6uuqFCCCHkRySLEEII5YpkEUIIoVypJwtJZ0la\nI2nTgnNRGyqEEDIk1WQhqTVwGD4bquRcW6AP0BboDtykZN/VvKruYpjaEnHWrIiz5uQhRshPnFWR\ndsviWuCcUud6UcdqQ+XlFyjirFkRZ83JQ4yQnzirIs1FeT2Bd81sRqmbtgHeLTiO2lAhhJCytGpD\nXQRcgHdBhRBCyLhUFuVJ+j6+T8VyPIG0xlsQHYABAGZ2RXLfddaGqq2YQwihLqnsorxMrOCW9DbQ\n3sw+Ltj8aF+8+2k8a9n8KIQQQu1IdT+LAoa3MKI2VAghZFAmWhYhhBCyLe2ps1UmqZukNyTNSTZP\nygRJt0laLOnVgnMtJI2TNFvSWEkbpxxja0kTJb0uaYak0zIaZ1NJz0p6OYlzYBbjLCGpgaSXJI1O\njjMXp6R3JL2SvKfPZTjOjSWNTBbmvi5p36zFKWmX5H18Kfn3U0mnZTDOMyS9JulVSUOT7R8qHWMu\nk4WkBsANQFdgD+B4SbulG9XX7sDjKnQeMMHMdgUmAufXelTftgo408z2AH4M/DZ5/zIVp5l9CRxs\nZnsDewHdJXUgY3EWOB3vPi2RxTjXAJ3MbG8zK1m/lMU4rwMeNbO2wJ74fjeZitPM5iTvY3tgH2AZ\n8AAZilNSK3wX0vZm1g4feji+SjGaWe4uwH7AmILj84Bz046rIJ42wKsFx28AWyXXWwJvpB1jqXgf\nBDpnOU6gOfAC8KMsxonP6BsPdAJGZ/XnDrwNbFbqXKbiBDYC5pVxPlNxloqtC/Bk1uIEWuEVMlok\niWJ0Vf/Wc9my4LsL9xaS7YV7W5rZYgAz+wDYMuV4vibpe/i39mn4L0+m4ky6dl4GPgDGm9nzZDBO\nvqlGUDgImMU4DRgv6XlJJyfnshbn9sASSXckXTy3SGpO9uIsdBwwLLmemTjNbBFwDbAAX57wqZlN\nqEqMeU0WeZeJWQWSNgDuA043sy/4blypx2lma8y7oVoDHSTtQcbilHQ4sNjMppPM6luL1N9PoKN5\nt0kPvPvxADL2fuLfgNsDNyaxLsN7D7IWJwCSGgM9gZHJqczEKWkTvIRSG7yVsb6kn5cRU7kx5jVZ\nvAdsV3BcsqgvqxZL2gpAUkvgvynHg6RGeKIYYmajktOZi7OEmX0GTAa6kb04OwI9Jb0FDAcOkTQE\n+CBjcWJm7yf/foh3P3Yge+/nQrwU0AvJ8f148shanCW6Ay+a2ZLkOEtxdgbeMrOlZrYaH1P5SVVi\nzGuyeB7YSVIbSU2AvnhfXFaIb3/DHA30T673A0aVfkAKbgdmmtl1BecyFaekzUtmaUhqhpeHmUXG\n4jSzC8xsOzPbAf9dnGhmvwAeIkNxSmqetCaRtD7ezz6D7L2fi4F3Je2SnDoUeJ2MxVngePxLQoks\nxbkA2E/SepKEv5czqUqMaQ8MVWPgphswG69Ke17a8RTENQxYBHyZ/KBOwgeXJiTxjgM2STnGjsBq\nYDrwMvBS8n5umrE4f5DENh14FbgwOZ+pOEvFfBDfDHBnKk58LKDkZz6j5O8ma3EmMe2JfymcDvwH\n2DijcTYHPgQ2LDiXqTiBgfiXrFeBu4DGVYkxFuWFEEIoV167oUIIIdSiSBYhhBDKFckihBBCuSJZ\nhBBCKFckixBCCOWKZBFCCKFckSxCqKSkfPZv0o4jhNoUySKEymsBnFKROya1eULIvUgWIVTeX4Ad\nkoqog8u574OSHpR0pKSGtRFcCMUQK7hDqCRJbYCHzDeTqcj9DwR+ie/DMhK4w8zmFTHEEGpctCxC\nKDIze8LM+gE/TE69IenoNGMKobIapR1ACHkm6TLgcHw/gB8CLybXR5vZoOQ+6wFHAwPwgnin4rvq\nhZAb0Q0VQiVJ2hTfv2D7Ctx3MHAs8Ahwm5m9Uuz4QiiGSBYhVIGke4B2+F7w567jft3w/S2+qrXg\nQiiCSBYhhBDKFQPcIYQQyhXJIoQQQrkiWYQQQihXJIsQQgjlimQRQgihXJEsQgghlCuSRQghhHJF\nsgghhFCu/w+ip4Q9s46W/wAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fa23c83bc50>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "from math import sin\n",
-    "from math import pi\n",
-    "\n",
-    "VZ1=20;                   #Assumed zener voltage, V\n",
-    "VF1=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-    "VZ2=20;                   #Assumed zener voltage, V\n",
-    "VF2=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "for t in range(0,(int)(2*pi*10)):                  #time interval from 0s to 151s\n",
-    "    Vin.append(30*sin(t/10.0));\n",
-    "    \n",
-    "plt.subplot(211)\n",
-    "plt.plot(Vin);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=-(VZ1+VF2)):\n",
-    "        vout.append(-(VZ1+VF2));               #Zener diode forward biased, \n",
-    "    elif(v>=VZ2+VF1):\n",
-    "        vout.append(VZ2+VF1);        #Input voltage exceeds zener voltage\n",
-    "    else:\n",
-    "        vout.append(v);           #Zener diode reverse biased\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout); \n",
-    "plt.xlim([0,80])\n",
-    "plt.ylim([-40,40])\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_5.ipynb
deleted file mode 100755
index 5cc7f4af..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_5.ipynb
+++ /dev/null
@@ -1,804 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 18 : SOLID-STATE SWITCHING CIRCUITS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.1 : Page number 472"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Input voltage required to saturate the transistor switch=5.4V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10;             #Supply voltage, V\n",
-    "RC=1.0;               #Collector resistor, kΩ\n",
-    "RB=47.0;              #Base resistor, kΩ\n",
-    "beta=100.0;           #Base current amplification factor\n",
-    "VBE=0.7;            #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "IC_sat=VCC/RC;              #Collector saturation current, mA\n",
-    "IB=IC_sat/beta;             #Base current, mA\n",
-    "V=IB*RB+VBE;                #Input voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Input voltage required to saturate the transistor switch=%.1fV.\"%V);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.2 : Page number 475"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The collector emitter voltage at cut-off=9.99V.\n",
-      "(ii) The collector emitter voltage at saturation=0.7V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10;             #Supply voltage, V\n",
-    "RC=1.0;               #Collector resistor, kΩ\n",
-    "ICBO=10.0;            #Collector leakage current, μA\n",
-    "V_knee=0.7;         #Knee voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IC=ICBO;                    #Collector current, μA\n",
-    "VCE=VCC-(ICBO/1000)*RC;            #Collector-emitter voltage, V\n",
-    "\n",
-    "print(\"(i) The collector emitter voltage at cut-off=%.2fV.\"%VCE);\n",
-    "\n",
-    "#(ii)\n",
-    "#Since, saturation current=IC_sat=(VCC-V_knee)/RC;         \n",
-    "VCE=V_knee;                    #Collector-emitter voltage, V\n",
-    "\n",
-    "print(\"(ii) The collector emitter voltage at saturation=%.1fV.\"%VCE);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.3 : Page number 475-476"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Minimum β=19.4.\n",
-      "(ii) The transistor will not be saturated.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=10;             #Supply voltage, V\n",
-    "RC=1;               #Collector resistor, kΩ\n",
-    "VBB=2;              #Supply voltage to base, V\n",
-    "RB=2.7;             #Base resistor, kΩ\n",
-    "V_knee=0.7;         #Knee voltage, V\n",
-    "VBE=0.7;            #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IB=round((VBB-VBE)/RB,2);           #Base current, mA\n",
-    "Ic_sat=(VCC-V_knee)/RC;             #Collector saturation current, mA\n",
-    "beta_min=Ic_sat/IB;                 #Minimum value of base current amplification factor\n",
-    "print(\"(i) Minimum β=%.1f.\"%beta_min);\n",
-    "\n",
-    "#(ii)\n",
-    "VBB=1;                       #Supply voltage to base(changed), V\n",
-    "beta=50;                     #Base current amplification factor\n",
-    "IB=(VBB-VBE)/RB;             #Base current, mA\n",
-    "IC=beta*IB;                  #Collector current,mA\n",
-    "\n",
-    "if(IC<Ic_sat):\n",
-    "    print(\"(ii) The transistor will not be saturated.\");\n",
-    "else:\n",
-    "    print(\"(ii) The transistor will be saturated.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.4 : Page number 480"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Time period of the square wave=0.14 m sec.\n",
-      "Time frequency of the square wave=7 kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R2=10;                  #Resistor R2, kΩ\n",
-    "R3=10;                  #Resistor R3, kΩ\n",
-    "C1=0.01;                #Capacitor of 1st transistor, μF\n",
-    "C2=0.01;                #Capacitor of 2nd transistor, μF\n",
-    "\n",
-    "#Calculation\n",
-    "R=R2*1000;                   #Resistance, Ω\n",
-    "C=C1*10**-6;                 #Capacitance, F\n",
-    "T=round((1.4*R*C)*1000,2);   #Time period,m sec\n",
-    "f=1/(T*10**-3);              #Frequency, Hz\n",
-    "f=f/1000;                    #Frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Time period of the square wave=%.2f m sec.\"%T);\n",
-    "print(\"Time frequency of the square wave=%d kHz.\"%f);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.6 : Page number 485"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output voltage=0.55V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=10;               #Resistance in differentiating circuit, kΩ\n",
-    "C=2.2;              #Capacitance in differentiating circuit, μF\n",
-    "d_ei=10;            #Change in input voltage, V\n",
-    "dt=0.4;             #Time in which change occurs, s\n",
-    "\n",
-    "#Calculation\n",
-    "eo=R*1000*C*10**-6*d_ei/dt\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output voltage=%.2fV.\"%eo);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.7 : Page number 489"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The peak output voltage=11.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vin_peak=12;                #Peak value of input voltage, V\n",
-    "V_D=0.7;                    #Forward bias voltage of diode, V\n",
-    "\n",
-    "#Calculation\n",
-    "Vout_peak=Vin_peak-V_D;         #Peak value of output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The peak output voltage=%.1fV.\"%Vout_peak);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.8 : Page number 489"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The peak output voltage=8V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vin_peak=10;                #Peak value of input voltage, V\n",
-    "R=1;                        #Input resistor, kΩ\n",
-    "RL=4;                       #Load resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "Vout_peak=(Vin_peak*RL)/(R+RL);         #Peak output voltage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The peak output voltage=%dV.\"%Vout_peak);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.9 : Page number 490"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The diode will be forward biased for the negative half-cycle of input signal.\n",
-      "The output voltage=-0.7V.\n",
-      "The voltage across R=-9.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Vin=-10;                #Input voltage, V\n",
-    "V_D=0.7;                #Forward bias voltage of the diode, V\n",
-    "R=1;                    #Resistance, kΩ\n",
-    "\n",
-    "\n",
-    "print(\"The diode will be forward biased for the negative half-cycle of input signal.\");\n",
-    "Vout=-V_D;              #Output voltage, V\n",
-    "V_R=Vin-(-V_D);         #Voltage across resistor R, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output voltage=%.1fV.\"%Vout);\n",
-    "print(\"The voltage across R=%.1fV.\"%V_R);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.10 : Page number 490-491"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "During the positive half cycle, the diode is foward biased and can be replaced by battery of 0.7V.\n",
-      "Therefore, Vout=0.7V.\n",
-      "During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\n",
-      "Therefore, Vout_peak=-8.33V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_F=0.7;                        #Forward bias voltage of diode, V\n",
-    "R=200.0;                          #Input resistor of the circuit,  Ω\n",
-    "RL=1.0;                           #Load resistor, kΩ\n",
-    "Vin_peak=10.0;                    #Peak input voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#Positive half-cycle:\n",
-    "print(\"During the positive half cycle, the diode is foward biased and can be replaced by battery of %.1fV.\"%V_F);\n",
-    "print(\"Therefore, Vout=%.1fV.\"%V_F);\n",
-    "\n",
-    "#Negative half-cycle:\n",
-    "print(\"During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\");\n",
-    "Vout_peak=RL*(-Vin_peak)/(R/1000+RL);\n",
-    "print(\"Therefore, Vout_peak=%.2fV.\"%Vout_peak);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.12 : Page number 491"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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Nkq5P7h8J7AA8TahyaidJhN6Mo4DNJH1LKDUhqZekdyUtSNqatss478fJ9SYB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\n0dVmhIEDB/54v6SkhJKSkix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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fa23cd37e80>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "from math import sin\n",
-    "from math import pi\n",
-    "\n",
-    "V_biasing=10.0;                        #Biasing voltage, V\n",
-    "vin=[30*sin(t/10.0) for t in range(0,(int)(2*pi*10))]      #input voltage  waveform, V\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.plot(vin);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "vout=[];                             #Output voltage waveform, V\n",
-    "for v in vin[:]:\n",
-    "    if(v-V_biasing)>0 :              #Diode is forward biased.\n",
-    "        vout.append(v-V_biasing);\n",
-    "    else:                            #Diode is reverse biased.\n",
-    "        vout.append(0);\n",
-    "\n",
-    "plt.subplot(212)        \n",
-    "plt.plot(vout);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.13 : Page number 492"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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ZldfuMY6rJX1X0h6S1ii86eaSPiPpBmBCq29uZmbN0+tVVZIOAI4AdgNGAW+Q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\n5CKPkAAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fa23dcfd668>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "t1=50;         #Assumed time interval, s\n",
-    "t2=100;         #Assumed time interval, s\n",
-    "V_biasing=10;  #Biasing voltage, V\n",
-    "for t in range(0,151):                  #time interval from 0s to 151s\n",
-    "    if(t<=t1):                      \n",
-    "        Vin.append(15);               #Value of input voltage for time 0 to t1 seconds \n",
-    "    elif(t<=t2 and t>t1):\n",
-    "        Vin.append(-30);             #Value of input voltage for time t1 to t2 seconds\n",
-    "    else :\n",
-    "        Vin.append(15);             #Value of input voltage after t2 seconds\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.plot(Vin);\n",
-    "plt.xlim([0,160])\n",
-    "plt.ylim([-35,20])\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=0):\n",
-    "        vout.append(0);               #Diode reverse biased\n",
-    "    else:\n",
-    "        vout.append(v-V_biasing);      #Diode forward biased\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout);\n",
-    "plt.xlim(0,160)\n",
-    "plt.ylim(-35,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.14 : Page number 492-493"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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2yMNXAZA0RtLXJN0KjG1W4SSNzaOvZuRncZuVnofP2mDUa80iIvaWtD9wFLCb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\ng4WZmdXkYGFmZjU5WJiZWU3/Hz1pk1JFdhIyAAAAAElFTkSuQmCC\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fa23c928d30>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "t1=50;         #Assumed time interval, s\n",
-    "t2=100;         #Assumed time interval, s\n",
-    "V_biasing=5;  #Biasing voltage, V\n",
-    "for t in range(0,151):                  #time interval from 0s to 151s\n",
-    "    if(t<=t1):                      \n",
-    "        Vin.append(10);               #Value of input voltage for time 0 to t1 seconds \n",
-    "    elif(t<=t2 and t>t1):\n",
-    "        Vin.append(-10);             #Value of input voltage for time t1 to t2 seconds\n",
-    "    else :\n",
-    "        Vin.append(0);\n",
-    "\n",
-    "plt.subplot(211)        \n",
-    "plt.plot(Vin);\n",
-    "plt.xlim(0,101)\n",
-    "plt.ylim(-20,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=0):\n",
-    "        vout.append(v);               #Diode reverse biased\n",
-    "    else:\n",
-    "        vout.append(v-V_biasing);      #Diode forward biased\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout);\n",
-    "plt.xlim(0,101)\n",
-    "plt.ylim(-20,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.15 : Page number 493"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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xH0DSNGAiMLvmnInAj7NY7pM0StKYiFjWoJjMcrkGYVVT7zDXX9WzbwC2BhbW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\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fa23cd370b8>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "t1=50;         #Assumed time interval, s\n",
-    "t2=100;         #Assumed time interval, s\n",
-    "V_D1=0.6;          #Forward Biasing voltage of the 1st diode, V\n",
-    "V_D2=0.6;          #Forward Biasing voltage of the 2nd diode, V\n",
-    "for t in range(0,151):                  #time interval from 0s to 151s\n",
-    "    if(t<=t1):                      \n",
-    "        Vin.append(10);               #Value of input voltage for time 0 to t1 seconds \n",
-    "    elif(t<=t2 and t>t1):\n",
-    "        Vin.append(-10);             #Value of input voltage for time t1 to t2 seconds\n",
-    "    else :\n",
-    "        Vin.append(0);\n",
-    "\n",
-    "plt.subplot(211);\n",
-    "plt.plot(Vin);\n",
-    "plt.xlim(0,110)\n",
-    "plt.ylim(-20,20)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=0):\n",
-    "        vout.append(-V_D1);               #Diode D1 forward biased, \n",
-    "    else:\n",
-    "        vout.append(V_D2);      #Diode D2 forward biased\n",
-    "\n",
-    "plt.subplot(212)        \n",
-    "plt.plot(vout);\n",
-    "plt.xlim(0,110)\n",
-    "plt.ylim(-1,1)\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.16 : Page number 493-494"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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XvVPSHdHt44HfAi8Rupx2kSTCbMYxwFaSviO0mpB0lKQPJC2Nxpp2zbju59Hr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\nTQAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fa23cac80f0>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "from math import sin\n",
-    "from math import pi\n",
-    "\n",
-    "VZ=20;                   #Assumed zener voltage, V\n",
-    "VF=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "for t in range(0,(int)(2*pi*10)):                  #time interval from 0s to 151s\n",
-    "    Vin.append(30*sin(t/10.0));\n",
-    "\n",
-    "plt.subplot(211)\n",
-    "plt.plot(Vin);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=-VF):\n",
-    "        vout.append(-VF);               #Zener diode forward biased, \n",
-    "    elif(v>=VZ):\n",
-    "        vout.append(VZ);        #Input voltage exceeds zener voltage\n",
-    "    else:\n",
-    "        vout.append(v);           #Zener diode reverse biased\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout);\n",
-    "plt.xlim([0,80])\n",
-    "plt.ylim([-1,40])\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 18.17 : Page number 494"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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2Bx7Fu5x2lSR8NuM4oJWkz/BWE5J6SnpN0tJkrGm3gud9O3m9V4AvJDVMzp2d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\nsgghhFCu/w+ip4Q9s46W/wAAAABJRU5ErkJggg==\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fa23c83bc50>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "from math import sin\n",
-    "from math import pi\n",
-    "\n",
-    "VZ1=20;                   #Assumed zener voltage, V\n",
-    "VF1=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-    "VZ2=20;                   #Assumed zener voltage, V\n",
-    "VF2=0.7;          #Assumed forward biasing voltage of the zener diode, V\n",
-    "Vin=[];      #Input voltage waveform, V\n",
-    "for t in range(0,(int)(2*pi*10)):                  #time interval from 0s to 151s\n",
-    "    Vin.append(30*sin(t/10.0));\n",
-    "    \n",
-    "plt.subplot(211)\n",
-    "plt.plot(Vin);\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vin(V)');\n",
-    "plt.title('Input waveform');\n",
-    "\n",
-    "\n",
-    "vout=[];                 #Output voltage waveform, V\n",
-    "for v in Vin[:]:                  #Loop iterating input voltage \n",
-    "    if(v<=-(VZ1+VF2)):\n",
-    "        vout.append(-(VZ1+VF2));               #Zener diode forward biased, \n",
-    "    elif(v>=VZ2+VF1):\n",
-    "        vout.append(VZ2+VF1);        #Input voltage exceeds zener voltage\n",
-    "    else:\n",
-    "        vout.append(v);           #Zener diode reverse biased\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout); \n",
-    "plt.xlim([0,80])\n",
-    "plt.ylim([-40,40])\n",
-    "plt.xlabel('t-->');\n",
-    "plt.ylabel('Vout(V)');\n",
-    "plt.title('Output waveform');\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19.ipynb
deleted file mode 100755
index 5c74053c..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19.ipynb
+++ /dev/null
@@ -1,1666 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e80dba2d461291c68fb06b32e702542441b6f13e4e124da0b4826c1be4471e9c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 19 : FIELD EFFECT TRANSISTORS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.1 : Page number 515\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_DSS=12.0;                                 #Shorted gate drain current, mA\n",
-      "V_GS_off=-5.0;                              #Gate-source cut-off voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"ID=%d[1 + VGS/%d]\u00b2mA.\"%(I_DSS,abs(V_GS_off)));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ID=12[1 + VGS/5]\u00b2mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.2 : Page number 515\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_DSS=32.0;                                 #Shorted gate drain current, mA\n",
-      "V_GS_off=-8.0;                              #Gate-source cut-off voltage, V\n",
-      "V_GS=-4.5;                                  #Gate-source voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The drain current=%.2fmA.\"%I_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drain current=6.12mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.3 : Page number  515\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_DSS=10.0;                                 #Shorted gate drain current, mA\n",
-      "V_GS_off=-6.0;                              #Gate-source cut-off voltage, V\n",
-      "I_D=5.0;                                    #Drain current mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, I_D=I_DSS*[1 - (V_GS/V_GS_off)]\u00b2\n",
-      "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS));              #Gate-source voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_P=-V_GS_off;                                  #Pinch-off voltage, V        \n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) VGS=%.2fV.\"%V_GS);\n",
-      "print(\"(ii) VP=%dV\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) VGS=-1.76V.\n",
-        "(ii) VP=6V\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.4 : Page number 515-516\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GS_off=-4.0;                              #Gate-source cut-off voltage, V\n",
-      "I_DSS=12.0;                                 #Shorted gate drain current, mA\n",
-      "R_D=560.0;                                  #Drain resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "V_P=-V_GS_off;          #Pinch-off voltage, V\n",
-      "V_DS=V_P;               #Minimum drain-source voltage for JFET to be in constant current region, V\n",
-      "I_D=I_DSS;              #Maximum drain current, mA (V_GS=0)\n",
-      "V_RD=(I_D/1000)*R_D;    #Voltage across drain resistor, V (OHM's LAW)\n",
-      "V_DD=V_DS+V_RD;         #Minimum value of supply voltage to drain, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum value of VDD required =%.2fV.\"%V_DD);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum value of VDD required =10.72V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.5 : Page number 516\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_DSS=3.0;                                 #Shorted gate drain current, mA\n",
-      "V_GS_off=-6.0;                              #Gate-source cut-off voltage, V\n",
-      "V_GS=-2.0;                                  #Gate-source voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The drain current=%.2fmA.\"%I_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drain current=1.33mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.6 : Page number 516"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VGS_off=4;                #Gate-source cut-off voltage, V\n",
-      "VGS=6;                    #Gate source voltage, V\n",
-      "\n",
-      "print(\"p-channel JFET requires a positive gate-to-source voltage to pass drain current.\");\n",
-      "print(\"More positive voltage, the less the drain current. \");\n",
-      "print(\"Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\");"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "p-channel JFET requires a positive gate-to-source voltage to pass drain current.\n",
-        "More positive voltage, the less the drain current. \n",
-        "Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.7 : Page number 517-518\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GS=15.0;              #Gate-source voltage, V\n",
-      "I_G=1e-03;               #Gate current, \u03bcA\n",
-      "\n",
-      "#Calculation\n",
-      "R_GS=(V_GS/(I_G*10**-6))/10**6;          #Gate to source resistance, M\u03a9 (OHM's LAW)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The gate to source resistance=%dM\u03a9.\"%R_GS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The gate to source resistance=15000M\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 47
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.8 : Page  number 518\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "V_GS_max=-3.1;                      #Maximum gate to source voltage, V\n",
-      "V_GS_min=-3.0;                      #Minimum gate to source voltage, V\n",
-      "I_D_max=1.3;                      #Maximum drain current, mA\n",
-      "I_D_min=1.0;                      #Minimum drain current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "delta_V_GS=abs(V_GS_max-V_GS_min);              #Change in gate to source voltage, V\n",
-      "delta_I_D=I_D_max-I_D_min;                      #Change in drain current, mA\n",
-      "g_fs=(delta_I_D/delta_V_GS)*1000;               #Transconductance, \u03bc mho\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Transconductance=%.0f \u03bc mho\"%g_fs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Transconductance=3000 \u03bc mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.9 : Page number 518\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GS=[0,0,-0.2];            #Readings of Gate-source voltage, V\n",
-      "V_DS=[7,15,15];             #Readings of Drain-source voltage, V\n",
-      "ID=[10,10.25,9.65];         #Readings of drain current, mA\n",
-      "\n",
-      "\n",
-      "#Displaying the readings:\n",
-      "print(\"VGS= %dV     %dV        %.1fV\"%(V_GS[0],V_GS[1],V_GS[2]));\n",
-      "print(\"VDS= %dV     %dV        %dV\"%(V_DS[0],V_DS[1],V_DS[2]));\n",
-      "print(\"ID = %dV    %.2fV     %.2fV\"%(ID[0],ID[1],ID[2]));\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "#V_GS constant at 0V,\n",
-      "delta_VDS=V_DS[1]-V_DS[0];          #Change in drain-source voltage, V\n",
-      "delta_ID=ID[1]-ID[0];               #Change in drain current, mA\n",
-      "rd=delta_VDS/delta_ID;              #a.c drain resistance, k\u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "#V_DS constant at 15V,\n",
-      "delta_VGS=V_GS[2]-V_GS[1];                  #Change in gate-source voltage, V\n",
-      "delta_ID=ID[2]-ID[1];                       #Change in drain current, mA\n",
-      "g_fs=round((delta_ID/delta_VGS)*1000,);     #Transconductance, \u03bc mho\n",
-      "\n",
-      "#(iii)\n",
-      "amplification_factor=rd*1000*g_fs*10**-6;                          #Amplification factor\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The a.c drain resistance=%dk\u03a9.\"%rd);\n",
-      "print(\"(ii)  The transconductance=%d \u03bc mho.\"%g_fs);\n",
-      "print(\"(iii) The amplification factor=%d.\"%amplification_factor );\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VGS= 0V     0V        -0.2V\n",
-        "VDS= 7V     15V        15V\n",
-        "ID = 10V    10.25V     9.65V\n",
-        "(i)   The a.c drain resistance=32k\u03a9.\n",
-        "(ii)  The transconductance=3000 \u03bc mho.\n",
-        "(iii) The amplification factor=96.\n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.10 : Page number 519\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "g_mo=4000.0;                        #Maximum transconductance, \u03bcS\n",
-      "V_GS=-3.0;                          #Gate to source voltage, V\n",
-      "V_GS_off=-8.0;                      #Gate-source cut-off voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "g_m=g_mo*(1-(V_GS/V_GS_off));               #Transconductance, \u03bcS\n",
-      "\n",
-      "#Result\n",
-      "print(\"The transconductance=%d \u03bcS.\"%g_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The transconductance=2500 \u03bcS.\n"
-       ]
-      }
-     ],
-     "prompt_number": 54
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.11 : Page number 519\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "g_mo=5000.0;                        #Maximum transconductance, \u03bcS\n",
-      "V_GS=-4.0;                          #Gate to source voltage, V\n",
-      "V_GS_off=-6.0;                      #Gate-source cut-off voltage, V\n",
-      "I_DSS=3.0;                          #Shorted-gate drain current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "g_m=g_mo*(1-(V_GS/V_GS_off));               #Transconductance, \u03bcS\n",
-      "I_D=(I_DSS*(1-(V_GS/V_GS_off))**2)*1000;                    #Drain current \u03bcA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The transconductance=%.0f \u03bcS.\"%g_m);\n",
-      "print(\"The drain current=%d \u03bcA.\"%I_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The transconductance=1667 \u03bcS.\n",
-        "The drain current=333 \u03bcA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.12 : Page number 520\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GS_off=-8.0;                      #Gate-source cut-off voltage, V\n",
-      "I_DSS=16.0;                         #Shorted-gate drain current, mA\n",
-      "R_D=2.2;                            #Drain resistor, k\u03a9\n",
-      "R_G=1.0;                            #Gate resistor, M\u03a9\n",
-      "V_DD=10.0;                          #Drain supply voltage, V\n",
-      "V_GG=-5.0;                           #Gate supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_GS=V_GG;                                           #Gate-source voltage, V\n",
-      "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current \u03bcA\n",
-      "V_DS=V_DD-I_D*R_D;                                   #Drain-source voltage, V (Kirchhoff's voltage law)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The gate-source voltage=%dV.\"%V_GS);\n",
-      "print(\"The drain current=%.2fmA.\"%I_D);\n",
-      "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The gate-source voltage=-5V.\n",
-        "The drain current=2.25mA.\n",
-        "The drain-source voltage=5.05V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.13 : Page number 521\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_D=5.0;                   #Drain current mA\n",
-      "V_DD=15.0;                 #Drain supply voltage, V\n",
-      "V_G=0;                     #Gate voltage, V\n",
-      "R_D=1.0;                   #Drain resistor, k\u03a9\n",
-      "R_S=470.0;                 #Source resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_S=(I_D/1000)*R_S;                 #Source voltage, V (OHM's LAW)\n",
-      "V_D=V_DD-I_D*R_D;                   #Drain voltage, V (Kirchhoff's voltage law)\n",
-      "V_DS=V_D-V_S;                       #Drain-source voltage, V\n",
-      "V_GS=V_G-V_S;                       #Gate-source voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n",
-      "print(\"The gate-source voltage=%.2fV.\"%V_GS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drain-source voltage=7.65V.\n",
-        "The gate-source voltage=-2.35V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 57
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.14 : Page number 521\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GS=-5.0;                     #Gate-source voltage, V\n",
-      "I_D=6.25;                      #Drain current mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "R_S=abs(V_GS/(I_D/1000));              #Required source resistor, \u03a9 (OHM's LAW)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required source resistor=%d \u03a9.\"%R_S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required source resistor=800 \u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.15 : Page number : 521\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_DSS=25.0;                                 #Shorted gate drain current, mA\n",
-      "V_GS_off=15.0;                              #Gate-source cut-off voltage, V\n",
-      "V_GS=5.0;                                   #Gate-source voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;               #Drain current mA\n",
-      "R_S=V_GS/(I_D/1000);                            #Required source resistor, \u03a9 (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The source resistance=%.0f\u03a9.\"%R_S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The source resistance=450\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 59
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.16 : Page number 522\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_DSS=15.0;                                 #Shorted gate drain current, mA\n",
-      "V_GS_off=-8.0;                              #Gate-source cut-off voltage, V\n",
-      "V_DD=12.0;                                  #Drain supply voltage,V\n",
-      "V_D=V_DD/2;                                 #Drain voltage(half of V_DD), V\n",
-      "\n",
-      "#Calculation\n",
-      "I_D=I_DSS/2;                                #Drain current(approximately half of I_DSS), mA\n",
-      "V_GS=V_GS_off/3.4;                          #Gate-source voltage, V\n",
-      "R_S=abs(V_GS/(I_D/1000));                   #Source resistor, \u03a9 (OHM's LAW)\n",
-      "#Since,V_D=V_DD-I_D*R_D;                      \n",
-      "R_D=(V_DD-V_D)/(I_D/1000);                         #Drain resistor, \u03a9 (OHM's LAW)\n",
-      "\n",
-      "#Result\n",
-      "print(\" RS=%d \u03a9 and RD=%d \u03a9.\"%(R_S,R_D));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " RS=313 \u03a9 and RD=800 \u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.17 : Page number 522\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_DSS=5.0;                       #Shorted gate drain current, mA\n",
-      "V_GS_off=-2.0;                   #Gate-source cut-off voltage, V\n",
-      "V_DS=10.0;                       #Drain-source voltage,V\n",
-      "I_D=1.5;                         #Drain current, mA\n",
-      "V_DD=20.0;                       #Drain supply voltage,V\n",
-      "V_G=0;                           #Gate voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since,  Drain current, I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    \n",
-      "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS));              #Gate-source voltage, V\n",
-      "\n",
-      "#Since, V_GS=V_G-V_S,\n",
-      "V_S=V_G-V_GS;                           #Source voltage, V\n",
-      "\n",
-      "R_S=V_S/I_D;                            #Source resistor, k\u03a9\n",
-      "\n",
-      "#Since, V_DD=I_D*R_D +V_DS+ I_D*R_S,\n",
-      "R_D=(V_DD-I_D*R_S-V_DS)/I_D;                    #Drain resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"The source resistance=%.1f k\u03a9\"%R_S);\n",
-      "print(\"The drain resistance=%d k\u03a9.\"%R_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The source resistance=0.6 k\u03a9\n",
-        "The drain resistance=6 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 61
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.18 : Page number 522-523\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_DD=30.0;                  #Drain supply voltage, V\n",
-      "R_D=5.0;                    #Drain resistor, k\u03a9\n",
-      "I_D=2.5;                    #Drain current, mA\n",
-      "R_S=200.0;                  #Source resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_DS=V_DD-I_D*(R_D+(R_S/1000));                 #Drain-source voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_GS=-(I_D/1000)*R_S;                          #Gate-source voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The drain-source voltage=%dV.\"%V_DS);\n",
-      "print(\"The gate-source voltage=%.1fV.\"%V_GS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drain-source voltage=17V.\n",
-        "The gate-source voltage=-0.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 62
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.19 : Page number 523"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ID_1=2.15;                  #First stage drain current, mA\n",
-      "ID_2=9.15;                  #Second stage drain current, mA\n",
-      "VDD=30;                     #Drain supply voltage, V\n",
-      "RS_1=0.68;                  #Source resistance of 1st stage, k\u03a9\n",
-      "RS_2=0.22;                  #Source resistance of 2nd stage, k\u03a9\n",
-      "RD_1=8.2;                   #Drain resistor of 1st stage, k\u03a9\n",
-      "RD_2=2;                     #Drain resistor of 2nd stage, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "V_RD1=ID_1*RD_1;                #Voltage drop across 8.2k\u03a9\n",
-      "VD_1=VDD-V_RD1;                 #Drain voltage of 1st stage, V\n",
-      "VS_1=ID_1*RS_1;                 #D.C potential of source of first stage, V\n",
-      "V_RD2=ID_2*RD_2;                #Voltage drop across 2k\u03a9\n",
-      "VD_2=VDD-V_RD2;                 #Drain voltage of 2nd stage, V\n",
-      "VS_2=ID_2*RS_2;                 #D.C potential of source of 2nd stage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"drain voltage of 1st stage=%.2fV.\"%VD_1);\n",
-      "print(\"Source voltage of 1st stage=%.2fV.\"%VS_1);\n",
-      "print(\"drain voltage of 2nd stage=%.1fV.\"%VD_2);\n",
-      "print(\"Source voltage of 2nd stage=%.2fV.\"%VS_2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "drain voltage of 1st stage=12.37V.\n",
-        "Source voltage of 1st stage=1.46V.\n",
-        "drain voltage of 2nd stage=11.7V.\n",
-        "Source voltage of 2nd stage=2.01V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 63
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.20 : Page number 524"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=12;                     #Drain supply voltage, V\n",
-      "VD=7;                       #Drain voltage, V\n",
-      "R1=6.8;                     #Resistor R1, M\u03a9\n",
-      "R2=1;                       #Resistor R2, M\u03a9\n",
-      "RS=1.8;                     #Source resistance, k\u03a9\n",
-      "RD=3.3;                     #Drain resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "ID=(VDD-VD)/RD;                  #Second stage drain current, mA\n",
-      "VS=ID*RS;                        #Source voltage, V\n",
-      "VG=VDD*R2/(R1+R2);               #Drain voltage, V\n",
-      "VGS=VG-VS;                       #Drain-source voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"Drain current=%.2fmA.\"%ID);\n",
-      "print(\"Gate-source voltage=%.1fV.\"%VGS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain current=1.52mA.\n",
-        "Gate-source voltage=-1.2V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 65
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.21 : Page number 524-525"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VDD=30;                     #Drain supply voltage, V\n",
-      "ID=2.5;                     #Drain current, mA\n",
-      "VDS=8;                      #Drain-source voltage, V\n",
-      "VGS_off=-5;                 #Gate-source cutoff voltage, V\n",
-      "R1=1;                       #Resistor R1, M\u03a9\n",
-      "R2=500;                     #Resistor R2, k\u03a9\n",
-      "IDSS=10;                    #Shorted gate drain current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#ID=IDSS*square_of(1-(VGS/VGS_off))\n",
-      "VGS=VGS_off*(1-sqrt(ID/IDSS));           #Gate-source voltage, V\n",
-      "V2=VDD*R2/(R1*1000+R2);                       #Voltage across R2, V\n",
-      "\n",
-      "\n",
-      "#V2=VGS+ID*RS\n",
-      "RS=(V2-VGS)/ID;                         #Source resistor, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"Source resistor, RS=%dk\u03a9.\"%RS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Source resistor, RS=5k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 66
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "\n",
-      "Example 19.22 : Page number 528-529\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=20.0;                 #Drain supply voltage, V\n",
-      "RS=50.0;                  #Source resistor, \u03a9\n",
-      "RD=150.0;                 #Drain resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "VDS_max=VDD;                           #Maximum drain source voltage, V\n",
-      "ID_max=(VDD/(RD+RS))*1000;             #Maximum drain current, mA\n",
-      "\n",
-      "\n",
-      "#plot\n",
-      "x=[i for i in range(0,(int)(VDS_max+1))];          #Plot variable for V_DS\n",
-      "y=[(i/(RD+RS))*1000 for i in reversed(x[:])];      #Plot variable for ID\n",
-      "p=plot(x,y);\n",
-      "xlabel(\"VDS(V)\");\n",
-      "ylabel(\"ID(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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4wRkbG+s8d+6cJ340oFEoB0BSYmKi3nnnHV2+fFlS+amOT5w4UeVU8ZJ0ww03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\nTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7fafb8158f50>"
-       ]
-      }
-     ],
-     "prompt_number": 68
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.23 : Page number 529"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=20;                 #Drain supply voltage, V\n",
-      "RD=500.0;                 #Drain resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "VDS_max=VDD;                           #Maximum drain source voltage, v \n",
-      "ID_max=(VDD/RD)*1000;                  #Maximum drain current, mA\n",
-      "\n",
-      "#Plot\n",
-      "x=[i for i in range(0,(int)(VDS_max+1))];          #Plot variable for V_DS\n",
-      "y=[(i/RD)*1000 for i in reversed(x[:])];      #Plot variable for ID\n",
-      "p=plot(x,y);\n",
-      "xlabel(\"VDS(V)\");\n",
-      "ylabel(\"ID(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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7MASIawVEtlPU5iCivuAeRETSYwiQ4rWvFaSnc62AyNEYAqQKggAsX85WQORoDAFSFe5B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\nEWnY/wGORYC0smnRPgAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7fafb804f310>"
-       ]
-      }
-     ],
-     "prompt_number": 71
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.24 : Page number 530-531"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=20;                 #Drain supply voltage, V\n",
-      "RD=12.0;                  #Drain resistor, k\u03a9\n",
-      "RL=8.0;                   #Load resistor, k\u03a9\n",
-      "RG=1.0;                   #Gate resistor, M\u03a9\n",
-      "gm=1.0;                   #transconductance, mA/V\n",
-      "\n",
-      "#Calculation\n",
-      "gm=gm*10**-3;                   #transconductance, mho\n",
-      "RAC=(RD*RL)/(RD+RL);            #Total a.c load, k\u03a9\n",
-      "Av=gm*RAC*1000;                 #Voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage gain=%.1f.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain=4.8.\n"
-       ]
-      }
-     ],
-     "prompt_number": 73
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.25 : Page number 531"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "gm=3000;                   #transconductance, \u03bcmho\n",
-      "RD=10;                     #Drain resistance, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Av=gm*10**-6*RD*1000;       #Voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage gain=%d.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain=30.\n"
-       ]
-      }
-     ],
-     "prompt_number": 74
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.26 : Page number 531"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IDSS=8;                     #Shorted gate drain current, mA\n",
-      "VGS_off=-10;                #Gate-source cut-off voltage, V\n",
-      "ID=1.9;                     #Drain current, mA\n",
-      "RD=3.3;                     #Drain resistance, k\u03a9\n",
-      "RS=2.7;                     #Source resistor, k\u03a9\n",
-      "vin=100;                    #Input voltage, mV\n",
-      "\n",
-      "#Calculation\n",
-      "VGS=-ID*RS;                      #Gate-source voltage, V\n",
-      "gmo=2*IDSS*10**-3/abs(VGS_off);  #Maximum transconductance, S\n",
-      "gm=gmo*(1-(VGS/VGS_off));        #Transconductance, S\n",
-      "Av=gm*RD*1000;                   #Voltage gain\n",
-      "vout=Av*vin;                     #Output voltage, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%dmV(r.m.s).\"%vout);\n",
-      "                \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=257mV(r.m.s).\n"
-       ]
-      }
-     ],
-     "prompt_number": 76
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.27 : Page number 531-532"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=4.7;                    #Load resistor, \u03a9\n",
-      "RD=3.3;                     #Drain resistance, k\u03a9\n",
-      "gm=779*10**-6;              #Transconductance, S\n",
-      "vin=100;                    #Input voltage, mV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "RAC=RD*RL/(RD+RL);          #Total a.c drain resistance, k\u03a9\n",
-      "Av=gm*RAC*1000;             #Voltage gain\n",
-      "vout=Av*vin;                     #Output voltage, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%dmV(r.m.s).\"%vout);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=151mV(r.m.s).\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.28 : Page number 532-533"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RD=1.5;            #Drain resistance, k\u03a9\n",
-      "gm=4;              #Transconductance, mS\n",
-      "RS=560;            #Source resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Av=gm*10**-3*RD*1000/(1+gm*10**-3*RS);\n",
-      "print(\"Voltage gain=%.2f.\"%Av);\n",
-      "\n",
-      "#If RS is bypassed by a capacitor\n",
-      "Av=gm*10**-3*RD*1000;\n",
-      "print(\"Voltage gain, if RS resistor is bypassed=%d.\"%Av);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain=1.85.\n",
-        "Voltage gain, if RS resistor is bypassed=6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.29 : Page number 533"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "from math import sqrt\n",
-      "\n",
-      "IDSS=10;                     #Shorted gate drain current, mA\n",
-      "VGS_off=-3.5;                #Gate-source cut-off voltage, V\n",
-      "RD=1.5;            #Drain resistance, k\u03a9\n",
-      "RS=750;            #Source resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#From d.c biasing\n",
-      "ID=2.3;                                     #Drain current, mA\n",
-      "VGS=round(VGS_off*(1-sqrt(ID/IDSS)),1);              #Gate-source voltage, V\n",
-      "gm=round(round((2*IDSS/abs(VGS_off)),1)*round((1-(VGS/VGS_off)),3),2);           #Transconductance, mS\n",
-      "\n",
-      "\n",
-      "#(i)\n",
-      "Av=gm*RD;                         #Voltage gain with RS resistor bypassed\n",
-      "print(\"(i) Voltage gain with RS bypassed=%.3f.\"%Av);\n",
-      "\n",
-      "#(ii)\n",
-      "Av=Av/(1+gm*(RS/1000.0));\n",
-      "print(\"(ii) Voltage gain with RS unbypassed=%.2f.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Voltage gain with RS bypassed=4.155.\n",
-        "(ii) Voltage gain with RS unbypassed=1.35.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.30 : Page number 539-540"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IDSS=10.0;                     #Shorted gate drain current, mA\n",
-      "VGS_off=-8.0;                  #Gate-source cut-off voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "if(VGS_off<0):\n",
-      "    print(\"(i)  n-channel D-MOSFET\");\n",
-      "else:\n",
-      "    print(\"(i)  p-channel D-MOSFET\");\n",
-      "    \n",
-      "\n",
-      "#(ii)\n",
-      "VGS=-3.0;                                #Gate-source voltage, V\n",
-      "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-      "print(\"(ii) Drain current=%.2fmA\"%ID);\n",
-      "\n",
-      "#(iii)\n",
-      "VGS=3.0;                                #Gate-source voltage, V\n",
-      "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-      "print(\"(iii) Drain current=%.1fmA\"%ID);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  n-channel D-MOSFET\n",
-        "(ii) Drain current=3.91mA\n",
-        "(iii) Drain current=18.9mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.31 : Page number 540\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IDSS=1.0;                     #Shorted gate drain current, mA\n",
-      "VGS_off=-6.0;                  #Gate-source cut-off voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Point 1\n",
-      "VGS=0;                  #Gate source voltage, V               \n",
-      "ID=IDSS;                #Drain current, mA\n",
-      "print(\"Point 1: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n",
-      "\n",
-      "#Point 2\n",
-      "VGS=VGS_off;                  #Gate source voltage, V               \n",
-      "ID=0;                        #Drain current, mA\n",
-      "print(\"Point 2: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n",
-      "\n",
-      "#locating more points by changing VG values\n",
-      "VGS=-3;                  #Gate source voltage, V               \n",
-      "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-      "print(\"Point 3: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n",
-      "\n",
-      "VGS=-1;                  #Gate source voltage, V               \n",
-      "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-      "print(\"Point 4: VGS=%dV and ID=%.3fmA.\"%(VGS,ID));\n",
-      "\n",
-      "VGS=1;                  #Gate source voltage, V               \n",
-      "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-      "print(\"Point 5: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n",
-      "\n",
-      "VGS=3;                  #Gate source voltage, V               \n",
-      "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-      "print(\"Point 6: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Point 1: VGS=0V and ID=1mA.\n",
-        "Point 2: VGS=-6V and ID=0mA.\n",
-        "Point 3: VGS=-3V and ID=0.25mA.\n",
-        "Point 4: VGS=-1V and ID=0.694mA.\n",
-        "Point 5: VGS=1V and ID=1.36mA.\n",
-        "Point 6: VGS=3V and ID=2.25mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.32 : Page number 541"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=18;          #Drain supply voltage, V\n",
-      "RD=620.0;           #Drain resistor, \u03a9\n",
-      "IDSS=12.0;        #Shorted gate drain current, mA\n",
-      "VGS_off=-8.0;      #Gate-source cut-off voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "ID=IDSS;               #Drain current, mA\n",
-      "VDS=VDD-IDSS*(RD/1000);       #Drain source voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Drain source voltage=%.1fV.\"%VDS);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain source voltage=10.6V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.33 : Page number 542"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=15;           #Drain supply voltage\n",
-      "RD=620.0;          #Drain resistor, \u03a9\n",
-      "RL=8.2;          #Load resistor, k\u03a9\n",
-      "vin=500.0;         #Input voltage, V\n",
-      "IDSS=12.0;        #Shorted gate drain current, mA\n",
-      "gm=3.2;            #Transconductance, mS\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "VDS=VDD-IDSS*(RD/1000.0);       #Drain source voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "RAC=RD*RL*1000/(RD+RL*1000);  #Total a.c drain resistace, \u03a9\n",
-      "vout=(gm/1000.0)*RAC*vin;     #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  Drain source voltage=%.2fV.\"%VDS);\n",
-      "print(\"(ii) Output voltage=%dmV\"%vout);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Drain source voltage=7.56V.\n",
-        "(ii) Output voltage=922mV\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.34 : Page number 545"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ID_on=500.0;             #Drain current for MOSFET ON, mA\n",
-      "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-      "VGS_th=1.0;              #Threshold value of gate-source voltage, V\n",
-      "VGS=5;                 #Gate-source voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "K=round(ID_on/(VGS_on-VGS_th)**2,2);   #Constant for a E-MOSFET, mA/V\u00b2\n",
-      "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Drain current=%.1fmA\"%ID);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain current=98.7mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.35 : Page number 545"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ID_on=3.0;             #Drain current for MOSFET ON, mA\n",
-      "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-      "VGS_th=3.0;              #Threshold value of gate-source voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "K=round((ID_on/(VGS_on-VGS_th)**2),3);   #Constant for a E-MOSFET, mA/V\u00b2\n",
-      "print(\"K=%.3fe-03A/V\u00b2.\"%K);\n",
-      "\n",
-      "#Determining different points for plotting\n",
-      "VGS=5;                                 #Gate-source voltage, V\n",
-      "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-      "print(\"For VGS=5V, Drain current=%.3fmA\"%ID);\n",
-      "VGS=8;                                 #Gate-source voltage, V\n",
-      "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-      "print(\"For VGS=8V, Drain current=%.3fmA\"%ID);\n",
-      "VGS=10;                                 #Gate-source voltage, V\n",
-      "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-      "print(\"For VGS=10V, Drain current=%.dmA\"%ID);\n",
-      "VGS=12;                                 #Gate-source voltage, V\n",
-      "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-      "print(\"For VGS=12V, Drain current=%.2fmA\"%ID);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "K=0.061e-03A/V\u00b2.\n",
-        "For VGS=5V, Drain current=0.244mA\n",
-        "For VGS=8V, Drain current=1.525mA\n",
-        "For VGS=10V, Drain current=2mA\n",
-        "For VGS=12V, Drain current=4.94mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.36 : Page number 546-547"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=24.0;             #Drain supply voltage, V\n",
-      "RD=470.0;             #Drain resistor, \u03a9\n",
-      "R1=100.0;             #Resistor R1, k\u03a9\n",
-      "R2=15.0;              #Resistor R2, k\u03a9\n",
-      "ID_on=500.0;             #Drain current for MOSFET ON, mA\n",
-      "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-      "VGS_th=1.0;              #Threshold value of gate-source voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "VGS=VDD*R2/(R1+R2);                      #Gate-source voltage, V (Voltage divider rule)\n",
-      "K=round((ID_on/(VGS_on-VGS_th)**2),2);   #Constant for a E-MOSFET, mA/V\u00b2\n",
-      "ID=K*(VGS-VGS_th)**2;                    #Drain current, mA\n",
-      "VDS=VDD-(ID/1000)*RD;                           #Drain-source voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Drain-source voltage=%.1fV.\"%VDS);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain-source voltage=10.8V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.37 : Page number 547"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=20.0;             #Drain supply voltage, V\n",
-      "RD=1.0;               #Drain resistor, k\u03a9\n",
-      "RG=5.0;               #Gate resistor , M\u03a9\n",
-      "ID_on=10.0;             #Drain current for MOSFET ON, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#since, VGS=VDS\n",
-      "ID=ID_on;              #Drain current, mA\n",
-      "VDS=VDD-ID*RD;         #Drain-source voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Drain current=%dmA.\"%ID);\n",
-      "print(\"Drain-source voltage=%dV.\"%VDS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain current=10mA.\n",
-        "Drain-source voltage=10V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.38 : Page number 547-548"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=10.0;             #Drain supply voltage, V\n",
-      "RD=3.0;             #Drain resistor, k\u03a9\n",
-      "R1=1.0;             #Resistor R1, M\u03a9\n",
-      "R2=1.0;              #Resistor R2, M\u03a9\n",
-      "ID_on=10.0;             #Drain current for MOSFET ON, mA\n",
-      "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-      "VGS_th=1.5;              #Threshold value of gate-source voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "K=round((ID_on/(VGS_on-VGS_th)**2),3);   #Constant for a E-MOSFET, mA/V\u00b2\n",
-      "VGS=VDD*R2/(R1+R2);                      #Gate-source voltage, V (Voltage divider rule)\n",
-      "ID=K*(VGS-VGS_th)**2;                    #Drain current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Drain current=%.2fmA.\"%ID);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain current=1.69mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_1.ipynb
deleted file mode 100755
index 5c74053c..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_1.ipynb
+++ /dev/null
@@ -1,1666 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e80dba2d461291c68fb06b32e702542441b6f13e4e124da0b4826c1be4471e9c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 19 : FIELD EFFECT TRANSISTORS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.1 : Page number 515\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_DSS=12.0;                                 #Shorted gate drain current, mA\n",
-      "V_GS_off=-5.0;                              #Gate-source cut-off voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"ID=%d[1 + VGS/%d]\u00b2mA.\"%(I_DSS,abs(V_GS_off)));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ID=12[1 + VGS/5]\u00b2mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.2 : Page number 515\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_DSS=32.0;                                 #Shorted gate drain current, mA\n",
-      "V_GS_off=-8.0;                              #Gate-source cut-off voltage, V\n",
-      "V_GS=-4.5;                                  #Gate-source voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The drain current=%.2fmA.\"%I_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drain current=6.12mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.3 : Page number  515\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_DSS=10.0;                                 #Shorted gate drain current, mA\n",
-      "V_GS_off=-6.0;                              #Gate-source cut-off voltage, V\n",
-      "I_D=5.0;                                    #Drain current mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, I_D=I_DSS*[1 - (V_GS/V_GS_off)]\u00b2\n",
-      "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS));              #Gate-source voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_P=-V_GS_off;                                  #Pinch-off voltage, V        \n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) VGS=%.2fV.\"%V_GS);\n",
-      "print(\"(ii) VP=%dV\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) VGS=-1.76V.\n",
-        "(ii) VP=6V\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.4 : Page number 515-516\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GS_off=-4.0;                              #Gate-source cut-off voltage, V\n",
-      "I_DSS=12.0;                                 #Shorted gate drain current, mA\n",
-      "R_D=560.0;                                  #Drain resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "V_P=-V_GS_off;          #Pinch-off voltage, V\n",
-      "V_DS=V_P;               #Minimum drain-source voltage for JFET to be in constant current region, V\n",
-      "I_D=I_DSS;              #Maximum drain current, mA (V_GS=0)\n",
-      "V_RD=(I_D/1000)*R_D;    #Voltage across drain resistor, V (OHM's LAW)\n",
-      "V_DD=V_DS+V_RD;         #Minimum value of supply voltage to drain, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum value of VDD required =%.2fV.\"%V_DD);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum value of VDD required =10.72V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.5 : Page number 516\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_DSS=3.0;                                 #Shorted gate drain current, mA\n",
-      "V_GS_off=-6.0;                              #Gate-source cut-off voltage, V\n",
-      "V_GS=-2.0;                                  #Gate-source voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The drain current=%.2fmA.\"%I_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drain current=1.33mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.6 : Page number 516"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VGS_off=4;                #Gate-source cut-off voltage, V\n",
-      "VGS=6;                    #Gate source voltage, V\n",
-      "\n",
-      "print(\"p-channel JFET requires a positive gate-to-source voltage to pass drain current.\");\n",
-      "print(\"More positive voltage, the less the drain current. \");\n",
-      "print(\"Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\");"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "p-channel JFET requires a positive gate-to-source voltage to pass drain current.\n",
-        "More positive voltage, the less the drain current. \n",
-        "Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.7 : Page number 517-518\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GS=15.0;              #Gate-source voltage, V\n",
-      "I_G=1e-03;               #Gate current, \u03bcA\n",
-      "\n",
-      "#Calculation\n",
-      "R_GS=(V_GS/(I_G*10**-6))/10**6;          #Gate to source resistance, M\u03a9 (OHM's LAW)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The gate to source resistance=%dM\u03a9.\"%R_GS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The gate to source resistance=15000M\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 47
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.8 : Page  number 518\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "V_GS_max=-3.1;                      #Maximum gate to source voltage, V\n",
-      "V_GS_min=-3.0;                      #Minimum gate to source voltage, V\n",
-      "I_D_max=1.3;                      #Maximum drain current, mA\n",
-      "I_D_min=1.0;                      #Minimum drain current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "delta_V_GS=abs(V_GS_max-V_GS_min);              #Change in gate to source voltage, V\n",
-      "delta_I_D=I_D_max-I_D_min;                      #Change in drain current, mA\n",
-      "g_fs=(delta_I_D/delta_V_GS)*1000;               #Transconductance, \u03bc mho\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Transconductance=%.0f \u03bc mho\"%g_fs);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Transconductance=3000 \u03bc mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.9 : Page number 518\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GS=[0,0,-0.2];            #Readings of Gate-source voltage, V\n",
-      "V_DS=[7,15,15];             #Readings of Drain-source voltage, V\n",
-      "ID=[10,10.25,9.65];         #Readings of drain current, mA\n",
-      "\n",
-      "\n",
-      "#Displaying the readings:\n",
-      "print(\"VGS= %dV     %dV        %.1fV\"%(V_GS[0],V_GS[1],V_GS[2]));\n",
-      "print(\"VDS= %dV     %dV        %dV\"%(V_DS[0],V_DS[1],V_DS[2]));\n",
-      "print(\"ID = %dV    %.2fV     %.2fV\"%(ID[0],ID[1],ID[2]));\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "#V_GS constant at 0V,\n",
-      "delta_VDS=V_DS[1]-V_DS[0];          #Change in drain-source voltage, V\n",
-      "delta_ID=ID[1]-ID[0];               #Change in drain current, mA\n",
-      "rd=delta_VDS/delta_ID;              #a.c drain resistance, k\u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "#V_DS constant at 15V,\n",
-      "delta_VGS=V_GS[2]-V_GS[1];                  #Change in gate-source voltage, V\n",
-      "delta_ID=ID[2]-ID[1];                       #Change in drain current, mA\n",
-      "g_fs=round((delta_ID/delta_VGS)*1000,);     #Transconductance, \u03bc mho\n",
-      "\n",
-      "#(iii)\n",
-      "amplification_factor=rd*1000*g_fs*10**-6;                          #Amplification factor\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   The a.c drain resistance=%dk\u03a9.\"%rd);\n",
-      "print(\"(ii)  The transconductance=%d \u03bc mho.\"%g_fs);\n",
-      "print(\"(iii) The amplification factor=%d.\"%amplification_factor );\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VGS= 0V     0V        -0.2V\n",
-        "VDS= 7V     15V        15V\n",
-        "ID = 10V    10.25V     9.65V\n",
-        "(i)   The a.c drain resistance=32k\u03a9.\n",
-        "(ii)  The transconductance=3000 \u03bc mho.\n",
-        "(iii) The amplification factor=96.\n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.10 : Page number 519\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "g_mo=4000.0;                        #Maximum transconductance, \u03bcS\n",
-      "V_GS=-3.0;                          #Gate to source voltage, V\n",
-      "V_GS_off=-8.0;                      #Gate-source cut-off voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "g_m=g_mo*(1-(V_GS/V_GS_off));               #Transconductance, \u03bcS\n",
-      "\n",
-      "#Result\n",
-      "print(\"The transconductance=%d \u03bcS.\"%g_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The transconductance=2500 \u03bcS.\n"
-       ]
-      }
-     ],
-     "prompt_number": 54
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.11 : Page number 519\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "g_mo=5000.0;                        #Maximum transconductance, \u03bcS\n",
-      "V_GS=-4.0;                          #Gate to source voltage, V\n",
-      "V_GS_off=-6.0;                      #Gate-source cut-off voltage, V\n",
-      "I_DSS=3.0;                          #Shorted-gate drain current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "g_m=g_mo*(1-(V_GS/V_GS_off));               #Transconductance, \u03bcS\n",
-      "I_D=(I_DSS*(1-(V_GS/V_GS_off))**2)*1000;                    #Drain current \u03bcA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The transconductance=%.0f \u03bcS.\"%g_m);\n",
-      "print(\"The drain current=%d \u03bcA.\"%I_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The transconductance=1667 \u03bcS.\n",
-        "The drain current=333 \u03bcA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.12 : Page number 520\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GS_off=-8.0;                      #Gate-source cut-off voltage, V\n",
-      "I_DSS=16.0;                         #Shorted-gate drain current, mA\n",
-      "R_D=2.2;                            #Drain resistor, k\u03a9\n",
-      "R_G=1.0;                            #Gate resistor, M\u03a9\n",
-      "V_DD=10.0;                          #Drain supply voltage, V\n",
-      "V_GG=-5.0;                           #Gate supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_GS=V_GG;                                           #Gate-source voltage, V\n",
-      "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current \u03bcA\n",
-      "V_DS=V_DD-I_D*R_D;                                   #Drain-source voltage, V (Kirchhoff's voltage law)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The gate-source voltage=%dV.\"%V_GS);\n",
-      "print(\"The drain current=%.2fmA.\"%I_D);\n",
-      "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The gate-source voltage=-5V.\n",
-        "The drain current=2.25mA.\n",
-        "The drain-source voltage=5.05V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.13 : Page number 521\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_D=5.0;                   #Drain current mA\n",
-      "V_DD=15.0;                 #Drain supply voltage, V\n",
-      "V_G=0;                     #Gate voltage, V\n",
-      "R_D=1.0;                   #Drain resistor, k\u03a9\n",
-      "R_S=470.0;                 #Source resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_S=(I_D/1000)*R_S;                 #Source voltage, V (OHM's LAW)\n",
-      "V_D=V_DD-I_D*R_D;                   #Drain voltage, V (Kirchhoff's voltage law)\n",
-      "V_DS=V_D-V_S;                       #Drain-source voltage, V\n",
-      "V_GS=V_G-V_S;                       #Gate-source voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n",
-      "print(\"The gate-source voltage=%.2fV.\"%V_GS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drain-source voltage=7.65V.\n",
-        "The gate-source voltage=-2.35V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 57
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.14 : Page number 521\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GS=-5.0;                     #Gate-source voltage, V\n",
-      "I_D=6.25;                      #Drain current mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "R_S=abs(V_GS/(I_D/1000));              #Required source resistor, \u03a9 (OHM's LAW)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required source resistor=%d \u03a9.\"%R_S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required source resistor=800 \u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.15 : Page number : 521\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_DSS=25.0;                                 #Shorted gate drain current, mA\n",
-      "V_GS_off=15.0;                              #Gate-source cut-off voltage, V\n",
-      "V_GS=5.0;                                   #Gate-source voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;               #Drain current mA\n",
-      "R_S=V_GS/(I_D/1000);                            #Required source resistor, \u03a9 (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The source resistance=%.0f\u03a9.\"%R_S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The source resistance=450\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 59
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.16 : Page number 522\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_DSS=15.0;                                 #Shorted gate drain current, mA\n",
-      "V_GS_off=-8.0;                              #Gate-source cut-off voltage, V\n",
-      "V_DD=12.0;                                  #Drain supply voltage,V\n",
-      "V_D=V_DD/2;                                 #Drain voltage(half of V_DD), V\n",
-      "\n",
-      "#Calculation\n",
-      "I_D=I_DSS/2;                                #Drain current(approximately half of I_DSS), mA\n",
-      "V_GS=V_GS_off/3.4;                          #Gate-source voltage, V\n",
-      "R_S=abs(V_GS/(I_D/1000));                   #Source resistor, \u03a9 (OHM's LAW)\n",
-      "#Since,V_D=V_DD-I_D*R_D;                      \n",
-      "R_D=(V_DD-V_D)/(I_D/1000);                         #Drain resistor, \u03a9 (OHM's LAW)\n",
-      "\n",
-      "#Result\n",
-      "print(\" RS=%d \u03a9 and RD=%d \u03a9.\"%(R_S,R_D));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " RS=313 \u03a9 and RD=800 \u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.17 : Page number 522\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_DSS=5.0;                       #Shorted gate drain current, mA\n",
-      "V_GS_off=-2.0;                   #Gate-source cut-off voltage, V\n",
-      "V_DS=10.0;                       #Drain-source voltage,V\n",
-      "I_D=1.5;                         #Drain current, mA\n",
-      "V_DD=20.0;                       #Drain supply voltage,V\n",
-      "V_G=0;                           #Gate voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since,  Drain current, I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    \n",
-      "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS));              #Gate-source voltage, V\n",
-      "\n",
-      "#Since, V_GS=V_G-V_S,\n",
-      "V_S=V_G-V_GS;                           #Source voltage, V\n",
-      "\n",
-      "R_S=V_S/I_D;                            #Source resistor, k\u03a9\n",
-      "\n",
-      "#Since, V_DD=I_D*R_D +V_DS+ I_D*R_S,\n",
-      "R_D=(V_DD-I_D*R_S-V_DS)/I_D;                    #Drain resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"The source resistance=%.1f k\u03a9\"%R_S);\n",
-      "print(\"The drain resistance=%d k\u03a9.\"%R_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The source resistance=0.6 k\u03a9\n",
-        "The drain resistance=6 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 61
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.18 : Page number 522-523\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_DD=30.0;                  #Drain supply voltage, V\n",
-      "R_D=5.0;                    #Drain resistor, k\u03a9\n",
-      "I_D=2.5;                    #Drain current, mA\n",
-      "R_S=200.0;                  #Source resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_DS=V_DD-I_D*(R_D+(R_S/1000));                 #Drain-source voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_GS=-(I_D/1000)*R_S;                          #Gate-source voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The drain-source voltage=%dV.\"%V_DS);\n",
-      "print(\"The gate-source voltage=%.1fV.\"%V_GS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The drain-source voltage=17V.\n",
-        "The gate-source voltage=-0.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 62
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.19 : Page number 523"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ID_1=2.15;                  #First stage drain current, mA\n",
-      "ID_2=9.15;                  #Second stage drain current, mA\n",
-      "VDD=30;                     #Drain supply voltage, V\n",
-      "RS_1=0.68;                  #Source resistance of 1st stage, k\u03a9\n",
-      "RS_2=0.22;                  #Source resistance of 2nd stage, k\u03a9\n",
-      "RD_1=8.2;                   #Drain resistor of 1st stage, k\u03a9\n",
-      "RD_2=2;                     #Drain resistor of 2nd stage, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "V_RD1=ID_1*RD_1;                #Voltage drop across 8.2k\u03a9\n",
-      "VD_1=VDD-V_RD1;                 #Drain voltage of 1st stage, V\n",
-      "VS_1=ID_1*RS_1;                 #D.C potential of source of first stage, V\n",
-      "V_RD2=ID_2*RD_2;                #Voltage drop across 2k\u03a9\n",
-      "VD_2=VDD-V_RD2;                 #Drain voltage of 2nd stage, V\n",
-      "VS_2=ID_2*RS_2;                 #D.C potential of source of 2nd stage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"drain voltage of 1st stage=%.2fV.\"%VD_1);\n",
-      "print(\"Source voltage of 1st stage=%.2fV.\"%VS_1);\n",
-      "print(\"drain voltage of 2nd stage=%.1fV.\"%VD_2);\n",
-      "print(\"Source voltage of 2nd stage=%.2fV.\"%VS_2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "drain voltage of 1st stage=12.37V.\n",
-        "Source voltage of 1st stage=1.46V.\n",
-        "drain voltage of 2nd stage=11.7V.\n",
-        "Source voltage of 2nd stage=2.01V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 63
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.20 : Page number 524"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=12;                     #Drain supply voltage, V\n",
-      "VD=7;                       #Drain voltage, V\n",
-      "R1=6.8;                     #Resistor R1, M\u03a9\n",
-      "R2=1;                       #Resistor R2, M\u03a9\n",
-      "RS=1.8;                     #Source resistance, k\u03a9\n",
-      "RD=3.3;                     #Drain resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "ID=(VDD-VD)/RD;                  #Second stage drain current, mA\n",
-      "VS=ID*RS;                        #Source voltage, V\n",
-      "VG=VDD*R2/(R1+R2);               #Drain voltage, V\n",
-      "VGS=VG-VS;                       #Drain-source voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"Drain current=%.2fmA.\"%ID);\n",
-      "print(\"Gate-source voltage=%.1fV.\"%VGS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain current=1.52mA.\n",
-        "Gate-source voltage=-1.2V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 65
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.21 : Page number 524-525"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VDD=30;                     #Drain supply voltage, V\n",
-      "ID=2.5;                     #Drain current, mA\n",
-      "VDS=8;                      #Drain-source voltage, V\n",
-      "VGS_off=-5;                 #Gate-source cutoff voltage, V\n",
-      "R1=1;                       #Resistor R1, M\u03a9\n",
-      "R2=500;                     #Resistor R2, k\u03a9\n",
-      "IDSS=10;                    #Shorted gate drain current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#ID=IDSS*square_of(1-(VGS/VGS_off))\n",
-      "VGS=VGS_off*(1-sqrt(ID/IDSS));           #Gate-source voltage, V\n",
-      "V2=VDD*R2/(R1*1000+R2);                       #Voltage across R2, V\n",
-      "\n",
-      "\n",
-      "#V2=VGS+ID*RS\n",
-      "RS=(V2-VGS)/ID;                         #Source resistor, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"Source resistor, RS=%dk\u03a9.\"%RS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Source resistor, RS=5k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 66
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "\n",
-      "Example 19.22 : Page number 528-529\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=20.0;                 #Drain supply voltage, V\n",
-      "RS=50.0;                  #Source resistor, \u03a9\n",
-      "RD=150.0;                 #Drain resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "VDS_max=VDD;                           #Maximum drain source voltage, V\n",
-      "ID_max=(VDD/(RD+RS))*1000;             #Maximum drain current, mA\n",
-      "\n",
-      "\n",
-      "#plot\n",
-      "x=[i for i in range(0,(int)(VDS_max+1))];          #Plot variable for V_DS\n",
-      "y=[(i/(RD+RS))*1000 for i in reversed(x[:])];      #Plot variable for ID\n",
-      "p=plot(x,y);\n",
-      "xlabel(\"VDS(V)\");\n",
-      "ylabel(\"ID(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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4wRkbG+s8d+6cJ340oFEoB0BSYmKi3nnnHV2+fFlS+amOT5w4UeVU8ZJ0ww03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\nTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7fafb8158f50>"
-       ]
-      }
-     ],
-     "prompt_number": 68
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.23 : Page number 529"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=20;                 #Drain supply voltage, V\n",
-      "RD=500.0;                 #Drain resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "VDS_max=VDD;                           #Maximum drain source voltage, v \n",
-      "ID_max=(VDD/RD)*1000;                  #Maximum drain current, mA\n",
-      "\n",
-      "#Plot\n",
-      "x=[i for i in range(0,(int)(VDS_max+1))];          #Plot variable for V_DS\n",
-      "y=[(i/RD)*1000 for i in reversed(x[:])];      #Plot variable for ID\n",
-      "p=plot(x,y);\n",
-      "xlabel(\"VDS(V)\");\n",
-      "ylabel(\"ID(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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7MASIawVEtlPU5iCivuAeRETSYwiQ4rWvFaSnc62AyNEYAqQKggAsX85WQORoDAFSFe5B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\nEWnY/wGORYC0smnRPgAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7fafb804f310>"
-       ]
-      }
-     ],
-     "prompt_number": 71
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.24 : Page number 530-531"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=20;                 #Drain supply voltage, V\n",
-      "RD=12.0;                  #Drain resistor, k\u03a9\n",
-      "RL=8.0;                   #Load resistor, k\u03a9\n",
-      "RG=1.0;                   #Gate resistor, M\u03a9\n",
-      "gm=1.0;                   #transconductance, mA/V\n",
-      "\n",
-      "#Calculation\n",
-      "gm=gm*10**-3;                   #transconductance, mho\n",
-      "RAC=(RD*RL)/(RD+RL);            #Total a.c load, k\u03a9\n",
-      "Av=gm*RAC*1000;                 #Voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage gain=%.1f.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain=4.8.\n"
-       ]
-      }
-     ],
-     "prompt_number": 73
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.25 : Page number 531"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "gm=3000;                   #transconductance, \u03bcmho\n",
-      "RD=10;                     #Drain resistance, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Av=gm*10**-6*RD*1000;       #Voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage gain=%d.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain=30.\n"
-       ]
-      }
-     ],
-     "prompt_number": 74
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.26 : Page number 531"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IDSS=8;                     #Shorted gate drain current, mA\n",
-      "VGS_off=-10;                #Gate-source cut-off voltage, V\n",
-      "ID=1.9;                     #Drain current, mA\n",
-      "RD=3.3;                     #Drain resistance, k\u03a9\n",
-      "RS=2.7;                     #Source resistor, k\u03a9\n",
-      "vin=100;                    #Input voltage, mV\n",
-      "\n",
-      "#Calculation\n",
-      "VGS=-ID*RS;                      #Gate-source voltage, V\n",
-      "gmo=2*IDSS*10**-3/abs(VGS_off);  #Maximum transconductance, S\n",
-      "gm=gmo*(1-(VGS/VGS_off));        #Transconductance, S\n",
-      "Av=gm*RD*1000;                   #Voltage gain\n",
-      "vout=Av*vin;                     #Output voltage, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%dmV(r.m.s).\"%vout);\n",
-      "                \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=257mV(r.m.s).\n"
-       ]
-      }
-     ],
-     "prompt_number": 76
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.27 : Page number 531-532"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RL=4.7;                    #Load resistor, \u03a9\n",
-      "RD=3.3;                     #Drain resistance, k\u03a9\n",
-      "gm=779*10**-6;              #Transconductance, S\n",
-      "vin=100;                    #Input voltage, mV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "RAC=RD*RL/(RD+RL);          #Total a.c drain resistance, k\u03a9\n",
-      "Av=gm*RAC*1000;             #Voltage gain\n",
-      "vout=Av*vin;                     #Output voltage, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%dmV(r.m.s).\"%vout);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=151mV(r.m.s).\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.28 : Page number 532-533"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RD=1.5;            #Drain resistance, k\u03a9\n",
-      "gm=4;              #Transconductance, mS\n",
-      "RS=560;            #Source resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Av=gm*10**-3*RD*1000/(1+gm*10**-3*RS);\n",
-      "print(\"Voltage gain=%.2f.\"%Av);\n",
-      "\n",
-      "#If RS is bypassed by a capacitor\n",
-      "Av=gm*10**-3*RD*1000;\n",
-      "print(\"Voltage gain, if RS resistor is bypassed=%d.\"%Av);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain=1.85.\n",
-        "Voltage gain, if RS resistor is bypassed=6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.29 : Page number 533"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "from math import sqrt\n",
-      "\n",
-      "IDSS=10;                     #Shorted gate drain current, mA\n",
-      "VGS_off=-3.5;                #Gate-source cut-off voltage, V\n",
-      "RD=1.5;            #Drain resistance, k\u03a9\n",
-      "RS=750;            #Source resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#From d.c biasing\n",
-      "ID=2.3;                                     #Drain current, mA\n",
-      "VGS=round(VGS_off*(1-sqrt(ID/IDSS)),1);              #Gate-source voltage, V\n",
-      "gm=round(round((2*IDSS/abs(VGS_off)),1)*round((1-(VGS/VGS_off)),3),2);           #Transconductance, mS\n",
-      "\n",
-      "\n",
-      "#(i)\n",
-      "Av=gm*RD;                         #Voltage gain with RS resistor bypassed\n",
-      "print(\"(i) Voltage gain with RS bypassed=%.3f.\"%Av);\n",
-      "\n",
-      "#(ii)\n",
-      "Av=Av/(1+gm*(RS/1000.0));\n",
-      "print(\"(ii) Voltage gain with RS unbypassed=%.2f.\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Voltage gain with RS bypassed=4.155.\n",
-        "(ii) Voltage gain with RS unbypassed=1.35.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.30 : Page number 539-540"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IDSS=10.0;                     #Shorted gate drain current, mA\n",
-      "VGS_off=-8.0;                  #Gate-source cut-off voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "if(VGS_off<0):\n",
-      "    print(\"(i)  n-channel D-MOSFET\");\n",
-      "else:\n",
-      "    print(\"(i)  p-channel D-MOSFET\");\n",
-      "    \n",
-      "\n",
-      "#(ii)\n",
-      "VGS=-3.0;                                #Gate-source voltage, V\n",
-      "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-      "print(\"(ii) Drain current=%.2fmA\"%ID);\n",
-      "\n",
-      "#(iii)\n",
-      "VGS=3.0;                                #Gate-source voltage, V\n",
-      "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-      "print(\"(iii) Drain current=%.1fmA\"%ID);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  n-channel D-MOSFET\n",
-        "(ii) Drain current=3.91mA\n",
-        "(iii) Drain current=18.9mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.31 : Page number 540\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "IDSS=1.0;                     #Shorted gate drain current, mA\n",
-      "VGS_off=-6.0;                  #Gate-source cut-off voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Point 1\n",
-      "VGS=0;                  #Gate source voltage, V               \n",
-      "ID=IDSS;                #Drain current, mA\n",
-      "print(\"Point 1: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n",
-      "\n",
-      "#Point 2\n",
-      "VGS=VGS_off;                  #Gate source voltage, V               \n",
-      "ID=0;                        #Drain current, mA\n",
-      "print(\"Point 2: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n",
-      "\n",
-      "#locating more points by changing VG values\n",
-      "VGS=-3;                  #Gate source voltage, V               \n",
-      "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-      "print(\"Point 3: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n",
-      "\n",
-      "VGS=-1;                  #Gate source voltage, V               \n",
-      "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-      "print(\"Point 4: VGS=%dV and ID=%.3fmA.\"%(VGS,ID));\n",
-      "\n",
-      "VGS=1;                  #Gate source voltage, V               \n",
-      "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-      "print(\"Point 5: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n",
-      "\n",
-      "VGS=3;                  #Gate source voltage, V               \n",
-      "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-      "print(\"Point 6: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Point 1: VGS=0V and ID=1mA.\n",
-        "Point 2: VGS=-6V and ID=0mA.\n",
-        "Point 3: VGS=-3V and ID=0.25mA.\n",
-        "Point 4: VGS=-1V and ID=0.694mA.\n",
-        "Point 5: VGS=1V and ID=1.36mA.\n",
-        "Point 6: VGS=3V and ID=2.25mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.32 : Page number 541"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=18;          #Drain supply voltage, V\n",
-      "RD=620.0;           #Drain resistor, \u03a9\n",
-      "IDSS=12.0;        #Shorted gate drain current, mA\n",
-      "VGS_off=-8.0;      #Gate-source cut-off voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "ID=IDSS;               #Drain current, mA\n",
-      "VDS=VDD-IDSS*(RD/1000);       #Drain source voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Drain source voltage=%.1fV.\"%VDS);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain source voltage=10.6V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.33 : Page number 542"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=15;           #Drain supply voltage\n",
-      "RD=620.0;          #Drain resistor, \u03a9\n",
-      "RL=8.2;          #Load resistor, k\u03a9\n",
-      "vin=500.0;         #Input voltage, V\n",
-      "IDSS=12.0;        #Shorted gate drain current, mA\n",
-      "gm=3.2;            #Transconductance, mS\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "VDS=VDD-IDSS*(RD/1000.0);       #Drain source voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "RAC=RD*RL*1000/(RD+RL*1000);  #Total a.c drain resistace, \u03a9\n",
-      "vout=(gm/1000.0)*RAC*vin;     #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  Drain source voltage=%.2fV.\"%VDS);\n",
-      "print(\"(ii) Output voltage=%dmV\"%vout);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Drain source voltage=7.56V.\n",
-        "(ii) Output voltage=922mV\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.34 : Page number 545"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ID_on=500.0;             #Drain current for MOSFET ON, mA\n",
-      "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-      "VGS_th=1.0;              #Threshold value of gate-source voltage, V\n",
-      "VGS=5;                 #Gate-source voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "K=round(ID_on/(VGS_on-VGS_th)**2,2);   #Constant for a E-MOSFET, mA/V\u00b2\n",
-      "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Drain current=%.1fmA\"%ID);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain current=98.7mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.35 : Page number 545"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ID_on=3.0;             #Drain current for MOSFET ON, mA\n",
-      "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-      "VGS_th=3.0;              #Threshold value of gate-source voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "K=round((ID_on/(VGS_on-VGS_th)**2),3);   #Constant for a E-MOSFET, mA/V\u00b2\n",
-      "print(\"K=%.3fe-03A/V\u00b2.\"%K);\n",
-      "\n",
-      "#Determining different points for plotting\n",
-      "VGS=5;                                 #Gate-source voltage, V\n",
-      "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-      "print(\"For VGS=5V, Drain current=%.3fmA\"%ID);\n",
-      "VGS=8;                                 #Gate-source voltage, V\n",
-      "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-      "print(\"For VGS=8V, Drain current=%.3fmA\"%ID);\n",
-      "VGS=10;                                 #Gate-source voltage, V\n",
-      "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-      "print(\"For VGS=10V, Drain current=%.dmA\"%ID);\n",
-      "VGS=12;                                 #Gate-source voltage, V\n",
-      "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-      "print(\"For VGS=12V, Drain current=%.2fmA\"%ID);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "K=0.061e-03A/V\u00b2.\n",
-        "For VGS=5V, Drain current=0.244mA\n",
-        "For VGS=8V, Drain current=1.525mA\n",
-        "For VGS=10V, Drain current=2mA\n",
-        "For VGS=12V, Drain current=4.94mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.36 : Page number 546-547"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=24.0;             #Drain supply voltage, V\n",
-      "RD=470.0;             #Drain resistor, \u03a9\n",
-      "R1=100.0;             #Resistor R1, k\u03a9\n",
-      "R2=15.0;              #Resistor R2, k\u03a9\n",
-      "ID_on=500.0;             #Drain current for MOSFET ON, mA\n",
-      "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-      "VGS_th=1.0;              #Threshold value of gate-source voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "VGS=VDD*R2/(R1+R2);                      #Gate-source voltage, V (Voltage divider rule)\n",
-      "K=round((ID_on/(VGS_on-VGS_th)**2),2);   #Constant for a E-MOSFET, mA/V\u00b2\n",
-      "ID=K*(VGS-VGS_th)**2;                    #Drain current, mA\n",
-      "VDS=VDD-(ID/1000)*RD;                           #Drain-source voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Drain-source voltage=%.1fV.\"%VDS);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain-source voltage=10.8V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.37 : Page number 547"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=20.0;             #Drain supply voltage, V\n",
-      "RD=1.0;               #Drain resistor, k\u03a9\n",
-      "RG=5.0;               #Gate resistor , M\u03a9\n",
-      "ID_on=10.0;             #Drain current for MOSFET ON, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#since, VGS=VDS\n",
-      "ID=ID_on;              #Drain current, mA\n",
-      "VDS=VDD-ID*RD;         #Drain-source voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Drain current=%dmA.\"%ID);\n",
-      "print(\"Drain-source voltage=%dV.\"%VDS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain current=10mA.\n",
-        "Drain-source voltage=10V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 19.38 : Page number 547-548"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VDD=10.0;             #Drain supply voltage, V\n",
-      "RD=3.0;             #Drain resistor, k\u03a9\n",
-      "R1=1.0;             #Resistor R1, M\u03a9\n",
-      "R2=1.0;              #Resistor R2, M\u03a9\n",
-      "ID_on=10.0;             #Drain current for MOSFET ON, mA\n",
-      "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-      "VGS_th=1.5;              #Threshold value of gate-source voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "K=round((ID_on/(VGS_on-VGS_th)**2),3);   #Constant for a E-MOSFET, mA/V\u00b2\n",
-      "VGS=VDD*R2/(R1+R2);                      #Gate-source voltage, V (Voltage divider rule)\n",
-      "ID=K*(VGS-VGS_th)**2;                    #Drain current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Drain current=%.2fmA.\"%ID);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Drain current=1.69mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_2.ipynb
deleted file mode 100755
index 36c07b9b..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_2.ipynb
+++ /dev/null
@@ -1,1660 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 19 : FIELD EFFECT TRANSISTORS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.1 : Page number 515"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "ID=12[1 + VGS/5]²mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=12.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-5.0;                              #Gate-source cut-off voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"ID=%d[1 + VGS/%d]²mA.\"%(I_DSS,abs(V_GS_off)));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.2 : Page number 515"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain current=6.12mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=32.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-8.0;                              #Gate-source cut-off voltage, V\n",
-    "V_GS=-4.5;                                  #Gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain current=%.2fmA.\"%I_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.3 : Page number  515"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) VGS=-1.76V.\n",
-      "(ii) VP=6V\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "\n",
-    "#Variable declaration\n",
-    "I_DSS=10.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-6.0;                              #Gate-source cut-off voltage, V\n",
-    "I_D=5.0;                                    #Drain current mA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "#Since, I_D=I_DSS*[1 - (V_GS/V_GS_off)]²\n",
-    "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS));              #Gate-source voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "V_P=-V_GS_off;                                  #Pinch-off voltage, V        \n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) VGS=%.2fV.\"%V_GS);\n",
-    "print(\"(ii) VP=%dV\"%V_P);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.4 : Page number 515-516"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The minimum value of VDD required =10.72V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS_off=-4.0;                              #Gate-source cut-off voltage, V\n",
-    "I_DSS=12.0;                                 #Shorted gate drain current, mA\n",
-    "R_D=560.0;                                  #Drain resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "V_P=-V_GS_off;          #Pinch-off voltage, V\n",
-    "V_DS=V_P;               #Minimum drain-source voltage for JFET to be in constant current region, V\n",
-    "I_D=I_DSS;              #Maximum drain current, mA (V_GS=0)\n",
-    "V_RD=(I_D/1000)*R_D;    #Voltage across drain resistor, V (OHM's LAW)\n",
-    "V_DD=V_DS+V_RD;         #Minimum value of supply voltage to drain, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The minimum value of VDD required =%.2fV.\"%V_DD);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.5 : Page number 516"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain current=1.33mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=3.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-6.0;                              #Gate-source cut-off voltage, V\n",
-    "V_GS=-2.0;                                  #Gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain current=%.2fmA.\"%I_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.6 : Page number 516"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "p-channel JFET requires a positive gate-to-source voltage to pass drain current.\n",
-      "More positive voltage, the less the drain current. \n",
-      "Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VGS_off=4;                #Gate-source cut-off voltage, V\n",
-    "VGS=6;                    #Gate source voltage, V\n",
-    "\n",
-    "print(\"p-channel JFET requires a positive gate-to-source voltage to pass drain current.\");\n",
-    "print(\"More positive voltage, the less the drain current. \");\n",
-    "print(\"Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\");"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.7 : Page number 517-518"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The gate to source resistance=15000MΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS=15.0;              #Gate-source voltage, V\n",
-    "I_G=1e-03;               #Gate current, μA\n",
-    "\n",
-    "#Calculation\n",
-    "R_GS=(V_GS/(I_G*10**-6))/10**6;          #Gate to source resistance, MΩ (OHM's LAW)\n",
-    "\n",
-    "#Result\n",
-    "print(\"The gate to source resistance=%dMΩ.\"%R_GS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.8 : Page  number 518"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Transconductance=3000 μ mho\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "V_GS_max=-3.1;                      #Maximum gate to source voltage, V\n",
-    "V_GS_min=-3.0;                      #Minimum gate to source voltage, V\n",
-    "I_D_max=1.3;                      #Maximum drain current, mA\n",
-    "I_D_min=1.0;                      #Minimum drain current, mA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "delta_V_GS=abs(V_GS_max-V_GS_min);              #Change in gate to source voltage, V\n",
-    "delta_I_D=I_D_max-I_D_min;                      #Change in drain current, mA\n",
-    "g_fs=(delta_I_D/delta_V_GS)*1000;               #Transconductance, μ mho\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Transconductance=%.0f μ mho\"%g_fs);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.9 : Page number 518"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VGS= 0V     0V        -0.2V\n",
-      "VDS= 7V     15V        15V\n",
-      "ID = 10V    10.25V     9.65V\n",
-      "(i)   The a.c drain resistance=32kΩ.\n",
-      "(ii)  The transconductance=3000 μ mho.\n",
-      "(iii) The amplification factor=96.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS=[0,0,-0.2];            #Readings of Gate-source voltage, V\n",
-    "V_DS=[7,15,15];             #Readings of Drain-source voltage, V\n",
-    "ID=[10,10.25,9.65];         #Readings of drain current, mA\n",
-    "\n",
-    "\n",
-    "#Displaying the readings:\n",
-    "print(\"VGS= %dV     %dV        %.1fV\"%(V_GS[0],V_GS[1],V_GS[2]));\n",
-    "print(\"VDS= %dV     %dV        %dV\"%(V_DS[0],V_DS[1],V_DS[2]));\n",
-    "print(\"ID = %dV    %.2fV     %.2fV\"%(ID[0],ID[1],ID[2]));\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "#V_GS constant at 0V,\n",
-    "delta_VDS=V_DS[1]-V_DS[0];          #Change in drain-source voltage, V\n",
-    "delta_ID=ID[1]-ID[0];               #Change in drain current, mA\n",
-    "rd=delta_VDS/delta_ID;              #a.c drain resistance, kΩ\n",
-    "\n",
-    "#(ii)\n",
-    "#V_DS constant at 15V,\n",
-    "delta_VGS=V_GS[2]-V_GS[1];                  #Change in gate-source voltage, V\n",
-    "delta_ID=ID[2]-ID[1];                       #Change in drain current, mA\n",
-    "g_fs=round((delta_ID/delta_VGS)*1000,);     #Transconductance, μ mho\n",
-    "\n",
-    "#(iii)\n",
-    "amplification_factor=rd*1000*g_fs*10**-6;                          #Amplification factor\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)   The a.c drain resistance=%dkΩ.\"%rd);\n",
-    "print(\"(ii)  The transconductance=%d μ mho.\"%g_fs);\n",
-    "print(\"(iii) The amplification factor=%d.\"%amplification_factor );\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.10 : Page number 519"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The transconductance=2500 μS.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "g_mo=4000.0;                        #Maximum transconductance, μS\n",
-    "V_GS=-3.0;                          #Gate to source voltage, V\n",
-    "V_GS_off=-8.0;                      #Gate-source cut-off voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "g_m=g_mo*(1-(V_GS/V_GS_off));               #Transconductance, μS\n",
-    "\n",
-    "#Result\n",
-    "print(\"The transconductance=%d μS.\"%g_m);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.11 : Page number 519"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The transconductance=1667 μS.\n",
-      "The drain current=333 μA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "g_mo=5000.0;                        #Maximum transconductance, μS\n",
-    "V_GS=-4.0;                          #Gate to source voltage, V\n",
-    "V_GS_off=-6.0;                      #Gate-source cut-off voltage, V\n",
-    "I_DSS=3.0;                          #Shorted-gate drain current, mA\n",
-    "\n",
-    "#Calculation\n",
-    "g_m=g_mo*(1-(V_GS/V_GS_off));               #Transconductance, μS\n",
-    "I_D=(I_DSS*(1-(V_GS/V_GS_off))**2)*1000;                    #Drain current μA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The transconductance=%.0f μS.\"%g_m);\n",
-    "print(\"The drain current=%d μA.\"%I_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.12 : Page number 520"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The gate-source voltage=-5V.\n",
-      "The drain current=2.25mA.\n",
-      "The drain-source voltage=5.05V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS_off=-8.0;                      #Gate-source cut-off voltage, V\n",
-    "I_DSS=16.0;                         #Shorted-gate drain current, mA\n",
-    "R_D=2.2;                            #Drain resistor, kΩ\n",
-    "R_G=1.0;                            #Gate resistor, MΩ\n",
-    "V_DD=10.0;                          #Drain supply voltage, V\n",
-    "V_GG=-5.0;                           #Gate supply voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_GS=V_GG;                                           #Gate-source voltage, V\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current μA\n",
-    "V_DS=V_DD-I_D*R_D;                                   #Drain-source voltage, V (Kirchhoff's voltage law)\n",
-    "\n",
-    "#Result\n",
-    "print(\"The gate-source voltage=%dV.\"%V_GS);\n",
-    "print(\"The drain current=%.2fmA.\"%I_D);\n",
-    "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.13 : Page number 521"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain-source voltage=7.65V.\n",
-      "The gate-source voltage=-2.35V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_D=5.0;                   #Drain current mA\n",
-    "V_DD=15.0;                 #Drain supply voltage, V\n",
-    "V_G=0;                     #Gate voltage, V\n",
-    "R_D=1.0;                   #Drain resistor, kΩ\n",
-    "R_S=470.0;                 #Source resistor, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_S=(I_D/1000)*R_S;                 #Source voltage, V (OHM's LAW)\n",
-    "V_D=V_DD-I_D*R_D;                   #Drain voltage, V (Kirchhoff's voltage law)\n",
-    "V_DS=V_D-V_S;                       #Drain-source voltage, V\n",
-    "V_GS=V_G-V_S;                       #Gate-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n",
-    "print(\"The gate-source voltage=%.2fV.\"%V_GS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.14 : Page number 521"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The required source resistor=800 Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS=-5.0;                     #Gate-source voltage, V\n",
-    "I_D=6.25;                      #Drain current mA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "R_S=abs(V_GS/(I_D/1000));              #Required source resistor, Ω (OHM's LAW)\n",
-    "\n",
-    "#Result\n",
-    "print(\"The required source resistor=%d Ω.\"%R_S);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.15 : Page number : 521"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The source resistance=450Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=25.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=15.0;                              #Gate-source cut-off voltage, V\n",
-    "V_GS=5.0;                                   #Gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;               #Drain current mA\n",
-    "R_S=V_GS/(I_D/1000);                            #Required source resistor, Ω (OHM's LAW)\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The source resistance=%.0fΩ.\"%R_S);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.16 : Page number 522"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      " RS=313 Ω and RD=800 Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=15.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-8.0;                              #Gate-source cut-off voltage, V\n",
-    "V_DD=12.0;                                  #Drain supply voltage,V\n",
-    "V_D=V_DD/2;                                 #Drain voltage(half of V_DD), V\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS/2;                                #Drain current(approximately half of I_DSS), mA\n",
-    "V_GS=V_GS_off/3.4;                          #Gate-source voltage, V\n",
-    "R_S=abs(V_GS/(I_D/1000));                   #Source resistor, Ω (OHM's LAW)\n",
-    "#Since,V_D=V_DD-I_D*R_D;                      \n",
-    "R_D=(V_DD-V_D)/(I_D/1000);                         #Drain resistor, Ω (OHM's LAW)\n",
-    "\n",
-    "#Result\n",
-    "print(\" RS=%d Ω and RD=%d Ω.\"%(R_S,R_D));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.17 : Page number 522"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The source resistance=0.6 kΩ\n",
-      "The drain resistance=6 kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "\n",
-    "#Variable declaration\n",
-    "I_DSS=5.0;                       #Shorted gate drain current, mA\n",
-    "V_GS_off=-2.0;                   #Gate-source cut-off voltage, V\n",
-    "V_DS=10.0;                       #Drain-source voltage,V\n",
-    "I_D=1.5;                         #Drain current, mA\n",
-    "V_DD=20.0;                       #Drain supply voltage,V\n",
-    "V_G=0;                           #Gate voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since,  Drain current, I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    \n",
-    "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS));              #Gate-source voltage, V\n",
-    "\n",
-    "#Since, V_GS=V_G-V_S,\n",
-    "V_S=V_G-V_GS;                           #Source voltage, V\n",
-    "\n",
-    "R_S=V_S/I_D;                            #Source resistor, kΩ\n",
-    "\n",
-    "#Since, V_DD=I_D*R_D +V_DS+ I_D*R_S,\n",
-    "R_D=(V_DD-I_D*R_S-V_DS)/I_D;                    #Drain resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "print(\"The source resistance=%.1f kΩ\"%R_S);\n",
-    "print(\"The drain resistance=%d kΩ.\"%R_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.18 : Page number 522-523"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain-source voltage=17V.\n",
-      "The gate-source voltage=-0.5V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_DD=30.0;                  #Drain supply voltage, V\n",
-    "R_D=5.0;                    #Drain resistor, kΩ\n",
-    "I_D=2.5;                    #Drain current, mA\n",
-    "R_S=200.0;                  #Source resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "V_DS=V_DD-I_D*(R_D+(R_S/1000));                 #Drain-source voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "V_GS=-(I_D/1000)*R_S;                          #Gate-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain-source voltage=%dV.\"%V_DS);\n",
-    "print(\"The gate-source voltage=%.1fV.\"%V_GS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.19 : Page number 523"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "drain voltage of 1st stage=12.37V.\n",
-      "Source voltage of 1st stage=1.46V.\n",
-      "drain voltage of 2nd stage=11.7V.\n",
-      "Source voltage of 2nd stage=2.01V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "ID_1=2.15;                  #First stage drain current, mA\n",
-    "ID_2=9.15;                  #Second stage drain current, mA\n",
-    "VDD=30;                     #Drain supply voltage, V\n",
-    "RS_1=0.68;                  #Source resistance of 1st stage, kΩ\n",
-    "RS_2=0.22;                  #Source resistance of 2nd stage, kΩ\n",
-    "RD_1=8.2;                   #Drain resistor of 1st stage, kΩ\n",
-    "RD_2=2;                     #Drain resistor of 2nd stage, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "V_RD1=ID_1*RD_1;                #Voltage drop across 8.2kΩ\n",
-    "VD_1=VDD-V_RD1;                 #Drain voltage of 1st stage, V\n",
-    "VS_1=ID_1*RS_1;                 #D.C potential of source of first stage, V\n",
-    "V_RD2=ID_2*RD_2;                #Voltage drop across 2kΩ\n",
-    "VD_2=VDD-V_RD2;                 #Drain voltage of 2nd stage, V\n",
-    "VS_2=ID_2*RS_2;                 #D.C potential of source of 2nd stage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"drain voltage of 1st stage=%.2fV.\"%VD_1);\n",
-    "print(\"Source voltage of 1st stage=%.2fV.\"%VS_1);\n",
-    "print(\"drain voltage of 2nd stage=%.1fV.\"%VD_2);\n",
-    "print(\"Source voltage of 2nd stage=%.2fV.\"%VS_2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.20 : Page number 524"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=1.52mA.\n",
-      "Gate-source voltage=-1.2V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=12;                     #Drain supply voltage, V\n",
-    "VD=7;                       #Drain voltage, V\n",
-    "R1=6.8;                     #Resistor R1, MΩ\n",
-    "R2=1;                       #Resistor R2, MΩ\n",
-    "RS=1.8;                     #Source resistance, kΩ\n",
-    "RD=3.3;                     #Drain resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "ID=(VDD-VD)/RD;                  #Second stage drain current, mA\n",
-    "VS=ID*RS;                        #Source voltage, V\n",
-    "VG=VDD*R2/(R1+R2);               #Drain voltage, V\n",
-    "VGS=VG-VS;                       #Drain-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "print(\"Drain current=%.2fmA.\"%ID);\n",
-    "print(\"Gate-source voltage=%.1fV.\"%VGS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.21 : Page number 524-525"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Source resistor, RS=5kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VDD=30;                     #Drain supply voltage, V\n",
-    "ID=2.5;                     #Drain current, mA\n",
-    "VDS=8;                      #Drain-source voltage, V\n",
-    "VGS_off=-5;                 #Gate-source cutoff voltage, V\n",
-    "R1=1;                       #Resistor R1, MΩ\n",
-    "R2=500;                     #Resistor R2, kΩ\n",
-    "IDSS=10;                    #Shorted gate drain current, mA\n",
-    "\n",
-    "#Calculation\n",
-    "#ID=IDSS*square_of(1-(VGS/VGS_off))\n",
-    "VGS=VGS_off*(1-sqrt(ID/IDSS));           #Gate-source voltage, V\n",
-    "V2=VDD*R2/(R1*1000+R2);                       #Voltage across R2, V\n",
-    "\n",
-    "\n",
-    "#V2=VGS+ID*RS\n",
-    "RS=(V2-VGS)/ID;                         #Source resistor, kΩ\n",
-    "\n",
-    "#Result\n",
-    "print(\"Source resistor, RS=%dkΩ.\"%RS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##  Example 19.22 : Page number 528-529"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fbbbf169d68>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VDD=20.0;                 #Drain supply voltage, V\n",
-    "RS=50.0;                  #Source resistor, Ω\n",
-    "RD=150.0;                 #Drain resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "VDS_max=VDD;                           #Maximum drain source voltage, V\n",
-    "ID_max=(VDD/(RD+RS))*1000;             #Maximum drain current, mA\n",
-    "\n",
-    "\n",
-    "#plot\n",
-    "x=[i for i in range(0,(int)(VDS_max+1))];          #Plot variable for V_DS\n",
-    "y=[(i/(RD+RS))*1000 for i in reversed(x[:])];      #Plot variable for ID\n",
-    "\n",
-    "\n",
-    "plt.plot(x,y);\n",
-    "plt.xlabel(\"VDS(V)\");\n",
-    "plt.ylabel(\"ID(mA)\");\n",
-    "plt.title(\"d.c load line\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.23 : Page number 529"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fbbbf188710>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VDD=20;                 #Drain supply voltage, V\n",
-    "RD=500.0;                 #Drain resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "VDS_max=VDD;                           #Maximum drain source voltage, v \n",
-    "ID_max=(VDD/RD)*1000;                  #Maximum drain current, mA\n",
-    "\n",
-    "#Plot\n",
-    "x=[i for i in range(0,(int)(VDS_max+1))];          #Plot variable for V_DS\n",
-    "y=[(i/RD)*1000 for i in reversed(x[:])];      #Plot variable for ID\n",
-    "\n",
-    "plt.plot(x,y);\n",
-    "plt.xlabel(\"VDS(V)\");\n",
-    "plt.ylabel(\"ID(mA)\");\n",
-    "plt.title(\"d.c load line\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.24 : Page number 530-531"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain=4.8.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=20;                 #Drain supply voltage, V\n",
-    "RD=12.0;                  #Drain resistor, kΩ\n",
-    "RL=8.0;                   #Load resistor, kΩ\n",
-    "RG=1.0;                   #Gate resistor, MΩ\n",
-    "gm=1.0;                   #transconductance, mA/V\n",
-    "\n",
-    "#Calculation\n",
-    "gm=gm*10**-3;                   #transconductance, mho\n",
-    "RAC=(RD*RL)/(RD+RL);            #Total a.c load, kΩ\n",
-    "Av=gm*RAC*1000;                 #Voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Voltage gain=%.1f.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.25 : Page number 531"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain=30.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "gm=3000;                   #transconductance, μmho\n",
-    "RD=10;                     #Drain resistance, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "Av=gm*10**-6*RD*1000;       #Voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"Voltage gain=%d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.26 : Page number 531"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=257mV(r.m.s).\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IDSS=8;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-10;                #Gate-source cut-off voltage, V\n",
-    "ID=1.9;                     #Drain current, mA\n",
-    "RD=3.3;                     #Drain resistance, kΩ\n",
-    "RS=2.7;                     #Source resistor, kΩ\n",
-    "vin=100;                    #Input voltage, mV\n",
-    "\n",
-    "#Calculation\n",
-    "VGS=-ID*RS;                      #Gate-source voltage, V\n",
-    "gmo=2*IDSS*10**-3/abs(VGS_off);  #Maximum transconductance, S\n",
-    "gm=gmo*(1-(VGS/VGS_off));        #Transconductance, S\n",
-    "Av=gm*RD*1000;                   #Voltage gain\n",
-    "vout=Av*vin;                     #Output voltage, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dmV(r.m.s).\"%vout);\n",
-    "                \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.27 : Page number 531-532"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 28,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=151mV(r.m.s).\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RL=4.7;                    #Load resistor, Ω\n",
-    "RD=3.3;                     #Drain resistance, kΩ\n",
-    "gm=779*10**-6;              #Transconductance, S\n",
-    "vin=100;                    #Input voltage, mV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "RAC=RD*RL/(RD+RL);          #Total a.c drain resistance, kΩ\n",
-    "Av=gm*RAC*1000;             #Voltage gain\n",
-    "vout=Av*vin;                     #Output voltage, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dmV(r.m.s).\"%vout);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.28 : Page number 532-533"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 29,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain=1.85.\n",
-      "Voltage gain, if RS resistor is bypassed=6.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RD=1.5;            #Drain resistance, kΩ\n",
-    "gm=4;              #Transconductance, mS\n",
-    "RS=560;            #Source resistance, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "Av=gm*10**-3*RD*1000/(1+gm*10**-3*RS);\n",
-    "print(\"Voltage gain=%.2f.\"%Av);\n",
-    "\n",
-    "#If RS is bypassed by a capacitor\n",
-    "Av=gm*10**-3*RD*1000;\n",
-    "print(\"Voltage gain, if RS resistor is bypassed=%d.\"%Av);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.29 : Page number 533"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 30,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Voltage gain with RS bypassed=4.155.\n",
-      "(ii) Voltage gain with RS unbypassed=1.35.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "from math import sqrt\n",
-    "\n",
-    "IDSS=10;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-3.5;                #Gate-source cut-off voltage, V\n",
-    "RD=1.5;            #Drain resistance, kΩ\n",
-    "RS=750;            #Source resistance, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#From d.c biasing\n",
-    "ID=2.3;                                     #Drain current, mA\n",
-    "VGS=round(VGS_off*(1-sqrt(ID/IDSS)),1);              #Gate-source voltage, V\n",
-    "gm=round(round((2*IDSS/abs(VGS_off)),1)*round((1-(VGS/VGS_off)),3),2);           #Transconductance, mS\n",
-    "\n",
-    "\n",
-    "#(i)\n",
-    "Av=gm*RD;                         #Voltage gain with RS resistor bypassed\n",
-    "print(\"(i) Voltage gain with RS bypassed=%.3f.\"%Av);\n",
-    "\n",
-    "#(ii)\n",
-    "Av=Av/(1+gm*(RS/1000.0));\n",
-    "print(\"(ii) Voltage gain with RS unbypassed=%.2f.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.30 : Page number 539-540"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  n-channel D-MOSFET\n",
-      "(ii) Drain current=3.91mA\n",
-      "(iii) Drain current=18.9mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IDSS=10.0;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-8.0;                  #Gate-source cut-off voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "if(VGS_off<0):\n",
-    "    print(\"(i)  n-channel D-MOSFET\");\n",
-    "else:\n",
-    "    print(\"(i)  p-channel D-MOSFET\");\n",
-    "    \n",
-    "\n",
-    "#(ii)\n",
-    "VGS=-3.0;                                #Gate-source voltage, V\n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"(ii) Drain current=%.2fmA\"%ID);\n",
-    "\n",
-    "#(iii)\n",
-    "VGS=3.0;                                #Gate-source voltage, V\n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"(iii) Drain current=%.1fmA\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.31 : Page number 540"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 32,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Point 1: VGS=0V and ID=1mA.\n",
-      "Point 2: VGS=-6V and ID=0mA.\n",
-      "Point 3: VGS=-3V and ID=0.25mA.\n",
-      "Point 4: VGS=-1V and ID=0.694mA.\n",
-      "Point 5: VGS=1V and ID=1.36mA.\n",
-      "Point 6: VGS=3V and ID=2.25mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IDSS=1.0;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-6.0;                  #Gate-source cut-off voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Point 1\n",
-    "VGS=0;                  #Gate source voltage, V               \n",
-    "ID=IDSS;                #Drain current, mA\n",
-    "print(\"Point 1: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n",
-    "\n",
-    "#Point 2\n",
-    "VGS=VGS_off;                  #Gate source voltage, V               \n",
-    "ID=0;                        #Drain current, mA\n",
-    "print(\"Point 2: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n",
-    "\n",
-    "#locating more points by changing VG values\n",
-    "VGS=-3;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 3: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n",
-    "\n",
-    "VGS=-1;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 4: VGS=%dV and ID=%.3fmA.\"%(VGS,ID));\n",
-    "\n",
-    "VGS=1;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 5: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n",
-    "\n",
-    "VGS=3;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 6: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.32 : Page number 541"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 33,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain source voltage=10.6V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=18;          #Drain supply voltage, V\n",
-    "RD=620.0;           #Drain resistor, Ω\n",
-    "IDSS=12.0;        #Shorted gate drain current, mA\n",
-    "VGS_off=-8.0;      #Gate-source cut-off voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "ID=IDSS;               #Drain current, mA\n",
-    "VDS=VDD-IDSS*(RD/1000);       #Drain source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain source voltage=%.1fV.\"%VDS);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.33 : Page number 542"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Drain source voltage=7.56V.\n",
-      "(ii) Output voltage=922mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=15;           #Drain supply voltage\n",
-    "RD=620.0;          #Drain resistor, Ω\n",
-    "RL=8.2;          #Load resistor, kΩ\n",
-    "vin=500.0;         #Input voltage, V\n",
-    "IDSS=12.0;        #Shorted gate drain current, mA\n",
-    "gm=3.2;            #Transconductance, mS\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "VDS=VDD-IDSS*(RD/1000.0);       #Drain source voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "RAC=RD*RL*1000/(RD+RL*1000);  #Total a.c drain resistace, Ω\n",
-    "vout=(gm/1000.0)*RAC*vin;     #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  Drain source voltage=%.2fV.\"%VDS);\n",
-    "print(\"(ii) Output voltage=%dmV\"%vout);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.34 : Page number 545"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 35,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=98.7mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "ID_on=500.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=1.0;              #Threshold value of gate-source voltage, V\n",
-    "VGS=5;                 #Gate-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "K=round(ID_on/(VGS_on-VGS_th)**2,2);   #Constant for a E-MOSFET, mA/V²\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain current=%.1fmA\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.35 : Page number 545"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 36,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "K=0.061e-03A/V².\n",
-      "For VGS=5V, Drain current=0.244mA\n",
-      "For VGS=8V, Drain current=1.525mA\n",
-      "For VGS=10V, Drain current=2mA\n",
-      "For VGS=12V, Drain current=4.94mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "ID_on=3.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=3.0;              #Threshold value of gate-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "K=round((ID_on/(VGS_on-VGS_th)**2),3);   #Constant for a E-MOSFET, mA/V²\n",
-    "print(\"K=%.3fe-03A/V².\"%K);\n",
-    "\n",
-    "#Determining different points for plotting\n",
-    "VGS=5;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=5V, Drain current=%.3fmA\"%ID);\n",
-    "VGS=8;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=8V, Drain current=%.3fmA\"%ID);\n",
-    "VGS=10;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=10V, Drain current=%.dmA\"%ID);\n",
-    "VGS=12;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=12V, Drain current=%.2fmA\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.36 : Page number 546-547"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 37,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain-source voltage=10.8V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=24.0;             #Drain supply voltage, V\n",
-    "RD=470.0;             #Drain resistor, Ω\n",
-    "R1=100.0;             #Resistor R1, kΩ\n",
-    "R2=15.0;              #Resistor R2, kΩ\n",
-    "ID_on=500.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=1.0;              #Threshold value of gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "VGS=VDD*R2/(R1+R2);                      #Gate-source voltage, V (Voltage divider rule)\n",
-    "K=round((ID_on/(VGS_on-VGS_th)**2),2);   #Constant for a E-MOSFET, mA/V²\n",
-    "ID=K*(VGS-VGS_th)**2;                    #Drain current, mA\n",
-    "VDS=VDD-(ID/1000)*RD;                           #Drain-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain-source voltage=%.1fV.\"%VDS);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.37 : Page number 547"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 38,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=10mA.\n",
-      "Drain-source voltage=10V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=20.0;             #Drain supply voltage, V\n",
-    "RD=1.0;               #Drain resistor, kΩ\n",
-    "RG=5.0;               #Gate resistor , MΩ\n",
-    "ID_on=10.0;             #Drain current for MOSFET ON, mA\n",
-    "\n",
-    "#Calculation\n",
-    "#since, VGS=VDS\n",
-    "ID=ID_on;              #Drain current, mA\n",
-    "VDS=VDD-ID*RD;         #Drain-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain current=%dmA.\"%ID);\n",
-    "print(\"Drain-source voltage=%dV.\"%VDS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.38 : Page number 547-548"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 39,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=1.69mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=10.0;             #Drain supply voltage, V\n",
-    "RD=3.0;             #Drain resistor, kΩ\n",
-    "R1=1.0;             #Resistor R1, MΩ\n",
-    "R2=1.0;              #Resistor R2, MΩ\n",
-    "ID_on=10.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=1.5;              #Threshold value of gate-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "K=round((ID_on/(VGS_on-VGS_th)**2),3);   #Constant for a E-MOSFET, mA/V²\n",
-    "VGS=VDD*R2/(R1+R2);                      #Gate-source voltage, V (Voltage divider rule)\n",
-    "ID=K*(VGS-VGS_th)**2;                    #Drain current, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain current=%.2fmA.\"%ID);\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_3.ipynb
deleted file mode 100755
index 2da379bc..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_3.ipynb
+++ /dev/null
@@ -1,1671 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 19 : FIELD EFFECT TRANSISTORS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.1 : Page number 515"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "ID=12[1 + VGS/5]²mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=12.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-5.0;                              #Gate-source cut-off voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"ID=%d[1 + VGS/%d]²mA.\"%(I_DSS,abs(V_GS_off)));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.2 : Page number 515"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain current=6.12mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=32.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-8.0;                              #Gate-source cut-off voltage, V\n",
-    "V_GS=-4.5;                                  #Gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain current=%.2fmA.\"%I_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.3 : Page number  515"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) VGS=-1.76V.\n",
-      "(ii) VP=6V\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "\n",
-    "#Variable declaration\n",
-    "I_DSS=10.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-6.0;                              #Gate-source cut-off voltage, V\n",
-    "I_D=5.0;                                    #Drain current mA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "#Since, I_D=I_DSS*[1 - (V_GS/V_GS_off)]²\n",
-    "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS));              #Gate-source voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "V_P=-V_GS_off;                                  #Pinch-off voltage, V        \n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) VGS=%.2fV.\"%V_GS);\n",
-    "print(\"(ii) VP=%dV\"%V_P);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.4 : Page number 515-516"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The minimum value of VDD required =10.72V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS_off=-4.0;                              #Gate-source cut-off voltage, V\n",
-    "I_DSS=12.0;                                 #Shorted gate drain current, mA\n",
-    "R_D=560.0;                                  #Drain resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "V_P=-V_GS_off;          #Pinch-off voltage, V\n",
-    "V_DS=V_P;               #Minimum drain-source voltage for JFET to be in constant current region, V\n",
-    "I_D=I_DSS;              #Maximum drain current, mA (V_GS=0)\n",
-    "V_RD=(I_D/1000)*R_D;    #Voltage across drain resistor, V (OHM's LAW)\n",
-    "V_DD=V_DS+V_RD;         #Minimum value of supply voltage to drain, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The minimum value of VDD required =%.2fV.\"%V_DD);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.5 : Page number 516"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain current=1.33mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=3.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-6.0;                              #Gate-source cut-off voltage, V\n",
-    "V_GS=-2.0;                                  #Gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain current=%.2fmA.\"%I_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.6 : Page number 516"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "p-channel JFET requires a positive gate-to-source voltage to pass drain current.\n",
-      "More positive voltage, the less the drain current. \n",
-      "Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VGS_off=4;                #Gate-source cut-off voltage, V\n",
-    "VGS=6;                    #Gate source voltage, V\n",
-    "\n",
-    "print(\"p-channel JFET requires a positive gate-to-source voltage to pass drain current.\");\n",
-    "print(\"More positive voltage, the less the drain current. \");\n",
-    "print(\"Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\");"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.7 : Page number 517-518"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The gate to source resistance=15000MΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS=15.0;              #Gate-source voltage, V\n",
-    "I_G=1e-03;               #Gate current, μA\n",
-    "\n",
-    "#Calculation\n",
-    "R_GS=(V_GS/(I_G*10**-6))/10**6;          #Gate to source resistance, MΩ (OHM's LAW)\n",
-    "\n",
-    "#Result\n",
-    "print(\"The gate to source resistance=%dMΩ.\"%R_GS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.8 : Page  number 518"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Transconductance=3000 μ mho\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "V_GS_max=-3.1;                      #Maximum gate to source voltage, V\n",
-    "V_GS_min=-3.0;                      #Minimum gate to source voltage, V\n",
-    "I_D_max=1.3;                      #Maximum drain current, mA\n",
-    "I_D_min=1.0;                      #Minimum drain current, mA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "delta_V_GS=abs(V_GS_max-V_GS_min);              #Change in gate to source voltage, V\n",
-    "delta_I_D=I_D_max-I_D_min;                      #Change in drain current, mA\n",
-    "g_fs=(delta_I_D/delta_V_GS)*1000;               #Transconductance, μ mho\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Transconductance=%.0f μ mho\"%g_fs);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.9 : Page number 518"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VGS= 0V     0V        -0.2V\n",
-      "VDS= 7V     15V        15V\n",
-      "ID = 10V    10.25V     9.65V\n",
-      "(i)   The a.c drain resistance=32kΩ.\n",
-      "(ii)  The transconductance=3000 μ mho.\n",
-      "(iii) The amplification factor=96.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS=[0,0,-0.2];            #Readings of Gate-source voltage, V\n",
-    "V_DS=[7,15,15];             #Readings of Drain-source voltage, V\n",
-    "ID=[10,10.25,9.65];         #Readings of drain current, mA\n",
-    "\n",
-    "\n",
-    "#Displaying the readings:\n",
-    "print(\"VGS= %dV     %dV        %.1fV\"%(V_GS[0],V_GS[1],V_GS[2]));\n",
-    "print(\"VDS= %dV     %dV        %dV\"%(V_DS[0],V_DS[1],V_DS[2]));\n",
-    "print(\"ID = %dV    %.2fV     %.2fV\"%(ID[0],ID[1],ID[2]));\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "#V_GS constant at 0V,\n",
-    "delta_VDS=V_DS[1]-V_DS[0];          #Change in drain-source voltage, V\n",
-    "delta_ID=ID[1]-ID[0];               #Change in drain current, mA\n",
-    "rd=delta_VDS/delta_ID;              #a.c drain resistance, kΩ\n",
-    "\n",
-    "#(ii)\n",
-    "#V_DS constant at 15V,\n",
-    "delta_VGS=V_GS[2]-V_GS[1];                  #Change in gate-source voltage, V\n",
-    "delta_ID=ID[2]-ID[1];                       #Change in drain current, mA\n",
-    "g_fs=round((delta_ID/delta_VGS)*1000,);     #Transconductance, μ mho\n",
-    "\n",
-    "#(iii)\n",
-    "amplification_factor=rd*1000*g_fs*10**-6;                          #Amplification factor\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)   The a.c drain resistance=%dkΩ.\"%rd);\n",
-    "print(\"(ii)  The transconductance=%d μ mho.\"%g_fs);\n",
-    "print(\"(iii) The amplification factor=%d.\"%amplification_factor );\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.10 : Page number 519"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The transconductance=2500 μS.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "g_mo=4000.0;                        #Maximum transconductance, μS\n",
-    "V_GS=-3.0;                          #Gate to source voltage, V\n",
-    "V_GS_off=-8.0;                      #Gate-source cut-off voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "g_m=g_mo*(1-(V_GS/V_GS_off));               #Transconductance, μS\n",
-    "\n",
-    "#Result\n",
-    "print(\"The transconductance=%d μS.\"%g_m);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.11 : Page number 519"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The transconductance=1667 μS.\n",
-      "The drain current=333 μA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "g_mo=5000.0;                        #Maximum transconductance, μS\n",
-    "V_GS=-4.0;                          #Gate to source voltage, V\n",
-    "V_GS_off=-6.0;                      #Gate-source cut-off voltage, V\n",
-    "I_DSS=3.0;                          #Shorted-gate drain current, mA\n",
-    "\n",
-    "#Calculation\n",
-    "g_m=g_mo*(1-(V_GS/V_GS_off));               #Transconductance, μS\n",
-    "I_D=(I_DSS*(1-(V_GS/V_GS_off))**2)*1000;                    #Drain current μA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The transconductance=%.0f μS.\"%g_m);\n",
-    "print(\"The drain current=%d μA.\"%I_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.12 : Page number 520"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The gate-source voltage=-5V.\n",
-      "The drain current=2.25mA.\n",
-      "The drain-source voltage=5.05V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS_off=-8.0;                      #Gate-source cut-off voltage, V\n",
-    "I_DSS=16.0;                         #Shorted-gate drain current, mA\n",
-    "R_D=2.2;                            #Drain resistor, kΩ\n",
-    "R_G=1.0;                            #Gate resistor, MΩ\n",
-    "V_DD=10.0;                          #Drain supply voltage, V\n",
-    "V_GG=-5.0;                           #Gate supply voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_GS=V_GG;                                           #Gate-source voltage, V\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current μA\n",
-    "V_DS=V_DD-I_D*R_D;                                   #Drain-source voltage, V (Kirchhoff's voltage law)\n",
-    "\n",
-    "#Result\n",
-    "print(\"The gate-source voltage=%dV.\"%V_GS);\n",
-    "print(\"The drain current=%.2fmA.\"%I_D);\n",
-    "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.13 : Page number 521"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain-source voltage=7.65V.\n",
-      "The gate-source voltage=-2.35V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_D=5.0;                   #Drain current mA\n",
-    "V_DD=15.0;                 #Drain supply voltage, V\n",
-    "V_G=0;                     #Gate voltage, V\n",
-    "R_D=1.0;                   #Drain resistor, kΩ\n",
-    "R_S=470.0;                 #Source resistor, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_S=(I_D/1000)*R_S;                 #Source voltage, V (OHM's LAW)\n",
-    "V_D=V_DD-I_D*R_D;                   #Drain voltage, V (Kirchhoff's voltage law)\n",
-    "V_DS=V_D-V_S;                       #Drain-source voltage, V\n",
-    "V_GS=V_G-V_S;                       #Gate-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n",
-    "print(\"The gate-source voltage=%.2fV.\"%V_GS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.14 : Page number 521"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The required source resistor=800 Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS=-5.0;                     #Gate-source voltage, V\n",
-    "I_D=6.25;                      #Drain current mA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "R_S=abs(V_GS/(I_D/1000));              #Required source resistor, Ω (OHM's LAW)\n",
-    "\n",
-    "#Result\n",
-    "print(\"The required source resistor=%d Ω.\"%R_S);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.15 : Page number : 521"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The source resistance=450Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=25.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=15.0;                              #Gate-source cut-off voltage, V\n",
-    "V_GS=5.0;                                   #Gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;               #Drain current mA\n",
-    "R_S=V_GS/(I_D/1000);                            #Required source resistor, Ω (OHM's LAW)\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The source resistance=%.0fΩ.\"%R_S);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.16 : Page number 522"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      " RS=313 Ω and RD=800 Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=15.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-8.0;                              #Gate-source cut-off voltage, V\n",
-    "V_DD=12.0;                                  #Drain supply voltage,V\n",
-    "V_D=V_DD/2;                                 #Drain voltage(half of V_DD), V\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS/2;                                #Drain current(approximately half of I_DSS), mA\n",
-    "V_GS=V_GS_off/3.4;                          #Gate-source voltage, V\n",
-    "R_S=abs(V_GS/(I_D/1000));                   #Source resistor, Ω (OHM's LAW)\n",
-    "#Since,V_D=V_DD-I_D*R_D;                      \n",
-    "R_D=(V_DD-V_D)/(I_D/1000);                         #Drain resistor, Ω (OHM's LAW)\n",
-    "\n",
-    "#Result\n",
-    "print(\" RS=%d Ω and RD=%d Ω.\"%(R_S,R_D));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.17 : Page number 522"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The source resistance=0.6 kΩ\n",
-      "The drain resistance=6 kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "\n",
-    "#Variable declaration\n",
-    "I_DSS=5.0;                       #Shorted gate drain current, mA\n",
-    "V_GS_off=-2.0;                   #Gate-source cut-off voltage, V\n",
-    "V_DS=10.0;                       #Drain-source voltage,V\n",
-    "I_D=1.5;                         #Drain current, mA\n",
-    "V_DD=20.0;                       #Drain supply voltage,V\n",
-    "V_G=0;                           #Gate voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since,  Drain current, I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    \n",
-    "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS));              #Gate-source voltage, V\n",
-    "\n",
-    "#Since, V_GS=V_G-V_S,\n",
-    "V_S=V_G-V_GS;                           #Source voltage, V\n",
-    "\n",
-    "R_S=V_S/I_D;                            #Source resistor, kΩ\n",
-    "\n",
-    "#Since, V_DD=I_D*R_D +V_DS+ I_D*R_S,\n",
-    "R_D=(V_DD-I_D*R_S-V_DS)/I_D;                    #Drain resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "print(\"The source resistance=%.1f kΩ\"%R_S);\n",
-    "print(\"The drain resistance=%d kΩ.\"%R_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.18 : Page number 522-523"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain-source voltage=17V.\n",
-      "The gate-source voltage=-0.5V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_DD=30.0;                  #Drain supply voltage, V\n",
-    "R_D=5.0;                    #Drain resistor, kΩ\n",
-    "I_D=2.5;                    #Drain current, mA\n",
-    "R_S=200.0;                  #Source resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "V_DS=V_DD-I_D*(R_D+(R_S/1000));                 #Drain-source voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "V_GS=-(I_D/1000)*R_S;                          #Gate-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain-source voltage=%dV.\"%V_DS);\n",
-    "print(\"The gate-source voltage=%.1fV.\"%V_GS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.19 : Page number 523"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "drain voltage of 1st stage=12.37V.\n",
-      "Source voltage of 1st stage=1.46V.\n",
-      "drain voltage of 2nd stage=11.7V.\n",
-      "Source voltage of 2nd stage=2.01V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "ID_1=2.15;                  #First stage drain current, mA\n",
-    "ID_2=9.15;                  #Second stage drain current, mA\n",
-    "VDD=30;                     #Drain supply voltage, V\n",
-    "RS_1=0.68;                  #Source resistance of 1st stage, kΩ\n",
-    "RS_2=0.22;                  #Source resistance of 2nd stage, kΩ\n",
-    "RD_1=8.2;                   #Drain resistor of 1st stage, kΩ\n",
-    "RD_2=2;                     #Drain resistor of 2nd stage, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "V_RD1=ID_1*RD_1;                #Voltage drop across 8.2kΩ\n",
-    "VD_1=VDD-V_RD1;                 #Drain voltage of 1st stage, V\n",
-    "VS_1=ID_1*RS_1;                 #D.C potential of source of first stage, V\n",
-    "V_RD2=ID_2*RD_2;                #Voltage drop across 2kΩ\n",
-    "VD_2=VDD-V_RD2;                 #Drain voltage of 2nd stage, V\n",
-    "VS_2=ID_2*RS_2;                 #D.C potential of source of 2nd stage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"drain voltage of 1st stage=%.2fV.\"%VD_1);\n",
-    "print(\"Source voltage of 1st stage=%.2fV.\"%VS_1);\n",
-    "print(\"drain voltage of 2nd stage=%.1fV.\"%VD_2);\n",
-    "print(\"Source voltage of 2nd stage=%.2fV.\"%VS_2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.20 : Page number 524"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=1.52mA.\n",
-      "Gate-source voltage=-1.2V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=12;                     #Drain supply voltage, V\n",
-    "VD=7;                       #Drain voltage, V\n",
-    "R1=6.8;                     #Resistor R1, MΩ\n",
-    "R2=1;                       #Resistor R2, MΩ\n",
-    "RS=1.8;                     #Source resistance, kΩ\n",
-    "RD=3.3;                     #Drain resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "ID=(VDD-VD)/RD;                  #Second stage drain current, mA\n",
-    "VS=ID*RS;                        #Source voltage, V\n",
-    "VG=VDD*R2/(R1+R2);               #Drain voltage, V\n",
-    "VGS=VG-VS;                       #Drain-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "print(\"Drain current=%.2fmA.\"%ID);\n",
-    "print(\"Gate-source voltage=%.1fV.\"%VGS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.21 : Page number 524-525"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Source resistor, RS=5kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VDD=30;                     #Drain supply voltage, V\n",
-    "ID=2.5;                     #Drain current, mA\n",
-    "VDS=8;                      #Drain-source voltage, V\n",
-    "VGS_off=-5;                 #Gate-source cutoff voltage, V\n",
-    "R1=1;                       #Resistor R1, MΩ\n",
-    "R2=500;                     #Resistor R2, kΩ\n",
-    "IDSS=10;                    #Shorted gate drain current, mA\n",
-    "\n",
-    "#Calculation\n",
-    "#ID=IDSS*square_of(1-(VGS/VGS_off))\n",
-    "VGS=VGS_off*(1-sqrt(ID/IDSS));           #Gate-source voltage, V\n",
-    "V2=VDD*R2/(R1*1000+R2);                       #Voltage across R2, V\n",
-    "\n",
-    "\n",
-    "#V2=VGS+ID*RS\n",
-    "RS=(V2-VGS)/ID;                         #Source resistor, kΩ\n",
-    "\n",
-    "#Result\n",
-    "print(\"Source resistor, RS=%dkΩ.\"%RS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##  Example 19.22 : Page number 528-529"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fb05356ef60>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline \n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VDD=20.0;                 #Drain supply voltage, V\n",
-    "RS=50.0;                  #Source resistor, Ω\n",
-    "RD=150.0;                 #Drain resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "VDS_max=VDD;                           #Maximum drain source voltage, V\n",
-    "ID_max=(VDD/(RD+RS))*1000;             #Maximum drain current, mA\n",
-    "\n",
-    "\n",
-    "#plot\n",
-    "x=[i for i in range(0,(int)(VDS_max+1))];          #Plot variable for V_DS\n",
-    "y=[(i/(RD+RS))*1000 for i in reversed(x[:])];      #Plot variable for ID\n",
-    "\n",
-    "\n",
-    "plt.plot(x,y);\n",
-    "plt.xlabel(\"VDS(V)\");\n",
-    "plt.ylabel(\"ID(mA)\");\n",
-    "plt.title(\"d.c load line\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.23 : Page number 529"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7fb05356db70>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline \n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VDD=20;                 #Drain supply voltage, V\n",
-    "RD=500.0;                 #Drain resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "VDS_max=VDD;                           #Maximum drain source voltage, v \n",
-    "ID_max=(VDD/RD)*1000;                  #Maximum drain current, mA\n",
-    "\n",
-    "#Plot\n",
-    "x=[i for i in range(0,(int)(VDS_max+1))];          #Plot variable for V_DS\n",
-    "y=[(i/RD)*1000 for i in reversed(x[:])];      #Plot variable for ID\n",
-    "\n",
-    "plt.plot(x,y);\n",
-    "plt.xlabel(\"VDS(V)\");\n",
-    "plt.ylabel(\"ID(mA)\");\n",
-    "plt.title(\"d.c load line\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.24 : Page number 530-531"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain=4.8.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=20;                 #Drain supply voltage, V\n",
-    "RD=12.0;                  #Drain resistor, kΩ\n",
-    "RL=8.0;                   #Load resistor, kΩ\n",
-    "RG=1.0;                   #Gate resistor, MΩ\n",
-    "gm=1.0;                   #transconductance, mA/V\n",
-    "\n",
-    "#Calculation\n",
-    "gm=gm*10**-3;                   #transconductance, mho\n",
-    "RAC=(RD*RL)/(RD+RL);            #Total a.c load, kΩ\n",
-    "Av=gm*RAC*1000;                 #Voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Voltage gain=%.1f.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.25 : Page number 531"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain=30.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "gm=3000;                   #transconductance, μmho\n",
-    "RD=10;                     #Drain resistance, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "Av=gm*10**-6*RD*1000;       #Voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"Voltage gain=%d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.26 : Page number 531"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=257mV(r.m.s).\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IDSS=8;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-10;                #Gate-source cut-off voltage, V\n",
-    "ID=1.9;                     #Drain current, mA\n",
-    "RD=3.3;                     #Drain resistance, kΩ\n",
-    "RS=2.7;                     #Source resistor, kΩ\n",
-    "vin=100;                    #Input voltage, mV\n",
-    "\n",
-    "#Calculation\n",
-    "VGS=-ID*RS;                      #Gate-source voltage, V\n",
-    "gmo=2*IDSS*10**-3/abs(VGS_off);  #Maximum transconductance, S\n",
-    "gm=gmo*(1-(VGS/VGS_off));        #Transconductance, S\n",
-    "Av=gm*RD*1000;                   #Voltage gain\n",
-    "vout=Av*vin;                     #Output voltage, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dmV(r.m.s).\"%vout);\n",
-    "                \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.27 : Page number 531-532"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 28,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=151mV(r.m.s).\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RL=4.7;                    #Load resistor, Ω\n",
-    "RD=3.3;                     #Drain resistance, kΩ\n",
-    "gm=779*10**-6;              #Transconductance, S\n",
-    "vin=100;                    #Input voltage, mV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "RAC=RD*RL/(RD+RL);          #Total a.c drain resistance, kΩ\n",
-    "Av=gm*RAC*1000;             #Voltage gain\n",
-    "vout=Av*vin;                     #Output voltage, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dmV(r.m.s).\"%vout);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.28 : Page number 532-533"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 29,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain=1.85.\n",
-      "Voltage gain, if RS resistor is bypassed=6.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RD=1.5;            #Drain resistance, kΩ\n",
-    "gm=4;              #Transconductance, mS\n",
-    "RS=560;            #Source resistance, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "Av=gm*10**-3*RD*1000/(1+gm*10**-3*RS);\n",
-    "print(\"Voltage gain=%.2f.\"%Av);\n",
-    "\n",
-    "#If RS is bypassed by a capacitor\n",
-    "Av=gm*10**-3*RD*1000;\n",
-    "print(\"Voltage gain, if RS resistor is bypassed=%d.\"%Av);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.29 : Page number 533"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 30,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Voltage gain with RS bypassed=4.155.\n",
-      "(ii) Voltage gain with RS unbypassed=1.35.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "from math import sqrt\n",
-    "\n",
-    "IDSS=10;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-3.5;                #Gate-source cut-off voltage, V\n",
-    "RD=1.5;            #Drain resistance, kΩ\n",
-    "RS=750;            #Source resistance, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#From d.c biasing\n",
-    "ID=2.3;                                     #Drain current, mA\n",
-    "VGS=round(VGS_off*(1-sqrt(ID/IDSS)),1);              #Gate-source voltage, V\n",
-    "gm=round(round((2*IDSS/abs(VGS_off)),1)*round((1-(VGS/VGS_off)),3),2);           #Transconductance, mS\n",
-    "\n",
-    "\n",
-    "#(i)\n",
-    "Av=gm*RD;                         #Voltage gain with RS resistor bypassed\n",
-    "print(\"(i) Voltage gain with RS bypassed=%.3f.\"%Av);\n",
-    "\n",
-    "#(ii)\n",
-    "Av=Av/(1+gm*(RS/1000.0));\n",
-    "print(\"(ii) Voltage gain with RS unbypassed=%.2f.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.30 : Page number 539-540"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  n-channel D-MOSFET\n",
-      "(ii) Drain current=3.91mA\n",
-      "(iii) Drain current=18.9mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IDSS=10.0;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-8.0;                  #Gate-source cut-off voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "if(VGS_off<0):\n",
-    "    print(\"(i)  n-channel D-MOSFET\");\n",
-    "else:\n",
-    "    print(\"(i)  p-channel D-MOSFET\");\n",
-    "    \n",
-    "\n",
-    "#(ii)\n",
-    "VGS=-3.0;                                #Gate-source voltage, V\n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"(ii) Drain current=%.2fmA\"%ID);\n",
-    "\n",
-    "#(iii)\n",
-    "VGS=3.0;                                #Gate-source voltage, V\n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"(iii) Drain current=%.1fmA\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.31 : Page number 540"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 32,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Point 1: VGS=0V and ID=1mA.\n",
-      "Point 2: VGS=-6V and ID=0mA.\n",
-      "Point 3: VGS=-3V and ID=0.25mA.\n",
-      "Point 4: VGS=-1V and ID=0.694mA.\n",
-      "Point 5: VGS=1V and ID=1.36mA.\n",
-      "Point 6: VGS=3V and ID=2.25mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IDSS=1.0;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-6.0;                  #Gate-source cut-off voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Point 1\n",
-    "VGS=0;                  #Gate source voltage, V               \n",
-    "ID=IDSS;                #Drain current, mA\n",
-    "print(\"Point 1: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n",
-    "\n",
-    "#Point 2\n",
-    "VGS=VGS_off;                  #Gate source voltage, V               \n",
-    "ID=0;                        #Drain current, mA\n",
-    "print(\"Point 2: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n",
-    "\n",
-    "#locating more points by changing VG values\n",
-    "VGS=-3;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 3: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n",
-    "\n",
-    "VGS=-1;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 4: VGS=%dV and ID=%.3fmA.\"%(VGS,ID));\n",
-    "\n",
-    "VGS=1;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 5: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n",
-    "\n",
-    "VGS=3;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 6: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.32 : Page number 541"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 33,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain source voltage=10.6V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=18;          #Drain supply voltage, V\n",
-    "RD=620.0;           #Drain resistor, Ω\n",
-    "IDSS=12.0;        #Shorted gate drain current, mA\n",
-    "VGS_off=-8.0;      #Gate-source cut-off voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "ID=IDSS;               #Drain current, mA\n",
-    "VDS=VDD-IDSS*(RD/1000);       #Drain source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain source voltage=%.1fV.\"%VDS);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.33 : Page number 542"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Drain source voltage=7.56V.\n",
-      "(ii) Output voltage=922mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=15;           #Drain supply voltage\n",
-    "RD=620.0;          #Drain resistor, Ω\n",
-    "RL=8.2;          #Load resistor, kΩ\n",
-    "vin=500.0;         #Input voltage, V\n",
-    "IDSS=12.0;        #Shorted gate drain current, mA\n",
-    "gm=3.2;            #Transconductance, mS\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "VDS=VDD-IDSS*(RD/1000.0);       #Drain source voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "RAC=RD*RL*1000/(RD+RL*1000);  #Total a.c drain resistace, Ω\n",
-    "vout=(gm/1000.0)*RAC*vin;     #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  Drain source voltage=%.2fV.\"%VDS);\n",
-    "print(\"(ii) Output voltage=%dmV\"%vout);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.34 : Page number 545"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 35,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=98.7mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "ID_on=500.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=1.0;              #Threshold value of gate-source voltage, V\n",
-    "VGS=5;                 #Gate-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "K=round(ID_on/(VGS_on-VGS_th)**2,2);   #Constant for a E-MOSFET, mA/V²\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain current=%.1fmA\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.35 : Page number 545"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 36,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "K=0.061e-03A/V².\n",
-      "For VGS=5V, Drain current=0.244mA\n",
-      "For VGS=8V, Drain current=1.525mA\n",
-      "For VGS=10V, Drain current=2mA\n",
-      "For VGS=12V, Drain current=4.94mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "ID_on=3.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=3.0;              #Threshold value of gate-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "K=round((ID_on/(VGS_on-VGS_th)**2),3);   #Constant for a E-MOSFET, mA/V²\n",
-    "print(\"K=%.3fe-03A/V².\"%K);\n",
-    "\n",
-    "#Determining different points for plotting\n",
-    "VGS=5;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=5V, Drain current=%.3fmA\"%ID);\n",
-    "VGS=8;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=8V, Drain current=%.3fmA\"%ID);\n",
-    "VGS=10;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=10V, Drain current=%.dmA\"%ID);\n",
-    "VGS=12;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=12V, Drain current=%.2fmA\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.36 : Page number 546-547"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 37,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain-source voltage=10.8V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=24.0;             #Drain supply voltage, V\n",
-    "RD=470.0;             #Drain resistor, Ω\n",
-    "R1=100.0;             #Resistor R1, kΩ\n",
-    "R2=15.0;              #Resistor R2, kΩ\n",
-    "ID_on=500.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=1.0;              #Threshold value of gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "VGS=VDD*R2/(R1+R2);                      #Gate-source voltage, V (Voltage divider rule)\n",
-    "K=round((ID_on/(VGS_on-VGS_th)**2),2);   #Constant for a E-MOSFET, mA/V²\n",
-    "ID=K*(VGS-VGS_th)**2;                    #Drain current, mA\n",
-    "VDS=VDD-(ID/1000)*RD;                           #Drain-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain-source voltage=%.1fV.\"%VDS);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.37 : Page number 547"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 38,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=10mA.\n",
-      "Drain-source voltage=10V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=20.0;             #Drain supply voltage, V\n",
-    "RD=1.0;               #Drain resistor, kΩ\n",
-    "RG=5.0;               #Gate resistor , MΩ\n",
-    "ID_on=10.0;             #Drain current for MOSFET ON, mA\n",
-    "\n",
-    "#Calculation\n",
-    "#since, VGS=VDS\n",
-    "ID=ID_on;              #Drain current, mA\n",
-    "VDS=VDD-ID*RD;         #Drain-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain current=%dmA.\"%ID);\n",
-    "print(\"Drain-source voltage=%dV.\"%VDS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.38 : Page number 547-548"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 39,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=1.69mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=10.0;             #Drain supply voltage, V\n",
-    "RD=3.0;             #Drain resistor, kΩ\n",
-    "R1=1.0;             #Resistor R1, MΩ\n",
-    "R2=1.0;              #Resistor R2, MΩ\n",
-    "ID_on=10.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=1.5;              #Threshold value of gate-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "K=round((ID_on/(VGS_on-VGS_th)**2),3);   #Constant for a E-MOSFET, mA/V²\n",
-    "VGS=VDD*R2/(R1+R2);                      #Gate-source voltage, V (Voltage divider rule)\n",
-    "ID=K*(VGS-VGS_th)**2;                    #Drain current, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain current=%.2fmA.\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_4.ipynb
deleted file mode 100755
index 65f7561a..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_4.ipynb
+++ /dev/null
@@ -1,1671 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 19 : FIELD EFFECT TRANSISTORS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.1 : Page number 515"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "ID=12[1 + VGS/5]²mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=12.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-5.0;                              #Gate-source cut-off voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"ID=%d[1 + VGS/%d]²mA.\"%(I_DSS,abs(V_GS_off)));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.2 : Page number 515"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain current=6.12mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=32.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-8.0;                              #Gate-source cut-off voltage, V\n",
-    "V_GS=-4.5;                                  #Gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain current=%.2fmA.\"%I_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.3 : Page number  515"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) VGS=-1.76V.\n",
-      "(ii) VP=6V\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "\n",
-    "#Variable declaration\n",
-    "I_DSS=10.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-6.0;                              #Gate-source cut-off voltage, V\n",
-    "I_D=5.0;                                    #Drain current mA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "#Since, I_D=I_DSS*[1 - (V_GS/V_GS_off)]²\n",
-    "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS));              #Gate-source voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "V_P=-V_GS_off;                                  #Pinch-off voltage, V        \n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) VGS=%.2fV.\"%V_GS);\n",
-    "print(\"(ii) VP=%dV\"%V_P);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.4 : Page number 515-516"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The minimum value of VDD required =10.72V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS_off=-4.0;                              #Gate-source cut-off voltage, V\n",
-    "I_DSS=12.0;                                 #Shorted gate drain current, mA\n",
-    "R_D=560.0;                                  #Drain resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "V_P=-V_GS_off;          #Pinch-off voltage, V\n",
-    "V_DS=V_P;               #Minimum drain-source voltage for JFET to be in constant current region, V\n",
-    "I_D=I_DSS;              #Maximum drain current, mA (V_GS=0)\n",
-    "V_RD=(I_D/1000)*R_D;    #Voltage across drain resistor, V (OHM's LAW)\n",
-    "V_DD=V_DS+V_RD;         #Minimum value of supply voltage to drain, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The minimum value of VDD required =%.2fV.\"%V_DD);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.5 : Page number 516"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain current=1.33mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=3.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-6.0;                              #Gate-source cut-off voltage, V\n",
-    "V_GS=-2.0;                                  #Gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain current=%.2fmA.\"%I_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.6 : Page number 516"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "p-channel JFET requires a positive gate-to-source voltage to pass drain current.\n",
-      "More positive voltage, the less the drain current. \n",
-      "Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VGS_off=4;                #Gate-source cut-off voltage, V\n",
-    "VGS=6;                    #Gate source voltage, V\n",
-    "\n",
-    "print(\"p-channel JFET requires a positive gate-to-source voltage to pass drain current.\");\n",
-    "print(\"More positive voltage, the less the drain current. \");\n",
-    "print(\"Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\");"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.7 : Page number 517-518"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The gate to source resistance=15000MΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS=15.0;              #Gate-source voltage, V\n",
-    "I_G=1e-03;               #Gate current, μA\n",
-    "\n",
-    "#Calculation\n",
-    "R_GS=(V_GS/(I_G*10**-6))/10**6;          #Gate to source resistance, MΩ (OHM's LAW)\n",
-    "\n",
-    "#Result\n",
-    "print(\"The gate to source resistance=%dMΩ.\"%R_GS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.8 : Page  number 518"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Transconductance=3000 μ mho\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "V_GS_max=-3.1;                      #Maximum gate to source voltage, V\n",
-    "V_GS_min=-3.0;                      #Minimum gate to source voltage, V\n",
-    "I_D_max=1.3;                      #Maximum drain current, mA\n",
-    "I_D_min=1.0;                      #Minimum drain current, mA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "delta_V_GS=abs(V_GS_max-V_GS_min);              #Change in gate to source voltage, V\n",
-    "delta_I_D=I_D_max-I_D_min;                      #Change in drain current, mA\n",
-    "g_fs=(delta_I_D/delta_V_GS)*1000;               #Transconductance, μ mho\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Transconductance=%.0f μ mho\"%g_fs);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.9 : Page number 518"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VGS= 0V     0V        -0.2V\n",
-      "VDS= 7V     15V        15V\n",
-      "ID = 10V    10.25V     9.65V\n",
-      "(i)   The a.c drain resistance=32kΩ.\n",
-      "(ii)  The transconductance=3000 μ mho.\n",
-      "(iii) The amplification factor=96.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS=[0,0,-0.2];            #Readings of Gate-source voltage, V\n",
-    "V_DS=[7,15,15];             #Readings of Drain-source voltage, V\n",
-    "ID=[10,10.25,9.65];         #Readings of drain current, mA\n",
-    "\n",
-    "\n",
-    "#Displaying the readings:\n",
-    "print(\"VGS= %dV     %dV        %.1fV\"%(V_GS[0],V_GS[1],V_GS[2]));\n",
-    "print(\"VDS= %dV     %dV        %dV\"%(V_DS[0],V_DS[1],V_DS[2]));\n",
-    "print(\"ID = %dV    %.2fV     %.2fV\"%(ID[0],ID[1],ID[2]));\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "#V_GS constant at 0V,\n",
-    "delta_VDS=V_DS[1]-V_DS[0];          #Change in drain-source voltage, V\n",
-    "delta_ID=ID[1]-ID[0];               #Change in drain current, mA\n",
-    "rd=delta_VDS/delta_ID;              #a.c drain resistance, kΩ\n",
-    "\n",
-    "#(ii)\n",
-    "#V_DS constant at 15V,\n",
-    "delta_VGS=V_GS[2]-V_GS[1];                  #Change in gate-source voltage, V\n",
-    "delta_ID=ID[2]-ID[1];                       #Change in drain current, mA\n",
-    "g_fs=round((delta_ID/delta_VGS)*1000,);     #Transconductance, μ mho\n",
-    "\n",
-    "#(iii)\n",
-    "amplification_factor=rd*1000*g_fs*10**-6;                          #Amplification factor\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)   The a.c drain resistance=%dkΩ.\"%rd);\n",
-    "print(\"(ii)  The transconductance=%d μ mho.\"%g_fs);\n",
-    "print(\"(iii) The amplification factor=%d.\"%amplification_factor );\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.10 : Page number 519"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The transconductance=2500 μS.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "g_mo=4000.0;                        #Maximum transconductance, μS\n",
-    "V_GS=-3.0;                          #Gate to source voltage, V\n",
-    "V_GS_off=-8.0;                      #Gate-source cut-off voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "g_m=g_mo*(1-(V_GS/V_GS_off));               #Transconductance, μS\n",
-    "\n",
-    "#Result\n",
-    "print(\"The transconductance=%d μS.\"%g_m);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.11 : Page number 519"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The transconductance=1667 μS.\n",
-      "The drain current=333 μA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "g_mo=5000.0;                        #Maximum transconductance, μS\n",
-    "V_GS=-4.0;                          #Gate to source voltage, V\n",
-    "V_GS_off=-6.0;                      #Gate-source cut-off voltage, V\n",
-    "I_DSS=3.0;                          #Shorted-gate drain current, mA\n",
-    "\n",
-    "#Calculation\n",
-    "g_m=g_mo*(1-(V_GS/V_GS_off));               #Transconductance, μS\n",
-    "I_D=(I_DSS*(1-(V_GS/V_GS_off))**2)*1000;                    #Drain current μA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The transconductance=%.0f μS.\"%g_m);\n",
-    "print(\"The drain current=%d μA.\"%I_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.12 : Page number 520"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The gate-source voltage=-5V.\n",
-      "The drain current=2.25mA.\n",
-      "The drain-source voltage=5.05V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS_off=-8.0;                      #Gate-source cut-off voltage, V\n",
-    "I_DSS=16.0;                         #Shorted-gate drain current, mA\n",
-    "R_D=2.2;                            #Drain resistor, kΩ\n",
-    "R_G=1.0;                            #Gate resistor, MΩ\n",
-    "V_DD=10.0;                          #Drain supply voltage, V\n",
-    "V_GG=-5.0;                           #Gate supply voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_GS=V_GG;                                           #Gate-source voltage, V\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current μA\n",
-    "V_DS=V_DD-I_D*R_D;                                   #Drain-source voltage, V (Kirchhoff's voltage law)\n",
-    "\n",
-    "#Result\n",
-    "print(\"The gate-source voltage=%dV.\"%V_GS);\n",
-    "print(\"The drain current=%.2fmA.\"%I_D);\n",
-    "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.13 : Page number 521"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain-source voltage=7.65V.\n",
-      "The gate-source voltage=-2.35V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_D=5.0;                   #Drain current mA\n",
-    "V_DD=15.0;                 #Drain supply voltage, V\n",
-    "V_G=0;                     #Gate voltage, V\n",
-    "R_D=1.0;                   #Drain resistor, kΩ\n",
-    "R_S=470.0;                 #Source resistor, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_S=(I_D/1000)*R_S;                 #Source voltage, V (OHM's LAW)\n",
-    "V_D=V_DD-I_D*R_D;                   #Drain voltage, V (Kirchhoff's voltage law)\n",
-    "V_DS=V_D-V_S;                       #Drain-source voltage, V\n",
-    "V_GS=V_G-V_S;                       #Gate-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n",
-    "print(\"The gate-source voltage=%.2fV.\"%V_GS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.14 : Page number 521"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The required source resistor=800 Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS=-5.0;                     #Gate-source voltage, V\n",
-    "I_D=6.25;                      #Drain current mA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "R_S=abs(V_GS/(I_D/1000));              #Required source resistor, Ω (OHM's LAW)\n",
-    "\n",
-    "#Result\n",
-    "print(\"The required source resistor=%d Ω.\"%R_S);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.15 : Page number : 521"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The source resistance=450Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=25.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=15.0;                              #Gate-source cut-off voltage, V\n",
-    "V_GS=5.0;                                   #Gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;               #Drain current mA\n",
-    "R_S=V_GS/(I_D/1000);                            #Required source resistor, Ω (OHM's LAW)\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The source resistance=%.0fΩ.\"%R_S);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.16 : Page number 522"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      " RS=313 Ω and RD=800 Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=15.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-8.0;                              #Gate-source cut-off voltage, V\n",
-    "V_DD=12.0;                                  #Drain supply voltage,V\n",
-    "V_D=V_DD/2;                                 #Drain voltage(half of V_DD), V\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS/2;                                #Drain current(approximately half of I_DSS), mA\n",
-    "V_GS=V_GS_off/3.4;                          #Gate-source voltage, V\n",
-    "R_S=abs(V_GS/(I_D/1000));                   #Source resistor, Ω (OHM's LAW)\n",
-    "#Since,V_D=V_DD-I_D*R_D;                      \n",
-    "R_D=(V_DD-V_D)/(I_D/1000);                         #Drain resistor, Ω (OHM's LAW)\n",
-    "\n",
-    "#Result\n",
-    "print(\" RS=%d Ω and RD=%d Ω.\"%(R_S,R_D));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.17 : Page number 522"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The source resistance=0.6 kΩ\n",
-      "The drain resistance=6 kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "\n",
-    "#Variable declaration\n",
-    "I_DSS=5.0;                       #Shorted gate drain current, mA\n",
-    "V_GS_off=-2.0;                   #Gate-source cut-off voltage, V\n",
-    "V_DS=10.0;                       #Drain-source voltage,V\n",
-    "I_D=1.5;                         #Drain current, mA\n",
-    "V_DD=20.0;                       #Drain supply voltage,V\n",
-    "V_G=0;                           #Gate voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since,  Drain current, I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    \n",
-    "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS));              #Gate-source voltage, V\n",
-    "\n",
-    "#Since, V_GS=V_G-V_S,\n",
-    "V_S=V_G-V_GS;                           #Source voltage, V\n",
-    "\n",
-    "R_S=V_S/I_D;                            #Source resistor, kΩ\n",
-    "\n",
-    "#Since, V_DD=I_D*R_D +V_DS+ I_D*R_S,\n",
-    "R_D=(V_DD-I_D*R_S-V_DS)/I_D;                    #Drain resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "print(\"The source resistance=%.1f kΩ\"%R_S);\n",
-    "print(\"The drain resistance=%d kΩ.\"%R_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.18 : Page number 522-523"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain-source voltage=17V.\n",
-      "The gate-source voltage=-0.5V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_DD=30.0;                  #Drain supply voltage, V\n",
-    "R_D=5.0;                    #Drain resistor, kΩ\n",
-    "I_D=2.5;                    #Drain current, mA\n",
-    "R_S=200.0;                  #Source resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "V_DS=V_DD-I_D*(R_D+(R_S/1000));                 #Drain-source voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "V_GS=-(I_D/1000)*R_S;                          #Gate-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain-source voltage=%dV.\"%V_DS);\n",
-    "print(\"The gate-source voltage=%.1fV.\"%V_GS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.19 : Page number 523"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "drain voltage of 1st stage=12.37V.\n",
-      "Source voltage of 1st stage=1.46V.\n",
-      "drain voltage of 2nd stage=11.7V.\n",
-      "Source voltage of 2nd stage=2.01V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "ID_1=2.15;                  #First stage drain current, mA\n",
-    "ID_2=9.15;                  #Second stage drain current, mA\n",
-    "VDD=30;                     #Drain supply voltage, V\n",
-    "RS_1=0.68;                  #Source resistance of 1st stage, kΩ\n",
-    "RS_2=0.22;                  #Source resistance of 2nd stage, kΩ\n",
-    "RD_1=8.2;                   #Drain resistor of 1st stage, kΩ\n",
-    "RD_2=2;                     #Drain resistor of 2nd stage, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "V_RD1=ID_1*RD_1;                #Voltage drop across 8.2kΩ\n",
-    "VD_1=VDD-V_RD1;                 #Drain voltage of 1st stage, V\n",
-    "VS_1=ID_1*RS_1;                 #D.C potential of source of first stage, V\n",
-    "V_RD2=ID_2*RD_2;                #Voltage drop across 2kΩ\n",
-    "VD_2=VDD-V_RD2;                 #Drain voltage of 2nd stage, V\n",
-    "VS_2=ID_2*RS_2;                 #D.C potential of source of 2nd stage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"drain voltage of 1st stage=%.2fV.\"%VD_1);\n",
-    "print(\"Source voltage of 1st stage=%.2fV.\"%VS_1);\n",
-    "print(\"drain voltage of 2nd stage=%.1fV.\"%VD_2);\n",
-    "print(\"Source voltage of 2nd stage=%.2fV.\"%VS_2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.20 : Page number 524"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=1.52mA.\n",
-      "Gate-source voltage=-1.2V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=12;                     #Drain supply voltage, V\n",
-    "VD=7;                       #Drain voltage, V\n",
-    "R1=6.8;                     #Resistor R1, MΩ\n",
-    "R2=1;                       #Resistor R2, MΩ\n",
-    "RS=1.8;                     #Source resistance, kΩ\n",
-    "RD=3.3;                     #Drain resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "ID=(VDD-VD)/RD;                  #Second stage drain current, mA\n",
-    "VS=ID*RS;                        #Source voltage, V\n",
-    "VG=VDD*R2/(R1+R2);               #Drain voltage, V\n",
-    "VGS=VG-VS;                       #Drain-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "print(\"Drain current=%.2fmA.\"%ID);\n",
-    "print(\"Gate-source voltage=%.1fV.\"%VGS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.21 : Page number 524-525"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Source resistor, RS=5kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VDD=30;                     #Drain supply voltage, V\n",
-    "ID=2.5;                     #Drain current, mA\n",
-    "VDS=8;                      #Drain-source voltage, V\n",
-    "VGS_off=-5;                 #Gate-source cutoff voltage, V\n",
-    "R1=1;                       #Resistor R1, MΩ\n",
-    "R2=500;                     #Resistor R2, kΩ\n",
-    "IDSS=10;                    #Shorted gate drain current, mA\n",
-    "\n",
-    "#Calculation\n",
-    "#ID=IDSS*square_of(1-(VGS/VGS_off))\n",
-    "VGS=VGS_off*(1-sqrt(ID/IDSS));           #Gate-source voltage, V\n",
-    "V2=VDD*R2/(R1*1000+R2);                       #Voltage across R2, V\n",
-    "\n",
-    "\n",
-    "#V2=VGS+ID*RS\n",
-    "RS=(V2-VGS)/ID;                         #Source resistor, kΩ\n",
-    "\n",
-    "#Result\n",
-    "print(\"Source resistor, RS=%dkΩ.\"%RS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##  Example 19.22 : Page number 528-529"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f70cf070f28>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline \n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VDD=20.0;                 #Drain supply voltage, V\n",
-    "RS=50.0;                  #Source resistor, Ω\n",
-    "RD=150.0;                 #Drain resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "VDS_max=VDD;                           #Maximum drain source voltage, V\n",
-    "ID_max=(VDD/(RD+RS))*1000;             #Maximum drain current, mA\n",
-    "\n",
-    "\n",
-    "#plot\n",
-    "x=[i for i in range(0,(int)(VDS_max+1))];          #Plot variable for V_DS\n",
-    "y=[(i/(RD+RS))*1000 for i in reversed(x[:])];      #Plot variable for ID\n",
-    "\n",
-    "\n",
-    "plt.plot(x,y);\n",
-    "plt.xlabel(\"VDS(V)\");\n",
-    "plt.ylabel(\"ID(mA)\");\n",
-    "plt.title(\"d.c load line\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.23 : Page number 529"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f70cf06ab38>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline \n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VDD=20;                 #Drain supply voltage, V\n",
-    "RD=500.0;                 #Drain resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "VDS_max=VDD;                           #Maximum drain source voltage, v \n",
-    "ID_max=(VDD/RD)*1000;                  #Maximum drain current, mA\n",
-    "\n",
-    "#Plot\n",
-    "x=[i for i in range(0,(int)(VDS_max+1))];          #Plot variable for V_DS\n",
-    "y=[(i/RD)*1000 for i in reversed(x[:])];      #Plot variable for ID\n",
-    "\n",
-    "plt.plot(x,y);\n",
-    "plt.xlabel(\"VDS(V)\");\n",
-    "plt.ylabel(\"ID(mA)\");\n",
-    "plt.title(\"d.c load line\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.24 : Page number 530-531"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain=4.8.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=20;                 #Drain supply voltage, V\n",
-    "RD=12.0;                  #Drain resistor, kΩ\n",
-    "RL=8.0;                   #Load resistor, kΩ\n",
-    "RG=1.0;                   #Gate resistor, MΩ\n",
-    "gm=1.0;                   #transconductance, mA/V\n",
-    "\n",
-    "#Calculation\n",
-    "gm=gm*10**-3;                   #transconductance, mho\n",
-    "RAC=(RD*RL)/(RD+RL);            #Total a.c load, kΩ\n",
-    "Av=gm*RAC*1000;                 #Voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Voltage gain=%.1f.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.25 : Page number 531"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain=30.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "gm=3000;                   #transconductance, μmho\n",
-    "RD=10;                     #Drain resistance, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "Av=gm*10**-6*RD*1000;       #Voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"Voltage gain=%d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.26 : Page number 531"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=257mV(r.m.s).\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IDSS=8;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-10;                #Gate-source cut-off voltage, V\n",
-    "ID=1.9;                     #Drain current, mA\n",
-    "RD=3.3;                     #Drain resistance, kΩ\n",
-    "RS=2.7;                     #Source resistor, kΩ\n",
-    "vin=100;                    #Input voltage, mV\n",
-    "\n",
-    "#Calculation\n",
-    "VGS=-ID*RS;                      #Gate-source voltage, V\n",
-    "gmo=2*IDSS*10**-3/abs(VGS_off);  #Maximum transconductance, S\n",
-    "gm=gmo*(1-(VGS/VGS_off));        #Transconductance, S\n",
-    "Av=gm*RD*1000;                   #Voltage gain\n",
-    "vout=Av*vin;                     #Output voltage, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dmV(r.m.s).\"%vout);\n",
-    "                \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.27 : Page number 531-532"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 28,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=151mV(r.m.s).\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RL=4.7;                    #Load resistor, Ω\n",
-    "RD=3.3;                     #Drain resistance, kΩ\n",
-    "gm=779*10**-6;              #Transconductance, S\n",
-    "vin=100;                    #Input voltage, mV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "RAC=RD*RL/(RD+RL);          #Total a.c drain resistance, kΩ\n",
-    "Av=gm*RAC*1000;             #Voltage gain\n",
-    "vout=Av*vin;                     #Output voltage, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dmV(r.m.s).\"%vout);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.28 : Page number 532-533"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 29,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain=1.85.\n",
-      "Voltage gain, if RS resistor is bypassed=6.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RD=1.5;            #Drain resistance, kΩ\n",
-    "gm=4;              #Transconductance, mS\n",
-    "RS=560;            #Source resistance, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "Av=gm*10**-3*RD*1000/(1+gm*10**-3*RS);\n",
-    "print(\"Voltage gain=%.2f.\"%Av);\n",
-    "\n",
-    "#If RS is bypassed by a capacitor\n",
-    "Av=gm*10**-3*RD*1000;\n",
-    "print(\"Voltage gain, if RS resistor is bypassed=%d.\"%Av);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.29 : Page number 533"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 30,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Voltage gain with RS bypassed=4.155.\n",
-      "(ii) Voltage gain with RS unbypassed=1.35.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "from math import sqrt\n",
-    "\n",
-    "IDSS=10;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-3.5;                #Gate-source cut-off voltage, V\n",
-    "RD=1.5;            #Drain resistance, kΩ\n",
-    "RS=750;            #Source resistance, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#From d.c biasing\n",
-    "ID=2.3;                                     #Drain current, mA\n",
-    "VGS=round(VGS_off*(1-sqrt(ID/IDSS)),1);              #Gate-source voltage, V\n",
-    "gm=round(round((2*IDSS/abs(VGS_off)),1)*round((1-(VGS/VGS_off)),3),2);           #Transconductance, mS\n",
-    "\n",
-    "\n",
-    "#(i)\n",
-    "Av=gm*RD;                         #Voltage gain with RS resistor bypassed\n",
-    "print(\"(i) Voltage gain with RS bypassed=%.3f.\"%Av);\n",
-    "\n",
-    "#(ii)\n",
-    "Av=Av/(1+gm*(RS/1000.0));\n",
-    "print(\"(ii) Voltage gain with RS unbypassed=%.2f.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.30 : Page number 539-540"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  n-channel D-MOSFET\n",
-      "(ii) Drain current=3.91mA\n",
-      "(iii) Drain current=18.9mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IDSS=10.0;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-8.0;                  #Gate-source cut-off voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "if(VGS_off<0):\n",
-    "    print(\"(i)  n-channel D-MOSFET\");\n",
-    "else:\n",
-    "    print(\"(i)  p-channel D-MOSFET\");\n",
-    "    \n",
-    "\n",
-    "#(ii)\n",
-    "VGS=-3.0;                                #Gate-source voltage, V\n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"(ii) Drain current=%.2fmA\"%ID);\n",
-    "\n",
-    "#(iii)\n",
-    "VGS=3.0;                                #Gate-source voltage, V\n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"(iii) Drain current=%.1fmA\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.31 : Page number 540"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 32,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Point 1: VGS=0V and ID=1mA.\n",
-      "Point 2: VGS=-6V and ID=0mA.\n",
-      "Point 3: VGS=-3V and ID=0.25mA.\n",
-      "Point 4: VGS=-1V and ID=0.694mA.\n",
-      "Point 5: VGS=1V and ID=1.36mA.\n",
-      "Point 6: VGS=3V and ID=2.25mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IDSS=1.0;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-6.0;                  #Gate-source cut-off voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Point 1\n",
-    "VGS=0;                  #Gate source voltage, V               \n",
-    "ID=IDSS;                #Drain current, mA\n",
-    "print(\"Point 1: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n",
-    "\n",
-    "#Point 2\n",
-    "VGS=VGS_off;                  #Gate source voltage, V               \n",
-    "ID=0;                        #Drain current, mA\n",
-    "print(\"Point 2: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n",
-    "\n",
-    "#locating more points by changing VG values\n",
-    "VGS=-3;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 3: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n",
-    "\n",
-    "VGS=-1;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 4: VGS=%dV and ID=%.3fmA.\"%(VGS,ID));\n",
-    "\n",
-    "VGS=1;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 5: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n",
-    "\n",
-    "VGS=3;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 6: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.32 : Page number 541"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 33,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain source voltage=10.6V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=18;          #Drain supply voltage, V\n",
-    "RD=620.0;           #Drain resistor, Ω\n",
-    "IDSS=12.0;        #Shorted gate drain current, mA\n",
-    "VGS_off=-8.0;      #Gate-source cut-off voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "ID=IDSS;               #Drain current, mA\n",
-    "VDS=VDD-IDSS*(RD/1000);       #Drain source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain source voltage=%.1fV.\"%VDS);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.33 : Page number 542"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Drain source voltage=7.56V.\n",
-      "(ii) Output voltage=922mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=15;           #Drain supply voltage\n",
-    "RD=620.0;          #Drain resistor, Ω\n",
-    "RL=8.2;          #Load resistor, kΩ\n",
-    "vin=500.0;         #Input voltage, V\n",
-    "IDSS=12.0;        #Shorted gate drain current, mA\n",
-    "gm=3.2;            #Transconductance, mS\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "VDS=VDD-IDSS*(RD/1000.0);       #Drain source voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "RAC=RD*RL*1000/(RD+RL*1000);  #Total a.c drain resistace, Ω\n",
-    "vout=(gm/1000.0)*RAC*vin;     #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  Drain source voltage=%.2fV.\"%VDS);\n",
-    "print(\"(ii) Output voltage=%dmV\"%vout);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.34 : Page number 545"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 35,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=98.7mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "ID_on=500.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=1.0;              #Threshold value of gate-source voltage, V\n",
-    "VGS=5;                 #Gate-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "K=round(ID_on/(VGS_on-VGS_th)**2,2);   #Constant for a E-MOSFET, mA/V²\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain current=%.1fmA\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.35 : Page number 545"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 36,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "K=0.061e-03A/V².\n",
-      "For VGS=5V, Drain current=0.244mA\n",
-      "For VGS=8V, Drain current=1.525mA\n",
-      "For VGS=10V, Drain current=2mA\n",
-      "For VGS=12V, Drain current=4.94mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "ID_on=3.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=3.0;              #Threshold value of gate-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "K=round((ID_on/(VGS_on-VGS_th)**2),3);   #Constant for a E-MOSFET, mA/V²\n",
-    "print(\"K=%.3fe-03A/V².\"%K);\n",
-    "\n",
-    "#Determining different points for plotting\n",
-    "VGS=5;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=5V, Drain current=%.3fmA\"%ID);\n",
-    "VGS=8;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=8V, Drain current=%.3fmA\"%ID);\n",
-    "VGS=10;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=10V, Drain current=%.dmA\"%ID);\n",
-    "VGS=12;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=12V, Drain current=%.2fmA\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.36 : Page number 546-547"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 37,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain-source voltage=10.8V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=24.0;             #Drain supply voltage, V\n",
-    "RD=470.0;             #Drain resistor, Ω\n",
-    "R1=100.0;             #Resistor R1, kΩ\n",
-    "R2=15.0;              #Resistor R2, kΩ\n",
-    "ID_on=500.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=1.0;              #Threshold value of gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "VGS=VDD*R2/(R1+R2);                      #Gate-source voltage, V (Voltage divider rule)\n",
-    "K=round((ID_on/(VGS_on-VGS_th)**2),2);   #Constant for a E-MOSFET, mA/V²\n",
-    "ID=K*(VGS-VGS_th)**2;                    #Drain current, mA\n",
-    "VDS=VDD-(ID/1000)*RD;                           #Drain-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain-source voltage=%.1fV.\"%VDS);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.37 : Page number 547"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 38,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=10mA.\n",
-      "Drain-source voltage=10V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=20.0;             #Drain supply voltage, V\n",
-    "RD=1.0;               #Drain resistor, kΩ\n",
-    "RG=5.0;               #Gate resistor , MΩ\n",
-    "ID_on=10.0;             #Drain current for MOSFET ON, mA\n",
-    "\n",
-    "#Calculation\n",
-    "#since, VGS=VDS\n",
-    "ID=ID_on;              #Drain current, mA\n",
-    "VDS=VDD-ID*RD;         #Drain-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain current=%dmA.\"%ID);\n",
-    "print(\"Drain-source voltage=%dV.\"%VDS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.38 : Page number 547-548"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 39,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=1.69mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=10.0;             #Drain supply voltage, V\n",
-    "RD=3.0;             #Drain resistor, kΩ\n",
-    "R1=1.0;             #Resistor R1, MΩ\n",
-    "R2=1.0;              #Resistor R2, MΩ\n",
-    "ID_on=10.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=1.5;              #Threshold value of gate-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "K=round((ID_on/(VGS_on-VGS_th)**2),3);   #Constant for a E-MOSFET, mA/V²\n",
-    "VGS=VDD*R2/(R1+R2);                      #Gate-source voltage, V (Voltage divider rule)\n",
-    "ID=K*(VGS-VGS_th)**2;                    #Drain current, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain current=%.2fmA.\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_5.ipynb
deleted file mode 100755
index 65f7561a..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_5.ipynb
+++ /dev/null
@@ -1,1671 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 19 : FIELD EFFECT TRANSISTORS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.1 : Page number 515"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "ID=12[1 + VGS/5]²mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=12.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-5.0;                              #Gate-source cut-off voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"ID=%d[1 + VGS/%d]²mA.\"%(I_DSS,abs(V_GS_off)));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.2 : Page number 515"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain current=6.12mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=32.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-8.0;                              #Gate-source cut-off voltage, V\n",
-    "V_GS=-4.5;                                  #Gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain current=%.2fmA.\"%I_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.3 : Page number  515"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) VGS=-1.76V.\n",
-      "(ii) VP=6V\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "\n",
-    "#Variable declaration\n",
-    "I_DSS=10.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-6.0;                              #Gate-source cut-off voltage, V\n",
-    "I_D=5.0;                                    #Drain current mA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "#Since, I_D=I_DSS*[1 - (V_GS/V_GS_off)]²\n",
-    "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS));              #Gate-source voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "V_P=-V_GS_off;                                  #Pinch-off voltage, V        \n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) VGS=%.2fV.\"%V_GS);\n",
-    "print(\"(ii) VP=%dV\"%V_P);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.4 : Page number 515-516"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The minimum value of VDD required =10.72V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS_off=-4.0;                              #Gate-source cut-off voltage, V\n",
-    "I_DSS=12.0;                                 #Shorted gate drain current, mA\n",
-    "R_D=560.0;                                  #Drain resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "V_P=-V_GS_off;          #Pinch-off voltage, V\n",
-    "V_DS=V_P;               #Minimum drain-source voltage for JFET to be in constant current region, V\n",
-    "I_D=I_DSS;              #Maximum drain current, mA (V_GS=0)\n",
-    "V_RD=(I_D/1000)*R_D;    #Voltage across drain resistor, V (OHM's LAW)\n",
-    "V_DD=V_DS+V_RD;         #Minimum value of supply voltage to drain, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The minimum value of VDD required =%.2fV.\"%V_DD);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.5 : Page number 516"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain current=1.33mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=3.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-6.0;                              #Gate-source cut-off voltage, V\n",
-    "V_GS=-2.0;                                  #Gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain current=%.2fmA.\"%I_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.6 : Page number 516"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "p-channel JFET requires a positive gate-to-source voltage to pass drain current.\n",
-      "More positive voltage, the less the drain current. \n",
-      "Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VGS_off=4;                #Gate-source cut-off voltage, V\n",
-    "VGS=6;                    #Gate source voltage, V\n",
-    "\n",
-    "print(\"p-channel JFET requires a positive gate-to-source voltage to pass drain current.\");\n",
-    "print(\"More positive voltage, the less the drain current. \");\n",
-    "print(\"Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\");"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.7 : Page number 517-518"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The gate to source resistance=15000MΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS=15.0;              #Gate-source voltage, V\n",
-    "I_G=1e-03;               #Gate current, μA\n",
-    "\n",
-    "#Calculation\n",
-    "R_GS=(V_GS/(I_G*10**-6))/10**6;          #Gate to source resistance, MΩ (OHM's LAW)\n",
-    "\n",
-    "#Result\n",
-    "print(\"The gate to source resistance=%dMΩ.\"%R_GS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.8 : Page  number 518"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Transconductance=3000 μ mho\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "V_GS_max=-3.1;                      #Maximum gate to source voltage, V\n",
-    "V_GS_min=-3.0;                      #Minimum gate to source voltage, V\n",
-    "I_D_max=1.3;                      #Maximum drain current, mA\n",
-    "I_D_min=1.0;                      #Minimum drain current, mA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "delta_V_GS=abs(V_GS_max-V_GS_min);              #Change in gate to source voltage, V\n",
-    "delta_I_D=I_D_max-I_D_min;                      #Change in drain current, mA\n",
-    "g_fs=(delta_I_D/delta_V_GS)*1000;               #Transconductance, μ mho\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Transconductance=%.0f μ mho\"%g_fs);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.9 : Page number 518"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VGS= 0V     0V        -0.2V\n",
-      "VDS= 7V     15V        15V\n",
-      "ID = 10V    10.25V     9.65V\n",
-      "(i)   The a.c drain resistance=32kΩ.\n",
-      "(ii)  The transconductance=3000 μ mho.\n",
-      "(iii) The amplification factor=96.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS=[0,0,-0.2];            #Readings of Gate-source voltage, V\n",
-    "V_DS=[7,15,15];             #Readings of Drain-source voltage, V\n",
-    "ID=[10,10.25,9.65];         #Readings of drain current, mA\n",
-    "\n",
-    "\n",
-    "#Displaying the readings:\n",
-    "print(\"VGS= %dV     %dV        %.1fV\"%(V_GS[0],V_GS[1],V_GS[2]));\n",
-    "print(\"VDS= %dV     %dV        %dV\"%(V_DS[0],V_DS[1],V_DS[2]));\n",
-    "print(\"ID = %dV    %.2fV     %.2fV\"%(ID[0],ID[1],ID[2]));\n",
-    "\n",
-    "#Calculations\n",
-    "#(i)\n",
-    "#V_GS constant at 0V,\n",
-    "delta_VDS=V_DS[1]-V_DS[0];          #Change in drain-source voltage, V\n",
-    "delta_ID=ID[1]-ID[0];               #Change in drain current, mA\n",
-    "rd=delta_VDS/delta_ID;              #a.c drain resistance, kΩ\n",
-    "\n",
-    "#(ii)\n",
-    "#V_DS constant at 15V,\n",
-    "delta_VGS=V_GS[2]-V_GS[1];                  #Change in gate-source voltage, V\n",
-    "delta_ID=ID[2]-ID[1];                       #Change in drain current, mA\n",
-    "g_fs=round((delta_ID/delta_VGS)*1000,);     #Transconductance, μ mho\n",
-    "\n",
-    "#(iii)\n",
-    "amplification_factor=rd*1000*g_fs*10**-6;                          #Amplification factor\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)   The a.c drain resistance=%dkΩ.\"%rd);\n",
-    "print(\"(ii)  The transconductance=%d μ mho.\"%g_fs);\n",
-    "print(\"(iii) The amplification factor=%d.\"%amplification_factor );\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.10 : Page number 519"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The transconductance=2500 μS.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "g_mo=4000.0;                        #Maximum transconductance, μS\n",
-    "V_GS=-3.0;                          #Gate to source voltage, V\n",
-    "V_GS_off=-8.0;                      #Gate-source cut-off voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "g_m=g_mo*(1-(V_GS/V_GS_off));               #Transconductance, μS\n",
-    "\n",
-    "#Result\n",
-    "print(\"The transconductance=%d μS.\"%g_m);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.11 : Page number 519"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The transconductance=1667 μS.\n",
-      "The drain current=333 μA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "g_mo=5000.0;                        #Maximum transconductance, μS\n",
-    "V_GS=-4.0;                          #Gate to source voltage, V\n",
-    "V_GS_off=-6.0;                      #Gate-source cut-off voltage, V\n",
-    "I_DSS=3.0;                          #Shorted-gate drain current, mA\n",
-    "\n",
-    "#Calculation\n",
-    "g_m=g_mo*(1-(V_GS/V_GS_off));               #Transconductance, μS\n",
-    "I_D=(I_DSS*(1-(V_GS/V_GS_off))**2)*1000;                    #Drain current μA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The transconductance=%.0f μS.\"%g_m);\n",
-    "print(\"The drain current=%d μA.\"%I_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.12 : Page number 520"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The gate-source voltage=-5V.\n",
-      "The drain current=2.25mA.\n",
-      "The drain-source voltage=5.05V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS_off=-8.0;                      #Gate-source cut-off voltage, V\n",
-    "I_DSS=16.0;                         #Shorted-gate drain current, mA\n",
-    "R_D=2.2;                            #Drain resistor, kΩ\n",
-    "R_G=1.0;                            #Gate resistor, MΩ\n",
-    "V_DD=10.0;                          #Drain supply voltage, V\n",
-    "V_GG=-5.0;                           #Gate supply voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_GS=V_GG;                                           #Gate-source voltage, V\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    #Drain current μA\n",
-    "V_DS=V_DD-I_D*R_D;                                   #Drain-source voltage, V (Kirchhoff's voltage law)\n",
-    "\n",
-    "#Result\n",
-    "print(\"The gate-source voltage=%dV.\"%V_GS);\n",
-    "print(\"The drain current=%.2fmA.\"%I_D);\n",
-    "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.13 : Page number 521"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain-source voltage=7.65V.\n",
-      "The gate-source voltage=-2.35V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_D=5.0;                   #Drain current mA\n",
-    "V_DD=15.0;                 #Drain supply voltage, V\n",
-    "V_G=0;                     #Gate voltage, V\n",
-    "R_D=1.0;                   #Drain resistor, kΩ\n",
-    "R_S=470.0;                 #Source resistor, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_S=(I_D/1000)*R_S;                 #Source voltage, V (OHM's LAW)\n",
-    "V_D=V_DD-I_D*R_D;                   #Drain voltage, V (Kirchhoff's voltage law)\n",
-    "V_DS=V_D-V_S;                       #Drain-source voltage, V\n",
-    "V_GS=V_G-V_S;                       #Gate-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n",
-    "print(\"The gate-source voltage=%.2fV.\"%V_GS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.14 : Page number 521"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The required source resistor=800 Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_GS=-5.0;                     #Gate-source voltage, V\n",
-    "I_D=6.25;                      #Drain current mA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "R_S=abs(V_GS/(I_D/1000));              #Required source resistor, Ω (OHM's LAW)\n",
-    "\n",
-    "#Result\n",
-    "print(\"The required source resistor=%d Ω.\"%R_S);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.15 : Page number : 521"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The source resistance=450Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=25.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=15.0;                              #Gate-source cut-off voltage, V\n",
-    "V_GS=5.0;                                   #Gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;               #Drain current mA\n",
-    "R_S=V_GS/(I_D/1000);                            #Required source resistor, Ω (OHM's LAW)\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The source resistance=%.0fΩ.\"%R_S);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.16 : Page number 522"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      " RS=313 Ω and RD=800 Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_DSS=15.0;                                 #Shorted gate drain current, mA\n",
-    "V_GS_off=-8.0;                              #Gate-source cut-off voltage, V\n",
-    "V_DD=12.0;                                  #Drain supply voltage,V\n",
-    "V_D=V_DD/2;                                 #Drain voltage(half of V_DD), V\n",
-    "\n",
-    "#Calculation\n",
-    "I_D=I_DSS/2;                                #Drain current(approximately half of I_DSS), mA\n",
-    "V_GS=V_GS_off/3.4;                          #Gate-source voltage, V\n",
-    "R_S=abs(V_GS/(I_D/1000));                   #Source resistor, Ω (OHM's LAW)\n",
-    "#Since,V_D=V_DD-I_D*R_D;                      \n",
-    "R_D=(V_DD-V_D)/(I_D/1000);                         #Drain resistor, Ω (OHM's LAW)\n",
-    "\n",
-    "#Result\n",
-    "print(\" RS=%d Ω and RD=%d Ω.\"%(R_S,R_D));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.17 : Page number 522"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The source resistance=0.6 kΩ\n",
-      "The drain resistance=6 kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "\n",
-    "#Variable declaration\n",
-    "I_DSS=5.0;                       #Shorted gate drain current, mA\n",
-    "V_GS_off=-2.0;                   #Gate-source cut-off voltage, V\n",
-    "V_DS=10.0;                       #Drain-source voltage,V\n",
-    "I_D=1.5;                         #Drain current, mA\n",
-    "V_DD=20.0;                       #Drain supply voltage,V\n",
-    "V_G=0;                           #Gate voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since,  Drain current, I_D=I_DSS*(1-(V_GS/V_GS_off))**2;                    \n",
-    "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS));              #Gate-source voltage, V\n",
-    "\n",
-    "#Since, V_GS=V_G-V_S,\n",
-    "V_S=V_G-V_GS;                           #Source voltage, V\n",
-    "\n",
-    "R_S=V_S/I_D;                            #Source resistor, kΩ\n",
-    "\n",
-    "#Since, V_DD=I_D*R_D +V_DS+ I_D*R_S,\n",
-    "R_D=(V_DD-I_D*R_S-V_DS)/I_D;                    #Drain resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "print(\"The source resistance=%.1f kΩ\"%R_S);\n",
-    "print(\"The drain resistance=%d kΩ.\"%R_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.18 : Page number 522-523"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The drain-source voltage=17V.\n",
-      "The gate-source voltage=-0.5V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_DD=30.0;                  #Drain supply voltage, V\n",
-    "R_D=5.0;                    #Drain resistor, kΩ\n",
-    "I_D=2.5;                    #Drain current, mA\n",
-    "R_S=200.0;                  #Source resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "V_DS=V_DD-I_D*(R_D+(R_S/1000));                 #Drain-source voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "V_GS=-(I_D/1000)*R_S;                          #Gate-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The drain-source voltage=%dV.\"%V_DS);\n",
-    "print(\"The gate-source voltage=%.1fV.\"%V_GS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.19 : Page number 523"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "drain voltage of 1st stage=12.37V.\n",
-      "Source voltage of 1st stage=1.46V.\n",
-      "drain voltage of 2nd stage=11.7V.\n",
-      "Source voltage of 2nd stage=2.01V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "ID_1=2.15;                  #First stage drain current, mA\n",
-    "ID_2=9.15;                  #Second stage drain current, mA\n",
-    "VDD=30;                     #Drain supply voltage, V\n",
-    "RS_1=0.68;                  #Source resistance of 1st stage, kΩ\n",
-    "RS_2=0.22;                  #Source resistance of 2nd stage, kΩ\n",
-    "RD_1=8.2;                   #Drain resistor of 1st stage, kΩ\n",
-    "RD_2=2;                     #Drain resistor of 2nd stage, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "V_RD1=ID_1*RD_1;                #Voltage drop across 8.2kΩ\n",
-    "VD_1=VDD-V_RD1;                 #Drain voltage of 1st stage, V\n",
-    "VS_1=ID_1*RS_1;                 #D.C potential of source of first stage, V\n",
-    "V_RD2=ID_2*RD_2;                #Voltage drop across 2kΩ\n",
-    "VD_2=VDD-V_RD2;                 #Drain voltage of 2nd stage, V\n",
-    "VS_2=ID_2*RS_2;                 #D.C potential of source of 2nd stage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"drain voltage of 1st stage=%.2fV.\"%VD_1);\n",
-    "print(\"Source voltage of 1st stage=%.2fV.\"%VS_1);\n",
-    "print(\"drain voltage of 2nd stage=%.1fV.\"%VD_2);\n",
-    "print(\"Source voltage of 2nd stage=%.2fV.\"%VS_2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.20 : Page number 524"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=1.52mA.\n",
-      "Gate-source voltage=-1.2V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=12;                     #Drain supply voltage, V\n",
-    "VD=7;                       #Drain voltage, V\n",
-    "R1=6.8;                     #Resistor R1, MΩ\n",
-    "R2=1;                       #Resistor R2, MΩ\n",
-    "RS=1.8;                     #Source resistance, kΩ\n",
-    "RD=3.3;                     #Drain resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "ID=(VDD-VD)/RD;                  #Second stage drain current, mA\n",
-    "VS=ID*RS;                        #Source voltage, V\n",
-    "VG=VDD*R2/(R1+R2);               #Drain voltage, V\n",
-    "VGS=VG-VS;                       #Drain-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "print(\"Drain current=%.2fmA.\"%ID);\n",
-    "print(\"Gate-source voltage=%.1fV.\"%VGS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.21 : Page number 524-525"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Source resistor, RS=5kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VDD=30;                     #Drain supply voltage, V\n",
-    "ID=2.5;                     #Drain current, mA\n",
-    "VDS=8;                      #Drain-source voltage, V\n",
-    "VGS_off=-5;                 #Gate-source cutoff voltage, V\n",
-    "R1=1;                       #Resistor R1, MΩ\n",
-    "R2=500;                     #Resistor R2, kΩ\n",
-    "IDSS=10;                    #Shorted gate drain current, mA\n",
-    "\n",
-    "#Calculation\n",
-    "#ID=IDSS*square_of(1-(VGS/VGS_off))\n",
-    "VGS=VGS_off*(1-sqrt(ID/IDSS));           #Gate-source voltage, V\n",
-    "V2=VDD*R2/(R1*1000+R2);                       #Voltage across R2, V\n",
-    "\n",
-    "\n",
-    "#V2=VGS+ID*RS\n",
-    "RS=(V2-VGS)/ID;                         #Source resistor, kΩ\n",
-    "\n",
-    "#Result\n",
-    "print(\"Source resistor, RS=%dkΩ.\"%RS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##  Example 19.22 : Page number 528-529"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f70cf070f28>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline \n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VDD=20.0;                 #Drain supply voltage, V\n",
-    "RS=50.0;                  #Source resistor, Ω\n",
-    "RD=150.0;                 #Drain resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "VDS_max=VDD;                           #Maximum drain source voltage, V\n",
-    "ID_max=(VDD/(RD+RS))*1000;             #Maximum drain current, mA\n",
-    "\n",
-    "\n",
-    "#plot\n",
-    "x=[i for i in range(0,(int)(VDS_max+1))];          #Plot variable for V_DS\n",
-    "y=[(i/(RD+RS))*1000 for i in reversed(x[:])];      #Plot variable for ID\n",
-    "\n",
-    "\n",
-    "plt.plot(x,y);\n",
-    "plt.xlabel(\"VDS(V)\");\n",
-    "plt.ylabel(\"ID(mA)\");\n",
-    "plt.title(\"d.c load line\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.23 : Page number 529"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f70cf06ab38>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline \n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VDD=20;                 #Drain supply voltage, V\n",
-    "RD=500.0;                 #Drain resistor, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "VDS_max=VDD;                           #Maximum drain source voltage, v \n",
-    "ID_max=(VDD/RD)*1000;                  #Maximum drain current, mA\n",
-    "\n",
-    "#Plot\n",
-    "x=[i for i in range(0,(int)(VDS_max+1))];          #Plot variable for V_DS\n",
-    "y=[(i/RD)*1000 for i in reversed(x[:])];      #Plot variable for ID\n",
-    "\n",
-    "plt.plot(x,y);\n",
-    "plt.xlabel(\"VDS(V)\");\n",
-    "plt.ylabel(\"ID(mA)\");\n",
-    "plt.title(\"d.c load line\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.24 : Page number 530-531"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain=4.8.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=20;                 #Drain supply voltage, V\n",
-    "RD=12.0;                  #Drain resistor, kΩ\n",
-    "RL=8.0;                   #Load resistor, kΩ\n",
-    "RG=1.0;                   #Gate resistor, MΩ\n",
-    "gm=1.0;                   #transconductance, mA/V\n",
-    "\n",
-    "#Calculation\n",
-    "gm=gm*10**-3;                   #transconductance, mho\n",
-    "RAC=(RD*RL)/(RD+RL);            #Total a.c load, kΩ\n",
-    "Av=gm*RAC*1000;                 #Voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Voltage gain=%.1f.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.25 : Page number 531"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain=30.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "gm=3000;                   #transconductance, μmho\n",
-    "RD=10;                     #Drain resistance, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "Av=gm*10**-6*RD*1000;       #Voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"Voltage gain=%d.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.26 : Page number 531"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=257mV(r.m.s).\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IDSS=8;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-10;                #Gate-source cut-off voltage, V\n",
-    "ID=1.9;                     #Drain current, mA\n",
-    "RD=3.3;                     #Drain resistance, kΩ\n",
-    "RS=2.7;                     #Source resistor, kΩ\n",
-    "vin=100;                    #Input voltage, mV\n",
-    "\n",
-    "#Calculation\n",
-    "VGS=-ID*RS;                      #Gate-source voltage, V\n",
-    "gmo=2*IDSS*10**-3/abs(VGS_off);  #Maximum transconductance, S\n",
-    "gm=gmo*(1-(VGS/VGS_off));        #Transconductance, S\n",
-    "Av=gm*RD*1000;                   #Voltage gain\n",
-    "vout=Av*vin;                     #Output voltage, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dmV(r.m.s).\"%vout);\n",
-    "                \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.27 : Page number 531-532"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 28,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=151mV(r.m.s).\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RL=4.7;                    #Load resistor, Ω\n",
-    "RD=3.3;                     #Drain resistance, kΩ\n",
-    "gm=779*10**-6;              #Transconductance, S\n",
-    "vin=100;                    #Input voltage, mV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "RAC=RD*RL/(RD+RL);          #Total a.c drain resistance, kΩ\n",
-    "Av=gm*RAC*1000;             #Voltage gain\n",
-    "vout=Av*vin;                     #Output voltage, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dmV(r.m.s).\"%vout);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.28 : Page number 532-533"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 29,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Voltage gain=1.85.\n",
-      "Voltage gain, if RS resistor is bypassed=6.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RD=1.5;            #Drain resistance, kΩ\n",
-    "gm=4;              #Transconductance, mS\n",
-    "RS=560;            #Source resistance, Ω\n",
-    "\n",
-    "#Calculation\n",
-    "Av=gm*10**-3*RD*1000/(1+gm*10**-3*RS);\n",
-    "print(\"Voltage gain=%.2f.\"%Av);\n",
-    "\n",
-    "#If RS is bypassed by a capacitor\n",
-    "Av=gm*10**-3*RD*1000;\n",
-    "print(\"Voltage gain, if RS resistor is bypassed=%d.\"%Av);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.29 : Page number 533"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 30,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Voltage gain with RS bypassed=4.155.\n",
-      "(ii) Voltage gain with RS unbypassed=1.35.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "from math import sqrt\n",
-    "\n",
-    "IDSS=10;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-3.5;                #Gate-source cut-off voltage, V\n",
-    "RD=1.5;            #Drain resistance, kΩ\n",
-    "RS=750;            #Source resistance, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#From d.c biasing\n",
-    "ID=2.3;                                     #Drain current, mA\n",
-    "VGS=round(VGS_off*(1-sqrt(ID/IDSS)),1);              #Gate-source voltage, V\n",
-    "gm=round(round((2*IDSS/abs(VGS_off)),1)*round((1-(VGS/VGS_off)),3),2);           #Transconductance, mS\n",
-    "\n",
-    "\n",
-    "#(i)\n",
-    "Av=gm*RD;                         #Voltage gain with RS resistor bypassed\n",
-    "print(\"(i) Voltage gain with RS bypassed=%.3f.\"%Av);\n",
-    "\n",
-    "#(ii)\n",
-    "Av=Av/(1+gm*(RS/1000.0));\n",
-    "print(\"(ii) Voltage gain with RS unbypassed=%.2f.\"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.30 : Page number 539-540"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  n-channel D-MOSFET\n",
-      "(ii) Drain current=3.91mA\n",
-      "(iii) Drain current=18.9mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IDSS=10.0;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-8.0;                  #Gate-source cut-off voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "if(VGS_off<0):\n",
-    "    print(\"(i)  n-channel D-MOSFET\");\n",
-    "else:\n",
-    "    print(\"(i)  p-channel D-MOSFET\");\n",
-    "    \n",
-    "\n",
-    "#(ii)\n",
-    "VGS=-3.0;                                #Gate-source voltage, V\n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"(ii) Drain current=%.2fmA\"%ID);\n",
-    "\n",
-    "#(iii)\n",
-    "VGS=3.0;                                #Gate-source voltage, V\n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"(iii) Drain current=%.1fmA\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.31 : Page number 540"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 32,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Point 1: VGS=0V and ID=1mA.\n",
-      "Point 2: VGS=-6V and ID=0mA.\n",
-      "Point 3: VGS=-3V and ID=0.25mA.\n",
-      "Point 4: VGS=-1V and ID=0.694mA.\n",
-      "Point 5: VGS=1V and ID=1.36mA.\n",
-      "Point 6: VGS=3V and ID=2.25mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "IDSS=1.0;                     #Shorted gate drain current, mA\n",
-    "VGS_off=-6.0;                  #Gate-source cut-off voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Point 1\n",
-    "VGS=0;                  #Gate source voltage, V               \n",
-    "ID=IDSS;                #Drain current, mA\n",
-    "print(\"Point 1: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n",
-    "\n",
-    "#Point 2\n",
-    "VGS=VGS_off;                  #Gate source voltage, V               \n",
-    "ID=0;                        #Drain current, mA\n",
-    "print(\"Point 2: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n",
-    "\n",
-    "#locating more points by changing VG values\n",
-    "VGS=-3;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 3: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n",
-    "\n",
-    "VGS=-1;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 4: VGS=%dV and ID=%.3fmA.\"%(VGS,ID));\n",
-    "\n",
-    "VGS=1;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 5: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n",
-    "\n",
-    "VGS=3;                  #Gate source voltage, V               \n",
-    "ID=IDSS*(1-(VGS/VGS_off))**2;          #Drain current mA\n",
-    "print(\"Point 6: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.32 : Page number 541"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 33,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain source voltage=10.6V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=18;          #Drain supply voltage, V\n",
-    "RD=620.0;           #Drain resistor, Ω\n",
-    "IDSS=12.0;        #Shorted gate drain current, mA\n",
-    "VGS_off=-8.0;      #Gate-source cut-off voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "ID=IDSS;               #Drain current, mA\n",
-    "VDS=VDD-IDSS*(RD/1000);       #Drain source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain source voltage=%.1fV.\"%VDS);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.33 : Page number 542"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Drain source voltage=7.56V.\n",
-      "(ii) Output voltage=922mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=15;           #Drain supply voltage\n",
-    "RD=620.0;          #Drain resistor, Ω\n",
-    "RL=8.2;          #Load resistor, kΩ\n",
-    "vin=500.0;         #Input voltage, V\n",
-    "IDSS=12.0;        #Shorted gate drain current, mA\n",
-    "gm=3.2;            #Transconductance, mS\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "VDS=VDD-IDSS*(RD/1000.0);       #Drain source voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "RAC=RD*RL*1000/(RD+RL*1000);  #Total a.c drain resistace, Ω\n",
-    "vout=(gm/1000.0)*RAC*vin;     #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  Drain source voltage=%.2fV.\"%VDS);\n",
-    "print(\"(ii) Output voltage=%dmV\"%vout);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.34 : Page number 545"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 35,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=98.7mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "ID_on=500.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=1.0;              #Threshold value of gate-source voltage, V\n",
-    "VGS=5;                 #Gate-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "K=round(ID_on/(VGS_on-VGS_th)**2,2);   #Constant for a E-MOSFET, mA/V²\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain current=%.1fmA\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.35 : Page number 545"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 36,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "K=0.061e-03A/V².\n",
-      "For VGS=5V, Drain current=0.244mA\n",
-      "For VGS=8V, Drain current=1.525mA\n",
-      "For VGS=10V, Drain current=2mA\n",
-      "For VGS=12V, Drain current=4.94mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "ID_on=3.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=3.0;              #Threshold value of gate-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "K=round((ID_on/(VGS_on-VGS_th)**2),3);   #Constant for a E-MOSFET, mA/V²\n",
-    "print(\"K=%.3fe-03A/V².\"%K);\n",
-    "\n",
-    "#Determining different points for plotting\n",
-    "VGS=5;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=5V, Drain current=%.3fmA\"%ID);\n",
-    "VGS=8;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=8V, Drain current=%.3fmA\"%ID);\n",
-    "VGS=10;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=10V, Drain current=%.dmA\"%ID);\n",
-    "VGS=12;                                 #Gate-source voltage, V\n",
-    "ID=K*(VGS-VGS_th)**2;                  #Drain current, mA\n",
-    "print(\"For VGS=12V, Drain current=%.2fmA\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.36 : Page number 546-547"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 37,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain-source voltage=10.8V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=24.0;             #Drain supply voltage, V\n",
-    "RD=470.0;             #Drain resistor, Ω\n",
-    "R1=100.0;             #Resistor R1, kΩ\n",
-    "R2=15.0;              #Resistor R2, kΩ\n",
-    "ID_on=500.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=1.0;              #Threshold value of gate-source voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "VGS=VDD*R2/(R1+R2);                      #Gate-source voltage, V (Voltage divider rule)\n",
-    "K=round((ID_on/(VGS_on-VGS_th)**2),2);   #Constant for a E-MOSFET, mA/V²\n",
-    "ID=K*(VGS-VGS_th)**2;                    #Drain current, mA\n",
-    "VDS=VDD-(ID/1000)*RD;                           #Drain-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain-source voltage=%.1fV.\"%VDS);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.37 : Page number 547"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 38,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=10mA.\n",
-      "Drain-source voltage=10V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=20.0;             #Drain supply voltage, V\n",
-    "RD=1.0;               #Drain resistor, kΩ\n",
-    "RG=5.0;               #Gate resistor , MΩ\n",
-    "ID_on=10.0;             #Drain current for MOSFET ON, mA\n",
-    "\n",
-    "#Calculation\n",
-    "#since, VGS=VDS\n",
-    "ID=ID_on;              #Drain current, mA\n",
-    "VDS=VDD-ID*RD;         #Drain-source voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain current=%dmA.\"%ID);\n",
-    "print(\"Drain-source voltage=%dV.\"%VDS);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 19.38 : Page number 547-548"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 39,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Drain current=1.69mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VDD=10.0;             #Drain supply voltage, V\n",
-    "RD=3.0;             #Drain resistor, kΩ\n",
-    "R1=1.0;             #Resistor R1, MΩ\n",
-    "R2=1.0;              #Resistor R2, MΩ\n",
-    "ID_on=10.0;             #Drain current for MOSFET ON, mA\n",
-    "VGS_on=10.0;             #Gate-source voltage for MOSFET ON, V\n",
-    "VGS_th=1.5;              #Threshold value of gate-source voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "K=round((ID_on/(VGS_on-VGS_th)**2),3);   #Constant for a E-MOSFET, mA/V²\n",
-    "VGS=VDD*R2/(R1+R2);                      #Gate-source voltage, V (Voltage divider rule)\n",
-    "ID=K*(VGS-VGS_th)**2;                    #Drain current, mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Drain current=%.2fmA.\"%ID);\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_1.ipynb
deleted file mode 100755
index 49519941..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_1.ipynb
+++ /dev/null
@@ -1,646 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0ac98582dd0b2497034e459e869a2a3bd28001d0d4c4b37a61a8ed5d05f228e3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 1: INTRODUCTION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.1: Page Number 8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "Eg=24.0;                     #Generated voltage in V\n",
-      "Ri=0.01;                   #Internal Resistance in \u03a9\n",
-      "P=100;                     #Power supplied in watts\n",
-      "\n",
-      "#Calculations\n",
-      "# (i)\n",
-      "I=P/Eg;                         #Load current in A\n",
-      "V_Ri=I*Ri;                      #Voltage drop in internal resistance\n",
-      "\n",
-      "# (ii)\n",
-      "V=Eg-(I*Ri);                     #Terminal Voltage\n",
-      "\n",
-      "#Results\n",
-      "print (\"The voltage drop in internal resistance is %.4f V\"%V_Ri);\n",
-      "print (\"The terminal voltage is %.2f V\"%V);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage drop in internal resistance is 0.0417 V\n",
-        "The terminal voltage is 23.96 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.2: Page number 10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Eg=500.0;                 #Generated voltage in V\n",
-      "Ri=1000.0;                #Internal Resistance in \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "# (i)\n",
-      "RL=10;                    #Load resistance of case 1 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "\n",
-      "print(\"The load current for RL=10\u03a9 is %.3f A\"%I);\n",
-      "\n",
-      "# (ii)\n",
-      "RL=50;                    #Load resistance of case 2 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "\n",
-      "print(\"The load current for RL=50\u03a9 is %.3f A\"%I);\n",
-      "\n",
-      "# (iii)\n",
-      "RL=100;                    #Load resistance of case 3 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "I=round(I,3);\n",
-      "\n",
-      "print(\"The load current for RL=100\u03a9 is %.3f A\"%I);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load current for RL=10\u03a9 is 0.495 A\n",
-        "The load current for RL=50\u03a9 is 0.476 A\n",
-        "The load current for RL=100\u03a9 is 0.455 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3: Page Number 11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "E=10.0;                #voltage of voltage source in V\n",
-      "Ri=10.0;               #Internal Resistance of the voltage source in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Isc=E/Ri;                #short circuit current in A\n",
-      "I=Isc;                   #Current value of current source in A\n",
-      "R=Ri;                    #Internal Resistence of the current source in \u03a9\n",
-      "\n",
-      "#Results\n",
-      "print(\"The current value of the current source= %d A\"%Isc);\n",
-      "print(\"The internal resistance of the current source =%d \u03a9 \"%R);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current value of the current source= 1 A\n",
-        "The internal resistance of the current source =10 \u03a9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.4: Page number 11-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I=6.0;                    # current value of current source in mA\n",
-      "Ri=2000.0;              #Internal Resistance of the current source in \u03a9\n",
-      "\n",
-      "#Calcultion\n",
-      "V=(I/1000)*Ri;                      #Voltage of voltage source in V\n",
-      "R=Ri;                         #Internal resistance of voltage source in \u03a9\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage of voltage source is %d V\"%V);\n",
-      "print(\"The internal resistance of the voltage source is %d \u03a9\"%R);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage of voltage source is 12 V\n",
-        "The internal resistance of the voltage source is 2000 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5: Page number 13\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "E=200.0;                    #Generated voltage in V\n",
-      "Ri=100.0;                   #Internal Resistance of generator in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "RL=100;                      #Load resistance for 1st case in \u03a9\n",
-      "I=E/(RL+Ri);                 #Load current in 1st case A\n",
-      "P=(I*I)*RL;                  #Power delivered to load of 2nd case in watts\n",
-      "Pt=(I*I)*(Ri+RL);            #Total power generated in watts\n",
-      "\n",
-      "print(\"Power delivered for RL=100\u03a9 is %d watts\"%P);\n",
-      "print(\"Total power generated for RL=100\u03a9 is %d watts\"%Pt);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "RL=300;                      #Load resistance for 2nd case in \u03a9\n",
-      "I=E/(RL+Ri);                 #Load current in 2nd case in A\n",
-      "P=(I*I)*RL;                  #Power delivered to load of 2nd case in watts\n",
-      "Pt=(I*I)*(Ri+RL);            #Total power generated in watts\n",
-      "\n",
-      "print(\"Power delivered for RL=300\u03a9 is %d watts\"%P);\n",
-      "print(\"Total power generated for RL=300\u03a9 is %d watts\"%Pt);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power delivered for RL=100\u03a9 is 100 watts\n",
-        "Total power generated for RL=100\u03a9 is 200 watts\n",
-        "Power delivered for RL=300\u03a9 is 75 watts\n",
-        "Total power generated for RL=300\u03a9 is 100 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6: Page number 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=12.0;                  #Output from amplifier in V\n",
-      "R_out_eq=15;           #Equivalent resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "RL=R_out_eq;              #Load resistance in \u03a9\n",
-      "Rt=RL+R_out_eq;           #Total resistance in \u03a9\n",
-      "I=V/Rt;                   #Circuit current in A\n",
-      "PL=pow(I,2)*RL;           #Power delivered to load in W\n",
-      "\n",
-      "#Results\n",
-      "print(\"Load resistance required is = %d \u03a9\"%RL);\n",
-      "print(\"Power delivered to load = %.1f W\"%PL);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load resistance required is = 15 \u03a9\n",
-        "Power delivered to load = 2.4 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7, Page number 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=50.0;                      #voltage from ac generator in V\n",
-      "R=100.0;                     #Resistance of internal impedance in \u03a9\n",
-      "XL=50.0;                     #inductive reactance of internal impedance in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Zi=100+(50j);               #Internal impedance in complex form (\u03a9)\n",
-      "ZL=conjugate(Zi);          #Load impedance (conjugate of internal impedance ) in \u03a9\n",
-      "Zt=Zi+ZL;                  #Total impedance in \u03a9\n",
-      "I=real(V/Zt);              #Circuit current in A\n",
-      "\n",
-      "Max_Power=pow(I,2)*R;      #Maximum power transferred to the load in watts\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print (\"Load impedance %d %dj \u03a9\"%(real(ZL),imag(ZL)));\n",
-      "print(\"Maximum power transferred to the load =%.2f W\"%Max_Power);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load impedance 100 -50j \u03a9\n",
-        "Maximum power transferred to the load =6.25 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.8: Page number 16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pR(R1,R2):\n",
-      "    return((R1*R2)/(R1+R2));\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "E=100.0;                              #Source voltage in V\n",
-      "R1=10.0;                              #Resistance of resistor 1 in \u03a9\n",
-      "R2=20.0;                              #Resistance of resistor 2 in \u03a9\n",
-      "R3=12.0;                              #Resistance of resistor 3 in \u03a9\n",
-      "R4=8.0;                               #Resistance of resistor 4 in \u03a9\n",
-      "RL=100.0;                             #Resistance of load in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Req=R1+pR(R3+R4,R2);                 #Equivalent resistance after removing RL ,in \u03a9\n",
-      "I=E/Req;                             #Total circuit current in A\n",
-      "I8=I*R2/(R2+R3+R4);\n",
-      "\n",
-      "#Thevenin's equivalent circuit's parameters\n",
-      "E0=I8*R4;                                         #Thevenin voltage V\n",
-      "R0=pR(pR(R1,R2)+R3,R4);                           #Thevenin resistance   \n",
-      "I_RL=E0/(R0+RL);                                  #Load current in A  \n",
-      "\n",
-      "#Result \n",
-      "print (\"Current through load = %.2f A.\"%I_RL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through load = 0.19 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.9: Page number 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pR(R1,R2):\n",
-      "    return((R1*R2)/(R1+R2));\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "V=20.0;                           #Voltage source in V\n",
-      "R1=1000.0;                        #resistance of resistor 1 in \u03a9\n",
-      "R2=1000.0;                        #resistance of resistor 2 in \u03a9\n",
-      "R3=1000.0;                        #resistance of resistor 3 in \u03a9\n",
-      "\n",
-      "#calculation\n",
-      "#parameter for Thevenin's equivalent circuit\n",
-      "E0=(V*R3)/(R1+R3);                         #thevenin voltage in V\n",
-      "R0=pR(R1,R3)+R2;                     #Thevenins resistance in \u03a9\n",
-      "\n",
-      "#result\n",
-      "print(\"The thevenin voltage = %d V\"%E0);\n",
-      "print(\"The thevenin resistance = %d \u03a9\"%R0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The thevenin voltage = 10 V\n",
-        "The thevenin resistance = 1500 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.10: Page number 18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=120.0;                          #Supply voltage in V\n",
-      "R1=40.0;                          #Resistor 1's resistance in \u03a9\n",
-      "R2=20.0;                          #Resistor 2's resistance in \u03a9\n",
-      "R3=60.0;                          #Resistor 3's resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, Thevenin's voltage and resistance are calculated\n",
-      "E0=(V*R2)/(R1+R2);                  #Thevenin voltage (voltage across the load resistance RL, after removing RL)in V\n",
-      "R0=(R1*R2)/(R1+R2) + R3;                 #Thevenin's resistance (Resistance between the terminals of load RL, with RL removed and source voltage shorted)in \u03a9                \n",
-      "RL=R0;                              #Value of load resistance to be connected for maximum power transfer in \u03a9\n",
-      "Pmax=pow(E0,2)/(4*RL);              #Maximum power transferred to load in watts\n",
-      "\n",
-      "#Results\n",
-      "print(\"The value of load resistance RL to which maximum power will be transferred = %.2f \u03a9.\"%RL);\n",
-      "print(\"The maximum power transferred to load =%.2f W.\"%Pmax);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of load resistance RL to which maximum power will be transferred = 73.33 \u03a9.\n",
-        "The maximum power transferred to load =5.45 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.11: Page number 18-19-20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=80.0;                           #Supply voltage in V\n",
-      "R1=100.0;                         #Resistor 1's resistance in \u03a9\n",
-      "R2=100.0;                         #Resistor 2's resistance in \u03a9\n",
-      "R3=30.0;                          #Resistor 3's resistance in \u03a9\n",
-      "R4=80.0;                          #Resistor 4's resistance in \u03a9\n",
-      "R5=20.0;                          #Resistor 5's resistance in \u03a9\n",
-      "R6=60.0;                          #Resistor 6's resistance in \u03a9\n",
-      "R7=20.0;                          #Resistor 7's resistance in \u03a9\n",
-      "R8=50.0;                          #Resistor 8's resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem,\n",
-      "E0=(V*R2)/(R1+R2);              #Thevenin's voltage for the circuit containing V, R1, R2 in V.\n",
-      "R0=(R1*R2)/(R1+R2);             #Thevenin's resistance for R1, R2 in \u03a9.\n",
-      "\n",
-      "#Using Thevenin's theorem again on E0, R0 and rest of the circuit resistors.\n",
-      "E0_1=(E0*R4)/(R0+R3+R4);                #Thevenin's voltage for the cicruit containing E0, R0, R3, R4 in V\n",
-      "R0_1=((R0+R3)*R4)/(R0+R3+R4);           #Thevenin's resistance of R0,R3,R4 (R0 and R3 in series and both in parallel with R4), in \u03a9 \n",
-      "\n",
-      "#Using Thevenin's theorem again on E0_1, R0_1, and rest of the circuit resistors.\n",
-      "E0_2=(E0_1*R6)/(R0_1+R5+R6);                       #Thevenin's voltage for the circuit containing E0_1, R0_1, R5, R6 in V\n",
-      "R0_2=((R0_1+R5)*R6)/(R0_1+R5+R6);                  #Thevenin's resistance of R0_1,R5,R6 (R0 and R3 in series and both in parallel with R4), in \u03a9\n",
-      "\n",
-      "\n",
-      "I_50=E0_2/(R0_2+R7+R8);                           #Current through the 50 \u03a9 resistor in A\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The current through the 50 \u03a9 resistor =%.1f A.\"%I_50);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the 50 \u03a9 resistor =0.1 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.12: Page number 22\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import floor\n",
-      "#Variable declaration\n",
-      "V=40.0;                       #Voltage supply in V\n",
-      "R1=4.0;                       #Resistor 1's resistance in \u03a9\n",
-      "R2=6.0;                       #Resistor 2's resistance in \u03a9\n",
-      "R3=5.0;                       #Resistor 3's resistance in \u03a9\n",
-      "R4=8.0;                       #Resistor 4's resistance in \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Using Norton's theorem,\n",
-      "#calculating Norton current by removing the load resistance R4 and short circuiting those two terminals of the circuit\n",
-      "R=R1 + (R2*R3)/(R2+R3);         #Load on source after removing R4 resistor, in  \u03a9\n",
-      "I=V/R;                          #Source current in A\n",
-      "\n",
-      "#Using current dividing rule ,calculating the short circuit current.\n",
-      "I_N=(I*R2)/(R2+R3);             #Norton's equivalent current or the short circuit current in A\n",
-      "\n",
-      "R_N=R3 + (R1*R2)/(R1+R2);       #Norton's equivalent resistance in \u03a9\n",
-      "\n",
-      "I_8=(I_N*R_N)/(R_N+R4);         #Current through the 8 \u03a9 resistance in A\n",
-      "\n",
-      " \n",
-      "\n",
-      "#Results\n",
-      "print(\"The current through the 8\u03a9 resistance =%.2f A.\"%I_8);\n",
-      "\n",
-      "#Note: The answer in the book is 1.55 A, but in the above code the approximate value is obtained, i.e not 1.55A but 1.56A\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the 8\u03a9 resistance =1.56 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.13 :Page number 23\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V1=30.0;                #Voltage source 1, V\n",
-      "V2=18.0;                #Voltage source 2, V\n",
-      "R1=20.0;                #1st resistor, \u03a9\n",
-      "R2=10.0;                #2nd resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Finding Thevenin's Equivalent circuit\n",
-      "I=(V1-V2)/(R1+R2);              #Current in the circuit, A\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to 1st loop of the circuit,\n",
-      "#V1-I*R1-E0=0, where E0 is the voltage across the points X-Y.\n",
-      "E0=V1-I*R1;                     #Thevenin's voltage source, V\n",
-      "\n",
-      "R0=R1*R2/(R1+R2);               #Thevenin's resistance, \u03a9\n",
-      "\n",
-      "#Finding Norton's equivalent circuit\n",
-      "IN=E0/R0;                   #Norton's equivalent current source, A\n",
-      "RN=R0;                      #Norton's equivanlent resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"IN=%.1fA and RN=%.2f \u03a9\"%(IN,RN));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IN=3.3A and RN=6.67 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_2.ipynb
deleted file mode 100755
index 49519941..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_2.ipynb
+++ /dev/null
@@ -1,646 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0ac98582dd0b2497034e459e869a2a3bd28001d0d4c4b37a61a8ed5d05f228e3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 1: INTRODUCTION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.1: Page Number 8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "Eg=24.0;                     #Generated voltage in V\n",
-      "Ri=0.01;                   #Internal Resistance in \u03a9\n",
-      "P=100;                     #Power supplied in watts\n",
-      "\n",
-      "#Calculations\n",
-      "# (i)\n",
-      "I=P/Eg;                         #Load current in A\n",
-      "V_Ri=I*Ri;                      #Voltage drop in internal resistance\n",
-      "\n",
-      "# (ii)\n",
-      "V=Eg-(I*Ri);                     #Terminal Voltage\n",
-      "\n",
-      "#Results\n",
-      "print (\"The voltage drop in internal resistance is %.4f V\"%V_Ri);\n",
-      "print (\"The terminal voltage is %.2f V\"%V);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage drop in internal resistance is 0.0417 V\n",
-        "The terminal voltage is 23.96 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.2: Page number 10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Eg=500.0;                 #Generated voltage in V\n",
-      "Ri=1000.0;                #Internal Resistance in \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "# (i)\n",
-      "RL=10;                    #Load resistance of case 1 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "\n",
-      "print(\"The load current for RL=10\u03a9 is %.3f A\"%I);\n",
-      "\n",
-      "# (ii)\n",
-      "RL=50;                    #Load resistance of case 2 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "\n",
-      "print(\"The load current for RL=50\u03a9 is %.3f A\"%I);\n",
-      "\n",
-      "# (iii)\n",
-      "RL=100;                    #Load resistance of case 3 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "I=round(I,3);\n",
-      "\n",
-      "print(\"The load current for RL=100\u03a9 is %.3f A\"%I);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load current for RL=10\u03a9 is 0.495 A\n",
-        "The load current for RL=50\u03a9 is 0.476 A\n",
-        "The load current for RL=100\u03a9 is 0.455 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3: Page Number 11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "E=10.0;                #voltage of voltage source in V\n",
-      "Ri=10.0;               #Internal Resistance of the voltage source in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Isc=E/Ri;                #short circuit current in A\n",
-      "I=Isc;                   #Current value of current source in A\n",
-      "R=Ri;                    #Internal Resistence of the current source in \u03a9\n",
-      "\n",
-      "#Results\n",
-      "print(\"The current value of the current source= %d A\"%Isc);\n",
-      "print(\"The internal resistance of the current source =%d \u03a9 \"%R);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current value of the current source= 1 A\n",
-        "The internal resistance of the current source =10 \u03a9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.4: Page number 11-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I=6.0;                    # current value of current source in mA\n",
-      "Ri=2000.0;              #Internal Resistance of the current source in \u03a9\n",
-      "\n",
-      "#Calcultion\n",
-      "V=(I/1000)*Ri;                      #Voltage of voltage source in V\n",
-      "R=Ri;                         #Internal resistance of voltage source in \u03a9\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage of voltage source is %d V\"%V);\n",
-      "print(\"The internal resistance of the voltage source is %d \u03a9\"%R);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage of voltage source is 12 V\n",
-        "The internal resistance of the voltage source is 2000 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5: Page number 13\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "E=200.0;                    #Generated voltage in V\n",
-      "Ri=100.0;                   #Internal Resistance of generator in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "RL=100;                      #Load resistance for 1st case in \u03a9\n",
-      "I=E/(RL+Ri);                 #Load current in 1st case A\n",
-      "P=(I*I)*RL;                  #Power delivered to load of 2nd case in watts\n",
-      "Pt=(I*I)*(Ri+RL);            #Total power generated in watts\n",
-      "\n",
-      "print(\"Power delivered for RL=100\u03a9 is %d watts\"%P);\n",
-      "print(\"Total power generated for RL=100\u03a9 is %d watts\"%Pt);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "RL=300;                      #Load resistance for 2nd case in \u03a9\n",
-      "I=E/(RL+Ri);                 #Load current in 2nd case in A\n",
-      "P=(I*I)*RL;                  #Power delivered to load of 2nd case in watts\n",
-      "Pt=(I*I)*(Ri+RL);            #Total power generated in watts\n",
-      "\n",
-      "print(\"Power delivered for RL=300\u03a9 is %d watts\"%P);\n",
-      "print(\"Total power generated for RL=300\u03a9 is %d watts\"%Pt);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power delivered for RL=100\u03a9 is 100 watts\n",
-        "Total power generated for RL=100\u03a9 is 200 watts\n",
-        "Power delivered for RL=300\u03a9 is 75 watts\n",
-        "Total power generated for RL=300\u03a9 is 100 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6: Page number 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=12.0;                  #Output from amplifier in V\n",
-      "R_out_eq=15;           #Equivalent resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "RL=R_out_eq;              #Load resistance in \u03a9\n",
-      "Rt=RL+R_out_eq;           #Total resistance in \u03a9\n",
-      "I=V/Rt;                   #Circuit current in A\n",
-      "PL=pow(I,2)*RL;           #Power delivered to load in W\n",
-      "\n",
-      "#Results\n",
-      "print(\"Load resistance required is = %d \u03a9\"%RL);\n",
-      "print(\"Power delivered to load = %.1f W\"%PL);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load resistance required is = 15 \u03a9\n",
-        "Power delivered to load = 2.4 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7, Page number 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=50.0;                      #voltage from ac generator in V\n",
-      "R=100.0;                     #Resistance of internal impedance in \u03a9\n",
-      "XL=50.0;                     #inductive reactance of internal impedance in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Zi=100+(50j);               #Internal impedance in complex form (\u03a9)\n",
-      "ZL=conjugate(Zi);          #Load impedance (conjugate of internal impedance ) in \u03a9\n",
-      "Zt=Zi+ZL;                  #Total impedance in \u03a9\n",
-      "I=real(V/Zt);              #Circuit current in A\n",
-      "\n",
-      "Max_Power=pow(I,2)*R;      #Maximum power transferred to the load in watts\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print (\"Load impedance %d %dj \u03a9\"%(real(ZL),imag(ZL)));\n",
-      "print(\"Maximum power transferred to the load =%.2f W\"%Max_Power);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load impedance 100 -50j \u03a9\n",
-        "Maximum power transferred to the load =6.25 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.8: Page number 16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pR(R1,R2):\n",
-      "    return((R1*R2)/(R1+R2));\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "E=100.0;                              #Source voltage in V\n",
-      "R1=10.0;                              #Resistance of resistor 1 in \u03a9\n",
-      "R2=20.0;                              #Resistance of resistor 2 in \u03a9\n",
-      "R3=12.0;                              #Resistance of resistor 3 in \u03a9\n",
-      "R4=8.0;                               #Resistance of resistor 4 in \u03a9\n",
-      "RL=100.0;                             #Resistance of load in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Req=R1+pR(R3+R4,R2);                 #Equivalent resistance after removing RL ,in \u03a9\n",
-      "I=E/Req;                             #Total circuit current in A\n",
-      "I8=I*R2/(R2+R3+R4);\n",
-      "\n",
-      "#Thevenin's equivalent circuit's parameters\n",
-      "E0=I8*R4;                                         #Thevenin voltage V\n",
-      "R0=pR(pR(R1,R2)+R3,R4);                           #Thevenin resistance   \n",
-      "I_RL=E0/(R0+RL);                                  #Load current in A  \n",
-      "\n",
-      "#Result \n",
-      "print (\"Current through load = %.2f A.\"%I_RL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through load = 0.19 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.9: Page number 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pR(R1,R2):\n",
-      "    return((R1*R2)/(R1+R2));\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "V=20.0;                           #Voltage source in V\n",
-      "R1=1000.0;                        #resistance of resistor 1 in \u03a9\n",
-      "R2=1000.0;                        #resistance of resistor 2 in \u03a9\n",
-      "R3=1000.0;                        #resistance of resistor 3 in \u03a9\n",
-      "\n",
-      "#calculation\n",
-      "#parameter for Thevenin's equivalent circuit\n",
-      "E0=(V*R3)/(R1+R3);                         #thevenin voltage in V\n",
-      "R0=pR(R1,R3)+R2;                     #Thevenins resistance in \u03a9\n",
-      "\n",
-      "#result\n",
-      "print(\"The thevenin voltage = %d V\"%E0);\n",
-      "print(\"The thevenin resistance = %d \u03a9\"%R0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The thevenin voltage = 10 V\n",
-        "The thevenin resistance = 1500 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.10: Page number 18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=120.0;                          #Supply voltage in V\n",
-      "R1=40.0;                          #Resistor 1's resistance in \u03a9\n",
-      "R2=20.0;                          #Resistor 2's resistance in \u03a9\n",
-      "R3=60.0;                          #Resistor 3's resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, Thevenin's voltage and resistance are calculated\n",
-      "E0=(V*R2)/(R1+R2);                  #Thevenin voltage (voltage across the load resistance RL, after removing RL)in V\n",
-      "R0=(R1*R2)/(R1+R2) + R3;                 #Thevenin's resistance (Resistance between the terminals of load RL, with RL removed and source voltage shorted)in \u03a9                \n",
-      "RL=R0;                              #Value of load resistance to be connected for maximum power transfer in \u03a9\n",
-      "Pmax=pow(E0,2)/(4*RL);              #Maximum power transferred to load in watts\n",
-      "\n",
-      "#Results\n",
-      "print(\"The value of load resistance RL to which maximum power will be transferred = %.2f \u03a9.\"%RL);\n",
-      "print(\"The maximum power transferred to load =%.2f W.\"%Pmax);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of load resistance RL to which maximum power will be transferred = 73.33 \u03a9.\n",
-        "The maximum power transferred to load =5.45 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.11: Page number 18-19-20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=80.0;                           #Supply voltage in V\n",
-      "R1=100.0;                         #Resistor 1's resistance in \u03a9\n",
-      "R2=100.0;                         #Resistor 2's resistance in \u03a9\n",
-      "R3=30.0;                          #Resistor 3's resistance in \u03a9\n",
-      "R4=80.0;                          #Resistor 4's resistance in \u03a9\n",
-      "R5=20.0;                          #Resistor 5's resistance in \u03a9\n",
-      "R6=60.0;                          #Resistor 6's resistance in \u03a9\n",
-      "R7=20.0;                          #Resistor 7's resistance in \u03a9\n",
-      "R8=50.0;                          #Resistor 8's resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem,\n",
-      "E0=(V*R2)/(R1+R2);              #Thevenin's voltage for the circuit containing V, R1, R2 in V.\n",
-      "R0=(R1*R2)/(R1+R2);             #Thevenin's resistance for R1, R2 in \u03a9.\n",
-      "\n",
-      "#Using Thevenin's theorem again on E0, R0 and rest of the circuit resistors.\n",
-      "E0_1=(E0*R4)/(R0+R3+R4);                #Thevenin's voltage for the cicruit containing E0, R0, R3, R4 in V\n",
-      "R0_1=((R0+R3)*R4)/(R0+R3+R4);           #Thevenin's resistance of R0,R3,R4 (R0 and R3 in series and both in parallel with R4), in \u03a9 \n",
-      "\n",
-      "#Using Thevenin's theorem again on E0_1, R0_1, and rest of the circuit resistors.\n",
-      "E0_2=(E0_1*R6)/(R0_1+R5+R6);                       #Thevenin's voltage for the circuit containing E0_1, R0_1, R5, R6 in V\n",
-      "R0_2=((R0_1+R5)*R6)/(R0_1+R5+R6);                  #Thevenin's resistance of R0_1,R5,R6 (R0 and R3 in series and both in parallel with R4), in \u03a9\n",
-      "\n",
-      "\n",
-      "I_50=E0_2/(R0_2+R7+R8);                           #Current through the 50 \u03a9 resistor in A\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The current through the 50 \u03a9 resistor =%.1f A.\"%I_50);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the 50 \u03a9 resistor =0.1 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.12: Page number 22\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import floor\n",
-      "#Variable declaration\n",
-      "V=40.0;                       #Voltage supply in V\n",
-      "R1=4.0;                       #Resistor 1's resistance in \u03a9\n",
-      "R2=6.0;                       #Resistor 2's resistance in \u03a9\n",
-      "R3=5.0;                       #Resistor 3's resistance in \u03a9\n",
-      "R4=8.0;                       #Resistor 4's resistance in \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Using Norton's theorem,\n",
-      "#calculating Norton current by removing the load resistance R4 and short circuiting those two terminals of the circuit\n",
-      "R=R1 + (R2*R3)/(R2+R3);         #Load on source after removing R4 resistor, in  \u03a9\n",
-      "I=V/R;                          #Source current in A\n",
-      "\n",
-      "#Using current dividing rule ,calculating the short circuit current.\n",
-      "I_N=(I*R2)/(R2+R3);             #Norton's equivalent current or the short circuit current in A\n",
-      "\n",
-      "R_N=R3 + (R1*R2)/(R1+R2);       #Norton's equivalent resistance in \u03a9\n",
-      "\n",
-      "I_8=(I_N*R_N)/(R_N+R4);         #Current through the 8 \u03a9 resistance in A\n",
-      "\n",
-      " \n",
-      "\n",
-      "#Results\n",
-      "print(\"The current through the 8\u03a9 resistance =%.2f A.\"%I_8);\n",
-      "\n",
-      "#Note: The answer in the book is 1.55 A, but in the above code the approximate value is obtained, i.e not 1.55A but 1.56A\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the 8\u03a9 resistance =1.56 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.13 :Page number 23\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V1=30.0;                #Voltage source 1, V\n",
-      "V2=18.0;                #Voltage source 2, V\n",
-      "R1=20.0;                #1st resistor, \u03a9\n",
-      "R2=10.0;                #2nd resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Finding Thevenin's Equivalent circuit\n",
-      "I=(V1-V2)/(R1+R2);              #Current in the circuit, A\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to 1st loop of the circuit,\n",
-      "#V1-I*R1-E0=0, where E0 is the voltage across the points X-Y.\n",
-      "E0=V1-I*R1;                     #Thevenin's voltage source, V\n",
-      "\n",
-      "R0=R1*R2/(R1+R2);               #Thevenin's resistance, \u03a9\n",
-      "\n",
-      "#Finding Norton's equivalent circuit\n",
-      "IN=E0/R0;                   #Norton's equivalent current source, A\n",
-      "RN=R0;                      #Norton's equivanlent resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"IN=%.1fA and RN=%.2f \u03a9\"%(IN,RN));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IN=3.3A and RN=6.67 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_3.ipynb
deleted file mode 100755
index 49519941..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_3.ipynb
+++ /dev/null
@@ -1,646 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0ac98582dd0b2497034e459e869a2a3bd28001d0d4c4b37a61a8ed5d05f228e3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 1: INTRODUCTION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.1: Page Number 8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "Eg=24.0;                     #Generated voltage in V\n",
-      "Ri=0.01;                   #Internal Resistance in \u03a9\n",
-      "P=100;                     #Power supplied in watts\n",
-      "\n",
-      "#Calculations\n",
-      "# (i)\n",
-      "I=P/Eg;                         #Load current in A\n",
-      "V_Ri=I*Ri;                      #Voltage drop in internal resistance\n",
-      "\n",
-      "# (ii)\n",
-      "V=Eg-(I*Ri);                     #Terminal Voltage\n",
-      "\n",
-      "#Results\n",
-      "print (\"The voltage drop in internal resistance is %.4f V\"%V_Ri);\n",
-      "print (\"The terminal voltage is %.2f V\"%V);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage drop in internal resistance is 0.0417 V\n",
-        "The terminal voltage is 23.96 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.2: Page number 10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Eg=500.0;                 #Generated voltage in V\n",
-      "Ri=1000.0;                #Internal Resistance in \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "# (i)\n",
-      "RL=10;                    #Load resistance of case 1 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "\n",
-      "print(\"The load current for RL=10\u03a9 is %.3f A\"%I);\n",
-      "\n",
-      "# (ii)\n",
-      "RL=50;                    #Load resistance of case 2 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "\n",
-      "print(\"The load current for RL=50\u03a9 is %.3f A\"%I);\n",
-      "\n",
-      "# (iii)\n",
-      "RL=100;                    #Load resistance of case 3 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "I=round(I,3);\n",
-      "\n",
-      "print(\"The load current for RL=100\u03a9 is %.3f A\"%I);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load current for RL=10\u03a9 is 0.495 A\n",
-        "The load current for RL=50\u03a9 is 0.476 A\n",
-        "The load current for RL=100\u03a9 is 0.455 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3: Page Number 11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "E=10.0;                #voltage of voltage source in V\n",
-      "Ri=10.0;               #Internal Resistance of the voltage source in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Isc=E/Ri;                #short circuit current in A\n",
-      "I=Isc;                   #Current value of current source in A\n",
-      "R=Ri;                    #Internal Resistence of the current source in \u03a9\n",
-      "\n",
-      "#Results\n",
-      "print(\"The current value of the current source= %d A\"%Isc);\n",
-      "print(\"The internal resistance of the current source =%d \u03a9 \"%R);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current value of the current source= 1 A\n",
-        "The internal resistance of the current source =10 \u03a9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.4: Page number 11-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I=6.0;                    # current value of current source in mA\n",
-      "Ri=2000.0;              #Internal Resistance of the current source in \u03a9\n",
-      "\n",
-      "#Calcultion\n",
-      "V=(I/1000)*Ri;                      #Voltage of voltage source in V\n",
-      "R=Ri;                         #Internal resistance of voltage source in \u03a9\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage of voltage source is %d V\"%V);\n",
-      "print(\"The internal resistance of the voltage source is %d \u03a9\"%R);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage of voltage source is 12 V\n",
-        "The internal resistance of the voltage source is 2000 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5: Page number 13\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "E=200.0;                    #Generated voltage in V\n",
-      "Ri=100.0;                   #Internal Resistance of generator in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "RL=100;                      #Load resistance for 1st case in \u03a9\n",
-      "I=E/(RL+Ri);                 #Load current in 1st case A\n",
-      "P=(I*I)*RL;                  #Power delivered to load of 2nd case in watts\n",
-      "Pt=(I*I)*(Ri+RL);            #Total power generated in watts\n",
-      "\n",
-      "print(\"Power delivered for RL=100\u03a9 is %d watts\"%P);\n",
-      "print(\"Total power generated for RL=100\u03a9 is %d watts\"%Pt);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "RL=300;                      #Load resistance for 2nd case in \u03a9\n",
-      "I=E/(RL+Ri);                 #Load current in 2nd case in A\n",
-      "P=(I*I)*RL;                  #Power delivered to load of 2nd case in watts\n",
-      "Pt=(I*I)*(Ri+RL);            #Total power generated in watts\n",
-      "\n",
-      "print(\"Power delivered for RL=300\u03a9 is %d watts\"%P);\n",
-      "print(\"Total power generated for RL=300\u03a9 is %d watts\"%Pt);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power delivered for RL=100\u03a9 is 100 watts\n",
-        "Total power generated for RL=100\u03a9 is 200 watts\n",
-        "Power delivered for RL=300\u03a9 is 75 watts\n",
-        "Total power generated for RL=300\u03a9 is 100 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6: Page number 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=12.0;                  #Output from amplifier in V\n",
-      "R_out_eq=15;           #Equivalent resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "RL=R_out_eq;              #Load resistance in \u03a9\n",
-      "Rt=RL+R_out_eq;           #Total resistance in \u03a9\n",
-      "I=V/Rt;                   #Circuit current in A\n",
-      "PL=pow(I,2)*RL;           #Power delivered to load in W\n",
-      "\n",
-      "#Results\n",
-      "print(\"Load resistance required is = %d \u03a9\"%RL);\n",
-      "print(\"Power delivered to load = %.1f W\"%PL);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load resistance required is = 15 \u03a9\n",
-        "Power delivered to load = 2.4 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7, Page number 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=50.0;                      #voltage from ac generator in V\n",
-      "R=100.0;                     #Resistance of internal impedance in \u03a9\n",
-      "XL=50.0;                     #inductive reactance of internal impedance in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Zi=100+(50j);               #Internal impedance in complex form (\u03a9)\n",
-      "ZL=conjugate(Zi);          #Load impedance (conjugate of internal impedance ) in \u03a9\n",
-      "Zt=Zi+ZL;                  #Total impedance in \u03a9\n",
-      "I=real(V/Zt);              #Circuit current in A\n",
-      "\n",
-      "Max_Power=pow(I,2)*R;      #Maximum power transferred to the load in watts\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print (\"Load impedance %d %dj \u03a9\"%(real(ZL),imag(ZL)));\n",
-      "print(\"Maximum power transferred to the load =%.2f W\"%Max_Power);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load impedance 100 -50j \u03a9\n",
-        "Maximum power transferred to the load =6.25 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.8: Page number 16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pR(R1,R2):\n",
-      "    return((R1*R2)/(R1+R2));\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "E=100.0;                              #Source voltage in V\n",
-      "R1=10.0;                              #Resistance of resistor 1 in \u03a9\n",
-      "R2=20.0;                              #Resistance of resistor 2 in \u03a9\n",
-      "R3=12.0;                              #Resistance of resistor 3 in \u03a9\n",
-      "R4=8.0;                               #Resistance of resistor 4 in \u03a9\n",
-      "RL=100.0;                             #Resistance of load in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Req=R1+pR(R3+R4,R2);                 #Equivalent resistance after removing RL ,in \u03a9\n",
-      "I=E/Req;                             #Total circuit current in A\n",
-      "I8=I*R2/(R2+R3+R4);\n",
-      "\n",
-      "#Thevenin's equivalent circuit's parameters\n",
-      "E0=I8*R4;                                         #Thevenin voltage V\n",
-      "R0=pR(pR(R1,R2)+R3,R4);                           #Thevenin resistance   \n",
-      "I_RL=E0/(R0+RL);                                  #Load current in A  \n",
-      "\n",
-      "#Result \n",
-      "print (\"Current through load = %.2f A.\"%I_RL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through load = 0.19 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.9: Page number 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pR(R1,R2):\n",
-      "    return((R1*R2)/(R1+R2));\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "V=20.0;                           #Voltage source in V\n",
-      "R1=1000.0;                        #resistance of resistor 1 in \u03a9\n",
-      "R2=1000.0;                        #resistance of resistor 2 in \u03a9\n",
-      "R3=1000.0;                        #resistance of resistor 3 in \u03a9\n",
-      "\n",
-      "#calculation\n",
-      "#parameter for Thevenin's equivalent circuit\n",
-      "E0=(V*R3)/(R1+R3);                         #thevenin voltage in V\n",
-      "R0=pR(R1,R3)+R2;                     #Thevenins resistance in \u03a9\n",
-      "\n",
-      "#result\n",
-      "print(\"The thevenin voltage = %d V\"%E0);\n",
-      "print(\"The thevenin resistance = %d \u03a9\"%R0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The thevenin voltage = 10 V\n",
-        "The thevenin resistance = 1500 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.10: Page number 18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=120.0;                          #Supply voltage in V\n",
-      "R1=40.0;                          #Resistor 1's resistance in \u03a9\n",
-      "R2=20.0;                          #Resistor 2's resistance in \u03a9\n",
-      "R3=60.0;                          #Resistor 3's resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, Thevenin's voltage and resistance are calculated\n",
-      "E0=(V*R2)/(R1+R2);                  #Thevenin voltage (voltage across the load resistance RL, after removing RL)in V\n",
-      "R0=(R1*R2)/(R1+R2) + R3;                 #Thevenin's resistance (Resistance between the terminals of load RL, with RL removed and source voltage shorted)in \u03a9                \n",
-      "RL=R0;                              #Value of load resistance to be connected for maximum power transfer in \u03a9\n",
-      "Pmax=pow(E0,2)/(4*RL);              #Maximum power transferred to load in watts\n",
-      "\n",
-      "#Results\n",
-      "print(\"The value of load resistance RL to which maximum power will be transferred = %.2f \u03a9.\"%RL);\n",
-      "print(\"The maximum power transferred to load =%.2f W.\"%Pmax);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of load resistance RL to which maximum power will be transferred = 73.33 \u03a9.\n",
-        "The maximum power transferred to load =5.45 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.11: Page number 18-19-20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=80.0;                           #Supply voltage in V\n",
-      "R1=100.0;                         #Resistor 1's resistance in \u03a9\n",
-      "R2=100.0;                         #Resistor 2's resistance in \u03a9\n",
-      "R3=30.0;                          #Resistor 3's resistance in \u03a9\n",
-      "R4=80.0;                          #Resistor 4's resistance in \u03a9\n",
-      "R5=20.0;                          #Resistor 5's resistance in \u03a9\n",
-      "R6=60.0;                          #Resistor 6's resistance in \u03a9\n",
-      "R7=20.0;                          #Resistor 7's resistance in \u03a9\n",
-      "R8=50.0;                          #Resistor 8's resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem,\n",
-      "E0=(V*R2)/(R1+R2);              #Thevenin's voltage for the circuit containing V, R1, R2 in V.\n",
-      "R0=(R1*R2)/(R1+R2);             #Thevenin's resistance for R1, R2 in \u03a9.\n",
-      "\n",
-      "#Using Thevenin's theorem again on E0, R0 and rest of the circuit resistors.\n",
-      "E0_1=(E0*R4)/(R0+R3+R4);                #Thevenin's voltage for the cicruit containing E0, R0, R3, R4 in V\n",
-      "R0_1=((R0+R3)*R4)/(R0+R3+R4);           #Thevenin's resistance of R0,R3,R4 (R0 and R3 in series and both in parallel with R4), in \u03a9 \n",
-      "\n",
-      "#Using Thevenin's theorem again on E0_1, R0_1, and rest of the circuit resistors.\n",
-      "E0_2=(E0_1*R6)/(R0_1+R5+R6);                       #Thevenin's voltage for the circuit containing E0_1, R0_1, R5, R6 in V\n",
-      "R0_2=((R0_1+R5)*R6)/(R0_1+R5+R6);                  #Thevenin's resistance of R0_1,R5,R6 (R0 and R3 in series and both in parallel with R4), in \u03a9\n",
-      "\n",
-      "\n",
-      "I_50=E0_2/(R0_2+R7+R8);                           #Current through the 50 \u03a9 resistor in A\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The current through the 50 \u03a9 resistor =%.1f A.\"%I_50);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the 50 \u03a9 resistor =0.1 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.12: Page number 22\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import floor\n",
-      "#Variable declaration\n",
-      "V=40.0;                       #Voltage supply in V\n",
-      "R1=4.0;                       #Resistor 1's resistance in \u03a9\n",
-      "R2=6.0;                       #Resistor 2's resistance in \u03a9\n",
-      "R3=5.0;                       #Resistor 3's resistance in \u03a9\n",
-      "R4=8.0;                       #Resistor 4's resistance in \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Using Norton's theorem,\n",
-      "#calculating Norton current by removing the load resistance R4 and short circuiting those two terminals of the circuit\n",
-      "R=R1 + (R2*R3)/(R2+R3);         #Load on source after removing R4 resistor, in  \u03a9\n",
-      "I=V/R;                          #Source current in A\n",
-      "\n",
-      "#Using current dividing rule ,calculating the short circuit current.\n",
-      "I_N=(I*R2)/(R2+R3);             #Norton's equivalent current or the short circuit current in A\n",
-      "\n",
-      "R_N=R3 + (R1*R2)/(R1+R2);       #Norton's equivalent resistance in \u03a9\n",
-      "\n",
-      "I_8=(I_N*R_N)/(R_N+R4);         #Current through the 8 \u03a9 resistance in A\n",
-      "\n",
-      " \n",
-      "\n",
-      "#Results\n",
-      "print(\"The current through the 8\u03a9 resistance =%.2f A.\"%I_8);\n",
-      "\n",
-      "#Note: The answer in the book is 1.55 A, but in the above code the approximate value is obtained, i.e not 1.55A but 1.56A\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the 8\u03a9 resistance =1.56 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.13 :Page number 23\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V1=30.0;                #Voltage source 1, V\n",
-      "V2=18.0;                #Voltage source 2, V\n",
-      "R1=20.0;                #1st resistor, \u03a9\n",
-      "R2=10.0;                #2nd resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Finding Thevenin's Equivalent circuit\n",
-      "I=(V1-V2)/(R1+R2);              #Current in the circuit, A\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to 1st loop of the circuit,\n",
-      "#V1-I*R1-E0=0, where E0 is the voltage across the points X-Y.\n",
-      "E0=V1-I*R1;                     #Thevenin's voltage source, V\n",
-      "\n",
-      "R0=R1*R2/(R1+R2);               #Thevenin's resistance, \u03a9\n",
-      "\n",
-      "#Finding Norton's equivalent circuit\n",
-      "IN=E0/R0;                   #Norton's equivalent current source, A\n",
-      "RN=R0;                      #Norton's equivanlent resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"IN=%.1fA and RN=%.2f \u03a9\"%(IN,RN));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IN=3.3A and RN=6.67 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_4.ipynb
deleted file mode 100755
index 49519941..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_4.ipynb
+++ /dev/null
@@ -1,646 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0ac98582dd0b2497034e459e869a2a3bd28001d0d4c4b37a61a8ed5d05f228e3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 1: INTRODUCTION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.1: Page Number 8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "Eg=24.0;                     #Generated voltage in V\n",
-      "Ri=0.01;                   #Internal Resistance in \u03a9\n",
-      "P=100;                     #Power supplied in watts\n",
-      "\n",
-      "#Calculations\n",
-      "# (i)\n",
-      "I=P/Eg;                         #Load current in A\n",
-      "V_Ri=I*Ri;                      #Voltage drop in internal resistance\n",
-      "\n",
-      "# (ii)\n",
-      "V=Eg-(I*Ri);                     #Terminal Voltage\n",
-      "\n",
-      "#Results\n",
-      "print (\"The voltage drop in internal resistance is %.4f V\"%V_Ri);\n",
-      "print (\"The terminal voltage is %.2f V\"%V);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage drop in internal resistance is 0.0417 V\n",
-        "The terminal voltage is 23.96 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.2: Page number 10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Eg=500.0;                 #Generated voltage in V\n",
-      "Ri=1000.0;                #Internal Resistance in \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "# (i)\n",
-      "RL=10;                    #Load resistance of case 1 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "\n",
-      "print(\"The load current for RL=10\u03a9 is %.3f A\"%I);\n",
-      "\n",
-      "# (ii)\n",
-      "RL=50;                    #Load resistance of case 2 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "\n",
-      "print(\"The load current for RL=50\u03a9 is %.3f A\"%I);\n",
-      "\n",
-      "# (iii)\n",
-      "RL=100;                    #Load resistance of case 3 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "I=round(I,3);\n",
-      "\n",
-      "print(\"The load current for RL=100\u03a9 is %.3f A\"%I);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load current for RL=10\u03a9 is 0.495 A\n",
-        "The load current for RL=50\u03a9 is 0.476 A\n",
-        "The load current for RL=100\u03a9 is 0.455 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3: Page Number 11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "E=10.0;                #voltage of voltage source in V\n",
-      "Ri=10.0;               #Internal Resistance of the voltage source in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Isc=E/Ri;                #short circuit current in A\n",
-      "I=Isc;                   #Current value of current source in A\n",
-      "R=Ri;                    #Internal Resistence of the current source in \u03a9\n",
-      "\n",
-      "#Results\n",
-      "print(\"The current value of the current source= %d A\"%Isc);\n",
-      "print(\"The internal resistance of the current source =%d \u03a9 \"%R);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current value of the current source= 1 A\n",
-        "The internal resistance of the current source =10 \u03a9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.4: Page number 11-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I=6.0;                    # current value of current source in mA\n",
-      "Ri=2000.0;              #Internal Resistance of the current source in \u03a9\n",
-      "\n",
-      "#Calcultion\n",
-      "V=(I/1000)*Ri;                      #Voltage of voltage source in V\n",
-      "R=Ri;                         #Internal resistance of voltage source in \u03a9\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage of voltage source is %d V\"%V);\n",
-      "print(\"The internal resistance of the voltage source is %d \u03a9\"%R);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage of voltage source is 12 V\n",
-        "The internal resistance of the voltage source is 2000 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5: Page number 13\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "E=200.0;                    #Generated voltage in V\n",
-      "Ri=100.0;                   #Internal Resistance of generator in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "RL=100;                      #Load resistance for 1st case in \u03a9\n",
-      "I=E/(RL+Ri);                 #Load current in 1st case A\n",
-      "P=(I*I)*RL;                  #Power delivered to load of 2nd case in watts\n",
-      "Pt=(I*I)*(Ri+RL);            #Total power generated in watts\n",
-      "\n",
-      "print(\"Power delivered for RL=100\u03a9 is %d watts\"%P);\n",
-      "print(\"Total power generated for RL=100\u03a9 is %d watts\"%Pt);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "RL=300;                      #Load resistance for 2nd case in \u03a9\n",
-      "I=E/(RL+Ri);                 #Load current in 2nd case in A\n",
-      "P=(I*I)*RL;                  #Power delivered to load of 2nd case in watts\n",
-      "Pt=(I*I)*(Ri+RL);            #Total power generated in watts\n",
-      "\n",
-      "print(\"Power delivered for RL=300\u03a9 is %d watts\"%P);\n",
-      "print(\"Total power generated for RL=300\u03a9 is %d watts\"%Pt);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power delivered for RL=100\u03a9 is 100 watts\n",
-        "Total power generated for RL=100\u03a9 is 200 watts\n",
-        "Power delivered for RL=300\u03a9 is 75 watts\n",
-        "Total power generated for RL=300\u03a9 is 100 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6: Page number 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=12.0;                  #Output from amplifier in V\n",
-      "R_out_eq=15;           #Equivalent resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "RL=R_out_eq;              #Load resistance in \u03a9\n",
-      "Rt=RL+R_out_eq;           #Total resistance in \u03a9\n",
-      "I=V/Rt;                   #Circuit current in A\n",
-      "PL=pow(I,2)*RL;           #Power delivered to load in W\n",
-      "\n",
-      "#Results\n",
-      "print(\"Load resistance required is = %d \u03a9\"%RL);\n",
-      "print(\"Power delivered to load = %.1f W\"%PL);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load resistance required is = 15 \u03a9\n",
-        "Power delivered to load = 2.4 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7, Page number 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=50.0;                      #voltage from ac generator in V\n",
-      "R=100.0;                     #Resistance of internal impedance in \u03a9\n",
-      "XL=50.0;                     #inductive reactance of internal impedance in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Zi=100+(50j);               #Internal impedance in complex form (\u03a9)\n",
-      "ZL=conjugate(Zi);          #Load impedance (conjugate of internal impedance ) in \u03a9\n",
-      "Zt=Zi+ZL;                  #Total impedance in \u03a9\n",
-      "I=real(V/Zt);              #Circuit current in A\n",
-      "\n",
-      "Max_Power=pow(I,2)*R;      #Maximum power transferred to the load in watts\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print (\"Load impedance %d %dj \u03a9\"%(real(ZL),imag(ZL)));\n",
-      "print(\"Maximum power transferred to the load =%.2f W\"%Max_Power);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load impedance 100 -50j \u03a9\n",
-        "Maximum power transferred to the load =6.25 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.8: Page number 16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pR(R1,R2):\n",
-      "    return((R1*R2)/(R1+R2));\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "E=100.0;                              #Source voltage in V\n",
-      "R1=10.0;                              #Resistance of resistor 1 in \u03a9\n",
-      "R2=20.0;                              #Resistance of resistor 2 in \u03a9\n",
-      "R3=12.0;                              #Resistance of resistor 3 in \u03a9\n",
-      "R4=8.0;                               #Resistance of resistor 4 in \u03a9\n",
-      "RL=100.0;                             #Resistance of load in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Req=R1+pR(R3+R4,R2);                 #Equivalent resistance after removing RL ,in \u03a9\n",
-      "I=E/Req;                             #Total circuit current in A\n",
-      "I8=I*R2/(R2+R3+R4);\n",
-      "\n",
-      "#Thevenin's equivalent circuit's parameters\n",
-      "E0=I8*R4;                                         #Thevenin voltage V\n",
-      "R0=pR(pR(R1,R2)+R3,R4);                           #Thevenin resistance   \n",
-      "I_RL=E0/(R0+RL);                                  #Load current in A  \n",
-      "\n",
-      "#Result \n",
-      "print (\"Current through load = %.2f A.\"%I_RL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through load = 0.19 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.9: Page number 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pR(R1,R2):\n",
-      "    return((R1*R2)/(R1+R2));\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "V=20.0;                           #Voltage source in V\n",
-      "R1=1000.0;                        #resistance of resistor 1 in \u03a9\n",
-      "R2=1000.0;                        #resistance of resistor 2 in \u03a9\n",
-      "R3=1000.0;                        #resistance of resistor 3 in \u03a9\n",
-      "\n",
-      "#calculation\n",
-      "#parameter for Thevenin's equivalent circuit\n",
-      "E0=(V*R3)/(R1+R3);                         #thevenin voltage in V\n",
-      "R0=pR(R1,R3)+R2;                     #Thevenins resistance in \u03a9\n",
-      "\n",
-      "#result\n",
-      "print(\"The thevenin voltage = %d V\"%E0);\n",
-      "print(\"The thevenin resistance = %d \u03a9\"%R0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The thevenin voltage = 10 V\n",
-        "The thevenin resistance = 1500 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.10: Page number 18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=120.0;                          #Supply voltage in V\n",
-      "R1=40.0;                          #Resistor 1's resistance in \u03a9\n",
-      "R2=20.0;                          #Resistor 2's resistance in \u03a9\n",
-      "R3=60.0;                          #Resistor 3's resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, Thevenin's voltage and resistance are calculated\n",
-      "E0=(V*R2)/(R1+R2);                  #Thevenin voltage (voltage across the load resistance RL, after removing RL)in V\n",
-      "R0=(R1*R2)/(R1+R2) + R3;                 #Thevenin's resistance (Resistance between the terminals of load RL, with RL removed and source voltage shorted)in \u03a9                \n",
-      "RL=R0;                              #Value of load resistance to be connected for maximum power transfer in \u03a9\n",
-      "Pmax=pow(E0,2)/(4*RL);              #Maximum power transferred to load in watts\n",
-      "\n",
-      "#Results\n",
-      "print(\"The value of load resistance RL to which maximum power will be transferred = %.2f \u03a9.\"%RL);\n",
-      "print(\"The maximum power transferred to load =%.2f W.\"%Pmax);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of load resistance RL to which maximum power will be transferred = 73.33 \u03a9.\n",
-        "The maximum power transferred to load =5.45 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.11: Page number 18-19-20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=80.0;                           #Supply voltage in V\n",
-      "R1=100.0;                         #Resistor 1's resistance in \u03a9\n",
-      "R2=100.0;                         #Resistor 2's resistance in \u03a9\n",
-      "R3=30.0;                          #Resistor 3's resistance in \u03a9\n",
-      "R4=80.0;                          #Resistor 4's resistance in \u03a9\n",
-      "R5=20.0;                          #Resistor 5's resistance in \u03a9\n",
-      "R6=60.0;                          #Resistor 6's resistance in \u03a9\n",
-      "R7=20.0;                          #Resistor 7's resistance in \u03a9\n",
-      "R8=50.0;                          #Resistor 8's resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem,\n",
-      "E0=(V*R2)/(R1+R2);              #Thevenin's voltage for the circuit containing V, R1, R2 in V.\n",
-      "R0=(R1*R2)/(R1+R2);             #Thevenin's resistance for R1, R2 in \u03a9.\n",
-      "\n",
-      "#Using Thevenin's theorem again on E0, R0 and rest of the circuit resistors.\n",
-      "E0_1=(E0*R4)/(R0+R3+R4);                #Thevenin's voltage for the cicruit containing E0, R0, R3, R4 in V\n",
-      "R0_1=((R0+R3)*R4)/(R0+R3+R4);           #Thevenin's resistance of R0,R3,R4 (R0 and R3 in series and both in parallel with R4), in \u03a9 \n",
-      "\n",
-      "#Using Thevenin's theorem again on E0_1, R0_1, and rest of the circuit resistors.\n",
-      "E0_2=(E0_1*R6)/(R0_1+R5+R6);                       #Thevenin's voltage for the circuit containing E0_1, R0_1, R5, R6 in V\n",
-      "R0_2=((R0_1+R5)*R6)/(R0_1+R5+R6);                  #Thevenin's resistance of R0_1,R5,R6 (R0 and R3 in series and both in parallel with R4), in \u03a9\n",
-      "\n",
-      "\n",
-      "I_50=E0_2/(R0_2+R7+R8);                           #Current through the 50 \u03a9 resistor in A\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The current through the 50 \u03a9 resistor =%.1f A.\"%I_50);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the 50 \u03a9 resistor =0.1 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.12: Page number 22\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import floor\n",
-      "#Variable declaration\n",
-      "V=40.0;                       #Voltage supply in V\n",
-      "R1=4.0;                       #Resistor 1's resistance in \u03a9\n",
-      "R2=6.0;                       #Resistor 2's resistance in \u03a9\n",
-      "R3=5.0;                       #Resistor 3's resistance in \u03a9\n",
-      "R4=8.0;                       #Resistor 4's resistance in \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Using Norton's theorem,\n",
-      "#calculating Norton current by removing the load resistance R4 and short circuiting those two terminals of the circuit\n",
-      "R=R1 + (R2*R3)/(R2+R3);         #Load on source after removing R4 resistor, in  \u03a9\n",
-      "I=V/R;                          #Source current in A\n",
-      "\n",
-      "#Using current dividing rule ,calculating the short circuit current.\n",
-      "I_N=(I*R2)/(R2+R3);             #Norton's equivalent current or the short circuit current in A\n",
-      "\n",
-      "R_N=R3 + (R1*R2)/(R1+R2);       #Norton's equivalent resistance in \u03a9\n",
-      "\n",
-      "I_8=(I_N*R_N)/(R_N+R4);         #Current through the 8 \u03a9 resistance in A\n",
-      "\n",
-      " \n",
-      "\n",
-      "#Results\n",
-      "print(\"The current through the 8\u03a9 resistance =%.2f A.\"%I_8);\n",
-      "\n",
-      "#Note: The answer in the book is 1.55 A, but in the above code the approximate value is obtained, i.e not 1.55A but 1.56A\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the 8\u03a9 resistance =1.56 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.13 :Page number 23\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V1=30.0;                #Voltage source 1, V\n",
-      "V2=18.0;                #Voltage source 2, V\n",
-      "R1=20.0;                #1st resistor, \u03a9\n",
-      "R2=10.0;                #2nd resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Finding Thevenin's Equivalent circuit\n",
-      "I=(V1-V2)/(R1+R2);              #Current in the circuit, A\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to 1st loop of the circuit,\n",
-      "#V1-I*R1-E0=0, where E0 is the voltage across the points X-Y.\n",
-      "E0=V1-I*R1;                     #Thevenin's voltage source, V\n",
-      "\n",
-      "R0=R1*R2/(R1+R2);               #Thevenin's resistance, \u03a9\n",
-      "\n",
-      "#Finding Norton's equivalent circuit\n",
-      "IN=E0/R0;                   #Norton's equivalent current source, A\n",
-      "RN=R0;                      #Norton's equivanlent resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"IN=%.1fA and RN=%.2f \u03a9\"%(IN,RN));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IN=3.3A and RN=6.67 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_5.ipynb
deleted file mode 100755
index 49519941..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_5.ipynb
+++ /dev/null
@@ -1,646 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0ac98582dd0b2497034e459e869a2a3bd28001d0d4c4b37a61a8ed5d05f228e3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 1: INTRODUCTION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.1: Page Number 8"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "Eg=24.0;                     #Generated voltage in V\n",
-      "Ri=0.01;                   #Internal Resistance in \u03a9\n",
-      "P=100;                     #Power supplied in watts\n",
-      "\n",
-      "#Calculations\n",
-      "# (i)\n",
-      "I=P/Eg;                         #Load current in A\n",
-      "V_Ri=I*Ri;                      #Voltage drop in internal resistance\n",
-      "\n",
-      "# (ii)\n",
-      "V=Eg-(I*Ri);                     #Terminal Voltage\n",
-      "\n",
-      "#Results\n",
-      "print (\"The voltage drop in internal resistance is %.4f V\"%V_Ri);\n",
-      "print (\"The terminal voltage is %.2f V\"%V);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage drop in internal resistance is 0.0417 V\n",
-        "The terminal voltage is 23.96 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.2: Page number 10"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Eg=500.0;                 #Generated voltage in V\n",
-      "Ri=1000.0;                #Internal Resistance in \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "# (i)\n",
-      "RL=10;                    #Load resistance of case 1 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "\n",
-      "print(\"The load current for RL=10\u03a9 is %.3f A\"%I);\n",
-      "\n",
-      "# (ii)\n",
-      "RL=50;                    #Load resistance of case 2 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "\n",
-      "print(\"The load current for RL=50\u03a9 is %.3f A\"%I);\n",
-      "\n",
-      "# (iii)\n",
-      "RL=100;                    #Load resistance of case 3 in \u03a9 \n",
-      "I= Eg/(RL+Ri);            #Load current in A\n",
-      "I=round(I,3);\n",
-      "\n",
-      "print(\"The load current for RL=100\u03a9 is %.3f A\"%I);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load current for RL=10\u03a9 is 0.495 A\n",
-        "The load current for RL=50\u03a9 is 0.476 A\n",
-        "The load current for RL=100\u03a9 is 0.455 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.3: Page Number 11"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "E=10.0;                #voltage of voltage source in V\n",
-      "Ri=10.0;               #Internal Resistance of the voltage source in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Isc=E/Ri;                #short circuit current in A\n",
-      "I=Isc;                   #Current value of current source in A\n",
-      "R=Ri;                    #Internal Resistence of the current source in \u03a9\n",
-      "\n",
-      "#Results\n",
-      "print(\"The current value of the current source= %d A\"%Isc);\n",
-      "print(\"The internal resistance of the current source =%d \u03a9 \"%R);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current value of the current source= 1 A\n",
-        "The internal resistance of the current source =10 \u03a9 \n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "EXAMPLE 1.4: Page number 11-12"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I=6.0;                    # current value of current source in mA\n",
-      "Ri=2000.0;              #Internal Resistance of the current source in \u03a9\n",
-      "\n",
-      "#Calcultion\n",
-      "V=(I/1000)*Ri;                      #Voltage of voltage source in V\n",
-      "R=Ri;                         #Internal resistance of voltage source in \u03a9\n",
-      "\n",
-      "#Results\n",
-      "print(\"The voltage of voltage source is %d V\"%V);\n",
-      "print(\"The internal resistance of the voltage source is %d \u03a9\"%R);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage of voltage source is 12 V\n",
-        "The internal resistance of the voltage source is 2000 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.5: Page number 13\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "E=200.0;                    #Generated voltage in V\n",
-      "Ri=100.0;                   #Internal Resistance of generator in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#(i)\n",
-      "RL=100;                      #Load resistance for 1st case in \u03a9\n",
-      "I=E/(RL+Ri);                 #Load current in 1st case A\n",
-      "P=(I*I)*RL;                  #Power delivered to load of 2nd case in watts\n",
-      "Pt=(I*I)*(Ri+RL);            #Total power generated in watts\n",
-      "\n",
-      "print(\"Power delivered for RL=100\u03a9 is %d watts\"%P);\n",
-      "print(\"Total power generated for RL=100\u03a9 is %d watts\"%Pt);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "RL=300;                      #Load resistance for 2nd case in \u03a9\n",
-      "I=E/(RL+Ri);                 #Load current in 2nd case in A\n",
-      "P=(I*I)*RL;                  #Power delivered to load of 2nd case in watts\n",
-      "Pt=(I*I)*(Ri+RL);            #Total power generated in watts\n",
-      "\n",
-      "print(\"Power delivered for RL=300\u03a9 is %d watts\"%P);\n",
-      "print(\"Total power generated for RL=300\u03a9 is %d watts\"%Pt);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power delivered for RL=100\u03a9 is 100 watts\n",
-        "Total power generated for RL=100\u03a9 is 200 watts\n",
-        "Power delivered for RL=300\u03a9 is 75 watts\n",
-        "Total power generated for RL=300\u03a9 is 100 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6: Page number 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=12.0;                  #Output from amplifier in V\n",
-      "R_out_eq=15;           #Equivalent resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "RL=R_out_eq;              #Load resistance in \u03a9\n",
-      "Rt=RL+R_out_eq;           #Total resistance in \u03a9\n",
-      "I=V/Rt;                   #Circuit current in A\n",
-      "PL=pow(I,2)*RL;           #Power delivered to load in W\n",
-      "\n",
-      "#Results\n",
-      "print(\"Load resistance required is = %d \u03a9\"%RL);\n",
-      "print(\"Power delivered to load = %.1f W\"%PL);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load resistance required is = 15 \u03a9\n",
-        "Power delivered to load = 2.4 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.7, Page number 14"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=50.0;                      #voltage from ac generator in V\n",
-      "R=100.0;                     #Resistance of internal impedance in \u03a9\n",
-      "XL=50.0;                     #inductive reactance of internal impedance in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Zi=100+(50j);               #Internal impedance in complex form (\u03a9)\n",
-      "ZL=conjugate(Zi);          #Load impedance (conjugate of internal impedance ) in \u03a9\n",
-      "Zt=Zi+ZL;                  #Total impedance in \u03a9\n",
-      "I=real(V/Zt);              #Circuit current in A\n",
-      "\n",
-      "Max_Power=pow(I,2)*R;      #Maximum power transferred to the load in watts\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print (\"Load impedance %d %dj \u03a9\"%(real(ZL),imag(ZL)));\n",
-      "print(\"Maximum power transferred to the load =%.2f W\"%Max_Power);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load impedance 100 -50j \u03a9\n",
-        "Maximum power transferred to the load =6.25 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.8: Page number 16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pR(R1,R2):\n",
-      "    return((R1*R2)/(R1+R2));\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "E=100.0;                              #Source voltage in V\n",
-      "R1=10.0;                              #Resistance of resistor 1 in \u03a9\n",
-      "R2=20.0;                              #Resistance of resistor 2 in \u03a9\n",
-      "R3=12.0;                              #Resistance of resistor 3 in \u03a9\n",
-      "R4=8.0;                               #Resistance of resistor 4 in \u03a9\n",
-      "RL=100.0;                             #Resistance of load in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Req=R1+pR(R3+R4,R2);                 #Equivalent resistance after removing RL ,in \u03a9\n",
-      "I=E/Req;                             #Total circuit current in A\n",
-      "I8=I*R2/(R2+R3+R4);\n",
-      "\n",
-      "#Thevenin's equivalent circuit's parameters\n",
-      "E0=I8*R4;                                         #Thevenin voltage V\n",
-      "R0=pR(pR(R1,R2)+R3,R4);                           #Thevenin resistance   \n",
-      "I_RL=E0/(R0+RL);                                  #Load current in A  \n",
-      "\n",
-      "#Result \n",
-      "print (\"Current through load = %.2f A.\"%I_RL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through load = 0.19 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.9: Page number 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pR(R1,R2):\n",
-      "    return((R1*R2)/(R1+R2));\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "V=20.0;                           #Voltage source in V\n",
-      "R1=1000.0;                        #resistance of resistor 1 in \u03a9\n",
-      "R2=1000.0;                        #resistance of resistor 2 in \u03a9\n",
-      "R3=1000.0;                        #resistance of resistor 3 in \u03a9\n",
-      "\n",
-      "#calculation\n",
-      "#parameter for Thevenin's equivalent circuit\n",
-      "E0=(V*R3)/(R1+R3);                         #thevenin voltage in V\n",
-      "R0=pR(R1,R3)+R2;                     #Thevenins resistance in \u03a9\n",
-      "\n",
-      "#result\n",
-      "print(\"The thevenin voltage = %d V\"%E0);\n",
-      "print(\"The thevenin resistance = %d \u03a9\"%R0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The thevenin voltage = 10 V\n",
-        "The thevenin resistance = 1500 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.10: Page number 18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=120.0;                          #Supply voltage in V\n",
-      "R1=40.0;                          #Resistor 1's resistance in \u03a9\n",
-      "R2=20.0;                          #Resistor 2's resistance in \u03a9\n",
-      "R3=60.0;                          #Resistor 3's resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, Thevenin's voltage and resistance are calculated\n",
-      "E0=(V*R2)/(R1+R2);                  #Thevenin voltage (voltage across the load resistance RL, after removing RL)in V\n",
-      "R0=(R1*R2)/(R1+R2) + R3;                 #Thevenin's resistance (Resistance between the terminals of load RL, with RL removed and source voltage shorted)in \u03a9                \n",
-      "RL=R0;                              #Value of load resistance to be connected for maximum power transfer in \u03a9\n",
-      "Pmax=pow(E0,2)/(4*RL);              #Maximum power transferred to load in watts\n",
-      "\n",
-      "#Results\n",
-      "print(\"The value of load resistance RL to which maximum power will be transferred = %.2f \u03a9.\"%RL);\n",
-      "print(\"The maximum power transferred to load =%.2f W.\"%Pmax);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of load resistance RL to which maximum power will be transferred = 73.33 \u03a9.\n",
-        "The maximum power transferred to load =5.45 W.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.11: Page number 18-19-20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=80.0;                           #Supply voltage in V\n",
-      "R1=100.0;                         #Resistor 1's resistance in \u03a9\n",
-      "R2=100.0;                         #Resistor 2's resistance in \u03a9\n",
-      "R3=30.0;                          #Resistor 3's resistance in \u03a9\n",
-      "R4=80.0;                          #Resistor 4's resistance in \u03a9\n",
-      "R5=20.0;                          #Resistor 5's resistance in \u03a9\n",
-      "R6=60.0;                          #Resistor 6's resistance in \u03a9\n",
-      "R7=20.0;                          #Resistor 7's resistance in \u03a9\n",
-      "R8=50.0;                          #Resistor 8's resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem,\n",
-      "E0=(V*R2)/(R1+R2);              #Thevenin's voltage for the circuit containing V, R1, R2 in V.\n",
-      "R0=(R1*R2)/(R1+R2);             #Thevenin's resistance for R1, R2 in \u03a9.\n",
-      "\n",
-      "#Using Thevenin's theorem again on E0, R0 and rest of the circuit resistors.\n",
-      "E0_1=(E0*R4)/(R0+R3+R4);                #Thevenin's voltage for the cicruit containing E0, R0, R3, R4 in V\n",
-      "R0_1=((R0+R3)*R4)/(R0+R3+R4);           #Thevenin's resistance of R0,R3,R4 (R0 and R3 in series and both in parallel with R4), in \u03a9 \n",
-      "\n",
-      "#Using Thevenin's theorem again on E0_1, R0_1, and rest of the circuit resistors.\n",
-      "E0_2=(E0_1*R6)/(R0_1+R5+R6);                       #Thevenin's voltage for the circuit containing E0_1, R0_1, R5, R6 in V\n",
-      "R0_2=((R0_1+R5)*R6)/(R0_1+R5+R6);                  #Thevenin's resistance of R0_1,R5,R6 (R0 and R3 in series and both in parallel with R4), in \u03a9\n",
-      "\n",
-      "\n",
-      "I_50=E0_2/(R0_2+R7+R8);                           #Current through the 50 \u03a9 resistor in A\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The current through the 50 \u03a9 resistor =%.1f A.\"%I_50);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the 50 \u03a9 resistor =0.1 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.12: Page number 22\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "from math import floor\n",
-      "#Variable declaration\n",
-      "V=40.0;                       #Voltage supply in V\n",
-      "R1=4.0;                       #Resistor 1's resistance in \u03a9\n",
-      "R2=6.0;                       #Resistor 2's resistance in \u03a9\n",
-      "R3=5.0;                       #Resistor 3's resistance in \u03a9\n",
-      "R4=8.0;                       #Resistor 4's resistance in \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Using Norton's theorem,\n",
-      "#calculating Norton current by removing the load resistance R4 and short circuiting those two terminals of the circuit\n",
-      "R=R1 + (R2*R3)/(R2+R3);         #Load on source after removing R4 resistor, in  \u03a9\n",
-      "I=V/R;                          #Source current in A\n",
-      "\n",
-      "#Using current dividing rule ,calculating the short circuit current.\n",
-      "I_N=(I*R2)/(R2+R3);             #Norton's equivalent current or the short circuit current in A\n",
-      "\n",
-      "R_N=R3 + (R1*R2)/(R1+R2);       #Norton's equivalent resistance in \u03a9\n",
-      "\n",
-      "I_8=(I_N*R_N)/(R_N+R4);         #Current through the 8 \u03a9 resistance in A\n",
-      "\n",
-      " \n",
-      "\n",
-      "#Results\n",
-      "print(\"The current through the 8\u03a9 resistance =%.2f A.\"%I_8);\n",
-      "\n",
-      "#Note: The answer in the book is 1.55 A, but in the above code the approximate value is obtained, i.e not 1.55A but 1.56A\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the 8\u03a9 resistance =1.56 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.13 :Page number 23\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V1=30.0;                #Voltage source 1, V\n",
-      "V2=18.0;                #Voltage source 2, V\n",
-      "R1=20.0;                #1st resistor, \u03a9\n",
-      "R2=10.0;                #2nd resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Finding Thevenin's Equivalent circuit\n",
-      "I=(V1-V2)/(R1+R2);              #Current in the circuit, A\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to 1st loop of the circuit,\n",
-      "#V1-I*R1-E0=0, where E0 is the voltage across the points X-Y.\n",
-      "E0=V1-I*R1;                     #Thevenin's voltage source, V\n",
-      "\n",
-      "R0=R1*R2/(R1+R2);               #Thevenin's resistance, \u03a9\n",
-      "\n",
-      "#Finding Norton's equivalent circuit\n",
-      "IN=E0/R0;                   #Norton's equivalent current source, A\n",
-      "RN=R0;                      #Norton's equivanlent resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"IN=%.1fA and RN=%.2f \u03a9\"%(IN,RN));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IN=3.3A and RN=6.67 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2.ipynb
deleted file mode 100755
index 06a555a6..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2.ipynb
+++ /dev/null
@@ -1,125 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c0ff4f67576afe73a11c06eedd0a50709b7f5831737f83db1cd640098e3e9740"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 2 : ELECTRONIC EMISSION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1: Page number 31\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "\n",
-      "from math import exp\n",
-      "from math import pi\n",
-      "\n",
-      "l=5.0;                        #length of tungsten filament in cm\n",
-      "d=0.01;                       #diameter of the filament in cm\n",
-      "T=2500.0;                     #operating temperature in K\n",
-      "A=60.2*pow(10,4);             #constant, depending upon the type of thermionic emitter, in amp/m\u00b2/K\u00b2\n",
-      "phi=4.517;                    #work function of emitter in eV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "b=round(11600*phi,-1);                       #constant for a metal, in K\n",
-      "Js=round(A*T*T*exp(-b/T),-2);                #Emission current density in amp/m\u00b2\n",
-      "a=pi*(d/100)*(l/100);                        #Surface area of the cathode in m\u00b2\n",
-      "E_I=Js*a;                                     #Emission current in A\n",
-      "\n",
-      "#Result\n",
-      "print(\"emission current =%.3f A\"%E_I);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "emission current =0.047 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2:Page number 31\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "Js=0.1;                   #Emission current density in amp/cm\u00b2\n",
-      "A=60.2;                   #Constant depending upon the type of thermionic emitter, in amp/cm\u00b2/K\u00b2\n",
-      "T=1900.0;                 #Absolute temperature in K\n",
-      "\n",
-      "\n",
-      "#calculations\n",
-      "#Calculating b according to the formula Js=A*T\u00b2*exp(-b/T) for emission current density\n",
-      "b=-T*(log(Js/(A*T*T)));          #constant for emitter, in K\n",
-      "phi= round(b/11600,2);           # work function in eV\n",
-      "\n",
-      "print (\"Work function of the tungsten wire = %.2f eV\"%phi);\n",
-      "\n",
-      "if(phi==4.52):\n",
-      "\tprint(\"Given sample is pure Tungsten\");\n",
-      "elif(phi!=4.52 and phi>=2.63 and phi<=4.52):\n",
-      "\tprint (\"The sample is not pure Tungsten\");\n",
-      "    \n",
-      "#Note : In the text book, the work function has been approximated to 3.56eV, but in the code it calculates as  3.52eV\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Work function of the tungsten wire = 3.52 eV\n",
-        "The sample is not pure Tungsten\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20.ipynb
deleted file mode 100755
index cad31534..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20.ipynb
+++ /dev/null
@@ -1,677 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:369e36634d005b832372dcae6796c76b979f32b499d8baadce951517f2201533"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 20 : SILICON CONTROLLED RECTIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.2 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I=50.0;                             #Surge current, A\n",
-      "t=12.0;                             #Time for which surge current lasts, ms\n",
-      "circuit_fusing_rating_max=90;       #Maximum circuit fusing rating, A\u00b2s\n",
-      "\n",
-      "#Calculation\n",
-      "circuit_fusing_rating=I**2*(t*10**-3);              #Circuit fusing rating, A\u00b2s\n",
-      "\n",
-      "#Result\n",
-      "if(circuit_fusing_rating<circuit_fusing_rating_max):\n",
-      "    print(\"The device will not be destroyed.\");\n",
-      "else:\n",
-      "    print(\"The device will be destroyed.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The device will not be destroyed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.3 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I2_t_rating=50.0;                   #circuit fuse rating, A\u00b2s\n",
-      "Is=100.0;                           #Surge current, A\n",
-      "\n",
-      "#Calculation\n",
-      "t_max=(I2_t_rating/Is**2)*1000;            #Maximum allowable duration, ms\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable duration =%dms\"%t_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable duration =5ms\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.4 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "R=220.0;                    #Gate resistor, \u03a9\n",
-      "I_G=7.0;                    #Gate current, mA\n",
-      "V_GK=0.7;                   #Junction voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_in=V_GK+(I_G/1000)*R;                    #Input voltage, V (Kirchhoff's voltage law)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required input voltage=%.2fV.\"%V_in);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required input voltage=2.24V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.5 : Page number 564\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "from math import cos\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_G=1.0;                #Gate current, mA\n",
-      "V_m=200.0;              #Peak value of input sinusoidal voltage, V\n",
-      "v=100.0;              #Forward breakdown voltage of SCR, V\n",
-      "R_L=100.0;              #Load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#v=Vm*sin(theta)\n",
-      "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n",
-      "theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad\n",
-      "theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees\n",
-      "\n",
-      "alpha=round(theta,0);                    #Firing angle, degrees\n",
-      "\n",
-      "#(ii)\n",
-      "phi=180-alpha;                  #Conduction angle, degrees\n",
-      "\n",
-      "#(iii)\n",
-      "V_avg=(V_m/(2*pi))*(1+cos(alpha*pi/180));           #Average voltage, V\n",
-      "I_avg=V_avg/R_L;                                    #Average current, A\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n",
-      "print(\"(ii) The conduction angle=%.0f\u00b0\"%phi);\n",
-      "print(\"(iii) The average current=%.4fA \"%I_avg);\n",
-      "\n",
-      "#Note: In the text book has approximated the average current to 0.5925A but in the code it gets approximated to 0.5940A.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The firing agle=30\u00b0\n",
-        "(ii) The conduction angle=150\u00b0\n",
-        "(iii) The average current=0.5940A \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.6 : Page number 564\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "from math import cos\n",
-      "from math import pi\n",
-      "from math import floor\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_G=1.0;                #Gate current, mA\n",
-      "V_m=400.0;              #Peak value of input sinusoidal voltage, V\n",
-      "v=150.0;                #Forward breakdown voltage of SCR, V\n",
-      "R_L=200.0;              #Load resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#v=Vm*sin(theta)\n",
-      "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n",
-      "theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad\n",
-      "theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees\n",
-      "\n",
-      "alpha=theta;                    #Firing angle, degrees\n",
-      "\n",
-      "#(ii)\n",
-      "V_av=floor((V_m/(2*pi))*(1+cos(alpha*pi/180))*10)/10;           #Average voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "I_av=V_av/R_L;                                    #Average current, A\n",
-      "\n",
-      "#(iv)\n",
-      "P_out=V_av*I_av;                                #Output power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n",
-      "print(\"(ii) The average output voltage=%.1f V\"%V_av);\n",
-      "print(\"(iii) The average current=%.3fA \"%I_av);\n",
-      "print(\"(iv) The output power=%.2f W\"%P_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The firing agle=22\u00b0\n",
-        "(ii) The average output voltage=122.6 V\n",
-        "(iii) The average current=0.613A \n",
-        "(iv) The output power=75.15 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.7 : Page number 564-565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "\n",
-      "#Variable declaration\n",
-      "v=180.0;                    #Forward breakdown voltage, V\n",
-      "V_m=240.0;                  #Peak value of input voltage, V\n",
-      "w=314.0;                    #Angular frequency of input ,rad/s\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#v=Vm*sin(w*t)\n",
-      "#So, t=asin(v/Vm)/w\n",
-      "t=(asin(v/V_m)/w)*1000;                        #Time for which SCR remains off, ms\n",
-      "\n",
-      "#Result\n",
-      "print(\"The SCR remains off for %.1f ms.\"%t);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The SCR remains off for 2.7 ms.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.8 : Page number 565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import cos\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_dc=1.0;                   #d.c load current, A\n",
-      "alpha=30.0;                 #Firing angle, \u00b0\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_av=I_dc;                                                 #Average current(= d.c current), A\n",
-      "\n",
-      "#Since, Iav=(Vm/(2*pi*RL))*(1+cos(alpha)) and Im=Vm/RL\n",
-      "I_m=floor((2*pi*I_av/(1+cos(alpha*pi/180)))*100)/100;      #Peak-load current, A\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Peak-load current=%.2f A.\"%I_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Peak-load current=3.36 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.9: Page number 565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "from math import sin\n",
-      "\n",
-      "#Variable declaration\n",
-      "alpha=60.0;                 #Firing angle, \u00b0\n",
-      "P=100.0;                    #Power rating of tungsten lamp, W\n",
-      "V=110.0;                    #Voltage rating of tungsten lamp, V\n",
-      "V_ac=110.0;                 #a.c supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_m=V_ac*sqrt(2);               #Peak value of input voltage, V\n",
-      "\n",
-      "alpha=alpha*pi/180;             #firing angle, rad\n",
-      "\n",
-      "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n",
-      "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "E_rms=round(V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)));            #r.m.s voltage developed in the lamp, V\n",
-      "\n",
-      "\n",
-      "RL=V**2/P;                      #Load resistance, \u03a9\n",
-      "\n",
-      "I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The r.m.s voltage developed in the lamp=%d V.\"%E_rms);\n",
-      "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The r.m.s voltage developed in the lamp=70 V.\n",
-        "The r.m.s current developed in the lamp=0.58 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.10 : Page number 567\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import cos\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=100.0;                   #Load resistance, \u03a9\n",
-      "V_m=200.0;                  #Peak a.c voltage, V\n",
-      "alpha=60;                   #firing angle, \u00b0\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_av=(V_m/pi)*(1+cos(alpha*pi/180));                #D.C output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "I_av=V_av/RL;                   #Load current, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  d.c output voltage=%.1f V.\"%V_av);\n",
-      "print(\"(ii) Load current=%.3f A\"%I_av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  d.c output voltage=95.5 V.\n",
-        "(ii) Load current=0.955 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.11: Page number 567\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "from math import sin\n",
-      "\n",
-      "#Variable declaration\n",
-      "alpha=60.0;                 #Firing angle, \u00b0\n",
-      "P=100.0;                    #Power rating of tungsten lamp, W\n",
-      "V=110.0;                    #Voltage rating of tungsten lamp, V\n",
-      "V_ac=110.0;                 #a.c supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_m=round(V_ac*sqrt(2));               #Peak value of input voltage, V\n",
-      "\n",
-      "alpha=alpha*pi/180;             #firing angle, rad\n",
-      "\n",
-      "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n",
-      "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "E_rms=V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(4*pi));            #r.m.s voltage developed in the lamp, V\n",
-      "\n",
-      "RL=V**2/P;                      #Load resistance, \u03a9\n",
-      "\n",
-      "I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The r.m.s voltage developed in the lamp=%.1f V.\"%E_rms);\n",
-      "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The r.m.s voltage developed in the lamp=98.9 V.\n",
-        "The r.m.s current developed in the lamp=0.82 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.12 : Page number 572\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15;                     #Suuply voltage, V\n",
-      "V_T=0.7;                    #Gate trigger voltage, V\n",
-      "I_T=7.0;                    #Gate trigger current, mA\n",
-      "I_H=6.0;                    #Holding current. mA\n",
-      "R_Vin=1;                    #Resistance at Vin, k\u03a9\n",
-      "R_VCC=100;                  #Resistance at Vcc, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) when SCR is off, there is no current, therefore no voltage drop across the resistor\n",
-      "V_out=VCC;                      #Output voltage, when SCR is off, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_in=V_T+I_T*R_Vin;             #Input voltage required to trigger the SCR, V\n",
-      "\n",
-      "#(iii)\n",
-      "#Since, I_H=(Vcc-VT)/R_Vin;\n",
-      "VCC_SCR_open=(I_H/1000)*R_VCC+V_T;                 #Decreased value of supply voltage at which SCR opens, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output voltage when SCR is off=%dV.\"%V_out);\n",
-      "print(\"(ii) The input voltage required to trigger the SCR=%.1f V.\"%V_in);\n",
-      "print(\"(iii) The decreased supply voltage at which SCR opens=%.1f V.\"%VCC_SCR_open);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage when SCR is off=15V.\n",
-        "(ii) The input voltage required to trigger the SCR=7.7 V.\n",
-        "(iii) The decreased supply voltage at which SCR opens=1.3 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.13 : Page number 572-573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=5.6;                 #zener voltage, V\n",
-      "V_T=0.7;                #Trigger voltage of SCR, V\n",
-      "\n",
-      "#Calculation\n",
-      "VCC=Vz+V_T;                 #Required supply voltage to turn on the crowbar, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required supply voltage to turn on the crowbar=%.1fV.\"%VCC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required supply voltage to turn on the crowbar=6.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.14 : Page number 573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=12;                  #Zener breakdown voltage, V\n",
-      "V_T=1.5;                #Trigger voltage, V\n",
-      "tolerance_z=10.0;         #Tolerance of zener diode, %\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vz_max=Vz*(1+tolerance_z/100);          #Maximum value of zener breakdown, V\n",
-      "Vz_min=Vz*(1-tolerance_z/100);          #Minimum value of zener breakdown, V\n",
-      "V_crowbar=Vz_max+V_T;                   #Maximum value of supply voltage for crowbarring, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum value of supply voltage for crowbarring=%.1fV\"%V_crowbar);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of supply voltage for crowbarring=14.7V\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.15 : Page number 573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=25.0;               #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#When brights light is on, LASCR conducts and thus gets short circuited to ground, hence,\n",
-      "V_out=0;                #Output voltage, V\n",
-      "\n",
-      "print(\"Output voltage when bright light is on=%dV\"%V_out);\n",
-      "\n",
-      "\n",
-      "#When brights light is off, LASCR stops conducting and thus no current through resistor, hence,\n",
-      "V_out=VCC;                #Output voltage, V\n",
-      "print(\"Output voltage when bright light is off=%dV\"%V_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage when bright light is on=0V\n",
-        "Output voltage when bright light is off=25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_1.ipynb
deleted file mode 100755
index cad31534..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_1.ipynb
+++ /dev/null
@@ -1,677 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:369e36634d005b832372dcae6796c76b979f32b499d8baadce951517f2201533"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 20 : SILICON CONTROLLED RECTIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.2 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I=50.0;                             #Surge current, A\n",
-      "t=12.0;                             #Time for which surge current lasts, ms\n",
-      "circuit_fusing_rating_max=90;       #Maximum circuit fusing rating, A\u00b2s\n",
-      "\n",
-      "#Calculation\n",
-      "circuit_fusing_rating=I**2*(t*10**-3);              #Circuit fusing rating, A\u00b2s\n",
-      "\n",
-      "#Result\n",
-      "if(circuit_fusing_rating<circuit_fusing_rating_max):\n",
-      "    print(\"The device will not be destroyed.\");\n",
-      "else:\n",
-      "    print(\"The device will be destroyed.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The device will not be destroyed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.3 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I2_t_rating=50.0;                   #circuit fuse rating, A\u00b2s\n",
-      "Is=100.0;                           #Surge current, A\n",
-      "\n",
-      "#Calculation\n",
-      "t_max=(I2_t_rating/Is**2)*1000;            #Maximum allowable duration, ms\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable duration =%dms\"%t_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable duration =5ms\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.4 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "R=220.0;                    #Gate resistor, \u03a9\n",
-      "I_G=7.0;                    #Gate current, mA\n",
-      "V_GK=0.7;                   #Junction voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_in=V_GK+(I_G/1000)*R;                    #Input voltage, V (Kirchhoff's voltage law)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required input voltage=%.2fV.\"%V_in);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required input voltage=2.24V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.5 : Page number 564\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "from math import cos\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_G=1.0;                #Gate current, mA\n",
-      "V_m=200.0;              #Peak value of input sinusoidal voltage, V\n",
-      "v=100.0;              #Forward breakdown voltage of SCR, V\n",
-      "R_L=100.0;              #Load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#v=Vm*sin(theta)\n",
-      "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n",
-      "theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad\n",
-      "theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees\n",
-      "\n",
-      "alpha=round(theta,0);                    #Firing angle, degrees\n",
-      "\n",
-      "#(ii)\n",
-      "phi=180-alpha;                  #Conduction angle, degrees\n",
-      "\n",
-      "#(iii)\n",
-      "V_avg=(V_m/(2*pi))*(1+cos(alpha*pi/180));           #Average voltage, V\n",
-      "I_avg=V_avg/R_L;                                    #Average current, A\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n",
-      "print(\"(ii) The conduction angle=%.0f\u00b0\"%phi);\n",
-      "print(\"(iii) The average current=%.4fA \"%I_avg);\n",
-      "\n",
-      "#Note: In the text book has approximated the average current to 0.5925A but in the code it gets approximated to 0.5940A.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The firing agle=30\u00b0\n",
-        "(ii) The conduction angle=150\u00b0\n",
-        "(iii) The average current=0.5940A \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.6 : Page number 564\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "from math import cos\n",
-      "from math import pi\n",
-      "from math import floor\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_G=1.0;                #Gate current, mA\n",
-      "V_m=400.0;              #Peak value of input sinusoidal voltage, V\n",
-      "v=150.0;                #Forward breakdown voltage of SCR, V\n",
-      "R_L=200.0;              #Load resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#v=Vm*sin(theta)\n",
-      "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n",
-      "theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad\n",
-      "theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees\n",
-      "\n",
-      "alpha=theta;                    #Firing angle, degrees\n",
-      "\n",
-      "#(ii)\n",
-      "V_av=floor((V_m/(2*pi))*(1+cos(alpha*pi/180))*10)/10;           #Average voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "I_av=V_av/R_L;                                    #Average current, A\n",
-      "\n",
-      "#(iv)\n",
-      "P_out=V_av*I_av;                                #Output power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n",
-      "print(\"(ii) The average output voltage=%.1f V\"%V_av);\n",
-      "print(\"(iii) The average current=%.3fA \"%I_av);\n",
-      "print(\"(iv) The output power=%.2f W\"%P_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The firing agle=22\u00b0\n",
-        "(ii) The average output voltage=122.6 V\n",
-        "(iii) The average current=0.613A \n",
-        "(iv) The output power=75.15 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.7 : Page number 564-565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "\n",
-      "#Variable declaration\n",
-      "v=180.0;                    #Forward breakdown voltage, V\n",
-      "V_m=240.0;                  #Peak value of input voltage, V\n",
-      "w=314.0;                    #Angular frequency of input ,rad/s\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#v=Vm*sin(w*t)\n",
-      "#So, t=asin(v/Vm)/w\n",
-      "t=(asin(v/V_m)/w)*1000;                        #Time for which SCR remains off, ms\n",
-      "\n",
-      "#Result\n",
-      "print(\"The SCR remains off for %.1f ms.\"%t);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The SCR remains off for 2.7 ms.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.8 : Page number 565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import cos\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_dc=1.0;                   #d.c load current, A\n",
-      "alpha=30.0;                 #Firing angle, \u00b0\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_av=I_dc;                                                 #Average current(= d.c current), A\n",
-      "\n",
-      "#Since, Iav=(Vm/(2*pi*RL))*(1+cos(alpha)) and Im=Vm/RL\n",
-      "I_m=floor((2*pi*I_av/(1+cos(alpha*pi/180)))*100)/100;      #Peak-load current, A\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Peak-load current=%.2f A.\"%I_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Peak-load current=3.36 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.9: Page number 565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "from math import sin\n",
-      "\n",
-      "#Variable declaration\n",
-      "alpha=60.0;                 #Firing angle, \u00b0\n",
-      "P=100.0;                    #Power rating of tungsten lamp, W\n",
-      "V=110.0;                    #Voltage rating of tungsten lamp, V\n",
-      "V_ac=110.0;                 #a.c supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_m=V_ac*sqrt(2);               #Peak value of input voltage, V\n",
-      "\n",
-      "alpha=alpha*pi/180;             #firing angle, rad\n",
-      "\n",
-      "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n",
-      "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "E_rms=round(V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)));            #r.m.s voltage developed in the lamp, V\n",
-      "\n",
-      "\n",
-      "RL=V**2/P;                      #Load resistance, \u03a9\n",
-      "\n",
-      "I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The r.m.s voltage developed in the lamp=%d V.\"%E_rms);\n",
-      "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The r.m.s voltage developed in the lamp=70 V.\n",
-        "The r.m.s current developed in the lamp=0.58 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.10 : Page number 567\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import cos\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=100.0;                   #Load resistance, \u03a9\n",
-      "V_m=200.0;                  #Peak a.c voltage, V\n",
-      "alpha=60;                   #firing angle, \u00b0\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_av=(V_m/pi)*(1+cos(alpha*pi/180));                #D.C output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "I_av=V_av/RL;                   #Load current, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  d.c output voltage=%.1f V.\"%V_av);\n",
-      "print(\"(ii) Load current=%.3f A\"%I_av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  d.c output voltage=95.5 V.\n",
-        "(ii) Load current=0.955 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.11: Page number 567\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "from math import sin\n",
-      "\n",
-      "#Variable declaration\n",
-      "alpha=60.0;                 #Firing angle, \u00b0\n",
-      "P=100.0;                    #Power rating of tungsten lamp, W\n",
-      "V=110.0;                    #Voltage rating of tungsten lamp, V\n",
-      "V_ac=110.0;                 #a.c supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_m=round(V_ac*sqrt(2));               #Peak value of input voltage, V\n",
-      "\n",
-      "alpha=alpha*pi/180;             #firing angle, rad\n",
-      "\n",
-      "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n",
-      "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "E_rms=V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(4*pi));            #r.m.s voltage developed in the lamp, V\n",
-      "\n",
-      "RL=V**2/P;                      #Load resistance, \u03a9\n",
-      "\n",
-      "I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The r.m.s voltage developed in the lamp=%.1f V.\"%E_rms);\n",
-      "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The r.m.s voltage developed in the lamp=98.9 V.\n",
-        "The r.m.s current developed in the lamp=0.82 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.12 : Page number 572\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15;                     #Suuply voltage, V\n",
-      "V_T=0.7;                    #Gate trigger voltage, V\n",
-      "I_T=7.0;                    #Gate trigger current, mA\n",
-      "I_H=6.0;                    #Holding current. mA\n",
-      "R_Vin=1;                    #Resistance at Vin, k\u03a9\n",
-      "R_VCC=100;                  #Resistance at Vcc, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) when SCR is off, there is no current, therefore no voltage drop across the resistor\n",
-      "V_out=VCC;                      #Output voltage, when SCR is off, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_in=V_T+I_T*R_Vin;             #Input voltage required to trigger the SCR, V\n",
-      "\n",
-      "#(iii)\n",
-      "#Since, I_H=(Vcc-VT)/R_Vin;\n",
-      "VCC_SCR_open=(I_H/1000)*R_VCC+V_T;                 #Decreased value of supply voltage at which SCR opens, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output voltage when SCR is off=%dV.\"%V_out);\n",
-      "print(\"(ii) The input voltage required to trigger the SCR=%.1f V.\"%V_in);\n",
-      "print(\"(iii) The decreased supply voltage at which SCR opens=%.1f V.\"%VCC_SCR_open);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage when SCR is off=15V.\n",
-        "(ii) The input voltage required to trigger the SCR=7.7 V.\n",
-        "(iii) The decreased supply voltage at which SCR opens=1.3 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.13 : Page number 572-573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=5.6;                 #zener voltage, V\n",
-      "V_T=0.7;                #Trigger voltage of SCR, V\n",
-      "\n",
-      "#Calculation\n",
-      "VCC=Vz+V_T;                 #Required supply voltage to turn on the crowbar, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required supply voltage to turn on the crowbar=%.1fV.\"%VCC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required supply voltage to turn on the crowbar=6.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.14 : Page number 573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=12;                  #Zener breakdown voltage, V\n",
-      "V_T=1.5;                #Trigger voltage, V\n",
-      "tolerance_z=10.0;         #Tolerance of zener diode, %\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vz_max=Vz*(1+tolerance_z/100);          #Maximum value of zener breakdown, V\n",
-      "Vz_min=Vz*(1-tolerance_z/100);          #Minimum value of zener breakdown, V\n",
-      "V_crowbar=Vz_max+V_T;                   #Maximum value of supply voltage for crowbarring, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum value of supply voltage for crowbarring=%.1fV\"%V_crowbar);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of supply voltage for crowbarring=14.7V\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.15 : Page number 573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=25.0;               #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#When brights light is on, LASCR conducts and thus gets short circuited to ground, hence,\n",
-      "V_out=0;                #Output voltage, V\n",
-      "\n",
-      "print(\"Output voltage when bright light is on=%dV\"%V_out);\n",
-      "\n",
-      "\n",
-      "#When brights light is off, LASCR stops conducting and thus no current through resistor, hence,\n",
-      "V_out=VCC;                #Output voltage, V\n",
-      "print(\"Output voltage when bright light is off=%dV\"%V_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage when bright light is on=0V\n",
-        "Output voltage when bright light is off=25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_2.ipynb
deleted file mode 100755
index cad31534..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_2.ipynb
+++ /dev/null
@@ -1,677 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:369e36634d005b832372dcae6796c76b979f32b499d8baadce951517f2201533"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 20 : SILICON CONTROLLED RECTIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.2 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I=50.0;                             #Surge current, A\n",
-      "t=12.0;                             #Time for which surge current lasts, ms\n",
-      "circuit_fusing_rating_max=90;       #Maximum circuit fusing rating, A\u00b2s\n",
-      "\n",
-      "#Calculation\n",
-      "circuit_fusing_rating=I**2*(t*10**-3);              #Circuit fusing rating, A\u00b2s\n",
-      "\n",
-      "#Result\n",
-      "if(circuit_fusing_rating<circuit_fusing_rating_max):\n",
-      "    print(\"The device will not be destroyed.\");\n",
-      "else:\n",
-      "    print(\"The device will be destroyed.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The device will not be destroyed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.3 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I2_t_rating=50.0;                   #circuit fuse rating, A\u00b2s\n",
-      "Is=100.0;                           #Surge current, A\n",
-      "\n",
-      "#Calculation\n",
-      "t_max=(I2_t_rating/Is**2)*1000;            #Maximum allowable duration, ms\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable duration =%dms\"%t_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable duration =5ms\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.4 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "R=220.0;                    #Gate resistor, \u03a9\n",
-      "I_G=7.0;                    #Gate current, mA\n",
-      "V_GK=0.7;                   #Junction voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_in=V_GK+(I_G/1000)*R;                    #Input voltage, V (Kirchhoff's voltage law)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required input voltage=%.2fV.\"%V_in);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required input voltage=2.24V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.5 : Page number 564\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "from math import cos\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_G=1.0;                #Gate current, mA\n",
-      "V_m=200.0;              #Peak value of input sinusoidal voltage, V\n",
-      "v=100.0;              #Forward breakdown voltage of SCR, V\n",
-      "R_L=100.0;              #Load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#v=Vm*sin(theta)\n",
-      "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n",
-      "theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad\n",
-      "theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees\n",
-      "\n",
-      "alpha=round(theta,0);                    #Firing angle, degrees\n",
-      "\n",
-      "#(ii)\n",
-      "phi=180-alpha;                  #Conduction angle, degrees\n",
-      "\n",
-      "#(iii)\n",
-      "V_avg=(V_m/(2*pi))*(1+cos(alpha*pi/180));           #Average voltage, V\n",
-      "I_avg=V_avg/R_L;                                    #Average current, A\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n",
-      "print(\"(ii) The conduction angle=%.0f\u00b0\"%phi);\n",
-      "print(\"(iii) The average current=%.4fA \"%I_avg);\n",
-      "\n",
-      "#Note: In the text book has approximated the average current to 0.5925A but in the code it gets approximated to 0.5940A.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The firing agle=30\u00b0\n",
-        "(ii) The conduction angle=150\u00b0\n",
-        "(iii) The average current=0.5940A \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.6 : Page number 564\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "from math import cos\n",
-      "from math import pi\n",
-      "from math import floor\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_G=1.0;                #Gate current, mA\n",
-      "V_m=400.0;              #Peak value of input sinusoidal voltage, V\n",
-      "v=150.0;                #Forward breakdown voltage of SCR, V\n",
-      "R_L=200.0;              #Load resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#v=Vm*sin(theta)\n",
-      "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n",
-      "theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad\n",
-      "theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees\n",
-      "\n",
-      "alpha=theta;                    #Firing angle, degrees\n",
-      "\n",
-      "#(ii)\n",
-      "V_av=floor((V_m/(2*pi))*(1+cos(alpha*pi/180))*10)/10;           #Average voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "I_av=V_av/R_L;                                    #Average current, A\n",
-      "\n",
-      "#(iv)\n",
-      "P_out=V_av*I_av;                                #Output power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n",
-      "print(\"(ii) The average output voltage=%.1f V\"%V_av);\n",
-      "print(\"(iii) The average current=%.3fA \"%I_av);\n",
-      "print(\"(iv) The output power=%.2f W\"%P_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The firing agle=22\u00b0\n",
-        "(ii) The average output voltage=122.6 V\n",
-        "(iii) The average current=0.613A \n",
-        "(iv) The output power=75.15 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.7 : Page number 564-565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "\n",
-      "#Variable declaration\n",
-      "v=180.0;                    #Forward breakdown voltage, V\n",
-      "V_m=240.0;                  #Peak value of input voltage, V\n",
-      "w=314.0;                    #Angular frequency of input ,rad/s\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#v=Vm*sin(w*t)\n",
-      "#So, t=asin(v/Vm)/w\n",
-      "t=(asin(v/V_m)/w)*1000;                        #Time for which SCR remains off, ms\n",
-      "\n",
-      "#Result\n",
-      "print(\"The SCR remains off for %.1f ms.\"%t);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The SCR remains off for 2.7 ms.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.8 : Page number 565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import cos\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_dc=1.0;                   #d.c load current, A\n",
-      "alpha=30.0;                 #Firing angle, \u00b0\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_av=I_dc;                                                 #Average current(= d.c current), A\n",
-      "\n",
-      "#Since, Iav=(Vm/(2*pi*RL))*(1+cos(alpha)) and Im=Vm/RL\n",
-      "I_m=floor((2*pi*I_av/(1+cos(alpha*pi/180)))*100)/100;      #Peak-load current, A\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Peak-load current=%.2f A.\"%I_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Peak-load current=3.36 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.9: Page number 565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "from math import sin\n",
-      "\n",
-      "#Variable declaration\n",
-      "alpha=60.0;                 #Firing angle, \u00b0\n",
-      "P=100.0;                    #Power rating of tungsten lamp, W\n",
-      "V=110.0;                    #Voltage rating of tungsten lamp, V\n",
-      "V_ac=110.0;                 #a.c supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_m=V_ac*sqrt(2);               #Peak value of input voltage, V\n",
-      "\n",
-      "alpha=alpha*pi/180;             #firing angle, rad\n",
-      "\n",
-      "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n",
-      "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "E_rms=round(V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)));            #r.m.s voltage developed in the lamp, V\n",
-      "\n",
-      "\n",
-      "RL=V**2/P;                      #Load resistance, \u03a9\n",
-      "\n",
-      "I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The r.m.s voltage developed in the lamp=%d V.\"%E_rms);\n",
-      "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The r.m.s voltage developed in the lamp=70 V.\n",
-        "The r.m.s current developed in the lamp=0.58 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.10 : Page number 567\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import cos\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=100.0;                   #Load resistance, \u03a9\n",
-      "V_m=200.0;                  #Peak a.c voltage, V\n",
-      "alpha=60;                   #firing angle, \u00b0\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_av=(V_m/pi)*(1+cos(alpha*pi/180));                #D.C output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "I_av=V_av/RL;                   #Load current, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  d.c output voltage=%.1f V.\"%V_av);\n",
-      "print(\"(ii) Load current=%.3f A\"%I_av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  d.c output voltage=95.5 V.\n",
-        "(ii) Load current=0.955 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.11: Page number 567\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "from math import sin\n",
-      "\n",
-      "#Variable declaration\n",
-      "alpha=60.0;                 #Firing angle, \u00b0\n",
-      "P=100.0;                    #Power rating of tungsten lamp, W\n",
-      "V=110.0;                    #Voltage rating of tungsten lamp, V\n",
-      "V_ac=110.0;                 #a.c supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_m=round(V_ac*sqrt(2));               #Peak value of input voltage, V\n",
-      "\n",
-      "alpha=alpha*pi/180;             #firing angle, rad\n",
-      "\n",
-      "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n",
-      "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "E_rms=V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(4*pi));            #r.m.s voltage developed in the lamp, V\n",
-      "\n",
-      "RL=V**2/P;                      #Load resistance, \u03a9\n",
-      "\n",
-      "I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The r.m.s voltage developed in the lamp=%.1f V.\"%E_rms);\n",
-      "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The r.m.s voltage developed in the lamp=98.9 V.\n",
-        "The r.m.s current developed in the lamp=0.82 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.12 : Page number 572\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15;                     #Suuply voltage, V\n",
-      "V_T=0.7;                    #Gate trigger voltage, V\n",
-      "I_T=7.0;                    #Gate trigger current, mA\n",
-      "I_H=6.0;                    #Holding current. mA\n",
-      "R_Vin=1;                    #Resistance at Vin, k\u03a9\n",
-      "R_VCC=100;                  #Resistance at Vcc, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) when SCR is off, there is no current, therefore no voltage drop across the resistor\n",
-      "V_out=VCC;                      #Output voltage, when SCR is off, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_in=V_T+I_T*R_Vin;             #Input voltage required to trigger the SCR, V\n",
-      "\n",
-      "#(iii)\n",
-      "#Since, I_H=(Vcc-VT)/R_Vin;\n",
-      "VCC_SCR_open=(I_H/1000)*R_VCC+V_T;                 #Decreased value of supply voltage at which SCR opens, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output voltage when SCR is off=%dV.\"%V_out);\n",
-      "print(\"(ii) The input voltage required to trigger the SCR=%.1f V.\"%V_in);\n",
-      "print(\"(iii) The decreased supply voltage at which SCR opens=%.1f V.\"%VCC_SCR_open);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage when SCR is off=15V.\n",
-        "(ii) The input voltage required to trigger the SCR=7.7 V.\n",
-        "(iii) The decreased supply voltage at which SCR opens=1.3 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.13 : Page number 572-573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=5.6;                 #zener voltage, V\n",
-      "V_T=0.7;                #Trigger voltage of SCR, V\n",
-      "\n",
-      "#Calculation\n",
-      "VCC=Vz+V_T;                 #Required supply voltage to turn on the crowbar, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required supply voltage to turn on the crowbar=%.1fV.\"%VCC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required supply voltage to turn on the crowbar=6.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.14 : Page number 573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=12;                  #Zener breakdown voltage, V\n",
-      "V_T=1.5;                #Trigger voltage, V\n",
-      "tolerance_z=10.0;         #Tolerance of zener diode, %\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vz_max=Vz*(1+tolerance_z/100);          #Maximum value of zener breakdown, V\n",
-      "Vz_min=Vz*(1-tolerance_z/100);          #Minimum value of zener breakdown, V\n",
-      "V_crowbar=Vz_max+V_T;                   #Maximum value of supply voltage for crowbarring, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum value of supply voltage for crowbarring=%.1fV\"%V_crowbar);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of supply voltage for crowbarring=14.7V\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.15 : Page number 573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=25.0;               #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#When brights light is on, LASCR conducts and thus gets short circuited to ground, hence,\n",
-      "V_out=0;                #Output voltage, V\n",
-      "\n",
-      "print(\"Output voltage when bright light is on=%dV\"%V_out);\n",
-      "\n",
-      "\n",
-      "#When brights light is off, LASCR stops conducting and thus no current through resistor, hence,\n",
-      "V_out=VCC;                #Output voltage, V\n",
-      "print(\"Output voltage when bright light is off=%dV\"%V_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage when bright light is on=0V\n",
-        "Output voltage when bright light is off=25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_3.ipynb
deleted file mode 100755
index cad31534..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_3.ipynb
+++ /dev/null
@@ -1,677 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:369e36634d005b832372dcae6796c76b979f32b499d8baadce951517f2201533"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 20 : SILICON CONTROLLED RECTIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.2 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I=50.0;                             #Surge current, A\n",
-      "t=12.0;                             #Time for which surge current lasts, ms\n",
-      "circuit_fusing_rating_max=90;       #Maximum circuit fusing rating, A\u00b2s\n",
-      "\n",
-      "#Calculation\n",
-      "circuit_fusing_rating=I**2*(t*10**-3);              #Circuit fusing rating, A\u00b2s\n",
-      "\n",
-      "#Result\n",
-      "if(circuit_fusing_rating<circuit_fusing_rating_max):\n",
-      "    print(\"The device will not be destroyed.\");\n",
-      "else:\n",
-      "    print(\"The device will be destroyed.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The device will not be destroyed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.3 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I2_t_rating=50.0;                   #circuit fuse rating, A\u00b2s\n",
-      "Is=100.0;                           #Surge current, A\n",
-      "\n",
-      "#Calculation\n",
-      "t_max=(I2_t_rating/Is**2)*1000;            #Maximum allowable duration, ms\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable duration =%dms\"%t_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable duration =5ms\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.4 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "R=220.0;                    #Gate resistor, \u03a9\n",
-      "I_G=7.0;                    #Gate current, mA\n",
-      "V_GK=0.7;                   #Junction voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_in=V_GK+(I_G/1000)*R;                    #Input voltage, V (Kirchhoff's voltage law)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required input voltage=%.2fV.\"%V_in);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required input voltage=2.24V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.5 : Page number 564\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "from math import cos\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_G=1.0;                #Gate current, mA\n",
-      "V_m=200.0;              #Peak value of input sinusoidal voltage, V\n",
-      "v=100.0;              #Forward breakdown voltage of SCR, V\n",
-      "R_L=100.0;              #Load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#v=Vm*sin(theta)\n",
-      "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n",
-      "theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad\n",
-      "theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees\n",
-      "\n",
-      "alpha=round(theta,0);                    #Firing angle, degrees\n",
-      "\n",
-      "#(ii)\n",
-      "phi=180-alpha;                  #Conduction angle, degrees\n",
-      "\n",
-      "#(iii)\n",
-      "V_avg=(V_m/(2*pi))*(1+cos(alpha*pi/180));           #Average voltage, V\n",
-      "I_avg=V_avg/R_L;                                    #Average current, A\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n",
-      "print(\"(ii) The conduction angle=%.0f\u00b0\"%phi);\n",
-      "print(\"(iii) The average current=%.4fA \"%I_avg);\n",
-      "\n",
-      "#Note: In the text book has approximated the average current to 0.5925A but in the code it gets approximated to 0.5940A.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The firing agle=30\u00b0\n",
-        "(ii) The conduction angle=150\u00b0\n",
-        "(iii) The average current=0.5940A \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.6 : Page number 564\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "from math import cos\n",
-      "from math import pi\n",
-      "from math import floor\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_G=1.0;                #Gate current, mA\n",
-      "V_m=400.0;              #Peak value of input sinusoidal voltage, V\n",
-      "v=150.0;                #Forward breakdown voltage of SCR, V\n",
-      "R_L=200.0;              #Load resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#v=Vm*sin(theta)\n",
-      "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n",
-      "theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad\n",
-      "theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees\n",
-      "\n",
-      "alpha=theta;                    #Firing angle, degrees\n",
-      "\n",
-      "#(ii)\n",
-      "V_av=floor((V_m/(2*pi))*(1+cos(alpha*pi/180))*10)/10;           #Average voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "I_av=V_av/R_L;                                    #Average current, A\n",
-      "\n",
-      "#(iv)\n",
-      "P_out=V_av*I_av;                                #Output power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n",
-      "print(\"(ii) The average output voltage=%.1f V\"%V_av);\n",
-      "print(\"(iii) The average current=%.3fA \"%I_av);\n",
-      "print(\"(iv) The output power=%.2f W\"%P_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The firing agle=22\u00b0\n",
-        "(ii) The average output voltage=122.6 V\n",
-        "(iii) The average current=0.613A \n",
-        "(iv) The output power=75.15 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.7 : Page number 564-565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "\n",
-      "#Variable declaration\n",
-      "v=180.0;                    #Forward breakdown voltage, V\n",
-      "V_m=240.0;                  #Peak value of input voltage, V\n",
-      "w=314.0;                    #Angular frequency of input ,rad/s\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#v=Vm*sin(w*t)\n",
-      "#So, t=asin(v/Vm)/w\n",
-      "t=(asin(v/V_m)/w)*1000;                        #Time for which SCR remains off, ms\n",
-      "\n",
-      "#Result\n",
-      "print(\"The SCR remains off for %.1f ms.\"%t);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The SCR remains off for 2.7 ms.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.8 : Page number 565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import cos\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_dc=1.0;                   #d.c load current, A\n",
-      "alpha=30.0;                 #Firing angle, \u00b0\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_av=I_dc;                                                 #Average current(= d.c current), A\n",
-      "\n",
-      "#Since, Iav=(Vm/(2*pi*RL))*(1+cos(alpha)) and Im=Vm/RL\n",
-      "I_m=floor((2*pi*I_av/(1+cos(alpha*pi/180)))*100)/100;      #Peak-load current, A\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Peak-load current=%.2f A.\"%I_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Peak-load current=3.36 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.9: Page number 565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "from math import sin\n",
-      "\n",
-      "#Variable declaration\n",
-      "alpha=60.0;                 #Firing angle, \u00b0\n",
-      "P=100.0;                    #Power rating of tungsten lamp, W\n",
-      "V=110.0;                    #Voltage rating of tungsten lamp, V\n",
-      "V_ac=110.0;                 #a.c supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_m=V_ac*sqrt(2);               #Peak value of input voltage, V\n",
-      "\n",
-      "alpha=alpha*pi/180;             #firing angle, rad\n",
-      "\n",
-      "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n",
-      "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "E_rms=round(V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)));            #r.m.s voltage developed in the lamp, V\n",
-      "\n",
-      "\n",
-      "RL=V**2/P;                      #Load resistance, \u03a9\n",
-      "\n",
-      "I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The r.m.s voltage developed in the lamp=%d V.\"%E_rms);\n",
-      "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The r.m.s voltage developed in the lamp=70 V.\n",
-        "The r.m.s current developed in the lamp=0.58 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.10 : Page number 567\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import cos\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=100.0;                   #Load resistance, \u03a9\n",
-      "V_m=200.0;                  #Peak a.c voltage, V\n",
-      "alpha=60;                   #firing angle, \u00b0\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_av=(V_m/pi)*(1+cos(alpha*pi/180));                #D.C output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "I_av=V_av/RL;                   #Load current, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  d.c output voltage=%.1f V.\"%V_av);\n",
-      "print(\"(ii) Load current=%.3f A\"%I_av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  d.c output voltage=95.5 V.\n",
-        "(ii) Load current=0.955 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.11: Page number 567\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "from math import sin\n",
-      "\n",
-      "#Variable declaration\n",
-      "alpha=60.0;                 #Firing angle, \u00b0\n",
-      "P=100.0;                    #Power rating of tungsten lamp, W\n",
-      "V=110.0;                    #Voltage rating of tungsten lamp, V\n",
-      "V_ac=110.0;                 #a.c supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_m=round(V_ac*sqrt(2));               #Peak value of input voltage, V\n",
-      "\n",
-      "alpha=alpha*pi/180;             #firing angle, rad\n",
-      "\n",
-      "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n",
-      "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "E_rms=V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(4*pi));            #r.m.s voltage developed in the lamp, V\n",
-      "\n",
-      "RL=V**2/P;                      #Load resistance, \u03a9\n",
-      "\n",
-      "I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The r.m.s voltage developed in the lamp=%.1f V.\"%E_rms);\n",
-      "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The r.m.s voltage developed in the lamp=98.9 V.\n",
-        "The r.m.s current developed in the lamp=0.82 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.12 : Page number 572\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15;                     #Suuply voltage, V\n",
-      "V_T=0.7;                    #Gate trigger voltage, V\n",
-      "I_T=7.0;                    #Gate trigger current, mA\n",
-      "I_H=6.0;                    #Holding current. mA\n",
-      "R_Vin=1;                    #Resistance at Vin, k\u03a9\n",
-      "R_VCC=100;                  #Resistance at Vcc, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) when SCR is off, there is no current, therefore no voltage drop across the resistor\n",
-      "V_out=VCC;                      #Output voltage, when SCR is off, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_in=V_T+I_T*R_Vin;             #Input voltage required to trigger the SCR, V\n",
-      "\n",
-      "#(iii)\n",
-      "#Since, I_H=(Vcc-VT)/R_Vin;\n",
-      "VCC_SCR_open=(I_H/1000)*R_VCC+V_T;                 #Decreased value of supply voltage at which SCR opens, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output voltage when SCR is off=%dV.\"%V_out);\n",
-      "print(\"(ii) The input voltage required to trigger the SCR=%.1f V.\"%V_in);\n",
-      "print(\"(iii) The decreased supply voltage at which SCR opens=%.1f V.\"%VCC_SCR_open);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage when SCR is off=15V.\n",
-        "(ii) The input voltage required to trigger the SCR=7.7 V.\n",
-        "(iii) The decreased supply voltage at which SCR opens=1.3 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.13 : Page number 572-573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=5.6;                 #zener voltage, V\n",
-      "V_T=0.7;                #Trigger voltage of SCR, V\n",
-      "\n",
-      "#Calculation\n",
-      "VCC=Vz+V_T;                 #Required supply voltage to turn on the crowbar, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required supply voltage to turn on the crowbar=%.1fV.\"%VCC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required supply voltage to turn on the crowbar=6.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.14 : Page number 573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=12;                  #Zener breakdown voltage, V\n",
-      "V_T=1.5;                #Trigger voltage, V\n",
-      "tolerance_z=10.0;         #Tolerance of zener diode, %\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vz_max=Vz*(1+tolerance_z/100);          #Maximum value of zener breakdown, V\n",
-      "Vz_min=Vz*(1-tolerance_z/100);          #Minimum value of zener breakdown, V\n",
-      "V_crowbar=Vz_max+V_T;                   #Maximum value of supply voltage for crowbarring, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum value of supply voltage for crowbarring=%.1fV\"%V_crowbar);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of supply voltage for crowbarring=14.7V\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.15 : Page number 573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=25.0;               #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#When brights light is on, LASCR conducts and thus gets short circuited to ground, hence,\n",
-      "V_out=0;                #Output voltage, V\n",
-      "\n",
-      "print(\"Output voltage when bright light is on=%dV\"%V_out);\n",
-      "\n",
-      "\n",
-      "#When brights light is off, LASCR stops conducting and thus no current through resistor, hence,\n",
-      "V_out=VCC;                #Output voltage, V\n",
-      "print(\"Output voltage when bright light is off=%dV\"%V_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage when bright light is on=0V\n",
-        "Output voltage when bright light is off=25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_4.ipynb
deleted file mode 100755
index cad31534..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_4.ipynb
+++ /dev/null
@@ -1,677 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:369e36634d005b832372dcae6796c76b979f32b499d8baadce951517f2201533"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 20 : SILICON CONTROLLED RECTIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.2 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I=50.0;                             #Surge current, A\n",
-      "t=12.0;                             #Time for which surge current lasts, ms\n",
-      "circuit_fusing_rating_max=90;       #Maximum circuit fusing rating, A\u00b2s\n",
-      "\n",
-      "#Calculation\n",
-      "circuit_fusing_rating=I**2*(t*10**-3);              #Circuit fusing rating, A\u00b2s\n",
-      "\n",
-      "#Result\n",
-      "if(circuit_fusing_rating<circuit_fusing_rating_max):\n",
-      "    print(\"The device will not be destroyed.\");\n",
-      "else:\n",
-      "    print(\"The device will be destroyed.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The device will not be destroyed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.3 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I2_t_rating=50.0;                   #circuit fuse rating, A\u00b2s\n",
-      "Is=100.0;                           #Surge current, A\n",
-      "\n",
-      "#Calculation\n",
-      "t_max=(I2_t_rating/Is**2)*1000;            #Maximum allowable duration, ms\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable duration =%dms\"%t_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable duration =5ms\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.4 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "R=220.0;                    #Gate resistor, \u03a9\n",
-      "I_G=7.0;                    #Gate current, mA\n",
-      "V_GK=0.7;                   #Junction voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_in=V_GK+(I_G/1000)*R;                    #Input voltage, V (Kirchhoff's voltage law)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required input voltage=%.2fV.\"%V_in);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required input voltage=2.24V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.5 : Page number 564\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "from math import cos\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_G=1.0;                #Gate current, mA\n",
-      "V_m=200.0;              #Peak value of input sinusoidal voltage, V\n",
-      "v=100.0;              #Forward breakdown voltage of SCR, V\n",
-      "R_L=100.0;              #Load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#v=Vm*sin(theta)\n",
-      "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n",
-      "theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad\n",
-      "theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees\n",
-      "\n",
-      "alpha=round(theta,0);                    #Firing angle, degrees\n",
-      "\n",
-      "#(ii)\n",
-      "phi=180-alpha;                  #Conduction angle, degrees\n",
-      "\n",
-      "#(iii)\n",
-      "V_avg=(V_m/(2*pi))*(1+cos(alpha*pi/180));           #Average voltage, V\n",
-      "I_avg=V_avg/R_L;                                    #Average current, A\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n",
-      "print(\"(ii) The conduction angle=%.0f\u00b0\"%phi);\n",
-      "print(\"(iii) The average current=%.4fA \"%I_avg);\n",
-      "\n",
-      "#Note: In the text book has approximated the average current to 0.5925A but in the code it gets approximated to 0.5940A.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The firing agle=30\u00b0\n",
-        "(ii) The conduction angle=150\u00b0\n",
-        "(iii) The average current=0.5940A \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.6 : Page number 564\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "from math import cos\n",
-      "from math import pi\n",
-      "from math import floor\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_G=1.0;                #Gate current, mA\n",
-      "V_m=400.0;              #Peak value of input sinusoidal voltage, V\n",
-      "v=150.0;                #Forward breakdown voltage of SCR, V\n",
-      "R_L=200.0;              #Load resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#v=Vm*sin(theta)\n",
-      "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n",
-      "theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad\n",
-      "theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees\n",
-      "\n",
-      "alpha=theta;                    #Firing angle, degrees\n",
-      "\n",
-      "#(ii)\n",
-      "V_av=floor((V_m/(2*pi))*(1+cos(alpha*pi/180))*10)/10;           #Average voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "I_av=V_av/R_L;                                    #Average current, A\n",
-      "\n",
-      "#(iv)\n",
-      "P_out=V_av*I_av;                                #Output power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n",
-      "print(\"(ii) The average output voltage=%.1f V\"%V_av);\n",
-      "print(\"(iii) The average current=%.3fA \"%I_av);\n",
-      "print(\"(iv) The output power=%.2f W\"%P_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The firing agle=22\u00b0\n",
-        "(ii) The average output voltage=122.6 V\n",
-        "(iii) The average current=0.613A \n",
-        "(iv) The output power=75.15 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.7 : Page number 564-565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "\n",
-      "#Variable declaration\n",
-      "v=180.0;                    #Forward breakdown voltage, V\n",
-      "V_m=240.0;                  #Peak value of input voltage, V\n",
-      "w=314.0;                    #Angular frequency of input ,rad/s\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#v=Vm*sin(w*t)\n",
-      "#So, t=asin(v/Vm)/w\n",
-      "t=(asin(v/V_m)/w)*1000;                        #Time for which SCR remains off, ms\n",
-      "\n",
-      "#Result\n",
-      "print(\"The SCR remains off for %.1f ms.\"%t);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The SCR remains off for 2.7 ms.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.8 : Page number 565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import cos\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_dc=1.0;                   #d.c load current, A\n",
-      "alpha=30.0;                 #Firing angle, \u00b0\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_av=I_dc;                                                 #Average current(= d.c current), A\n",
-      "\n",
-      "#Since, Iav=(Vm/(2*pi*RL))*(1+cos(alpha)) and Im=Vm/RL\n",
-      "I_m=floor((2*pi*I_av/(1+cos(alpha*pi/180)))*100)/100;      #Peak-load current, A\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Peak-load current=%.2f A.\"%I_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Peak-load current=3.36 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.9: Page number 565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "from math import sin\n",
-      "\n",
-      "#Variable declaration\n",
-      "alpha=60.0;                 #Firing angle, \u00b0\n",
-      "P=100.0;                    #Power rating of tungsten lamp, W\n",
-      "V=110.0;                    #Voltage rating of tungsten lamp, V\n",
-      "V_ac=110.0;                 #a.c supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_m=V_ac*sqrt(2);               #Peak value of input voltage, V\n",
-      "\n",
-      "alpha=alpha*pi/180;             #firing angle, rad\n",
-      "\n",
-      "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n",
-      "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "E_rms=round(V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)));            #r.m.s voltage developed in the lamp, V\n",
-      "\n",
-      "\n",
-      "RL=V**2/P;                      #Load resistance, \u03a9\n",
-      "\n",
-      "I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The r.m.s voltage developed in the lamp=%d V.\"%E_rms);\n",
-      "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The r.m.s voltage developed in the lamp=70 V.\n",
-        "The r.m.s current developed in the lamp=0.58 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.10 : Page number 567\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import cos\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=100.0;                   #Load resistance, \u03a9\n",
-      "V_m=200.0;                  #Peak a.c voltage, V\n",
-      "alpha=60;                   #firing angle, \u00b0\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_av=(V_m/pi)*(1+cos(alpha*pi/180));                #D.C output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "I_av=V_av/RL;                   #Load current, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  d.c output voltage=%.1f V.\"%V_av);\n",
-      "print(\"(ii) Load current=%.3f A\"%I_av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  d.c output voltage=95.5 V.\n",
-        "(ii) Load current=0.955 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.11: Page number 567\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "from math import sin\n",
-      "\n",
-      "#Variable declaration\n",
-      "alpha=60.0;                 #Firing angle, \u00b0\n",
-      "P=100.0;                    #Power rating of tungsten lamp, W\n",
-      "V=110.0;                    #Voltage rating of tungsten lamp, V\n",
-      "V_ac=110.0;                 #a.c supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_m=round(V_ac*sqrt(2));               #Peak value of input voltage, V\n",
-      "\n",
-      "alpha=alpha*pi/180;             #firing angle, rad\n",
-      "\n",
-      "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n",
-      "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "E_rms=V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(4*pi));            #r.m.s voltage developed in the lamp, V\n",
-      "\n",
-      "RL=V**2/P;                      #Load resistance, \u03a9\n",
-      "\n",
-      "I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The r.m.s voltage developed in the lamp=%.1f V.\"%E_rms);\n",
-      "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The r.m.s voltage developed in the lamp=98.9 V.\n",
-        "The r.m.s current developed in the lamp=0.82 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.12 : Page number 572\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15;                     #Suuply voltage, V\n",
-      "V_T=0.7;                    #Gate trigger voltage, V\n",
-      "I_T=7.0;                    #Gate trigger current, mA\n",
-      "I_H=6.0;                    #Holding current. mA\n",
-      "R_Vin=1;                    #Resistance at Vin, k\u03a9\n",
-      "R_VCC=100;                  #Resistance at Vcc, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) when SCR is off, there is no current, therefore no voltage drop across the resistor\n",
-      "V_out=VCC;                      #Output voltage, when SCR is off, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_in=V_T+I_T*R_Vin;             #Input voltage required to trigger the SCR, V\n",
-      "\n",
-      "#(iii)\n",
-      "#Since, I_H=(Vcc-VT)/R_Vin;\n",
-      "VCC_SCR_open=(I_H/1000)*R_VCC+V_T;                 #Decreased value of supply voltage at which SCR opens, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output voltage when SCR is off=%dV.\"%V_out);\n",
-      "print(\"(ii) The input voltage required to trigger the SCR=%.1f V.\"%V_in);\n",
-      "print(\"(iii) The decreased supply voltage at which SCR opens=%.1f V.\"%VCC_SCR_open);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage when SCR is off=15V.\n",
-        "(ii) The input voltage required to trigger the SCR=7.7 V.\n",
-        "(iii) The decreased supply voltage at which SCR opens=1.3 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.13 : Page number 572-573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=5.6;                 #zener voltage, V\n",
-      "V_T=0.7;                #Trigger voltage of SCR, V\n",
-      "\n",
-      "#Calculation\n",
-      "VCC=Vz+V_T;                 #Required supply voltage to turn on the crowbar, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required supply voltage to turn on the crowbar=%.1fV.\"%VCC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required supply voltage to turn on the crowbar=6.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.14 : Page number 573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=12;                  #Zener breakdown voltage, V\n",
-      "V_T=1.5;                #Trigger voltage, V\n",
-      "tolerance_z=10.0;         #Tolerance of zener diode, %\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vz_max=Vz*(1+tolerance_z/100);          #Maximum value of zener breakdown, V\n",
-      "Vz_min=Vz*(1-tolerance_z/100);          #Minimum value of zener breakdown, V\n",
-      "V_crowbar=Vz_max+V_T;                   #Maximum value of supply voltage for crowbarring, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum value of supply voltage for crowbarring=%.1fV\"%V_crowbar);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of supply voltage for crowbarring=14.7V\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.15 : Page number 573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=25.0;               #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#When brights light is on, LASCR conducts and thus gets short circuited to ground, hence,\n",
-      "V_out=0;                #Output voltage, V\n",
-      "\n",
-      "print(\"Output voltage when bright light is on=%dV\"%V_out);\n",
-      "\n",
-      "\n",
-      "#When brights light is off, LASCR stops conducting and thus no current through resistor, hence,\n",
-      "V_out=VCC;                #Output voltage, V\n",
-      "print(\"Output voltage when bright light is off=%dV\"%V_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage when bright light is on=0V\n",
-        "Output voltage when bright light is off=25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_5.ipynb
deleted file mode 100755
index cad31534..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_5.ipynb
+++ /dev/null
@@ -1,677 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:369e36634d005b832372dcae6796c76b979f32b499d8baadce951517f2201533"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 20 : SILICON CONTROLLED RECTIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.2 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I=50.0;                             #Surge current, A\n",
-      "t=12.0;                             #Time for which surge current lasts, ms\n",
-      "circuit_fusing_rating_max=90;       #Maximum circuit fusing rating, A\u00b2s\n",
-      "\n",
-      "#Calculation\n",
-      "circuit_fusing_rating=I**2*(t*10**-3);              #Circuit fusing rating, A\u00b2s\n",
-      "\n",
-      "#Result\n",
-      "if(circuit_fusing_rating<circuit_fusing_rating_max):\n",
-      "    print(\"The device will not be destroyed.\");\n",
-      "else:\n",
-      "    print(\"The device will be destroyed.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The device will not be destroyed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.3 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I2_t_rating=50.0;                   #circuit fuse rating, A\u00b2s\n",
-      "Is=100.0;                           #Surge current, A\n",
-      "\n",
-      "#Calculation\n",
-      "t_max=(I2_t_rating/Is**2)*1000;            #Maximum allowable duration, ms\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable duration =%dms\"%t_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable duration =5ms\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.4 : Page number 559\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "R=220.0;                    #Gate resistor, \u03a9\n",
-      "I_G=7.0;                    #Gate current, mA\n",
-      "V_GK=0.7;                   #Junction voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_in=V_GK+(I_G/1000)*R;                    #Input voltage, V (Kirchhoff's voltage law)\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required input voltage=%.2fV.\"%V_in);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required input voltage=2.24V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.5 : Page number 564\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "from math import cos\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_G=1.0;                #Gate current, mA\n",
-      "V_m=200.0;              #Peak value of input sinusoidal voltage, V\n",
-      "v=100.0;              #Forward breakdown voltage of SCR, V\n",
-      "R_L=100.0;              #Load resistance, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#v=Vm*sin(theta)\n",
-      "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n",
-      "theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad\n",
-      "theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees\n",
-      "\n",
-      "alpha=round(theta,0);                    #Firing angle, degrees\n",
-      "\n",
-      "#(ii)\n",
-      "phi=180-alpha;                  #Conduction angle, degrees\n",
-      "\n",
-      "#(iii)\n",
-      "V_avg=(V_m/(2*pi))*(1+cos(alpha*pi/180));           #Average voltage, V\n",
-      "I_avg=V_avg/R_L;                                    #Average current, A\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n",
-      "print(\"(ii) The conduction angle=%.0f\u00b0\"%phi);\n",
-      "print(\"(iii) The average current=%.4fA \"%I_avg);\n",
-      "\n",
-      "#Note: In the text book has approximated the average current to 0.5925A but in the code it gets approximated to 0.5940A.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The firing agle=30\u00b0\n",
-        "(ii) The conduction angle=150\u00b0\n",
-        "(iii) The average current=0.5940A \n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.6 : Page number 564\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "from math import cos\n",
-      "from math import pi\n",
-      "from math import floor\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_G=1.0;                #Gate current, mA\n",
-      "V_m=400.0;              #Peak value of input sinusoidal voltage, V\n",
-      "v=150.0;                #Forward breakdown voltage of SCR, V\n",
-      "R_L=200.0;              #Load resistance, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#v=Vm*sin(theta)\n",
-      "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n",
-      "theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad\n",
-      "theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees\n",
-      "\n",
-      "alpha=theta;                    #Firing angle, degrees\n",
-      "\n",
-      "#(ii)\n",
-      "V_av=floor((V_m/(2*pi))*(1+cos(alpha*pi/180))*10)/10;           #Average voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "I_av=V_av/R_L;                                    #Average current, A\n",
-      "\n",
-      "#(iv)\n",
-      "P_out=V_av*I_av;                                #Output power, W\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n",
-      "print(\"(ii) The average output voltage=%.1f V\"%V_av);\n",
-      "print(\"(iii) The average current=%.3fA \"%I_av);\n",
-      "print(\"(iv) The output power=%.2f W\"%P_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The firing agle=22\u00b0\n",
-        "(ii) The average output voltage=122.6 V\n",
-        "(iii) The average current=0.613A \n",
-        "(iv) The output power=75.15 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.7 : Page number 564-565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import asin\n",
-      "\n",
-      "#Variable declaration\n",
-      "v=180.0;                    #Forward breakdown voltage, V\n",
-      "V_m=240.0;                  #Peak value of input voltage, V\n",
-      "w=314.0;                    #Angular frequency of input ,rad/s\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#v=Vm*sin(w*t)\n",
-      "#So, t=asin(v/Vm)/w\n",
-      "t=(asin(v/V_m)/w)*1000;                        #Time for which SCR remains off, ms\n",
-      "\n",
-      "#Result\n",
-      "print(\"The SCR remains off for %.1f ms.\"%t);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The SCR remains off for 2.7 ms.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.8 : Page number 565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import cos\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_dc=1.0;                   #d.c load current, A\n",
-      "alpha=30.0;                 #Firing angle, \u00b0\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_av=I_dc;                                                 #Average current(= d.c current), A\n",
-      "\n",
-      "#Since, Iav=(Vm/(2*pi*RL))*(1+cos(alpha)) and Im=Vm/RL\n",
-      "I_m=floor((2*pi*I_av/(1+cos(alpha*pi/180)))*100)/100;      #Peak-load current, A\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Peak-load current=%.2f A.\"%I_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Peak-load current=3.36 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.9: Page number 565\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "from math import sin\n",
-      "\n",
-      "#Variable declaration\n",
-      "alpha=60.0;                 #Firing angle, \u00b0\n",
-      "P=100.0;                    #Power rating of tungsten lamp, W\n",
-      "V=110.0;                    #Voltage rating of tungsten lamp, V\n",
-      "V_ac=110.0;                 #a.c supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_m=V_ac*sqrt(2);               #Peak value of input voltage, V\n",
-      "\n",
-      "alpha=alpha*pi/180;             #firing angle, rad\n",
-      "\n",
-      "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n",
-      "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "E_rms=round(V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)));            #r.m.s voltage developed in the lamp, V\n",
-      "\n",
-      "\n",
-      "RL=V**2/P;                      #Load resistance, \u03a9\n",
-      "\n",
-      "I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The r.m.s voltage developed in the lamp=%d V.\"%E_rms);\n",
-      "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The r.m.s voltage developed in the lamp=70 V.\n",
-        "The r.m.s current developed in the lamp=0.58 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.10 : Page number 567\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import cos\n",
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=100.0;                   #Load resistance, \u03a9\n",
-      "V_m=200.0;                  #Peak a.c voltage, V\n",
-      "alpha=60;                   #firing angle, \u00b0\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_av=(V_m/pi)*(1+cos(alpha*pi/180));                #D.C output voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "I_av=V_av/RL;                   #Load current, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  d.c output voltage=%.1f V.\"%V_av);\n",
-      "print(\"(ii) Load current=%.3f A\"%I_av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  d.c output voltage=95.5 V.\n",
-        "(ii) Load current=0.955 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.11: Page number 567\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import pi\n",
-      "from math import sin\n",
-      "\n",
-      "#Variable declaration\n",
-      "alpha=60.0;                 #Firing angle, \u00b0\n",
-      "P=100.0;                    #Power rating of tungsten lamp, W\n",
-      "V=110.0;                    #Voltage rating of tungsten lamp, V\n",
-      "V_ac=110.0;                 #a.c supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_m=round(V_ac*sqrt(2));               #Peak value of input voltage, V\n",
-      "\n",
-      "alpha=alpha*pi/180;             #firing angle, rad\n",
-      "\n",
-      "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n",
-      "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n",
-      "E_rms=V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(4*pi));            #r.m.s voltage developed in the lamp, V\n",
-      "\n",
-      "RL=V**2/P;                      #Load resistance, \u03a9\n",
-      "\n",
-      "I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The r.m.s voltage developed in the lamp=%.1f V.\"%E_rms);\n",
-      "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The r.m.s voltage developed in the lamp=98.9 V.\n",
-        "The r.m.s current developed in the lamp=0.82 A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.12 : Page number 572\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15;                     #Suuply voltage, V\n",
-      "V_T=0.7;                    #Gate trigger voltage, V\n",
-      "I_T=7.0;                    #Gate trigger current, mA\n",
-      "I_H=6.0;                    #Holding current. mA\n",
-      "R_Vin=1;                    #Resistance at Vin, k\u03a9\n",
-      "R_VCC=100;                  #Resistance at Vcc, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) when SCR is off, there is no current, therefore no voltage drop across the resistor\n",
-      "V_out=VCC;                      #Output voltage, when SCR is off, V\n",
-      "\n",
-      "#(ii)\n",
-      "V_in=V_T+I_T*R_Vin;             #Input voltage required to trigger the SCR, V\n",
-      "\n",
-      "#(iii)\n",
-      "#Since, I_H=(Vcc-VT)/R_Vin;\n",
-      "VCC_SCR_open=(I_H/1000)*R_VCC+V_T;                 #Decreased value of supply voltage at which SCR opens, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The output voltage when SCR is off=%dV.\"%V_out);\n",
-      "print(\"(ii) The input voltage required to trigger the SCR=%.1f V.\"%V_in);\n",
-      "print(\"(iii) The decreased supply voltage at which SCR opens=%.1f V.\"%VCC_SCR_open);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage when SCR is off=15V.\n",
-        "(ii) The input voltage required to trigger the SCR=7.7 V.\n",
-        "(iii) The decreased supply voltage at which SCR opens=1.3 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.13 : Page number 572-573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=5.6;                 #zener voltage, V\n",
-      "V_T=0.7;                #Trigger voltage of SCR, V\n",
-      "\n",
-      "#Calculation\n",
-      "VCC=Vz+V_T;                 #Required supply voltage to turn on the crowbar, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The required supply voltage to turn on the crowbar=%.1fV.\"%VCC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required supply voltage to turn on the crowbar=6.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.14 : Page number 573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=12;                  #Zener breakdown voltage, V\n",
-      "V_T=1.5;                #Trigger voltage, V\n",
-      "tolerance_z=10.0;         #Tolerance of zener diode, %\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vz_max=Vz*(1+tolerance_z/100);          #Maximum value of zener breakdown, V\n",
-      "Vz_min=Vz*(1-tolerance_z/100);          #Minimum value of zener breakdown, V\n",
-      "V_crowbar=Vz_max+V_T;                   #Maximum value of supply voltage for crowbarring, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum value of supply voltage for crowbarring=%.1fV\"%V_crowbar);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum value of supply voltage for crowbarring=14.7V\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 20.15 : Page number 573\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=25.0;               #Supply voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#When brights light is on, LASCR conducts and thus gets short circuited to ground, hence,\n",
-      "V_out=0;                #Output voltage, V\n",
-      "\n",
-      "print(\"Output voltage when bright light is on=%dV\"%V_out);\n",
-      "\n",
-      "\n",
-      "#When brights light is off, LASCR stops conducting and thus no current through resistor, hence,\n",
-      "V_out=VCC;                #Output voltage, V\n",
-      "print(\"Output voltage when bright light is off=%dV\"%V_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage when bright light is on=0V\n",
-        "Output voltage when bright light is off=25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21.ipynb
deleted file mode 100755
index acca0cfa..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21.ipynb
+++ /dev/null
@@ -1,467 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:412bf04e25192c77f9fa9664d995cc0ae6446a81f631fb5e0e755ebfa36436bf"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 21 : POWER ELECTRONICS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.3: Page number 585\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GT=2;                 #Gate triggering voltage, V\n",
-      "V_F=0.7;               #Forward voltage for diode D1\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)Triggering only by a positive gate voltage,\n",
-      "#A diode is connected at the gatewith the n-side connected to thegate of the device,\n",
-      "V_A=V_F+V_GT;                   #Required voltage to trigger the device, V\n",
-      "\n",
-      "print(\"The  required voltage to trigger the device only by positive voltage=%.1fV.\"%V_A);\n",
-      "\n",
-      "#(ii)\n",
-      "print(\"In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The  required voltage to trigger the device only by positive voltage=2.7V.\n",
-        "In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.4 : Page number 585-586\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=50.0;                   #Resitor, \u03a9\n",
-      "V=50.0;                   #Supply voltage, V\n",
-      "V_drop=1.0;               #Drop across the triac in conduction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Ideal triac\n",
-      "#Since the triac is ideal, voltage drop across it is zero,\n",
-      "I=V/R;                  #Current through the 50 \u03a9 resistor, A\n",
-      "\n",
-      "print(\"(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=%dA.\"%I);\n",
-      "\n",
-      "#(ii) Triac has a drop of 1V\n",
-      "I=(V-V_drop)/R;                     #Current through the 50 \u03a9 resistor, A\n",
-      "\n",
-      "print(\"(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=%.2fA.\"%I);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=1A.\n",
-        "(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=0.98A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.5 : Page number 588-589\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GT=2;             #Gate triggering voltage, V\n",
-      "V_BO=20;            #Breakover voltage,V\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"The triggering level is raised by using a diac.\");\n",
-      "V_A=V_BO+V_GT;                  #Gate trigger signal, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"In order to turn on the triac, the gate trigger signal=%dV.\"%V_A);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The triggering level is raised by using a diac.\n",
-        "In order to turn on the triac, the gate trigger signal=22V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.6 : Page number 589\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BO=30;            #Breakover voltage of diac, V\n",
-      "V_GT=1;             #Trigger voltage of the triac, V\n",
-      "I_T=10;             #Trigger current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_A=V_BO+V_GT;              #Voltage required for triggering the triac, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum capacitor voltage that will trigger the triac=%d V.\"%V_A);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum capacitor voltage that will trigger the triac=31 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.7 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "eta=0.6;                        #Intrinsic stand-off ratio for UJT\n",
-      "R_BB=10;                        #Inter-base resistance, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, RBB=RB1+RB2 and eta=RB1/(RB1+RB2),\n",
-      "#eta=RB1/RBB.\n",
-      "R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, k\u03a9\n",
-      "R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, k\u03a9 \n",
-      "\n",
-      "#Result\n",
-      "print(\"Resistance of the bar between B1 and emitter junction=%d k\u03a9.\"%R_B1);\n",
-      "print(\"Resistance of the bar between B2 and emitter junction=%d k\u03a9.\"%R_B2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Resistance of the bar between B1 and emitter junction=6 k\u03a9.\n",
-        "Resistance of the bar between B2 and emitter junction=4 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.8 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BB=10;                #Interbase voltage, V\n",
-      "eta=0.65;                #Intrinsic stand-off ratio for UJT\n",
-      "V_D=0.7;                #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_stand_off=eta*V_BB;               #Stand off voltage, V\n",
-      "V_P=V_stand_off+V_D;                #Peak-point voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Stand off voltage=%.1f V.\"%V_stand_off);\n",
-      "print(\"Peak-point voltage=%.1f V.\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Stand off voltage=6.5 V.\n",
-        "Peak-point voltage=7.2 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.9 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BB=25;                #Interbase voltage, V\n",
-      "eta_max=0.86;           #Maximum intrinsic stand-off ratio for UJT\n",
-      "eta_min=0.74;           #Minimum intrinsic stand-off ratio for UJT\n",
-      "V_D=0.7;                #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_P_max=eta_max*V_BB+V_D;           #Maximum peak-point, V\n",
-      "V_P_min=eta_min*V_BB+V_D;           #Minimum peak-point, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum peak-point voltage=%.1fV\"%V_P_max);\n",
-      "print(\"Minimum peak-point voltage=%.1fV\"%V_P_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum peak-point voltage=22.2V\n",
-        "Minimum peak-point voltage=19.2V\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.10 : Page number 593-594\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "eta=0.65;                        #Intrinsic stand-off ratio for UJT\n",
-      "R_BB=7.0;                          #Inter-base resistance, k\u03a9\n",
-      "R1=100.0;                          #Resistor R1, \u03a9\n",
-      "R2=400.0;                          #Resistor R2, \u03a9\n",
-      "V_S=12.0;                          #Source voltage, V\n",
-      "V_D=0.7;                         #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, eta=RB1/RBB,\n",
-      "R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, k\u03a9\n",
-      "R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, k\u03a9 \n",
-      "\n",
-      "print(\"(i)  Resistance of the bar between B1 and emitter junction=%.2f k\u03a9.\"%R_B1);\n",
-      "print(\"     Resistance of the bar between B2 and emitter junction=%.2f k\u03a9.\"%R_B2);\n",
-      "\n",
-      "#(ii)\n",
-      "V_B2_B1=V_S*R_BB/(R_BB + (R1/1000) + (R2/1000));          #Voltage  across B2-B1, V (voltage divider rule)\n",
-      "V_P=eta*V_B2_B1+V_D;                        #Peak-point voltage, V\n",
-      "\n",
-      "print(\"(ii) The voltage across the base B2-B1=%.1fV.\"%V_B2_B1);\n",
-      "print(\"     Peak-point voltage=%.2fV\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Resistance of the bar between B1 and emitter junction=4.55 k\u03a9.\n",
-        "     Resistance of the bar between B2 and emitter junction=2.45 k\u03a9.\n",
-        "(ii) The voltage across the base B2-B1=11.2V.\n",
-        "     Peak-point voltage=7.98V\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.11 : Page number  596\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "RE_initial=5;                   #Initial value of emitter resistor, k\u03a9\n",
-      "RE_adjusted=10;                 #Adjusted value of emitter resistor, k\u03a9\n",
-      "C=0.2;                          #Capacitance, \u03bcF\n",
-      "eta=0.54;                         #intrinsic stand-off ratio\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "t=round((RE_initial*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                             #frequency, Hz\n",
-      "\n",
-      "print(\"Frequency for 5k\u03a9 setting=%dHz.\"%f);\n",
-      "\n",
-      "#(i)\n",
-      "t=round((RE_adjusted*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                             #frequency, Hz\n",
-      "\n",
-      "print(\"Frequency for 10k\u03a9 setting=%dHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Frequency for 5k\u03a9 setting=1282Hz.\n",
-        "Frequency for 10k\u03a9 setting=645Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.12 : Page number 596-597\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "V_S=12;                         #Supply voltage, V\n",
-      "R_BB=5;                         #Interbase resistance, k\u03a9\n",
-      "R_1=50;                         #Resistor R1, k\u03a9\n",
-      "R_2=0.1;                        #Resistor R2, k\u03a9\n",
-      "C=0.1;                          #Capacitance, \u03bcF\n",
-      "eta=0.6;                          #intrinsic stand-off ratio\n",
-      "V_D=0.7;                        #Voltage drop across pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, \u03b7=R_B1/R_BB,\n",
-      "R_B1=eta*R_BB;                            #Resitance between base B1 and emitter junction, k\u03a9\n",
-      "\n",
-      "#Since, R_BB=R_B1+R_B2\n",
-      "R_B2=R_BB-R_B1;                         #Resitance between base B2 and emitter junction, k\u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "V_RB1_R2=V_S*(R_B1+R_2)/(R_BB+R_2);           #Voltage drop across R_B1 and R_2 resistors, V\n",
-      "V_P=V_D+V_RB1_R2;                           #Peak-point voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "t=round((R_1*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                                    #frequency, Hz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   R_B1=%dk\u03a9 and R_B2=%dk\u03a9\"%(R_B1,R_B2));\n",
-      "print(\"(ii)  The peak-point voltage to turn on the UJT=%.0fV.\"%V_P);\n",
-      "print(\"(iii) Frequency of oscillations=%dHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   R_B1=3k\u03a9 and R_B2=2k\u03a9\n",
-        "(ii)  The peak-point voltage to turn on the UJT=8V.\n",
-        "(iii) Frequency of oscillations=218Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_1.ipynb
deleted file mode 100755
index acca0cfa..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_1.ipynb
+++ /dev/null
@@ -1,467 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:412bf04e25192c77f9fa9664d995cc0ae6446a81f631fb5e0e755ebfa36436bf"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 21 : POWER ELECTRONICS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.3: Page number 585\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GT=2;                 #Gate triggering voltage, V\n",
-      "V_F=0.7;               #Forward voltage for diode D1\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)Triggering only by a positive gate voltage,\n",
-      "#A diode is connected at the gatewith the n-side connected to thegate of the device,\n",
-      "V_A=V_F+V_GT;                   #Required voltage to trigger the device, V\n",
-      "\n",
-      "print(\"The  required voltage to trigger the device only by positive voltage=%.1fV.\"%V_A);\n",
-      "\n",
-      "#(ii)\n",
-      "print(\"In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The  required voltage to trigger the device only by positive voltage=2.7V.\n",
-        "In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.4 : Page number 585-586\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=50.0;                   #Resitor, \u03a9\n",
-      "V=50.0;                   #Supply voltage, V\n",
-      "V_drop=1.0;               #Drop across the triac in conduction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Ideal triac\n",
-      "#Since the triac is ideal, voltage drop across it is zero,\n",
-      "I=V/R;                  #Current through the 50 \u03a9 resistor, A\n",
-      "\n",
-      "print(\"(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=%dA.\"%I);\n",
-      "\n",
-      "#(ii) Triac has a drop of 1V\n",
-      "I=(V-V_drop)/R;                     #Current through the 50 \u03a9 resistor, A\n",
-      "\n",
-      "print(\"(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=%.2fA.\"%I);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=1A.\n",
-        "(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=0.98A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.5 : Page number 588-589\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GT=2;             #Gate triggering voltage, V\n",
-      "V_BO=20;            #Breakover voltage,V\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"The triggering level is raised by using a diac.\");\n",
-      "V_A=V_BO+V_GT;                  #Gate trigger signal, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"In order to turn on the triac, the gate trigger signal=%dV.\"%V_A);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The triggering level is raised by using a diac.\n",
-        "In order to turn on the triac, the gate trigger signal=22V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.6 : Page number 589\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BO=30;            #Breakover voltage of diac, V\n",
-      "V_GT=1;             #Trigger voltage of the triac, V\n",
-      "I_T=10;             #Trigger current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_A=V_BO+V_GT;              #Voltage required for triggering the triac, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum capacitor voltage that will trigger the triac=%d V.\"%V_A);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum capacitor voltage that will trigger the triac=31 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.7 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "eta=0.6;                        #Intrinsic stand-off ratio for UJT\n",
-      "R_BB=10;                        #Inter-base resistance, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, RBB=RB1+RB2 and eta=RB1/(RB1+RB2),\n",
-      "#eta=RB1/RBB.\n",
-      "R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, k\u03a9\n",
-      "R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, k\u03a9 \n",
-      "\n",
-      "#Result\n",
-      "print(\"Resistance of the bar between B1 and emitter junction=%d k\u03a9.\"%R_B1);\n",
-      "print(\"Resistance of the bar between B2 and emitter junction=%d k\u03a9.\"%R_B2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Resistance of the bar between B1 and emitter junction=6 k\u03a9.\n",
-        "Resistance of the bar between B2 and emitter junction=4 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.8 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BB=10;                #Interbase voltage, V\n",
-      "eta=0.65;                #Intrinsic stand-off ratio for UJT\n",
-      "V_D=0.7;                #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_stand_off=eta*V_BB;               #Stand off voltage, V\n",
-      "V_P=V_stand_off+V_D;                #Peak-point voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Stand off voltage=%.1f V.\"%V_stand_off);\n",
-      "print(\"Peak-point voltage=%.1f V.\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Stand off voltage=6.5 V.\n",
-        "Peak-point voltage=7.2 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.9 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BB=25;                #Interbase voltage, V\n",
-      "eta_max=0.86;           #Maximum intrinsic stand-off ratio for UJT\n",
-      "eta_min=0.74;           #Minimum intrinsic stand-off ratio for UJT\n",
-      "V_D=0.7;                #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_P_max=eta_max*V_BB+V_D;           #Maximum peak-point, V\n",
-      "V_P_min=eta_min*V_BB+V_D;           #Minimum peak-point, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum peak-point voltage=%.1fV\"%V_P_max);\n",
-      "print(\"Minimum peak-point voltage=%.1fV\"%V_P_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum peak-point voltage=22.2V\n",
-        "Minimum peak-point voltage=19.2V\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.10 : Page number 593-594\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "eta=0.65;                        #Intrinsic stand-off ratio for UJT\n",
-      "R_BB=7.0;                          #Inter-base resistance, k\u03a9\n",
-      "R1=100.0;                          #Resistor R1, \u03a9\n",
-      "R2=400.0;                          #Resistor R2, \u03a9\n",
-      "V_S=12.0;                          #Source voltage, V\n",
-      "V_D=0.7;                         #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, eta=RB1/RBB,\n",
-      "R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, k\u03a9\n",
-      "R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, k\u03a9 \n",
-      "\n",
-      "print(\"(i)  Resistance of the bar between B1 and emitter junction=%.2f k\u03a9.\"%R_B1);\n",
-      "print(\"     Resistance of the bar between B2 and emitter junction=%.2f k\u03a9.\"%R_B2);\n",
-      "\n",
-      "#(ii)\n",
-      "V_B2_B1=V_S*R_BB/(R_BB + (R1/1000) + (R2/1000));          #Voltage  across B2-B1, V (voltage divider rule)\n",
-      "V_P=eta*V_B2_B1+V_D;                        #Peak-point voltage, V\n",
-      "\n",
-      "print(\"(ii) The voltage across the base B2-B1=%.1fV.\"%V_B2_B1);\n",
-      "print(\"     Peak-point voltage=%.2fV\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Resistance of the bar between B1 and emitter junction=4.55 k\u03a9.\n",
-        "     Resistance of the bar between B2 and emitter junction=2.45 k\u03a9.\n",
-        "(ii) The voltage across the base B2-B1=11.2V.\n",
-        "     Peak-point voltage=7.98V\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.11 : Page number  596\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "RE_initial=5;                   #Initial value of emitter resistor, k\u03a9\n",
-      "RE_adjusted=10;                 #Adjusted value of emitter resistor, k\u03a9\n",
-      "C=0.2;                          #Capacitance, \u03bcF\n",
-      "eta=0.54;                         #intrinsic stand-off ratio\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "t=round((RE_initial*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                             #frequency, Hz\n",
-      "\n",
-      "print(\"Frequency for 5k\u03a9 setting=%dHz.\"%f);\n",
-      "\n",
-      "#(i)\n",
-      "t=round((RE_adjusted*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                             #frequency, Hz\n",
-      "\n",
-      "print(\"Frequency for 10k\u03a9 setting=%dHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Frequency for 5k\u03a9 setting=1282Hz.\n",
-        "Frequency for 10k\u03a9 setting=645Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.12 : Page number 596-597\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "V_S=12;                         #Supply voltage, V\n",
-      "R_BB=5;                         #Interbase resistance, k\u03a9\n",
-      "R_1=50;                         #Resistor R1, k\u03a9\n",
-      "R_2=0.1;                        #Resistor R2, k\u03a9\n",
-      "C=0.1;                          #Capacitance, \u03bcF\n",
-      "eta=0.6;                          #intrinsic stand-off ratio\n",
-      "V_D=0.7;                        #Voltage drop across pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, \u03b7=R_B1/R_BB,\n",
-      "R_B1=eta*R_BB;                            #Resitance between base B1 and emitter junction, k\u03a9\n",
-      "\n",
-      "#Since, R_BB=R_B1+R_B2\n",
-      "R_B2=R_BB-R_B1;                         #Resitance between base B2 and emitter junction, k\u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "V_RB1_R2=V_S*(R_B1+R_2)/(R_BB+R_2);           #Voltage drop across R_B1 and R_2 resistors, V\n",
-      "V_P=V_D+V_RB1_R2;                           #Peak-point voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "t=round((R_1*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                                    #frequency, Hz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   R_B1=%dk\u03a9 and R_B2=%dk\u03a9\"%(R_B1,R_B2));\n",
-      "print(\"(ii)  The peak-point voltage to turn on the UJT=%.0fV.\"%V_P);\n",
-      "print(\"(iii) Frequency of oscillations=%dHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   R_B1=3k\u03a9 and R_B2=2k\u03a9\n",
-        "(ii)  The peak-point voltage to turn on the UJT=8V.\n",
-        "(iii) Frequency of oscillations=218Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_2.ipynb
deleted file mode 100755
index acca0cfa..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_2.ipynb
+++ /dev/null
@@ -1,467 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:412bf04e25192c77f9fa9664d995cc0ae6446a81f631fb5e0e755ebfa36436bf"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 21 : POWER ELECTRONICS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.3: Page number 585\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GT=2;                 #Gate triggering voltage, V\n",
-      "V_F=0.7;               #Forward voltage for diode D1\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)Triggering only by a positive gate voltage,\n",
-      "#A diode is connected at the gatewith the n-side connected to thegate of the device,\n",
-      "V_A=V_F+V_GT;                   #Required voltage to trigger the device, V\n",
-      "\n",
-      "print(\"The  required voltage to trigger the device only by positive voltage=%.1fV.\"%V_A);\n",
-      "\n",
-      "#(ii)\n",
-      "print(\"In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The  required voltage to trigger the device only by positive voltage=2.7V.\n",
-        "In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.4 : Page number 585-586\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=50.0;                   #Resitor, \u03a9\n",
-      "V=50.0;                   #Supply voltage, V\n",
-      "V_drop=1.0;               #Drop across the triac in conduction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Ideal triac\n",
-      "#Since the triac is ideal, voltage drop across it is zero,\n",
-      "I=V/R;                  #Current through the 50 \u03a9 resistor, A\n",
-      "\n",
-      "print(\"(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=%dA.\"%I);\n",
-      "\n",
-      "#(ii) Triac has a drop of 1V\n",
-      "I=(V-V_drop)/R;                     #Current through the 50 \u03a9 resistor, A\n",
-      "\n",
-      "print(\"(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=%.2fA.\"%I);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=1A.\n",
-        "(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=0.98A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.5 : Page number 588-589\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GT=2;             #Gate triggering voltage, V\n",
-      "V_BO=20;            #Breakover voltage,V\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"The triggering level is raised by using a diac.\");\n",
-      "V_A=V_BO+V_GT;                  #Gate trigger signal, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"In order to turn on the triac, the gate trigger signal=%dV.\"%V_A);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The triggering level is raised by using a diac.\n",
-        "In order to turn on the triac, the gate trigger signal=22V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.6 : Page number 589\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BO=30;            #Breakover voltage of diac, V\n",
-      "V_GT=1;             #Trigger voltage of the triac, V\n",
-      "I_T=10;             #Trigger current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_A=V_BO+V_GT;              #Voltage required for triggering the triac, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum capacitor voltage that will trigger the triac=%d V.\"%V_A);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum capacitor voltage that will trigger the triac=31 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.7 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "eta=0.6;                        #Intrinsic stand-off ratio for UJT\n",
-      "R_BB=10;                        #Inter-base resistance, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, RBB=RB1+RB2 and eta=RB1/(RB1+RB2),\n",
-      "#eta=RB1/RBB.\n",
-      "R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, k\u03a9\n",
-      "R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, k\u03a9 \n",
-      "\n",
-      "#Result\n",
-      "print(\"Resistance of the bar between B1 and emitter junction=%d k\u03a9.\"%R_B1);\n",
-      "print(\"Resistance of the bar between B2 and emitter junction=%d k\u03a9.\"%R_B2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Resistance of the bar between B1 and emitter junction=6 k\u03a9.\n",
-        "Resistance of the bar between B2 and emitter junction=4 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.8 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BB=10;                #Interbase voltage, V\n",
-      "eta=0.65;                #Intrinsic stand-off ratio for UJT\n",
-      "V_D=0.7;                #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_stand_off=eta*V_BB;               #Stand off voltage, V\n",
-      "V_P=V_stand_off+V_D;                #Peak-point voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Stand off voltage=%.1f V.\"%V_stand_off);\n",
-      "print(\"Peak-point voltage=%.1f V.\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Stand off voltage=6.5 V.\n",
-        "Peak-point voltage=7.2 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.9 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BB=25;                #Interbase voltage, V\n",
-      "eta_max=0.86;           #Maximum intrinsic stand-off ratio for UJT\n",
-      "eta_min=0.74;           #Minimum intrinsic stand-off ratio for UJT\n",
-      "V_D=0.7;                #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_P_max=eta_max*V_BB+V_D;           #Maximum peak-point, V\n",
-      "V_P_min=eta_min*V_BB+V_D;           #Minimum peak-point, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum peak-point voltage=%.1fV\"%V_P_max);\n",
-      "print(\"Minimum peak-point voltage=%.1fV\"%V_P_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum peak-point voltage=22.2V\n",
-        "Minimum peak-point voltage=19.2V\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.10 : Page number 593-594\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "eta=0.65;                        #Intrinsic stand-off ratio for UJT\n",
-      "R_BB=7.0;                          #Inter-base resistance, k\u03a9\n",
-      "R1=100.0;                          #Resistor R1, \u03a9\n",
-      "R2=400.0;                          #Resistor R2, \u03a9\n",
-      "V_S=12.0;                          #Source voltage, V\n",
-      "V_D=0.7;                         #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, eta=RB1/RBB,\n",
-      "R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, k\u03a9\n",
-      "R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, k\u03a9 \n",
-      "\n",
-      "print(\"(i)  Resistance of the bar between B1 and emitter junction=%.2f k\u03a9.\"%R_B1);\n",
-      "print(\"     Resistance of the bar between B2 and emitter junction=%.2f k\u03a9.\"%R_B2);\n",
-      "\n",
-      "#(ii)\n",
-      "V_B2_B1=V_S*R_BB/(R_BB + (R1/1000) + (R2/1000));          #Voltage  across B2-B1, V (voltage divider rule)\n",
-      "V_P=eta*V_B2_B1+V_D;                        #Peak-point voltage, V\n",
-      "\n",
-      "print(\"(ii) The voltage across the base B2-B1=%.1fV.\"%V_B2_B1);\n",
-      "print(\"     Peak-point voltage=%.2fV\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Resistance of the bar between B1 and emitter junction=4.55 k\u03a9.\n",
-        "     Resistance of the bar between B2 and emitter junction=2.45 k\u03a9.\n",
-        "(ii) The voltage across the base B2-B1=11.2V.\n",
-        "     Peak-point voltage=7.98V\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.11 : Page number  596\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "RE_initial=5;                   #Initial value of emitter resistor, k\u03a9\n",
-      "RE_adjusted=10;                 #Adjusted value of emitter resistor, k\u03a9\n",
-      "C=0.2;                          #Capacitance, \u03bcF\n",
-      "eta=0.54;                         #intrinsic stand-off ratio\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "t=round((RE_initial*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                             #frequency, Hz\n",
-      "\n",
-      "print(\"Frequency for 5k\u03a9 setting=%dHz.\"%f);\n",
-      "\n",
-      "#(i)\n",
-      "t=round((RE_adjusted*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                             #frequency, Hz\n",
-      "\n",
-      "print(\"Frequency for 10k\u03a9 setting=%dHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Frequency for 5k\u03a9 setting=1282Hz.\n",
-        "Frequency for 10k\u03a9 setting=645Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.12 : Page number 596-597\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "V_S=12;                         #Supply voltage, V\n",
-      "R_BB=5;                         #Interbase resistance, k\u03a9\n",
-      "R_1=50;                         #Resistor R1, k\u03a9\n",
-      "R_2=0.1;                        #Resistor R2, k\u03a9\n",
-      "C=0.1;                          #Capacitance, \u03bcF\n",
-      "eta=0.6;                          #intrinsic stand-off ratio\n",
-      "V_D=0.7;                        #Voltage drop across pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, \u03b7=R_B1/R_BB,\n",
-      "R_B1=eta*R_BB;                            #Resitance between base B1 and emitter junction, k\u03a9\n",
-      "\n",
-      "#Since, R_BB=R_B1+R_B2\n",
-      "R_B2=R_BB-R_B1;                         #Resitance between base B2 and emitter junction, k\u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "V_RB1_R2=V_S*(R_B1+R_2)/(R_BB+R_2);           #Voltage drop across R_B1 and R_2 resistors, V\n",
-      "V_P=V_D+V_RB1_R2;                           #Peak-point voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "t=round((R_1*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                                    #frequency, Hz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   R_B1=%dk\u03a9 and R_B2=%dk\u03a9\"%(R_B1,R_B2));\n",
-      "print(\"(ii)  The peak-point voltage to turn on the UJT=%.0fV.\"%V_P);\n",
-      "print(\"(iii) Frequency of oscillations=%dHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   R_B1=3k\u03a9 and R_B2=2k\u03a9\n",
-        "(ii)  The peak-point voltage to turn on the UJT=8V.\n",
-        "(iii) Frequency of oscillations=218Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_3.ipynb
deleted file mode 100755
index acca0cfa..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_3.ipynb
+++ /dev/null
@@ -1,467 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:412bf04e25192c77f9fa9664d995cc0ae6446a81f631fb5e0e755ebfa36436bf"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 21 : POWER ELECTRONICS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.3: Page number 585\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GT=2;                 #Gate triggering voltage, V\n",
-      "V_F=0.7;               #Forward voltage for diode D1\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)Triggering only by a positive gate voltage,\n",
-      "#A diode is connected at the gatewith the n-side connected to thegate of the device,\n",
-      "V_A=V_F+V_GT;                   #Required voltage to trigger the device, V\n",
-      "\n",
-      "print(\"The  required voltage to trigger the device only by positive voltage=%.1fV.\"%V_A);\n",
-      "\n",
-      "#(ii)\n",
-      "print(\"In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The  required voltage to trigger the device only by positive voltage=2.7V.\n",
-        "In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.4 : Page number 585-586\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=50.0;                   #Resitor, \u03a9\n",
-      "V=50.0;                   #Supply voltage, V\n",
-      "V_drop=1.0;               #Drop across the triac in conduction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Ideal triac\n",
-      "#Since the triac is ideal, voltage drop across it is zero,\n",
-      "I=V/R;                  #Current through the 50 \u03a9 resistor, A\n",
-      "\n",
-      "print(\"(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=%dA.\"%I);\n",
-      "\n",
-      "#(ii) Triac has a drop of 1V\n",
-      "I=(V-V_drop)/R;                     #Current through the 50 \u03a9 resistor, A\n",
-      "\n",
-      "print(\"(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=%.2fA.\"%I);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=1A.\n",
-        "(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=0.98A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.5 : Page number 588-589\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GT=2;             #Gate triggering voltage, V\n",
-      "V_BO=20;            #Breakover voltage,V\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"The triggering level is raised by using a diac.\");\n",
-      "V_A=V_BO+V_GT;                  #Gate trigger signal, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"In order to turn on the triac, the gate trigger signal=%dV.\"%V_A);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The triggering level is raised by using a diac.\n",
-        "In order to turn on the triac, the gate trigger signal=22V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.6 : Page number 589\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BO=30;            #Breakover voltage of diac, V\n",
-      "V_GT=1;             #Trigger voltage of the triac, V\n",
-      "I_T=10;             #Trigger current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_A=V_BO+V_GT;              #Voltage required for triggering the triac, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum capacitor voltage that will trigger the triac=%d V.\"%V_A);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum capacitor voltage that will trigger the triac=31 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.7 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "eta=0.6;                        #Intrinsic stand-off ratio for UJT\n",
-      "R_BB=10;                        #Inter-base resistance, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, RBB=RB1+RB2 and eta=RB1/(RB1+RB2),\n",
-      "#eta=RB1/RBB.\n",
-      "R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, k\u03a9\n",
-      "R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, k\u03a9 \n",
-      "\n",
-      "#Result\n",
-      "print(\"Resistance of the bar between B1 and emitter junction=%d k\u03a9.\"%R_B1);\n",
-      "print(\"Resistance of the bar between B2 and emitter junction=%d k\u03a9.\"%R_B2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Resistance of the bar between B1 and emitter junction=6 k\u03a9.\n",
-        "Resistance of the bar between B2 and emitter junction=4 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.8 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BB=10;                #Interbase voltage, V\n",
-      "eta=0.65;                #Intrinsic stand-off ratio for UJT\n",
-      "V_D=0.7;                #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_stand_off=eta*V_BB;               #Stand off voltage, V\n",
-      "V_P=V_stand_off+V_D;                #Peak-point voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Stand off voltage=%.1f V.\"%V_stand_off);\n",
-      "print(\"Peak-point voltage=%.1f V.\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Stand off voltage=6.5 V.\n",
-        "Peak-point voltage=7.2 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.9 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BB=25;                #Interbase voltage, V\n",
-      "eta_max=0.86;           #Maximum intrinsic stand-off ratio for UJT\n",
-      "eta_min=0.74;           #Minimum intrinsic stand-off ratio for UJT\n",
-      "V_D=0.7;                #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_P_max=eta_max*V_BB+V_D;           #Maximum peak-point, V\n",
-      "V_P_min=eta_min*V_BB+V_D;           #Minimum peak-point, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum peak-point voltage=%.1fV\"%V_P_max);\n",
-      "print(\"Minimum peak-point voltage=%.1fV\"%V_P_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum peak-point voltage=22.2V\n",
-        "Minimum peak-point voltage=19.2V\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.10 : Page number 593-594\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "eta=0.65;                        #Intrinsic stand-off ratio for UJT\n",
-      "R_BB=7.0;                          #Inter-base resistance, k\u03a9\n",
-      "R1=100.0;                          #Resistor R1, \u03a9\n",
-      "R2=400.0;                          #Resistor R2, \u03a9\n",
-      "V_S=12.0;                          #Source voltage, V\n",
-      "V_D=0.7;                         #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, eta=RB1/RBB,\n",
-      "R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, k\u03a9\n",
-      "R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, k\u03a9 \n",
-      "\n",
-      "print(\"(i)  Resistance of the bar between B1 and emitter junction=%.2f k\u03a9.\"%R_B1);\n",
-      "print(\"     Resistance of the bar between B2 and emitter junction=%.2f k\u03a9.\"%R_B2);\n",
-      "\n",
-      "#(ii)\n",
-      "V_B2_B1=V_S*R_BB/(R_BB + (R1/1000) + (R2/1000));          #Voltage  across B2-B1, V (voltage divider rule)\n",
-      "V_P=eta*V_B2_B1+V_D;                        #Peak-point voltage, V\n",
-      "\n",
-      "print(\"(ii) The voltage across the base B2-B1=%.1fV.\"%V_B2_B1);\n",
-      "print(\"     Peak-point voltage=%.2fV\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Resistance of the bar between B1 and emitter junction=4.55 k\u03a9.\n",
-        "     Resistance of the bar between B2 and emitter junction=2.45 k\u03a9.\n",
-        "(ii) The voltage across the base B2-B1=11.2V.\n",
-        "     Peak-point voltage=7.98V\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.11 : Page number  596\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "RE_initial=5;                   #Initial value of emitter resistor, k\u03a9\n",
-      "RE_adjusted=10;                 #Adjusted value of emitter resistor, k\u03a9\n",
-      "C=0.2;                          #Capacitance, \u03bcF\n",
-      "eta=0.54;                         #intrinsic stand-off ratio\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "t=round((RE_initial*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                             #frequency, Hz\n",
-      "\n",
-      "print(\"Frequency for 5k\u03a9 setting=%dHz.\"%f);\n",
-      "\n",
-      "#(i)\n",
-      "t=round((RE_adjusted*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                             #frequency, Hz\n",
-      "\n",
-      "print(\"Frequency for 10k\u03a9 setting=%dHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Frequency for 5k\u03a9 setting=1282Hz.\n",
-        "Frequency for 10k\u03a9 setting=645Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.12 : Page number 596-597\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "V_S=12;                         #Supply voltage, V\n",
-      "R_BB=5;                         #Interbase resistance, k\u03a9\n",
-      "R_1=50;                         #Resistor R1, k\u03a9\n",
-      "R_2=0.1;                        #Resistor R2, k\u03a9\n",
-      "C=0.1;                          #Capacitance, \u03bcF\n",
-      "eta=0.6;                          #intrinsic stand-off ratio\n",
-      "V_D=0.7;                        #Voltage drop across pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, \u03b7=R_B1/R_BB,\n",
-      "R_B1=eta*R_BB;                            #Resitance between base B1 and emitter junction, k\u03a9\n",
-      "\n",
-      "#Since, R_BB=R_B1+R_B2\n",
-      "R_B2=R_BB-R_B1;                         #Resitance between base B2 and emitter junction, k\u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "V_RB1_R2=V_S*(R_B1+R_2)/(R_BB+R_2);           #Voltage drop across R_B1 and R_2 resistors, V\n",
-      "V_P=V_D+V_RB1_R2;                           #Peak-point voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "t=round((R_1*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                                    #frequency, Hz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   R_B1=%dk\u03a9 and R_B2=%dk\u03a9\"%(R_B1,R_B2));\n",
-      "print(\"(ii)  The peak-point voltage to turn on the UJT=%.0fV.\"%V_P);\n",
-      "print(\"(iii) Frequency of oscillations=%dHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   R_B1=3k\u03a9 and R_B2=2k\u03a9\n",
-        "(ii)  The peak-point voltage to turn on the UJT=8V.\n",
-        "(iii) Frequency of oscillations=218Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_4.ipynb
deleted file mode 100755
index acca0cfa..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_4.ipynb
+++ /dev/null
@@ -1,467 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:412bf04e25192c77f9fa9664d995cc0ae6446a81f631fb5e0e755ebfa36436bf"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 21 : POWER ELECTRONICS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.3: Page number 585\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GT=2;                 #Gate triggering voltage, V\n",
-      "V_F=0.7;               #Forward voltage for diode D1\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)Triggering only by a positive gate voltage,\n",
-      "#A diode is connected at the gatewith the n-side connected to thegate of the device,\n",
-      "V_A=V_F+V_GT;                   #Required voltage to trigger the device, V\n",
-      "\n",
-      "print(\"The  required voltage to trigger the device only by positive voltage=%.1fV.\"%V_A);\n",
-      "\n",
-      "#(ii)\n",
-      "print(\"In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The  required voltage to trigger the device only by positive voltage=2.7V.\n",
-        "In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.4 : Page number 585-586\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=50.0;                   #Resitor, \u03a9\n",
-      "V=50.0;                   #Supply voltage, V\n",
-      "V_drop=1.0;               #Drop across the triac in conduction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Ideal triac\n",
-      "#Since the triac is ideal, voltage drop across it is zero,\n",
-      "I=V/R;                  #Current through the 50 \u03a9 resistor, A\n",
-      "\n",
-      "print(\"(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=%dA.\"%I);\n",
-      "\n",
-      "#(ii) Triac has a drop of 1V\n",
-      "I=(V-V_drop)/R;                     #Current through the 50 \u03a9 resistor, A\n",
-      "\n",
-      "print(\"(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=%.2fA.\"%I);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=1A.\n",
-        "(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=0.98A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.5 : Page number 588-589\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GT=2;             #Gate triggering voltage, V\n",
-      "V_BO=20;            #Breakover voltage,V\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"The triggering level is raised by using a diac.\");\n",
-      "V_A=V_BO+V_GT;                  #Gate trigger signal, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"In order to turn on the triac, the gate trigger signal=%dV.\"%V_A);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The triggering level is raised by using a diac.\n",
-        "In order to turn on the triac, the gate trigger signal=22V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.6 : Page number 589\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BO=30;            #Breakover voltage of diac, V\n",
-      "V_GT=1;             #Trigger voltage of the triac, V\n",
-      "I_T=10;             #Trigger current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_A=V_BO+V_GT;              #Voltage required for triggering the triac, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum capacitor voltage that will trigger the triac=%d V.\"%V_A);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum capacitor voltage that will trigger the triac=31 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.7 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "eta=0.6;                        #Intrinsic stand-off ratio for UJT\n",
-      "R_BB=10;                        #Inter-base resistance, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, RBB=RB1+RB2 and eta=RB1/(RB1+RB2),\n",
-      "#eta=RB1/RBB.\n",
-      "R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, k\u03a9\n",
-      "R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, k\u03a9 \n",
-      "\n",
-      "#Result\n",
-      "print(\"Resistance of the bar between B1 and emitter junction=%d k\u03a9.\"%R_B1);\n",
-      "print(\"Resistance of the bar between B2 and emitter junction=%d k\u03a9.\"%R_B2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Resistance of the bar between B1 and emitter junction=6 k\u03a9.\n",
-        "Resistance of the bar between B2 and emitter junction=4 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.8 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BB=10;                #Interbase voltage, V\n",
-      "eta=0.65;                #Intrinsic stand-off ratio for UJT\n",
-      "V_D=0.7;                #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_stand_off=eta*V_BB;               #Stand off voltage, V\n",
-      "V_P=V_stand_off+V_D;                #Peak-point voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Stand off voltage=%.1f V.\"%V_stand_off);\n",
-      "print(\"Peak-point voltage=%.1f V.\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Stand off voltage=6.5 V.\n",
-        "Peak-point voltage=7.2 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.9 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BB=25;                #Interbase voltage, V\n",
-      "eta_max=0.86;           #Maximum intrinsic stand-off ratio for UJT\n",
-      "eta_min=0.74;           #Minimum intrinsic stand-off ratio for UJT\n",
-      "V_D=0.7;                #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_P_max=eta_max*V_BB+V_D;           #Maximum peak-point, V\n",
-      "V_P_min=eta_min*V_BB+V_D;           #Minimum peak-point, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum peak-point voltage=%.1fV\"%V_P_max);\n",
-      "print(\"Minimum peak-point voltage=%.1fV\"%V_P_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum peak-point voltage=22.2V\n",
-        "Minimum peak-point voltage=19.2V\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.10 : Page number 593-594\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "eta=0.65;                        #Intrinsic stand-off ratio for UJT\n",
-      "R_BB=7.0;                          #Inter-base resistance, k\u03a9\n",
-      "R1=100.0;                          #Resistor R1, \u03a9\n",
-      "R2=400.0;                          #Resistor R2, \u03a9\n",
-      "V_S=12.0;                          #Source voltage, V\n",
-      "V_D=0.7;                         #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, eta=RB1/RBB,\n",
-      "R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, k\u03a9\n",
-      "R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, k\u03a9 \n",
-      "\n",
-      "print(\"(i)  Resistance of the bar between B1 and emitter junction=%.2f k\u03a9.\"%R_B1);\n",
-      "print(\"     Resistance of the bar between B2 and emitter junction=%.2f k\u03a9.\"%R_B2);\n",
-      "\n",
-      "#(ii)\n",
-      "V_B2_B1=V_S*R_BB/(R_BB + (R1/1000) + (R2/1000));          #Voltage  across B2-B1, V (voltage divider rule)\n",
-      "V_P=eta*V_B2_B1+V_D;                        #Peak-point voltage, V\n",
-      "\n",
-      "print(\"(ii) The voltage across the base B2-B1=%.1fV.\"%V_B2_B1);\n",
-      "print(\"     Peak-point voltage=%.2fV\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Resistance of the bar between B1 and emitter junction=4.55 k\u03a9.\n",
-        "     Resistance of the bar between B2 and emitter junction=2.45 k\u03a9.\n",
-        "(ii) The voltage across the base B2-B1=11.2V.\n",
-        "     Peak-point voltage=7.98V\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.11 : Page number  596\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "RE_initial=5;                   #Initial value of emitter resistor, k\u03a9\n",
-      "RE_adjusted=10;                 #Adjusted value of emitter resistor, k\u03a9\n",
-      "C=0.2;                          #Capacitance, \u03bcF\n",
-      "eta=0.54;                         #intrinsic stand-off ratio\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "t=round((RE_initial*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                             #frequency, Hz\n",
-      "\n",
-      "print(\"Frequency for 5k\u03a9 setting=%dHz.\"%f);\n",
-      "\n",
-      "#(i)\n",
-      "t=round((RE_adjusted*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                             #frequency, Hz\n",
-      "\n",
-      "print(\"Frequency for 10k\u03a9 setting=%dHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Frequency for 5k\u03a9 setting=1282Hz.\n",
-        "Frequency for 10k\u03a9 setting=645Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.12 : Page number 596-597\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "V_S=12;                         #Supply voltage, V\n",
-      "R_BB=5;                         #Interbase resistance, k\u03a9\n",
-      "R_1=50;                         #Resistor R1, k\u03a9\n",
-      "R_2=0.1;                        #Resistor R2, k\u03a9\n",
-      "C=0.1;                          #Capacitance, \u03bcF\n",
-      "eta=0.6;                          #intrinsic stand-off ratio\n",
-      "V_D=0.7;                        #Voltage drop across pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, \u03b7=R_B1/R_BB,\n",
-      "R_B1=eta*R_BB;                            #Resitance between base B1 and emitter junction, k\u03a9\n",
-      "\n",
-      "#Since, R_BB=R_B1+R_B2\n",
-      "R_B2=R_BB-R_B1;                         #Resitance between base B2 and emitter junction, k\u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "V_RB1_R2=V_S*(R_B1+R_2)/(R_BB+R_2);           #Voltage drop across R_B1 and R_2 resistors, V\n",
-      "V_P=V_D+V_RB1_R2;                           #Peak-point voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "t=round((R_1*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                                    #frequency, Hz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   R_B1=%dk\u03a9 and R_B2=%dk\u03a9\"%(R_B1,R_B2));\n",
-      "print(\"(ii)  The peak-point voltage to turn on the UJT=%.0fV.\"%V_P);\n",
-      "print(\"(iii) Frequency of oscillations=%dHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   R_B1=3k\u03a9 and R_B2=2k\u03a9\n",
-        "(ii)  The peak-point voltage to turn on the UJT=8V.\n",
-        "(iii) Frequency of oscillations=218Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_5.ipynb
deleted file mode 100755
index acca0cfa..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_5.ipynb
+++ /dev/null
@@ -1,467 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:412bf04e25192c77f9fa9664d995cc0ae6446a81f631fb5e0e755ebfa36436bf"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 21 : POWER ELECTRONICS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.3: Page number 585\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GT=2;                 #Gate triggering voltage, V\n",
-      "V_F=0.7;               #Forward voltage for diode D1\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)Triggering only by a positive gate voltage,\n",
-      "#A diode is connected at the gatewith the n-side connected to thegate of the device,\n",
-      "V_A=V_F+V_GT;                   #Required voltage to trigger the device, V\n",
-      "\n",
-      "print(\"The  required voltage to trigger the device only by positive voltage=%.1fV.\"%V_A);\n",
-      "\n",
-      "#(ii)\n",
-      "print(\"In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The  required voltage to trigger the device only by positive voltage=2.7V.\n",
-        "In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.4 : Page number 585-586\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=50.0;                   #Resitor, \u03a9\n",
-      "V=50.0;                   #Supply voltage, V\n",
-      "V_drop=1.0;               #Drop across the triac in conduction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) Ideal triac\n",
-      "#Since the triac is ideal, voltage drop across it is zero,\n",
-      "I=V/R;                  #Current through the 50 \u03a9 resistor, A\n",
-      "\n",
-      "print(\"(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=%dA.\"%I);\n",
-      "\n",
-      "#(ii) Triac has a drop of 1V\n",
-      "I=(V-V_drop)/R;                     #Current through the 50 \u03a9 resistor, A\n",
-      "\n",
-      "print(\"(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=%.2fA.\"%I);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=1A.\n",
-        "(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=0.98A.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.5 : Page number 588-589\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_GT=2;             #Gate triggering voltage, V\n",
-      "V_BO=20;            #Breakover voltage,V\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"The triggering level is raised by using a diac.\");\n",
-      "V_A=V_BO+V_GT;                  #Gate trigger signal, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"In order to turn on the triac, the gate trigger signal=%dV.\"%V_A);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The triggering level is raised by using a diac.\n",
-        "In order to turn on the triac, the gate trigger signal=22V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.6 : Page number 589\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BO=30;            #Breakover voltage of diac, V\n",
-      "V_GT=1;             #Trigger voltage of the triac, V\n",
-      "I_T=10;             #Trigger current, mA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_A=V_BO+V_GT;              #Voltage required for triggering the triac, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The minimum capacitor voltage that will trigger the triac=%d V.\"%V_A);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum capacitor voltage that will trigger the triac=31 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.7 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "eta=0.6;                        #Intrinsic stand-off ratio for UJT\n",
-      "R_BB=10;                        #Inter-base resistance, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, RBB=RB1+RB2 and eta=RB1/(RB1+RB2),\n",
-      "#eta=RB1/RBB.\n",
-      "R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, k\u03a9\n",
-      "R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, k\u03a9 \n",
-      "\n",
-      "#Result\n",
-      "print(\"Resistance of the bar between B1 and emitter junction=%d k\u03a9.\"%R_B1);\n",
-      "print(\"Resistance of the bar between B2 and emitter junction=%d k\u03a9.\"%R_B2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Resistance of the bar between B1 and emitter junction=6 k\u03a9.\n",
-        "Resistance of the bar between B2 and emitter junction=4 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.8 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BB=10;                #Interbase voltage, V\n",
-      "eta=0.65;                #Intrinsic stand-off ratio for UJT\n",
-      "V_D=0.7;                #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_stand_off=eta*V_BB;               #Stand off voltage, V\n",
-      "V_P=V_stand_off+V_D;                #Peak-point voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Stand off voltage=%.1f V.\"%V_stand_off);\n",
-      "print(\"Peak-point voltage=%.1f V.\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Stand off voltage=6.5 V.\n",
-        "Peak-point voltage=7.2 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.9 : Page number 593\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BB=25;                #Interbase voltage, V\n",
-      "eta_max=0.86;           #Maximum intrinsic stand-off ratio for UJT\n",
-      "eta_min=0.74;           #Minimum intrinsic stand-off ratio for UJT\n",
-      "V_D=0.7;                #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_P_max=eta_max*V_BB+V_D;           #Maximum peak-point, V\n",
-      "V_P_min=eta_min*V_BB+V_D;           #Minimum peak-point, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum peak-point voltage=%.1fV\"%V_P_max);\n",
-      "print(\"Minimum peak-point voltage=%.1fV\"%V_P_min);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum peak-point voltage=22.2V\n",
-        "Minimum peak-point voltage=19.2V\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.10 : Page number 593-594\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "eta=0.65;                        #Intrinsic stand-off ratio for UJT\n",
-      "R_BB=7.0;                          #Inter-base resistance, k\u03a9\n",
-      "R1=100.0;                          #Resistor R1, \u03a9\n",
-      "R2=400.0;                          #Resistor R2, \u03a9\n",
-      "V_S=12.0;                          #Source voltage, V\n",
-      "V_D=0.7;                         #Voltage drop in the pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, eta=RB1/RBB,\n",
-      "R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, k\u03a9\n",
-      "R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, k\u03a9 \n",
-      "\n",
-      "print(\"(i)  Resistance of the bar between B1 and emitter junction=%.2f k\u03a9.\"%R_B1);\n",
-      "print(\"     Resistance of the bar between B2 and emitter junction=%.2f k\u03a9.\"%R_B2);\n",
-      "\n",
-      "#(ii)\n",
-      "V_B2_B1=V_S*R_BB/(R_BB + (R1/1000) + (R2/1000));          #Voltage  across B2-B1, V (voltage divider rule)\n",
-      "V_P=eta*V_B2_B1+V_D;                        #Peak-point voltage, V\n",
-      "\n",
-      "print(\"(ii) The voltage across the base B2-B1=%.1fV.\"%V_B2_B1);\n",
-      "print(\"     Peak-point voltage=%.2fV\"%V_P);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Resistance of the bar between B1 and emitter junction=4.55 k\u03a9.\n",
-        "     Resistance of the bar between B2 and emitter junction=2.45 k\u03a9.\n",
-        "(ii) The voltage across the base B2-B1=11.2V.\n",
-        "     Peak-point voltage=7.98V\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.11 : Page number  596\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "RE_initial=5;                   #Initial value of emitter resistor, k\u03a9\n",
-      "RE_adjusted=10;                 #Adjusted value of emitter resistor, k\u03a9\n",
-      "C=0.2;                          #Capacitance, \u03bcF\n",
-      "eta=0.54;                         #intrinsic stand-off ratio\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "t=round((RE_initial*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                             #frequency, Hz\n",
-      "\n",
-      "print(\"Frequency for 5k\u03a9 setting=%dHz.\"%f);\n",
-      "\n",
-      "#(i)\n",
-      "t=round((RE_adjusted*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                             #frequency, Hz\n",
-      "\n",
-      "print(\"Frequency for 10k\u03a9 setting=%dHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Frequency for 5k\u03a9 setting=1282Hz.\n",
-        "Frequency for 10k\u03a9 setting=645Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 21.12 : Page number 596-597\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "V_S=12;                         #Supply voltage, V\n",
-      "R_BB=5;                         #Interbase resistance, k\u03a9\n",
-      "R_1=50;                         #Resistor R1, k\u03a9\n",
-      "R_2=0.1;                        #Resistor R2, k\u03a9\n",
-      "C=0.1;                          #Capacitance, \u03bcF\n",
-      "eta=0.6;                          #intrinsic stand-off ratio\n",
-      "V_D=0.7;                        #Voltage drop across pn junction, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "#Since, \u03b7=R_B1/R_BB,\n",
-      "R_B1=eta*R_BB;                            #Resitance between base B1 and emitter junction, k\u03a9\n",
-      "\n",
-      "#Since, R_BB=R_B1+R_B2\n",
-      "R_B2=R_BB-R_B1;                         #Resitance between base B2 and emitter junction, k\u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "V_RB1_R2=V_S*(R_B1+R_2)/(R_BB+R_2);           #Voltage drop across R_B1 and R_2 resistors, V\n",
-      "V_P=V_D+V_RB1_R2;                           #Peak-point voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "t=round((R_1*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms\n",
-      "f=(1/t)*1000;                                                    #frequency, Hz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   R_B1=%dk\u03a9 and R_B2=%dk\u03a9\"%(R_B1,R_B2));\n",
-      "print(\"(ii)  The peak-point voltage to turn on the UJT=%.0fV.\"%V_P);\n",
-      "print(\"(iii) Frequency of oscillations=%dHz.\"%f);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   R_B1=3k\u03a9 and R_B2=2k\u03a9\n",
-        "(ii)  The peak-point voltage to turn on the UJT=8V.\n",
-        "(iii) Frequency of oscillations=218Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22.ipynb
deleted file mode 100755
index 5f13ea0e..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22.ipynb
+++ /dev/null
@@ -1,668 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a688629536ad6915939234eacc1ed3eaaf36e0aaa88de5df5b6a309da4d2c64d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 22: ELECTRONIC INSTRUMENTS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.1 : Page number 606\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_g=1;                  #Full scale deflection current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "MS=1/(I_g/1000.0);        #Multimeter sensitivity, \u03a9 per volt\n",
-      "\n",
-      "#Result\n",
-      "print(\"The multimeter sensitivity=%d \u03a9 per volt.\"%MS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The multimeter sensitivity=1000 \u03a9 per volt.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.2 : Page number 606-607\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=1000.0;                  #Meter sensitivity, \u03a9 per volt\n",
-      "V_full_scale=50.0;                        #Full scale volts\n",
-      "R=50000.0;                                #Resistance to be measured, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "meter_resistance=V_full_scale*meter_sensitivity;            #Meter resistance, \u03a9\n",
-      "R_p=R*meter_resistance/(R+meter_resistance);                #Parallel resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"When the meter is used to measure the voltage across the resistance %d\u03a9, total resistance =%d\u03a9.\"%(R,R_p));\n",
-      "print(\"\u2234 Meter will give highly incorrect reading.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "When the meter is used to measure the voltage across the resistance 50000\u03a9, total resistance =25000\u03a9.\n",
-        "\u2234 Meter will give highly incorrect reading.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.3 : Page number 607\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=4.0;                #Meter sensitivity, k\u03a9/V\n",
-      "R_1=10.0;                             #Resistance across which voltage is to be measured, k\u03a9\n",
-      "R_2=10.0;                             #Resistance, k\u03a9\n",
-      "range_max=10.0;                       #Maximum range of the meter, V\n",
-      "range_min=0;                        #Minimum range of the meter, V\n",
-      "V=20.0;                               #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "R_meter=meter_sensitivity*range_max;                       #Resistance of the meter, k\u03a9\n",
-      "R_T=(R_meter*R_1)/(R_1+R_meter) +  R_2;                     #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T,2);                                            #Circuit current, mA\n",
-      "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter));       #Voltage read by multimeter, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage read by multimeter=8.88V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.4 : Page number 607-608\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=20.0;                #Meter sensitivity, k\u03a9/V\n",
-      "R_1=10.0;                             #Resistance across which voltage is to be measured, k\u03a9\n",
-      "R_2=10.0;                             #Resistance, k\u03a9\n",
-      "range_max=10.0;                       #Maximum range of the meter, V\n",
-      "range_min=0;                        #Minimum range of the meter, V\n",
-      "V=20.0;                               #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "R_meter=meter_sensitivity*range_max;                       #Resistance of the meter, k\u03a9\n",
-      "R_T=round((R_meter*R_1)/(R_1+R_meter) +  R_2,1);                     #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T,2);                                           #Circuit current, mA\n",
-      "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter));       #Voltage read by multimeter, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n",
-      "\n",
-      "\n",
-      "#Note: The circuit current=1.0256mA, has been approximated in the text  as 1.04mA. But, in the code 1.03 mA has been used. Therefore, the final answer is obtained as 9.81V and not 9.88V.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage read by multimeter=9.81V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.5 : Page number 608-609\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "R_A=20.0;             #Resistance after point A, k\u03a9\n",
-      "R_B=20.0;             #Resistance after point B, k\u03a9\n",
-      "R_C=30.0;             #Resistance after point C, k\u03a9\n",
-      "R_D=30.0;             #Resistance after point D, k\u03a9\n",
-      "R_meter=60.0;         #Resistance of the meter, k\u03a9\n",
-      "V=100.0;              #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) When meter is not connected:\n",
-      "R_T=R_A+R_B+R_C+R_D;                    #Total circuit resistance, k\u03a9\n",
-      "I_circuit=V/R_T;                        #Circuit current, mA\n",
-      "V_A=V;                                  #Voltage at point A, V\n",
-      "V_B=V-(I_circuit*R_A);                  #Voltage at point B, V\n",
-      "V_C=V-(I_circuit*(R_A+R_B));            #Voltage at point C, V\n",
-      "V_D=V-(I_circuit*(R_T-R_D));            #Voltage at point D, V\n",
-      "\n",
-      "print(\"(i) When meter is not connected:\");\n",
-      "print(\"    Voltage at point A=%dV.\"%V_A);\n",
-      "print(\"    Voltage at point B=%dV.\"%V_B);\n",
-      "print(\"    Voltage at point C=%dV.\"%V_C);\n",
-      "print(\"    Voltage at point D=%dV.\"%V_D);\n",
-      "\n",
-      "\n",
-      "#(ii) When meter is connected:\n",
-      "#(a) Since, point A is directly connected to the source, voltage at point A is equal to source voltage.\n",
-      "V_A=V;                                      #Voltage at point A, V\n",
-      "\n",
-      "#(b)\n",
-      "R_T_B=R_A  + round((R_T-R_A)*R_meter/(R_meter + (R_T-R_A)),2);             #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T_B,2);                                                #Circuit current, mA\n",
-      "V_B=I_circuit*(R_T-R_A)*R_meter/(R_meter + (R_T-R_A));                   #Voltage at point B, V\n",
-      "\n",
-      "\n",
-      "#(c)\n",
-      "R_T_C=(R_A+R_B)  + (R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B));               #Total circuit resistance, k\u03a9\n",
-      "I_circuit=V/R_T_C;                                                                #Circuit current, mA\n",
-      "V_C=floor((I_circuit*(R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B)))*10)/10;     #Voltage at point C, V\n",
-      "\n",
-      "\n",
-      "\n",
-      "#(c)\n",
-      "R_T_D=(R_T-R_D)  + R_D*R_meter/(R_meter + R_D);             #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T_D,2);                                 #Circuit current, mA\n",
-      "V_D=I_circuit*(R_D*R_meter)/(R_meter + R_D);                #Voltage at point D, V\n",
-      "\n",
-      "\n",
-      "print(\"(ii) When meter is connected:\");\n",
-      "print(\"     Voltage at point A=%dV.\"%V_A);\n",
-      "print(\"     Voltage at point B=%dV.\"%V_B);\n",
-      "print(\"     Voltage at point C=%.1fV.\"%V_C);\n",
-      "print(\"     Voltage at point D=%.1fV.\"%V_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) When meter is not connected:\n",
-        "    Voltage at point A=100V.\n",
-        "    Voltage at point B=80V.\n",
-        "    Voltage at point C=60V.\n",
-        "    Voltage at point D=30V.\n",
-        "(ii) When meter is connected:\n",
-        "     Voltage at point A=100V.\n",
-        "     Voltage at point B=63V.\n",
-        "     Voltage at point C=42.8V.\n",
-        "     Voltage at point D=22.2V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.6 : Page number 614\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                     #Supply voltage, V\n",
-      "R_m=1.0;                      #Meter resistance, k\u03a9\n",
-      "I_m_fsd=2.0;                  #Full scale deflection of meter current, mA\n",
-      "beta=80.0;                    #Base current amplification factor\n",
-      "E=5.0;                        #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_E=E-V_BE;                 #Emitter voltage, V\n",
-      "\n",
-      "#(i)\n",
-      "#I_m_fsd=V_E/(R_s+R_m), (OHM's LAW)\n",
-      "R_s=((V_E/I_m_fsd)-R_m)*1000;                  #Multiplier resistor, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "IB=I_m_fsd/beta;     \t\t\t\t#Base current, mA\n",
-      "R_i=E/IB;                    \t\t\t#Input resistance of voltmeter, k\u03a9\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The multiplier resistor=%d\u03a9.\"%R_s);\n",
-      "print(\"(ii) The voltmeter input resistance=%dk\u03a9\"%R_i);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The multiplier resistor=1150\u03a9.\n",
-        "(ii) The voltmeter input resistance=200k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.7 : Page number 614\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20;                     #Supply voltage, V\n",
-      "Rs_Rm=9.3;                  #Sum of multipier resistance and meter resistance, k\u03a9\n",
-      "I_m=1;                      #Meter current, mA\n",
-      "beta=100;                   #Base current amplification factor\n",
-      "E=10;                       #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_E=E-V_BE;             #Emitter voltage, V\n",
-      "I_m=V_E/Rs_Rm;          #Meter current, mA\n",
-      "\n",
-      "#(ii)\n",
-      "I_B=I_m/beta;               #Base current, mA\n",
-      "R_i_T=(E/I_B)/1000;         #Input resistance of voltmeter, with transistor, M\u03a9\n",
-      "R_i_WT=Rs_Rm;               #Input resistance of voltmeter, without transistor, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The meter current=%dmA\"%I_m);\n",
-      "print(\"(ii) The input resistance of voltmeter with transistor=%dM\u03a9.\"%R_i_T);\n",
-      "print(\"     The input resistance of voltmeter without transistor=%.1fk\u03a9.\"%R_i_WT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The meter current=1mA\n",
-        "(ii) The input resistance of voltmeter with transistor=1M\u03a9.\n",
-        "     The input resistance of voltmeter without transistor=9.3k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.8 : Page number 614-615\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20;                     #Supply voltage, V\n",
-      "Rs_Rm=9.3;                  #Sum of multipier resistance and meter resistance, k\u03a9\n",
-      "I_m=1;                      #Meter current, mA\n",
-      "beta=100;                   #Base current amplification factor\n",
-      "E=5;                        #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_m=(E-V_BE)/Rs_Rm;          #Meter current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The meter current=%.2fmA\"%I_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The meter current=0.46mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.9 : Page number 616\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_m_fsd=100.0;                  #Full scale deflection of meter current, \u03bcA\n",
-      "R_m=1.0;                        #Meter resistance, k\u03a9\n",
-      "V_rms=100.0;                    #r.m.s voltage to be measured, V\n",
-      "V_F=0.7;                      #Forward voltage drop of rectifier diode, V \n",
-      "\n",
-      "#Calculation\n",
-      "V_m=round(sqrt(2)*V_rms,1);                                                          #Peak value of applied voltage, V\n",
-      "V_rectifier_drop=2*V_F;                                                              #Total rectifier drop, V\n",
-      "I_peak=round(I_m_fsd/0.637,2);                                                       #Peak f.s.d current, \u03bcA\n",
-      "R_s=floor(((((V_m-V_rectifier_drop)/(I_peak*10**-6))-(R_m*1000))/1000)*10)/10;       #Multiplier resistance, k\u03a9 (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The multiplier resistance=%.1fk\u03a9.\"%R_s);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The multiplier resistance=890.7k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.10 : Page number 616\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_av=75;                  #Full scale deflection of meter current, \u03bcA\n",
-      "R_s=708;                     #Multiplier resistor, k\u03a9\n",
-      "R_m=900;                     #Meter coil resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "I_peak=I_av*10**-6/0.637;                   #Peak f.s.d meter current, A\n",
-      "R_T=R_s*1000+R_m;                                #Total circuit resistance, \u03a9\n",
-      "\n",
-      "#I_peak=(Vm-V_drop)/R_T; (OHM's LAW)\n",
-      "#And, Vm=sqrt(2)*Vrms\n",
-      "V_rms=(I_peak*R_T+(2*0.7))/sqrt(2) ;             #applied r.m.s voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The applied r.m.s voltage=%dV\"%V_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The applied r.m.s voltage=60V\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.11 : Page number 618\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection_sensitivity=0.01;            #Deflection sensitivity, mm/V\n",
-      "V=400;                                  #Applied voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "spot_shift=V*deflection_sensitivity;         #Spot shift produced, mm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The shift produced in the spot=%dmm.\"%spot_shift);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The shift produced in the spot=4mm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.12 : Page number 618-619\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection_sensitivity=0.03;            #Deflection sensitivity, mm/V\n",
-      "spot_shift=3;                           #Spot shift produced, mm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, spot_shift=Applied_Voltage*deflection_sensitivity,\n",
-      "V=spot_shift/deflection_sensitivity;    #Applied voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Applied voltage=%dV.\"%V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Applied voltage=100V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.13 : Page number  622\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection=2;                            #Deflection produced by applied voltage, cm\n",
-      "V=200;                                   #Applied voltage, V\n",
-      "deflection_by_another_voltage=3;         #Deflection by another voltage, cm\n",
-      "\n",
-      "#Calculation\n",
-      "deflection_sensitivity=V/deflection;                                #deflection sensitivity, V/cm\n",
-      "V_unknown=deflection_sensitivity*deflection_by_another_voltage;     #Unknown voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The unknown voltage=%dV.\"%V_unknown);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The unknown voltage=300V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.14 : Page number 622\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "f_H=1000;                   #Frequency applied to horizontal plates, Hz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Loops_H=1;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(i)   Unknown frequency=%dHz.\"%f_V);\n",
-      "\n",
-      "#(ii)\n",
-      "Loops_H=2;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(ii)  Unknown frequency=%dHz.\"%f_V);\n",
-      "\n",
-      "#(iii)\n",
-      "Loops_H=6;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(iii) Unknown frequency=%dHz.\"%f_V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Unknown frequency=1000Hz.\n",
-        "(ii)  Unknown frequency=2000Hz.\n",
-        "(iii) Unknown frequency=6000Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_1.ipynb
deleted file mode 100755
index 5f13ea0e..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_1.ipynb
+++ /dev/null
@@ -1,668 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a688629536ad6915939234eacc1ed3eaaf36e0aaa88de5df5b6a309da4d2c64d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 22: ELECTRONIC INSTRUMENTS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.1 : Page number 606\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_g=1;                  #Full scale deflection current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "MS=1/(I_g/1000.0);        #Multimeter sensitivity, \u03a9 per volt\n",
-      "\n",
-      "#Result\n",
-      "print(\"The multimeter sensitivity=%d \u03a9 per volt.\"%MS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The multimeter sensitivity=1000 \u03a9 per volt.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.2 : Page number 606-607\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=1000.0;                  #Meter sensitivity, \u03a9 per volt\n",
-      "V_full_scale=50.0;                        #Full scale volts\n",
-      "R=50000.0;                                #Resistance to be measured, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "meter_resistance=V_full_scale*meter_sensitivity;            #Meter resistance, \u03a9\n",
-      "R_p=R*meter_resistance/(R+meter_resistance);                #Parallel resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"When the meter is used to measure the voltage across the resistance %d\u03a9, total resistance =%d\u03a9.\"%(R,R_p));\n",
-      "print(\"\u2234 Meter will give highly incorrect reading.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "When the meter is used to measure the voltage across the resistance 50000\u03a9, total resistance =25000\u03a9.\n",
-        "\u2234 Meter will give highly incorrect reading.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.3 : Page number 607\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=4.0;                #Meter sensitivity, k\u03a9/V\n",
-      "R_1=10.0;                             #Resistance across which voltage is to be measured, k\u03a9\n",
-      "R_2=10.0;                             #Resistance, k\u03a9\n",
-      "range_max=10.0;                       #Maximum range of the meter, V\n",
-      "range_min=0;                        #Minimum range of the meter, V\n",
-      "V=20.0;                               #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "R_meter=meter_sensitivity*range_max;                       #Resistance of the meter, k\u03a9\n",
-      "R_T=(R_meter*R_1)/(R_1+R_meter) +  R_2;                     #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T,2);                                            #Circuit current, mA\n",
-      "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter));       #Voltage read by multimeter, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage read by multimeter=8.88V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.4 : Page number 607-608\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=20.0;                #Meter sensitivity, k\u03a9/V\n",
-      "R_1=10.0;                             #Resistance across which voltage is to be measured, k\u03a9\n",
-      "R_2=10.0;                             #Resistance, k\u03a9\n",
-      "range_max=10.0;                       #Maximum range of the meter, V\n",
-      "range_min=0;                        #Minimum range of the meter, V\n",
-      "V=20.0;                               #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "R_meter=meter_sensitivity*range_max;                       #Resistance of the meter, k\u03a9\n",
-      "R_T=round((R_meter*R_1)/(R_1+R_meter) +  R_2,1);                     #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T,2);                                           #Circuit current, mA\n",
-      "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter));       #Voltage read by multimeter, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n",
-      "\n",
-      "\n",
-      "#Note: The circuit current=1.0256mA, has been approximated in the text  as 1.04mA. But, in the code 1.03 mA has been used. Therefore, the final answer is obtained as 9.81V and not 9.88V.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage read by multimeter=9.81V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.5 : Page number 608-609\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "R_A=20.0;             #Resistance after point A, k\u03a9\n",
-      "R_B=20.0;             #Resistance after point B, k\u03a9\n",
-      "R_C=30.0;             #Resistance after point C, k\u03a9\n",
-      "R_D=30.0;             #Resistance after point D, k\u03a9\n",
-      "R_meter=60.0;         #Resistance of the meter, k\u03a9\n",
-      "V=100.0;              #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) When meter is not connected:\n",
-      "R_T=R_A+R_B+R_C+R_D;                    #Total circuit resistance, k\u03a9\n",
-      "I_circuit=V/R_T;                        #Circuit current, mA\n",
-      "V_A=V;                                  #Voltage at point A, V\n",
-      "V_B=V-(I_circuit*R_A);                  #Voltage at point B, V\n",
-      "V_C=V-(I_circuit*(R_A+R_B));            #Voltage at point C, V\n",
-      "V_D=V-(I_circuit*(R_T-R_D));            #Voltage at point D, V\n",
-      "\n",
-      "print(\"(i) When meter is not connected:\");\n",
-      "print(\"    Voltage at point A=%dV.\"%V_A);\n",
-      "print(\"    Voltage at point B=%dV.\"%V_B);\n",
-      "print(\"    Voltage at point C=%dV.\"%V_C);\n",
-      "print(\"    Voltage at point D=%dV.\"%V_D);\n",
-      "\n",
-      "\n",
-      "#(ii) When meter is connected:\n",
-      "#(a) Since, point A is directly connected to the source, voltage at point A is equal to source voltage.\n",
-      "V_A=V;                                      #Voltage at point A, V\n",
-      "\n",
-      "#(b)\n",
-      "R_T_B=R_A  + round((R_T-R_A)*R_meter/(R_meter + (R_T-R_A)),2);             #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T_B,2);                                                #Circuit current, mA\n",
-      "V_B=I_circuit*(R_T-R_A)*R_meter/(R_meter + (R_T-R_A));                   #Voltage at point B, V\n",
-      "\n",
-      "\n",
-      "#(c)\n",
-      "R_T_C=(R_A+R_B)  + (R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B));               #Total circuit resistance, k\u03a9\n",
-      "I_circuit=V/R_T_C;                                                                #Circuit current, mA\n",
-      "V_C=floor((I_circuit*(R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B)))*10)/10;     #Voltage at point C, V\n",
-      "\n",
-      "\n",
-      "\n",
-      "#(c)\n",
-      "R_T_D=(R_T-R_D)  + R_D*R_meter/(R_meter + R_D);             #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T_D,2);                                 #Circuit current, mA\n",
-      "V_D=I_circuit*(R_D*R_meter)/(R_meter + R_D);                #Voltage at point D, V\n",
-      "\n",
-      "\n",
-      "print(\"(ii) When meter is connected:\");\n",
-      "print(\"     Voltage at point A=%dV.\"%V_A);\n",
-      "print(\"     Voltage at point B=%dV.\"%V_B);\n",
-      "print(\"     Voltage at point C=%.1fV.\"%V_C);\n",
-      "print(\"     Voltage at point D=%.1fV.\"%V_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) When meter is not connected:\n",
-        "    Voltage at point A=100V.\n",
-        "    Voltage at point B=80V.\n",
-        "    Voltage at point C=60V.\n",
-        "    Voltage at point D=30V.\n",
-        "(ii) When meter is connected:\n",
-        "     Voltage at point A=100V.\n",
-        "     Voltage at point B=63V.\n",
-        "     Voltage at point C=42.8V.\n",
-        "     Voltage at point D=22.2V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.6 : Page number 614\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                     #Supply voltage, V\n",
-      "R_m=1.0;                      #Meter resistance, k\u03a9\n",
-      "I_m_fsd=2.0;                  #Full scale deflection of meter current, mA\n",
-      "beta=80.0;                    #Base current amplification factor\n",
-      "E=5.0;                        #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_E=E-V_BE;                 #Emitter voltage, V\n",
-      "\n",
-      "#(i)\n",
-      "#I_m_fsd=V_E/(R_s+R_m), (OHM's LAW)\n",
-      "R_s=((V_E/I_m_fsd)-R_m)*1000;                  #Multiplier resistor, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "IB=I_m_fsd/beta;     \t\t\t\t#Base current, mA\n",
-      "R_i=E/IB;                    \t\t\t#Input resistance of voltmeter, k\u03a9\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The multiplier resistor=%d\u03a9.\"%R_s);\n",
-      "print(\"(ii) The voltmeter input resistance=%dk\u03a9\"%R_i);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The multiplier resistor=1150\u03a9.\n",
-        "(ii) The voltmeter input resistance=200k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.7 : Page number 614\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20;                     #Supply voltage, V\n",
-      "Rs_Rm=9.3;                  #Sum of multipier resistance and meter resistance, k\u03a9\n",
-      "I_m=1;                      #Meter current, mA\n",
-      "beta=100;                   #Base current amplification factor\n",
-      "E=10;                       #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_E=E-V_BE;             #Emitter voltage, V\n",
-      "I_m=V_E/Rs_Rm;          #Meter current, mA\n",
-      "\n",
-      "#(ii)\n",
-      "I_B=I_m/beta;               #Base current, mA\n",
-      "R_i_T=(E/I_B)/1000;         #Input resistance of voltmeter, with transistor, M\u03a9\n",
-      "R_i_WT=Rs_Rm;               #Input resistance of voltmeter, without transistor, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The meter current=%dmA\"%I_m);\n",
-      "print(\"(ii) The input resistance of voltmeter with transistor=%dM\u03a9.\"%R_i_T);\n",
-      "print(\"     The input resistance of voltmeter without transistor=%.1fk\u03a9.\"%R_i_WT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The meter current=1mA\n",
-        "(ii) The input resistance of voltmeter with transistor=1M\u03a9.\n",
-        "     The input resistance of voltmeter without transistor=9.3k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.8 : Page number 614-615\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20;                     #Supply voltage, V\n",
-      "Rs_Rm=9.3;                  #Sum of multipier resistance and meter resistance, k\u03a9\n",
-      "I_m=1;                      #Meter current, mA\n",
-      "beta=100;                   #Base current amplification factor\n",
-      "E=5;                        #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_m=(E-V_BE)/Rs_Rm;          #Meter current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The meter current=%.2fmA\"%I_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The meter current=0.46mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.9 : Page number 616\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_m_fsd=100.0;                  #Full scale deflection of meter current, \u03bcA\n",
-      "R_m=1.0;                        #Meter resistance, k\u03a9\n",
-      "V_rms=100.0;                    #r.m.s voltage to be measured, V\n",
-      "V_F=0.7;                      #Forward voltage drop of rectifier diode, V \n",
-      "\n",
-      "#Calculation\n",
-      "V_m=round(sqrt(2)*V_rms,1);                                                          #Peak value of applied voltage, V\n",
-      "V_rectifier_drop=2*V_F;                                                              #Total rectifier drop, V\n",
-      "I_peak=round(I_m_fsd/0.637,2);                                                       #Peak f.s.d current, \u03bcA\n",
-      "R_s=floor(((((V_m-V_rectifier_drop)/(I_peak*10**-6))-(R_m*1000))/1000)*10)/10;       #Multiplier resistance, k\u03a9 (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The multiplier resistance=%.1fk\u03a9.\"%R_s);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The multiplier resistance=890.7k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.10 : Page number 616\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_av=75;                  #Full scale deflection of meter current, \u03bcA\n",
-      "R_s=708;                     #Multiplier resistor, k\u03a9\n",
-      "R_m=900;                     #Meter coil resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "I_peak=I_av*10**-6/0.637;                   #Peak f.s.d meter current, A\n",
-      "R_T=R_s*1000+R_m;                                #Total circuit resistance, \u03a9\n",
-      "\n",
-      "#I_peak=(Vm-V_drop)/R_T; (OHM's LAW)\n",
-      "#And, Vm=sqrt(2)*Vrms\n",
-      "V_rms=(I_peak*R_T+(2*0.7))/sqrt(2) ;             #applied r.m.s voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The applied r.m.s voltage=%dV\"%V_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The applied r.m.s voltage=60V\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.11 : Page number 618\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection_sensitivity=0.01;            #Deflection sensitivity, mm/V\n",
-      "V=400;                                  #Applied voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "spot_shift=V*deflection_sensitivity;         #Spot shift produced, mm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The shift produced in the spot=%dmm.\"%spot_shift);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The shift produced in the spot=4mm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.12 : Page number 618-619\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection_sensitivity=0.03;            #Deflection sensitivity, mm/V\n",
-      "spot_shift=3;                           #Spot shift produced, mm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, spot_shift=Applied_Voltage*deflection_sensitivity,\n",
-      "V=spot_shift/deflection_sensitivity;    #Applied voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Applied voltage=%dV.\"%V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Applied voltage=100V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.13 : Page number  622\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection=2;                            #Deflection produced by applied voltage, cm\n",
-      "V=200;                                   #Applied voltage, V\n",
-      "deflection_by_another_voltage=3;         #Deflection by another voltage, cm\n",
-      "\n",
-      "#Calculation\n",
-      "deflection_sensitivity=V/deflection;                                #deflection sensitivity, V/cm\n",
-      "V_unknown=deflection_sensitivity*deflection_by_another_voltage;     #Unknown voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The unknown voltage=%dV.\"%V_unknown);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The unknown voltage=300V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.14 : Page number 622\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "f_H=1000;                   #Frequency applied to horizontal plates, Hz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Loops_H=1;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(i)   Unknown frequency=%dHz.\"%f_V);\n",
-      "\n",
-      "#(ii)\n",
-      "Loops_H=2;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(ii)  Unknown frequency=%dHz.\"%f_V);\n",
-      "\n",
-      "#(iii)\n",
-      "Loops_H=6;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(iii) Unknown frequency=%dHz.\"%f_V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Unknown frequency=1000Hz.\n",
-        "(ii)  Unknown frequency=2000Hz.\n",
-        "(iii) Unknown frequency=6000Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_2.ipynb
deleted file mode 100755
index 5f13ea0e..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_2.ipynb
+++ /dev/null
@@ -1,668 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a688629536ad6915939234eacc1ed3eaaf36e0aaa88de5df5b6a309da4d2c64d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 22: ELECTRONIC INSTRUMENTS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.1 : Page number 606\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_g=1;                  #Full scale deflection current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "MS=1/(I_g/1000.0);        #Multimeter sensitivity, \u03a9 per volt\n",
-      "\n",
-      "#Result\n",
-      "print(\"The multimeter sensitivity=%d \u03a9 per volt.\"%MS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The multimeter sensitivity=1000 \u03a9 per volt.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.2 : Page number 606-607\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=1000.0;                  #Meter sensitivity, \u03a9 per volt\n",
-      "V_full_scale=50.0;                        #Full scale volts\n",
-      "R=50000.0;                                #Resistance to be measured, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "meter_resistance=V_full_scale*meter_sensitivity;            #Meter resistance, \u03a9\n",
-      "R_p=R*meter_resistance/(R+meter_resistance);                #Parallel resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"When the meter is used to measure the voltage across the resistance %d\u03a9, total resistance =%d\u03a9.\"%(R,R_p));\n",
-      "print(\"\u2234 Meter will give highly incorrect reading.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "When the meter is used to measure the voltage across the resistance 50000\u03a9, total resistance =25000\u03a9.\n",
-        "\u2234 Meter will give highly incorrect reading.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.3 : Page number 607\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=4.0;                #Meter sensitivity, k\u03a9/V\n",
-      "R_1=10.0;                             #Resistance across which voltage is to be measured, k\u03a9\n",
-      "R_2=10.0;                             #Resistance, k\u03a9\n",
-      "range_max=10.0;                       #Maximum range of the meter, V\n",
-      "range_min=0;                        #Minimum range of the meter, V\n",
-      "V=20.0;                               #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "R_meter=meter_sensitivity*range_max;                       #Resistance of the meter, k\u03a9\n",
-      "R_T=(R_meter*R_1)/(R_1+R_meter) +  R_2;                     #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T,2);                                            #Circuit current, mA\n",
-      "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter));       #Voltage read by multimeter, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage read by multimeter=8.88V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.4 : Page number 607-608\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=20.0;                #Meter sensitivity, k\u03a9/V\n",
-      "R_1=10.0;                             #Resistance across which voltage is to be measured, k\u03a9\n",
-      "R_2=10.0;                             #Resistance, k\u03a9\n",
-      "range_max=10.0;                       #Maximum range of the meter, V\n",
-      "range_min=0;                        #Minimum range of the meter, V\n",
-      "V=20.0;                               #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "R_meter=meter_sensitivity*range_max;                       #Resistance of the meter, k\u03a9\n",
-      "R_T=round((R_meter*R_1)/(R_1+R_meter) +  R_2,1);                     #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T,2);                                           #Circuit current, mA\n",
-      "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter));       #Voltage read by multimeter, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n",
-      "\n",
-      "\n",
-      "#Note: The circuit current=1.0256mA, has been approximated in the text  as 1.04mA. But, in the code 1.03 mA has been used. Therefore, the final answer is obtained as 9.81V and not 9.88V.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage read by multimeter=9.81V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.5 : Page number 608-609\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "R_A=20.0;             #Resistance after point A, k\u03a9\n",
-      "R_B=20.0;             #Resistance after point B, k\u03a9\n",
-      "R_C=30.0;             #Resistance after point C, k\u03a9\n",
-      "R_D=30.0;             #Resistance after point D, k\u03a9\n",
-      "R_meter=60.0;         #Resistance of the meter, k\u03a9\n",
-      "V=100.0;              #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) When meter is not connected:\n",
-      "R_T=R_A+R_B+R_C+R_D;                    #Total circuit resistance, k\u03a9\n",
-      "I_circuit=V/R_T;                        #Circuit current, mA\n",
-      "V_A=V;                                  #Voltage at point A, V\n",
-      "V_B=V-(I_circuit*R_A);                  #Voltage at point B, V\n",
-      "V_C=V-(I_circuit*(R_A+R_B));            #Voltage at point C, V\n",
-      "V_D=V-(I_circuit*(R_T-R_D));            #Voltage at point D, V\n",
-      "\n",
-      "print(\"(i) When meter is not connected:\");\n",
-      "print(\"    Voltage at point A=%dV.\"%V_A);\n",
-      "print(\"    Voltage at point B=%dV.\"%V_B);\n",
-      "print(\"    Voltage at point C=%dV.\"%V_C);\n",
-      "print(\"    Voltage at point D=%dV.\"%V_D);\n",
-      "\n",
-      "\n",
-      "#(ii) When meter is connected:\n",
-      "#(a) Since, point A is directly connected to the source, voltage at point A is equal to source voltage.\n",
-      "V_A=V;                                      #Voltage at point A, V\n",
-      "\n",
-      "#(b)\n",
-      "R_T_B=R_A  + round((R_T-R_A)*R_meter/(R_meter + (R_T-R_A)),2);             #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T_B,2);                                                #Circuit current, mA\n",
-      "V_B=I_circuit*(R_T-R_A)*R_meter/(R_meter + (R_T-R_A));                   #Voltage at point B, V\n",
-      "\n",
-      "\n",
-      "#(c)\n",
-      "R_T_C=(R_A+R_B)  + (R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B));               #Total circuit resistance, k\u03a9\n",
-      "I_circuit=V/R_T_C;                                                                #Circuit current, mA\n",
-      "V_C=floor((I_circuit*(R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B)))*10)/10;     #Voltage at point C, V\n",
-      "\n",
-      "\n",
-      "\n",
-      "#(c)\n",
-      "R_T_D=(R_T-R_D)  + R_D*R_meter/(R_meter + R_D);             #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T_D,2);                                 #Circuit current, mA\n",
-      "V_D=I_circuit*(R_D*R_meter)/(R_meter + R_D);                #Voltage at point D, V\n",
-      "\n",
-      "\n",
-      "print(\"(ii) When meter is connected:\");\n",
-      "print(\"     Voltage at point A=%dV.\"%V_A);\n",
-      "print(\"     Voltage at point B=%dV.\"%V_B);\n",
-      "print(\"     Voltage at point C=%.1fV.\"%V_C);\n",
-      "print(\"     Voltage at point D=%.1fV.\"%V_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) When meter is not connected:\n",
-        "    Voltage at point A=100V.\n",
-        "    Voltage at point B=80V.\n",
-        "    Voltage at point C=60V.\n",
-        "    Voltage at point D=30V.\n",
-        "(ii) When meter is connected:\n",
-        "     Voltage at point A=100V.\n",
-        "     Voltage at point B=63V.\n",
-        "     Voltage at point C=42.8V.\n",
-        "     Voltage at point D=22.2V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.6 : Page number 614\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                     #Supply voltage, V\n",
-      "R_m=1.0;                      #Meter resistance, k\u03a9\n",
-      "I_m_fsd=2.0;                  #Full scale deflection of meter current, mA\n",
-      "beta=80.0;                    #Base current amplification factor\n",
-      "E=5.0;                        #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_E=E-V_BE;                 #Emitter voltage, V\n",
-      "\n",
-      "#(i)\n",
-      "#I_m_fsd=V_E/(R_s+R_m), (OHM's LAW)\n",
-      "R_s=((V_E/I_m_fsd)-R_m)*1000;                  #Multiplier resistor, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "IB=I_m_fsd/beta;     \t\t\t\t#Base current, mA\n",
-      "R_i=E/IB;                    \t\t\t#Input resistance of voltmeter, k\u03a9\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The multiplier resistor=%d\u03a9.\"%R_s);\n",
-      "print(\"(ii) The voltmeter input resistance=%dk\u03a9\"%R_i);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The multiplier resistor=1150\u03a9.\n",
-        "(ii) The voltmeter input resistance=200k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.7 : Page number 614\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20;                     #Supply voltage, V\n",
-      "Rs_Rm=9.3;                  #Sum of multipier resistance and meter resistance, k\u03a9\n",
-      "I_m=1;                      #Meter current, mA\n",
-      "beta=100;                   #Base current amplification factor\n",
-      "E=10;                       #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_E=E-V_BE;             #Emitter voltage, V\n",
-      "I_m=V_E/Rs_Rm;          #Meter current, mA\n",
-      "\n",
-      "#(ii)\n",
-      "I_B=I_m/beta;               #Base current, mA\n",
-      "R_i_T=(E/I_B)/1000;         #Input resistance of voltmeter, with transistor, M\u03a9\n",
-      "R_i_WT=Rs_Rm;               #Input resistance of voltmeter, without transistor, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The meter current=%dmA\"%I_m);\n",
-      "print(\"(ii) The input resistance of voltmeter with transistor=%dM\u03a9.\"%R_i_T);\n",
-      "print(\"     The input resistance of voltmeter without transistor=%.1fk\u03a9.\"%R_i_WT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The meter current=1mA\n",
-        "(ii) The input resistance of voltmeter with transistor=1M\u03a9.\n",
-        "     The input resistance of voltmeter without transistor=9.3k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.8 : Page number 614-615\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20;                     #Supply voltage, V\n",
-      "Rs_Rm=9.3;                  #Sum of multipier resistance and meter resistance, k\u03a9\n",
-      "I_m=1;                      #Meter current, mA\n",
-      "beta=100;                   #Base current amplification factor\n",
-      "E=5;                        #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_m=(E-V_BE)/Rs_Rm;          #Meter current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The meter current=%.2fmA\"%I_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The meter current=0.46mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.9 : Page number 616\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_m_fsd=100.0;                  #Full scale deflection of meter current, \u03bcA\n",
-      "R_m=1.0;                        #Meter resistance, k\u03a9\n",
-      "V_rms=100.0;                    #r.m.s voltage to be measured, V\n",
-      "V_F=0.7;                      #Forward voltage drop of rectifier diode, V \n",
-      "\n",
-      "#Calculation\n",
-      "V_m=round(sqrt(2)*V_rms,1);                                                          #Peak value of applied voltage, V\n",
-      "V_rectifier_drop=2*V_F;                                                              #Total rectifier drop, V\n",
-      "I_peak=round(I_m_fsd/0.637,2);                                                       #Peak f.s.d current, \u03bcA\n",
-      "R_s=floor(((((V_m-V_rectifier_drop)/(I_peak*10**-6))-(R_m*1000))/1000)*10)/10;       #Multiplier resistance, k\u03a9 (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The multiplier resistance=%.1fk\u03a9.\"%R_s);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The multiplier resistance=890.7k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.10 : Page number 616\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_av=75;                  #Full scale deflection of meter current, \u03bcA\n",
-      "R_s=708;                     #Multiplier resistor, k\u03a9\n",
-      "R_m=900;                     #Meter coil resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "I_peak=I_av*10**-6/0.637;                   #Peak f.s.d meter current, A\n",
-      "R_T=R_s*1000+R_m;                                #Total circuit resistance, \u03a9\n",
-      "\n",
-      "#I_peak=(Vm-V_drop)/R_T; (OHM's LAW)\n",
-      "#And, Vm=sqrt(2)*Vrms\n",
-      "V_rms=(I_peak*R_T+(2*0.7))/sqrt(2) ;             #applied r.m.s voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The applied r.m.s voltage=%dV\"%V_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The applied r.m.s voltage=60V\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.11 : Page number 618\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection_sensitivity=0.01;            #Deflection sensitivity, mm/V\n",
-      "V=400;                                  #Applied voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "spot_shift=V*deflection_sensitivity;         #Spot shift produced, mm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The shift produced in the spot=%dmm.\"%spot_shift);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The shift produced in the spot=4mm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.12 : Page number 618-619\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection_sensitivity=0.03;            #Deflection sensitivity, mm/V\n",
-      "spot_shift=3;                           #Spot shift produced, mm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, spot_shift=Applied_Voltage*deflection_sensitivity,\n",
-      "V=spot_shift/deflection_sensitivity;    #Applied voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Applied voltage=%dV.\"%V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Applied voltage=100V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.13 : Page number  622\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection=2;                            #Deflection produced by applied voltage, cm\n",
-      "V=200;                                   #Applied voltage, V\n",
-      "deflection_by_another_voltage=3;         #Deflection by another voltage, cm\n",
-      "\n",
-      "#Calculation\n",
-      "deflection_sensitivity=V/deflection;                                #deflection sensitivity, V/cm\n",
-      "V_unknown=deflection_sensitivity*deflection_by_another_voltage;     #Unknown voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The unknown voltage=%dV.\"%V_unknown);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The unknown voltage=300V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.14 : Page number 622\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "f_H=1000;                   #Frequency applied to horizontal plates, Hz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Loops_H=1;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(i)   Unknown frequency=%dHz.\"%f_V);\n",
-      "\n",
-      "#(ii)\n",
-      "Loops_H=2;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(ii)  Unknown frequency=%dHz.\"%f_V);\n",
-      "\n",
-      "#(iii)\n",
-      "Loops_H=6;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(iii) Unknown frequency=%dHz.\"%f_V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Unknown frequency=1000Hz.\n",
-        "(ii)  Unknown frequency=2000Hz.\n",
-        "(iii) Unknown frequency=6000Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_3.ipynb
deleted file mode 100755
index 5f13ea0e..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_3.ipynb
+++ /dev/null
@@ -1,668 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a688629536ad6915939234eacc1ed3eaaf36e0aaa88de5df5b6a309da4d2c64d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 22: ELECTRONIC INSTRUMENTS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.1 : Page number 606\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_g=1;                  #Full scale deflection current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "MS=1/(I_g/1000.0);        #Multimeter sensitivity, \u03a9 per volt\n",
-      "\n",
-      "#Result\n",
-      "print(\"The multimeter sensitivity=%d \u03a9 per volt.\"%MS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The multimeter sensitivity=1000 \u03a9 per volt.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.2 : Page number 606-607\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=1000.0;                  #Meter sensitivity, \u03a9 per volt\n",
-      "V_full_scale=50.0;                        #Full scale volts\n",
-      "R=50000.0;                                #Resistance to be measured, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "meter_resistance=V_full_scale*meter_sensitivity;            #Meter resistance, \u03a9\n",
-      "R_p=R*meter_resistance/(R+meter_resistance);                #Parallel resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"When the meter is used to measure the voltage across the resistance %d\u03a9, total resistance =%d\u03a9.\"%(R,R_p));\n",
-      "print(\"\u2234 Meter will give highly incorrect reading.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "When the meter is used to measure the voltage across the resistance 50000\u03a9, total resistance =25000\u03a9.\n",
-        "\u2234 Meter will give highly incorrect reading.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.3 : Page number 607\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=4.0;                #Meter sensitivity, k\u03a9/V\n",
-      "R_1=10.0;                             #Resistance across which voltage is to be measured, k\u03a9\n",
-      "R_2=10.0;                             #Resistance, k\u03a9\n",
-      "range_max=10.0;                       #Maximum range of the meter, V\n",
-      "range_min=0;                        #Minimum range of the meter, V\n",
-      "V=20.0;                               #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "R_meter=meter_sensitivity*range_max;                       #Resistance of the meter, k\u03a9\n",
-      "R_T=(R_meter*R_1)/(R_1+R_meter) +  R_2;                     #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T,2);                                            #Circuit current, mA\n",
-      "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter));       #Voltage read by multimeter, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage read by multimeter=8.88V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.4 : Page number 607-608\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=20.0;                #Meter sensitivity, k\u03a9/V\n",
-      "R_1=10.0;                             #Resistance across which voltage is to be measured, k\u03a9\n",
-      "R_2=10.0;                             #Resistance, k\u03a9\n",
-      "range_max=10.0;                       #Maximum range of the meter, V\n",
-      "range_min=0;                        #Minimum range of the meter, V\n",
-      "V=20.0;                               #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "R_meter=meter_sensitivity*range_max;                       #Resistance of the meter, k\u03a9\n",
-      "R_T=round((R_meter*R_1)/(R_1+R_meter) +  R_2,1);                     #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T,2);                                           #Circuit current, mA\n",
-      "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter));       #Voltage read by multimeter, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n",
-      "\n",
-      "\n",
-      "#Note: The circuit current=1.0256mA, has been approximated in the text  as 1.04mA. But, in the code 1.03 mA has been used. Therefore, the final answer is obtained as 9.81V and not 9.88V.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage read by multimeter=9.81V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.5 : Page number 608-609\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "R_A=20.0;             #Resistance after point A, k\u03a9\n",
-      "R_B=20.0;             #Resistance after point B, k\u03a9\n",
-      "R_C=30.0;             #Resistance after point C, k\u03a9\n",
-      "R_D=30.0;             #Resistance after point D, k\u03a9\n",
-      "R_meter=60.0;         #Resistance of the meter, k\u03a9\n",
-      "V=100.0;              #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) When meter is not connected:\n",
-      "R_T=R_A+R_B+R_C+R_D;                    #Total circuit resistance, k\u03a9\n",
-      "I_circuit=V/R_T;                        #Circuit current, mA\n",
-      "V_A=V;                                  #Voltage at point A, V\n",
-      "V_B=V-(I_circuit*R_A);                  #Voltage at point B, V\n",
-      "V_C=V-(I_circuit*(R_A+R_B));            #Voltage at point C, V\n",
-      "V_D=V-(I_circuit*(R_T-R_D));            #Voltage at point D, V\n",
-      "\n",
-      "print(\"(i) When meter is not connected:\");\n",
-      "print(\"    Voltage at point A=%dV.\"%V_A);\n",
-      "print(\"    Voltage at point B=%dV.\"%V_B);\n",
-      "print(\"    Voltage at point C=%dV.\"%V_C);\n",
-      "print(\"    Voltage at point D=%dV.\"%V_D);\n",
-      "\n",
-      "\n",
-      "#(ii) When meter is connected:\n",
-      "#(a) Since, point A is directly connected to the source, voltage at point A is equal to source voltage.\n",
-      "V_A=V;                                      #Voltage at point A, V\n",
-      "\n",
-      "#(b)\n",
-      "R_T_B=R_A  + round((R_T-R_A)*R_meter/(R_meter + (R_T-R_A)),2);             #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T_B,2);                                                #Circuit current, mA\n",
-      "V_B=I_circuit*(R_T-R_A)*R_meter/(R_meter + (R_T-R_A));                   #Voltage at point B, V\n",
-      "\n",
-      "\n",
-      "#(c)\n",
-      "R_T_C=(R_A+R_B)  + (R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B));               #Total circuit resistance, k\u03a9\n",
-      "I_circuit=V/R_T_C;                                                                #Circuit current, mA\n",
-      "V_C=floor((I_circuit*(R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B)))*10)/10;     #Voltage at point C, V\n",
-      "\n",
-      "\n",
-      "\n",
-      "#(c)\n",
-      "R_T_D=(R_T-R_D)  + R_D*R_meter/(R_meter + R_D);             #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T_D,2);                                 #Circuit current, mA\n",
-      "V_D=I_circuit*(R_D*R_meter)/(R_meter + R_D);                #Voltage at point D, V\n",
-      "\n",
-      "\n",
-      "print(\"(ii) When meter is connected:\");\n",
-      "print(\"     Voltage at point A=%dV.\"%V_A);\n",
-      "print(\"     Voltage at point B=%dV.\"%V_B);\n",
-      "print(\"     Voltage at point C=%.1fV.\"%V_C);\n",
-      "print(\"     Voltage at point D=%.1fV.\"%V_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) When meter is not connected:\n",
-        "    Voltage at point A=100V.\n",
-        "    Voltage at point B=80V.\n",
-        "    Voltage at point C=60V.\n",
-        "    Voltage at point D=30V.\n",
-        "(ii) When meter is connected:\n",
-        "     Voltage at point A=100V.\n",
-        "     Voltage at point B=63V.\n",
-        "     Voltage at point C=42.8V.\n",
-        "     Voltage at point D=22.2V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.6 : Page number 614\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                     #Supply voltage, V\n",
-      "R_m=1.0;                      #Meter resistance, k\u03a9\n",
-      "I_m_fsd=2.0;                  #Full scale deflection of meter current, mA\n",
-      "beta=80.0;                    #Base current amplification factor\n",
-      "E=5.0;                        #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_E=E-V_BE;                 #Emitter voltage, V\n",
-      "\n",
-      "#(i)\n",
-      "#I_m_fsd=V_E/(R_s+R_m), (OHM's LAW)\n",
-      "R_s=((V_E/I_m_fsd)-R_m)*1000;                  #Multiplier resistor, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "IB=I_m_fsd/beta;     \t\t\t\t#Base current, mA\n",
-      "R_i=E/IB;                    \t\t\t#Input resistance of voltmeter, k\u03a9\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The multiplier resistor=%d\u03a9.\"%R_s);\n",
-      "print(\"(ii) The voltmeter input resistance=%dk\u03a9\"%R_i);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The multiplier resistor=1150\u03a9.\n",
-        "(ii) The voltmeter input resistance=200k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.7 : Page number 614\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20;                     #Supply voltage, V\n",
-      "Rs_Rm=9.3;                  #Sum of multipier resistance and meter resistance, k\u03a9\n",
-      "I_m=1;                      #Meter current, mA\n",
-      "beta=100;                   #Base current amplification factor\n",
-      "E=10;                       #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_E=E-V_BE;             #Emitter voltage, V\n",
-      "I_m=V_E/Rs_Rm;          #Meter current, mA\n",
-      "\n",
-      "#(ii)\n",
-      "I_B=I_m/beta;               #Base current, mA\n",
-      "R_i_T=(E/I_B)/1000;         #Input resistance of voltmeter, with transistor, M\u03a9\n",
-      "R_i_WT=Rs_Rm;               #Input resistance of voltmeter, without transistor, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The meter current=%dmA\"%I_m);\n",
-      "print(\"(ii) The input resistance of voltmeter with transistor=%dM\u03a9.\"%R_i_T);\n",
-      "print(\"     The input resistance of voltmeter without transistor=%.1fk\u03a9.\"%R_i_WT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The meter current=1mA\n",
-        "(ii) The input resistance of voltmeter with transistor=1M\u03a9.\n",
-        "     The input resistance of voltmeter without transistor=9.3k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.8 : Page number 614-615\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20;                     #Supply voltage, V\n",
-      "Rs_Rm=9.3;                  #Sum of multipier resistance and meter resistance, k\u03a9\n",
-      "I_m=1;                      #Meter current, mA\n",
-      "beta=100;                   #Base current amplification factor\n",
-      "E=5;                        #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_m=(E-V_BE)/Rs_Rm;          #Meter current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The meter current=%.2fmA\"%I_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The meter current=0.46mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.9 : Page number 616\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_m_fsd=100.0;                  #Full scale deflection of meter current, \u03bcA\n",
-      "R_m=1.0;                        #Meter resistance, k\u03a9\n",
-      "V_rms=100.0;                    #r.m.s voltage to be measured, V\n",
-      "V_F=0.7;                      #Forward voltage drop of rectifier diode, V \n",
-      "\n",
-      "#Calculation\n",
-      "V_m=round(sqrt(2)*V_rms,1);                                                          #Peak value of applied voltage, V\n",
-      "V_rectifier_drop=2*V_F;                                                              #Total rectifier drop, V\n",
-      "I_peak=round(I_m_fsd/0.637,2);                                                       #Peak f.s.d current, \u03bcA\n",
-      "R_s=floor(((((V_m-V_rectifier_drop)/(I_peak*10**-6))-(R_m*1000))/1000)*10)/10;       #Multiplier resistance, k\u03a9 (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The multiplier resistance=%.1fk\u03a9.\"%R_s);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The multiplier resistance=890.7k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.10 : Page number 616\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_av=75;                  #Full scale deflection of meter current, \u03bcA\n",
-      "R_s=708;                     #Multiplier resistor, k\u03a9\n",
-      "R_m=900;                     #Meter coil resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "I_peak=I_av*10**-6/0.637;                   #Peak f.s.d meter current, A\n",
-      "R_T=R_s*1000+R_m;                                #Total circuit resistance, \u03a9\n",
-      "\n",
-      "#I_peak=(Vm-V_drop)/R_T; (OHM's LAW)\n",
-      "#And, Vm=sqrt(2)*Vrms\n",
-      "V_rms=(I_peak*R_T+(2*0.7))/sqrt(2) ;             #applied r.m.s voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The applied r.m.s voltage=%dV\"%V_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The applied r.m.s voltage=60V\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.11 : Page number 618\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection_sensitivity=0.01;            #Deflection sensitivity, mm/V\n",
-      "V=400;                                  #Applied voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "spot_shift=V*deflection_sensitivity;         #Spot shift produced, mm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The shift produced in the spot=%dmm.\"%spot_shift);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The shift produced in the spot=4mm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.12 : Page number 618-619\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection_sensitivity=0.03;            #Deflection sensitivity, mm/V\n",
-      "spot_shift=3;                           #Spot shift produced, mm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, spot_shift=Applied_Voltage*deflection_sensitivity,\n",
-      "V=spot_shift/deflection_sensitivity;    #Applied voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Applied voltage=%dV.\"%V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Applied voltage=100V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.13 : Page number  622\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection=2;                            #Deflection produced by applied voltage, cm\n",
-      "V=200;                                   #Applied voltage, V\n",
-      "deflection_by_another_voltage=3;         #Deflection by another voltage, cm\n",
-      "\n",
-      "#Calculation\n",
-      "deflection_sensitivity=V/deflection;                                #deflection sensitivity, V/cm\n",
-      "V_unknown=deflection_sensitivity*deflection_by_another_voltage;     #Unknown voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The unknown voltage=%dV.\"%V_unknown);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The unknown voltage=300V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.14 : Page number 622\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "f_H=1000;                   #Frequency applied to horizontal plates, Hz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Loops_H=1;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(i)   Unknown frequency=%dHz.\"%f_V);\n",
-      "\n",
-      "#(ii)\n",
-      "Loops_H=2;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(ii)  Unknown frequency=%dHz.\"%f_V);\n",
-      "\n",
-      "#(iii)\n",
-      "Loops_H=6;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(iii) Unknown frequency=%dHz.\"%f_V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Unknown frequency=1000Hz.\n",
-        "(ii)  Unknown frequency=2000Hz.\n",
-        "(iii) Unknown frequency=6000Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_4.ipynb
deleted file mode 100755
index 5f13ea0e..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_4.ipynb
+++ /dev/null
@@ -1,668 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a688629536ad6915939234eacc1ed3eaaf36e0aaa88de5df5b6a309da4d2c64d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 22: ELECTRONIC INSTRUMENTS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.1 : Page number 606\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_g=1;                  #Full scale deflection current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "MS=1/(I_g/1000.0);        #Multimeter sensitivity, \u03a9 per volt\n",
-      "\n",
-      "#Result\n",
-      "print(\"The multimeter sensitivity=%d \u03a9 per volt.\"%MS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The multimeter sensitivity=1000 \u03a9 per volt.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.2 : Page number 606-607\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=1000.0;                  #Meter sensitivity, \u03a9 per volt\n",
-      "V_full_scale=50.0;                        #Full scale volts\n",
-      "R=50000.0;                                #Resistance to be measured, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "meter_resistance=V_full_scale*meter_sensitivity;            #Meter resistance, \u03a9\n",
-      "R_p=R*meter_resistance/(R+meter_resistance);                #Parallel resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"When the meter is used to measure the voltage across the resistance %d\u03a9, total resistance =%d\u03a9.\"%(R,R_p));\n",
-      "print(\"\u2234 Meter will give highly incorrect reading.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "When the meter is used to measure the voltage across the resistance 50000\u03a9, total resistance =25000\u03a9.\n",
-        "\u2234 Meter will give highly incorrect reading.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.3 : Page number 607\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=4.0;                #Meter sensitivity, k\u03a9/V\n",
-      "R_1=10.0;                             #Resistance across which voltage is to be measured, k\u03a9\n",
-      "R_2=10.0;                             #Resistance, k\u03a9\n",
-      "range_max=10.0;                       #Maximum range of the meter, V\n",
-      "range_min=0;                        #Minimum range of the meter, V\n",
-      "V=20.0;                               #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "R_meter=meter_sensitivity*range_max;                       #Resistance of the meter, k\u03a9\n",
-      "R_T=(R_meter*R_1)/(R_1+R_meter) +  R_2;                     #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T,2);                                            #Circuit current, mA\n",
-      "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter));       #Voltage read by multimeter, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage read by multimeter=8.88V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.4 : Page number 607-608\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=20.0;                #Meter sensitivity, k\u03a9/V\n",
-      "R_1=10.0;                             #Resistance across which voltage is to be measured, k\u03a9\n",
-      "R_2=10.0;                             #Resistance, k\u03a9\n",
-      "range_max=10.0;                       #Maximum range of the meter, V\n",
-      "range_min=0;                        #Minimum range of the meter, V\n",
-      "V=20.0;                               #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "R_meter=meter_sensitivity*range_max;                       #Resistance of the meter, k\u03a9\n",
-      "R_T=round((R_meter*R_1)/(R_1+R_meter) +  R_2,1);                     #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T,2);                                           #Circuit current, mA\n",
-      "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter));       #Voltage read by multimeter, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n",
-      "\n",
-      "\n",
-      "#Note: The circuit current=1.0256mA, has been approximated in the text  as 1.04mA. But, in the code 1.03 mA has been used. Therefore, the final answer is obtained as 9.81V and not 9.88V.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage read by multimeter=9.81V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.5 : Page number 608-609\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "R_A=20.0;             #Resistance after point A, k\u03a9\n",
-      "R_B=20.0;             #Resistance after point B, k\u03a9\n",
-      "R_C=30.0;             #Resistance after point C, k\u03a9\n",
-      "R_D=30.0;             #Resistance after point D, k\u03a9\n",
-      "R_meter=60.0;         #Resistance of the meter, k\u03a9\n",
-      "V=100.0;              #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) When meter is not connected:\n",
-      "R_T=R_A+R_B+R_C+R_D;                    #Total circuit resistance, k\u03a9\n",
-      "I_circuit=V/R_T;                        #Circuit current, mA\n",
-      "V_A=V;                                  #Voltage at point A, V\n",
-      "V_B=V-(I_circuit*R_A);                  #Voltage at point B, V\n",
-      "V_C=V-(I_circuit*(R_A+R_B));            #Voltage at point C, V\n",
-      "V_D=V-(I_circuit*(R_T-R_D));            #Voltage at point D, V\n",
-      "\n",
-      "print(\"(i) When meter is not connected:\");\n",
-      "print(\"    Voltage at point A=%dV.\"%V_A);\n",
-      "print(\"    Voltage at point B=%dV.\"%V_B);\n",
-      "print(\"    Voltage at point C=%dV.\"%V_C);\n",
-      "print(\"    Voltage at point D=%dV.\"%V_D);\n",
-      "\n",
-      "\n",
-      "#(ii) When meter is connected:\n",
-      "#(a) Since, point A is directly connected to the source, voltage at point A is equal to source voltage.\n",
-      "V_A=V;                                      #Voltage at point A, V\n",
-      "\n",
-      "#(b)\n",
-      "R_T_B=R_A  + round((R_T-R_A)*R_meter/(R_meter + (R_T-R_A)),2);             #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T_B,2);                                                #Circuit current, mA\n",
-      "V_B=I_circuit*(R_T-R_A)*R_meter/(R_meter + (R_T-R_A));                   #Voltage at point B, V\n",
-      "\n",
-      "\n",
-      "#(c)\n",
-      "R_T_C=(R_A+R_B)  + (R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B));               #Total circuit resistance, k\u03a9\n",
-      "I_circuit=V/R_T_C;                                                                #Circuit current, mA\n",
-      "V_C=floor((I_circuit*(R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B)))*10)/10;     #Voltage at point C, V\n",
-      "\n",
-      "\n",
-      "\n",
-      "#(c)\n",
-      "R_T_D=(R_T-R_D)  + R_D*R_meter/(R_meter + R_D);             #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T_D,2);                                 #Circuit current, mA\n",
-      "V_D=I_circuit*(R_D*R_meter)/(R_meter + R_D);                #Voltage at point D, V\n",
-      "\n",
-      "\n",
-      "print(\"(ii) When meter is connected:\");\n",
-      "print(\"     Voltage at point A=%dV.\"%V_A);\n",
-      "print(\"     Voltage at point B=%dV.\"%V_B);\n",
-      "print(\"     Voltage at point C=%.1fV.\"%V_C);\n",
-      "print(\"     Voltage at point D=%.1fV.\"%V_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) When meter is not connected:\n",
-        "    Voltage at point A=100V.\n",
-        "    Voltage at point B=80V.\n",
-        "    Voltage at point C=60V.\n",
-        "    Voltage at point D=30V.\n",
-        "(ii) When meter is connected:\n",
-        "     Voltage at point A=100V.\n",
-        "     Voltage at point B=63V.\n",
-        "     Voltage at point C=42.8V.\n",
-        "     Voltage at point D=22.2V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.6 : Page number 614\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                     #Supply voltage, V\n",
-      "R_m=1.0;                      #Meter resistance, k\u03a9\n",
-      "I_m_fsd=2.0;                  #Full scale deflection of meter current, mA\n",
-      "beta=80.0;                    #Base current amplification factor\n",
-      "E=5.0;                        #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_E=E-V_BE;                 #Emitter voltage, V\n",
-      "\n",
-      "#(i)\n",
-      "#I_m_fsd=V_E/(R_s+R_m), (OHM's LAW)\n",
-      "R_s=((V_E/I_m_fsd)-R_m)*1000;                  #Multiplier resistor, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "IB=I_m_fsd/beta;     \t\t\t\t#Base current, mA\n",
-      "R_i=E/IB;                    \t\t\t#Input resistance of voltmeter, k\u03a9\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The multiplier resistor=%d\u03a9.\"%R_s);\n",
-      "print(\"(ii) The voltmeter input resistance=%dk\u03a9\"%R_i);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The multiplier resistor=1150\u03a9.\n",
-        "(ii) The voltmeter input resistance=200k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.7 : Page number 614\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20;                     #Supply voltage, V\n",
-      "Rs_Rm=9.3;                  #Sum of multipier resistance and meter resistance, k\u03a9\n",
-      "I_m=1;                      #Meter current, mA\n",
-      "beta=100;                   #Base current amplification factor\n",
-      "E=10;                       #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_E=E-V_BE;             #Emitter voltage, V\n",
-      "I_m=V_E/Rs_Rm;          #Meter current, mA\n",
-      "\n",
-      "#(ii)\n",
-      "I_B=I_m/beta;               #Base current, mA\n",
-      "R_i_T=(E/I_B)/1000;         #Input resistance of voltmeter, with transistor, M\u03a9\n",
-      "R_i_WT=Rs_Rm;               #Input resistance of voltmeter, without transistor, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The meter current=%dmA\"%I_m);\n",
-      "print(\"(ii) The input resistance of voltmeter with transistor=%dM\u03a9.\"%R_i_T);\n",
-      "print(\"     The input resistance of voltmeter without transistor=%.1fk\u03a9.\"%R_i_WT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The meter current=1mA\n",
-        "(ii) The input resistance of voltmeter with transistor=1M\u03a9.\n",
-        "     The input resistance of voltmeter without transistor=9.3k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.8 : Page number 614-615\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20;                     #Supply voltage, V\n",
-      "Rs_Rm=9.3;                  #Sum of multipier resistance and meter resistance, k\u03a9\n",
-      "I_m=1;                      #Meter current, mA\n",
-      "beta=100;                   #Base current amplification factor\n",
-      "E=5;                        #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_m=(E-V_BE)/Rs_Rm;          #Meter current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The meter current=%.2fmA\"%I_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The meter current=0.46mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.9 : Page number 616\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_m_fsd=100.0;                  #Full scale deflection of meter current, \u03bcA\n",
-      "R_m=1.0;                        #Meter resistance, k\u03a9\n",
-      "V_rms=100.0;                    #r.m.s voltage to be measured, V\n",
-      "V_F=0.7;                      #Forward voltage drop of rectifier diode, V \n",
-      "\n",
-      "#Calculation\n",
-      "V_m=round(sqrt(2)*V_rms,1);                                                          #Peak value of applied voltage, V\n",
-      "V_rectifier_drop=2*V_F;                                                              #Total rectifier drop, V\n",
-      "I_peak=round(I_m_fsd/0.637,2);                                                       #Peak f.s.d current, \u03bcA\n",
-      "R_s=floor(((((V_m-V_rectifier_drop)/(I_peak*10**-6))-(R_m*1000))/1000)*10)/10;       #Multiplier resistance, k\u03a9 (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The multiplier resistance=%.1fk\u03a9.\"%R_s);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The multiplier resistance=890.7k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.10 : Page number 616\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_av=75;                  #Full scale deflection of meter current, \u03bcA\n",
-      "R_s=708;                     #Multiplier resistor, k\u03a9\n",
-      "R_m=900;                     #Meter coil resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "I_peak=I_av*10**-6/0.637;                   #Peak f.s.d meter current, A\n",
-      "R_T=R_s*1000+R_m;                                #Total circuit resistance, \u03a9\n",
-      "\n",
-      "#I_peak=(Vm-V_drop)/R_T; (OHM's LAW)\n",
-      "#And, Vm=sqrt(2)*Vrms\n",
-      "V_rms=(I_peak*R_T+(2*0.7))/sqrt(2) ;             #applied r.m.s voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The applied r.m.s voltage=%dV\"%V_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The applied r.m.s voltage=60V\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.11 : Page number 618\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection_sensitivity=0.01;            #Deflection sensitivity, mm/V\n",
-      "V=400;                                  #Applied voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "spot_shift=V*deflection_sensitivity;         #Spot shift produced, mm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The shift produced in the spot=%dmm.\"%spot_shift);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The shift produced in the spot=4mm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.12 : Page number 618-619\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection_sensitivity=0.03;            #Deflection sensitivity, mm/V\n",
-      "spot_shift=3;                           #Spot shift produced, mm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, spot_shift=Applied_Voltage*deflection_sensitivity,\n",
-      "V=spot_shift/deflection_sensitivity;    #Applied voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Applied voltage=%dV.\"%V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Applied voltage=100V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.13 : Page number  622\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection=2;                            #Deflection produced by applied voltage, cm\n",
-      "V=200;                                   #Applied voltage, V\n",
-      "deflection_by_another_voltage=3;         #Deflection by another voltage, cm\n",
-      "\n",
-      "#Calculation\n",
-      "deflection_sensitivity=V/deflection;                                #deflection sensitivity, V/cm\n",
-      "V_unknown=deflection_sensitivity*deflection_by_another_voltage;     #Unknown voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The unknown voltage=%dV.\"%V_unknown);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The unknown voltage=300V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.14 : Page number 622\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "f_H=1000;                   #Frequency applied to horizontal plates, Hz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Loops_H=1;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(i)   Unknown frequency=%dHz.\"%f_V);\n",
-      "\n",
-      "#(ii)\n",
-      "Loops_H=2;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(ii)  Unknown frequency=%dHz.\"%f_V);\n",
-      "\n",
-      "#(iii)\n",
-      "Loops_H=6;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(iii) Unknown frequency=%dHz.\"%f_V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Unknown frequency=1000Hz.\n",
-        "(ii)  Unknown frequency=2000Hz.\n",
-        "(iii) Unknown frequency=6000Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_5.ipynb
deleted file mode 100755
index 5f13ea0e..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_5.ipynb
+++ /dev/null
@@ -1,668 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a688629536ad6915939234eacc1ed3eaaf36e0aaa88de5df5b6a309da4d2c64d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 22: ELECTRONIC INSTRUMENTS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.1 : Page number 606\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_g=1;                  #Full scale deflection current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "MS=1/(I_g/1000.0);        #Multimeter sensitivity, \u03a9 per volt\n",
-      "\n",
-      "#Result\n",
-      "print(\"The multimeter sensitivity=%d \u03a9 per volt.\"%MS);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The multimeter sensitivity=1000 \u03a9 per volt.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.2 : Page number 606-607\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=1000.0;                  #Meter sensitivity, \u03a9 per volt\n",
-      "V_full_scale=50.0;                        #Full scale volts\n",
-      "R=50000.0;                                #Resistance to be measured, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "meter_resistance=V_full_scale*meter_sensitivity;            #Meter resistance, \u03a9\n",
-      "R_p=R*meter_resistance/(R+meter_resistance);                #Parallel resistance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"When the meter is used to measure the voltage across the resistance %d\u03a9, total resistance =%d\u03a9.\"%(R,R_p));\n",
-      "print(\"\u2234 Meter will give highly incorrect reading.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "When the meter is used to measure the voltage across the resistance 50000\u03a9, total resistance =25000\u03a9.\n",
-        "\u2234 Meter will give highly incorrect reading.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.3 : Page number 607\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=4.0;                #Meter sensitivity, k\u03a9/V\n",
-      "R_1=10.0;                             #Resistance across which voltage is to be measured, k\u03a9\n",
-      "R_2=10.0;                             #Resistance, k\u03a9\n",
-      "range_max=10.0;                       #Maximum range of the meter, V\n",
-      "range_min=0;                        #Minimum range of the meter, V\n",
-      "V=20.0;                               #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "R_meter=meter_sensitivity*range_max;                       #Resistance of the meter, k\u03a9\n",
-      "R_T=(R_meter*R_1)/(R_1+R_meter) +  R_2;                     #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T,2);                                            #Circuit current, mA\n",
-      "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter));       #Voltage read by multimeter, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage read by multimeter=8.88V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.4 : Page number 607-608\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "meter_sensitivity=20.0;                #Meter sensitivity, k\u03a9/V\n",
-      "R_1=10.0;                             #Resistance across which voltage is to be measured, k\u03a9\n",
-      "R_2=10.0;                             #Resistance, k\u03a9\n",
-      "range_max=10.0;                       #Maximum range of the meter, V\n",
-      "range_min=0;                        #Minimum range of the meter, V\n",
-      "V=20.0;                               #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "R_meter=meter_sensitivity*range_max;                       #Resistance of the meter, k\u03a9\n",
-      "R_T=round((R_meter*R_1)/(R_1+R_meter) +  R_2,1);                     #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T,2);                                           #Circuit current, mA\n",
-      "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter));       #Voltage read by multimeter, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n",
-      "\n",
-      "\n",
-      "#Note: The circuit current=1.0256mA, has been approximated in the text  as 1.04mA. But, in the code 1.03 mA has been used. Therefore, the final answer is obtained as 9.81V and not 9.88V.\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage read by multimeter=9.81V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.5 : Page number 608-609\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "R_A=20.0;             #Resistance after point A, k\u03a9\n",
-      "R_B=20.0;             #Resistance after point B, k\u03a9\n",
-      "R_C=30.0;             #Resistance after point C, k\u03a9\n",
-      "R_D=30.0;             #Resistance after point D, k\u03a9\n",
-      "R_meter=60.0;         #Resistance of the meter, k\u03a9\n",
-      "V=100.0;              #Battery voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) When meter is not connected:\n",
-      "R_T=R_A+R_B+R_C+R_D;                    #Total circuit resistance, k\u03a9\n",
-      "I_circuit=V/R_T;                        #Circuit current, mA\n",
-      "V_A=V;                                  #Voltage at point A, V\n",
-      "V_B=V-(I_circuit*R_A);                  #Voltage at point B, V\n",
-      "V_C=V-(I_circuit*(R_A+R_B));            #Voltage at point C, V\n",
-      "V_D=V-(I_circuit*(R_T-R_D));            #Voltage at point D, V\n",
-      "\n",
-      "print(\"(i) When meter is not connected:\");\n",
-      "print(\"    Voltage at point A=%dV.\"%V_A);\n",
-      "print(\"    Voltage at point B=%dV.\"%V_B);\n",
-      "print(\"    Voltage at point C=%dV.\"%V_C);\n",
-      "print(\"    Voltage at point D=%dV.\"%V_D);\n",
-      "\n",
-      "\n",
-      "#(ii) When meter is connected:\n",
-      "#(a) Since, point A is directly connected to the source, voltage at point A is equal to source voltage.\n",
-      "V_A=V;                                      #Voltage at point A, V\n",
-      "\n",
-      "#(b)\n",
-      "R_T_B=R_A  + round((R_T-R_A)*R_meter/(R_meter + (R_T-R_A)),2);             #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T_B,2);                                                #Circuit current, mA\n",
-      "V_B=I_circuit*(R_T-R_A)*R_meter/(R_meter + (R_T-R_A));                   #Voltage at point B, V\n",
-      "\n",
-      "\n",
-      "#(c)\n",
-      "R_T_C=(R_A+R_B)  + (R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B));               #Total circuit resistance, k\u03a9\n",
-      "I_circuit=V/R_T_C;                                                                #Circuit current, mA\n",
-      "V_C=floor((I_circuit*(R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B)))*10)/10;     #Voltage at point C, V\n",
-      "\n",
-      "\n",
-      "\n",
-      "#(c)\n",
-      "R_T_D=(R_T-R_D)  + R_D*R_meter/(R_meter + R_D);             #Total circuit resistance, k\u03a9\n",
-      "I_circuit=round(V/R_T_D,2);                                 #Circuit current, mA\n",
-      "V_D=I_circuit*(R_D*R_meter)/(R_meter + R_D);                #Voltage at point D, V\n",
-      "\n",
-      "\n",
-      "print(\"(ii) When meter is connected:\");\n",
-      "print(\"     Voltage at point A=%dV.\"%V_A);\n",
-      "print(\"     Voltage at point B=%dV.\"%V_B);\n",
-      "print(\"     Voltage at point C=%.1fV.\"%V_C);\n",
-      "print(\"     Voltage at point D=%.1fV.\"%V_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) When meter is not connected:\n",
-        "    Voltage at point A=100V.\n",
-        "    Voltage at point B=80V.\n",
-        "    Voltage at point C=60V.\n",
-        "    Voltage at point D=30V.\n",
-        "(ii) When meter is connected:\n",
-        "     Voltage at point A=100V.\n",
-        "     Voltage at point B=63V.\n",
-        "     Voltage at point C=42.8V.\n",
-        "     Voltage at point D=22.2V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.6 : Page number 614\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                     #Supply voltage, V\n",
-      "R_m=1.0;                      #Meter resistance, k\u03a9\n",
-      "I_m_fsd=2.0;                  #Full scale deflection of meter current, mA\n",
-      "beta=80.0;                    #Base current amplification factor\n",
-      "E=5.0;                        #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_E=E-V_BE;                 #Emitter voltage, V\n",
-      "\n",
-      "#(i)\n",
-      "#I_m_fsd=V_E/(R_s+R_m), (OHM's LAW)\n",
-      "R_s=((V_E/I_m_fsd)-R_m)*1000;                  #Multiplier resistor, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "IB=I_m_fsd/beta;     \t\t\t\t#Base current, mA\n",
-      "R_i=E/IB;                    \t\t\t#Input resistance of voltmeter, k\u03a9\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The multiplier resistor=%d\u03a9.\"%R_s);\n",
-      "print(\"(ii) The voltmeter input resistance=%dk\u03a9\"%R_i);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The multiplier resistor=1150\u03a9.\n",
-        "(ii) The voltmeter input resistance=200k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.7 : Page number 614\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20;                     #Supply voltage, V\n",
-      "Rs_Rm=9.3;                  #Sum of multipier resistance and meter resistance, k\u03a9\n",
-      "I_m=1;                      #Meter current, mA\n",
-      "beta=100;                   #Base current amplification factor\n",
-      "E=10;                       #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "V_E=E-V_BE;             #Emitter voltage, V\n",
-      "I_m=V_E/Rs_Rm;          #Meter current, mA\n",
-      "\n",
-      "#(ii)\n",
-      "I_B=I_m/beta;               #Base current, mA\n",
-      "R_i_T=(E/I_B)/1000;         #Input resistance of voltmeter, with transistor, M\u03a9\n",
-      "R_i_WT=Rs_Rm;               #Input resistance of voltmeter, without transistor, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The meter current=%dmA\"%I_m);\n",
-      "print(\"(ii) The input resistance of voltmeter with transistor=%dM\u03a9.\"%R_i_T);\n",
-      "print(\"     The input resistance of voltmeter without transistor=%.1fk\u03a9.\"%R_i_WT);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The meter current=1mA\n",
-        "(ii) The input resistance of voltmeter with transistor=1M\u03a9.\n",
-        "     The input resistance of voltmeter without transistor=9.3k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.8 : Page number 614-615\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20;                     #Supply voltage, V\n",
-      "Rs_Rm=9.3;                  #Sum of multipier resistance and meter resistance, k\u03a9\n",
-      "I_m=1;                      #Meter current, mA\n",
-      "beta=100;                   #Base current amplification factor\n",
-      "E=5;                        #Voltage to be measured, V\n",
-      "V_BE=0.7;                   #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_m=(E-V_BE)/Rs_Rm;          #Meter current, mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The meter current=%.2fmA\"%I_m);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The meter current=0.46mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.9 : Page number 616\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_m_fsd=100.0;                  #Full scale deflection of meter current, \u03bcA\n",
-      "R_m=1.0;                        #Meter resistance, k\u03a9\n",
-      "V_rms=100.0;                    #r.m.s voltage to be measured, V\n",
-      "V_F=0.7;                      #Forward voltage drop of rectifier diode, V \n",
-      "\n",
-      "#Calculation\n",
-      "V_m=round(sqrt(2)*V_rms,1);                                                          #Peak value of applied voltage, V\n",
-      "V_rectifier_drop=2*V_F;                                                              #Total rectifier drop, V\n",
-      "I_peak=round(I_m_fsd/0.637,2);                                                       #Peak f.s.d current, \u03bcA\n",
-      "R_s=floor(((((V_m-V_rectifier_drop)/(I_peak*10**-6))-(R_m*1000))/1000)*10)/10;       #Multiplier resistance, k\u03a9 (OHM's LAW)\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The multiplier resistance=%.1fk\u03a9.\"%R_s);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The multiplier resistance=890.7k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.10 : Page number 616\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "I_av=75;                  #Full scale deflection of meter current, \u03bcA\n",
-      "R_s=708;                     #Multiplier resistor, k\u03a9\n",
-      "R_m=900;                     #Meter coil resistor, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "I_peak=I_av*10**-6/0.637;                   #Peak f.s.d meter current, A\n",
-      "R_T=R_s*1000+R_m;                                #Total circuit resistance, \u03a9\n",
-      "\n",
-      "#I_peak=(Vm-V_drop)/R_T; (OHM's LAW)\n",
-      "#And, Vm=sqrt(2)*Vrms\n",
-      "V_rms=(I_peak*R_T+(2*0.7))/sqrt(2) ;             #applied r.m.s voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The applied r.m.s voltage=%dV\"%V_rms);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The applied r.m.s voltage=60V\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.11 : Page number 618\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection_sensitivity=0.01;            #Deflection sensitivity, mm/V\n",
-      "V=400;                                  #Applied voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "spot_shift=V*deflection_sensitivity;         #Spot shift produced, mm\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The shift produced in the spot=%dmm.\"%spot_shift);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The shift produced in the spot=4mm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.12 : Page number 618-619\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection_sensitivity=0.03;            #Deflection sensitivity, mm/V\n",
-      "spot_shift=3;                           #Spot shift produced, mm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, spot_shift=Applied_Voltage*deflection_sensitivity,\n",
-      "V=spot_shift/deflection_sensitivity;    #Applied voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Applied voltage=%dV.\"%V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Applied voltage=100V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.13 : Page number  622\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "deflection=2;                            #Deflection produced by applied voltage, cm\n",
-      "V=200;                                   #Applied voltage, V\n",
-      "deflection_by_another_voltage=3;         #Deflection by another voltage, cm\n",
-      "\n",
-      "#Calculation\n",
-      "deflection_sensitivity=V/deflection;                                #deflection sensitivity, V/cm\n",
-      "V_unknown=deflection_sensitivity*deflection_by_another_voltage;     #Unknown voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The unknown voltage=%dV.\"%V_unknown);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The unknown voltage=300V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 22.14 : Page number 622\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "f_H=1000;                   #Frequency applied to horizontal plates, Hz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Loops_H=1;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(i)   Unknown frequency=%dHz.\"%f_V);\n",
-      "\n",
-      "#(ii)\n",
-      "Loops_H=2;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(ii)  Unknown frequency=%dHz.\"%f_V);\n",
-      "\n",
-      "#(iii)\n",
-      "Loops_H=6;                        #Number of loops cut by horizontal line\n",
-      "Loops_V=1;                        #Number of loops cut by vertical line\n",
-      "f_V=f_H*(Loops_H/Loops_V);        #Unknown frequency, Hz\n",
-      "\n",
-      "print(\"(iii) Unknown frequency=%dHz.\"%f_V);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Unknown frequency=1000Hz.\n",
-        "(ii)  Unknown frequency=2000Hz.\n",
-        "(iii) Unknown frequency=6000Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23.ipynb
deleted file mode 100755
index 19741354..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23.ipynb
+++ /dev/null
@@ -1,133 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:87bd5c8d9448f5bb2e75909f89934a6fb2b64e65e6ea37b085ce080a58026071"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 23 : INTEGRATED CIRCUITS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.1: Page number 637"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=240;                     #Adjusted resistance of R2 resistor of LM317 voltage regulator, in kilo ohm\n",
-      "R2=2.4;                     #Fixed value of R1 resistor of LM317 voltage regulator, in ohm\n",
-      "\n",
-      "#Calculations\n",
-      "#Output voltage of LM317 voltage regulator IC = 1.25(R2/R1 +1)\n",
-      "Vout=1.25*((R2*1000)/R1 + 1);                      #Regulated d.c output voltage for the circuit in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The regulated d.c output voltage = %.2fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated d.c output voltage = 13.75V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.2 : Page number 638"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=1.2;                  #Value of resistance of monostable multivibrator in kilo ohm\n",
-      "C=0.1;                  #Value of capacitance of monostable multivibrator in microfarad\n",
-      "\n",
-      "#Calculations\n",
-      "T=1.1*(R*1000)*C;              #Time for which the circuit is ON, in microseconds\n",
-      "\n",
-      "#Results\n",
-      "print(\"Time for which the circuit is ON = %d microseconds.\"%T);                \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time for which the circuit is ON = 132 microseconds.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.3 : Page number 639\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=3.0;                       #Resistance of R1 resistor of 555 timer circuit in kilo ohm\n",
-      "R2=2.7;                       #Resistance of R2 resistor of 555 timer circuit in kilo ohm\n",
-      "C=0.033;                      #Capacitance of the capacitor of 555 timer circuit in microfarad\n",
-      "\n",
-      "#Calculations\n",
-      "f=1.44/(((R1*1000) + 2*(R2*1000))*(C*pow(10,-6)));                   #Frequency of the circuit in Hz\n",
-      "f=f/1000;                                                            #Frequency of the circuit in kHz\n",
-      "\n",
-      "#Results\n",
-      "print(\"The frequency of the circuit = %.2fkHz\"%f);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of the circuit = 5.19kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_1.ipynb
deleted file mode 100755
index 19741354..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_1.ipynb
+++ /dev/null
@@ -1,133 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:87bd5c8d9448f5bb2e75909f89934a6fb2b64e65e6ea37b085ce080a58026071"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 23 : INTEGRATED CIRCUITS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.1: Page number 637"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=240;                     #Adjusted resistance of R2 resistor of LM317 voltage regulator, in kilo ohm\n",
-      "R2=2.4;                     #Fixed value of R1 resistor of LM317 voltage regulator, in ohm\n",
-      "\n",
-      "#Calculations\n",
-      "#Output voltage of LM317 voltage regulator IC = 1.25(R2/R1 +1)\n",
-      "Vout=1.25*((R2*1000)/R1 + 1);                      #Regulated d.c output voltage for the circuit in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The regulated d.c output voltage = %.2fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated d.c output voltage = 13.75V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.2 : Page number 638"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=1.2;                  #Value of resistance of monostable multivibrator in kilo ohm\n",
-      "C=0.1;                  #Value of capacitance of monostable multivibrator in microfarad\n",
-      "\n",
-      "#Calculations\n",
-      "T=1.1*(R*1000)*C;              #Time for which the circuit is ON, in microseconds\n",
-      "\n",
-      "#Results\n",
-      "print(\"Time for which the circuit is ON = %d microseconds.\"%T);                \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time for which the circuit is ON = 132 microseconds.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.3 : Page number 639\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=3.0;                       #Resistance of R1 resistor of 555 timer circuit in kilo ohm\n",
-      "R2=2.7;                       #Resistance of R2 resistor of 555 timer circuit in kilo ohm\n",
-      "C=0.033;                      #Capacitance of the capacitor of 555 timer circuit in microfarad\n",
-      "\n",
-      "#Calculations\n",
-      "f=1.44/(((R1*1000) + 2*(R2*1000))*(C*pow(10,-6)));                   #Frequency of the circuit in Hz\n",
-      "f=f/1000;                                                            #Frequency of the circuit in kHz\n",
-      "\n",
-      "#Results\n",
-      "print(\"The frequency of the circuit = %.2fkHz\"%f);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of the circuit = 5.19kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_2.ipynb
deleted file mode 100755
index 19741354..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_2.ipynb
+++ /dev/null
@@ -1,133 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:87bd5c8d9448f5bb2e75909f89934a6fb2b64e65e6ea37b085ce080a58026071"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 23 : INTEGRATED CIRCUITS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.1: Page number 637"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=240;                     #Adjusted resistance of R2 resistor of LM317 voltage regulator, in kilo ohm\n",
-      "R2=2.4;                     #Fixed value of R1 resistor of LM317 voltage regulator, in ohm\n",
-      "\n",
-      "#Calculations\n",
-      "#Output voltage of LM317 voltage regulator IC = 1.25(R2/R1 +1)\n",
-      "Vout=1.25*((R2*1000)/R1 + 1);                      #Regulated d.c output voltage for the circuit in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The regulated d.c output voltage = %.2fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated d.c output voltage = 13.75V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.2 : Page number 638"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=1.2;                  #Value of resistance of monostable multivibrator in kilo ohm\n",
-      "C=0.1;                  #Value of capacitance of monostable multivibrator in microfarad\n",
-      "\n",
-      "#Calculations\n",
-      "T=1.1*(R*1000)*C;              #Time for which the circuit is ON, in microseconds\n",
-      "\n",
-      "#Results\n",
-      "print(\"Time for which the circuit is ON = %d microseconds.\"%T);                \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time for which the circuit is ON = 132 microseconds.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.3 : Page number 639\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=3.0;                       #Resistance of R1 resistor of 555 timer circuit in kilo ohm\n",
-      "R2=2.7;                       #Resistance of R2 resistor of 555 timer circuit in kilo ohm\n",
-      "C=0.033;                      #Capacitance of the capacitor of 555 timer circuit in microfarad\n",
-      "\n",
-      "#Calculations\n",
-      "f=1.44/(((R1*1000) + 2*(R2*1000))*(C*pow(10,-6)));                   #Frequency of the circuit in Hz\n",
-      "f=f/1000;                                                            #Frequency of the circuit in kHz\n",
-      "\n",
-      "#Results\n",
-      "print(\"The frequency of the circuit = %.2fkHz\"%f);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of the circuit = 5.19kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_3.ipynb
deleted file mode 100755
index 19741354..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_3.ipynb
+++ /dev/null
@@ -1,133 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:87bd5c8d9448f5bb2e75909f89934a6fb2b64e65e6ea37b085ce080a58026071"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 23 : INTEGRATED CIRCUITS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.1: Page number 637"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=240;                     #Adjusted resistance of R2 resistor of LM317 voltage regulator, in kilo ohm\n",
-      "R2=2.4;                     #Fixed value of R1 resistor of LM317 voltage regulator, in ohm\n",
-      "\n",
-      "#Calculations\n",
-      "#Output voltage of LM317 voltage regulator IC = 1.25(R2/R1 +1)\n",
-      "Vout=1.25*((R2*1000)/R1 + 1);                      #Regulated d.c output voltage for the circuit in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The regulated d.c output voltage = %.2fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated d.c output voltage = 13.75V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.2 : Page number 638"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=1.2;                  #Value of resistance of monostable multivibrator in kilo ohm\n",
-      "C=0.1;                  #Value of capacitance of monostable multivibrator in microfarad\n",
-      "\n",
-      "#Calculations\n",
-      "T=1.1*(R*1000)*C;              #Time for which the circuit is ON, in microseconds\n",
-      "\n",
-      "#Results\n",
-      "print(\"Time for which the circuit is ON = %d microseconds.\"%T);                \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time for which the circuit is ON = 132 microseconds.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.3 : Page number 639\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=3.0;                       #Resistance of R1 resistor of 555 timer circuit in kilo ohm\n",
-      "R2=2.7;                       #Resistance of R2 resistor of 555 timer circuit in kilo ohm\n",
-      "C=0.033;                      #Capacitance of the capacitor of 555 timer circuit in microfarad\n",
-      "\n",
-      "#Calculations\n",
-      "f=1.44/(((R1*1000) + 2*(R2*1000))*(C*pow(10,-6)));                   #Frequency of the circuit in Hz\n",
-      "f=f/1000;                                                            #Frequency of the circuit in kHz\n",
-      "\n",
-      "#Results\n",
-      "print(\"The frequency of the circuit = %.2fkHz\"%f);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of the circuit = 5.19kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_4.ipynb
deleted file mode 100755
index 19741354..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_4.ipynb
+++ /dev/null
@@ -1,133 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:87bd5c8d9448f5bb2e75909f89934a6fb2b64e65e6ea37b085ce080a58026071"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 23 : INTEGRATED CIRCUITS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.1: Page number 637"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=240;                     #Adjusted resistance of R2 resistor of LM317 voltage regulator, in kilo ohm\n",
-      "R2=2.4;                     #Fixed value of R1 resistor of LM317 voltage regulator, in ohm\n",
-      "\n",
-      "#Calculations\n",
-      "#Output voltage of LM317 voltage regulator IC = 1.25(R2/R1 +1)\n",
-      "Vout=1.25*((R2*1000)/R1 + 1);                      #Regulated d.c output voltage for the circuit in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The regulated d.c output voltage = %.2fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated d.c output voltage = 13.75V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.2 : Page number 638"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=1.2;                  #Value of resistance of monostable multivibrator in kilo ohm\n",
-      "C=0.1;                  #Value of capacitance of monostable multivibrator in microfarad\n",
-      "\n",
-      "#Calculations\n",
-      "T=1.1*(R*1000)*C;              #Time for which the circuit is ON, in microseconds\n",
-      "\n",
-      "#Results\n",
-      "print(\"Time for which the circuit is ON = %d microseconds.\"%T);                \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time for which the circuit is ON = 132 microseconds.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.3 : Page number 639\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=3.0;                       #Resistance of R1 resistor of 555 timer circuit in kilo ohm\n",
-      "R2=2.7;                       #Resistance of R2 resistor of 555 timer circuit in kilo ohm\n",
-      "C=0.033;                      #Capacitance of the capacitor of 555 timer circuit in microfarad\n",
-      "\n",
-      "#Calculations\n",
-      "f=1.44/(((R1*1000) + 2*(R2*1000))*(C*pow(10,-6)));                   #Frequency of the circuit in Hz\n",
-      "f=f/1000;                                                            #Frequency of the circuit in kHz\n",
-      "\n",
-      "#Results\n",
-      "print(\"The frequency of the circuit = %.2fkHz\"%f);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of the circuit = 5.19kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_5.ipynb
deleted file mode 100755
index 19741354..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_5.ipynb
+++ /dev/null
@@ -1,133 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:87bd5c8d9448f5bb2e75909f89934a6fb2b64e65e6ea37b085ce080a58026071"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 23 : INTEGRATED CIRCUITS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.1: Page number 637"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=240;                     #Adjusted resistance of R2 resistor of LM317 voltage regulator, in kilo ohm\n",
-      "R2=2.4;                     #Fixed value of R1 resistor of LM317 voltage regulator, in ohm\n",
-      "\n",
-      "#Calculations\n",
-      "#Output voltage of LM317 voltage regulator IC = 1.25(R2/R1 +1)\n",
-      "Vout=1.25*((R2*1000)/R1 + 1);                      #Regulated d.c output voltage for the circuit in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The regulated d.c output voltage = %.2fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulated d.c output voltage = 13.75V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.2 : Page number 638"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=1.2;                  #Value of resistance of monostable multivibrator in kilo ohm\n",
-      "C=0.1;                  #Value of capacitance of monostable multivibrator in microfarad\n",
-      "\n",
-      "#Calculations\n",
-      "T=1.1*(R*1000)*C;              #Time for which the circuit is ON, in microseconds\n",
-      "\n",
-      "#Results\n",
-      "print(\"Time for which the circuit is ON = %d microseconds.\"%T);                \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Time for which the circuit is ON = 132 microseconds.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 23.3 : Page number 639\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=3.0;                       #Resistance of R1 resistor of 555 timer circuit in kilo ohm\n",
-      "R2=2.7;                       #Resistance of R2 resistor of 555 timer circuit in kilo ohm\n",
-      "C=0.033;                      #Capacitance of the capacitor of 555 timer circuit in microfarad\n",
-      "\n",
-      "#Calculations\n",
-      "f=1.44/(((R1*1000) + 2*(R2*1000))*(C*pow(10,-6)));                   #Frequency of the circuit in Hz\n",
-      "f=f/1000;                                                            #Frequency of the circuit in kHz\n",
-      "\n",
-      "#Results\n",
-      "print(\"The frequency of the circuit = %.2fkHz\"%f);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of the circuit = 5.19kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24.ipynb
deleted file mode 100755
index e63c17a6..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24.ipynb
+++ /dev/null
@@ -1,604 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:947f358cf49d029c94d008f72a340051744678cf2e36ecc199100b78d31fcba5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 24 : HYBRID PARAMETERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.1 : Page number 644-645\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=10.0;                          #1st resistor, \u03a9\n",
-      "R2=5.0;                           #2nd resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"To find h11 and h21, output terminals are shorted.\");\n",
-      "h11=R1;                 #Input impedance with output shorted,  \u03a9\n",
-      "\n",
-      "print(\"h11=%d\u03a9.\"%h11);\n",
-      "\n",
-      "print(\"Output current flowing into the box= input current flowing out of the box.\");\n",
-      "print(\"i2=-i1\");                        #Output current flowing into the box= input current flowing out of the box.\n",
-      "print(\"h21=i2/i1 = -i1/i1= -1.\");       #Current gain with output shorted.\n",
-      "\n",
-      "\n",
-      "print(\"For finding h22 and h12, voltage source is connected at the output\");\n",
-      "#As, there will be no current through 10k\u03a9 resistor due to open circuited input,\n",
-      "print(\"v1=v2\");                         #Output voltage is equal to input voltage(equal to voltage drop across 5k\u03a9 resistor)\n",
-      "print(\"h12=v1/v2 = v2/v2 = 1\");         #Voltage feedback ratio with input terminals open\n",
-      "\n",
-      "h22=1/R2;                               #Output admittance, mho\n",
-      "print(\"h22=%.1f mho\"%h22);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To find h11 and h21, output terminals are shorted.\n",
-        "h11=10\u03a9.\n",
-        "Output current flowing into the box= input current flowing out of the box.\n",
-        "i2=-i1\n",
-        "h21=i2/i1 = -i1/i1= -1.\n",
-        "For finding h22 and h12, voltage source is connected at the output\n",
-        "v1=v2\n",
-        "h12=v1/v2 = v2/v2 = 1\n",
-        "h22=0.2 mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.2 : Page number 645-646\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=4.0;                          #1st resistor(at the input side), \u03a9\n",
-      "R2=4.0;                          #2nd resistor(at the middle), \u03a9\n",
-      "R3=4.0;                          #3rd resistor(at the output side), \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"To find h11 and h21, output terminals are shorted.\");\n",
-      "h11=R1 + (R2*R3/(R2+R3));                 #Input impedance with output shorted,  \u03a9\n",
-      "print(\"h11=%d\u03a9.\"%h11);\n",
-      "\n",
-      "#As the input current gets divided in half due to R2=R3.\n",
-      "print(\"Output current flowing into the box=negative of half of input current flowing out of the box.\");\n",
-      "print(\"i2=-i1/2 = -0.5i1\");                        \n",
-      "print(\"h21=i2/i1 = -0.5i1/i1= -0.5.\");       #Current gain with output shorted.\n",
-      "\n",
-      "print(\"For finding h22 and h12, voltage source is connected at the output\");\n",
-      "#As, there will be no current through the 1st 4k\u03a9 resistor due to open circuited input,\n",
-      "#Voltage gets equally divided across R2 and R3 resistor\n",
-      "print(\"v1=v2/2 = 0.5v2\");                    #Input voltage is equal to half of input voltage\n",
-      "print(\"h12=v1/v2 = 0.5v2/v2 = 0.5\");         #Voltage feedback ratio with input terminals open\n",
-      "\n",
-      "h22=1/(R2+R3);                               #Output admittance, mho\n",
-      "print(\"h22=%.3f mho\"%h22);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To find h11 and h21, output terminals are shorted.\n",
-        "h11=6\u03a9.\n",
-        "Output current flowing into the box=negative of half of input current flowing out of the box.\n",
-        "i2=-i1/2 = -0.5i1\n",
-        "h21=i2/i1 = -0.5i1/i1= -0.5.\n",
-        "For finding h22 and h12, voltage source is connected at the output\n",
-        "v1=v2/2 = 0.5v2\n",
-        "h12=v1/v2 = 0.5v2/v2 = 0.5\n",
-        "h22=0.125 mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.3 ; Page number 649-650\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=10.0;                  #Resistor at the input side, \u03a9\n",
-      "R2=5.0;                   #Resistor at the middle, \u03a9\n",
-      "rL=5.0;                   #Load resistor, \u03a9\n",
-      "\n",
-      "#h-parameter values from 24.1\n",
-      "h11=10.0;                     #Input impedance with output shorted,  \u03a9\n",
-      "h21=-1.0;                     #Current gain with output shorted\n",
-      "h12=1.0;                      #Voltage feedback ratio with input terminal open\n",
-      "h22=0.2;                    #Output admittance, mho\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=h11-(h12*h21/(h22+(1/rL)));                 #Input impedance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "Av=-h21/(Zin*(h22+(1/rL)));                     #voltage gain,\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The input impedance=%.1f\u03a9.\"%Zin );\n",
-      "print(\"(ii) The voltage gain=1/%d.\"%(1/Av));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The input impedance=12.5\u03a9.\n",
-        "(ii) The voltage gain=1/5.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.4 : Page number 652-653\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=10.0;                 #Collector-emitter voltage, V\n",
-      "IC=1.0;                   #Collector current, mA\n",
-      "rL=600.0;                 #a.c load seen by the transistor,\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=2000.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=10**-4;             #Output impedance, mho\n",
-      "hre=10**-3;             #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=hie - (hre*hfe/(hoe+(1/rL)));                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9. \\n As second term in the expression of Zin is small compared to first, Zin~hie=%d\u03a9.\"%(Zin,hie));\n",
-      "\n",
-      "#(ii)\n",
-      "Ai=hfe/(1+hoe*rL);                      #Current gain\n",
-      "print(\"Current gain=%d\"%Ai);\n",
-      "print(\"if hoe*rL<<1, then Ai~hfe=%d.\"%hfe);\n",
-      "\n",
-      "#(iii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/rL)));                   #Voltage gain\n",
-      "print(\"Voltage gain=%.1f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1972 \u03a9. \n",
-        " As second term in the expression of Zin is small compared to first, Zin~hie=2000\u03a9.\n",
-        "Current gain=47\n",
-        "if hoe*rL<<1, then Ai~hfe=50.\n",
-        "Voltage gain=-14.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.5 : Page number 653\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCE=5.0;                 #Collector-emitter voltage, V\n",
-      "IC=1.0;                   #Collector current, mA\n",
-      "rL=2.0;                 #a.c load seen by the transistor,\u03a9\n",
-      "\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1700.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=6*10**-6;           #Output impedance, mho\n",
-      "hre=1.3*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=38.0;                 #Current gain with output shorted\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=hie - (hre*hfe/(hoe+(1/(rL*1000))));                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n",
-      "\n",
-      "#(ii)\n",
-      "Ai=ceil((hfe/round((1+hoe*rL*1000),3))*10)/10;            #Current gain\n",
-      "print(\"Current gain=%.1f\"%Ai);\n",
-      "\n",
-      "#(iii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/(rL*1000))));                   #Voltage gain\n",
-      "print(\"Voltage gain=%.1f\"%abs(Av));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1690 \u03a9.\n",
-        "Current gain=37.6\n",
-        "Voltage gain=44.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.6 : Page number 653-654\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=10.0;                  #Collector resistance, k\u03a9\n",
-      "RL=30.0;                  #Load resistance, k\u03a9\n",
-      "R1=80.0;                  #Resistor R1, k\u03a9\n",
-      "R2=40.0;                  #Resistor R2, k\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1500.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=5*10**-5;           #Output impedance, mho\n",
-      "hre=4*10**-4;           #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=((RC*RL)/(RC+RL))*1000;             #a.c load as seen by resistance, \u03a9\n",
-      "\n",
-      "#(i)\n",
-      "Zin=round(hie - (hre*hfe/(hoe+(1/rL))),-1);                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n",
-      "\n",
-      "#Input impedance of stage=input impedance || bias resistors\n",
-      "Zin_stage=round(pr(pr(R1,R2)*1000,Zin),-1);        #\u03a9\n",
-      "print(\"Input impedance of the stage=%.0f \u03a9.\"%Zin_stage);\n",
-      "\n",
-      "#(ii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/rL)));                   #Voltage gain\n",
-      "print(\"Voltage gain=%d\"%Av);\n",
-      "print(\"The negative sign represents phase reversal.\");\n",
-      "\n",
-      "\n",
-      "#(iii)\n",
-      "Zout=(1/(hoe-(hfe*hre/hie)))/1000;                     #Output impedance of transistor, k\u03a9\n",
-      "Zout_stage=pr(Zout,pr(RL,RC));                    #Output impedance of the stage, k\u03a9\n",
-      "print(\"Output impedance=%.2f k\u03a9.\"%Zout);\n",
-      "print(\"Output impedance of the stage=%.2f k\u03a9.\"%Zout_stage);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1390 \u03a9.\n",
-        "Input impedance of the stage=1320 \u03a9.\n",
-        "Voltage gain=-196\n",
-        "The negative sign represents phase reversal.\n",
-        "Output impedance=27.27 k\u03a9.\n",
-        "Output impedance of the stage=5.88 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.7 : Page number 654\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=4.7;                  #Collector resistance, k\u03a9\n",
-      "RL=10.0;                   #Load resistance, k\u03a9\n",
-      "R1=33.0;                   #Resistor R1, k\u03a9\n",
-      "R2=10.0;                   #Resistor R2, k\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1;                  #Input impedance with output shorted, k\u03a9\n",
-      "hoe=25;                 #Output impedance, \u03bcS\n",
-      "hre=2.5*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=50;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=(RC*RL)/(RC+RL);             #a.c load as seen by resistance, k\u03a9\n",
-      "\n",
-      "Ai=hfe/(1+hoe*10**-6*rL*1000);            #Current gain\n",
-      "print(\"Current gain=%.1f\"%Ai);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current gain=46.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.8  : Page number 654-655\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R_S=100.0;                    #Series resistance, \u03a9 \n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1.0;                  #Input impedance with output shorted, k\u03a9\n",
-      "hoe=25.0;                 #Output impedance, \u03bcS\n",
-      "hre=2.5*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Zout=(1/(hoe*10**-6-(hfe*hre/(hie*1000+R_S))))/1000;                #Output impedance of transistor, k\u03a9\n",
-      "print(\"Output impedance=%.1f k\u03a9.\"%Zout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output impedance=73.3 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.9 : Page number 656\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=12.0;                  #Collector resistance, k\u03a9\n",
-      "RL=15.0;                  #Load resistance, k\u03a9\n",
-      "R1=50.0;                  #Resistor R1, k\u03a9\n",
-      "R2=5.0;                   #Resistor R2, k\u03a9\n",
-      "hie=1.94;               #Input impedance with output shorted, k\u03a9\n",
-      "hfe=71.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=(RC*RL)/(RC+RL);             #a.c load as seen by resistance, \u03a9\n",
-      "\n",
-      "#(i)\n",
-      "Zin_base=hie;                                           #Transistor input impedance, k\u03a9\n",
-      "Zin_circuit=floor(pr(Zin_base,pr(R1,R2))*100)/100;      #Circuit input impedance, k\u03a9\n",
-      "print(\"Circuit input impedance=%.2fk\u03a9\"%Zin_circuit);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "Av=hfe*rL/hie;                   #Voltage gain\n",
-      "print(\"Voltage gain=%.0f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Circuit input impedance=1.35k\u03a9\n",
-        "Voltage gain=244\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.10 : Page number 656\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "hie_min=600;            #Minimum input impedance with output shorted, \u03a9\n",
-      "hfe_min=110;            #Minimum current gain with output shorted\n",
-      "hie_max=800;            #Maximum input impedance with output shorted, \u03a9\n",
-      "hfe_max=140;            #Maximum current gain with output shorted\n",
-      "rL=460;                 #a.c collector load, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "hie=round(sqrt(hie_min*hie_max));          #Input impedance with output shorted, \u03a9\n",
-      "hfe=round(sqrt(hfe_min*hfe_max));          #Current gain with output shorted\n",
-      "Av=hfe*rL/hie;                      #Voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage gain=%.1f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain=82.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.11 : Page number 658-659\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#(a)Variable declaration\n",
-      "Ib=10;              #Base current, \u03bcA\n",
-      "Ic=1;               #Collector current, mA\n",
-      "Vbe=10;             #Base-emitter voltage, mV\n",
-      "\n",
-      "#Calculation\n",
-      "hie=Vbe*10**-3/(Ib*10**-6);           #Input impedance with output shorted, \u03a9\n",
-      "hfe=Ic*10**-3/(Ib*10**-6);            #Current gain with output shorted\n",
-      "\n",
-      "#(b) Variable declaration\n",
-      "Vbe=0.65;             #Base-emitter voltage, mV\n",
-      "Ic=60;                #Collector current, \u03bcA\n",
-      "Vce=1;                #Collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "hre=Vbe*10**-3/Vce;           #Voltage feedback ratio with input terminal open\n",
-      "hoe=Ic/Vce;                   #Output impedance, \u03bcmho\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"hie=%d\u03a9\"%hie);\n",
-      "print(\"hfe=%d\"%hfe);\n",
-      "print(\"hre=%.2fe\u201303\"%(hre*1000));\n",
-      "print(\"hoe=%d\u03bcmho\"%hoe);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hie=1000\u03a9\n",
-        "hfe=100\n",
-        "hre=0.65e\u201303\n",
-        "hoe=60\u03bcmho\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_1.ipynb
deleted file mode 100755
index e63c17a6..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_1.ipynb
+++ /dev/null
@@ -1,604 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:947f358cf49d029c94d008f72a340051744678cf2e36ecc199100b78d31fcba5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 24 : HYBRID PARAMETERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.1 : Page number 644-645\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=10.0;                          #1st resistor, \u03a9\n",
-      "R2=5.0;                           #2nd resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"To find h11 and h21, output terminals are shorted.\");\n",
-      "h11=R1;                 #Input impedance with output shorted,  \u03a9\n",
-      "\n",
-      "print(\"h11=%d\u03a9.\"%h11);\n",
-      "\n",
-      "print(\"Output current flowing into the box= input current flowing out of the box.\");\n",
-      "print(\"i2=-i1\");                        #Output current flowing into the box= input current flowing out of the box.\n",
-      "print(\"h21=i2/i1 = -i1/i1= -1.\");       #Current gain with output shorted.\n",
-      "\n",
-      "\n",
-      "print(\"For finding h22 and h12, voltage source is connected at the output\");\n",
-      "#As, there will be no current through 10k\u03a9 resistor due to open circuited input,\n",
-      "print(\"v1=v2\");                         #Output voltage is equal to input voltage(equal to voltage drop across 5k\u03a9 resistor)\n",
-      "print(\"h12=v1/v2 = v2/v2 = 1\");         #Voltage feedback ratio with input terminals open\n",
-      "\n",
-      "h22=1/R2;                               #Output admittance, mho\n",
-      "print(\"h22=%.1f mho\"%h22);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To find h11 and h21, output terminals are shorted.\n",
-        "h11=10\u03a9.\n",
-        "Output current flowing into the box= input current flowing out of the box.\n",
-        "i2=-i1\n",
-        "h21=i2/i1 = -i1/i1= -1.\n",
-        "For finding h22 and h12, voltage source is connected at the output\n",
-        "v1=v2\n",
-        "h12=v1/v2 = v2/v2 = 1\n",
-        "h22=0.2 mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.2 : Page number 645-646\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=4.0;                          #1st resistor(at the input side), \u03a9\n",
-      "R2=4.0;                          #2nd resistor(at the middle), \u03a9\n",
-      "R3=4.0;                          #3rd resistor(at the output side), \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"To find h11 and h21, output terminals are shorted.\");\n",
-      "h11=R1 + (R2*R3/(R2+R3));                 #Input impedance with output shorted,  \u03a9\n",
-      "print(\"h11=%d\u03a9.\"%h11);\n",
-      "\n",
-      "#As the input current gets divided in half due to R2=R3.\n",
-      "print(\"Output current flowing into the box=negative of half of input current flowing out of the box.\");\n",
-      "print(\"i2=-i1/2 = -0.5i1\");                        \n",
-      "print(\"h21=i2/i1 = -0.5i1/i1= -0.5.\");       #Current gain with output shorted.\n",
-      "\n",
-      "print(\"For finding h22 and h12, voltage source is connected at the output\");\n",
-      "#As, there will be no current through the 1st 4k\u03a9 resistor due to open circuited input,\n",
-      "#Voltage gets equally divided across R2 and R3 resistor\n",
-      "print(\"v1=v2/2 = 0.5v2\");                    #Input voltage is equal to half of input voltage\n",
-      "print(\"h12=v1/v2 = 0.5v2/v2 = 0.5\");         #Voltage feedback ratio with input terminals open\n",
-      "\n",
-      "h22=1/(R2+R3);                               #Output admittance, mho\n",
-      "print(\"h22=%.3f mho\"%h22);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To find h11 and h21, output terminals are shorted.\n",
-        "h11=6\u03a9.\n",
-        "Output current flowing into the box=negative of half of input current flowing out of the box.\n",
-        "i2=-i1/2 = -0.5i1\n",
-        "h21=i2/i1 = -0.5i1/i1= -0.5.\n",
-        "For finding h22 and h12, voltage source is connected at the output\n",
-        "v1=v2/2 = 0.5v2\n",
-        "h12=v1/v2 = 0.5v2/v2 = 0.5\n",
-        "h22=0.125 mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.3 ; Page number 649-650\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=10.0;                  #Resistor at the input side, \u03a9\n",
-      "R2=5.0;                   #Resistor at the middle, \u03a9\n",
-      "rL=5.0;                   #Load resistor, \u03a9\n",
-      "\n",
-      "#h-parameter values from 24.1\n",
-      "h11=10.0;                     #Input impedance with output shorted,  \u03a9\n",
-      "h21=-1.0;                     #Current gain with output shorted\n",
-      "h12=1.0;                      #Voltage feedback ratio with input terminal open\n",
-      "h22=0.2;                    #Output admittance, mho\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=h11-(h12*h21/(h22+(1/rL)));                 #Input impedance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "Av=-h21/(Zin*(h22+(1/rL)));                     #voltage gain,\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The input impedance=%.1f\u03a9.\"%Zin );\n",
-      "print(\"(ii) The voltage gain=1/%d.\"%(1/Av));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The input impedance=12.5\u03a9.\n",
-        "(ii) The voltage gain=1/5.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.4 : Page number 652-653\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=10.0;                 #Collector-emitter voltage, V\n",
-      "IC=1.0;                   #Collector current, mA\n",
-      "rL=600.0;                 #a.c load seen by the transistor,\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=2000.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=10**-4;             #Output impedance, mho\n",
-      "hre=10**-3;             #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=hie - (hre*hfe/(hoe+(1/rL)));                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9. \\n As second term in the expression of Zin is small compared to first, Zin~hie=%d\u03a9.\"%(Zin,hie));\n",
-      "\n",
-      "#(ii)\n",
-      "Ai=hfe/(1+hoe*rL);                      #Current gain\n",
-      "print(\"Current gain=%d\"%Ai);\n",
-      "print(\"if hoe*rL<<1, then Ai~hfe=%d.\"%hfe);\n",
-      "\n",
-      "#(iii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/rL)));                   #Voltage gain\n",
-      "print(\"Voltage gain=%.1f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1972 \u03a9. \n",
-        " As second term in the expression of Zin is small compared to first, Zin~hie=2000\u03a9.\n",
-        "Current gain=47\n",
-        "if hoe*rL<<1, then Ai~hfe=50.\n",
-        "Voltage gain=-14.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.5 : Page number 653\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCE=5.0;                 #Collector-emitter voltage, V\n",
-      "IC=1.0;                   #Collector current, mA\n",
-      "rL=2.0;                 #a.c load seen by the transistor,\u03a9\n",
-      "\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1700.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=6*10**-6;           #Output impedance, mho\n",
-      "hre=1.3*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=38.0;                 #Current gain with output shorted\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=hie - (hre*hfe/(hoe+(1/(rL*1000))));                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n",
-      "\n",
-      "#(ii)\n",
-      "Ai=ceil((hfe/round((1+hoe*rL*1000),3))*10)/10;            #Current gain\n",
-      "print(\"Current gain=%.1f\"%Ai);\n",
-      "\n",
-      "#(iii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/(rL*1000))));                   #Voltage gain\n",
-      "print(\"Voltage gain=%.1f\"%abs(Av));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1690 \u03a9.\n",
-        "Current gain=37.6\n",
-        "Voltage gain=44.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.6 : Page number 653-654\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=10.0;                  #Collector resistance, k\u03a9\n",
-      "RL=30.0;                  #Load resistance, k\u03a9\n",
-      "R1=80.0;                  #Resistor R1, k\u03a9\n",
-      "R2=40.0;                  #Resistor R2, k\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1500.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=5*10**-5;           #Output impedance, mho\n",
-      "hre=4*10**-4;           #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=((RC*RL)/(RC+RL))*1000;             #a.c load as seen by resistance, \u03a9\n",
-      "\n",
-      "#(i)\n",
-      "Zin=round(hie - (hre*hfe/(hoe+(1/rL))),-1);                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n",
-      "\n",
-      "#Input impedance of stage=input impedance || bias resistors\n",
-      "Zin_stage=round(pr(pr(R1,R2)*1000,Zin),-1);        #\u03a9\n",
-      "print(\"Input impedance of the stage=%.0f \u03a9.\"%Zin_stage);\n",
-      "\n",
-      "#(ii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/rL)));                   #Voltage gain\n",
-      "print(\"Voltage gain=%d\"%Av);\n",
-      "print(\"The negative sign represents phase reversal.\");\n",
-      "\n",
-      "\n",
-      "#(iii)\n",
-      "Zout=(1/(hoe-(hfe*hre/hie)))/1000;                     #Output impedance of transistor, k\u03a9\n",
-      "Zout_stage=pr(Zout,pr(RL,RC));                    #Output impedance of the stage, k\u03a9\n",
-      "print(\"Output impedance=%.2f k\u03a9.\"%Zout);\n",
-      "print(\"Output impedance of the stage=%.2f k\u03a9.\"%Zout_stage);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1390 \u03a9.\n",
-        "Input impedance of the stage=1320 \u03a9.\n",
-        "Voltage gain=-196\n",
-        "The negative sign represents phase reversal.\n",
-        "Output impedance=27.27 k\u03a9.\n",
-        "Output impedance of the stage=5.88 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.7 : Page number 654\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=4.7;                  #Collector resistance, k\u03a9\n",
-      "RL=10.0;                   #Load resistance, k\u03a9\n",
-      "R1=33.0;                   #Resistor R1, k\u03a9\n",
-      "R2=10.0;                   #Resistor R2, k\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1;                  #Input impedance with output shorted, k\u03a9\n",
-      "hoe=25;                 #Output impedance, \u03bcS\n",
-      "hre=2.5*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=50;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=(RC*RL)/(RC+RL);             #a.c load as seen by resistance, k\u03a9\n",
-      "\n",
-      "Ai=hfe/(1+hoe*10**-6*rL*1000);            #Current gain\n",
-      "print(\"Current gain=%.1f\"%Ai);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current gain=46.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.8  : Page number 654-655\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R_S=100.0;                    #Series resistance, \u03a9 \n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1.0;                  #Input impedance with output shorted, k\u03a9\n",
-      "hoe=25.0;                 #Output impedance, \u03bcS\n",
-      "hre=2.5*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Zout=(1/(hoe*10**-6-(hfe*hre/(hie*1000+R_S))))/1000;                #Output impedance of transistor, k\u03a9\n",
-      "print(\"Output impedance=%.1f k\u03a9.\"%Zout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output impedance=73.3 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.9 : Page number 656\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=12.0;                  #Collector resistance, k\u03a9\n",
-      "RL=15.0;                  #Load resistance, k\u03a9\n",
-      "R1=50.0;                  #Resistor R1, k\u03a9\n",
-      "R2=5.0;                   #Resistor R2, k\u03a9\n",
-      "hie=1.94;               #Input impedance with output shorted, k\u03a9\n",
-      "hfe=71.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=(RC*RL)/(RC+RL);             #a.c load as seen by resistance, \u03a9\n",
-      "\n",
-      "#(i)\n",
-      "Zin_base=hie;                                           #Transistor input impedance, k\u03a9\n",
-      "Zin_circuit=floor(pr(Zin_base,pr(R1,R2))*100)/100;      #Circuit input impedance, k\u03a9\n",
-      "print(\"Circuit input impedance=%.2fk\u03a9\"%Zin_circuit);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "Av=hfe*rL/hie;                   #Voltage gain\n",
-      "print(\"Voltage gain=%.0f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Circuit input impedance=1.35k\u03a9\n",
-        "Voltage gain=244\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.10 : Page number 656\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "hie_min=600;            #Minimum input impedance with output shorted, \u03a9\n",
-      "hfe_min=110;            #Minimum current gain with output shorted\n",
-      "hie_max=800;            #Maximum input impedance with output shorted, \u03a9\n",
-      "hfe_max=140;            #Maximum current gain with output shorted\n",
-      "rL=460;                 #a.c collector load, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "hie=round(sqrt(hie_min*hie_max));          #Input impedance with output shorted, \u03a9\n",
-      "hfe=round(sqrt(hfe_min*hfe_max));          #Current gain with output shorted\n",
-      "Av=hfe*rL/hie;                      #Voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage gain=%.1f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain=82.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.11 : Page number 658-659\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#(a)Variable declaration\n",
-      "Ib=10;              #Base current, \u03bcA\n",
-      "Ic=1;               #Collector current, mA\n",
-      "Vbe=10;             #Base-emitter voltage, mV\n",
-      "\n",
-      "#Calculation\n",
-      "hie=Vbe*10**-3/(Ib*10**-6);           #Input impedance with output shorted, \u03a9\n",
-      "hfe=Ic*10**-3/(Ib*10**-6);            #Current gain with output shorted\n",
-      "\n",
-      "#(b) Variable declaration\n",
-      "Vbe=0.65;             #Base-emitter voltage, mV\n",
-      "Ic=60;                #Collector current, \u03bcA\n",
-      "Vce=1;                #Collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "hre=Vbe*10**-3/Vce;           #Voltage feedback ratio with input terminal open\n",
-      "hoe=Ic/Vce;                   #Output impedance, \u03bcmho\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"hie=%d\u03a9\"%hie);\n",
-      "print(\"hfe=%d\"%hfe);\n",
-      "print(\"hre=%.2fe\u201303\"%(hre*1000));\n",
-      "print(\"hoe=%d\u03bcmho\"%hoe);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hie=1000\u03a9\n",
-        "hfe=100\n",
-        "hre=0.65e\u201303\n",
-        "hoe=60\u03bcmho\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_2.ipynb
deleted file mode 100755
index e63c17a6..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_2.ipynb
+++ /dev/null
@@ -1,604 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:947f358cf49d029c94d008f72a340051744678cf2e36ecc199100b78d31fcba5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 24 : HYBRID PARAMETERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.1 : Page number 644-645\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=10.0;                          #1st resistor, \u03a9\n",
-      "R2=5.0;                           #2nd resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"To find h11 and h21, output terminals are shorted.\");\n",
-      "h11=R1;                 #Input impedance with output shorted,  \u03a9\n",
-      "\n",
-      "print(\"h11=%d\u03a9.\"%h11);\n",
-      "\n",
-      "print(\"Output current flowing into the box= input current flowing out of the box.\");\n",
-      "print(\"i2=-i1\");                        #Output current flowing into the box= input current flowing out of the box.\n",
-      "print(\"h21=i2/i1 = -i1/i1= -1.\");       #Current gain with output shorted.\n",
-      "\n",
-      "\n",
-      "print(\"For finding h22 and h12, voltage source is connected at the output\");\n",
-      "#As, there will be no current through 10k\u03a9 resistor due to open circuited input,\n",
-      "print(\"v1=v2\");                         #Output voltage is equal to input voltage(equal to voltage drop across 5k\u03a9 resistor)\n",
-      "print(\"h12=v1/v2 = v2/v2 = 1\");         #Voltage feedback ratio with input terminals open\n",
-      "\n",
-      "h22=1/R2;                               #Output admittance, mho\n",
-      "print(\"h22=%.1f mho\"%h22);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To find h11 and h21, output terminals are shorted.\n",
-        "h11=10\u03a9.\n",
-        "Output current flowing into the box= input current flowing out of the box.\n",
-        "i2=-i1\n",
-        "h21=i2/i1 = -i1/i1= -1.\n",
-        "For finding h22 and h12, voltage source is connected at the output\n",
-        "v1=v2\n",
-        "h12=v1/v2 = v2/v2 = 1\n",
-        "h22=0.2 mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.2 : Page number 645-646\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=4.0;                          #1st resistor(at the input side), \u03a9\n",
-      "R2=4.0;                          #2nd resistor(at the middle), \u03a9\n",
-      "R3=4.0;                          #3rd resistor(at the output side), \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"To find h11 and h21, output terminals are shorted.\");\n",
-      "h11=R1 + (R2*R3/(R2+R3));                 #Input impedance with output shorted,  \u03a9\n",
-      "print(\"h11=%d\u03a9.\"%h11);\n",
-      "\n",
-      "#As the input current gets divided in half due to R2=R3.\n",
-      "print(\"Output current flowing into the box=negative of half of input current flowing out of the box.\");\n",
-      "print(\"i2=-i1/2 = -0.5i1\");                        \n",
-      "print(\"h21=i2/i1 = -0.5i1/i1= -0.5.\");       #Current gain with output shorted.\n",
-      "\n",
-      "print(\"For finding h22 and h12, voltage source is connected at the output\");\n",
-      "#As, there will be no current through the 1st 4k\u03a9 resistor due to open circuited input,\n",
-      "#Voltage gets equally divided across R2 and R3 resistor\n",
-      "print(\"v1=v2/2 = 0.5v2\");                    #Input voltage is equal to half of input voltage\n",
-      "print(\"h12=v1/v2 = 0.5v2/v2 = 0.5\");         #Voltage feedback ratio with input terminals open\n",
-      "\n",
-      "h22=1/(R2+R3);                               #Output admittance, mho\n",
-      "print(\"h22=%.3f mho\"%h22);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To find h11 and h21, output terminals are shorted.\n",
-        "h11=6\u03a9.\n",
-        "Output current flowing into the box=negative of half of input current flowing out of the box.\n",
-        "i2=-i1/2 = -0.5i1\n",
-        "h21=i2/i1 = -0.5i1/i1= -0.5.\n",
-        "For finding h22 and h12, voltage source is connected at the output\n",
-        "v1=v2/2 = 0.5v2\n",
-        "h12=v1/v2 = 0.5v2/v2 = 0.5\n",
-        "h22=0.125 mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.3 ; Page number 649-650\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=10.0;                  #Resistor at the input side, \u03a9\n",
-      "R2=5.0;                   #Resistor at the middle, \u03a9\n",
-      "rL=5.0;                   #Load resistor, \u03a9\n",
-      "\n",
-      "#h-parameter values from 24.1\n",
-      "h11=10.0;                     #Input impedance with output shorted,  \u03a9\n",
-      "h21=-1.0;                     #Current gain with output shorted\n",
-      "h12=1.0;                      #Voltage feedback ratio with input terminal open\n",
-      "h22=0.2;                    #Output admittance, mho\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=h11-(h12*h21/(h22+(1/rL)));                 #Input impedance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "Av=-h21/(Zin*(h22+(1/rL)));                     #voltage gain,\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The input impedance=%.1f\u03a9.\"%Zin );\n",
-      "print(\"(ii) The voltage gain=1/%d.\"%(1/Av));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The input impedance=12.5\u03a9.\n",
-        "(ii) The voltage gain=1/5.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.4 : Page number 652-653\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=10.0;                 #Collector-emitter voltage, V\n",
-      "IC=1.0;                   #Collector current, mA\n",
-      "rL=600.0;                 #a.c load seen by the transistor,\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=2000.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=10**-4;             #Output impedance, mho\n",
-      "hre=10**-3;             #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=hie - (hre*hfe/(hoe+(1/rL)));                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9. \\n As second term in the expression of Zin is small compared to first, Zin~hie=%d\u03a9.\"%(Zin,hie));\n",
-      "\n",
-      "#(ii)\n",
-      "Ai=hfe/(1+hoe*rL);                      #Current gain\n",
-      "print(\"Current gain=%d\"%Ai);\n",
-      "print(\"if hoe*rL<<1, then Ai~hfe=%d.\"%hfe);\n",
-      "\n",
-      "#(iii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/rL)));                   #Voltage gain\n",
-      "print(\"Voltage gain=%.1f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1972 \u03a9. \n",
-        " As second term in the expression of Zin is small compared to first, Zin~hie=2000\u03a9.\n",
-        "Current gain=47\n",
-        "if hoe*rL<<1, then Ai~hfe=50.\n",
-        "Voltage gain=-14.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.5 : Page number 653\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCE=5.0;                 #Collector-emitter voltage, V\n",
-      "IC=1.0;                   #Collector current, mA\n",
-      "rL=2.0;                 #a.c load seen by the transistor,\u03a9\n",
-      "\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1700.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=6*10**-6;           #Output impedance, mho\n",
-      "hre=1.3*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=38.0;                 #Current gain with output shorted\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=hie - (hre*hfe/(hoe+(1/(rL*1000))));                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n",
-      "\n",
-      "#(ii)\n",
-      "Ai=ceil((hfe/round((1+hoe*rL*1000),3))*10)/10;            #Current gain\n",
-      "print(\"Current gain=%.1f\"%Ai);\n",
-      "\n",
-      "#(iii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/(rL*1000))));                   #Voltage gain\n",
-      "print(\"Voltage gain=%.1f\"%abs(Av));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1690 \u03a9.\n",
-        "Current gain=37.6\n",
-        "Voltage gain=44.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.6 : Page number 653-654\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=10.0;                  #Collector resistance, k\u03a9\n",
-      "RL=30.0;                  #Load resistance, k\u03a9\n",
-      "R1=80.0;                  #Resistor R1, k\u03a9\n",
-      "R2=40.0;                  #Resistor R2, k\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1500.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=5*10**-5;           #Output impedance, mho\n",
-      "hre=4*10**-4;           #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=((RC*RL)/(RC+RL))*1000;             #a.c load as seen by resistance, \u03a9\n",
-      "\n",
-      "#(i)\n",
-      "Zin=round(hie - (hre*hfe/(hoe+(1/rL))),-1);                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n",
-      "\n",
-      "#Input impedance of stage=input impedance || bias resistors\n",
-      "Zin_stage=round(pr(pr(R1,R2)*1000,Zin),-1);        #\u03a9\n",
-      "print(\"Input impedance of the stage=%.0f \u03a9.\"%Zin_stage);\n",
-      "\n",
-      "#(ii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/rL)));                   #Voltage gain\n",
-      "print(\"Voltage gain=%d\"%Av);\n",
-      "print(\"The negative sign represents phase reversal.\");\n",
-      "\n",
-      "\n",
-      "#(iii)\n",
-      "Zout=(1/(hoe-(hfe*hre/hie)))/1000;                     #Output impedance of transistor, k\u03a9\n",
-      "Zout_stage=pr(Zout,pr(RL,RC));                    #Output impedance of the stage, k\u03a9\n",
-      "print(\"Output impedance=%.2f k\u03a9.\"%Zout);\n",
-      "print(\"Output impedance of the stage=%.2f k\u03a9.\"%Zout_stage);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1390 \u03a9.\n",
-        "Input impedance of the stage=1320 \u03a9.\n",
-        "Voltage gain=-196\n",
-        "The negative sign represents phase reversal.\n",
-        "Output impedance=27.27 k\u03a9.\n",
-        "Output impedance of the stage=5.88 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.7 : Page number 654\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=4.7;                  #Collector resistance, k\u03a9\n",
-      "RL=10.0;                   #Load resistance, k\u03a9\n",
-      "R1=33.0;                   #Resistor R1, k\u03a9\n",
-      "R2=10.0;                   #Resistor R2, k\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1;                  #Input impedance with output shorted, k\u03a9\n",
-      "hoe=25;                 #Output impedance, \u03bcS\n",
-      "hre=2.5*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=50;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=(RC*RL)/(RC+RL);             #a.c load as seen by resistance, k\u03a9\n",
-      "\n",
-      "Ai=hfe/(1+hoe*10**-6*rL*1000);            #Current gain\n",
-      "print(\"Current gain=%.1f\"%Ai);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current gain=46.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.8  : Page number 654-655\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R_S=100.0;                    #Series resistance, \u03a9 \n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1.0;                  #Input impedance with output shorted, k\u03a9\n",
-      "hoe=25.0;                 #Output impedance, \u03bcS\n",
-      "hre=2.5*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Zout=(1/(hoe*10**-6-(hfe*hre/(hie*1000+R_S))))/1000;                #Output impedance of transistor, k\u03a9\n",
-      "print(\"Output impedance=%.1f k\u03a9.\"%Zout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output impedance=73.3 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.9 : Page number 656\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=12.0;                  #Collector resistance, k\u03a9\n",
-      "RL=15.0;                  #Load resistance, k\u03a9\n",
-      "R1=50.0;                  #Resistor R1, k\u03a9\n",
-      "R2=5.0;                   #Resistor R2, k\u03a9\n",
-      "hie=1.94;               #Input impedance with output shorted, k\u03a9\n",
-      "hfe=71.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=(RC*RL)/(RC+RL);             #a.c load as seen by resistance, \u03a9\n",
-      "\n",
-      "#(i)\n",
-      "Zin_base=hie;                                           #Transistor input impedance, k\u03a9\n",
-      "Zin_circuit=floor(pr(Zin_base,pr(R1,R2))*100)/100;      #Circuit input impedance, k\u03a9\n",
-      "print(\"Circuit input impedance=%.2fk\u03a9\"%Zin_circuit);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "Av=hfe*rL/hie;                   #Voltage gain\n",
-      "print(\"Voltage gain=%.0f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Circuit input impedance=1.35k\u03a9\n",
-        "Voltage gain=244\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.10 : Page number 656\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "hie_min=600;            #Minimum input impedance with output shorted, \u03a9\n",
-      "hfe_min=110;            #Minimum current gain with output shorted\n",
-      "hie_max=800;            #Maximum input impedance with output shorted, \u03a9\n",
-      "hfe_max=140;            #Maximum current gain with output shorted\n",
-      "rL=460;                 #a.c collector load, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "hie=round(sqrt(hie_min*hie_max));          #Input impedance with output shorted, \u03a9\n",
-      "hfe=round(sqrt(hfe_min*hfe_max));          #Current gain with output shorted\n",
-      "Av=hfe*rL/hie;                      #Voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage gain=%.1f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain=82.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.11 : Page number 658-659\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#(a)Variable declaration\n",
-      "Ib=10;              #Base current, \u03bcA\n",
-      "Ic=1;               #Collector current, mA\n",
-      "Vbe=10;             #Base-emitter voltage, mV\n",
-      "\n",
-      "#Calculation\n",
-      "hie=Vbe*10**-3/(Ib*10**-6);           #Input impedance with output shorted, \u03a9\n",
-      "hfe=Ic*10**-3/(Ib*10**-6);            #Current gain with output shorted\n",
-      "\n",
-      "#(b) Variable declaration\n",
-      "Vbe=0.65;             #Base-emitter voltage, mV\n",
-      "Ic=60;                #Collector current, \u03bcA\n",
-      "Vce=1;                #Collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "hre=Vbe*10**-3/Vce;           #Voltage feedback ratio with input terminal open\n",
-      "hoe=Ic/Vce;                   #Output impedance, \u03bcmho\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"hie=%d\u03a9\"%hie);\n",
-      "print(\"hfe=%d\"%hfe);\n",
-      "print(\"hre=%.2fe\u201303\"%(hre*1000));\n",
-      "print(\"hoe=%d\u03bcmho\"%hoe);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hie=1000\u03a9\n",
-        "hfe=100\n",
-        "hre=0.65e\u201303\n",
-        "hoe=60\u03bcmho\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_3.ipynb
deleted file mode 100755
index e63c17a6..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_3.ipynb
+++ /dev/null
@@ -1,604 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:947f358cf49d029c94d008f72a340051744678cf2e36ecc199100b78d31fcba5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 24 : HYBRID PARAMETERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.1 : Page number 644-645\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=10.0;                          #1st resistor, \u03a9\n",
-      "R2=5.0;                           #2nd resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"To find h11 and h21, output terminals are shorted.\");\n",
-      "h11=R1;                 #Input impedance with output shorted,  \u03a9\n",
-      "\n",
-      "print(\"h11=%d\u03a9.\"%h11);\n",
-      "\n",
-      "print(\"Output current flowing into the box= input current flowing out of the box.\");\n",
-      "print(\"i2=-i1\");                        #Output current flowing into the box= input current flowing out of the box.\n",
-      "print(\"h21=i2/i1 = -i1/i1= -1.\");       #Current gain with output shorted.\n",
-      "\n",
-      "\n",
-      "print(\"For finding h22 and h12, voltage source is connected at the output\");\n",
-      "#As, there will be no current through 10k\u03a9 resistor due to open circuited input,\n",
-      "print(\"v1=v2\");                         #Output voltage is equal to input voltage(equal to voltage drop across 5k\u03a9 resistor)\n",
-      "print(\"h12=v1/v2 = v2/v2 = 1\");         #Voltage feedback ratio with input terminals open\n",
-      "\n",
-      "h22=1/R2;                               #Output admittance, mho\n",
-      "print(\"h22=%.1f mho\"%h22);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To find h11 and h21, output terminals are shorted.\n",
-        "h11=10\u03a9.\n",
-        "Output current flowing into the box= input current flowing out of the box.\n",
-        "i2=-i1\n",
-        "h21=i2/i1 = -i1/i1= -1.\n",
-        "For finding h22 and h12, voltage source is connected at the output\n",
-        "v1=v2\n",
-        "h12=v1/v2 = v2/v2 = 1\n",
-        "h22=0.2 mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.2 : Page number 645-646\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=4.0;                          #1st resistor(at the input side), \u03a9\n",
-      "R2=4.0;                          #2nd resistor(at the middle), \u03a9\n",
-      "R3=4.0;                          #3rd resistor(at the output side), \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"To find h11 and h21, output terminals are shorted.\");\n",
-      "h11=R1 + (R2*R3/(R2+R3));                 #Input impedance with output shorted,  \u03a9\n",
-      "print(\"h11=%d\u03a9.\"%h11);\n",
-      "\n",
-      "#As the input current gets divided in half due to R2=R3.\n",
-      "print(\"Output current flowing into the box=negative of half of input current flowing out of the box.\");\n",
-      "print(\"i2=-i1/2 = -0.5i1\");                        \n",
-      "print(\"h21=i2/i1 = -0.5i1/i1= -0.5.\");       #Current gain with output shorted.\n",
-      "\n",
-      "print(\"For finding h22 and h12, voltage source is connected at the output\");\n",
-      "#As, there will be no current through the 1st 4k\u03a9 resistor due to open circuited input,\n",
-      "#Voltage gets equally divided across R2 and R3 resistor\n",
-      "print(\"v1=v2/2 = 0.5v2\");                    #Input voltage is equal to half of input voltage\n",
-      "print(\"h12=v1/v2 = 0.5v2/v2 = 0.5\");         #Voltage feedback ratio with input terminals open\n",
-      "\n",
-      "h22=1/(R2+R3);                               #Output admittance, mho\n",
-      "print(\"h22=%.3f mho\"%h22);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To find h11 and h21, output terminals are shorted.\n",
-        "h11=6\u03a9.\n",
-        "Output current flowing into the box=negative of half of input current flowing out of the box.\n",
-        "i2=-i1/2 = -0.5i1\n",
-        "h21=i2/i1 = -0.5i1/i1= -0.5.\n",
-        "For finding h22 and h12, voltage source is connected at the output\n",
-        "v1=v2/2 = 0.5v2\n",
-        "h12=v1/v2 = 0.5v2/v2 = 0.5\n",
-        "h22=0.125 mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.3 ; Page number 649-650\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=10.0;                  #Resistor at the input side, \u03a9\n",
-      "R2=5.0;                   #Resistor at the middle, \u03a9\n",
-      "rL=5.0;                   #Load resistor, \u03a9\n",
-      "\n",
-      "#h-parameter values from 24.1\n",
-      "h11=10.0;                     #Input impedance with output shorted,  \u03a9\n",
-      "h21=-1.0;                     #Current gain with output shorted\n",
-      "h12=1.0;                      #Voltage feedback ratio with input terminal open\n",
-      "h22=0.2;                    #Output admittance, mho\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=h11-(h12*h21/(h22+(1/rL)));                 #Input impedance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "Av=-h21/(Zin*(h22+(1/rL)));                     #voltage gain,\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The input impedance=%.1f\u03a9.\"%Zin );\n",
-      "print(\"(ii) The voltage gain=1/%d.\"%(1/Av));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The input impedance=12.5\u03a9.\n",
-        "(ii) The voltage gain=1/5.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.4 : Page number 652-653\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=10.0;                 #Collector-emitter voltage, V\n",
-      "IC=1.0;                   #Collector current, mA\n",
-      "rL=600.0;                 #a.c load seen by the transistor,\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=2000.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=10**-4;             #Output impedance, mho\n",
-      "hre=10**-3;             #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=hie - (hre*hfe/(hoe+(1/rL)));                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9. \\n As second term in the expression of Zin is small compared to first, Zin~hie=%d\u03a9.\"%(Zin,hie));\n",
-      "\n",
-      "#(ii)\n",
-      "Ai=hfe/(1+hoe*rL);                      #Current gain\n",
-      "print(\"Current gain=%d\"%Ai);\n",
-      "print(\"if hoe*rL<<1, then Ai~hfe=%d.\"%hfe);\n",
-      "\n",
-      "#(iii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/rL)));                   #Voltage gain\n",
-      "print(\"Voltage gain=%.1f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1972 \u03a9. \n",
-        " As second term in the expression of Zin is small compared to first, Zin~hie=2000\u03a9.\n",
-        "Current gain=47\n",
-        "if hoe*rL<<1, then Ai~hfe=50.\n",
-        "Voltage gain=-14.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.5 : Page number 653\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCE=5.0;                 #Collector-emitter voltage, V\n",
-      "IC=1.0;                   #Collector current, mA\n",
-      "rL=2.0;                 #a.c load seen by the transistor,\u03a9\n",
-      "\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1700.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=6*10**-6;           #Output impedance, mho\n",
-      "hre=1.3*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=38.0;                 #Current gain with output shorted\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=hie - (hre*hfe/(hoe+(1/(rL*1000))));                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n",
-      "\n",
-      "#(ii)\n",
-      "Ai=ceil((hfe/round((1+hoe*rL*1000),3))*10)/10;            #Current gain\n",
-      "print(\"Current gain=%.1f\"%Ai);\n",
-      "\n",
-      "#(iii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/(rL*1000))));                   #Voltage gain\n",
-      "print(\"Voltage gain=%.1f\"%abs(Av));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1690 \u03a9.\n",
-        "Current gain=37.6\n",
-        "Voltage gain=44.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.6 : Page number 653-654\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=10.0;                  #Collector resistance, k\u03a9\n",
-      "RL=30.0;                  #Load resistance, k\u03a9\n",
-      "R1=80.0;                  #Resistor R1, k\u03a9\n",
-      "R2=40.0;                  #Resistor R2, k\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1500.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=5*10**-5;           #Output impedance, mho\n",
-      "hre=4*10**-4;           #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=((RC*RL)/(RC+RL))*1000;             #a.c load as seen by resistance, \u03a9\n",
-      "\n",
-      "#(i)\n",
-      "Zin=round(hie - (hre*hfe/(hoe+(1/rL))),-1);                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n",
-      "\n",
-      "#Input impedance of stage=input impedance || bias resistors\n",
-      "Zin_stage=round(pr(pr(R1,R2)*1000,Zin),-1);        #\u03a9\n",
-      "print(\"Input impedance of the stage=%.0f \u03a9.\"%Zin_stage);\n",
-      "\n",
-      "#(ii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/rL)));                   #Voltage gain\n",
-      "print(\"Voltage gain=%d\"%Av);\n",
-      "print(\"The negative sign represents phase reversal.\");\n",
-      "\n",
-      "\n",
-      "#(iii)\n",
-      "Zout=(1/(hoe-(hfe*hre/hie)))/1000;                     #Output impedance of transistor, k\u03a9\n",
-      "Zout_stage=pr(Zout,pr(RL,RC));                    #Output impedance of the stage, k\u03a9\n",
-      "print(\"Output impedance=%.2f k\u03a9.\"%Zout);\n",
-      "print(\"Output impedance of the stage=%.2f k\u03a9.\"%Zout_stage);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1390 \u03a9.\n",
-        "Input impedance of the stage=1320 \u03a9.\n",
-        "Voltage gain=-196\n",
-        "The negative sign represents phase reversal.\n",
-        "Output impedance=27.27 k\u03a9.\n",
-        "Output impedance of the stage=5.88 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.7 : Page number 654\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=4.7;                  #Collector resistance, k\u03a9\n",
-      "RL=10.0;                   #Load resistance, k\u03a9\n",
-      "R1=33.0;                   #Resistor R1, k\u03a9\n",
-      "R2=10.0;                   #Resistor R2, k\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1;                  #Input impedance with output shorted, k\u03a9\n",
-      "hoe=25;                 #Output impedance, \u03bcS\n",
-      "hre=2.5*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=50;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=(RC*RL)/(RC+RL);             #a.c load as seen by resistance, k\u03a9\n",
-      "\n",
-      "Ai=hfe/(1+hoe*10**-6*rL*1000);            #Current gain\n",
-      "print(\"Current gain=%.1f\"%Ai);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current gain=46.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.8  : Page number 654-655\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R_S=100.0;                    #Series resistance, \u03a9 \n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1.0;                  #Input impedance with output shorted, k\u03a9\n",
-      "hoe=25.0;                 #Output impedance, \u03bcS\n",
-      "hre=2.5*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Zout=(1/(hoe*10**-6-(hfe*hre/(hie*1000+R_S))))/1000;                #Output impedance of transistor, k\u03a9\n",
-      "print(\"Output impedance=%.1f k\u03a9.\"%Zout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output impedance=73.3 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.9 : Page number 656\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=12.0;                  #Collector resistance, k\u03a9\n",
-      "RL=15.0;                  #Load resistance, k\u03a9\n",
-      "R1=50.0;                  #Resistor R1, k\u03a9\n",
-      "R2=5.0;                   #Resistor R2, k\u03a9\n",
-      "hie=1.94;               #Input impedance with output shorted, k\u03a9\n",
-      "hfe=71.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=(RC*RL)/(RC+RL);             #a.c load as seen by resistance, \u03a9\n",
-      "\n",
-      "#(i)\n",
-      "Zin_base=hie;                                           #Transistor input impedance, k\u03a9\n",
-      "Zin_circuit=floor(pr(Zin_base,pr(R1,R2))*100)/100;      #Circuit input impedance, k\u03a9\n",
-      "print(\"Circuit input impedance=%.2fk\u03a9\"%Zin_circuit);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "Av=hfe*rL/hie;                   #Voltage gain\n",
-      "print(\"Voltage gain=%.0f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Circuit input impedance=1.35k\u03a9\n",
-        "Voltage gain=244\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.10 : Page number 656\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "hie_min=600;            #Minimum input impedance with output shorted, \u03a9\n",
-      "hfe_min=110;            #Minimum current gain with output shorted\n",
-      "hie_max=800;            #Maximum input impedance with output shorted, \u03a9\n",
-      "hfe_max=140;            #Maximum current gain with output shorted\n",
-      "rL=460;                 #a.c collector load, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "hie=round(sqrt(hie_min*hie_max));          #Input impedance with output shorted, \u03a9\n",
-      "hfe=round(sqrt(hfe_min*hfe_max));          #Current gain with output shorted\n",
-      "Av=hfe*rL/hie;                      #Voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage gain=%.1f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain=82.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.11 : Page number 658-659\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#(a)Variable declaration\n",
-      "Ib=10;              #Base current, \u03bcA\n",
-      "Ic=1;               #Collector current, mA\n",
-      "Vbe=10;             #Base-emitter voltage, mV\n",
-      "\n",
-      "#Calculation\n",
-      "hie=Vbe*10**-3/(Ib*10**-6);           #Input impedance with output shorted, \u03a9\n",
-      "hfe=Ic*10**-3/(Ib*10**-6);            #Current gain with output shorted\n",
-      "\n",
-      "#(b) Variable declaration\n",
-      "Vbe=0.65;             #Base-emitter voltage, mV\n",
-      "Ic=60;                #Collector current, \u03bcA\n",
-      "Vce=1;                #Collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "hre=Vbe*10**-3/Vce;           #Voltage feedback ratio with input terminal open\n",
-      "hoe=Ic/Vce;                   #Output impedance, \u03bcmho\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"hie=%d\u03a9\"%hie);\n",
-      "print(\"hfe=%d\"%hfe);\n",
-      "print(\"hre=%.2fe\u201303\"%(hre*1000));\n",
-      "print(\"hoe=%d\u03bcmho\"%hoe);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hie=1000\u03a9\n",
-        "hfe=100\n",
-        "hre=0.65e\u201303\n",
-        "hoe=60\u03bcmho\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_4.ipynb
deleted file mode 100755
index e63c17a6..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_4.ipynb
+++ /dev/null
@@ -1,604 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:947f358cf49d029c94d008f72a340051744678cf2e36ecc199100b78d31fcba5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 24 : HYBRID PARAMETERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.1 : Page number 644-645\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=10.0;                          #1st resistor, \u03a9\n",
-      "R2=5.0;                           #2nd resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"To find h11 and h21, output terminals are shorted.\");\n",
-      "h11=R1;                 #Input impedance with output shorted,  \u03a9\n",
-      "\n",
-      "print(\"h11=%d\u03a9.\"%h11);\n",
-      "\n",
-      "print(\"Output current flowing into the box= input current flowing out of the box.\");\n",
-      "print(\"i2=-i1\");                        #Output current flowing into the box= input current flowing out of the box.\n",
-      "print(\"h21=i2/i1 = -i1/i1= -1.\");       #Current gain with output shorted.\n",
-      "\n",
-      "\n",
-      "print(\"For finding h22 and h12, voltage source is connected at the output\");\n",
-      "#As, there will be no current through 10k\u03a9 resistor due to open circuited input,\n",
-      "print(\"v1=v2\");                         #Output voltage is equal to input voltage(equal to voltage drop across 5k\u03a9 resistor)\n",
-      "print(\"h12=v1/v2 = v2/v2 = 1\");         #Voltage feedback ratio with input terminals open\n",
-      "\n",
-      "h22=1/R2;                               #Output admittance, mho\n",
-      "print(\"h22=%.1f mho\"%h22);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To find h11 and h21, output terminals are shorted.\n",
-        "h11=10\u03a9.\n",
-        "Output current flowing into the box= input current flowing out of the box.\n",
-        "i2=-i1\n",
-        "h21=i2/i1 = -i1/i1= -1.\n",
-        "For finding h22 and h12, voltage source is connected at the output\n",
-        "v1=v2\n",
-        "h12=v1/v2 = v2/v2 = 1\n",
-        "h22=0.2 mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.2 : Page number 645-646\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=4.0;                          #1st resistor(at the input side), \u03a9\n",
-      "R2=4.0;                          #2nd resistor(at the middle), \u03a9\n",
-      "R3=4.0;                          #3rd resistor(at the output side), \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"To find h11 and h21, output terminals are shorted.\");\n",
-      "h11=R1 + (R2*R3/(R2+R3));                 #Input impedance with output shorted,  \u03a9\n",
-      "print(\"h11=%d\u03a9.\"%h11);\n",
-      "\n",
-      "#As the input current gets divided in half due to R2=R3.\n",
-      "print(\"Output current flowing into the box=negative of half of input current flowing out of the box.\");\n",
-      "print(\"i2=-i1/2 = -0.5i1\");                        \n",
-      "print(\"h21=i2/i1 = -0.5i1/i1= -0.5.\");       #Current gain with output shorted.\n",
-      "\n",
-      "print(\"For finding h22 and h12, voltage source is connected at the output\");\n",
-      "#As, there will be no current through the 1st 4k\u03a9 resistor due to open circuited input,\n",
-      "#Voltage gets equally divided across R2 and R3 resistor\n",
-      "print(\"v1=v2/2 = 0.5v2\");                    #Input voltage is equal to half of input voltage\n",
-      "print(\"h12=v1/v2 = 0.5v2/v2 = 0.5\");         #Voltage feedback ratio with input terminals open\n",
-      "\n",
-      "h22=1/(R2+R3);                               #Output admittance, mho\n",
-      "print(\"h22=%.3f mho\"%h22);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To find h11 and h21, output terminals are shorted.\n",
-        "h11=6\u03a9.\n",
-        "Output current flowing into the box=negative of half of input current flowing out of the box.\n",
-        "i2=-i1/2 = -0.5i1\n",
-        "h21=i2/i1 = -0.5i1/i1= -0.5.\n",
-        "For finding h22 and h12, voltage source is connected at the output\n",
-        "v1=v2/2 = 0.5v2\n",
-        "h12=v1/v2 = 0.5v2/v2 = 0.5\n",
-        "h22=0.125 mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.3 ; Page number 649-650\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=10.0;                  #Resistor at the input side, \u03a9\n",
-      "R2=5.0;                   #Resistor at the middle, \u03a9\n",
-      "rL=5.0;                   #Load resistor, \u03a9\n",
-      "\n",
-      "#h-parameter values from 24.1\n",
-      "h11=10.0;                     #Input impedance with output shorted,  \u03a9\n",
-      "h21=-1.0;                     #Current gain with output shorted\n",
-      "h12=1.0;                      #Voltage feedback ratio with input terminal open\n",
-      "h22=0.2;                    #Output admittance, mho\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=h11-(h12*h21/(h22+(1/rL)));                 #Input impedance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "Av=-h21/(Zin*(h22+(1/rL)));                     #voltage gain,\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The input impedance=%.1f\u03a9.\"%Zin );\n",
-      "print(\"(ii) The voltage gain=1/%d.\"%(1/Av));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The input impedance=12.5\u03a9.\n",
-        "(ii) The voltage gain=1/5.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.4 : Page number 652-653\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=10.0;                 #Collector-emitter voltage, V\n",
-      "IC=1.0;                   #Collector current, mA\n",
-      "rL=600.0;                 #a.c load seen by the transistor,\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=2000.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=10**-4;             #Output impedance, mho\n",
-      "hre=10**-3;             #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=hie - (hre*hfe/(hoe+(1/rL)));                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9. \\n As second term in the expression of Zin is small compared to first, Zin~hie=%d\u03a9.\"%(Zin,hie));\n",
-      "\n",
-      "#(ii)\n",
-      "Ai=hfe/(1+hoe*rL);                      #Current gain\n",
-      "print(\"Current gain=%d\"%Ai);\n",
-      "print(\"if hoe*rL<<1, then Ai~hfe=%d.\"%hfe);\n",
-      "\n",
-      "#(iii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/rL)));                   #Voltage gain\n",
-      "print(\"Voltage gain=%.1f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1972 \u03a9. \n",
-        " As second term in the expression of Zin is small compared to first, Zin~hie=2000\u03a9.\n",
-        "Current gain=47\n",
-        "if hoe*rL<<1, then Ai~hfe=50.\n",
-        "Voltage gain=-14.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.5 : Page number 653\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCE=5.0;                 #Collector-emitter voltage, V\n",
-      "IC=1.0;                   #Collector current, mA\n",
-      "rL=2.0;                 #a.c load seen by the transistor,\u03a9\n",
-      "\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1700.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=6*10**-6;           #Output impedance, mho\n",
-      "hre=1.3*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=38.0;                 #Current gain with output shorted\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=hie - (hre*hfe/(hoe+(1/(rL*1000))));                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n",
-      "\n",
-      "#(ii)\n",
-      "Ai=ceil((hfe/round((1+hoe*rL*1000),3))*10)/10;            #Current gain\n",
-      "print(\"Current gain=%.1f\"%Ai);\n",
-      "\n",
-      "#(iii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/(rL*1000))));                   #Voltage gain\n",
-      "print(\"Voltage gain=%.1f\"%abs(Av));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1690 \u03a9.\n",
-        "Current gain=37.6\n",
-        "Voltage gain=44.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.6 : Page number 653-654\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=10.0;                  #Collector resistance, k\u03a9\n",
-      "RL=30.0;                  #Load resistance, k\u03a9\n",
-      "R1=80.0;                  #Resistor R1, k\u03a9\n",
-      "R2=40.0;                  #Resistor R2, k\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1500.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=5*10**-5;           #Output impedance, mho\n",
-      "hre=4*10**-4;           #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=((RC*RL)/(RC+RL))*1000;             #a.c load as seen by resistance, \u03a9\n",
-      "\n",
-      "#(i)\n",
-      "Zin=round(hie - (hre*hfe/(hoe+(1/rL))),-1);                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n",
-      "\n",
-      "#Input impedance of stage=input impedance || bias resistors\n",
-      "Zin_stage=round(pr(pr(R1,R2)*1000,Zin),-1);        #\u03a9\n",
-      "print(\"Input impedance of the stage=%.0f \u03a9.\"%Zin_stage);\n",
-      "\n",
-      "#(ii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/rL)));                   #Voltage gain\n",
-      "print(\"Voltage gain=%d\"%Av);\n",
-      "print(\"The negative sign represents phase reversal.\");\n",
-      "\n",
-      "\n",
-      "#(iii)\n",
-      "Zout=(1/(hoe-(hfe*hre/hie)))/1000;                     #Output impedance of transistor, k\u03a9\n",
-      "Zout_stage=pr(Zout,pr(RL,RC));                    #Output impedance of the stage, k\u03a9\n",
-      "print(\"Output impedance=%.2f k\u03a9.\"%Zout);\n",
-      "print(\"Output impedance of the stage=%.2f k\u03a9.\"%Zout_stage);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1390 \u03a9.\n",
-        "Input impedance of the stage=1320 \u03a9.\n",
-        "Voltage gain=-196\n",
-        "The negative sign represents phase reversal.\n",
-        "Output impedance=27.27 k\u03a9.\n",
-        "Output impedance of the stage=5.88 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.7 : Page number 654\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=4.7;                  #Collector resistance, k\u03a9\n",
-      "RL=10.0;                   #Load resistance, k\u03a9\n",
-      "R1=33.0;                   #Resistor R1, k\u03a9\n",
-      "R2=10.0;                   #Resistor R2, k\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1;                  #Input impedance with output shorted, k\u03a9\n",
-      "hoe=25;                 #Output impedance, \u03bcS\n",
-      "hre=2.5*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=50;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=(RC*RL)/(RC+RL);             #a.c load as seen by resistance, k\u03a9\n",
-      "\n",
-      "Ai=hfe/(1+hoe*10**-6*rL*1000);            #Current gain\n",
-      "print(\"Current gain=%.1f\"%Ai);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current gain=46.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.8  : Page number 654-655\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R_S=100.0;                    #Series resistance, \u03a9 \n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1.0;                  #Input impedance with output shorted, k\u03a9\n",
-      "hoe=25.0;                 #Output impedance, \u03bcS\n",
-      "hre=2.5*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Zout=(1/(hoe*10**-6-(hfe*hre/(hie*1000+R_S))))/1000;                #Output impedance of transistor, k\u03a9\n",
-      "print(\"Output impedance=%.1f k\u03a9.\"%Zout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output impedance=73.3 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.9 : Page number 656\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=12.0;                  #Collector resistance, k\u03a9\n",
-      "RL=15.0;                  #Load resistance, k\u03a9\n",
-      "R1=50.0;                  #Resistor R1, k\u03a9\n",
-      "R2=5.0;                   #Resistor R2, k\u03a9\n",
-      "hie=1.94;               #Input impedance with output shorted, k\u03a9\n",
-      "hfe=71.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=(RC*RL)/(RC+RL);             #a.c load as seen by resistance, \u03a9\n",
-      "\n",
-      "#(i)\n",
-      "Zin_base=hie;                                           #Transistor input impedance, k\u03a9\n",
-      "Zin_circuit=floor(pr(Zin_base,pr(R1,R2))*100)/100;      #Circuit input impedance, k\u03a9\n",
-      "print(\"Circuit input impedance=%.2fk\u03a9\"%Zin_circuit);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "Av=hfe*rL/hie;                   #Voltage gain\n",
-      "print(\"Voltage gain=%.0f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Circuit input impedance=1.35k\u03a9\n",
-        "Voltage gain=244\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.10 : Page number 656\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "hie_min=600;            #Minimum input impedance with output shorted, \u03a9\n",
-      "hfe_min=110;            #Minimum current gain with output shorted\n",
-      "hie_max=800;            #Maximum input impedance with output shorted, \u03a9\n",
-      "hfe_max=140;            #Maximum current gain with output shorted\n",
-      "rL=460;                 #a.c collector load, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "hie=round(sqrt(hie_min*hie_max));          #Input impedance with output shorted, \u03a9\n",
-      "hfe=round(sqrt(hfe_min*hfe_max));          #Current gain with output shorted\n",
-      "Av=hfe*rL/hie;                      #Voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage gain=%.1f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain=82.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.11 : Page number 658-659\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#(a)Variable declaration\n",
-      "Ib=10;              #Base current, \u03bcA\n",
-      "Ic=1;               #Collector current, mA\n",
-      "Vbe=10;             #Base-emitter voltage, mV\n",
-      "\n",
-      "#Calculation\n",
-      "hie=Vbe*10**-3/(Ib*10**-6);           #Input impedance with output shorted, \u03a9\n",
-      "hfe=Ic*10**-3/(Ib*10**-6);            #Current gain with output shorted\n",
-      "\n",
-      "#(b) Variable declaration\n",
-      "Vbe=0.65;             #Base-emitter voltage, mV\n",
-      "Ic=60;                #Collector current, \u03bcA\n",
-      "Vce=1;                #Collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "hre=Vbe*10**-3/Vce;           #Voltage feedback ratio with input terminal open\n",
-      "hoe=Ic/Vce;                   #Output impedance, \u03bcmho\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"hie=%d\u03a9\"%hie);\n",
-      "print(\"hfe=%d\"%hfe);\n",
-      "print(\"hre=%.2fe\u201303\"%(hre*1000));\n",
-      "print(\"hoe=%d\u03bcmho\"%hoe);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hie=1000\u03a9\n",
-        "hfe=100\n",
-        "hre=0.65e\u201303\n",
-        "hoe=60\u03bcmho\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_5.ipynb
deleted file mode 100755
index e63c17a6..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_5.ipynb
+++ /dev/null
@@ -1,604 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:947f358cf49d029c94d008f72a340051744678cf2e36ecc199100b78d31fcba5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 24 : HYBRID PARAMETERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.1 : Page number 644-645\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=10.0;                          #1st resistor, \u03a9\n",
-      "R2=5.0;                           #2nd resistor, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"To find h11 and h21, output terminals are shorted.\");\n",
-      "h11=R1;                 #Input impedance with output shorted,  \u03a9\n",
-      "\n",
-      "print(\"h11=%d\u03a9.\"%h11);\n",
-      "\n",
-      "print(\"Output current flowing into the box= input current flowing out of the box.\");\n",
-      "print(\"i2=-i1\");                        #Output current flowing into the box= input current flowing out of the box.\n",
-      "print(\"h21=i2/i1 = -i1/i1= -1.\");       #Current gain with output shorted.\n",
-      "\n",
-      "\n",
-      "print(\"For finding h22 and h12, voltage source is connected at the output\");\n",
-      "#As, there will be no current through 10k\u03a9 resistor due to open circuited input,\n",
-      "print(\"v1=v2\");                         #Output voltage is equal to input voltage(equal to voltage drop across 5k\u03a9 resistor)\n",
-      "print(\"h12=v1/v2 = v2/v2 = 1\");         #Voltage feedback ratio with input terminals open\n",
-      "\n",
-      "h22=1/R2;                               #Output admittance, mho\n",
-      "print(\"h22=%.1f mho\"%h22);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To find h11 and h21, output terminals are shorted.\n",
-        "h11=10\u03a9.\n",
-        "Output current flowing into the box= input current flowing out of the box.\n",
-        "i2=-i1\n",
-        "h21=i2/i1 = -i1/i1= -1.\n",
-        "For finding h22 and h12, voltage source is connected at the output\n",
-        "v1=v2\n",
-        "h12=v1/v2 = v2/v2 = 1\n",
-        "h22=0.2 mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.2 : Page number 645-646\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=4.0;                          #1st resistor(at the input side), \u03a9\n",
-      "R2=4.0;                          #2nd resistor(at the middle), \u03a9\n",
-      "R3=4.0;                          #3rd resistor(at the output side), \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "print(\"To find h11 and h21, output terminals are shorted.\");\n",
-      "h11=R1 + (R2*R3/(R2+R3));                 #Input impedance with output shorted,  \u03a9\n",
-      "print(\"h11=%d\u03a9.\"%h11);\n",
-      "\n",
-      "#As the input current gets divided in half due to R2=R3.\n",
-      "print(\"Output current flowing into the box=negative of half of input current flowing out of the box.\");\n",
-      "print(\"i2=-i1/2 = -0.5i1\");                        \n",
-      "print(\"h21=i2/i1 = -0.5i1/i1= -0.5.\");       #Current gain with output shorted.\n",
-      "\n",
-      "print(\"For finding h22 and h12, voltage source is connected at the output\");\n",
-      "#As, there will be no current through the 1st 4k\u03a9 resistor due to open circuited input,\n",
-      "#Voltage gets equally divided across R2 and R3 resistor\n",
-      "print(\"v1=v2/2 = 0.5v2\");                    #Input voltage is equal to half of input voltage\n",
-      "print(\"h12=v1/v2 = 0.5v2/v2 = 0.5\");         #Voltage feedback ratio with input terminals open\n",
-      "\n",
-      "h22=1/(R2+R3);                               #Output admittance, mho\n",
-      "print(\"h22=%.3f mho\"%h22);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "To find h11 and h21, output terminals are shorted.\n",
-        "h11=6\u03a9.\n",
-        "Output current flowing into the box=negative of half of input current flowing out of the box.\n",
-        "i2=-i1/2 = -0.5i1\n",
-        "h21=i2/i1 = -0.5i1/i1= -0.5.\n",
-        "For finding h22 and h12, voltage source is connected at the output\n",
-        "v1=v2/2 = 0.5v2\n",
-        "h12=v1/v2 = 0.5v2/v2 = 0.5\n",
-        "h22=0.125 mho\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.3 ; Page number 649-650\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R1=10.0;                  #Resistor at the input side, \u03a9\n",
-      "R2=5.0;                   #Resistor at the middle, \u03a9\n",
-      "rL=5.0;                   #Load resistor, \u03a9\n",
-      "\n",
-      "#h-parameter values from 24.1\n",
-      "h11=10.0;                     #Input impedance with output shorted,  \u03a9\n",
-      "h21=-1.0;                     #Current gain with output shorted\n",
-      "h12=1.0;                      #Voltage feedback ratio with input terminal open\n",
-      "h22=0.2;                    #Output admittance, mho\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=h11-(h12*h21/(h22+(1/rL)));                 #Input impedance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "Av=-h21/(Zin*(h22+(1/rL)));                     #voltage gain,\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The input impedance=%.1f\u03a9.\"%Zin );\n",
-      "print(\"(ii) The voltage gain=1/%d.\"%(1/Av));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The input impedance=12.5\u03a9.\n",
-        "(ii) The voltage gain=1/5.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.4 : Page number 652-653\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=10.0;                 #Collector-emitter voltage, V\n",
-      "IC=1.0;                   #Collector current, mA\n",
-      "rL=600.0;                 #a.c load seen by the transistor,\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=2000.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=10**-4;             #Output impedance, mho\n",
-      "hre=10**-3;             #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=hie - (hre*hfe/(hoe+(1/rL)));                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9. \\n As second term in the expression of Zin is small compared to first, Zin~hie=%d\u03a9.\"%(Zin,hie));\n",
-      "\n",
-      "#(ii)\n",
-      "Ai=hfe/(1+hoe*rL);                      #Current gain\n",
-      "print(\"Current gain=%d\"%Ai);\n",
-      "print(\"if hoe*rL<<1, then Ai~hfe=%d.\"%hfe);\n",
-      "\n",
-      "#(iii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/rL)));                   #Voltage gain\n",
-      "print(\"Voltage gain=%.1f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1972 \u03a9. \n",
-        " As second term in the expression of Zin is small compared to first, Zin~hie=2000\u03a9.\n",
-        "Current gain=47\n",
-        "if hoe*rL<<1, then Ai~hfe=50.\n",
-        "Voltage gain=-14.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.5 : Page number 653\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import ceil\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCE=5.0;                 #Collector-emitter voltage, V\n",
-      "IC=1.0;                   #Collector current, mA\n",
-      "rL=2.0;                 #a.c load seen by the transistor,\u03a9\n",
-      "\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1700.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=6*10**-6;           #Output impedance, mho\n",
-      "hre=1.3*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=38.0;                 #Current gain with output shorted\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Zin=hie - (hre*hfe/(hoe+(1/(rL*1000))));                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n",
-      "\n",
-      "#(ii)\n",
-      "Ai=ceil((hfe/round((1+hoe*rL*1000),3))*10)/10;            #Current gain\n",
-      "print(\"Current gain=%.1f\"%Ai);\n",
-      "\n",
-      "#(iii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/(rL*1000))));                   #Voltage gain\n",
-      "print(\"Voltage gain=%.1f\"%abs(Av));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1690 \u03a9.\n",
-        "Current gain=37.6\n",
-        "Voltage gain=44.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.6 : Page number 653-654\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=10.0;                  #Collector resistance, k\u03a9\n",
-      "RL=30.0;                  #Load resistance, k\u03a9\n",
-      "R1=80.0;                  #Resistor R1, k\u03a9\n",
-      "R2=40.0;                  #Resistor R2, k\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1500.0;               #Input impedance with output shorted, \u03a9\n",
-      "hoe=5*10**-5;           #Output impedance, mho\n",
-      "hre=4*10**-4;           #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=((RC*RL)/(RC+RL))*1000;             #a.c load as seen by resistance, \u03a9\n",
-      "\n",
-      "#(i)\n",
-      "Zin=round(hie - (hre*hfe/(hoe+(1/rL))),-1);                  #Input impedance, \u03a9\n",
-      "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n",
-      "\n",
-      "#Input impedance of stage=input impedance || bias resistors\n",
-      "Zin_stage=round(pr(pr(R1,R2)*1000,Zin),-1);        #\u03a9\n",
-      "print(\"Input impedance of the stage=%.0f \u03a9.\"%Zin_stage);\n",
-      "\n",
-      "#(ii)\n",
-      "Av=-hfe/(Zin*(hoe+(1/rL)));                   #Voltage gain\n",
-      "print(\"Voltage gain=%d\"%Av);\n",
-      "print(\"The negative sign represents phase reversal.\");\n",
-      "\n",
-      "\n",
-      "#(iii)\n",
-      "Zout=(1/(hoe-(hfe*hre/hie)))/1000;                     #Output impedance of transistor, k\u03a9\n",
-      "Zout_stage=pr(Zout,pr(RL,RC));                    #Output impedance of the stage, k\u03a9\n",
-      "print(\"Output impedance=%.2f k\u03a9.\"%Zout);\n",
-      "print(\"Output impedance of the stage=%.2f k\u03a9.\"%Zout_stage);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input impedance=1390 \u03a9.\n",
-        "Input impedance of the stage=1320 \u03a9.\n",
-        "Voltage gain=-196\n",
-        "The negative sign represents phase reversal.\n",
-        "Output impedance=27.27 k\u03a9.\n",
-        "Output impedance of the stage=5.88 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.7 : Page number 654\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=4.7;                  #Collector resistance, k\u03a9\n",
-      "RL=10.0;                   #Load resistance, k\u03a9\n",
-      "R1=33.0;                   #Resistor R1, k\u03a9\n",
-      "R2=10.0;                   #Resistor R2, k\u03a9\n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1;                  #Input impedance with output shorted, k\u03a9\n",
-      "hoe=25;                 #Output impedance, \u03bcS\n",
-      "hre=2.5*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=50;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=(RC*RL)/(RC+RL);             #a.c load as seen by resistance, k\u03a9\n",
-      "\n",
-      "Ai=hfe/(1+hoe*10**-6*rL*1000);            #Current gain\n",
-      "print(\"Current gain=%.1f\"%Ai);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current gain=46.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.8  : Page number 654-655\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R_S=100.0;                    #Series resistance, \u03a9 \n",
-      "\n",
-      "#h-parameters\n",
-      "hie=1.0;                  #Input impedance with output shorted, k\u03a9\n",
-      "hoe=25.0;                 #Output impedance, \u03bcS\n",
-      "hre=2.5*10**-4;         #Voltage feedback ratio with input terminal open\n",
-      "hfe=50.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Zout=(1/(hoe*10**-6-(hfe*hre/(hie*1000+R_S))))/1000;                #Output impedance of transistor, k\u03a9\n",
-      "print(\"Output impedance=%.1f k\u03a9.\"%Zout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output impedance=73.3 k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.9 : Page number 656\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Function for calculating parallel resistance\n",
-      "def pr(r1,r2):\n",
-      "    return r1*r2/(r1+r2);\n",
-      "\n",
-      "#Variable declaration\n",
-      "RC=12.0;                  #Collector resistance, k\u03a9\n",
-      "RL=15.0;                  #Load resistance, k\u03a9\n",
-      "R1=50.0;                  #Resistor R1, k\u03a9\n",
-      "R2=5.0;                   #Resistor R2, k\u03a9\n",
-      "hie=1.94;               #Input impedance with output shorted, k\u03a9\n",
-      "hfe=71.0;                 #Current gain with output shorted\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "rL=(RC*RL)/(RC+RL);             #a.c load as seen by resistance, \u03a9\n",
-      "\n",
-      "#(i)\n",
-      "Zin_base=hie;                                           #Transistor input impedance, k\u03a9\n",
-      "Zin_circuit=floor(pr(Zin_base,pr(R1,R2))*100)/100;      #Circuit input impedance, k\u03a9\n",
-      "print(\"Circuit input impedance=%.2fk\u03a9\"%Zin_circuit);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "Av=hfe*rL/hie;                   #Voltage gain\n",
-      "print(\"Voltage gain=%.0f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Circuit input impedance=1.35k\u03a9\n",
-        "Voltage gain=244\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.10 : Page number 656\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "hie_min=600;            #Minimum input impedance with output shorted, \u03a9\n",
-      "hfe_min=110;            #Minimum current gain with output shorted\n",
-      "hie_max=800;            #Maximum input impedance with output shorted, \u03a9\n",
-      "hfe_max=140;            #Maximum current gain with output shorted\n",
-      "rL=460;                 #a.c collector load, \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "hie=round(sqrt(hie_min*hie_max));          #Input impedance with output shorted, \u03a9\n",
-      "hfe=round(sqrt(hfe_min*hfe_max));          #Current gain with output shorted\n",
-      "Av=hfe*rL/hie;                      #Voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"Voltage gain=%.1f\"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage gain=82.3\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 24.11 : Page number 658-659\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#(a)Variable declaration\n",
-      "Ib=10;              #Base current, \u03bcA\n",
-      "Ic=1;               #Collector current, mA\n",
-      "Vbe=10;             #Base-emitter voltage, mV\n",
-      "\n",
-      "#Calculation\n",
-      "hie=Vbe*10**-3/(Ib*10**-6);           #Input impedance with output shorted, \u03a9\n",
-      "hfe=Ic*10**-3/(Ib*10**-6);            #Current gain with output shorted\n",
-      "\n",
-      "#(b) Variable declaration\n",
-      "Vbe=0.65;             #Base-emitter voltage, mV\n",
-      "Ic=60;                #Collector current, \u03bcA\n",
-      "Vce=1;                #Collector-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "hre=Vbe*10**-3/Vce;           #Voltage feedback ratio with input terminal open\n",
-      "hoe=Ic/Vce;                   #Output impedance, \u03bcmho\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"hie=%d\u03a9\"%hie);\n",
-      "print(\"hfe=%d\"%hfe);\n",
-      "print(\"hre=%.2fe\u201303\"%(hre*1000));\n",
-      "print(\"hoe=%d\u03bcmho\"%hoe);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "hie=1000\u03a9\n",
-        "hfe=100\n",
-        "hre=0.65e\u201303\n",
-        "hoe=60\u03bcmho\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25.ipynb
deleted file mode 100755
index 6add7ba9..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25.ipynb
+++ /dev/null
@@ -1,2552 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:f21def3bd5ac5e7f3368a4e0e9a53b6961ecc94a2dca6b1917cdfda88271b542"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 25 : OPERATIONAL AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.1: Page number 664\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A=100.0;              #Open-circuit voltage gain of differential amplifier\n",
-      "V1=3.25;              #Input voltage to terminal 1 in V\n",
-      "V2=3.15;              #Input voltage to terminal 2 in V\n",
-      "\n",
-      "#Calculations\n",
-      "V0=A*(V1-V2);         #Output voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The output voltage of the differential amplifier = %dV\"%V0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage of the differential amplifier = 10V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.2: Page number 672\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_DM=2000.0;              #Differential mode voltage gain\n",
-      "A_CM=0.2;                 #Common mode voltage gain\n",
-      "\n",
-      "#Calculations\n",
-      "CMRR=A_DM/A_CM;           #Common mode rejection ratio\n",
-      "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The common mode rejection ratio = %d.\"%CMRR);\n",
-      "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The common mode rejection ratio = 10000.\n",
-        "The common mode rejection ratio in decibels= 80dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.3: Page number 672\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "VD_in=10.0;               #Differential mode input in mV\n",
-      "VD_out=1.0;               #Output for differential mode input in V\n",
-      "VC_in=10.0;                 #Common mode input in mV\n",
-      "VC_out=5.0;                 #Output for common mode input in mV\\\n",
-      "\n",
-      "#Calculations\n",
-      "A_DM=(VD_out*1000)/VD_in;       #Differntial mode voltage gain\n",
-      "A_CM=VC_out/VC_in;       #Common mode voltage gain\n",
-      "CMRR=A_DM/A_CM;           #Common mode rejection ratio\n",
-      "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The common mode rejection ratio in decibels= 46dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.4: Page number 672\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_DM=150.0;           #Differential mode voltage gain\n",
-      "CMRR_dB=90.0;         #Common mode rejection ratio\n",
-      "V1=100.0;             #Input voltage for terminal 1 in mV\n",
-      "V2=50.0;              #Input voltage for terminal 2 in mV\n",
-      "V_noise=1.0;          #Voltage of noise signal in mV\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "#Case(i)\n",
-      "V_out=A_DM*(V1-V2)/1000.0;            #Output voltage for differntial mode input, in V\n",
-      "\n",
-      "#Since CMRR_dB=20*log10(differential mode gain/common mode gain),\n",
-      "A_CM=A_DM/pow(10,(CMRR_dB/20));         #Common mode gain\n",
-      "V_OUT_noise=A_CM*(V_noise/1000);        #Noise on output in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Output voltage =%.1fV\"%V_out);\n",
-      "print(\"Noise on output = %.1fx10^-6V\"%(V_OUT_noise*pow(10,6)));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage =7.5V\n",
-        "Noise on output = 4.7x10^-6V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.5 : Page number 672-673\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_DM=2500.0;                    #Differential mode voltage gain\n",
-      "CMRR=30000.0;                   #Common mode rejection ratio\n",
-      "Input_signal=500.0;             #Single ended input r.m.s signal in microvolts\n",
-      "Interference=1.0;               #Interference signal, in V\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "A_CM=A_DM/CMRR;                 #Common mode gain\n",
-      "\n",
-      "#(ii)\n",
-      "CMRR_dB=20*log10(CMRR);         #Common mode rejection ratio in decibels\n",
-      "\n",
-      "#(iii)\n",
-      "V_out=A_DM*(Input_signal/pow(10,6)-0);        #r.m.s output signal  in V\n",
-      "\n",
-      "#(iv)\n",
-      "Interference_out=A_CM*Interference;          #r.m.s interference output in V\n",
-      "Interference_out=Interference_out*1000;      #r.m.s interference output in mV\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Common mode gain =%.3f\"%A_CM);\n",
-      "print(\"Common mode rejection ratio in decibels=%.1fdB\"%CMRR_dB);\n",
-      "print(\"r.m.s output signal =%.2fV\"%V_out);\n",
-      "print(\"r.m.s interfernce output voltage = %dmV\"%Interference_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Common mode gain =0.083\n",
-        "Common mode rejection ratio in decibels=89.5dB\n",
-        "r.m.s output signal =1.25V\n",
-        "r.m.s interfernce output voltage = 83mV\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.6 : Page number 674-675\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12;                 #Collector supply voltage, V\n",
-      "VEE=12;                 #Emitter supply voltage, V\n",
-      "RB=10;                  #Base resistor, k\u03a9\n",
-      "RC2=10;                 #Collector resistor, k\u03a9\n",
-      "RE=25;                  #Emitter resistor, k\u03a9\n",
-      "VBE=0.7;               #Base-emitter voltage, V\n",
-      "beta=100;               #Base amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-      "IE=(VEE-VBE)/RE;            #Tail current, mA\n",
-      "IE1=IE/2;                   #Emitter current of 1st transistor, mA\n",
-      "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-      "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-      "IC2=IC1;                    #Collector current of 2nd transistor, mA\n",
-      "IB1=(IC1/beta)*1000;        #Base current of 1st transistor, \u03bcA\n",
-      "IB2=IB1;                    #Base current of 2nd transistor, \u03bcA\n",
-      "VC1=VCC;                    #Collector voltage of 1st transistor, V\n",
-      "VC2=VCC-IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"VE=%.1fV\"%VE);\n",
-      "print(\"IE=%.3fmA\"%IE);\n",
-      "print(\"IE1=%.3fmA\"%IE1);\n",
-      "print(\"IE2=%.3fmA\"%IE2);\n",
-      "print(\"IC1=%.3fmA\"%IC1);\n",
-      "print(\"IC2=%.3fmA\"%IC2);\n",
-      "print(\"IB1=%.2f\u03bcA\"%IB1);\n",
-      "print(\"IB2=%.2f\u03bcA\"%IB2);\n",
-      "print(\"VC1=%dV\"%VC1);\n",
-      "print(\"VC2=%.1fV\"%VC2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VE=-0.7V\n",
-        "IE=0.452mA\n",
-        "IE1=0.226mA\n",
-        "IE2=0.226mA\n",
-        "IC1=0.226mA\n",
-        "IC2=0.226mA\n",
-        "IB1=2.26\u03bcA\n",
-        "IB2=2.26\u03bcA\n",
-        "VC1=12V\n",
-        "VC2=9.7V\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.7 : Page number 675\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15;                 #Collector supply voltage, V\n",
-      "VEE=15;                 #Emitter supply voltage, V\n",
-      "RB=33;                  #Base resistor, k\u03a9\n",
-      "RC=15;                  #Collector resistor, k\u03a9\n",
-      "RE=15;                  #Emitter resistor, k\u03a9\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IE_tail=(VEE-VBE)/RE;            #Tail current, mA\n",
-      "IE=round(IE_tail/2,3);                    #Emitter current in each transistor, mA\n",
-      "IC=IE;                           #Collector current(=emitter current), mA\n",
-      "Vout=VCC-IC*RC;                  #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage=%.2fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage=7.85V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.8 : Page number 675\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                 #Collector supply voltage, V\n",
-      "VEE=15.0;                 #Emitter supply voltage, V\n",
-      "RB=33.0;                  #Base resistor, k\u03a9\n",
-      "RC=15.0;                  #Collector resistor, k\u03a9\n",
-      "RE=15.0;                  #Emitter resistor, k\u03a9\n",
-      "VBE=0;                  #Base-emitter voltage, V\n",
-      "beta_dc_l=90.0;           #base current amplification factor for left transistor\n",
-      "beta_dc_r=110.0;          #base current amplification factor for right transistor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IE_tail=(VEE-VBE)/RE;             #Tail current, mA\n",
-      "IE=IE_tail/2;                      #Emitter current in each transistor, mA\n",
-      "IB1=(IE/beta_dc_l)*1000;          #Base current of 1st transistor, \u03bcA\n",
-      "IB2=(IE/beta_dc_r)*1000;          #Base current of 2nd transistor, \u03bcA\n",
-      "\n",
-      "#(ii)\n",
-      "VB1=-IB1/1000*RB;                    #Base voltage of 1st transistor, V\n",
-      "VB2=-IB2/1000*RB;                    #Base voltage of 1st transistor, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) IB1=%.2f\u03bcA\"%IB1);\n",
-      "print(\"    IB2=%.2f\u03bcA\"%IB2);\n",
-      "print(\"(ii) VB1=%.3fV\"%VB1);\n",
-      "print(\"     VB2=%.2fV\"%VB2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) IB1=5.56\u03bcA\n",
-        "    IB2=4.55\u03bcA\n",
-        "(ii) VB1=-0.183V\n",
-        "     VB2=-0.15V\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.9 : Page number 675-676\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                 #Collector supply voltage, V\n",
-      "VEE=15.0;                 #Emitter supply voltage, V\n",
-      "RB=10.0;                  #Base resistor, k\u03a9\n",
-      "RC1=10.0;                 #Collector resistor of 1st transistor, k\u03a9\n",
-      "RC2=10.0;                 #Collector resistor of 2nd transistor, k\u03a9\n",
-      "IE=1.0;                   #Tail current, mA\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-      "IE1=IE/2.0;                   #Emitter current of 1st transistor, mA\n",
-      "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-      "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-      "IC2=IE2;                    #Collector current of 2nd transistor, mA\n",
-      "VC1=VCC-IC1*RC1;            #Collector voltage of 1st transistor, V\n",
-      "VC2=VCC-IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"VE=%.1fV\"%VE);\n",
-      "print(\"Emitter current in each transistor=%.1fmA.\"%(IE/2.0));\n",
-      "print(\"IC1~IE1=%.1fmA and IC2~IE2=%.1fmA\"%(IE1,IE2));\n",
-      "print(\"VC1=VC2=%dV.\"%VC2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VE=-0.7V\n",
-        "Emitter current in each transistor=0.5mA.\n",
-        "IC1~IE1=0.5mA and IC2~IE2=0.5mA\n",
-        "VC1=VC2=10V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.10 : Page number 676-677\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage, V\n",
-      "VEE=12.0;                 #Emitter supply voltage, V\n",
-      "RC2=10.0;                 #Collector resistor of 2nd transistor, k\u03a9\n",
-      "RE=25.0;                  #Emitter current, k\u03a9\n",
-      "VBE=-0.7;               #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-      "IE=(VCC-VE)/RE;             #Tail current, mA\n",
-      "IE1=IE/2.0;                   #Emitter current of 1st transistor, mA\n",
-      "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-      "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-      "IC2=IE2;                    #Collector current of 2nd transistor, mA\n",
-      "VC1=-VEE;                   #Collector voltage of 1st transistor, V\n",
-      "VC2=-VEE+IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"VE=%.1fV\"%VE);\n",
-      "print(\"Tail current=%.3fmA.\"%IE);\n",
-      "print(\"Emitter current in each transistor=%.3fmA.\"%(IE/2.0));\n",
-      "print(\"IC1~IE1=%.3fmA and IC2~IE2=%.3fmA\"%(IC1,IC2));\n",
-      "print(\"VC1=%dV\"%VC1);\n",
-      "print(\"VC2=%.2fV\"%VC2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VE=0.7V\n",
-        "Tail current=0.452mA.\n",
-        "Emitter current in each transistor=0.226mA.\n",
-        "IC1~IE1=0.226mA and IC2~IE2=0.226mA\n",
-        "VC1=-12V\n",
-        "VC2=-9.74V\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.11 : Page number 679\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15;                 #Collector supply voltage, V\n",
-      "VEE=15;                 #Emitter supply voltage, V\n",
-      "RB=1;                   #Base resistor, M\u03a9\n",
-      "RC2=1;                  #Collector resistor, M\u03a9\n",
-      "RE=1;                   #Emitter resistor, M\u03a9\n",
-      "VBE=0;                  #Base-emitter voltage, V (Neglected)\n",
-      "beta_dc_l=90.0;           #base current amplification factor for left transistor\n",
-      "beta_dc_r=110.0;          #base current amplification factor for right transistor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IE=(VEE-VBE)/RE;                    #Tail current, \u03bcA\n",
-      "IE1=IE/2.0;                           #Emitter current of 1st transistor, \u03bcA\n",
-      "IE2=IE1;                            #Emitter current of 2nd transistor, \u03bcA\n",
-      "IB1=round((IE1/beta_dc_l)*1000,1);           #Base current of 1st transistor, nA\n",
-      "IB2=round((IE2/beta_dc_r)*1000,1);           #Base current of 2nd transistor, nA\n",
-      "I_in_offset=IB1-IB2;                 #Input offset current, nA\n",
-      "\n",
-      "#(ii)\n",
-      "I_in_bias=(IB1+IB2)/2;                  #Input bias current, nA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The input offset current=%.1fnA\"%I_in_offset);\n",
-      "print(\"(ii) The input bias current=%.1fnA\"%I_in_bias);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The input offset current=15.1nA\n",
-        "(ii) The input bias current=75.8nA\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.12 : Page number 679\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_in_offset=20;                 #Input offset current, nA\n",
-      "I_in_bias=80;                   #Input bias current, nA\n",
-      "\n",
-      "#Calculation\n",
-      "IB1=I_in_bias+I_in_offset/2;            #Base current in 1st transistor, nA\n",
-      "IB2=I_in_bias-I_in_offset/2;            #Base current in 2nd transistor, nA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The two base currents are: IB1=%dnA and IB2=%dnA.\"%(IB1,IB2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The two base currents are: IB1=90nA and IB2=70nA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.13 : Page number 679-680\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declration\n",
-      "I_in_offset=20;                 #Input offset current, nA\n",
-      "I_in_bias=80;                   #Input bias current, nA\n",
-      "A=150;                          #Voltage gain\n",
-      "RB=100;                         #Base resistor, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_io=(I_in_offset*10**-9*RB*1000)*1000;       #Input offset voltage, mV\n",
-      "V_out_offset=(A*V_io)/1000;                     #Output offset voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The input offset voltage=%dmV.\"%V_io);\n",
-      "print(\"The output offset voltage=%.1fV.\"%V_out_offset);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input offset voltage=2mV.\n",
-        "The output offset voltage=0.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.14 : Page number 682\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15;                 #Collector supply voltage, V\n",
-      "VEE=15;                 #Emitter supply voltage, V\n",
-      "RE=1;                   #Emitter resistor, M\u03a9\n",
-      "RC=1;                   #Collector resistor, M\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IE=VEE/RE;                  #Tail current, \u03bcA\n",
-      "IE1=IE/2.0;                   #Emitter current of 1st transistor, \u03bcA\n",
-      "IE2=IE1;                    #Emitter current of 2nd transistor, \u03bcA\n",
-      "re=25/IE1;                  #a.c emitter resistance, k\u03a9\n",
-      "A_DM=RC/(2.0*re);        #Differential voltage gain,\n",
-      "\n",
-      "#(i)\n",
-      "vin=1;                      #Input voltage, V\n",
-      "Vout=A_DM*vin;              #Output voltage, V\n",
-      "\n",
-      "print(\"(i) Output voltage=%.2fV.\"%Vout);\n",
-      "\n",
-      "#(ii)\n",
-      "vin=-1;                      #Input voltage, V\n",
-      "Vout=A_DM*vin;              #Output voltage, V;\n",
-      "print(\"(ii) Output voltage=%.2fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Output voltage=0.15V.\n",
-        "(ii) Output voltage=-0.15V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.15 : Page number 682-683\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12;                 #Collector supply voltage, V\n",
-      "VEE=12;                 #Emitter supply voltage, V\n",
-      "RE=100;                 #Emitter resistor, k\u03a9\n",
-      "RC1=120;                #Collector resistor of 1st transistor, k\u03a9\n",
-      "RC2=120;                #Collector resistor of 2nd transistor, k\u03a9\n",
-      "beta=220;               #Base amplification factor\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calcualtion\n",
-      "IE=((VEE-VBE)/RE)*1000;     #Tail current, \u03bcA\n",
-      "IE1=IE/2.0;                   #Emitter current of 1st transistor, \u03bcA\n",
-      "IE2=IE1;                    #Emitter current of 2nd transistor, \u03bcA\n",
-      "re=(25/IE1)*1000;           #a.c emitter resistance, \u03a9\n",
-      "Zin=2*beta*re/1000;         #Input impedance, k\u03a9\n",
-      "A_DM=RC1*1000/(2.0*re);             #Differential voltage gain,\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The input impedance=%dk\u03a9.\"%Zin);\n",
-      "print(\"(ii) The differential voltage gain=%.0f.\"%A_DM);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The input impedance=194k\u03a9.\n",
-        "(ii) The differential voltage gain=136.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.16: Page number 683-684\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12;                 #Collector supply voltage, V\n",
-      "VEE=12;                 #Emitter supply voltage, V\n",
-      "RE=200;                 #Emitter resistor, k\u03a9\n",
-      "RC=100;                 #Collector resistor, k\u03a9\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "IE=round((VEE-VBE)/RE,4);            #Tail current, mA\n",
-      "IE1=round(IE/2,4);                   #Emitter current of 1st transistor, mA\n",
-      "IE2=IE1;                             #Emitter current of 2nd transistor, mA\n",
-      "re=round(25/IE1,1);                  #a.c emitter resistance, \u03a9\n",
-      "A_DM=RC*1000/(2*re);                 #Differential voltage gain,\n",
-      "\n",
-      "#Result\n",
-      "print(\"Differential voltage gain=%.1f.\"%A_DM);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Differential voltage gain=56.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.17 : Page number 685-686\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "v1=0.5;               #Voltage in terminal 1, mV\n",
-      "v2=-0.5;               #Voltage in terminal 2, mV\n",
-      "vo=8.0;                   #Output voltage, V\n",
-      "vo_cm=12.0;               #Common mode output, mV\n",
-      "\n",
-      "#Calculation\n",
-      "vin=v1-v2;                  #Differential input, mV\n",
-      "A_DM=vo/(vin/1000.0);         #Differential mode gain,\n",
-      "vin_cm=1;                   #Common mode input, mV\n",
-      "A_CM=vo_cm/vin_cm;          #Common mode gain\n",
-      "CMRR=A_DM/A_CM;             #Common mode rejection ratio\n",
-      "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-      "\n",
-      "#Result\n",
-      "print(\"Common mode rejection ratio=%.1f.\"%CMRR)\n",
-      "print(\"Common mode rejection ratio in decibel=%.2fdB\"%CMRR_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Common mode rejection ratio=666.7.\n",
-        "Common mode rejection ratio in decibel=56.48dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.18 : Page number 686\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_DM=200000;                #Differential mode gain\n",
-      "CMRR_dB=90;                 #Common mode rejection ratio, dB\n",
-      "\n",
-      "#Calculation\n",
-      "CMRR=10**(CMRR_dB/20.0);              #Common mode rejection ratio\n",
-      "A_CM=A_DM/CMRR;                     #Common mode gain\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Common mode voltage gain=%.2f.\"%A_CM);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Common mode voltage gain=6.32.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.19 : Page number 686\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "vin_cm=3.2;                     #Common input voltage, V\n",
-      "vout=26;                        #Output voltage, V\n",
-      "A_DM=100;                       #Open-circuit voltage gain\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "A_CM=vout*10**-3/vin_cm;                #Common mode gain\n",
-      "\n",
-      "#(ii)\n",
-      "CMRR_dB=20*log10(A_DM/A_CM);            #Common mode rejection ratio, dB\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The Common mode gain=%.4f\"%A_CM);\n",
-      "print(\"(ii) The common mode rejection ratio=%.1fdB.\"%CMRR_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The Common mode gain=0.0081\n",
-        "(ii) The common mode rejection ratio=81.8dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.20 : Page number 686-687\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12;                 #Collector supply voltage, V\n",
-      "VEE=12;                 #Emitter supply voltage, V\n",
-      "RE=200.0;                 #Emitter resistor, k\u03a9\n",
-      "RC=100.0;                 #Collector resistor, k\u03a9\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "A_CM=round(RC/(2*RE),2);             #Common mode voltage gain\n",
-      "\n",
-      "#(ii)\n",
-      "IE=round((VEE-VBE)/RE,4);            #Tail current, mA\n",
-      "IE1=round(IE/2,4);                   #Emitter current of 1st transistor, mA\n",
-      "IE2=IE1;                             #Emitter current of 2nd transistor, mA\n",
-      "re=round(25/IE1,1);                  #a.c emitter resistance, \u03a9\n",
-      "A_DM=RC*1000/(2*re);                 #Differential voltage gain,\n",
-      "CMRR_dB=floor(20*log10(A_DM/A_CM)*100)/100;         #Common mode rejection ratio, dB\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) Common mode gain=%.2f\"%A_CM);\n",
-      "print(\"(ii)Common mode rejection ratio=%.2fdB\"%CMRR_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Common mode gain=0.25\n",
-        "(ii)Common mode rejection ratio=47.09dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.21 : Page number 691\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "ACL=500;                    #closed loop gain\n",
-      "f_unity=15;                 #frequency with cloased-loop unity gain, MHz\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "f2=f_unity*1000/500                 #Upper frequency of bandwidth,kHz\n",
-      "BW=f2-0;                            #Bandwidth, kHz\n",
-      "A_CL=f_unity*1000/200;                    #Maximum value of A_CL when f2=200kHz\n",
-      "A_CL_dB=20*log10(A_CL);             #Maximum value of A_CL in decibel\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"f2=%dkHz\"%f2);\n",
-      "print(\"ACL=%d or %.1fdB.\"%(A_CL,A_CL_dB));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "f2=30kHz\n",
-        "ACL=75 or 37.5dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.22 : Page number 691-692\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "GBW=1.5;                    #Gain-bandwidth, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) For A_CL=1;\n",
-      "A_CL=1;                         #Closed loop gain\n",
-      "BW=GBW/A_CL;                    #Bandwidth, MHz\n",
-      "\n",
-      "print(\"(i) Operating Bandwidth=%.1fMHz.\"%BW);\n",
-      "\n",
-      "#(ii) For A_CL=10;\n",
-      "A_CL=10;                         #Closed loop gain\n",
-      "BW=(GBW/A_CL)*1000;              #Bandwidth, kHz\n",
-      "\n",
-      "print(\"(ii) Operating Bandwidth=%dkHz.\"%BW);\n",
-      "\n",
-      "#(iii) For A_CL=100;\n",
-      "A_CL=100;                         #Closed loop gain\n",
-      "BW=(GBW/A_CL)*1000;               #Bandwidth, kHz\n",
-      "\n",
-      "print(\"(iii) Operating Bandwidth=%dkHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Operating Bandwidth=1.5MHz.\n",
-        "(ii) Operating Bandwidth=150kHz.\n",
-        "(iii) Operating Bandwidth=15kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.23 : Page number 692\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "slew_rate=0.5;                  #Slew rate, V/\u03bcs\n",
-      "V_supply=10;                     #Supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_sat=V_supply-2;                           #Saturation voltage, V\n",
-      "V_pk=V_sat;                                 #Maximum peak-output voltage, V\n",
-      "f_max=((slew_rate*10**6)/(2*pi*V_pk))/1000;       #Maximum operating frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum operating frequency=%.2fkHz.\"%f_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum operating frequency=9.95kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.24 : Page number 692\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "slew_rate=0.5;                  #Slew rate, V/\u03bcs\n",
-      "V_pk=100.0;                       #Peak-output voltage, mV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_pk=V_pk/1000.0;                                #Peak-output voltage, V\n",
-      "f_max=(slew_rate*10**6/(2*pi*V_pk))/1000.0;      #Maximum operating frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum operating frequency=%.0fkHz\"%f_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum operating frequency=796kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.25 : Page number 695-696\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_CL=-100;              #Closed-loop voltage gain\n",
-      "Ri=2.2;                 #Input resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, A_CL=-(Rf/Ri)\n",
-      "Rf=-A_CL*Ri;                #Feedback resistor, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"Feedback resistor=%dk\u03a9\"%Rf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Feedback resistor=220k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.26 : Page number 696\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "vin=2.5;                #Input voltage, mV\n",
-      "Rf=200;                 #Feedback resistor, k\u03a9\n",
-      "Ri=2;                   #Input resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-      "vout=A_CL*vin/1000;                  #Output voltage,V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%.2fV\"%vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=-0.25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.27 : Page number 696\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Varaiable declaration\n",
-      "Rf=1.0;                   #Feedback resistor, k\u03a9\n",
-      "Ri=1.0;                   #Input resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"Closed-loop voltage gain=%d\"%A_CL);\n",
-      "print(\"Therefore, output will have same amplitude but 180\u00b0 phase inversion.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Closed-loop voltage gain=-1\n",
-        "Therefore, output will have same amplitude but 180\u00b0 phase inversion.\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.28 : Page number 696-697\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=40;                   #Feedback resistor, k\u03a9\n",
-      "Ri=1;                   #Input resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Closed-loop voltage gain=%d\"%A_CL);\n",
-      "print(\"Supply voltage=\u00b115V, saturation voltage=\u00b113V. Since gain=-40, op-Amp will be driven to saturation.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Closed-loop voltage gain=-40\n",
-        "Supply voltage=\u00b115V, saturation voltage=\u00b113V. Since gain=-40, op-Amp will be driven to saturation.\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.29 : Page number 697\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "Rf=100;                   #Feedback resistor, k\u03a9\n",
-      "Ri=10;                    #Input resistor, k\u03a9\n",
-      "Vpp=1;                    #Input peak-peak voltage, V\n",
-      "slew_rate=0.5;            #Slew rate, V//\u03bcs\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-      "\n",
-      "#(ii)\n",
-      "Zi=Ri;                  #Input impedance(~ Input resistor), k\u03a9\n",
-      "\n",
-      "#(iii)\n",
-      "Vout=A_CL*Vpp;                           #Peak-to-peak voltage, V\n",
-      "Vpk=Vout/2;                              #Peak output voltage, V\n",
-      "f_max=(slew_rate*10**6/(2*pi*abs(Vpk)))/1000;        #Maximum operating frequency, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  A_CL=%d.\"%A_CL);\n",
-      "print(\"(ii) Zi=%dk\u03a9\"%Zi);\n",
-      "print(\"(iii) Maximum operating frequency=%.1fkHz.\"%f_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  A_CL=-10.\n",
-        "(ii) Zi=10k\u03a9\n",
-        "(iii) Maximum operating frequency=15.9kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.30 : Page number 697\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_CL=-4;                    #Closed loop voltage gain\n",
-      "R=[1.0,5.0,10.0,20.0];              #List of available resistors, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "for i in R[:]:\n",
-      "    for j in R[:]:\n",
-      "        if -(i/j)==A_CL :\n",
-      "            print(\"Rf=%dk\u03a9 and Ri=%dk\u03a9.\"%(i,j));\n",
-      "            break;\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rf=20k\u03a9 and Ri=5k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.31 : Page  number 697-698\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=100;                   #Feedback resistor, k\u03a9\n",
-      "Ri=1;                    #Input resistor, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_source=0;               #Source resistor, k\u03a9\n",
-      "A_CL=-Rf/(R_source+Ri);    #Closed-loop voltage gain\n",
-      "\n",
-      "print(\"(i) Closed loop voltage gain=%d.\"%A_CL);\n",
-      "\n",
-      "#(ii)\n",
-      "R_source=1;               #Source resistor, k\u03a9\n",
-      "A_CL=-Rf/(R_source+Ri);    #Closed-loop voltage gain\n",
-      "\n",
-      "print(\"(ii) Closed loop voltage gain=%d.\"%A_CL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Closed loop voltage gain=-100.\n",
-        "(ii) Closed loop voltage gain=-50.\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.32 : Page number 699-700\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=240;                   #Feedback resistor, k\u03a9\n",
-      "Ri=2.4;                    #Input resistor, k\u03a9\n",
-      "Vin=120;                    #Input voltage, \u03bcV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "A_CL=1+(Rf/Ri);             #Closed loop voltage gain\n",
-      "Vout=(A_CL*Vin)/1000;       #Output voltage, mV\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%.2fmV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=12.12mV\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.33 : Page number 700\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=10;                   #Feedback resistor, k\u03a9\n",
-      "Ri=1;                    #Input resistor, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "A_CL=1+(Rf/Ri);             #Closed loop voltage gain\n",
-      "#(i)\n",
-      "Vin=1;                      #Input voltage, V\n",
-      "Vout=A_CL*Vin;              #Output voltage, V\n",
-      "\n",
-      "print(\"(i)  Output voltage=%dV\"%Vout);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "Vin=-1;                      #Input voltage, V\n",
-      "Vout=A_CL*Vin;               #Output voltage, V\n",
-      "\n",
-      "print(\"(ii) Output voltage=%dV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Output voltage=11V\n",
-        "(ii) Output voltage=-11V\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.34 : Page number 700\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=5;                       #Feedback resistor, k\u03a9\n",
-      "Ri=1;                       #Input resistor, k\u03a9\n",
-      "Vin_max=1;                  #Maximum input voltage, V\n",
-      "Vin_min=-1;                 #Minimum input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_inpp=Vin_max-Vin_min;               #Peak-peak input voltage, V\n",
-      "A_CL=1+(Rf/Ri);                     #Closed loop voltage gain\n",
-      "Vout_pp=A_CL*V_inpp;                #Peak-peak output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Peak to peak output voltage=%dV\"%Vout_pp);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Peak to peak output voltage=12V\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.35 : Page number 700-701\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "Rf=100;                       #Feedback resistor, k\u03a9\n",
-      "Ri=10;                       #Input resistor, k\u03a9\n",
-      "Vpp=1;                    #Input peak-peak voltage, V\n",
-      "slew_rate=0.5;              #Slew rate, V/\u03bcs\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "A_CL=1+(Rf/Ri);                                      #Closed loop voltage gain\n",
-      "\n",
-      "#(ii)\n",
-      "Vout_pp=A_CL*Vpp;                                    #Peak-peak output voltage, V\n",
-      "Vpk=Vout_pp/2.0;                                          #Peak output voltage, V\n",
-      "f_max=((slew_rate*10**6)/(2*pi*Vpk))/1000.0;           #Maximum operating frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  Closed-loop voltage gain=%d\"%A_CL);\n",
-      "\n",
-      "print(\"(ii) Maximum operating frequency=%.2fkHz\"%f_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Closed-loop voltage gain=11\n",
-        "(ii) Maximum operating frequency=14.47kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.36 : Page number 701\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=220;                       #Feedback resistor, k\u03a9\n",
-      "Ri=3.3;                       #Input resistor, k\u03a9\n",
-      "unity_gain_BW=3;              #Unity gain bandwidth, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) For non-inverting amplifier\n",
-      "A_CL=1+(Rf/Ri);                    #Closed loop voltage gain\n",
-      "BW=unity_gain_BW*1000.0/A_CL;        #Bandwidth, kHz\n",
-      "\n",
-      "print(\"(i)  Bandwidth=%.1fkHz.\"%BW);\n",
-      "\n",
-      "#(ii) For inverting amplifier\n",
-      "Rf=47;                          #Feedback resistor, k\u03a9\n",
-      "Ri=1;                           #Input resistor, k\u03a9\n",
-      "A_CL=-(Rf/Ri);                    #Closed loop voltage gain\n",
-      "BW=unity_gain_BW*1000.0/abs(A_CL);        #Bandwidth, kHz\n",
-      "\n",
-      "print(\"(ii)  Bandwidth=%.1fkHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Bandwidth=44.3kHz.\n",
-        "(ii)  Bandwidth=63.8kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 47
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.37 : Page number 701-702\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#(i)\n",
-      "A_CL=1;                     #Closed loop voltage gain for voltage follower\n",
-      "print(\"(i) For voltage follower A_CL=1.\");\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "slew_rate=0.5;              #Slew rate, V/\u03bcs\n",
-      "V_inpp=6;                   #peak-peak input voltage, V\n",
-      "Vout=A_CL*V_inpp;           #Peak-peak output voltage, V\n",
-      "Vpk=Vout/2;                 #Peak output voltage, V\n",
-      "\n",
-      "f_max=(slew_rate*10**6/(2*pi*Vpk))/1000;       #Maximum operating frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) The maximum output frequency=%.2fkHz.\"%f_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) For voltage follower A_CL=1.\n",
-        "(ii) The maximum output frequency=26.53kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.38 : Page number 702\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=470.0;                 #Feedback resistor, k\u03a9\n",
-      "R1=4.3;                 #Input resistor of 1st op-Amp, k\u03a9\n",
-      "R2=33.0;                 #Input resistor of 2nd op-Amp, k\u03a9\n",
-      "R3=33.0;                 #Input resistor of 3rd op-Amp, k\u03a9\n",
-      "Vin=80.0;                 #Input voltage, \u03bcV.\n",
-      "\n",
-      "#Calculation\n",
-      "A1=1+Rf/R1;                 #Gain of first op-Amp\n",
-      "A2=-round(Rf/R2,1);         #Gain of second op-Amp\n",
-      "A3=-round(Rf/R3,1);         #Gain of third op-Amp\n",
-      "A=A1*A2*A3;                 #Overall gain\n",
-      "Vout=A*Vin*10**-6;          #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%.2fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=1.78V\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.39 : Page number 702-703\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A1=10;                  #Voltage gain of 1st op-Amp\n",
-      "A2=-18;                 #Voltage gain of 2nd op-Amp\n",
-      "A3=-27;                 #Voltage gain of 3rd op-Amp\n",
-      "Rf=270;                 #Feedback resistor, k\u03a9\n",
-      "Vin=150;                #Input voltage, \u03bcV  \n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "R1=Rf/(A1-1);                   #Input resistor of 1st op-Amp, k\u03a9\n",
-      "R2=-Rf/A2;                      #Input resistr of 2nd op-Amp, k\u03a9\n",
-      "R3=-Rf/A3;                      #Input resistor of 3rd op-Amp, k\u03a9\n",
-      "\n",
-      "A=A1*A2*A3;                     #overall gain,\n",
-      "Vout=Vin*10**-6*A;              #Output voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"R1=%dk\u03a9, R2=%dk\u03a9 and R3=%dk\u03a9.\"%(R1,R2,R3));\n",
-      "print(\"Output voltage=%.3fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R1=30k\u03a9, R2=15k\u03a9 and R3=10k\u03a9.\n",
-        "Output voltage=0.729V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.40 : Page number 703-704\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=500;                 #Feedback resistor, k\u03a9\n",
-      "A1=-10;                 #Gain of 1st op-Amp\n",
-      "A2=-20;                 #Gain of 2nd op-Amp\n",
-      "A3=-50;                 #Gain of 3rd op-Amp\n",
-      "\n",
-      "#Calculation\n",
-      "R1=-Rf/A1;                      #Input resistor of 1st op-Amp, k\u03a9\n",
-      "R2=-Rf/A2;                      #Input resistor of 2nd op-Amp, k\u03a9\n",
-      "R3=-Rf/A3;                      #Input resistor of 3rd op-Amp, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"R1=%dk\u03a9, R2=%dk\u03a9 and R3=%dk\u03a9.\"%(R1,R2,R3));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R1=50k\u03a9, R2=25k\u03a9 and R3=10k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 52
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.41 : Page number 705\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Zin=2.0;              #Input impedance of op-Amp, M\u03a9\n",
-      "Zout=75.0;            #Output impedance of op-Amp, \u03a9\n",
-      "A_OL=200000.0;        #Open-loop voltage gain\n",
-      "Rf=220.0;             #Feedback resistor, k\u03a9\n",
-      "Ri=10.0;              #Input resistor, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "mv=round(Ri/(Ri+Rf),3);       #Feedback fraction\n",
-      "Zin_NI=Zin*(1+(A_OL*mv));     #Input impedance, M\u03a9\n",
-      "Zout_NI=Zout/(1+A_OL*mv);     #Output impedance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "A_CL=1+Rf/Ri;               #Closed loop voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The input impedance=%dM\u03a9 and output impedance=%.1e\u03a9.\"%(Zin_NI,Zout_NI));\n",
-      "print(\"(ii) The closed loop voltage gain=%d.\"%A_CL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The input impedance=17202M\u03a9 and output impedance=8.7e-03\u03a9.\n",
-        "(ii) The closed loop voltage gain=23.\n"
-       ]
-      }
-     ],
-     "prompt_number": 54
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.42 : Page number 705-706\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "#For voltage follower,\n",
-      "mv=1.0;               #Feedback fraction\n",
-      "A_OL=200000.0;        #Open-loop voltage gain\n",
-      "Zin=2.0;              #Input impedance of op-Amp, M\u03a9\n",
-      "Zout=75.0;            #Output impedance of op-Amp, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Zin_VF=Zin*(1+(A_OL*mv));     #Input impedance, M\u03a9\n",
-      "Zout_VF=round(round(Zout/(1+A_OL*mv),6),5);     #Output impedance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The input impedance=%dM\u03a9 and output impedance=%.2fe-03\u03a9.\"%(Zin_VF,Zout_VF*1000));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input impedance=400002M\u03a9 and output impedance=0.38e-03\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 63
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.43 : Page number 706\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=100;             #Feedback resistor, k\u03a9\n",
-      "Ri=1.0;               #Input resistor, k\u03a9\n",
-      "Zin=4;              #Input impedance of op-Amp, M\u03a9\n",
-      "Zout=50;            #Output impedance of op-Amp, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Zin_I=Ri;                   #Input impedance, k\u03a9\n",
-      "Zout_I=Zout;                #Output impedance, \u03a9\n",
-      "A_CL=-(Rf/Ri);              #Closed loop voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The input impedance=%dk\u03a9 and output impedance=%d\u03a9.\"%(Zin_I,Zout_I));\n",
-      "print(\"Closed-loop voltage gain=%d\"%A_CL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input impedance=1k\u03a9 and output impedance=50\u03a9.\n",
-        "Closed-loop voltage gain=-100\n"
-       ]
-      }
-     ],
-     "prompt_number": 65
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.44 : Page number 709\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=10;               #Feedback resistor, k\u03a9\n",
-      "Ri=10;               #Input resistor, k\u03a9\n",
-      "V1=3;                #Input voltage 1st, V\n",
-      "V2=1;                #Input voltage 2nd, V\n",
-      "V3=8;                #Input voltage 3rd, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Rf=Ri, Vout=-(Rf/Ri)*(V1+V2+V3)= -(V1+V2+V3);\n",
-      "Vout=-(V1+V2+V3);           #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%dV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=-12V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 66
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.45 : Page number 709\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=10;               #Feedback resistor, k\u03a9\n",
-      "R1=1;                #Input resistor for input 1, k\u03a9\n",
-      "R2=1;                #Input resistor for input 2, k\u03a9\n",
-      "V1=0.2;              #Input voltage 1st, V\n",
-      "V2=0.5;              #Input voltage 2nd, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "R=R1;                       #Input resistor(=R1 or R2), k\u03a9\n",
-      "Vout=-(Rf/R)*(V1+V2);       #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%dV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=-7V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 67
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.46 : Page number 709-710\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=1;                #Feedback resistor, k\u03a9\n",
-      "Ri=10.0;               #Input resistor, k\u03a9\n",
-      "V1=10;                #Input voltage 1st, V\n",
-      "V2=8.0;                #Input voltage 2nd, V\n",
-      "V3=7.0;                #Input voltage 3rd, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Vout=-(Rf/Ri)*(V1+V2+V3);\n",
-      "Vout=-(Rf/Ri)*(V1+V2+V3);           #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%.1fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=-2.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 70
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.47 : Page number 710\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V1=0.6;                 #Input voltage to 1st input resistor, V\n",
-      "V2=-1.4;                #Input voltage to 2nd input resistor, V\n",
-      "Rf=200;                 #Feedback resistor, k\u03a9\n",
-      "R1=400;                 #Input resistor 1, k\u03a9\n",
-      "R2=100.0;                 #Input resistor 2, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=-Rf*(V1/R1 +V2/R2);                 #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%.1fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=2.5V\n"
-       ]
-      }
-     ],
-     "prompt_number": 71
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.48 : Page number 710-711\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=1.0;                 #Feedback resistor, k\u03a9\n",
-      "R1=1.0;                 #Input resistor 1, k\u03a9\n",
-      "R2=2.0;                 #Input resistor 2, k\u03a9\n",
-      "R3=4.0;                 #Input resistor 3, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Rf_R1=Rf/R1;                #Ratio of feedback resistor and 1st input resistor\n",
-      "Rf_R2=Rf/R2;                #Ratio of feedback resistor and 2nd input resistor\n",
-      "Rf_R3=Rf/R3;                #Ratio of feedback resistor and 3rd input resistor\n",
-      "\n",
-      "#(i) First input combination\n",
-      "V1=10;                                   #Input voltage to 1st input resistor, V\n",
-      "V2=0;                                    #Input voltage to 2nd input resistor, V\n",
-      "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-      "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-      "print(\"(i) The output voltage=%.1fV\"%Vout);\n",
-      "\n",
-      "#(i) First input combination\n",
-      "V1=0;                                   #Input voltage to 1st input resistor, V\n",
-      "V2=10;                                    #Input voltage to 2nd input resistor, V\n",
-      "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-      "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-      "print(\"(ii) The output voltage=%.1fV\"%Vout);\n",
-      "\n",
-      "\n",
-      "#(i) First input combination\n",
-      "V1=10;                                   #Input voltage to 1st input resistor, V\n",
-      "V2=10;                                    #Input voltage to 2nd input resistor, V\n",
-      "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-      "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-      "print(\"(iii) The output voltage=%.1fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage=-12.5V\n",
-        "(ii) The output voltage=-7.5V\n",
-        "(iii) The output voltage=-17.5V\n"
-       ]
-      }
-     ],
-     "prompt_number": 72
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.49 : Page number 711\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=330;                 #Feedback resistor, k\u03a9\n",
-      "R1=33.0;                  #Input resistor 1, k\u03a9\n",
-      "R2=10.0;                  #Input resistor 2, k\u03a9\n",
-      "V1_m=50;                #Peak voltage of 1st input, mV\n",
-      "V2_m=10;                #Peak voltage of 2nd input, mV\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Vout=-((Rf/R1)*V1 + (Rf/R2)*V2)\n",
-      "print(\"Vout=-[%.1fsin(1000t)+%.2fsin(3000t)]V\"%((V1_m/1000.0)*(Rf/R1),(V2_m/1000.0)*(Rf/R2)));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vout=-[0.5sin(1000t)+0.33sin(3000t)]V\n"
-       ]
-      }
-     ],
-     "prompt_number": 74
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.50 : Page number 715\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=100;                      #Input resistor, k\u03a9\n",
-      "C=10;                       #Feedback capacitor, \u03bcF\n",
-      "\n",
-      "#Calculation\n",
-      "RC=R*10**3*C*10**-6;        #product of input resistance and feedback capacitance, s\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Vo=-1*(1/RC)\u222bvi dt.\");\n",
-      "print(\"=>Vo=-1*(1/%d)\u222bvi dt\"%RC);\n",
-      "print(\"=>Vo=\u222bvi dt\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vo=-1*(1/RC)\u222bvi dt.\n",
-        "=>Vo=-1*(1/1)\u222bvi dt\n",
-        "=>Vo=\u222bvi dt\n"
-       ]
-      }
-     ],
-     "prompt_number": 75
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.51 : Page number 715-716\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "Rf=100;                      #Feedback resistor, k\u03a9\n",
-      "C=0.01;                      #Feedback capacitor, \u03bcF\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "fc=1/(2*pi*Rf*1000*C*10**-6);               #Crictical frequency, Hz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The critical frequency=%dHz.\"%fc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The critical frequency=159Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 76
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.52 : Page number 716\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "R=10.0;                      #Input resistor, k\u03a9\n",
-      "C=0.01;                    #Feedback capacitor, \u03bcF\n",
-      "vin=5;                     #Input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout_change_rate=-vin/(R*C);            #Rate of change of output voltage, V/\u03bcs   \n",
-      "print(\"(i) Vout=-1*(1/RC)\u222bvi dt.\");\n",
-      "print(\"    \u0394Vout/dt = -vin/RC = %dmV/\u03bcs.\"%Vout_change_rate);\n",
-      "\n",
-      "#(ii) Plotting the output waveform\n",
-      "vin_plot=[];                    #Plotting variable for input waveform, V\n",
-      "dt=100;                          #time between edges, \u03bcs\n",
-      "for i in range(0,3*dt+1):\n",
-      "    if i<dt or i>2*dt :\n",
-      "        vin_plot.append(0);\n",
-      "    else:\n",
-      "        vin_plot.append(5); \n",
-      "p=plot(vin_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,300])\n",
-      "limit.set_ylim([-5,10])\n",
-      "xlabel(\"t(microsecond)\");\n",
-      "ylabel(\"Vin(V)\");\n",
-      "title(\"Input waveform\");\n",
-      "show(p);\n",
-      "\n",
-      "        \n",
-      "vout_plot=[];                  #Plotting variable for output waveform, V\n",
-      "t=[i for i in range(0,301)];  #Time scale, \u03bcs\n",
-      "for i in t[:] :\n",
-      "    if i<dt:\n",
-      "        vout_plot.append(0);\n",
-      "    elif i>2*dt:\n",
-      "        vout_plot.append((Vout_change_rate/1000.0)*dt);\n",
-      "    else :\n",
-      "        vout_plot.append((-vin_plot[i]/(R*C))/1000*(i-dt));\n",
-      "\n",
-      "    \n",
-      "p=plot(vout_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,300])\n",
-      "limit.set_ylim([-5,5]);\n",
-      "xlabel('t(microsecond)');\n",
-      "ylabel(\"Vout(V)\");\n",
-      "title(\"output waveform\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Vout=-1*(1/RC)\u222bvi dt.\n",
-        "    \u0394Vout/dt = -vin/RC = -50mV/\u03bcs.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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Lz0tKSjB58uSbmsezzz6LTz/91NClqWLOnDlIT08HAMTHx+PMmTMqV0SmhMFA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\nSGNftWoAAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7fdd1397e4d0>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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lZGRY1nPgwAE1b95c3bt3l3TpKl+1atWSw+FQQkKCGjRoIOnSNTTGjh0rSYqP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\nLAgDAMDifwGSFOcKwPnWKgAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7fdd137a8390>"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.53 : Page number 716-717\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_supply=15;              #Supply voltage, V\n",
-      "R=10;                     #Input resistor, k\u03a9\n",
-      "C=0.2;                    #Feedback capacitor, \u03bcF\n",
-      "vin=10;                   #Input voltage, mV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vs=-V_supply+2;         #Saturation voltage, V\n",
-      "print(\"Vout=-1*(1/RC)\u222bvi dt.\");\n",
-      "print(\"Vout=%d*t volts\"%(-vin/(R*C)));\n",
-      "t=Vs/(-vin/(R*C));                      #Time required, seconds\n",
-      "print(\"Time required=%.1fseconds.\"%t);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vout=-1*(1/RC)\u222bvi dt.\n",
-        "Vout=-5*t volts\n",
-        "Time required=2.6seconds.\n"
-       ]
-      }
-     ],
-     "prompt_number": 78
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.54 : Page number 717-718\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=1;                     #Feedback resistor, k\u03a9\n",
-      "C=0.1;                   #Input capacitor, \u03bcF\n",
-      "Vin_change=5;            #Change in input voltage, V\n",
-      "t=0.1;                   #Time taken for change in input voltage, ms\n",
-      "\n",
-      "#Calcualtion\n",
-      "dvi_dt=Vin_change/(t/1000);        #Rate of change of input voltage, V/s\n",
-      "RC=R*1000*C*10**-6;                #Product of feedback resistance and input capacitance, s\n",
-      "#Since, Vo=-R*C*(dvi/dt);\n",
-      "Vo=-RC*dvi_dt;                      #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Vo=%dV.\"%Vo);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vo=-5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 79
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.55 : Page number 718\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=10;                     #Feedback resistor, k\u03a9\n",
-      "C=2.2;                   #Input capacitor, \u03bcF\n",
-      "Vin_change=10;            #Change in input voltage, V\n",
-      "t=0.4;                   #Time taken for change in input voltage, s\n",
-      "\n",
-      "#Calcualtin\n",
-      "dvi_dt=Vin_change/t;        #Rate of change of input voltage, V/s\n",
-      "RC=R*1000*C*10**-6;         #Product of feedback resistance and input capacitance, s\n",
-      "#Since, Vo=-R*C*(dvi/dt);\n",
-      "Vo=-RC*dvi_dt;                      #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Vo=%.2fV.\"%Vo);\n",
-      "print(\"The output voltage stays constant at %.2fV.\"%Vo);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vo=-0.55V.\n",
-        "The output voltage stays constant at -0.55V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 80
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.56 : Page number 718-719\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=100;                     #Feedback resistor, k\u03a9\n",
-      "C=10;                   #Input capacitor, \u03bcF\n",
-      "Vin_change=1;            #Change in input voltage, V\n",
-      "t=0.2;                   #Time taken for change in input voltage, s\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "RC=R*1000*C*10**-6;             #Product of feedback resistor and input capacitance, s\n",
-      "#(i)\n",
-      "print(\"vo=-%d*(dvi/dt).\"%RC);\n",
-      "\n",
-      "#(ii)\n",
-      "dvi_dt=Vin_change/t;                #Rate of change of input voltage, V\n",
-      "vo=-dvi_dt;                         #Output voltage, V\n",
-      "print(\"vo=%dV.\"%vo);\n",
-      "\n",
-      "print(\"Therefore, between 0 to 0.2s, the output voltage is constant at %dV.\"%vo);\n",
-      "print(\"For t>0.2s, the input is constant so that output voltage is zero.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "vo=-1*(dvi/dt).\n",
-        "vo=-5V.\n",
-        "Therefore, between 0 to 0.2s, the output voltage is constant at -5V.\n",
-        "For t>0.2s, the input is constant so that output voltage is zero.\n"
-       ]
-      }
-     ],
-     "prompt_number": 82
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_1.ipynb
deleted file mode 100755
index 6add7ba9..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_1.ipynb
+++ /dev/null
@@ -1,2552 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:f21def3bd5ac5e7f3368a4e0e9a53b6961ecc94a2dca6b1917cdfda88271b542"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 25 : OPERATIONAL AMPLIFIERS"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.1: Page number 664\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A=100.0;              #Open-circuit voltage gain of differential amplifier\n",
-      "V1=3.25;              #Input voltage to terminal 1 in V\n",
-      "V2=3.15;              #Input voltage to terminal 2 in V\n",
-      "\n",
-      "#Calculations\n",
-      "V0=A*(V1-V2);         #Output voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The output voltage of the differential amplifier = %dV\"%V0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage of the differential amplifier = 10V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.2: Page number 672\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_DM=2000.0;              #Differential mode voltage gain\n",
-      "A_CM=0.2;                 #Common mode voltage gain\n",
-      "\n",
-      "#Calculations\n",
-      "CMRR=A_DM/A_CM;           #Common mode rejection ratio\n",
-      "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The common mode rejection ratio = %d.\"%CMRR);\n",
-      "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The common mode rejection ratio = 10000.\n",
-        "The common mode rejection ratio in decibels= 80dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.3: Page number 672\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "VD_in=10.0;               #Differential mode input in mV\n",
-      "VD_out=1.0;               #Output for differential mode input in V\n",
-      "VC_in=10.0;                 #Common mode input in mV\n",
-      "VC_out=5.0;                 #Output for common mode input in mV\\\n",
-      "\n",
-      "#Calculations\n",
-      "A_DM=(VD_out*1000)/VD_in;       #Differntial mode voltage gain\n",
-      "A_CM=VC_out/VC_in;       #Common mode voltage gain\n",
-      "CMRR=A_DM/A_CM;           #Common mode rejection ratio\n",
-      "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-      "\n",
-      "#Results\n",
-      "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The common mode rejection ratio in decibels= 46dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.4: Page number 672\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_DM=150.0;           #Differential mode voltage gain\n",
-      "CMRR_dB=90.0;         #Common mode rejection ratio\n",
-      "V1=100.0;             #Input voltage for terminal 1 in mV\n",
-      "V2=50.0;              #Input voltage for terminal 2 in mV\n",
-      "V_noise=1.0;          #Voltage of noise signal in mV\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "#Case(i)\n",
-      "V_out=A_DM*(V1-V2)/1000.0;            #Output voltage for differntial mode input, in V\n",
-      "\n",
-      "#Since CMRR_dB=20*log10(differential mode gain/common mode gain),\n",
-      "A_CM=A_DM/pow(10,(CMRR_dB/20));         #Common mode gain\n",
-      "V_OUT_noise=A_CM*(V_noise/1000);        #Noise on output in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Output voltage =%.1fV\"%V_out);\n",
-      "print(\"Noise on output = %.1fx10^-6V\"%(V_OUT_noise*pow(10,6)));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage =7.5V\n",
-        "Noise on output = 4.7x10^-6V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.5 : Page number 672-673\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "A_DM=2500.0;                    #Differential mode voltage gain\n",
-      "CMRR=30000.0;                   #Common mode rejection ratio\n",
-      "Input_signal=500.0;             #Single ended input r.m.s signal in microvolts\n",
-      "Interference=1.0;               #Interference signal, in V\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "A_CM=A_DM/CMRR;                 #Common mode gain\n",
-      "\n",
-      "#(ii)\n",
-      "CMRR_dB=20*log10(CMRR);         #Common mode rejection ratio in decibels\n",
-      "\n",
-      "#(iii)\n",
-      "V_out=A_DM*(Input_signal/pow(10,6)-0);        #r.m.s output signal  in V\n",
-      "\n",
-      "#(iv)\n",
-      "Interference_out=A_CM*Interference;          #r.m.s interference output in V\n",
-      "Interference_out=Interference_out*1000;      #r.m.s interference output in mV\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Common mode gain =%.3f\"%A_CM);\n",
-      "print(\"Common mode rejection ratio in decibels=%.1fdB\"%CMRR_dB);\n",
-      "print(\"r.m.s output signal =%.2fV\"%V_out);\n",
-      "print(\"r.m.s interfernce output voltage = %dmV\"%Interference_out);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Common mode gain =0.083\n",
-        "Common mode rejection ratio in decibels=89.5dB\n",
-        "r.m.s output signal =1.25V\n",
-        "r.m.s interfernce output voltage = 83mV\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.6 : Page number 674-675\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12;                 #Collector supply voltage, V\n",
-      "VEE=12;                 #Emitter supply voltage, V\n",
-      "RB=10;                  #Base resistor, k\u03a9\n",
-      "RC2=10;                 #Collector resistor, k\u03a9\n",
-      "RE=25;                  #Emitter resistor, k\u03a9\n",
-      "VBE=0.7;               #Base-emitter voltage, V\n",
-      "beta=100;               #Base amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-      "IE=(VEE-VBE)/RE;            #Tail current, mA\n",
-      "IE1=IE/2;                   #Emitter current of 1st transistor, mA\n",
-      "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-      "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-      "IC2=IC1;                    #Collector current of 2nd transistor, mA\n",
-      "IB1=(IC1/beta)*1000;        #Base current of 1st transistor, \u03bcA\n",
-      "IB2=IB1;                    #Base current of 2nd transistor, \u03bcA\n",
-      "VC1=VCC;                    #Collector voltage of 1st transistor, V\n",
-      "VC2=VCC-IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"VE=%.1fV\"%VE);\n",
-      "print(\"IE=%.3fmA\"%IE);\n",
-      "print(\"IE1=%.3fmA\"%IE1);\n",
-      "print(\"IE2=%.3fmA\"%IE2);\n",
-      "print(\"IC1=%.3fmA\"%IC1);\n",
-      "print(\"IC2=%.3fmA\"%IC2);\n",
-      "print(\"IB1=%.2f\u03bcA\"%IB1);\n",
-      "print(\"IB2=%.2f\u03bcA\"%IB2);\n",
-      "print(\"VC1=%dV\"%VC1);\n",
-      "print(\"VC2=%.1fV\"%VC2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VE=-0.7V\n",
-        "IE=0.452mA\n",
-        "IE1=0.226mA\n",
-        "IE2=0.226mA\n",
-        "IC1=0.226mA\n",
-        "IC2=0.226mA\n",
-        "IB1=2.26\u03bcA\n",
-        "IB2=2.26\u03bcA\n",
-        "VC1=12V\n",
-        "VC2=9.7V\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.7 : Page number 675\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15;                 #Collector supply voltage, V\n",
-      "VEE=15;                 #Emitter supply voltage, V\n",
-      "RB=33;                  #Base resistor, k\u03a9\n",
-      "RC=15;                  #Collector resistor, k\u03a9\n",
-      "RE=15;                  #Emitter resistor, k\u03a9\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IE_tail=(VEE-VBE)/RE;            #Tail current, mA\n",
-      "IE=round(IE_tail/2,3);                    #Emitter current in each transistor, mA\n",
-      "IC=IE;                           #Collector current(=emitter current), mA\n",
-      "Vout=VCC-IC*RC;                  #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output voltage=%.2fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output voltage=7.85V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.8 : Page number 675\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                 #Collector supply voltage, V\n",
-      "VEE=15.0;                 #Emitter supply voltage, V\n",
-      "RB=33.0;                  #Base resistor, k\u03a9\n",
-      "RC=15.0;                  #Collector resistor, k\u03a9\n",
-      "RE=15.0;                  #Emitter resistor, k\u03a9\n",
-      "VBE=0;                  #Base-emitter voltage, V\n",
-      "beta_dc_l=90.0;           #base current amplification factor for left transistor\n",
-      "beta_dc_r=110.0;          #base current amplification factor for right transistor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IE_tail=(VEE-VBE)/RE;             #Tail current, mA\n",
-      "IE=IE_tail/2;                      #Emitter current in each transistor, mA\n",
-      "IB1=(IE/beta_dc_l)*1000;          #Base current of 1st transistor, \u03bcA\n",
-      "IB2=(IE/beta_dc_r)*1000;          #Base current of 2nd transistor, \u03bcA\n",
-      "\n",
-      "#(ii)\n",
-      "VB1=-IB1/1000*RB;                    #Base voltage of 1st transistor, V\n",
-      "VB2=-IB2/1000*RB;                    #Base voltage of 1st transistor, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) IB1=%.2f\u03bcA\"%IB1);\n",
-      "print(\"    IB2=%.2f\u03bcA\"%IB2);\n",
-      "print(\"(ii) VB1=%.3fV\"%VB1);\n",
-      "print(\"     VB2=%.2fV\"%VB2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) IB1=5.56\u03bcA\n",
-        "    IB2=4.55\u03bcA\n",
-        "(ii) VB1=-0.183V\n",
-        "     VB2=-0.15V\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.9 : Page number 675-676\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                 #Collector supply voltage, V\n",
-      "VEE=15.0;                 #Emitter supply voltage, V\n",
-      "RB=10.0;                  #Base resistor, k\u03a9\n",
-      "RC1=10.0;                 #Collector resistor of 1st transistor, k\u03a9\n",
-      "RC2=10.0;                 #Collector resistor of 2nd transistor, k\u03a9\n",
-      "IE=1.0;                   #Tail current, mA\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-      "IE1=IE/2.0;                   #Emitter current of 1st transistor, mA\n",
-      "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-      "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-      "IC2=IE2;                    #Collector current of 2nd transistor, mA\n",
-      "VC1=VCC-IC1*RC1;            #Collector voltage of 1st transistor, V\n",
-      "VC2=VCC-IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"VE=%.1fV\"%VE);\n",
-      "print(\"Emitter current in each transistor=%.1fmA.\"%(IE/2.0));\n",
-      "print(\"IC1~IE1=%.1fmA and IC2~IE2=%.1fmA\"%(IE1,IE2));\n",
-      "print(\"VC1=VC2=%dV.\"%VC2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VE=-0.7V\n",
-        "Emitter current in each transistor=0.5mA.\n",
-        "IC1~IE1=0.5mA and IC2~IE2=0.5mA\n",
-        "VC1=VC2=10V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.10 : Page number 676-677\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage, V\n",
-      "VEE=12.0;                 #Emitter supply voltage, V\n",
-      "RC2=10.0;                 #Collector resistor of 2nd transistor, k\u03a9\n",
-      "RE=25.0;                  #Emitter current, k\u03a9\n",
-      "VBE=-0.7;               #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-      "IE=(VCC-VE)/RE;             #Tail current, mA\n",
-      "IE1=IE/2.0;                   #Emitter current of 1st transistor, mA\n",
-      "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-      "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-      "IC2=IE2;                    #Collector current of 2nd transistor, mA\n",
-      "VC1=-VEE;                   #Collector voltage of 1st transistor, V\n",
-      "VC2=-VEE+IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"VE=%.1fV\"%VE);\n",
-      "print(\"Tail current=%.3fmA.\"%IE);\n",
-      "print(\"Emitter current in each transistor=%.3fmA.\"%(IE/2.0));\n",
-      "print(\"IC1~IE1=%.3fmA and IC2~IE2=%.3fmA\"%(IC1,IC2));\n",
-      "print(\"VC1=%dV\"%VC1);\n",
-      "print(\"VC2=%.2fV\"%VC2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VE=0.7V\n",
-        "Tail current=0.452mA.\n",
-        "Emitter current in each transistor=0.226mA.\n",
-        "IC1~IE1=0.226mA and IC2~IE2=0.226mA\n",
-        "VC1=-12V\n",
-        "VC2=-9.74V\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.11 : Page number 679\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15;                 #Collector supply voltage, V\n",
-      "VEE=15;                 #Emitter supply voltage, V\n",
-      "RB=1;                   #Base resistor, M\u03a9\n",
-      "RC2=1;                  #Collector resistor, M\u03a9\n",
-      "RE=1;                   #Emitter resistor, M\u03a9\n",
-      "VBE=0;                  #Base-emitter voltage, V (Neglected)\n",
-      "beta_dc_l=90.0;           #base current amplification factor for left transistor\n",
-      "beta_dc_r=110.0;          #base current amplification factor for right transistor\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "IE=(VEE-VBE)/RE;                    #Tail current, \u03bcA\n",
-      "IE1=IE/2.0;                           #Emitter current of 1st transistor, \u03bcA\n",
-      "IE2=IE1;                            #Emitter current of 2nd transistor, \u03bcA\n",
-      "IB1=round((IE1/beta_dc_l)*1000,1);           #Base current of 1st transistor, nA\n",
-      "IB2=round((IE2/beta_dc_r)*1000,1);           #Base current of 2nd transistor, nA\n",
-      "I_in_offset=IB1-IB2;                 #Input offset current, nA\n",
-      "\n",
-      "#(ii)\n",
-      "I_in_bias=(IB1+IB2)/2;                  #Input bias current, nA\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The input offset current=%.1fnA\"%I_in_offset);\n",
-      "print(\"(ii) The input bias current=%.1fnA\"%I_in_bias);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The input offset current=15.1nA\n",
-        "(ii) The input bias current=75.8nA\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.12 : Page number 679\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_in_offset=20;                 #Input offset current, nA\n",
-      "I_in_bias=80;                   #Input bias current, nA\n",
-      "\n",
-      "#Calculation\n",
-      "IB1=I_in_bias+I_in_offset/2;            #Base current in 1st transistor, nA\n",
-      "IB2=I_in_bias-I_in_offset/2;            #Base current in 2nd transistor, nA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The two base currents are: IB1=%dnA and IB2=%dnA.\"%(IB1,IB2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The two base currents are: IB1=90nA and IB2=70nA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.13 : Page number 679-680\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declration\n",
-      "I_in_offset=20;                 #Input offset current, nA\n",
-      "I_in_bias=80;                   #Input bias current, nA\n",
-      "A=150;                          #Voltage gain\n",
-      "RB=100;                         #Base resistor, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_io=(I_in_offset*10**-9*RB*1000)*1000;       #Input offset voltage, mV\n",
-      "V_out_offset=(A*V_io)/1000;                     #Output offset voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The input offset voltage=%dmV.\"%V_io);\n",
-      "print(\"The output offset voltage=%.1fV.\"%V_out_offset);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input offset voltage=2mV.\n",
-        "The output offset voltage=0.3V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.14 : Page number 682\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15;                 #Collector supply voltage, V\n",
-      "VEE=15;                 #Emitter supply voltage, V\n",
-      "RE=1;                   #Emitter resistor, M\u03a9\n",
-      "RC=1;                   #Collector resistor, M\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "IE=VEE/RE;                  #Tail current, \u03bcA\n",
-      "IE1=IE/2.0;                   #Emitter current of 1st transistor, \u03bcA\n",
-      "IE2=IE1;                    #Emitter current of 2nd transistor, \u03bcA\n",
-      "re=25/IE1;                  #a.c emitter resistance, k\u03a9\n",
-      "A_DM=RC/(2.0*re);        #Differential voltage gain,\n",
-      "\n",
-      "#(i)\n",
-      "vin=1;                      #Input voltage, V\n",
-      "Vout=A_DM*vin;              #Output voltage, V\n",
-      "\n",
-      "print(\"(i) Output voltage=%.2fV.\"%Vout);\n",
-      "\n",
-      "#(ii)\n",
-      "vin=-1;                      #Input voltage, V\n",
-      "Vout=A_DM*vin;              #Output voltage, V;\n",
-      "print(\"(ii) Output voltage=%.2fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Output voltage=0.15V.\n",
-        "(ii) Output voltage=-0.15V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.15 : Page number 682-683\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12;                 #Collector supply voltage, V\n",
-      "VEE=12;                 #Emitter supply voltage, V\n",
-      "RE=100;                 #Emitter resistor, k\u03a9\n",
-      "RC1=120;                #Collector resistor of 1st transistor, k\u03a9\n",
-      "RC2=120;                #Collector resistor of 2nd transistor, k\u03a9\n",
-      "beta=220;               #Base amplification factor\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calcualtion\n",
-      "IE=((VEE-VBE)/RE)*1000;     #Tail current, \u03bcA\n",
-      "IE1=IE/2.0;                   #Emitter current of 1st transistor, \u03bcA\n",
-      "IE2=IE1;                    #Emitter current of 2nd transistor, \u03bcA\n",
-      "re=(25/IE1)*1000;           #a.c emitter resistance, \u03a9\n",
-      "Zin=2*beta*re/1000;         #Input impedance, k\u03a9\n",
-      "A_DM=RC1*1000/(2.0*re);             #Differential voltage gain,\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The input impedance=%dk\u03a9.\"%Zin);\n",
-      "print(\"(ii) The differential voltage gain=%.0f.\"%A_DM);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The input impedance=194k\u03a9.\n",
-        "(ii) The differential voltage gain=136.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.16: Page number 683-684\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12;                 #Collector supply voltage, V\n",
-      "VEE=12;                 #Emitter supply voltage, V\n",
-      "RE=200;                 #Emitter resistor, k\u03a9\n",
-      "RC=100;                 #Collector resistor, k\u03a9\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "IE=round((VEE-VBE)/RE,4);            #Tail current, mA\n",
-      "IE1=round(IE/2,4);                   #Emitter current of 1st transistor, mA\n",
-      "IE2=IE1;                             #Emitter current of 2nd transistor, mA\n",
-      "re=round(25/IE1,1);                  #a.c emitter resistance, \u03a9\n",
-      "A_DM=RC*1000/(2*re);                 #Differential voltage gain,\n",
-      "\n",
-      "#Result\n",
-      "print(\"Differential voltage gain=%.1f.\"%A_DM);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Differential voltage gain=56.6.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.17 : Page number 685-686\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "v1=0.5;               #Voltage in terminal 1, mV\n",
-      "v2=-0.5;               #Voltage in terminal 2, mV\n",
-      "vo=8.0;                   #Output voltage, V\n",
-      "vo_cm=12.0;               #Common mode output, mV\n",
-      "\n",
-      "#Calculation\n",
-      "vin=v1-v2;                  #Differential input, mV\n",
-      "A_DM=vo/(vin/1000.0);         #Differential mode gain,\n",
-      "vin_cm=1;                   #Common mode input, mV\n",
-      "A_CM=vo_cm/vin_cm;          #Common mode gain\n",
-      "CMRR=A_DM/A_CM;             #Common mode rejection ratio\n",
-      "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-      "\n",
-      "#Result\n",
-      "print(\"Common mode rejection ratio=%.1f.\"%CMRR)\n",
-      "print(\"Common mode rejection ratio in decibel=%.2fdB\"%CMRR_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Common mode rejection ratio=666.7.\n",
-        "Common mode rejection ratio in decibel=56.48dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.18 : Page number 686\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_DM=200000;                #Differential mode gain\n",
-      "CMRR_dB=90;                 #Common mode rejection ratio, dB\n",
-      "\n",
-      "#Calculation\n",
-      "CMRR=10**(CMRR_dB/20.0);              #Common mode rejection ratio\n",
-      "A_CM=A_DM/CMRR;                     #Common mode gain\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Common mode voltage gain=%.2f.\"%A_CM);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Common mode voltage gain=6.32.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.19 : Page number 686\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "vin_cm=3.2;                     #Common input voltage, V\n",
-      "vout=26;                        #Output voltage, V\n",
-      "A_DM=100;                       #Open-circuit voltage gain\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "A_CM=vout*10**-3/vin_cm;                #Common mode gain\n",
-      "\n",
-      "#(ii)\n",
-      "CMRR_dB=20*log10(A_DM/A_CM);            #Common mode rejection ratio, dB\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  The Common mode gain=%.4f\"%A_CM);\n",
-      "print(\"(ii) The common mode rejection ratio=%.1fdB.\"%CMRR_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The Common mode gain=0.0081\n",
-        "(ii) The common mode rejection ratio=81.8dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.20 : Page number 686-687\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12;                 #Collector supply voltage, V\n",
-      "VEE=12;                 #Emitter supply voltage, V\n",
-      "RE=200.0;                 #Emitter resistor, k\u03a9\n",
-      "RC=100.0;                 #Collector resistor, k\u03a9\n",
-      "VBE=0.7;                #Base-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "A_CM=round(RC/(2*RE),2);             #Common mode voltage gain\n",
-      "\n",
-      "#(ii)\n",
-      "IE=round((VEE-VBE)/RE,4);            #Tail current, mA\n",
-      "IE1=round(IE/2,4);                   #Emitter current of 1st transistor, mA\n",
-      "IE2=IE1;                             #Emitter current of 2nd transistor, mA\n",
-      "re=round(25/IE1,1);                  #a.c emitter resistance, \u03a9\n",
-      "A_DM=RC*1000/(2*re);                 #Differential voltage gain,\n",
-      "CMRR_dB=floor(20*log10(A_DM/A_CM)*100)/100;         #Common mode rejection ratio, dB\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) Common mode gain=%.2f\"%A_CM);\n",
-      "print(\"(ii)Common mode rejection ratio=%.2fdB\"%CMRR_dB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Common mode gain=0.25\n",
-        "(ii)Common mode rejection ratio=47.09dB\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.21 : Page number 691\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log10\n",
-      "\n",
-      "#Variable declaration\n",
-      "ACL=500;                    #closed loop gain\n",
-      "f_unity=15;                 #frequency with cloased-loop unity gain, MHz\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "f2=f_unity*1000/500                 #Upper frequency of bandwidth,kHz\n",
-      "BW=f2-0;                            #Bandwidth, kHz\n",
-      "A_CL=f_unity*1000/200;                    #Maximum value of A_CL when f2=200kHz\n",
-      "A_CL_dB=20*log10(A_CL);             #Maximum value of A_CL in decibel\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"f2=%dkHz\"%f2);\n",
-      "print(\"ACL=%d or %.1fdB.\"%(A_CL,A_CL_dB));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "f2=30kHz\n",
-        "ACL=75 or 37.5dB.\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.22 : Page number 691-692\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "GBW=1.5;                    #Gain-bandwidth, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) For A_CL=1;\n",
-      "A_CL=1;                         #Closed loop gain\n",
-      "BW=GBW/A_CL;                    #Bandwidth, MHz\n",
-      "\n",
-      "print(\"(i) Operating Bandwidth=%.1fMHz.\"%BW);\n",
-      "\n",
-      "#(ii) For A_CL=10;\n",
-      "A_CL=10;                         #Closed loop gain\n",
-      "BW=(GBW/A_CL)*1000;              #Bandwidth, kHz\n",
-      "\n",
-      "print(\"(ii) Operating Bandwidth=%dkHz.\"%BW);\n",
-      "\n",
-      "#(iii) For A_CL=100;\n",
-      "A_CL=100;                         #Closed loop gain\n",
-      "BW=(GBW/A_CL)*1000;               #Bandwidth, kHz\n",
-      "\n",
-      "print(\"(iii) Operating Bandwidth=%dkHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Operating Bandwidth=1.5MHz.\n",
-        "(ii) Operating Bandwidth=150kHz.\n",
-        "(iii) Operating Bandwidth=15kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.23 : Page number 692\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "slew_rate=0.5;                  #Slew rate, V/\u03bcs\n",
-      "V_supply=10;                     #Supply voltage, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_sat=V_supply-2;                           #Saturation voltage, V\n",
-      "V_pk=V_sat;                                 #Maximum peak-output voltage, V\n",
-      "f_max=((slew_rate*10**6)/(2*pi*V_pk))/1000;       #Maximum operating frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum operating frequency=%.2fkHz.\"%f_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum operating frequency=9.95kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.24 : Page number 692\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "slew_rate=0.5;                  #Slew rate, V/\u03bcs\n",
-      "V_pk=100.0;                       #Peak-output voltage, mV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "V_pk=V_pk/1000.0;                                #Peak-output voltage, V\n",
-      "f_max=(slew_rate*10**6/(2*pi*V_pk))/1000.0;      #Maximum operating frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum operating frequency=%.0fkHz\"%f_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum operating frequency=796kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.25 : Page number 695-696\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_CL=-100;              #Closed-loop voltage gain\n",
-      "Ri=2.2;                 #Input resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, A_CL=-(Rf/Ri)\n",
-      "Rf=-A_CL*Ri;                #Feedback resistor, k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"Feedback resistor=%dk\u03a9\"%Rf);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Feedback resistor=220k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.26 : Page number 696\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "vin=2.5;                #Input voltage, mV\n",
-      "Rf=200;                 #Feedback resistor, k\u03a9\n",
-      "Ri=2;                   #Input resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-      "vout=A_CL*vin/1000;                  #Output voltage,V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%.2fV\"%vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=-0.25V\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.27 : Page number 696\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Varaiable declaration\n",
-      "Rf=1.0;                   #Feedback resistor, k\u03a9\n",
-      "Ri=1.0;                   #Input resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"Closed-loop voltage gain=%d\"%A_CL);\n",
-      "print(\"Therefore, output will have same amplitude but 180\u00b0 phase inversion.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Closed-loop voltage gain=-1\n",
-        "Therefore, output will have same amplitude but 180\u00b0 phase inversion.\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.28 : Page number 696-697\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=40;                   #Feedback resistor, k\u03a9\n",
-      "Ri=1;                   #Input resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Closed-loop voltage gain=%d\"%A_CL);\n",
-      "print(\"Supply voltage=\u00b115V, saturation voltage=\u00b113V. Since gain=-40, op-Amp will be driven to saturation.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Closed-loop voltage gain=-40\n",
-        "Supply voltage=\u00b115V, saturation voltage=\u00b113V. Since gain=-40, op-Amp will be driven to saturation.\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.29 : Page number 697\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "Rf=100;                   #Feedback resistor, k\u03a9\n",
-      "Ri=10;                    #Input resistor, k\u03a9\n",
-      "Vpp=1;                    #Input peak-peak voltage, V\n",
-      "slew_rate=0.5;            #Slew rate, V//\u03bcs\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-      "\n",
-      "#(ii)\n",
-      "Zi=Ri;                  #Input impedance(~ Input resistor), k\u03a9\n",
-      "\n",
-      "#(iii)\n",
-      "Vout=A_CL*Vpp;                           #Peak-to-peak voltage, V\n",
-      "Vpk=Vout/2;                              #Peak output voltage, V\n",
-      "f_max=(slew_rate*10**6/(2*pi*abs(Vpk)))/1000;        #Maximum operating frequency, kHz\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  A_CL=%d.\"%A_CL);\n",
-      "print(\"(ii) Zi=%dk\u03a9\"%Zi);\n",
-      "print(\"(iii) Maximum operating frequency=%.1fkHz.\"%f_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  A_CL=-10.\n",
-        "(ii) Zi=10k\u03a9\n",
-        "(iii) Maximum operating frequency=15.9kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.30 : Page number 697\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A_CL=-4;                    #Closed loop voltage gain\n",
-      "R=[1.0,5.0,10.0,20.0];              #List of available resistors, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "for i in R[:]:\n",
-      "    for j in R[:]:\n",
-      "        if -(i/j)==A_CL :\n",
-      "            print(\"Rf=%dk\u03a9 and Ri=%dk\u03a9.\"%(i,j));\n",
-      "            break;\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rf=20k\u03a9 and Ri=5k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.31 : Page  number 697-698\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=100;                   #Feedback resistor, k\u03a9\n",
-      "Ri=1;                    #Input resistor, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "R_source=0;               #Source resistor, k\u03a9\n",
-      "A_CL=-Rf/(R_source+Ri);    #Closed-loop voltage gain\n",
-      "\n",
-      "print(\"(i) Closed loop voltage gain=%d.\"%A_CL);\n",
-      "\n",
-      "#(ii)\n",
-      "R_source=1;               #Source resistor, k\u03a9\n",
-      "A_CL=-Rf/(R_source+Ri);    #Closed-loop voltage gain\n",
-      "\n",
-      "print(\"(ii) Closed loop voltage gain=%d.\"%A_CL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Closed loop voltage gain=-100.\n",
-        "(ii) Closed loop voltage gain=-50.\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.32 : Page number 699-700\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=240;                   #Feedback resistor, k\u03a9\n",
-      "Ri=2.4;                    #Input resistor, k\u03a9\n",
-      "Vin=120;                    #Input voltage, \u03bcV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "A_CL=1+(Rf/Ri);             #Closed loop voltage gain\n",
-      "Vout=(A_CL*Vin)/1000;       #Output voltage, mV\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%.2fmV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=12.12mV\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.33 : Page number 700\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=10;                   #Feedback resistor, k\u03a9\n",
-      "Ri=1;                    #Input resistor, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "A_CL=1+(Rf/Ri);             #Closed loop voltage gain\n",
-      "#(i)\n",
-      "Vin=1;                      #Input voltage, V\n",
-      "Vout=A_CL*Vin;              #Output voltage, V\n",
-      "\n",
-      "print(\"(i)  Output voltage=%dV\"%Vout);\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "Vin=-1;                      #Input voltage, V\n",
-      "Vout=A_CL*Vin;               #Output voltage, V\n",
-      "\n",
-      "print(\"(ii) Output voltage=%dV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Output voltage=11V\n",
-        "(ii) Output voltage=-11V\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.34 : Page number 700\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=5;                       #Feedback resistor, k\u03a9\n",
-      "Ri=1;                       #Input resistor, k\u03a9\n",
-      "Vin_max=1;                  #Maximum input voltage, V\n",
-      "Vin_min=-1;                 #Minimum input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "V_inpp=Vin_max-Vin_min;               #Peak-peak input voltage, V\n",
-      "A_CL=1+(Rf/Ri);                     #Closed loop voltage gain\n",
-      "Vout_pp=A_CL*V_inpp;                #Peak-peak output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Peak to peak output voltage=%dV\"%Vout_pp);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Peak to peak output voltage=12V\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.35 : Page number 700-701\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "Rf=100;                       #Feedback resistor, k\u03a9\n",
-      "Ri=10;                       #Input resistor, k\u03a9\n",
-      "Vpp=1;                    #Input peak-peak voltage, V\n",
-      "slew_rate=0.5;              #Slew rate, V/\u03bcs\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "A_CL=1+(Rf/Ri);                                      #Closed loop voltage gain\n",
-      "\n",
-      "#(ii)\n",
-      "Vout_pp=A_CL*Vpp;                                    #Peak-peak output voltage, V\n",
-      "Vpk=Vout_pp/2.0;                                          #Peak output voltage, V\n",
-      "f_max=((slew_rate*10**6)/(2*pi*Vpk))/1000.0;           #Maximum operating frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)  Closed-loop voltage gain=%d\"%A_CL);\n",
-      "\n",
-      "print(\"(ii) Maximum operating frequency=%.2fkHz\"%f_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Closed-loop voltage gain=11\n",
-        "(ii) Maximum operating frequency=14.47kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.36 : Page number 701\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=220;                       #Feedback resistor, k\u03a9\n",
-      "Ri=3.3;                       #Input resistor, k\u03a9\n",
-      "unity_gain_BW=3;              #Unity gain bandwidth, MHz\n",
-      "\n",
-      "#Calculation\n",
-      "#(i) For non-inverting amplifier\n",
-      "A_CL=1+(Rf/Ri);                    #Closed loop voltage gain\n",
-      "BW=unity_gain_BW*1000.0/A_CL;        #Bandwidth, kHz\n",
-      "\n",
-      "print(\"(i)  Bandwidth=%.1fkHz.\"%BW);\n",
-      "\n",
-      "#(ii) For inverting amplifier\n",
-      "Rf=47;                          #Feedback resistor, k\u03a9\n",
-      "Ri=1;                           #Input resistor, k\u03a9\n",
-      "A_CL=-(Rf/Ri);                    #Closed loop voltage gain\n",
-      "BW=unity_gain_BW*1000.0/abs(A_CL);        #Bandwidth, kHz\n",
-      "\n",
-      "print(\"(ii)  Bandwidth=%.1fkHz.\"%BW);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Bandwidth=44.3kHz.\n",
-        "(ii)  Bandwidth=63.8kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 47
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.37 : Page number 701-702\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#(i)\n",
-      "A_CL=1;                     #Closed loop voltage gain for voltage follower\n",
-      "print(\"(i) For voltage follower A_CL=1.\");\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "slew_rate=0.5;              #Slew rate, V/\u03bcs\n",
-      "V_inpp=6;                   #peak-peak input voltage, V\n",
-      "Vout=A_CL*V_inpp;           #Peak-peak output voltage, V\n",
-      "Vpk=Vout/2;                 #Peak output voltage, V\n",
-      "\n",
-      "f_max=(slew_rate*10**6/(2*pi*Vpk))/1000;       #Maximum operating frequency, kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) The maximum output frequency=%.2fkHz.\"%f_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) For voltage follower A_CL=1.\n",
-        "(ii) The maximum output frequency=26.53kHz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.38 : Page number 702\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=470.0;                 #Feedback resistor, k\u03a9\n",
-      "R1=4.3;                 #Input resistor of 1st op-Amp, k\u03a9\n",
-      "R2=33.0;                 #Input resistor of 2nd op-Amp, k\u03a9\n",
-      "R3=33.0;                 #Input resistor of 3rd op-Amp, k\u03a9\n",
-      "Vin=80.0;                 #Input voltage, \u03bcV.\n",
-      "\n",
-      "#Calculation\n",
-      "A1=1+Rf/R1;                 #Gain of first op-Amp\n",
-      "A2=-round(Rf/R2,1);         #Gain of second op-Amp\n",
-      "A3=-round(Rf/R3,1);         #Gain of third op-Amp\n",
-      "A=A1*A2*A3;                 #Overall gain\n",
-      "Vout=A*Vin*10**-6;          #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%.2fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=1.78V\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.39 : Page number 702-703\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "A1=10;                  #Voltage gain of 1st op-Amp\n",
-      "A2=-18;                 #Voltage gain of 2nd op-Amp\n",
-      "A3=-27;                 #Voltage gain of 3rd op-Amp\n",
-      "Rf=270;                 #Feedback resistor, k\u03a9\n",
-      "Vin=150;                #Input voltage, \u03bcV  \n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "R1=Rf/(A1-1);                   #Input resistor of 1st op-Amp, k\u03a9\n",
-      "R2=-Rf/A2;                      #Input resistr of 2nd op-Amp, k\u03a9\n",
-      "R3=-Rf/A3;                      #Input resistor of 3rd op-Amp, k\u03a9\n",
-      "\n",
-      "A=A1*A2*A3;                     #overall gain,\n",
-      "Vout=Vin*10**-6*A;              #Output voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"R1=%dk\u03a9, R2=%dk\u03a9 and R3=%dk\u03a9.\"%(R1,R2,R3));\n",
-      "print(\"Output voltage=%.3fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R1=30k\u03a9, R2=15k\u03a9 and R3=10k\u03a9.\n",
-        "Output voltage=0.729V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.40 : Page number 703-704\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=500;                 #Feedback resistor, k\u03a9\n",
-      "A1=-10;                 #Gain of 1st op-Amp\n",
-      "A2=-20;                 #Gain of 2nd op-Amp\n",
-      "A3=-50;                 #Gain of 3rd op-Amp\n",
-      "\n",
-      "#Calculation\n",
-      "R1=-Rf/A1;                      #Input resistor of 1st op-Amp, k\u03a9\n",
-      "R2=-Rf/A2;                      #Input resistor of 2nd op-Amp, k\u03a9\n",
-      "R3=-Rf/A3;                      #Input resistor of 3rd op-Amp, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"R1=%dk\u03a9, R2=%dk\u03a9 and R3=%dk\u03a9.\"%(R1,R2,R3));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R1=50k\u03a9, R2=25k\u03a9 and R3=10k\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 52
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.41 : Page number 705\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Zin=2.0;              #Input impedance of op-Amp, M\u03a9\n",
-      "Zout=75.0;            #Output impedance of op-Amp, \u03a9\n",
-      "A_OL=200000.0;        #Open-loop voltage gain\n",
-      "Rf=220.0;             #Feedback resistor, k\u03a9\n",
-      "Ri=10.0;              #Input resistor, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "mv=round(Ri/(Ri+Rf),3);       #Feedback fraction\n",
-      "Zin_NI=Zin*(1+(A_OL*mv));     #Input impedance, M\u03a9\n",
-      "Zout_NI=Zout/(1+A_OL*mv);     #Output impedance, \u03a9\n",
-      "\n",
-      "#(ii)\n",
-      "A_CL=1+Rf/Ri;               #Closed loop voltage gain\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The input impedance=%dM\u03a9 and output impedance=%.1e\u03a9.\"%(Zin_NI,Zout_NI));\n",
-      "print(\"(ii) The closed loop voltage gain=%d.\"%A_CL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The input impedance=17202M\u03a9 and output impedance=8.7e-03\u03a9.\n",
-        "(ii) The closed loop voltage gain=23.\n"
-       ]
-      }
-     ],
-     "prompt_number": 54
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.42 : Page number 705-706\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "#For voltage follower,\n",
-      "mv=1.0;               #Feedback fraction\n",
-      "A_OL=200000.0;        #Open-loop voltage gain\n",
-      "Zin=2.0;              #Input impedance of op-Amp, M\u03a9\n",
-      "Zout=75.0;            #Output impedance of op-Amp, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Zin_VF=Zin*(1+(A_OL*mv));     #Input impedance, M\u03a9\n",
-      "Zout_VF=round(round(Zout/(1+A_OL*mv),6),5);     #Output impedance, \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The input impedance=%dM\u03a9 and output impedance=%.2fe-03\u03a9.\"%(Zin_VF,Zout_VF*1000));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input impedance=400002M\u03a9 and output impedance=0.38e-03\u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 63
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.43 : Page number 706\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=100;             #Feedback resistor, k\u03a9\n",
-      "Ri=1.0;               #Input resistor, k\u03a9\n",
-      "Zin=4;              #Input impedance of op-Amp, M\u03a9\n",
-      "Zout=50;            #Output impedance of op-Amp, \u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Zin_I=Ri;                   #Input impedance, k\u03a9\n",
-      "Zout_I=Zout;                #Output impedance, \u03a9\n",
-      "A_CL=-(Rf/Ri);              #Closed loop voltage gain\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The input impedance=%dk\u03a9 and output impedance=%d\u03a9.\"%(Zin_I,Zout_I));\n",
-      "print(\"Closed-loop voltage gain=%d\"%A_CL);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The input impedance=1k\u03a9 and output impedance=50\u03a9.\n",
-        "Closed-loop voltage gain=-100\n"
-       ]
-      }
-     ],
-     "prompt_number": 65
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.44 : Page number 709\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=10;               #Feedback resistor, k\u03a9\n",
-      "Ri=10;               #Input resistor, k\u03a9\n",
-      "V1=3;                #Input voltage 1st, V\n",
-      "V2=1;                #Input voltage 2nd, V\n",
-      "V3=8;                #Input voltage 3rd, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Rf=Ri, Vout=-(Rf/Ri)*(V1+V2+V3)= -(V1+V2+V3);\n",
-      "Vout=-(V1+V2+V3);           #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%dV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=-12V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 66
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.45 : Page number 709\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=10;               #Feedback resistor, k\u03a9\n",
-      "R1=1;                #Input resistor for input 1, k\u03a9\n",
-      "R2=1;                #Input resistor for input 2, k\u03a9\n",
-      "V1=0.2;              #Input voltage 1st, V\n",
-      "V2=0.5;              #Input voltage 2nd, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "R=R1;                       #Input resistor(=R1 or R2), k\u03a9\n",
-      "Vout=-(Rf/R)*(V1+V2);       #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%dV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=-7V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 67
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.46 : Page number 709-710\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=1;                #Feedback resistor, k\u03a9\n",
-      "Ri=10.0;               #Input resistor, k\u03a9\n",
-      "V1=10;                #Input voltage 1st, V\n",
-      "V2=8.0;                #Input voltage 2nd, V\n",
-      "V3=7.0;                #Input voltage 3rd, V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Vout=-(Rf/Ri)*(V1+V2+V3);\n",
-      "Vout=-(Rf/Ri)*(V1+V2+V3);           #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%.1fV.\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=-2.5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 70
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.47 : Page number 710\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V1=0.6;                 #Input voltage to 1st input resistor, V\n",
-      "V2=-1.4;                #Input voltage to 2nd input resistor, V\n",
-      "Rf=200;                 #Feedback resistor, k\u03a9\n",
-      "R1=400;                 #Input resistor 1, k\u03a9\n",
-      "R2=100.0;                 #Input resistor 2, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "Vout=-Rf*(V1/R1 +V2/R2);                 #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Output voltage=%.1fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Output voltage=2.5V\n"
-       ]
-      }
-     ],
-     "prompt_number": 71
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.48 : Page number 710-711\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=1.0;                 #Feedback resistor, k\u03a9\n",
-      "R1=1.0;                 #Input resistor 1, k\u03a9\n",
-      "R2=2.0;                 #Input resistor 2, k\u03a9\n",
-      "R3=4.0;                 #Input resistor 3, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Rf_R1=Rf/R1;                #Ratio of feedback resistor and 1st input resistor\n",
-      "Rf_R2=Rf/R2;                #Ratio of feedback resistor and 2nd input resistor\n",
-      "Rf_R3=Rf/R3;                #Ratio of feedback resistor and 3rd input resistor\n",
-      "\n",
-      "#(i) First input combination\n",
-      "V1=10;                                   #Input voltage to 1st input resistor, V\n",
-      "V2=0;                                    #Input voltage to 2nd input resistor, V\n",
-      "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-      "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-      "print(\"(i) The output voltage=%.1fV\"%Vout);\n",
-      "\n",
-      "#(i) First input combination\n",
-      "V1=0;                                   #Input voltage to 1st input resistor, V\n",
-      "V2=10;                                    #Input voltage to 2nd input resistor, V\n",
-      "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-      "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-      "print(\"(ii) The output voltage=%.1fV\"%Vout);\n",
-      "\n",
-      "\n",
-      "#(i) First input combination\n",
-      "V1=10;                                   #Input voltage to 1st input resistor, V\n",
-      "V2=10;                                    #Input voltage to 2nd input resistor, V\n",
-      "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-      "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-      "print(\"(iii) The output voltage=%.1fV\"%Vout);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The output voltage=-12.5V\n",
-        "(ii) The output voltage=-7.5V\n",
-        "(iii) The output voltage=-17.5V\n"
-       ]
-      }
-     ],
-     "prompt_number": 72
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.49 : Page number 711\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Rf=330;                 #Feedback resistor, k\u03a9\n",
-      "R1=33.0;                  #Input resistor 1, k\u03a9\n",
-      "R2=10.0;                  #Input resistor 2, k\u03a9\n",
-      "V1_m=50;                #Peak voltage of 1st input, mV\n",
-      "V2_m=10;                #Peak voltage of 2nd input, mV\n",
-      "\n",
-      "#Calculation\n",
-      "#Since, Vout=-((Rf/R1)*V1 + (Rf/R2)*V2)\n",
-      "print(\"Vout=-[%.1fsin(1000t)+%.2fsin(3000t)]V\"%((V1_m/1000.0)*(Rf/R1),(V2_m/1000.0)*(Rf/R2)));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vout=-[0.5sin(1000t)+0.33sin(3000t)]V\n"
-       ]
-      }
-     ],
-     "prompt_number": 74
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.50 : Page number 715\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=100;                      #Input resistor, k\u03a9\n",
-      "C=10;                       #Feedback capacitor, \u03bcF\n",
-      "\n",
-      "#Calculation\n",
-      "RC=R*10**3*C*10**-6;        #product of input resistance and feedback capacitance, s\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Vo=-1*(1/RC)\u222bvi dt.\");\n",
-      "print(\"=>Vo=-1*(1/%d)\u222bvi dt\"%RC);\n",
-      "print(\"=>Vo=\u222bvi dt\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vo=-1*(1/RC)\u222bvi dt.\n",
-        "=>Vo=-1*(1/1)\u222bvi dt\n",
-        "=>Vo=\u222bvi dt\n"
-       ]
-      }
-     ],
-     "prompt_number": 75
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.51 : Page number 715-716\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "\n",
-      "#Variable declaration\n",
-      "Rf=100;                      #Feedback resistor, k\u03a9\n",
-      "C=0.01;                      #Feedback capacitor, \u03bcF\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "fc=1/(2*pi*Rf*1000*C*10**-6);               #Crictical frequency, Hz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The critical frequency=%dHz.\"%fc);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The critical frequency=159Hz.\n"
-       ]
-      }
-     ],
-     "prompt_number": 76
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.52 : Page number 716\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "R=10.0;                      #Input resistor, k\u03a9\n",
-      "C=0.01;                    #Feedback capacitor, \u03bcF\n",
-      "vin=5;                     #Input voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "Vout_change_rate=-vin/(R*C);            #Rate of change of output voltage, V/\u03bcs   \n",
-      "print(\"(i) Vout=-1*(1/RC)\u222bvi dt.\");\n",
-      "print(\"    \u0394Vout/dt = -vin/RC = %dmV/\u03bcs.\"%Vout_change_rate);\n",
-      "\n",
-      "#(ii) Plotting the output waveform\n",
-      "vin_plot=[];                    #Plotting variable for input waveform, V\n",
-      "dt=100;                          #time between edges, \u03bcs\n",
-      "for i in range(0,3*dt+1):\n",
-      "    if i<dt or i>2*dt :\n",
-      "        vin_plot.append(0);\n",
-      "    else:\n",
-      "        vin_plot.append(5); \n",
-      "p=plot(vin_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,300])\n",
-      "limit.set_ylim([-5,10])\n",
-      "xlabel(\"t(microsecond)\");\n",
-      "ylabel(\"Vin(V)\");\n",
-      "title(\"Input waveform\");\n",
-      "show(p);\n",
-      "\n",
-      "        \n",
-      "vout_plot=[];                  #Plotting variable for output waveform, V\n",
-      "t=[i for i in range(0,301)];  #Time scale, \u03bcs\n",
-      "for i in t[:] :\n",
-      "    if i<dt:\n",
-      "        vout_plot.append(0);\n",
-      "    elif i>2*dt:\n",
-      "        vout_plot.append((Vout_change_rate/1000.0)*dt);\n",
-      "    else :\n",
-      "        vout_plot.append((-vin_plot[i]/(R*C))/1000*(i-dt));\n",
-      "\n",
-      "    \n",
-      "p=plot(vout_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,300])\n",
-      "limit.set_ylim([-5,5]);\n",
-      "xlabel('t(microsecond)');\n",
-      "ylabel(\"Vout(V)\");\n",
-      "title(\"output waveform\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Vout=-1*(1/RC)\u222bvi dt.\n",
-        "    \u0394Vout/dt = -vin/RC = -50mV/\u03bcs.\n"
-       ]
-      },
-      {
-       "metadata": {},
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Lz0tKSjB58uSbmsezzz6LTz/91NClqWLOnDlIT08HAMTHx+PMmTMqV0SmhMFA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\nSGNftWoAAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7fdd1397e4d0>"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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lZGRY1nPgwAE1b95c3bt3l3TpKl+1atWSw+FQQkKCGjRoIOnSNTTGjh0rSYqP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\nLAgDAMDifwGSFOcKwPnWKgAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7fdd137a8390>"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.53 : Page number 716-717\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_supply=15;              #Supply voltage, V\n",
-      "R=10;                     #Input resistor, k\u03a9\n",
-      "C=0.2;                    #Feedback capacitor, \u03bcF\n",
-      "vin=10;                   #Input voltage, mV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vs=-V_supply+2;         #Saturation voltage, V\n",
-      "print(\"Vout=-1*(1/RC)\u222bvi dt.\");\n",
-      "print(\"Vout=%d*t volts\"%(-vin/(R*C)));\n",
-      "t=Vs/(-vin/(R*C));                      #Time required, seconds\n",
-      "print(\"Time required=%.1fseconds.\"%t);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vout=-1*(1/RC)\u222bvi dt.\n",
-        "Vout=-5*t volts\n",
-        "Time required=2.6seconds.\n"
-       ]
-      }
-     ],
-     "prompt_number": 78
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.54 : Page number 717-718\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=1;                     #Feedback resistor, k\u03a9\n",
-      "C=0.1;                   #Input capacitor, \u03bcF\n",
-      "Vin_change=5;            #Change in input voltage, V\n",
-      "t=0.1;                   #Time taken for change in input voltage, ms\n",
-      "\n",
-      "#Calcualtion\n",
-      "dvi_dt=Vin_change/(t/1000);        #Rate of change of input voltage, V/s\n",
-      "RC=R*1000*C*10**-6;                #Product of feedback resistance and input capacitance, s\n",
-      "#Since, Vo=-R*C*(dvi/dt);\n",
-      "Vo=-RC*dvi_dt;                      #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Vo=%dV.\"%Vo);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vo=-5V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 79
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.55 : Page number 718\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=10;                     #Feedback resistor, k\u03a9\n",
-      "C=2.2;                   #Input capacitor, \u03bcF\n",
-      "Vin_change=10;            #Change in input voltage, V\n",
-      "t=0.4;                   #Time taken for change in input voltage, s\n",
-      "\n",
-      "#Calcualtin\n",
-      "dvi_dt=Vin_change/t;        #Rate of change of input voltage, V/s\n",
-      "RC=R*1000*C*10**-6;         #Product of feedback resistance and input capacitance, s\n",
-      "#Since, Vo=-R*C*(dvi/dt);\n",
-      "Vo=-RC*dvi_dt;                      #Output voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Vo=%.2fV.\"%Vo);\n",
-      "print(\"The output voltage stays constant at %.2fV.\"%Vo);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vo=-0.55V.\n",
-        "The output voltage stays constant at -0.55V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 80
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 25.56 : Page number 718-719\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R=100;                     #Feedback resistor, k\u03a9\n",
-      "C=10;                   #Input capacitor, \u03bcF\n",
-      "Vin_change=1;            #Change in input voltage, V\n",
-      "t=0.2;                   #Time taken for change in input voltage, s\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "RC=R*1000*C*10**-6;             #Product of feedback resistor and input capacitance, s\n",
-      "#(i)\n",
-      "print(\"vo=-%d*(dvi/dt).\"%RC);\n",
-      "\n",
-      "#(ii)\n",
-      "dvi_dt=Vin_change/t;                #Rate of change of input voltage, V\n",
-      "vo=-dvi_dt;                         #Output voltage, V\n",
-      "print(\"vo=%dV.\"%vo);\n",
-      "\n",
-      "print(\"Therefore, between 0 to 0.2s, the output voltage is constant at %dV.\"%vo);\n",
-      "print(\"For t>0.2s, the input is constant so that output voltage is zero.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "vo=-1*(dvi/dt).\n",
-        "vo=-5V.\n",
-        "Therefore, between 0 to 0.2s, the output voltage is constant at -5V.\n",
-        "For t>0.2s, the input is constant so that output voltage is zero.\n"
-       ]
-      }
-     ],
-     "prompt_number": 82
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_2.ipynb
deleted file mode 100755
index d24cf79d..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_2.ipynb
+++ /dev/null
@@ -1,2512 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 25 : OPERATIONAL AMPLIFIERS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.1: Page number 664"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output voltage of the differential amplifier = 10V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A=100.0;              #Open-circuit voltage gain of differential amplifier\n",
-    "V1=3.25;              #Input voltage to terminal 1 in V\n",
-    "V2=3.15;              #Input voltage to terminal 2 in V\n",
-    "\n",
-    "#Calculations\n",
-    "V0=A*(V1-V2);         #Output voltage in V\n",
-    "\n",
-    "#Results\n",
-    "print(\"The output voltage of the differential amplifier = %dV\"%V0);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.2: Page number 672"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The common mode rejection ratio = 10000.\n",
-      "The common mode rejection ratio in decibels= 80dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "A_DM=2000.0;              #Differential mode voltage gain\n",
-    "A_CM=0.2;                 #Common mode voltage gain\n",
-    "\n",
-    "#Calculations\n",
-    "CMRR=A_DM/A_CM;           #Common mode rejection ratio\n",
-    "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"The common mode rejection ratio = %d.\"%CMRR);\n",
-    "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.3: Page number 672"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The common mode rejection ratio in decibels= 46dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "VD_in=10.0;               #Differential mode input in mV\n",
-    "VD_out=1.0;               #Output for differential mode input in V\n",
-    "VC_in=10.0;                 #Common mode input in mV\n",
-    "VC_out=5.0;                 #Output for common mode input in mV\\\n",
-    "\n",
-    "#Calculations\n",
-    "A_DM=(VD_out*1000)/VD_in;       #Differntial mode voltage gain\n",
-    "A_CM=VC_out/VC_in;       #Common mode voltage gain\n",
-    "CMRR=A_DM/A_CM;           #Common mode rejection ratio\n",
-    "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-    "\n",
-    "#Results\n",
-    "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.4: Page number 672"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage =7.5V\n",
-      "Noise on output = 4.7x10^-6V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_DM=150.0;           #Differential mode voltage gain\n",
-    "CMRR_dB=90.0;         #Common mode rejection ratio\n",
-    "V1=100.0;             #Input voltage for terminal 1 in mV\n",
-    "V2=50.0;              #Input voltage for terminal 2 in mV\n",
-    "V_noise=1.0;          #Voltage of noise signal in mV\n",
-    "\n",
-    "#Calculation\n",
-    "\n",
-    "#Case(i)\n",
-    "V_out=A_DM*(V1-V2)/1000.0;            #Output voltage for differntial mode input, in V\n",
-    "\n",
-    "#Since CMRR_dB=20*log10(differential mode gain/common mode gain),\n",
-    "A_CM=A_DM/pow(10,(CMRR_dB/20));         #Common mode gain\n",
-    "V_OUT_noise=A_CM*(V_noise/1000);        #Noise on output in V\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"Output voltage =%.1fV\"%V_out);\n",
-    "print(\"Noise on output = %.1fx10^-6V\"%(V_OUT_noise*pow(10,6)));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.5 : Page number 672-673"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Common mode gain =0.083\n",
-      "Common mode rejection ratio in decibels=89.5dB\n",
-      "r.m.s output signal =1.25V\n",
-      "r.m.s interfernce output voltage = 83mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "A_DM=2500.0;                    #Differential mode voltage gain\n",
-    "CMRR=30000.0;                   #Common mode rejection ratio\n",
-    "Input_signal=500.0;             #Single ended input r.m.s signal in microvolts\n",
-    "Interference=1.0;               #Interference signal, in V\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#(i)\n",
-    "A_CM=A_DM/CMRR;                 #Common mode gain\n",
-    "\n",
-    "#(ii)\n",
-    "CMRR_dB=20*log10(CMRR);         #Common mode rejection ratio in decibels\n",
-    "\n",
-    "#(iii)\n",
-    "V_out=A_DM*(Input_signal/pow(10,6)-0);        #r.m.s output signal  in V\n",
-    "\n",
-    "#(iv)\n",
-    "Interference_out=A_CM*Interference;          #r.m.s interference output in V\n",
-    "Interference_out=Interference_out*1000;      #r.m.s interference output in mV\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"Common mode gain =%.3f\"%A_CM);\n",
-    "print(\"Common mode rejection ratio in decibels=%.1fdB\"%CMRR_dB);\n",
-    "print(\"r.m.s output signal =%.2fV\"%V_out);\n",
-    "print(\"r.m.s interfernce output voltage = %dmV\"%Interference_out);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.6 : Page number 674-675"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VE=-0.7V\n",
-      "IE=0.452mA\n",
-      "IE1=0.226mA\n",
-      "IE2=0.226mA\n",
-      "IC1=0.226mA\n",
-      "IC2=0.226mA\n",
-      "IB1=2.26μA\n",
-      "IB2=2.26μA\n",
-      "VC1=12V\n",
-      "VC2=9.7V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RB=10;                  #Base resistor, kΩ\n",
-    "RC2=10;                 #Collector resistor, kΩ\n",
-    "RE=25;                  #Emitter resistor, kΩ\n",
-    "VBE=0.7;               #Base-emitter voltage, V\n",
-    "beta=100;               #Base amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-    "IE=(VEE-VBE)/RE;            #Tail current, mA\n",
-    "IE1=IE/2;                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-    "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-    "IC2=IC1;                    #Collector current of 2nd transistor, mA\n",
-    "IB1=(IC1/beta)*1000;        #Base current of 1st transistor, μA\n",
-    "IB2=IB1;                    #Base current of 2nd transistor, μA\n",
-    "VC1=VCC;                    #Collector voltage of 1st transistor, V\n",
-    "VC2=VCC-IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"VE=%.1fV\"%VE);\n",
-    "print(\"IE=%.3fmA\"%IE);\n",
-    "print(\"IE1=%.3fmA\"%IE1);\n",
-    "print(\"IE2=%.3fmA\"%IE2);\n",
-    "print(\"IC1=%.3fmA\"%IC1);\n",
-    "print(\"IC2=%.3fmA\"%IC2);\n",
-    "print(\"IB1=%.2fμA\"%IB1);\n",
-    "print(\"IB2=%.2fμA\"%IB2);\n",
-    "print(\"VC1=%dV\"%VC1);\n",
-    "print(\"VC2=%.1fV\"%VC2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.7 : Page number 675"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output voltage=7.85V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15;                 #Collector supply voltage, V\n",
-    "VEE=15;                 #Emitter supply voltage, V\n",
-    "RB=33;                  #Base resistor, kΩ\n",
-    "RC=15;                  #Collector resistor, kΩ\n",
-    "RE=15;                  #Emitter resistor, kΩ\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "IE_tail=(VEE-VBE)/RE;            #Tail current, mA\n",
-    "IE=round(IE_tail/2,3);                    #Emitter current in each transistor, mA\n",
-    "IC=IE;                           #Collector current(=emitter current), mA\n",
-    "Vout=VCC-IC*RC;                  #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output voltage=%.2fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.8 : Page number 675"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) IB1=5.56μA\n",
-      "    IB2=4.55μA\n",
-      "(ii) VB1=-0.183V\n",
-      "     VB2=-0.15V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15.0;                 #Collector supply voltage, V\n",
-    "VEE=15.0;                 #Emitter supply voltage, V\n",
-    "RB=33.0;                  #Base resistor, kΩ\n",
-    "RC=15.0;                  #Collector resistor, kΩ\n",
-    "RE=15.0;                  #Emitter resistor, kΩ\n",
-    "VBE=0;                  #Base-emitter voltage, V\n",
-    "beta_dc_l=90.0;           #base current amplification factor for left transistor\n",
-    "beta_dc_r=110.0;          #base current amplification factor for right transistor\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IE_tail=(VEE-VBE)/RE;             #Tail current, mA\n",
-    "IE=IE_tail/2;                      #Emitter current in each transistor, mA\n",
-    "IB1=(IE/beta_dc_l)*1000;          #Base current of 1st transistor, μA\n",
-    "IB2=(IE/beta_dc_r)*1000;          #Base current of 2nd transistor, μA\n",
-    "\n",
-    "#(ii)\n",
-    "VB1=-IB1/1000*RB;                    #Base voltage of 1st transistor, V\n",
-    "VB2=-IB2/1000*RB;                    #Base voltage of 1st transistor, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) IB1=%.2fμA\"%IB1);\n",
-    "print(\"    IB2=%.2fμA\"%IB2);\n",
-    "print(\"(ii) VB1=%.3fV\"%VB1);\n",
-    "print(\"     VB2=%.2fV\"%VB2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.9 : Page number 675-676"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VE=-0.7V\n",
-      "Emitter current in each transistor=0.5mA.\n",
-      "IC1~IE1=0.5mA and IC2~IE2=0.5mA\n",
-      "VC1=VC2=10V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15.0;                 #Collector supply voltage, V\n",
-    "VEE=15.0;                 #Emitter supply voltage, V\n",
-    "RB=10.0;                  #Base resistor, kΩ\n",
-    "RC1=10.0;                 #Collector resistor of 1st transistor, kΩ\n",
-    "RC2=10.0;                 #Collector resistor of 2nd transistor, kΩ\n",
-    "IE=1.0;                   #Tail current, mA\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-    "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-    "IC2=IE2;                    #Collector current of 2nd transistor, mA\n",
-    "VC1=VCC-IC1*RC1;            #Collector voltage of 1st transistor, V\n",
-    "VC2=VCC-IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"VE=%.1fV\"%VE);\n",
-    "print(\"Emitter current in each transistor=%.1fmA.\"%(IE/2.0));\n",
-    "print(\"IC1~IE1=%.1fmA and IC2~IE2=%.1fmA\"%(IE1,IE2));\n",
-    "print(\"VC1=VC2=%dV.\"%VC2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.10 : Page number 676-677"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VE=0.7V\n",
-      "Tail current=0.452mA.\n",
-      "Emitter current in each transistor=0.226mA.\n",
-      "IC1~IE1=0.226mA and IC2~IE2=0.226mA\n",
-      "VC1=-12V\n",
-      "VC2=-9.74V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12.0;                 #Collector supply voltage, V\n",
-    "VEE=12.0;                 #Emitter supply voltage, V\n",
-    "RC2=10.0;                 #Collector resistor of 2nd transistor, kΩ\n",
-    "RE=25.0;                  #Emitter current, kΩ\n",
-    "VBE=-0.7;               #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-    "IE=(VCC-VE)/RE;             #Tail current, mA\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-    "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-    "IC2=IE2;                    #Collector current of 2nd transistor, mA\n",
-    "VC1=-VEE;                   #Collector voltage of 1st transistor, V\n",
-    "VC2=-VEE+IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"VE=%.1fV\"%VE);\n",
-    "print(\"Tail current=%.3fmA.\"%IE);\n",
-    "print(\"Emitter current in each transistor=%.3fmA.\"%(IE/2.0));\n",
-    "print(\"IC1~IE1=%.3fmA and IC2~IE2=%.3fmA\"%(IC1,IC2));\n",
-    "print(\"VC1=%dV\"%VC1);\n",
-    "print(\"VC2=%.2fV\"%VC2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.11 : Page number 679"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The input offset current=15.1nA\n",
-      "(ii) The input bias current=75.8nA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15;                 #Collector supply voltage, V\n",
-    "VEE=15;                 #Emitter supply voltage, V\n",
-    "RB=1;                   #Base resistor, MΩ\n",
-    "RC2=1;                  #Collector resistor, MΩ\n",
-    "RE=1;                   #Emitter resistor, MΩ\n",
-    "VBE=0;                  #Base-emitter voltage, V (Neglected)\n",
-    "beta_dc_l=90.0;           #base current amplification factor for left transistor\n",
-    "beta_dc_r=110.0;          #base current amplification factor for right transistor\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IE=(VEE-VBE)/RE;                    #Tail current, μA\n",
-    "IE1=IE/2.0;                           #Emitter current of 1st transistor, μA\n",
-    "IE2=IE1;                            #Emitter current of 2nd transistor, μA\n",
-    "IB1=round((IE1/beta_dc_l)*1000,1);           #Base current of 1st transistor, nA\n",
-    "IB2=round((IE2/beta_dc_r)*1000,1);           #Base current of 2nd transistor, nA\n",
-    "I_in_offset=IB1-IB2;                 #Input offset current, nA\n",
-    "\n",
-    "#(ii)\n",
-    "I_in_bias=(IB1+IB2)/2;                  #Input bias current, nA\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  The input offset current=%.1fnA\"%I_in_offset);\n",
-    "print(\"(ii) The input bias current=%.1fnA\"%I_in_bias);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.12 : Page number 679"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The two base currents are: IB1=90nA and IB2=70nA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_in_offset=20;                 #Input offset current, nA\n",
-    "I_in_bias=80;                   #Input bias current, nA\n",
-    "\n",
-    "#Calculation\n",
-    "IB1=I_in_bias+I_in_offset/2;            #Base current in 1st transistor, nA\n",
-    "IB2=I_in_bias-I_in_offset/2;            #Base current in 2nd transistor, nA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The two base currents are: IB1=%dnA and IB2=%dnA.\"%(IB1,IB2));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.13 : Page number 679-680"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input offset voltage=2mV.\n",
-      "The output offset voltage=0.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declration\n",
-    "I_in_offset=20;                 #Input offset current, nA\n",
-    "I_in_bias=80;                   #Input bias current, nA\n",
-    "A=150;                          #Voltage gain\n",
-    "RB=100;                         #Base resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_io=(I_in_offset*10**-9*RB*1000)*1000;       #Input offset voltage, mV\n",
-    "V_out_offset=(A*V_io)/1000;                     #Output offset voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input offset voltage=%dmV.\"%V_io);\n",
-    "print(\"The output offset voltage=%.1fV.\"%V_out_offset);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.14 : Page number 682"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Output voltage=0.15V.\n",
-      "(ii) Output voltage=-0.15V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15;                 #Collector supply voltage, V\n",
-    "VEE=15;                 #Emitter supply voltage, V\n",
-    "RE=1;                   #Emitter resistor, MΩ\n",
-    "RC=1;                   #Collector resistor, MΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "IE=VEE/RE;                  #Tail current, μA\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, μA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, μA\n",
-    "re=25/IE1;                  #a.c emitter resistance, kΩ\n",
-    "A_DM=RC/(2.0*re);        #Differential voltage gain,\n",
-    "\n",
-    "#(i)\n",
-    "vin=1;                      #Input voltage, V\n",
-    "Vout=A_DM*vin;              #Output voltage, V\n",
-    "\n",
-    "print(\"(i) Output voltage=%.2fV.\"%Vout);\n",
-    "\n",
-    "#(ii)\n",
-    "vin=-1;                      #Input voltage, V\n",
-    "Vout=A_DM*vin;              #Output voltage, V;\n",
-    "print(\"(ii) Output voltage=%.2fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.15 : Page number 682-683"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The input impedance=194kΩ.\n",
-      "(ii) The differential voltage gain=136.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RE=100;                 #Emitter resistor, kΩ\n",
-    "RC1=120;                #Collector resistor of 1st transistor, kΩ\n",
-    "RC2=120;                #Collector resistor of 2nd transistor, kΩ\n",
-    "beta=220;               #Base amplification factor\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calcualtion\n",
-    "IE=((VEE-VBE)/RE)*1000;     #Tail current, μA\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, μA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, μA\n",
-    "re=(25/IE1)*1000;           #a.c emitter resistance, Ω\n",
-    "Zin=2*beta*re/1000;         #Input impedance, kΩ\n",
-    "A_DM=RC1*1000/(2.0*re);             #Differential voltage gain,\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) The input impedance=%dkΩ.\"%Zin);\n",
-    "print(\"(ii) The differential voltage gain=%.0f.\"%A_DM);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.16: Page number 683-684"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Differential voltage gain=56.6.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RE=200;                 #Emitter resistor, kΩ\n",
-    "RC=100;                 #Collector resistor, kΩ\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "IE=round((VEE-VBE)/RE,4);            #Tail current, mA\n",
-    "IE1=round(IE/2,4);                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                             #Emitter current of 2nd transistor, mA\n",
-    "re=round(25/IE1,1);                  #a.c emitter resistance, Ω\n",
-    "A_DM=RC*1000/(2*re);                 #Differential voltage gain,\n",
-    "\n",
-    "#Result\n",
-    "print(\"Differential voltage gain=%.1f.\"%A_DM);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.17 : Page number 685-686"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Common mode rejection ratio=666.7.\n",
-      "Common mode rejection ratio in decibel=56.48dB\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "v1=0.5;               #Voltage in terminal 1, mV\n",
-    "v2=-0.5;               #Voltage in terminal 2, mV\n",
-    "vo=8.0;                   #Output voltage, V\n",
-    "vo_cm=12.0;               #Common mode output, mV\n",
-    "\n",
-    "#Calculation\n",
-    "vin=v1-v2;                  #Differential input, mV\n",
-    "A_DM=vo/(vin/1000.0);         #Differential mode gain,\n",
-    "vin_cm=1;                   #Common mode input, mV\n",
-    "A_CM=vo_cm/vin_cm;          #Common mode gain\n",
-    "CMRR=A_DM/A_CM;             #Common mode rejection ratio\n",
-    "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-    "\n",
-    "#Result\n",
-    "print(\"Common mode rejection ratio=%.1f.\"%CMRR)\n",
-    "print(\"Common mode rejection ratio in decibel=%.2fdB\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.18 : Page number 686"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Common mode voltage gain=6.32.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_DM=200000;                #Differential mode gain\n",
-    "CMRR_dB=90;                 #Common mode rejection ratio, dB\n",
-    "\n",
-    "#Calculation\n",
-    "CMRR=10**(CMRR_dB/20.0);              #Common mode rejection ratio\n",
-    "A_CM=A_DM/CMRR;                     #Common mode gain\n",
-    "\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Common mode voltage gain=%.2f.\"%A_CM);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.19 : Page number 686"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The Common mode gain=0.0081\n",
-      "(ii) The common mode rejection ratio=81.8dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "vin_cm=3.2;                     #Common input voltage, V\n",
-    "vout=26;                        #Output voltage, V\n",
-    "A_DM=100;                       #Open-circuit voltage gain\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CM=vout*10**-3/vin_cm;                #Common mode gain\n",
-    "\n",
-    "#(ii)\n",
-    "CMRR_dB=20*log10(A_DM/A_CM);            #Common mode rejection ratio, dB\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  The Common mode gain=%.4f\"%A_CM);\n",
-    "print(\"(ii) The common mode rejection ratio=%.1fdB.\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.20 : Page number 686-687"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Common mode gain=0.25\n",
-      "(ii)Common mode rejection ratio=47.09dB\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "from math import floor\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RE=200.0;                 #Emitter resistor, kΩ\n",
-    "RC=100.0;                 #Collector resistor, kΩ\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CM=round(RC/(2*RE),2);             #Common mode voltage gain\n",
-    "\n",
-    "#(ii)\n",
-    "IE=round((VEE-VBE)/RE,4);            #Tail current, mA\n",
-    "IE1=round(IE/2,4);                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                             #Emitter current of 2nd transistor, mA\n",
-    "re=round(25/IE1,1);                  #a.c emitter resistance, Ω\n",
-    "A_DM=RC*1000/(2*re);                 #Differential voltage gain,\n",
-    "CMRR_dB=floor(20*log10(A_DM/A_CM)*100)/100;         #Common mode rejection ratio, dB\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) Common mode gain=%.2f\"%A_CM);\n",
-    "print(\"(ii)Common mode rejection ratio=%.2fdB\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.21 : Page number 691"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "f2=30kHz\n",
-      "ACL=75 or 37.5dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "ACL=500;                    #closed loop gain\n",
-    "f_unity=15;                 #frequency with cloased-loop unity gain, MHz\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "f2=f_unity*1000/500                 #Upper frequency of bandwidth,kHz\n",
-    "BW=f2-0;                            #Bandwidth, kHz\n",
-    "A_CL=f_unity*1000/200;                    #Maximum value of A_CL when f2=200kHz\n",
-    "A_CL_dB=20*log10(A_CL);             #Maximum value of A_CL in decibel\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"f2=%dkHz\"%f2);\n",
-    "print(\"ACL=%d or %.1fdB.\"%(A_CL,A_CL_dB));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.22 : Page number 691-692"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Operating Bandwidth=1.5MHz.\n",
-      "(ii) Operating Bandwidth=150kHz.\n",
-      "(iii) Operating Bandwidth=15kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "GBW=1.5;                    #Gain-bandwidth, MHz\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) For A_CL=1;\n",
-    "A_CL=1;                         #Closed loop gain\n",
-    "BW=GBW/A_CL;                    #Bandwidth, MHz\n",
-    "\n",
-    "print(\"(i) Operating Bandwidth=%.1fMHz.\"%BW);\n",
-    "\n",
-    "#(ii) For A_CL=10;\n",
-    "A_CL=10;                         #Closed loop gain\n",
-    "BW=(GBW/A_CL)*1000;              #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(ii) Operating Bandwidth=%dkHz.\"%BW);\n",
-    "\n",
-    "#(iii) For A_CL=100;\n",
-    "A_CL=100;                         #Closed loop gain\n",
-    "BW=(GBW/A_CL)*1000;               #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(iii) Operating Bandwidth=%dkHz.\"%BW);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.23 : Page number 692"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Maximum operating frequency=9.95kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "slew_rate=0.5;                  #Slew rate, V/μs\n",
-    "V_supply=10;                     #Supply voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_sat=V_supply-2;                           #Saturation voltage, V\n",
-    "V_pk=V_sat;                                 #Maximum peak-output voltage, V\n",
-    "f_max=((slew_rate*10**6)/(2*pi*V_pk))/1000;       #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Maximum operating frequency=%.2fkHz.\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.24 : Page number 692"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Maximum operating frequency=796kHz\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "slew_rate=0.5;                  #Slew rate, V/μs\n",
-    "V_pk=100.0;                       #Peak-output voltage, mV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_pk=V_pk/1000.0;                                #Peak-output voltage, V\n",
-    "f_max=(slew_rate*10**6/(2*pi*V_pk))/1000.0;      #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Maximum operating frequency=%.0fkHz\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.25 : Page number 695-696"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Feedback resistor=220kΩ\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_CL=-100;              #Closed-loop voltage gain\n",
-    "Ri=2.2;                 #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, A_CL=-(Rf/Ri)\n",
-    "Rf=-A_CL*Ri;                #Feedback resistor, kΩ\n",
-    "\n",
-    "#Result\n",
-    "print(\"Feedback resistor=%dkΩ\"%Rf);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.26 : Page number 696"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-0.25V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "vin=2.5;                #Input voltage, mV\n",
-    "Rf=200;                 #Feedback resistor, kΩ\n",
-    "Ri=2;                   #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "vout=A_CL*vin/1000;                  #Output voltage,V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.2fV\"%vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.27 : Page number 696"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 28,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Closed-loop voltage gain=-1\n",
-      "Therefore, output will have same amplitude but 180° phase inversion.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Varaiable declaration\n",
-    "Rf=1.0;                   #Feedback resistor, kΩ\n",
-    "Ri=1.0;                   #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"Closed-loop voltage gain=%d\"%A_CL);\n",
-    "print(\"Therefore, output will have same amplitude but 180° phase inversion.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.28 : Page number 696-697"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 29,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Closed-loop voltage gain=-40\n",
-      "Supply voltage=±15V, saturation voltage=±13V. Since gain=-40, op-Amp will be driven to saturation.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=40;                   #Feedback resistor, kΩ\n",
-    "Ri=1;                   #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Closed-loop voltage gain=%d\"%A_CL);\n",
-    "print(\"Supply voltage=±15V, saturation voltage=±13V. Since gain=-40, op-Amp will be driven to saturation.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.29 : Page number 697"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 30,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  A_CL=-10.\n",
-      "(ii) Zi=10kΩ\n",
-      "(iii) Maximum operating frequency=15.9kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "Rf=100;                   #Feedback resistor, kΩ\n",
-    "Ri=10;                    #Input resistor, kΩ\n",
-    "Vpp=1;                    #Input peak-peak voltage, V\n",
-    "slew_rate=0.5;            #Slew rate, V//μs\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "\n",
-    "#(ii)\n",
-    "Zi=Ri;                  #Input impedance(~ Input resistor), kΩ\n",
-    "\n",
-    "#(iii)\n",
-    "Vout=A_CL*Vpp;                           #Peak-to-peak voltage, V\n",
-    "Vpk=Vout/2;                              #Peak output voltage, V\n",
-    "f_max=(slew_rate*10**6/(2*pi*abs(Vpk)))/1000;        #Maximum operating frequency, kHz\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  A_CL=%d.\"%A_CL);\n",
-    "print(\"(ii) Zi=%dkΩ\"%Zi);\n",
-    "print(\"(iii) Maximum operating frequency=%.1fkHz.\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.30 : Page number 697"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Rf=20kΩ and Ri=5kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_CL=-4;                    #Closed loop voltage gain\n",
-    "R=[1.0,5.0,10.0,20.0];              #List of available resistors, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "for i in R[:]:\n",
-    "    for j in R[:]:\n",
-    "        if -(i/j)==A_CL :\n",
-    "            print(\"Rf=%dkΩ and Ri=%dkΩ.\"%(i,j));\n",
-    "            break;\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.31 : Page  number 697-698"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 32,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Closed loop voltage gain=-100.\n",
-      "(ii) Closed loop voltage gain=-50.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=100;                   #Feedback resistor, kΩ\n",
-    "Ri=1;                    #Input resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "R_source=0;               #Source resistor, kΩ\n",
-    "A_CL=-Rf/(R_source+Ri);    #Closed-loop voltage gain\n",
-    "\n",
-    "print(\"(i) Closed loop voltage gain=%d.\"%A_CL);\n",
-    "\n",
-    "#(ii)\n",
-    "R_source=1;               #Source resistor, kΩ\n",
-    "A_CL=-Rf/(R_source+Ri);    #Closed-loop voltage gain\n",
-    "\n",
-    "print(\"(ii) Closed loop voltage gain=%d.\"%A_CL);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.32 : Page number 699-700"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 33,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=12.12mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=240;                   #Feedback resistor, kΩ\n",
-    "Ri=2.4;                    #Input resistor, kΩ\n",
-    "Vin=120;                    #Input voltage, μV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=1+(Rf/Ri);             #Closed loop voltage gain\n",
-    "Vout=(A_CL*Vin)/1000;       #Output voltage, mV\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.2fmV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.33 : Page number 700"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Output voltage=11V\n",
-      "(ii) Output voltage=-11V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=10;                   #Feedback resistor, kΩ\n",
-    "Ri=1;                    #Input resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=1+(Rf/Ri);             #Closed loop voltage gain\n",
-    "#(i)\n",
-    "Vin=1;                      #Input voltage, V\n",
-    "Vout=A_CL*Vin;              #Output voltage, V\n",
-    "\n",
-    "print(\"(i)  Output voltage=%dV\"%Vout);\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "Vin=-1;                      #Input voltage, V\n",
-    "Vout=A_CL*Vin;               #Output voltage, V\n",
-    "\n",
-    "print(\"(ii) Output voltage=%dV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.34 : Page number 700"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 35,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Peak to peak output voltage=12V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=5;                       #Feedback resistor, kΩ\n",
-    "Ri=1;                       #Input resistor, kΩ\n",
-    "Vin_max=1;                  #Maximum input voltage, V\n",
-    "Vin_min=-1;                 #Minimum input voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "V_inpp=Vin_max-Vin_min;               #Peak-peak input voltage, V\n",
-    "A_CL=1+(Rf/Ri);                     #Closed loop voltage gain\n",
-    "Vout_pp=A_CL*V_inpp;                #Peak-peak output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Peak to peak output voltage=%dV\"%Vout_pp);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.35 : Page number 700-701"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 36,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Closed-loop voltage gain=11\n",
-      "(ii) Maximum operating frequency=14.47kHz\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "Rf=100;                       #Feedback resistor, kΩ\n",
-    "Ri=10;                       #Input resistor, kΩ\n",
-    "Vpp=1;                    #Input peak-peak voltage, V\n",
-    "slew_rate=0.5;              #Slew rate, V/μs\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CL=1+(Rf/Ri);                                      #Closed loop voltage gain\n",
-    "\n",
-    "#(ii)\n",
-    "Vout_pp=A_CL*Vpp;                                    #Peak-peak output voltage, V\n",
-    "Vpk=Vout_pp/2.0;                                          #Peak output voltage, V\n",
-    "f_max=((slew_rate*10**6)/(2*pi*Vpk))/1000.0;           #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  Closed-loop voltage gain=%d\"%A_CL);\n",
-    "\n",
-    "print(\"(ii) Maximum operating frequency=%.2fkHz\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.36 : Page number 701"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 37,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Bandwidth=44.3kHz.\n",
-      "(ii)  Bandwidth=63.8kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=220;                       #Feedback resistor, kΩ\n",
-    "Ri=3.3;                       #Input resistor, kΩ\n",
-    "unity_gain_BW=3;              #Unity gain bandwidth, MHz\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) For non-inverting amplifier\n",
-    "A_CL=1+(Rf/Ri);                    #Closed loop voltage gain\n",
-    "BW=unity_gain_BW*1000.0/A_CL;        #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(i)  Bandwidth=%.1fkHz.\"%BW);\n",
-    "\n",
-    "#(ii) For inverting amplifier\n",
-    "Rf=47;                          #Feedback resistor, kΩ\n",
-    "Ri=1;                           #Input resistor, kΩ\n",
-    "A_CL=-(Rf/Ri);                    #Closed loop voltage gain\n",
-    "BW=unity_gain_BW*1000.0/abs(A_CL);        #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(ii)  Bandwidth=%.1fkHz.\"%BW);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.37 : Page number 701-702"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 38,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) For voltage follower A_CL=1.\n",
-      "(ii) The maximum output frequency=26.53kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#(i)\n",
-    "A_CL=1;                     #Closed loop voltage gain for voltage follower\n",
-    "print(\"(i) For voltage follower A_CL=1.\");\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "slew_rate=0.5;              #Slew rate, V/μs\n",
-    "V_inpp=6;                   #peak-peak input voltage, V\n",
-    "Vout=A_CL*V_inpp;           #Peak-peak output voltage, V\n",
-    "Vpk=Vout/2;                 #Peak output voltage, V\n",
-    "\n",
-    "f_max=(slew_rate*10**6/(2*pi*Vpk))/1000;       #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"(ii) The maximum output frequency=%.2fkHz.\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.38 : Page number 702"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 39,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=1.78V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=470.0;                 #Feedback resistor, kΩ\n",
-    "R1=4.3;                 #Input resistor of 1st op-Amp, kΩ\n",
-    "R2=33.0;                 #Input resistor of 2nd op-Amp, kΩ\n",
-    "R3=33.0;                 #Input resistor of 3rd op-Amp, kΩ\n",
-    "Vin=80.0;                 #Input voltage, μV.\n",
-    "\n",
-    "#Calculation\n",
-    "A1=1+Rf/R1;                 #Gain of first op-Amp\n",
-    "A2=-round(Rf/R2,1);         #Gain of second op-Amp\n",
-    "A3=-round(Rf/R3,1);         #Gain of third op-Amp\n",
-    "A=A1*A2*A3;                 #Overall gain\n",
-    "Vout=A*Vin*10**-6;          #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.2fV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.39 : Page number 702-703"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "R1=30kΩ, R2=15kΩ and R3=10kΩ.\n",
-      "Output voltage=0.729V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A1=10;                  #Voltage gain of 1st op-Amp\n",
-    "A2=-18;                 #Voltage gain of 2nd op-Amp\n",
-    "A3=-27;                 #Voltage gain of 3rd op-Amp\n",
-    "Rf=270;                 #Feedback resistor, kΩ\n",
-    "Vin=150;                #Input voltage, μV  \n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "R1=Rf/(A1-1);                   #Input resistor of 1st op-Amp, kΩ\n",
-    "R2=-Rf/A2;                      #Input resistr of 2nd op-Amp, kΩ\n",
-    "R3=-Rf/A3;                      #Input resistor of 3rd op-Amp, kΩ\n",
-    "\n",
-    "A=A1*A2*A3;                     #overall gain,\n",
-    "Vout=Vin*10**-6*A;              #Output voltage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"R1=%dkΩ, R2=%dkΩ and R3=%dkΩ.\"%(R1,R2,R3));\n",
-    "print(\"Output voltage=%.3fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.40 : Page number 703-704"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 41,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "R1=50kΩ, R2=25kΩ and R3=10kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=500;                 #Feedback resistor, kΩ\n",
-    "A1=-10;                 #Gain of 1st op-Amp\n",
-    "A2=-20;                 #Gain of 2nd op-Amp\n",
-    "A3=-50;                 #Gain of 3rd op-Amp\n",
-    "\n",
-    "#Calculation\n",
-    "R1=-Rf/A1;                      #Input resistor of 1st op-Amp, kΩ\n",
-    "R2=-Rf/A2;                      #Input resistor of 2nd op-Amp, kΩ\n",
-    "R3=-Rf/A3;                      #Input resistor of 3rd op-Amp, kΩ\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"R1=%dkΩ, R2=%dkΩ and R3=%dkΩ.\"%(R1,R2,R3));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.41 : Page number 705"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 42,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The input impedance=17202MΩ and output impedance=8.7e-03Ω.\n",
-      "(ii) The closed loop voltage gain=23.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Zin=2.0;              #Input impedance of op-Amp, MΩ\n",
-    "Zout=75.0;            #Output impedance of op-Amp, Ω\n",
-    "A_OL=200000.0;        #Open-loop voltage gain\n",
-    "Rf=220.0;             #Feedback resistor, kΩ\n",
-    "Ri=10.0;              #Input resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "mv=round(Ri/(Ri+Rf),3);       #Feedback fraction\n",
-    "Zin_NI=Zin*(1+(A_OL*mv));     #Input impedance, MΩ\n",
-    "Zout_NI=Zout/(1+A_OL*mv);     #Output impedance, Ω\n",
-    "\n",
-    "#(ii)\n",
-    "A_CL=1+Rf/Ri;               #Closed loop voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) The input impedance=%dMΩ and output impedance=%.1eΩ.\"%(Zin_NI,Zout_NI));\n",
-    "print(\"(ii) The closed loop voltage gain=%d.\"%A_CL);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.42 : Page number 705-706"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 43,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance=400002MΩ and output impedance=0.38e-03Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "#For voltage follower,\n",
-    "mv=1.0;               #Feedback fraction\n",
-    "A_OL=200000.0;        #Open-loop voltage gain\n",
-    "Zin=2.0;              #Input impedance of op-Amp, MΩ\n",
-    "Zout=75.0;            #Output impedance of op-Amp, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Zin_VF=Zin*(1+(A_OL*mv));     #Input impedance, MΩ\n",
-    "Zout_VF=round(round(Zout/(1+A_OL*mv),6),5);     #Output impedance, Ω\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance=%dMΩ and output impedance=%.2fe-03Ω.\"%(Zin_VF,Zout_VF*1000));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.43 : Page number 706"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 44,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance=1kΩ and output impedance=50Ω.\n",
-      "Closed-loop voltage gain=-100\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=100;             #Feedback resistor, kΩ\n",
-    "Ri=1.0;               #Input resistor, kΩ\n",
-    "Zin=4;              #Input impedance of op-Amp, MΩ\n",
-    "Zout=50;            #Output impedance of op-Amp, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Zin_I=Ri;                   #Input impedance, kΩ\n",
-    "Zout_I=Zout;                #Output impedance, Ω\n",
-    "A_CL=-(Rf/Ri);              #Closed loop voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance=%dkΩ and output impedance=%dΩ.\"%(Zin_I,Zout_I));\n",
-    "print(\"Closed-loop voltage gain=%d\"%A_CL);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.44 : Page number 709"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-12V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=10;               #Feedback resistor, kΩ\n",
-    "Ri=10;               #Input resistor, kΩ\n",
-    "V1=3;                #Input voltage 1st, V\n",
-    "V2=1;                #Input voltage 2nd, V\n",
-    "V3=8;                #Input voltage 3rd, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Rf=Ri, Vout=-(Rf/Ri)*(V1+V2+V3)= -(V1+V2+V3);\n",
-    "Vout=-(V1+V2+V3);           #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.45 : Page number 709"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 46,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-7V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=10;               #Feedback resistor, kΩ\n",
-    "R1=1;                #Input resistor for input 1, kΩ\n",
-    "R2=1;                #Input resistor for input 2, kΩ\n",
-    "V1=0.2;              #Input voltage 1st, V\n",
-    "V2=0.5;              #Input voltage 2nd, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "R=R1;                       #Input resistor(=R1 or R2), kΩ\n",
-    "Vout=-(Rf/R)*(V1+V2);       #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.46 : Page number 709-710"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 47,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-2.5V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=1;                #Feedback resistor, kΩ\n",
-    "Ri=10.0;               #Input resistor, kΩ\n",
-    "V1=10;                #Input voltage 1st, V\n",
-    "V2=8.0;                #Input voltage 2nd, V\n",
-    "V3=7.0;                #Input voltage 3rd, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Vout=-(Rf/Ri)*(V1+V2+V3);\n",
-    "Vout=-(Rf/Ri)*(V1+V2+V3);           #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.1fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.47 : Page number 710"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 48,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=2.5V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V1=0.6;                 #Input voltage to 1st input resistor, V\n",
-    "V2=-1.4;                #Input voltage to 2nd input resistor, V\n",
-    "Rf=200;                 #Feedback resistor, kΩ\n",
-    "R1=400;                 #Input resistor 1, kΩ\n",
-    "R2=100.0;                 #Input resistor 2, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "Vout=-Rf*(V1/R1 +V2/R2);                 #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.1fV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.48 : Page number 710-711"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 49,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The output voltage=-12.5V\n",
-      "(ii) The output voltage=-7.5V\n",
-      "(iii) The output voltage=-17.5V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=1.0;                 #Feedback resistor, kΩ\n",
-    "R1=1.0;                 #Input resistor 1, kΩ\n",
-    "R2=2.0;                 #Input resistor 2, kΩ\n",
-    "R3=4.0;                 #Input resistor 3, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Rf_R1=Rf/R1;                #Ratio of feedback resistor and 1st input resistor\n",
-    "Rf_R2=Rf/R2;                #Ratio of feedback resistor and 2nd input resistor\n",
-    "Rf_R3=Rf/R3;                #Ratio of feedback resistor and 3rd input resistor\n",
-    "\n",
-    "#(i) First input combination\n",
-    "V1=10;                                   #Input voltage to 1st input resistor, V\n",
-    "V2=0;                                    #Input voltage to 2nd input resistor, V\n",
-    "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-    "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-    "print(\"(i) The output voltage=%.1fV\"%Vout);\n",
-    "\n",
-    "#(i) First input combination\n",
-    "V1=0;                                   #Input voltage to 1st input resistor, V\n",
-    "V2=10;                                    #Input voltage to 2nd input resistor, V\n",
-    "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-    "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-    "print(\"(ii) The output voltage=%.1fV\"%Vout);\n",
-    "\n",
-    "\n",
-    "#(i) First input combination\n",
-    "V1=10;                                   #Input voltage to 1st input resistor, V\n",
-    "V2=10;                                    #Input voltage to 2nd input resistor, V\n",
-    "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-    "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-    "print(\"(iii) The output voltage=%.1fV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.49 : Page number 711"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 50,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vout=-[0.5sin(1000t)+0.33sin(3000t)]V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=330;                 #Feedback resistor, kΩ\n",
-    "R1=33.0;                  #Input resistor 1, kΩ\n",
-    "R2=10.0;                  #Input resistor 2, kΩ\n",
-    "V1_m=50;                #Peak voltage of 1st input, mV\n",
-    "V2_m=10;                #Peak voltage of 2nd input, mV\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Vout=-((Rf/R1)*V1 + (Rf/R2)*V2)\n",
-    "print(\"Vout=-[%.1fsin(1000t)+%.2fsin(3000t)]V\"%((V1_m/1000.0)*(Rf/R1),(V2_m/1000.0)*(Rf/R2)));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.50 : Page number 715"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 51,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vo=-1*(1/RC)∫vi dt.\n",
-      "=>Vo=-1*(1/1)∫vi dt\n",
-      "=>Vo=∫vi dt\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=100;                      #Input resistor, kΩ\n",
-    "C=10;                       #Feedback capacitor, μF\n",
-    "\n",
-    "#Calculation\n",
-    "RC=R*10**3*C*10**-6;        #product of input resistance and feedback capacitance, s\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vo=-1*(1/RC)∫vi dt.\");\n",
-    "print(\"=>Vo=-1*(1/%d)∫vi dt\"%RC);\n",
-    "print(\"=>Vo=∫vi dt\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.51 : Page number 715-716"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 52,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The critical frequency=159Hz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "Rf=100;                      #Feedback resistor, kΩ\n",
-    "C=0.01;                      #Feedback capacitor, μF\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "fc=1/(2*pi*Rf*1000*C*10**-6);               #Crictical frequency, Hz\n",
-    "\n",
-    "#Result\n",
-    "print(\"The critical frequency=%dHz.\"%fc);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.52 : Page number 716"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 53,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Vout=-1*(1/RC)∫vi dt.\n",
-      "    ΔVout/dt = -vin/RC = -50mV/μs.\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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/w4w8f/Q0SQ0NhCDpTElzJc2W9FNJ6zWyP7NaJk2C73/fd2ibDUfdYzFJ2hmY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\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7ff5052a4080>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "R=10.0;                      #Input resistor, kΩ\n",
-    "C=0.01;                    #Feedback capacitor, μF\n",
-    "vin=5;                     #Input voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "Vout_change_rate=-vin/(R*C);            #Rate of change of output voltage, V/μs   \n",
-    "print(\"(i) Vout=-1*(1/RC)∫vi dt.\");\n",
-    "print(\"    ΔVout/dt = -vin/RC = %dmV/μs.\"%Vout_change_rate);\n",
-    "\n",
-    "#(ii) Plotting the output waveform\n",
-    "vin_plot=[];                    #Plotting variable for input waveform, V\n",
-    "dt=100;                          #time between edges, μs\n",
-    "for i in range(0,3*dt+1):\n",
-    "    if i<dt or i>2*dt :\n",
-    "        vin_plot.append(0);\n",
-    "    else:\n",
-    "        vin_plot.append(5); \n",
-    "\n",
-    "plt.subplot(211);\n",
-    "plt.plot(vin_plot);\n",
-    "plt.xlim([0,300])\n",
-    "plt.ylim([-5,10])\n",
-    "plt.xlabel(\"t(microsecond)\");\n",
-    "plt.ylabel(\"Vin(V)\");\n",
-    "plt.title(\"Input waveform\");\n",
-    "\n",
-    "        \n",
-    "vout_plot=[];                  #Plotting variable for output waveform, V\n",
-    "t=[i for i in range(0,301)];  #Time scale, μs\n",
-    "for i in t[:] :\n",
-    "    if i<dt:\n",
-    "        vout_plot.append(0);\n",
-    "    elif i>2*dt:\n",
-    "        vout_plot.append((Vout_change_rate/1000.0)*dt);\n",
-    "    else :\n",
-    "        vout_plot.append((-vin_plot[i]/(R*C))/1000*(i-dt));\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout_plot);\n",
-    "plt.xlim([0,300])\n",
-    "plt.ylim([-5,5]);\n",
-    "plt.xlabel('t(microsecond)');\n",
-    "plt.ylabel(\"Vout(V)\");\n",
-    "plt.title(\"output waveform\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.53 : Page number 716-717"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 54,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vout=-1*(1/RC)∫vi dt.\n",
-      "Vout=-5*t volts\n",
-      "Time required=2.6seconds.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_supply=15;              #Supply voltage, V\n",
-    "R=10;                     #Input resistor, kΩ\n",
-    "C=0.2;                    #Feedback capacitor, μF\n",
-    "vin=10;                   #Input voltage, mV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Vs=-V_supply+2;         #Saturation voltage, V\n",
-    "print(\"Vout=-1*(1/RC)∫vi dt.\");\n",
-    "print(\"Vout=%d*t volts\"%(-vin/(R*C)));\n",
-    "t=Vs/(-vin/(R*C));                      #Time required, seconds\n",
-    "print(\"Time required=%.1fseconds.\"%t);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.54 : Page number 717-718"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 55,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vo=-5V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=1;                     #Feedback resistor, kΩ\n",
-    "C=0.1;                   #Input capacitor, μF\n",
-    "Vin_change=5;            #Change in input voltage, V\n",
-    "t=0.1;                   #Time taken for change in input voltage, ms\n",
-    "\n",
-    "#Calcualtion\n",
-    "dvi_dt=Vin_change/(t/1000);        #Rate of change of input voltage, V/s\n",
-    "RC=R*1000*C*10**-6;                #Product of feedback resistance and input capacitance, s\n",
-    "#Since, Vo=-R*C*(dvi/dt);\n",
-    "Vo=-RC*dvi_dt;                      #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vo=%dV.\"%Vo);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.55 : Page number 718"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 56,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vo=-0.55V.\n",
-      "The output voltage stays constant at -0.55V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=10;                     #Feedback resistor, kΩ\n",
-    "C=2.2;                   #Input capacitor, μF\n",
-    "Vin_change=10;            #Change in input voltage, V\n",
-    "t=0.4;                   #Time taken for change in input voltage, s\n",
-    "\n",
-    "#Calcualtin\n",
-    "dvi_dt=Vin_change/t;        #Rate of change of input voltage, V/s\n",
-    "RC=R*1000*C*10**-6;         #Product of feedback resistance and input capacitance, s\n",
-    "#Since, Vo=-R*C*(dvi/dt);\n",
-    "Vo=-RC*dvi_dt;                      #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vo=%.2fV.\"%Vo);\n",
-    "print(\"The output voltage stays constant at %.2fV.\"%Vo);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.56 : Page number 718-719"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 57,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "vo=-1*(dvi/dt).\n",
-      "vo=-5V.\n",
-      "Therefore, between 0 to 0.2s, the output voltage is constant at -5V.\n",
-      "For t>0.2s, the input is constant so that output voltage is zero.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=100;                     #Feedback resistor, kΩ\n",
-    "C=10;                   #Input capacitor, μF\n",
-    "Vin_change=1;            #Change in input voltage, V\n",
-    "t=0.2;                   #Time taken for change in input voltage, s\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "RC=R*1000*C*10**-6;             #Product of feedback resistor and input capacitance, s\n",
-    "#(i)\n",
-    "print(\"vo=-%d*(dvi/dt).\"%RC);\n",
-    "\n",
-    "#(ii)\n",
-    "dvi_dt=Vin_change/t;                #Rate of change of input voltage, V\n",
-    "vo=-dvi_dt;                         #Output voltage, V\n",
-    "print(\"vo=%dV.\"%vo);\n",
-    "\n",
-    "print(\"Therefore, between 0 to 0.2s, the output voltage is constant at %dV.\"%vo);\n",
-    "print(\"For t>0.2s, the input is constant so that output voltage is zero.\");\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_3.ipynb
deleted file mode 100755
index 8b1cb626..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_3.ipynb
+++ /dev/null
@@ -1,2522 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 25 : OPERATIONAL AMPLIFIERS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.1: Page number 664"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output voltage of the differential amplifier = 10V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A=100.0;              #Open-circuit voltage gain of differential amplifier\n",
-    "V1=3.25;              #Input voltage to terminal 1 in V\n",
-    "V2=3.15;              #Input voltage to terminal 2 in V\n",
-    "\n",
-    "#Calculations\n",
-    "V0=A*(V1-V2);         #Output voltage in V\n",
-    "\n",
-    "#Results\n",
-    "print(\"The output voltage of the differential amplifier = %dV\"%V0);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.2: Page number 672"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The common mode rejection ratio = 10000.\n",
-      "The common mode rejection ratio in decibels= 80dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "A_DM=2000.0;              #Differential mode voltage gain\n",
-    "A_CM=0.2;                 #Common mode voltage gain\n",
-    "\n",
-    "#Calculations\n",
-    "CMRR=A_DM/A_CM;           #Common mode rejection ratio\n",
-    "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"The common mode rejection ratio = %d.\"%CMRR);\n",
-    "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.3: Page number 672"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The common mode rejection ratio in decibels= 46dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "VD_in=10.0;               #Differential mode input in mV\n",
-    "VD_out=1.0;               #Output for differential mode input in V\n",
-    "VC_in=10.0;                 #Common mode input in mV\n",
-    "VC_out=5.0;                 #Output for common mode input in mV\\\n",
-    "\n",
-    "#Calculations\n",
-    "A_DM=(VD_out*1000)/VD_in;       #Differntial mode voltage gain\n",
-    "A_CM=VC_out/VC_in;       #Common mode voltage gain\n",
-    "CMRR=A_DM/A_CM;           #Common mode rejection ratio\n",
-    "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-    "\n",
-    "#Results\n",
-    "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.4: Page number 672"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage =7.5V\n",
-      "Noise on output = 4.7x10^-6V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_DM=150.0;           #Differential mode voltage gain\n",
-    "CMRR_dB=90.0;         #Common mode rejection ratio\n",
-    "V1=100.0;             #Input voltage for terminal 1 in mV\n",
-    "V2=50.0;              #Input voltage for terminal 2 in mV\n",
-    "V_noise=1.0;          #Voltage of noise signal in mV\n",
-    "\n",
-    "#Calculation\n",
-    "\n",
-    "#Case(i)\n",
-    "V_out=A_DM*(V1-V2)/1000.0;            #Output voltage for differntial mode input, in V\n",
-    "\n",
-    "#Since CMRR_dB=20*log10(differential mode gain/common mode gain),\n",
-    "A_CM=A_DM/pow(10,(CMRR_dB/20));         #Common mode gain\n",
-    "V_OUT_noise=A_CM*(V_noise/1000);        #Noise on output in V\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"Output voltage =%.1fV\"%V_out);\n",
-    "print(\"Noise on output = %.1fx10^-6V\"%(V_OUT_noise*pow(10,6)));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.5 : Page number 672-673"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Common mode gain =0.083\n",
-      "Common mode rejection ratio in decibels=89.5dB\n",
-      "r.m.s output signal =1.25V\n",
-      "r.m.s interfernce output voltage = 83mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "A_DM=2500.0;                    #Differential mode voltage gain\n",
-    "CMRR=30000.0;                   #Common mode rejection ratio\n",
-    "Input_signal=500.0;             #Single ended input r.m.s signal in microvolts\n",
-    "Interference=1.0;               #Interference signal, in V\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#(i)\n",
-    "A_CM=A_DM/CMRR;                 #Common mode gain\n",
-    "\n",
-    "#(ii)\n",
-    "CMRR_dB=20*log10(CMRR);         #Common mode rejection ratio in decibels\n",
-    "\n",
-    "#(iii)\n",
-    "V_out=A_DM*(Input_signal/pow(10,6)-0);        #r.m.s output signal  in V\n",
-    "\n",
-    "#(iv)\n",
-    "Interference_out=A_CM*Interference;          #r.m.s interference output in V\n",
-    "Interference_out=Interference_out*1000;      #r.m.s interference output in mV\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"Common mode gain =%.3f\"%A_CM);\n",
-    "print(\"Common mode rejection ratio in decibels=%.1fdB\"%CMRR_dB);\n",
-    "print(\"r.m.s output signal =%.2fV\"%V_out);\n",
-    "print(\"r.m.s interfernce output voltage = %dmV\"%Interference_out);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.6 : Page number 674-675"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VE=-0.7V\n",
-      "IE=0.452mA\n",
-      "IE1=0.226mA\n",
-      "IE2=0.226mA\n",
-      "IC1=0.226mA\n",
-      "IC2=0.226mA\n",
-      "IB1=2.26μA\n",
-      "IB2=2.26μA\n",
-      "VC1=12V\n",
-      "VC2=9.7V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RB=10;                  #Base resistor, kΩ\n",
-    "RC2=10;                 #Collector resistor, kΩ\n",
-    "RE=25;                  #Emitter resistor, kΩ\n",
-    "VBE=0.7;               #Base-emitter voltage, V\n",
-    "beta=100;               #Base amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-    "IE=(VEE-VBE)/RE;            #Tail current, mA\n",
-    "IE1=IE/2;                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-    "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-    "IC2=IC1;                    #Collector current of 2nd transistor, mA\n",
-    "IB1=(IC1/beta)*1000;        #Base current of 1st transistor, μA\n",
-    "IB2=IB1;                    #Base current of 2nd transistor, μA\n",
-    "VC1=VCC;                    #Collector voltage of 1st transistor, V\n",
-    "VC2=VCC-IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"VE=%.1fV\"%VE);\n",
-    "print(\"IE=%.3fmA\"%IE);\n",
-    "print(\"IE1=%.3fmA\"%IE1);\n",
-    "print(\"IE2=%.3fmA\"%IE2);\n",
-    "print(\"IC1=%.3fmA\"%IC1);\n",
-    "print(\"IC2=%.3fmA\"%IC2);\n",
-    "print(\"IB1=%.2fμA\"%IB1);\n",
-    "print(\"IB2=%.2fμA\"%IB2);\n",
-    "print(\"VC1=%dV\"%VC1);\n",
-    "print(\"VC2=%.1fV\"%VC2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.7 : Page number 675"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output voltage=7.85V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15;                 #Collector supply voltage, V\n",
-    "VEE=15;                 #Emitter supply voltage, V\n",
-    "RB=33;                  #Base resistor, kΩ\n",
-    "RC=15;                  #Collector resistor, kΩ\n",
-    "RE=15;                  #Emitter resistor, kΩ\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "IE_tail=(VEE-VBE)/RE;            #Tail current, mA\n",
-    "IE=round(IE_tail/2,3);                    #Emitter current in each transistor, mA\n",
-    "IC=IE;                           #Collector current(=emitter current), mA\n",
-    "Vout=VCC-IC*RC;                  #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output voltage=%.2fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.8 : Page number 675"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) IB1=5.56μA\n",
-      "    IB2=4.55μA\n",
-      "(ii) VB1=-0.183V\n",
-      "     VB2=-0.15V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15.0;                 #Collector supply voltage, V\n",
-    "VEE=15.0;                 #Emitter supply voltage, V\n",
-    "RB=33.0;                  #Base resistor, kΩ\n",
-    "RC=15.0;                  #Collector resistor, kΩ\n",
-    "RE=15.0;                  #Emitter resistor, kΩ\n",
-    "VBE=0;                  #Base-emitter voltage, V\n",
-    "beta_dc_l=90.0;           #base current amplification factor for left transistor\n",
-    "beta_dc_r=110.0;          #base current amplification factor for right transistor\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IE_tail=(VEE-VBE)/RE;             #Tail current, mA\n",
-    "IE=IE_tail/2;                      #Emitter current in each transistor, mA\n",
-    "IB1=(IE/beta_dc_l)*1000;          #Base current of 1st transistor, μA\n",
-    "IB2=(IE/beta_dc_r)*1000;          #Base current of 2nd transistor, μA\n",
-    "\n",
-    "#(ii)\n",
-    "VB1=-IB1/1000*RB;                    #Base voltage of 1st transistor, V\n",
-    "VB2=-IB2/1000*RB;                    #Base voltage of 1st transistor, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) IB1=%.2fμA\"%IB1);\n",
-    "print(\"    IB2=%.2fμA\"%IB2);\n",
-    "print(\"(ii) VB1=%.3fV\"%VB1);\n",
-    "print(\"     VB2=%.2fV\"%VB2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.9 : Page number 675-676"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VE=-0.7V\n",
-      "Emitter current in each transistor=0.5mA.\n",
-      "IC1~IE1=0.5mA and IC2~IE2=0.5mA\n",
-      "VC1=VC2=10V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15.0;                 #Collector supply voltage, V\n",
-    "VEE=15.0;                 #Emitter supply voltage, V\n",
-    "RB=10.0;                  #Base resistor, kΩ\n",
-    "RC1=10.0;                 #Collector resistor of 1st transistor, kΩ\n",
-    "RC2=10.0;                 #Collector resistor of 2nd transistor, kΩ\n",
-    "IE=1.0;                   #Tail current, mA\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-    "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-    "IC2=IE2;                    #Collector current of 2nd transistor, mA\n",
-    "VC1=VCC-IC1*RC1;            #Collector voltage of 1st transistor, V\n",
-    "VC2=VCC-IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"VE=%.1fV\"%VE);\n",
-    "print(\"Emitter current in each transistor=%.1fmA.\"%(IE/2.0));\n",
-    "print(\"IC1~IE1=%.1fmA and IC2~IE2=%.1fmA\"%(IE1,IE2));\n",
-    "print(\"VC1=VC2=%dV.\"%VC2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.10 : Page number 676-677"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VE=0.7V\n",
-      "Tail current=0.452mA.\n",
-      "Emitter current in each transistor=0.226mA.\n",
-      "IC1~IE1=0.226mA and IC2~IE2=0.226mA\n",
-      "VC1=-12V\n",
-      "VC2=-9.74V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12.0;                 #Collector supply voltage, V\n",
-    "VEE=12.0;                 #Emitter supply voltage, V\n",
-    "RC2=10.0;                 #Collector resistor of 2nd transistor, kΩ\n",
-    "RE=25.0;                  #Emitter current, kΩ\n",
-    "VBE=-0.7;               #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-    "IE=(VCC-VE)/RE;             #Tail current, mA\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-    "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-    "IC2=IE2;                    #Collector current of 2nd transistor, mA\n",
-    "VC1=-VEE;                   #Collector voltage of 1st transistor, V\n",
-    "VC2=-VEE+IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"VE=%.1fV\"%VE);\n",
-    "print(\"Tail current=%.3fmA.\"%IE);\n",
-    "print(\"Emitter current in each transistor=%.3fmA.\"%(IE/2.0));\n",
-    "print(\"IC1~IE1=%.3fmA and IC2~IE2=%.3fmA\"%(IC1,IC2));\n",
-    "print(\"VC1=%dV\"%VC1);\n",
-    "print(\"VC2=%.2fV\"%VC2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.11 : Page number 679"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The input offset current=15.1nA\n",
-      "(ii) The input bias current=75.8nA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15;                 #Collector supply voltage, V\n",
-    "VEE=15;                 #Emitter supply voltage, V\n",
-    "RB=1;                   #Base resistor, MΩ\n",
-    "RC2=1;                  #Collector resistor, MΩ\n",
-    "RE=1;                   #Emitter resistor, MΩ\n",
-    "VBE=0;                  #Base-emitter voltage, V (Neglected)\n",
-    "beta_dc_l=90.0;           #base current amplification factor for left transistor\n",
-    "beta_dc_r=110.0;          #base current amplification factor for right transistor\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IE=(VEE-VBE)/RE;                    #Tail current, μA\n",
-    "IE1=IE/2.0;                           #Emitter current of 1st transistor, μA\n",
-    "IE2=IE1;                            #Emitter current of 2nd transistor, μA\n",
-    "IB1=round((IE1/beta_dc_l)*1000,1);           #Base current of 1st transistor, nA\n",
-    "IB2=round((IE2/beta_dc_r)*1000,1);           #Base current of 2nd transistor, nA\n",
-    "I_in_offset=IB1-IB2;                 #Input offset current, nA\n",
-    "\n",
-    "#(ii)\n",
-    "I_in_bias=(IB1+IB2)/2;                  #Input bias current, nA\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  The input offset current=%.1fnA\"%I_in_offset);\n",
-    "print(\"(ii) The input bias current=%.1fnA\"%I_in_bias);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.12 : Page number 679"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The two base currents are: IB1=90nA and IB2=70nA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_in_offset=20;                 #Input offset current, nA\n",
-    "I_in_bias=80;                   #Input bias current, nA\n",
-    "\n",
-    "#Calculation\n",
-    "IB1=I_in_bias+I_in_offset/2;            #Base current in 1st transistor, nA\n",
-    "IB2=I_in_bias-I_in_offset/2;            #Base current in 2nd transistor, nA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The two base currents are: IB1=%dnA and IB2=%dnA.\"%(IB1,IB2));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.13 : Page number 679-680"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input offset voltage=2mV.\n",
-      "The output offset voltage=0.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declration\n",
-    "I_in_offset=20;                 #Input offset current, nA\n",
-    "I_in_bias=80;                   #Input bias current, nA\n",
-    "A=150;                          #Voltage gain\n",
-    "RB=100;                         #Base resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_io=(I_in_offset*10**-9*RB*1000)*1000;       #Input offset voltage, mV\n",
-    "V_out_offset=(A*V_io)/1000;                     #Output offset voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input offset voltage=%dmV.\"%V_io);\n",
-    "print(\"The output offset voltage=%.1fV.\"%V_out_offset);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.14 : Page number 682"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Output voltage=0.15V.\n",
-      "(ii) Output voltage=-0.15V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15;                 #Collector supply voltage, V\n",
-    "VEE=15;                 #Emitter supply voltage, V\n",
-    "RE=1;                   #Emitter resistor, MΩ\n",
-    "RC=1;                   #Collector resistor, MΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "IE=VEE/RE;                  #Tail current, μA\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, μA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, μA\n",
-    "re=25/IE1;                  #a.c emitter resistance, kΩ\n",
-    "A_DM=RC/(2.0*re);        #Differential voltage gain,\n",
-    "\n",
-    "#(i)\n",
-    "vin=1;                      #Input voltage, V\n",
-    "Vout=A_DM*vin;              #Output voltage, V\n",
-    "\n",
-    "print(\"(i) Output voltage=%.2fV.\"%Vout);\n",
-    "\n",
-    "#(ii)\n",
-    "vin=-1;                      #Input voltage, V\n",
-    "Vout=A_DM*vin;              #Output voltage, V;\n",
-    "print(\"(ii) Output voltage=%.2fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.15 : Page number 682-683"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The input impedance=194kΩ.\n",
-      "(ii) The differential voltage gain=136.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RE=100;                 #Emitter resistor, kΩ\n",
-    "RC1=120;                #Collector resistor of 1st transistor, kΩ\n",
-    "RC2=120;                #Collector resistor of 2nd transistor, kΩ\n",
-    "beta=220;               #Base amplification factor\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calcualtion\n",
-    "IE=((VEE-VBE)/RE)*1000;     #Tail current, μA\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, μA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, μA\n",
-    "re=(25/IE1)*1000;           #a.c emitter resistance, Ω\n",
-    "Zin=2*beta*re/1000;         #Input impedance, kΩ\n",
-    "A_DM=RC1*1000/(2.0*re);             #Differential voltage gain,\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) The input impedance=%dkΩ.\"%Zin);\n",
-    "print(\"(ii) The differential voltage gain=%.0f.\"%A_DM);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.16: Page number 683-684"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Differential voltage gain=56.6.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RE=200;                 #Emitter resistor, kΩ\n",
-    "RC=100;                 #Collector resistor, kΩ\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "IE=round((VEE-VBE)/RE,4);            #Tail current, mA\n",
-    "IE1=round(IE/2,4);                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                             #Emitter current of 2nd transistor, mA\n",
-    "re=round(25/IE1,1);                  #a.c emitter resistance, Ω\n",
-    "A_DM=RC*1000/(2*re);                 #Differential voltage gain,\n",
-    "\n",
-    "#Result\n",
-    "print(\"Differential voltage gain=%.1f.\"%A_DM);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.17 : Page number 685-686"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Common mode rejection ratio=666.7.\n",
-      "Common mode rejection ratio in decibel=56.48dB\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "v1=0.5;               #Voltage in terminal 1, mV\n",
-    "v2=-0.5;               #Voltage in terminal 2, mV\n",
-    "vo=8.0;                   #Output voltage, V\n",
-    "vo_cm=12.0;               #Common mode output, mV\n",
-    "\n",
-    "#Calculation\n",
-    "vin=v1-v2;                  #Differential input, mV\n",
-    "A_DM=vo/(vin/1000.0);         #Differential mode gain,\n",
-    "vin_cm=1;                   #Common mode input, mV\n",
-    "A_CM=vo_cm/vin_cm;          #Common mode gain\n",
-    "CMRR=A_DM/A_CM;             #Common mode rejection ratio\n",
-    "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-    "\n",
-    "#Result\n",
-    "print(\"Common mode rejection ratio=%.1f.\"%CMRR)\n",
-    "print(\"Common mode rejection ratio in decibel=%.2fdB\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.18 : Page number 686"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Common mode voltage gain=6.32.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_DM=200000;                #Differential mode gain\n",
-    "CMRR_dB=90;                 #Common mode rejection ratio, dB\n",
-    "\n",
-    "#Calculation\n",
-    "CMRR=10**(CMRR_dB/20.0);              #Common mode rejection ratio\n",
-    "A_CM=A_DM/CMRR;                     #Common mode gain\n",
-    "\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Common mode voltage gain=%.2f.\"%A_CM);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.19 : Page number 686"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The Common mode gain=0.0081\n",
-      "(ii) The common mode rejection ratio=81.8dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "vin_cm=3.2;                     #Common input voltage, V\n",
-    "vout=26;                        #Output voltage, V\n",
-    "A_DM=100;                       #Open-circuit voltage gain\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CM=vout*10**-3/vin_cm;                #Common mode gain\n",
-    "\n",
-    "#(ii)\n",
-    "CMRR_dB=20*log10(A_DM/A_CM);            #Common mode rejection ratio, dB\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  The Common mode gain=%.4f\"%A_CM);\n",
-    "print(\"(ii) The common mode rejection ratio=%.1fdB.\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.20 : Page number 686-687"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Common mode gain=0.25\n",
-      "(ii)Common mode rejection ratio=47.09dB\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "from math import floor\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RE=200.0;                 #Emitter resistor, kΩ\n",
-    "RC=100.0;                 #Collector resistor, kΩ\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CM=round(RC/(2*RE),2);             #Common mode voltage gain\n",
-    "\n",
-    "#(ii)\n",
-    "IE=round((VEE-VBE)/RE,4);            #Tail current, mA\n",
-    "IE1=round(IE/2,4);                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                             #Emitter current of 2nd transistor, mA\n",
-    "re=round(25/IE1,1);                  #a.c emitter resistance, Ω\n",
-    "A_DM=RC*1000/(2*re);                 #Differential voltage gain,\n",
-    "CMRR_dB=floor(20*log10(A_DM/A_CM)*100)/100;         #Common mode rejection ratio, dB\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) Common mode gain=%.2f\"%A_CM);\n",
-    "print(\"(ii)Common mode rejection ratio=%.2fdB\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.21 : Page number 691"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "f2=30kHz\n",
-      "ACL=75 or 37.5dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "ACL=500;                    #closed loop gain\n",
-    "f_unity=15;                 #frequency with cloased-loop unity gain, MHz\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "f2=f_unity*1000/500                 #Upper frequency of bandwidth,kHz\n",
-    "BW=f2-0;                            #Bandwidth, kHz\n",
-    "A_CL=f_unity*1000/200;                    #Maximum value of A_CL when f2=200kHz\n",
-    "A_CL_dB=20*log10(A_CL);             #Maximum value of A_CL in decibel\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"f2=%dkHz\"%f2);\n",
-    "print(\"ACL=%d or %.1fdB.\"%(A_CL,A_CL_dB));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.22 : Page number 691-692"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Operating Bandwidth=1.5MHz.\n",
-      "(ii) Operating Bandwidth=150kHz.\n",
-      "(iii) Operating Bandwidth=15kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "GBW=1.5;                    #Gain-bandwidth, MHz\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) For A_CL=1;\n",
-    "A_CL=1;                         #Closed loop gain\n",
-    "BW=GBW/A_CL;                    #Bandwidth, MHz\n",
-    "\n",
-    "print(\"(i) Operating Bandwidth=%.1fMHz.\"%BW);\n",
-    "\n",
-    "#(ii) For A_CL=10;\n",
-    "A_CL=10;                         #Closed loop gain\n",
-    "BW=(GBW/A_CL)*1000;              #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(ii) Operating Bandwidth=%dkHz.\"%BW);\n",
-    "\n",
-    "#(iii) For A_CL=100;\n",
-    "A_CL=100;                         #Closed loop gain\n",
-    "BW=(GBW/A_CL)*1000;               #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(iii) Operating Bandwidth=%dkHz.\"%BW);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.23 : Page number 692"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Maximum operating frequency=9.95kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "slew_rate=0.5;                  #Slew rate, V/μs\n",
-    "V_supply=10;                     #Supply voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_sat=V_supply-2;                           #Saturation voltage, V\n",
-    "V_pk=V_sat;                                 #Maximum peak-output voltage, V\n",
-    "f_max=((slew_rate*10**6)/(2*pi*V_pk))/1000;       #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Maximum operating frequency=%.2fkHz.\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.24 : Page number 692"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Maximum operating frequency=796kHz\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "slew_rate=0.5;                  #Slew rate, V/μs\n",
-    "V_pk=100.0;                       #Peak-output voltage, mV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_pk=V_pk/1000.0;                                #Peak-output voltage, V\n",
-    "f_max=(slew_rate*10**6/(2*pi*V_pk))/1000.0;      #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Maximum operating frequency=%.0fkHz\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.25 : Page number 695-696"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Feedback resistor=220kΩ\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_CL=-100;              #Closed-loop voltage gain\n",
-    "Ri=2.2;                 #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, A_CL=-(Rf/Ri)\n",
-    "Rf=-A_CL*Ri;                #Feedback resistor, kΩ\n",
-    "\n",
-    "#Result\n",
-    "print(\"Feedback resistor=%dkΩ\"%Rf);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.26 : Page number 696"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-0.25V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "vin=2.5;                #Input voltage, mV\n",
-    "Rf=200;                 #Feedback resistor, kΩ\n",
-    "Ri=2;                   #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "vout=A_CL*vin/1000;                  #Output voltage,V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.2fV\"%vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.27 : Page number 696"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 28,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Closed-loop voltage gain=-1\n",
-      "Therefore, output will have same amplitude but 180° phase inversion.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Varaiable declaration\n",
-    "Rf=1.0;                   #Feedback resistor, kΩ\n",
-    "Ri=1.0;                   #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"Closed-loop voltage gain=%d\"%A_CL);\n",
-    "print(\"Therefore, output will have same amplitude but 180° phase inversion.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.28 : Page number 696-697"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 29,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Closed-loop voltage gain=-40\n",
-      "Supply voltage=±15V, saturation voltage=±13V. Since gain=-40, op-Amp will be driven to saturation.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=40;                   #Feedback resistor, kΩ\n",
-    "Ri=1;                   #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Closed-loop voltage gain=%d\"%A_CL);\n",
-    "print(\"Supply voltage=±15V, saturation voltage=±13V. Since gain=-40, op-Amp will be driven to saturation.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.29 : Page number 697"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 30,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  A_CL=-10.\n",
-      "(ii) Zi=10kΩ\n",
-      "(iii) Maximum operating frequency=15.9kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "Rf=100;                   #Feedback resistor, kΩ\n",
-    "Ri=10;                    #Input resistor, kΩ\n",
-    "Vpp=1;                    #Input peak-peak voltage, V\n",
-    "slew_rate=0.5;            #Slew rate, V//μs\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "\n",
-    "#(ii)\n",
-    "Zi=Ri;                  #Input impedance(~ Input resistor), kΩ\n",
-    "\n",
-    "#(iii)\n",
-    "Vout=A_CL*Vpp;                           #Peak-to-peak voltage, V\n",
-    "Vpk=Vout/2;                              #Peak output voltage, V\n",
-    "f_max=(slew_rate*10**6/(2*pi*abs(Vpk)))/1000;        #Maximum operating frequency, kHz\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  A_CL=%d.\"%A_CL);\n",
-    "print(\"(ii) Zi=%dkΩ\"%Zi);\n",
-    "print(\"(iii) Maximum operating frequency=%.1fkHz.\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.30 : Page number 697"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Rf=20kΩ and Ri=5kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_CL=-4;                    #Closed loop voltage gain\n",
-    "R=[1.0,5.0,10.0,20.0];              #List of available resistors, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "for i in R[:]:\n",
-    "    for j in R[:]:\n",
-    "        if -(i/j)==A_CL :\n",
-    "            print(\"Rf=%dkΩ and Ri=%dkΩ.\"%(i,j));\n",
-    "            break;\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.31 : Page  number 697-698"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 32,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Closed loop voltage gain=-100.\n",
-      "(ii) Closed loop voltage gain=-50.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=100;                   #Feedback resistor, kΩ\n",
-    "Ri=1;                    #Input resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "R_source=0;               #Source resistor, kΩ\n",
-    "A_CL=-Rf/(R_source+Ri);    #Closed-loop voltage gain\n",
-    "\n",
-    "print(\"(i) Closed loop voltage gain=%d.\"%A_CL);\n",
-    "\n",
-    "#(ii)\n",
-    "R_source=1;               #Source resistor, kΩ\n",
-    "A_CL=-Rf/(R_source+Ri);    #Closed-loop voltage gain\n",
-    "\n",
-    "print(\"(ii) Closed loop voltage gain=%d.\"%A_CL);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.32 : Page number 699-700"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 33,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=12.12mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=240;                   #Feedback resistor, kΩ\n",
-    "Ri=2.4;                    #Input resistor, kΩ\n",
-    "Vin=120;                    #Input voltage, μV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=1+(Rf/Ri);             #Closed loop voltage gain\n",
-    "Vout=(A_CL*Vin)/1000;       #Output voltage, mV\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.2fmV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.33 : Page number 700"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Output voltage=11V\n",
-      "(ii) Output voltage=-11V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=10;                   #Feedback resistor, kΩ\n",
-    "Ri=1;                    #Input resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=1+(Rf/Ri);             #Closed loop voltage gain\n",
-    "#(i)\n",
-    "Vin=1;                      #Input voltage, V\n",
-    "Vout=A_CL*Vin;              #Output voltage, V\n",
-    "\n",
-    "print(\"(i)  Output voltage=%dV\"%Vout);\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "Vin=-1;                      #Input voltage, V\n",
-    "Vout=A_CL*Vin;               #Output voltage, V\n",
-    "\n",
-    "print(\"(ii) Output voltage=%dV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.34 : Page number 700"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 35,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Peak to peak output voltage=12V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=5;                       #Feedback resistor, kΩ\n",
-    "Ri=1;                       #Input resistor, kΩ\n",
-    "Vin_max=1;                  #Maximum input voltage, V\n",
-    "Vin_min=-1;                 #Minimum input voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "V_inpp=Vin_max-Vin_min;               #Peak-peak input voltage, V\n",
-    "A_CL=1+(Rf/Ri);                     #Closed loop voltage gain\n",
-    "Vout_pp=A_CL*V_inpp;                #Peak-peak output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Peak to peak output voltage=%dV\"%Vout_pp);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.35 : Page number 700-701"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 36,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Closed-loop voltage gain=11\n",
-      "(ii) Maximum operating frequency=14.47kHz\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "Rf=100;                       #Feedback resistor, kΩ\n",
-    "Ri=10;                       #Input resistor, kΩ\n",
-    "Vpp=1;                    #Input peak-peak voltage, V\n",
-    "slew_rate=0.5;              #Slew rate, V/μs\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CL=1+(Rf/Ri);                                      #Closed loop voltage gain\n",
-    "\n",
-    "#(ii)\n",
-    "Vout_pp=A_CL*Vpp;                                    #Peak-peak output voltage, V\n",
-    "Vpk=Vout_pp/2.0;                                          #Peak output voltage, V\n",
-    "f_max=((slew_rate*10**6)/(2*pi*Vpk))/1000.0;           #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  Closed-loop voltage gain=%d\"%A_CL);\n",
-    "\n",
-    "print(\"(ii) Maximum operating frequency=%.2fkHz\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.36 : Page number 701"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 37,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Bandwidth=44.3kHz.\n",
-      "(ii)  Bandwidth=63.8kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=220;                       #Feedback resistor, kΩ\n",
-    "Ri=3.3;                       #Input resistor, kΩ\n",
-    "unity_gain_BW=3;              #Unity gain bandwidth, MHz\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) For non-inverting amplifier\n",
-    "A_CL=1+(Rf/Ri);                    #Closed loop voltage gain\n",
-    "BW=unity_gain_BW*1000.0/A_CL;        #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(i)  Bandwidth=%.1fkHz.\"%BW);\n",
-    "\n",
-    "#(ii) For inverting amplifier\n",
-    "Rf=47;                          #Feedback resistor, kΩ\n",
-    "Ri=1;                           #Input resistor, kΩ\n",
-    "A_CL=-(Rf/Ri);                    #Closed loop voltage gain\n",
-    "BW=unity_gain_BW*1000.0/abs(A_CL);        #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(ii)  Bandwidth=%.1fkHz.\"%BW);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.37 : Page number 701-702"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 38,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) For voltage follower A_CL=1.\n",
-      "(ii) The maximum output frequency=26.53kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#(i)\n",
-    "A_CL=1;                     #Closed loop voltage gain for voltage follower\n",
-    "print(\"(i) For voltage follower A_CL=1.\");\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "slew_rate=0.5;              #Slew rate, V/μs\n",
-    "V_inpp=6;                   #peak-peak input voltage, V\n",
-    "Vout=A_CL*V_inpp;           #Peak-peak output voltage, V\n",
-    "Vpk=Vout/2;                 #Peak output voltage, V\n",
-    "\n",
-    "f_max=(slew_rate*10**6/(2*pi*Vpk))/1000;       #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"(ii) The maximum output frequency=%.2fkHz.\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.38 : Page number 702"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 39,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=1.78V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=470.0;                 #Feedback resistor, kΩ\n",
-    "R1=4.3;                 #Input resistor of 1st op-Amp, kΩ\n",
-    "R2=33.0;                 #Input resistor of 2nd op-Amp, kΩ\n",
-    "R3=33.0;                 #Input resistor of 3rd op-Amp, kΩ\n",
-    "Vin=80.0;                 #Input voltage, μV.\n",
-    "\n",
-    "#Calculation\n",
-    "A1=1+Rf/R1;                 #Gain of first op-Amp\n",
-    "A2=-round(Rf/R2,1);         #Gain of second op-Amp\n",
-    "A3=-round(Rf/R3,1);         #Gain of third op-Amp\n",
-    "A=A1*A2*A3;                 #Overall gain\n",
-    "Vout=A*Vin*10**-6;          #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.2fV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.39 : Page number 702-703"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "R1=30kΩ, R2=15kΩ and R3=10kΩ.\n",
-      "Output voltage=0.729V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A1=10;                  #Voltage gain of 1st op-Amp\n",
-    "A2=-18;                 #Voltage gain of 2nd op-Amp\n",
-    "A3=-27;                 #Voltage gain of 3rd op-Amp\n",
-    "Rf=270;                 #Feedback resistor, kΩ\n",
-    "Vin=150;                #Input voltage, μV  \n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "R1=Rf/(A1-1);                   #Input resistor of 1st op-Amp, kΩ\n",
-    "R2=-Rf/A2;                      #Input resistr of 2nd op-Amp, kΩ\n",
-    "R3=-Rf/A3;                      #Input resistor of 3rd op-Amp, kΩ\n",
-    "\n",
-    "A=A1*A2*A3;                     #overall gain,\n",
-    "Vout=Vin*10**-6*A;              #Output voltage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"R1=%dkΩ, R2=%dkΩ and R3=%dkΩ.\"%(R1,R2,R3));\n",
-    "print(\"Output voltage=%.3fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.40 : Page number 703-704"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 41,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "R1=50kΩ, R2=25kΩ and R3=10kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=500;                 #Feedback resistor, kΩ\n",
-    "A1=-10;                 #Gain of 1st op-Amp\n",
-    "A2=-20;                 #Gain of 2nd op-Amp\n",
-    "A3=-50;                 #Gain of 3rd op-Amp\n",
-    "\n",
-    "#Calculation\n",
-    "R1=-Rf/A1;                      #Input resistor of 1st op-Amp, kΩ\n",
-    "R2=-Rf/A2;                      #Input resistor of 2nd op-Amp, kΩ\n",
-    "R3=-Rf/A3;                      #Input resistor of 3rd op-Amp, kΩ\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"R1=%dkΩ, R2=%dkΩ and R3=%dkΩ.\"%(R1,R2,R3));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.41 : Page number 705"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 42,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The input impedance=17202MΩ and output impedance=8.7e-03Ω.\n",
-      "(ii) The closed loop voltage gain=23.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Zin=2.0;              #Input impedance of op-Amp, MΩ\n",
-    "Zout=75.0;            #Output impedance of op-Amp, Ω\n",
-    "A_OL=200000.0;        #Open-loop voltage gain\n",
-    "Rf=220.0;             #Feedback resistor, kΩ\n",
-    "Ri=10.0;              #Input resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "mv=round(Ri/(Ri+Rf),3);       #Feedback fraction\n",
-    "Zin_NI=Zin*(1+(A_OL*mv));     #Input impedance, MΩ\n",
-    "Zout_NI=Zout/(1+A_OL*mv);     #Output impedance, Ω\n",
-    "\n",
-    "#(ii)\n",
-    "A_CL=1+Rf/Ri;               #Closed loop voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) The input impedance=%dMΩ and output impedance=%.1eΩ.\"%(Zin_NI,Zout_NI));\n",
-    "print(\"(ii) The closed loop voltage gain=%d.\"%A_CL);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.42 : Page number 705-706"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 43,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance=400002MΩ and output impedance=0.38e-03Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "#For voltage follower,\n",
-    "mv=1.0;               #Feedback fraction\n",
-    "A_OL=200000.0;        #Open-loop voltage gain\n",
-    "Zin=2.0;              #Input impedance of op-Amp, MΩ\n",
-    "Zout=75.0;            #Output impedance of op-Amp, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Zin_VF=Zin*(1+(A_OL*mv));     #Input impedance, MΩ\n",
-    "Zout_VF=round(round(Zout/(1+A_OL*mv),6),5);     #Output impedance, Ω\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance=%dMΩ and output impedance=%.2fe-03Ω.\"%(Zin_VF,Zout_VF*1000));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.43 : Page number 706"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 44,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance=1kΩ and output impedance=50Ω.\n",
-      "Closed-loop voltage gain=-100\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=100;             #Feedback resistor, kΩ\n",
-    "Ri=1.0;               #Input resistor, kΩ\n",
-    "Zin=4;              #Input impedance of op-Amp, MΩ\n",
-    "Zout=50;            #Output impedance of op-Amp, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Zin_I=Ri;                   #Input impedance, kΩ\n",
-    "Zout_I=Zout;                #Output impedance, Ω\n",
-    "A_CL=-(Rf/Ri);              #Closed loop voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance=%dkΩ and output impedance=%dΩ.\"%(Zin_I,Zout_I));\n",
-    "print(\"Closed-loop voltage gain=%d\"%A_CL);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.44 : Page number 709"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-12V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=10;               #Feedback resistor, kΩ\n",
-    "Ri=10;               #Input resistor, kΩ\n",
-    "V1=3;                #Input voltage 1st, V\n",
-    "V2=1;                #Input voltage 2nd, V\n",
-    "V3=8;                #Input voltage 3rd, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Rf=Ri, Vout=-(Rf/Ri)*(V1+V2+V3)= -(V1+V2+V3);\n",
-    "Vout=-(V1+V2+V3);           #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.45 : Page number 709"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 46,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-7V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=10;               #Feedback resistor, kΩ\n",
-    "R1=1;                #Input resistor for input 1, kΩ\n",
-    "R2=1;                #Input resistor for input 2, kΩ\n",
-    "V1=0.2;              #Input voltage 1st, V\n",
-    "V2=0.5;              #Input voltage 2nd, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "R=R1;                       #Input resistor(=R1 or R2), kΩ\n",
-    "Vout=-(Rf/R)*(V1+V2);       #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.46 : Page number 709-710"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 47,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-2.5V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=1;                #Feedback resistor, kΩ\n",
-    "Ri=10.0;               #Input resistor, kΩ\n",
-    "V1=10;                #Input voltage 1st, V\n",
-    "V2=8.0;                #Input voltage 2nd, V\n",
-    "V3=7.0;                #Input voltage 3rd, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Vout=-(Rf/Ri)*(V1+V2+V3);\n",
-    "Vout=-(Rf/Ri)*(V1+V2+V3);           #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.1fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.47 : Page number 710"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 48,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=2.5V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V1=0.6;                 #Input voltage to 1st input resistor, V\n",
-    "V2=-1.4;                #Input voltage to 2nd input resistor, V\n",
-    "Rf=200;                 #Feedback resistor, kΩ\n",
-    "R1=400;                 #Input resistor 1, kΩ\n",
-    "R2=100.0;                 #Input resistor 2, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "Vout=-Rf*(V1/R1 +V2/R2);                 #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.1fV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.48 : Page number 710-711"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 49,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The output voltage=-12.5V\n",
-      "(ii) The output voltage=-7.5V\n",
-      "(iii) The output voltage=-17.5V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=1.0;                 #Feedback resistor, kΩ\n",
-    "R1=1.0;                 #Input resistor 1, kΩ\n",
-    "R2=2.0;                 #Input resistor 2, kΩ\n",
-    "R3=4.0;                 #Input resistor 3, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Rf_R1=Rf/R1;                #Ratio of feedback resistor and 1st input resistor\n",
-    "Rf_R2=Rf/R2;                #Ratio of feedback resistor and 2nd input resistor\n",
-    "Rf_R3=Rf/R3;                #Ratio of feedback resistor and 3rd input resistor\n",
-    "\n",
-    "#(i) First input combination\n",
-    "V1=10;                                   #Input voltage to 1st input resistor, V\n",
-    "V2=0;                                    #Input voltage to 2nd input resistor, V\n",
-    "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-    "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-    "print(\"(i) The output voltage=%.1fV\"%Vout);\n",
-    "\n",
-    "#(i) First input combination\n",
-    "V1=0;                                   #Input voltage to 1st input resistor, V\n",
-    "V2=10;                                    #Input voltage to 2nd input resistor, V\n",
-    "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-    "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-    "print(\"(ii) The output voltage=%.1fV\"%Vout);\n",
-    "\n",
-    "\n",
-    "#(i) First input combination\n",
-    "V1=10;                                   #Input voltage to 1st input resistor, V\n",
-    "V2=10;                                    #Input voltage to 2nd input resistor, V\n",
-    "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-    "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-    "print(\"(iii) The output voltage=%.1fV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.49 : Page number 711"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 50,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vout=-[0.5sin(1000t)+0.33sin(3000t)]V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=330;                 #Feedback resistor, kΩ\n",
-    "R1=33.0;                  #Input resistor 1, kΩ\n",
-    "R2=10.0;                  #Input resistor 2, kΩ\n",
-    "V1_m=50;                #Peak voltage of 1st input, mV\n",
-    "V2_m=10;                #Peak voltage of 2nd input, mV\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Vout=-((Rf/R1)*V1 + (Rf/R2)*V2)\n",
-    "print(\"Vout=-[%.1fsin(1000t)+%.2fsin(3000t)]V\"%((V1_m/1000.0)*(Rf/R1),(V2_m/1000.0)*(Rf/R2)));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.50 : Page number 715"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 51,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vo=-1*(1/RC)∫vi dt.\n",
-      "=>Vo=-1*(1/1)∫vi dt\n",
-      "=>Vo=∫vi dt\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=100;                      #Input resistor, kΩ\n",
-    "C=10;                       #Feedback capacitor, μF\n",
-    "\n",
-    "#Calculation\n",
-    "RC=R*10**3*C*10**-6;        #product of input resistance and feedback capacitance, s\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vo=-1*(1/RC)∫vi dt.\");\n",
-    "print(\"=>Vo=-1*(1/%d)∫vi dt\"%RC);\n",
-    "print(\"=>Vo=∫vi dt\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.51 : Page number 715-716"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 52,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The critical frequency=159Hz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "Rf=100;                      #Feedback resistor, kΩ\n",
-    "C=0.01;                      #Feedback capacitor, μF\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "fc=1/(2*pi*Rf*1000*C*10**-6);               #Crictical frequency, Hz\n",
-    "\n",
-    "#Result\n",
-    "print(\"The critical frequency=%dHz.\"%fc);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.52 : Page number 716"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 53,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Vout=-1*(1/RC)∫vi dt.\n",
-      "    ΔVout/dt = -vin/RC = -50mV/μs.\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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/w4w8f/Q0SQ0NhCDpTElzJc2W9FNJ6zWyP7NaJk2C73/fd2ibDUfdYzFJ2hmY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\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7ff1b9cb68d0>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "R=10.0;                      #Input resistor, kΩ\n",
-    "C=0.01;                    #Feedback capacitor, μF\n",
-    "vin=5;                     #Input voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "Vout_change_rate=-vin/(R*C);            #Rate of change of output voltage, V/μs   \n",
-    "print(\"(i) Vout=-1*(1/RC)∫vi dt.\");\n",
-    "print(\"    ΔVout/dt = -vin/RC = %dmV/μs.\"%Vout_change_rate);\n",
-    "\n",
-    "#(ii) Plotting the output waveform\n",
-    "vin_plot=[];                    #Plotting variable for input waveform, V\n",
-    "dt=100;                          #time between edges, μs\n",
-    "for i in range(0,3*dt+1):\n",
-    "    if i<dt or i>2*dt :\n",
-    "        vin_plot.append(0);\n",
-    "    else:\n",
-    "        vin_plot.append(5); \n",
-    "\n",
-    "plt.subplot(211);\n",
-    "plt.plot(vin_plot);\n",
-    "plt.xlim([0,300])\n",
-    "plt.ylim([-5,10])\n",
-    "plt.xlabel(\"t(microsecond)\");\n",
-    "plt.ylabel(\"Vin(V)\");\n",
-    "plt.title(\"Input waveform\");\n",
-    "\n",
-    "        \n",
-    "vout_plot=[];                  #Plotting variable for output waveform, V\n",
-    "t=[i for i in range(0,301)];  #Time scale, μs\n",
-    "for i in t[:] :\n",
-    "    if i<dt:\n",
-    "        vout_plot.append(0);\n",
-    "    elif i>2*dt:\n",
-    "        vout_plot.append((Vout_change_rate/1000.0)*dt);\n",
-    "    else :\n",
-    "        vout_plot.append((-vin_plot[i]/(R*C))/1000*(i-dt));\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout_plot);\n",
-    "plt.xlim([0,300])\n",
-    "plt.ylim([-5,5]);\n",
-    "plt.xlabel('t(microsecond)');\n",
-    "plt.ylabel(\"Vout(V)\");\n",
-    "plt.title(\"output waveform\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.53 : Page number 716-717"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 54,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vout=-1*(1/RC)∫vi dt.\n",
-      "Vout=-5*t volts\n",
-      "Time required=2.6seconds.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_supply=15;              #Supply voltage, V\n",
-    "R=10;                     #Input resistor, kΩ\n",
-    "C=0.2;                    #Feedback capacitor, μF\n",
-    "vin=10;                   #Input voltage, mV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Vs=-V_supply+2;         #Saturation voltage, V\n",
-    "print(\"Vout=-1*(1/RC)∫vi dt.\");\n",
-    "print(\"Vout=%d*t volts\"%(-vin/(R*C)));\n",
-    "t=Vs/(-vin/(R*C));                      #Time required, seconds\n",
-    "print(\"Time required=%.1fseconds.\"%t);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.54 : Page number 717-718"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 55,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vo=-5V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=1;                     #Feedback resistor, kΩ\n",
-    "C=0.1;                   #Input capacitor, μF\n",
-    "Vin_change=5;            #Change in input voltage, V\n",
-    "t=0.1;                   #Time taken for change in input voltage, ms\n",
-    "\n",
-    "#Calcualtion\n",
-    "dvi_dt=Vin_change/(t/1000);        #Rate of change of input voltage, V/s\n",
-    "RC=R*1000*C*10**-6;                #Product of feedback resistance and input capacitance, s\n",
-    "#Since, Vo=-R*C*(dvi/dt);\n",
-    "Vo=-RC*dvi_dt;                      #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vo=%dV.\"%Vo);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.55 : Page number 718"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 56,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vo=-0.55V.\n",
-      "The output voltage stays constant at -0.55V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=10;                     #Feedback resistor, kΩ\n",
-    "C=2.2;                   #Input capacitor, μF\n",
-    "Vin_change=10;            #Change in input voltage, V\n",
-    "t=0.4;                   #Time taken for change in input voltage, s\n",
-    "\n",
-    "#Calcualtin\n",
-    "dvi_dt=Vin_change/t;        #Rate of change of input voltage, V/s\n",
-    "RC=R*1000*C*10**-6;         #Product of feedback resistance and input capacitance, s\n",
-    "#Since, Vo=-R*C*(dvi/dt);\n",
-    "Vo=-RC*dvi_dt;                      #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vo=%.2fV.\"%Vo);\n",
-    "print(\"The output voltage stays constant at %.2fV.\"%Vo);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.56 : Page number 718-719"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 58,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "vo=-1*(dvi/dt).\n",
-      "vo=-5V.\n",
-      "Therefore, between 0 to 0.2s, the output voltage is constant at -5V.\n",
-      "For t>0.2s, the input is constant so that output voltage is zero.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=100;                     #Feedback resistor, kΩ\n",
-    "C=10;                   #Input capacitor, μF\n",
-    "Vin_change=1;            #Change in input voltage, V\n",
-    "t=0.2;                   #Time taken for change in input voltage, s\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "RC=R*1000*C*10**-6;             #Product of feedback resistor and input capacitance, s\n",
-    "#(i)\n",
-    "print(\"vo=-%d*(dvi/dt).\"%RC);\n",
-    "\n",
-    "#(ii)\n",
-    "dvi_dt=Vin_change/t;                #Rate of change of input voltage, V\n",
-    "vo=-dvi_dt;                         #Output voltage, V\n",
-    "print(\"vo=%dV.\"%vo);\n",
-    "\n",
-    "print(\"Therefore, between 0 to 0.2s, the output voltage is constant at %dV.\"%vo);\n",
-    "print(\"For t>0.2s, the input is constant so that output voltage is zero.\");\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_4.ipynb
deleted file mode 100755
index 92d6f1ec..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_4.ipynb
+++ /dev/null
@@ -1,2522 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 25 : OPERATIONAL AMPLIFIERS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.1: Page number 664"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output voltage of the differential amplifier = 10V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A=100.0;              #Open-circuit voltage gain of differential amplifier\n",
-    "V1=3.25;              #Input voltage to terminal 1 in V\n",
-    "V2=3.15;              #Input voltage to terminal 2 in V\n",
-    "\n",
-    "#Calculations\n",
-    "V0=A*(V1-V2);         #Output voltage in V\n",
-    "\n",
-    "#Results\n",
-    "print(\"The output voltage of the differential amplifier = %dV\"%V0);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.2: Page number 672"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The common mode rejection ratio = 10000.\n",
-      "The common mode rejection ratio in decibels= 80dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "A_DM=2000.0;              #Differential mode voltage gain\n",
-    "A_CM=0.2;                 #Common mode voltage gain\n",
-    "\n",
-    "#Calculations\n",
-    "CMRR=A_DM/A_CM;           #Common mode rejection ratio\n",
-    "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"The common mode rejection ratio = %d.\"%CMRR);\n",
-    "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.3: Page number 672"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The common mode rejection ratio in decibels= 46dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "VD_in=10.0;               #Differential mode input in mV\n",
-    "VD_out=1.0;               #Output for differential mode input in V\n",
-    "VC_in=10.0;                 #Common mode input in mV\n",
-    "VC_out=5.0;                 #Output for common mode input in mV\\\n",
-    "\n",
-    "#Calculations\n",
-    "A_DM=(VD_out*1000)/VD_in;       #Differntial mode voltage gain\n",
-    "A_CM=VC_out/VC_in;       #Common mode voltage gain\n",
-    "CMRR=A_DM/A_CM;           #Common mode rejection ratio\n",
-    "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-    "\n",
-    "#Results\n",
-    "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.4: Page number 672"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage =7.5V\n",
-      "Noise on output = 4.7x10^-6V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_DM=150.0;           #Differential mode voltage gain\n",
-    "CMRR_dB=90.0;         #Common mode rejection ratio\n",
-    "V1=100.0;             #Input voltage for terminal 1 in mV\n",
-    "V2=50.0;              #Input voltage for terminal 2 in mV\n",
-    "V_noise=1.0;          #Voltage of noise signal in mV\n",
-    "\n",
-    "#Calculation\n",
-    "\n",
-    "#Case(i)\n",
-    "V_out=A_DM*(V1-V2)/1000.0;            #Output voltage for differntial mode input, in V\n",
-    "\n",
-    "#Since CMRR_dB=20*log10(differential mode gain/common mode gain),\n",
-    "A_CM=A_DM/pow(10,(CMRR_dB/20));         #Common mode gain\n",
-    "V_OUT_noise=A_CM*(V_noise/1000);        #Noise on output in V\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"Output voltage =%.1fV\"%V_out);\n",
-    "print(\"Noise on output = %.1fx10^-6V\"%(V_OUT_noise*pow(10,6)));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.5 : Page number 672-673"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Common mode gain =0.083\n",
-      "Common mode rejection ratio in decibels=89.5dB\n",
-      "r.m.s output signal =1.25V\n",
-      "r.m.s interfernce output voltage = 83mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "A_DM=2500.0;                    #Differential mode voltage gain\n",
-    "CMRR=30000.0;                   #Common mode rejection ratio\n",
-    "Input_signal=500.0;             #Single ended input r.m.s signal in microvolts\n",
-    "Interference=1.0;               #Interference signal, in V\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#(i)\n",
-    "A_CM=A_DM/CMRR;                 #Common mode gain\n",
-    "\n",
-    "#(ii)\n",
-    "CMRR_dB=20*log10(CMRR);         #Common mode rejection ratio in decibels\n",
-    "\n",
-    "#(iii)\n",
-    "V_out=A_DM*(Input_signal/pow(10,6)-0);        #r.m.s output signal  in V\n",
-    "\n",
-    "#(iv)\n",
-    "Interference_out=A_CM*Interference;          #r.m.s interference output in V\n",
-    "Interference_out=Interference_out*1000;      #r.m.s interference output in mV\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"Common mode gain =%.3f\"%A_CM);\n",
-    "print(\"Common mode rejection ratio in decibels=%.1fdB\"%CMRR_dB);\n",
-    "print(\"r.m.s output signal =%.2fV\"%V_out);\n",
-    "print(\"r.m.s interfernce output voltage = %dmV\"%Interference_out);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.6 : Page number 674-675"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VE=-0.7V\n",
-      "IE=0.452mA\n",
-      "IE1=0.226mA\n",
-      "IE2=0.226mA\n",
-      "IC1=0.226mA\n",
-      "IC2=0.226mA\n",
-      "IB1=2.26μA\n",
-      "IB2=2.26μA\n",
-      "VC1=12V\n",
-      "VC2=9.7V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RB=10;                  #Base resistor, kΩ\n",
-    "RC2=10;                 #Collector resistor, kΩ\n",
-    "RE=25;                  #Emitter resistor, kΩ\n",
-    "VBE=0.7;               #Base-emitter voltage, V\n",
-    "beta=100;               #Base amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-    "IE=(VEE-VBE)/RE;            #Tail current, mA\n",
-    "IE1=IE/2;                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-    "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-    "IC2=IC1;                    #Collector current of 2nd transistor, mA\n",
-    "IB1=(IC1/beta)*1000;        #Base current of 1st transistor, μA\n",
-    "IB2=IB1;                    #Base current of 2nd transistor, μA\n",
-    "VC1=VCC;                    #Collector voltage of 1st transistor, V\n",
-    "VC2=VCC-IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"VE=%.1fV\"%VE);\n",
-    "print(\"IE=%.3fmA\"%IE);\n",
-    "print(\"IE1=%.3fmA\"%IE1);\n",
-    "print(\"IE2=%.3fmA\"%IE2);\n",
-    "print(\"IC1=%.3fmA\"%IC1);\n",
-    "print(\"IC2=%.3fmA\"%IC2);\n",
-    "print(\"IB1=%.2fμA\"%IB1);\n",
-    "print(\"IB2=%.2fμA\"%IB2);\n",
-    "print(\"VC1=%dV\"%VC1);\n",
-    "print(\"VC2=%.1fV\"%VC2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.7 : Page number 675"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output voltage=7.85V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15;                 #Collector supply voltage, V\n",
-    "VEE=15;                 #Emitter supply voltage, V\n",
-    "RB=33;                  #Base resistor, kΩ\n",
-    "RC=15;                  #Collector resistor, kΩ\n",
-    "RE=15;                  #Emitter resistor, kΩ\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "IE_tail=(VEE-VBE)/RE;            #Tail current, mA\n",
-    "IE=round(IE_tail/2,3);                    #Emitter current in each transistor, mA\n",
-    "IC=IE;                           #Collector current(=emitter current), mA\n",
-    "Vout=VCC-IC*RC;                  #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output voltage=%.2fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.8 : Page number 675"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) IB1=5.56μA\n",
-      "    IB2=4.55μA\n",
-      "(ii) VB1=-0.183V\n",
-      "     VB2=-0.15V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15.0;                 #Collector supply voltage, V\n",
-    "VEE=15.0;                 #Emitter supply voltage, V\n",
-    "RB=33.0;                  #Base resistor, kΩ\n",
-    "RC=15.0;                  #Collector resistor, kΩ\n",
-    "RE=15.0;                  #Emitter resistor, kΩ\n",
-    "VBE=0;                  #Base-emitter voltage, V\n",
-    "beta_dc_l=90.0;           #base current amplification factor for left transistor\n",
-    "beta_dc_r=110.0;          #base current amplification factor for right transistor\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IE_tail=(VEE-VBE)/RE;             #Tail current, mA\n",
-    "IE=IE_tail/2;                      #Emitter current in each transistor, mA\n",
-    "IB1=(IE/beta_dc_l)*1000;          #Base current of 1st transistor, μA\n",
-    "IB2=(IE/beta_dc_r)*1000;          #Base current of 2nd transistor, μA\n",
-    "\n",
-    "#(ii)\n",
-    "VB1=-IB1/1000*RB;                    #Base voltage of 1st transistor, V\n",
-    "VB2=-IB2/1000*RB;                    #Base voltage of 1st transistor, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) IB1=%.2fμA\"%IB1);\n",
-    "print(\"    IB2=%.2fμA\"%IB2);\n",
-    "print(\"(ii) VB1=%.3fV\"%VB1);\n",
-    "print(\"     VB2=%.2fV\"%VB2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.9 : Page number 675-676"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VE=-0.7V\n",
-      "Emitter current in each transistor=0.5mA.\n",
-      "IC1~IE1=0.5mA and IC2~IE2=0.5mA\n",
-      "VC1=VC2=10V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15.0;                 #Collector supply voltage, V\n",
-    "VEE=15.0;                 #Emitter supply voltage, V\n",
-    "RB=10.0;                  #Base resistor, kΩ\n",
-    "RC1=10.0;                 #Collector resistor of 1st transistor, kΩ\n",
-    "RC2=10.0;                 #Collector resistor of 2nd transistor, kΩ\n",
-    "IE=1.0;                   #Tail current, mA\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-    "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-    "IC2=IE2;                    #Collector current of 2nd transistor, mA\n",
-    "VC1=VCC-IC1*RC1;            #Collector voltage of 1st transistor, V\n",
-    "VC2=VCC-IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"VE=%.1fV\"%VE);\n",
-    "print(\"Emitter current in each transistor=%.1fmA.\"%(IE/2.0));\n",
-    "print(\"IC1~IE1=%.1fmA and IC2~IE2=%.1fmA\"%(IE1,IE2));\n",
-    "print(\"VC1=VC2=%dV.\"%VC2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.10 : Page number 676-677"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VE=0.7V\n",
-      "Tail current=0.452mA.\n",
-      "Emitter current in each transistor=0.226mA.\n",
-      "IC1~IE1=0.226mA and IC2~IE2=0.226mA\n",
-      "VC1=-12V\n",
-      "VC2=-9.74V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12.0;                 #Collector supply voltage, V\n",
-    "VEE=12.0;                 #Emitter supply voltage, V\n",
-    "RC2=10.0;                 #Collector resistor of 2nd transistor, kΩ\n",
-    "RE=25.0;                  #Emitter current, kΩ\n",
-    "VBE=-0.7;               #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-    "IE=(VCC-VE)/RE;             #Tail current, mA\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-    "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-    "IC2=IE2;                    #Collector current of 2nd transistor, mA\n",
-    "VC1=-VEE;                   #Collector voltage of 1st transistor, V\n",
-    "VC2=-VEE+IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"VE=%.1fV\"%VE);\n",
-    "print(\"Tail current=%.3fmA.\"%IE);\n",
-    "print(\"Emitter current in each transistor=%.3fmA.\"%(IE/2.0));\n",
-    "print(\"IC1~IE1=%.3fmA and IC2~IE2=%.3fmA\"%(IC1,IC2));\n",
-    "print(\"VC1=%dV\"%VC1);\n",
-    "print(\"VC2=%.2fV\"%VC2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.11 : Page number 679"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The input offset current=15.1nA\n",
-      "(ii) The input bias current=75.8nA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15;                 #Collector supply voltage, V\n",
-    "VEE=15;                 #Emitter supply voltage, V\n",
-    "RB=1;                   #Base resistor, MΩ\n",
-    "RC2=1;                  #Collector resistor, MΩ\n",
-    "RE=1;                   #Emitter resistor, MΩ\n",
-    "VBE=0;                  #Base-emitter voltage, V (Neglected)\n",
-    "beta_dc_l=90.0;           #base current amplification factor for left transistor\n",
-    "beta_dc_r=110.0;          #base current amplification factor for right transistor\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IE=(VEE-VBE)/RE;                    #Tail current, μA\n",
-    "IE1=IE/2.0;                           #Emitter current of 1st transistor, μA\n",
-    "IE2=IE1;                            #Emitter current of 2nd transistor, μA\n",
-    "IB1=round((IE1/beta_dc_l)*1000,1);           #Base current of 1st transistor, nA\n",
-    "IB2=round((IE2/beta_dc_r)*1000,1);           #Base current of 2nd transistor, nA\n",
-    "I_in_offset=IB1-IB2;                 #Input offset current, nA\n",
-    "\n",
-    "#(ii)\n",
-    "I_in_bias=(IB1+IB2)/2;                  #Input bias current, nA\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  The input offset current=%.1fnA\"%I_in_offset);\n",
-    "print(\"(ii) The input bias current=%.1fnA\"%I_in_bias);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.12 : Page number 679"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The two base currents are: IB1=90nA and IB2=70nA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_in_offset=20;                 #Input offset current, nA\n",
-    "I_in_bias=80;                   #Input bias current, nA\n",
-    "\n",
-    "#Calculation\n",
-    "IB1=I_in_bias+I_in_offset/2;            #Base current in 1st transistor, nA\n",
-    "IB2=I_in_bias-I_in_offset/2;            #Base current in 2nd transistor, nA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The two base currents are: IB1=%dnA and IB2=%dnA.\"%(IB1,IB2));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.13 : Page number 679-680"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input offset voltage=2mV.\n",
-      "The output offset voltage=0.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declration\n",
-    "I_in_offset=20;                 #Input offset current, nA\n",
-    "I_in_bias=80;                   #Input bias current, nA\n",
-    "A=150;                          #Voltage gain\n",
-    "RB=100;                         #Base resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_io=(I_in_offset*10**-9*RB*1000)*1000;       #Input offset voltage, mV\n",
-    "V_out_offset=(A*V_io)/1000;                     #Output offset voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input offset voltage=%dmV.\"%V_io);\n",
-    "print(\"The output offset voltage=%.1fV.\"%V_out_offset);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.14 : Page number 682"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Output voltage=0.15V.\n",
-      "(ii) Output voltage=-0.15V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15;                 #Collector supply voltage, V\n",
-    "VEE=15;                 #Emitter supply voltage, V\n",
-    "RE=1;                   #Emitter resistor, MΩ\n",
-    "RC=1;                   #Collector resistor, MΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "IE=VEE/RE;                  #Tail current, μA\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, μA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, μA\n",
-    "re=25/IE1;                  #a.c emitter resistance, kΩ\n",
-    "A_DM=RC/(2.0*re);        #Differential voltage gain,\n",
-    "\n",
-    "#(i)\n",
-    "vin=1;                      #Input voltage, V\n",
-    "Vout=A_DM*vin;              #Output voltage, V\n",
-    "\n",
-    "print(\"(i) Output voltage=%.2fV.\"%Vout);\n",
-    "\n",
-    "#(ii)\n",
-    "vin=-1;                      #Input voltage, V\n",
-    "Vout=A_DM*vin;              #Output voltage, V;\n",
-    "print(\"(ii) Output voltage=%.2fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.15 : Page number 682-683"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The input impedance=194kΩ.\n",
-      "(ii) The differential voltage gain=136.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RE=100;                 #Emitter resistor, kΩ\n",
-    "RC1=120;                #Collector resistor of 1st transistor, kΩ\n",
-    "RC2=120;                #Collector resistor of 2nd transistor, kΩ\n",
-    "beta=220;               #Base amplification factor\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calcualtion\n",
-    "IE=((VEE-VBE)/RE)*1000;     #Tail current, μA\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, μA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, μA\n",
-    "re=(25/IE1)*1000;           #a.c emitter resistance, Ω\n",
-    "Zin=2*beta*re/1000;         #Input impedance, kΩ\n",
-    "A_DM=RC1*1000/(2.0*re);             #Differential voltage gain,\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) The input impedance=%dkΩ.\"%Zin);\n",
-    "print(\"(ii) The differential voltage gain=%.0f.\"%A_DM);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.16: Page number 683-684"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Differential voltage gain=56.6.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RE=200;                 #Emitter resistor, kΩ\n",
-    "RC=100;                 #Collector resistor, kΩ\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "IE=round((VEE-VBE)/RE,4);            #Tail current, mA\n",
-    "IE1=round(IE/2,4);                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                             #Emitter current of 2nd transistor, mA\n",
-    "re=round(25/IE1,1);                  #a.c emitter resistance, Ω\n",
-    "A_DM=RC*1000/(2*re);                 #Differential voltage gain,\n",
-    "\n",
-    "#Result\n",
-    "print(\"Differential voltage gain=%.1f.\"%A_DM);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.17 : Page number 685-686"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Common mode rejection ratio=666.7.\n",
-      "Common mode rejection ratio in decibel=56.48dB\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "v1=0.5;               #Voltage in terminal 1, mV\n",
-    "v2=-0.5;               #Voltage in terminal 2, mV\n",
-    "vo=8.0;                   #Output voltage, V\n",
-    "vo_cm=12.0;               #Common mode output, mV\n",
-    "\n",
-    "#Calculation\n",
-    "vin=v1-v2;                  #Differential input, mV\n",
-    "A_DM=vo/(vin/1000.0);         #Differential mode gain,\n",
-    "vin_cm=1;                   #Common mode input, mV\n",
-    "A_CM=vo_cm/vin_cm;          #Common mode gain\n",
-    "CMRR=A_DM/A_CM;             #Common mode rejection ratio\n",
-    "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-    "\n",
-    "#Result\n",
-    "print(\"Common mode rejection ratio=%.1f.\"%CMRR)\n",
-    "print(\"Common mode rejection ratio in decibel=%.2fdB\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.18 : Page number 686"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Common mode voltage gain=6.32.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_DM=200000;                #Differential mode gain\n",
-    "CMRR_dB=90;                 #Common mode rejection ratio, dB\n",
-    "\n",
-    "#Calculation\n",
-    "CMRR=10**(CMRR_dB/20.0);              #Common mode rejection ratio\n",
-    "A_CM=A_DM/CMRR;                     #Common mode gain\n",
-    "\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Common mode voltage gain=%.2f.\"%A_CM);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.19 : Page number 686"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The Common mode gain=0.0081\n",
-      "(ii) The common mode rejection ratio=81.8dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "vin_cm=3.2;                     #Common input voltage, V\n",
-    "vout=26;                        #Output voltage, V\n",
-    "A_DM=100;                       #Open-circuit voltage gain\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CM=vout*10**-3/vin_cm;                #Common mode gain\n",
-    "\n",
-    "#(ii)\n",
-    "CMRR_dB=20*log10(A_DM/A_CM);            #Common mode rejection ratio, dB\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  The Common mode gain=%.4f\"%A_CM);\n",
-    "print(\"(ii) The common mode rejection ratio=%.1fdB.\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.20 : Page number 686-687"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Common mode gain=0.25\n",
-      "(ii)Common mode rejection ratio=47.09dB\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "from math import floor\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RE=200.0;                 #Emitter resistor, kΩ\n",
-    "RC=100.0;                 #Collector resistor, kΩ\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CM=round(RC/(2*RE),2);             #Common mode voltage gain\n",
-    "\n",
-    "#(ii)\n",
-    "IE=round((VEE-VBE)/RE,4);            #Tail current, mA\n",
-    "IE1=round(IE/2,4);                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                             #Emitter current of 2nd transistor, mA\n",
-    "re=round(25/IE1,1);                  #a.c emitter resistance, Ω\n",
-    "A_DM=RC*1000/(2*re);                 #Differential voltage gain,\n",
-    "CMRR_dB=floor(20*log10(A_DM/A_CM)*100)/100;         #Common mode rejection ratio, dB\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) Common mode gain=%.2f\"%A_CM);\n",
-    "print(\"(ii)Common mode rejection ratio=%.2fdB\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.21 : Page number 691"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "f2=30kHz\n",
-      "ACL=75 or 37.5dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "ACL=500;                    #closed loop gain\n",
-    "f_unity=15;                 #frequency with cloased-loop unity gain, MHz\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "f2=f_unity*1000/500                 #Upper frequency of bandwidth,kHz\n",
-    "BW=f2-0;                            #Bandwidth, kHz\n",
-    "A_CL=f_unity*1000/200;                    #Maximum value of A_CL when f2=200kHz\n",
-    "A_CL_dB=20*log10(A_CL);             #Maximum value of A_CL in decibel\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"f2=%dkHz\"%f2);\n",
-    "print(\"ACL=%d or %.1fdB.\"%(A_CL,A_CL_dB));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.22 : Page number 691-692"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Operating Bandwidth=1.5MHz.\n",
-      "(ii) Operating Bandwidth=150kHz.\n",
-      "(iii) Operating Bandwidth=15kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "GBW=1.5;                    #Gain-bandwidth, MHz\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) For A_CL=1;\n",
-    "A_CL=1;                         #Closed loop gain\n",
-    "BW=GBW/A_CL;                    #Bandwidth, MHz\n",
-    "\n",
-    "print(\"(i) Operating Bandwidth=%.1fMHz.\"%BW);\n",
-    "\n",
-    "#(ii) For A_CL=10;\n",
-    "A_CL=10;                         #Closed loop gain\n",
-    "BW=(GBW/A_CL)*1000;              #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(ii) Operating Bandwidth=%dkHz.\"%BW);\n",
-    "\n",
-    "#(iii) For A_CL=100;\n",
-    "A_CL=100;                         #Closed loop gain\n",
-    "BW=(GBW/A_CL)*1000;               #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(iii) Operating Bandwidth=%dkHz.\"%BW);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.23 : Page number 692"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Maximum operating frequency=9.95kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "slew_rate=0.5;                  #Slew rate, V/μs\n",
-    "V_supply=10;                     #Supply voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_sat=V_supply-2;                           #Saturation voltage, V\n",
-    "V_pk=V_sat;                                 #Maximum peak-output voltage, V\n",
-    "f_max=((slew_rate*10**6)/(2*pi*V_pk))/1000;       #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Maximum operating frequency=%.2fkHz.\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.24 : Page number 692"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Maximum operating frequency=796kHz\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "slew_rate=0.5;                  #Slew rate, V/μs\n",
-    "V_pk=100.0;                       #Peak-output voltage, mV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_pk=V_pk/1000.0;                                #Peak-output voltage, V\n",
-    "f_max=(slew_rate*10**6/(2*pi*V_pk))/1000.0;      #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Maximum operating frequency=%.0fkHz\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.25 : Page number 695-696"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Feedback resistor=220kΩ\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_CL=-100;              #Closed-loop voltage gain\n",
-    "Ri=2.2;                 #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, A_CL=-(Rf/Ri)\n",
-    "Rf=-A_CL*Ri;                #Feedback resistor, kΩ\n",
-    "\n",
-    "#Result\n",
-    "print(\"Feedback resistor=%dkΩ\"%Rf);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.26 : Page number 696"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-0.25V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "vin=2.5;                #Input voltage, mV\n",
-    "Rf=200;                 #Feedback resistor, kΩ\n",
-    "Ri=2;                   #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "vout=A_CL*vin/1000;                  #Output voltage,V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.2fV\"%vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.27 : Page number 696"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 28,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Closed-loop voltage gain=-1\n",
-      "Therefore, output will have same amplitude but 180° phase inversion.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Varaiable declaration\n",
-    "Rf=1.0;                   #Feedback resistor, kΩ\n",
-    "Ri=1.0;                   #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"Closed-loop voltage gain=%d\"%A_CL);\n",
-    "print(\"Therefore, output will have same amplitude but 180° phase inversion.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.28 : Page number 696-697"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 29,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Closed-loop voltage gain=-40\n",
-      "Supply voltage=±15V, saturation voltage=±13V. Since gain=-40, op-Amp will be driven to saturation.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=40;                   #Feedback resistor, kΩ\n",
-    "Ri=1;                   #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Closed-loop voltage gain=%d\"%A_CL);\n",
-    "print(\"Supply voltage=±15V, saturation voltage=±13V. Since gain=-40, op-Amp will be driven to saturation.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.29 : Page number 697"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 30,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  A_CL=-10.\n",
-      "(ii) Zi=10kΩ\n",
-      "(iii) Maximum operating frequency=15.9kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "Rf=100;                   #Feedback resistor, kΩ\n",
-    "Ri=10;                    #Input resistor, kΩ\n",
-    "Vpp=1;                    #Input peak-peak voltage, V\n",
-    "slew_rate=0.5;            #Slew rate, V//μs\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "\n",
-    "#(ii)\n",
-    "Zi=Ri;                  #Input impedance(~ Input resistor), kΩ\n",
-    "\n",
-    "#(iii)\n",
-    "Vout=A_CL*Vpp;                           #Peak-to-peak voltage, V\n",
-    "Vpk=Vout/2;                              #Peak output voltage, V\n",
-    "f_max=(slew_rate*10**6/(2*pi*abs(Vpk)))/1000;        #Maximum operating frequency, kHz\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  A_CL=%d.\"%A_CL);\n",
-    "print(\"(ii) Zi=%dkΩ\"%Zi);\n",
-    "print(\"(iii) Maximum operating frequency=%.1fkHz.\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.30 : Page number 697"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Rf=20kΩ and Ri=5kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_CL=-4;                    #Closed loop voltage gain\n",
-    "R=[1.0,5.0,10.0,20.0];              #List of available resistors, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "for i in R[:]:\n",
-    "    for j in R[:]:\n",
-    "        if -(i/j)==A_CL :\n",
-    "            print(\"Rf=%dkΩ and Ri=%dkΩ.\"%(i,j));\n",
-    "            break;\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.31 : Page  number 697-698"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 32,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Closed loop voltage gain=-100.\n",
-      "(ii) Closed loop voltage gain=-50.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=100;                   #Feedback resistor, kΩ\n",
-    "Ri=1;                    #Input resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "R_source=0;               #Source resistor, kΩ\n",
-    "A_CL=-Rf/(R_source+Ri);    #Closed-loop voltage gain\n",
-    "\n",
-    "print(\"(i) Closed loop voltage gain=%d.\"%A_CL);\n",
-    "\n",
-    "#(ii)\n",
-    "R_source=1;               #Source resistor, kΩ\n",
-    "A_CL=-Rf/(R_source+Ri);    #Closed-loop voltage gain\n",
-    "\n",
-    "print(\"(ii) Closed loop voltage gain=%d.\"%A_CL);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.32 : Page number 699-700"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 33,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=12.12mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=240;                   #Feedback resistor, kΩ\n",
-    "Ri=2.4;                    #Input resistor, kΩ\n",
-    "Vin=120;                    #Input voltage, μV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=1+(Rf/Ri);             #Closed loop voltage gain\n",
-    "Vout=(A_CL*Vin)/1000;       #Output voltage, mV\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.2fmV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.33 : Page number 700"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Output voltage=11V\n",
-      "(ii) Output voltage=-11V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=10;                   #Feedback resistor, kΩ\n",
-    "Ri=1;                    #Input resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=1+(Rf/Ri);             #Closed loop voltage gain\n",
-    "#(i)\n",
-    "Vin=1;                      #Input voltage, V\n",
-    "Vout=A_CL*Vin;              #Output voltage, V\n",
-    "\n",
-    "print(\"(i)  Output voltage=%dV\"%Vout);\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "Vin=-1;                      #Input voltage, V\n",
-    "Vout=A_CL*Vin;               #Output voltage, V\n",
-    "\n",
-    "print(\"(ii) Output voltage=%dV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.34 : Page number 700"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 35,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Peak to peak output voltage=12V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=5;                       #Feedback resistor, kΩ\n",
-    "Ri=1;                       #Input resistor, kΩ\n",
-    "Vin_max=1;                  #Maximum input voltage, V\n",
-    "Vin_min=-1;                 #Minimum input voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "V_inpp=Vin_max-Vin_min;               #Peak-peak input voltage, V\n",
-    "A_CL=1+(Rf/Ri);                     #Closed loop voltage gain\n",
-    "Vout_pp=A_CL*V_inpp;                #Peak-peak output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Peak to peak output voltage=%dV\"%Vout_pp);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.35 : Page number 700-701"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 36,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Closed-loop voltage gain=11\n",
-      "(ii) Maximum operating frequency=14.47kHz\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "Rf=100;                       #Feedback resistor, kΩ\n",
-    "Ri=10;                       #Input resistor, kΩ\n",
-    "Vpp=1;                    #Input peak-peak voltage, V\n",
-    "slew_rate=0.5;              #Slew rate, V/μs\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CL=1+(Rf/Ri);                                      #Closed loop voltage gain\n",
-    "\n",
-    "#(ii)\n",
-    "Vout_pp=A_CL*Vpp;                                    #Peak-peak output voltage, V\n",
-    "Vpk=Vout_pp/2.0;                                          #Peak output voltage, V\n",
-    "f_max=((slew_rate*10**6)/(2*pi*Vpk))/1000.0;           #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  Closed-loop voltage gain=%d\"%A_CL);\n",
-    "\n",
-    "print(\"(ii) Maximum operating frequency=%.2fkHz\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.36 : Page number 701"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 37,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Bandwidth=44.3kHz.\n",
-      "(ii)  Bandwidth=63.8kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=220;                       #Feedback resistor, kΩ\n",
-    "Ri=3.3;                       #Input resistor, kΩ\n",
-    "unity_gain_BW=3;              #Unity gain bandwidth, MHz\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) For non-inverting amplifier\n",
-    "A_CL=1+(Rf/Ri);                    #Closed loop voltage gain\n",
-    "BW=unity_gain_BW*1000.0/A_CL;        #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(i)  Bandwidth=%.1fkHz.\"%BW);\n",
-    "\n",
-    "#(ii) For inverting amplifier\n",
-    "Rf=47;                          #Feedback resistor, kΩ\n",
-    "Ri=1;                           #Input resistor, kΩ\n",
-    "A_CL=-(Rf/Ri);                    #Closed loop voltage gain\n",
-    "BW=unity_gain_BW*1000.0/abs(A_CL);        #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(ii)  Bandwidth=%.1fkHz.\"%BW);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.37 : Page number 701-702"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 38,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) For voltage follower A_CL=1.\n",
-      "(ii) The maximum output frequency=26.53kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#(i)\n",
-    "A_CL=1;                     #Closed loop voltage gain for voltage follower\n",
-    "print(\"(i) For voltage follower A_CL=1.\");\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "slew_rate=0.5;              #Slew rate, V/μs\n",
-    "V_inpp=6;                   #peak-peak input voltage, V\n",
-    "Vout=A_CL*V_inpp;           #Peak-peak output voltage, V\n",
-    "Vpk=Vout/2;                 #Peak output voltage, V\n",
-    "\n",
-    "f_max=(slew_rate*10**6/(2*pi*Vpk))/1000;       #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"(ii) The maximum output frequency=%.2fkHz.\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.38 : Page number 702"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 39,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=1.78V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=470.0;                 #Feedback resistor, kΩ\n",
-    "R1=4.3;                 #Input resistor of 1st op-Amp, kΩ\n",
-    "R2=33.0;                 #Input resistor of 2nd op-Amp, kΩ\n",
-    "R3=33.0;                 #Input resistor of 3rd op-Amp, kΩ\n",
-    "Vin=80.0;                 #Input voltage, μV.\n",
-    "\n",
-    "#Calculation\n",
-    "A1=1+Rf/R1;                 #Gain of first op-Amp\n",
-    "A2=-round(Rf/R2,1);         #Gain of second op-Amp\n",
-    "A3=-round(Rf/R3,1);         #Gain of third op-Amp\n",
-    "A=A1*A2*A3;                 #Overall gain\n",
-    "Vout=A*Vin*10**-6;          #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.2fV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.39 : Page number 702-703"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "R1=30kΩ, R2=15kΩ and R3=10kΩ.\n",
-      "Output voltage=0.729V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A1=10;                  #Voltage gain of 1st op-Amp\n",
-    "A2=-18;                 #Voltage gain of 2nd op-Amp\n",
-    "A3=-27;                 #Voltage gain of 3rd op-Amp\n",
-    "Rf=270;                 #Feedback resistor, kΩ\n",
-    "Vin=150;                #Input voltage, μV  \n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "R1=Rf/(A1-1);                   #Input resistor of 1st op-Amp, kΩ\n",
-    "R2=-Rf/A2;                      #Input resistr of 2nd op-Amp, kΩ\n",
-    "R3=-Rf/A3;                      #Input resistor of 3rd op-Amp, kΩ\n",
-    "\n",
-    "A=A1*A2*A3;                     #overall gain,\n",
-    "Vout=Vin*10**-6*A;              #Output voltage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"R1=%dkΩ, R2=%dkΩ and R3=%dkΩ.\"%(R1,R2,R3));\n",
-    "print(\"Output voltage=%.3fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.40 : Page number 703-704"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 41,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "R1=50kΩ, R2=25kΩ and R3=10kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=500;                 #Feedback resistor, kΩ\n",
-    "A1=-10;                 #Gain of 1st op-Amp\n",
-    "A2=-20;                 #Gain of 2nd op-Amp\n",
-    "A3=-50;                 #Gain of 3rd op-Amp\n",
-    "\n",
-    "#Calculation\n",
-    "R1=-Rf/A1;                      #Input resistor of 1st op-Amp, kΩ\n",
-    "R2=-Rf/A2;                      #Input resistor of 2nd op-Amp, kΩ\n",
-    "R3=-Rf/A3;                      #Input resistor of 3rd op-Amp, kΩ\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"R1=%dkΩ, R2=%dkΩ and R3=%dkΩ.\"%(R1,R2,R3));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.41 : Page number 705"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 42,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The input impedance=17202MΩ and output impedance=8.7e-03Ω.\n",
-      "(ii) The closed loop voltage gain=23.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Zin=2.0;              #Input impedance of op-Amp, MΩ\n",
-    "Zout=75.0;            #Output impedance of op-Amp, Ω\n",
-    "A_OL=200000.0;        #Open-loop voltage gain\n",
-    "Rf=220.0;             #Feedback resistor, kΩ\n",
-    "Ri=10.0;              #Input resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "mv=round(Ri/(Ri+Rf),3);       #Feedback fraction\n",
-    "Zin_NI=Zin*(1+(A_OL*mv));     #Input impedance, MΩ\n",
-    "Zout_NI=Zout/(1+A_OL*mv);     #Output impedance, Ω\n",
-    "\n",
-    "#(ii)\n",
-    "A_CL=1+Rf/Ri;               #Closed loop voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) The input impedance=%dMΩ and output impedance=%.1eΩ.\"%(Zin_NI,Zout_NI));\n",
-    "print(\"(ii) The closed loop voltage gain=%d.\"%A_CL);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.42 : Page number 705-706"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 43,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance=400002MΩ and output impedance=0.38e-03Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "#For voltage follower,\n",
-    "mv=1.0;               #Feedback fraction\n",
-    "A_OL=200000.0;        #Open-loop voltage gain\n",
-    "Zin=2.0;              #Input impedance of op-Amp, MΩ\n",
-    "Zout=75.0;            #Output impedance of op-Amp, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Zin_VF=Zin*(1+(A_OL*mv));     #Input impedance, MΩ\n",
-    "Zout_VF=round(round(Zout/(1+A_OL*mv),6),5);     #Output impedance, Ω\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance=%dMΩ and output impedance=%.2fe-03Ω.\"%(Zin_VF,Zout_VF*1000));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.43 : Page number 706"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 44,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance=1kΩ and output impedance=50Ω.\n",
-      "Closed-loop voltage gain=-100\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=100;             #Feedback resistor, kΩ\n",
-    "Ri=1.0;               #Input resistor, kΩ\n",
-    "Zin=4;              #Input impedance of op-Amp, MΩ\n",
-    "Zout=50;            #Output impedance of op-Amp, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Zin_I=Ri;                   #Input impedance, kΩ\n",
-    "Zout_I=Zout;                #Output impedance, Ω\n",
-    "A_CL=-(Rf/Ri);              #Closed loop voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance=%dkΩ and output impedance=%dΩ.\"%(Zin_I,Zout_I));\n",
-    "print(\"Closed-loop voltage gain=%d\"%A_CL);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.44 : Page number 709"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-12V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=10;               #Feedback resistor, kΩ\n",
-    "Ri=10;               #Input resistor, kΩ\n",
-    "V1=3;                #Input voltage 1st, V\n",
-    "V2=1;                #Input voltage 2nd, V\n",
-    "V3=8;                #Input voltage 3rd, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Rf=Ri, Vout=-(Rf/Ri)*(V1+V2+V3)= -(V1+V2+V3);\n",
-    "Vout=-(V1+V2+V3);           #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.45 : Page number 709"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 46,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-7V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=10;               #Feedback resistor, kΩ\n",
-    "R1=1;                #Input resistor for input 1, kΩ\n",
-    "R2=1;                #Input resistor for input 2, kΩ\n",
-    "V1=0.2;              #Input voltage 1st, V\n",
-    "V2=0.5;              #Input voltage 2nd, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "R=R1;                       #Input resistor(=R1 or R2), kΩ\n",
-    "Vout=-(Rf/R)*(V1+V2);       #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.46 : Page number 709-710"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 47,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-2.5V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=1;                #Feedback resistor, kΩ\n",
-    "Ri=10.0;               #Input resistor, kΩ\n",
-    "V1=10;                #Input voltage 1st, V\n",
-    "V2=8.0;                #Input voltage 2nd, V\n",
-    "V3=7.0;                #Input voltage 3rd, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Vout=-(Rf/Ri)*(V1+V2+V3);\n",
-    "Vout=-(Rf/Ri)*(V1+V2+V3);           #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.1fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.47 : Page number 710"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 48,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=2.5V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V1=0.6;                 #Input voltage to 1st input resistor, V\n",
-    "V2=-1.4;                #Input voltage to 2nd input resistor, V\n",
-    "Rf=200;                 #Feedback resistor, kΩ\n",
-    "R1=400;                 #Input resistor 1, kΩ\n",
-    "R2=100.0;                 #Input resistor 2, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "Vout=-Rf*(V1/R1 +V2/R2);                 #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.1fV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.48 : Page number 710-711"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 49,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The output voltage=-12.5V\n",
-      "(ii) The output voltage=-7.5V\n",
-      "(iii) The output voltage=-17.5V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=1.0;                 #Feedback resistor, kΩ\n",
-    "R1=1.0;                 #Input resistor 1, kΩ\n",
-    "R2=2.0;                 #Input resistor 2, kΩ\n",
-    "R3=4.0;                 #Input resistor 3, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Rf_R1=Rf/R1;                #Ratio of feedback resistor and 1st input resistor\n",
-    "Rf_R2=Rf/R2;                #Ratio of feedback resistor and 2nd input resistor\n",
-    "Rf_R3=Rf/R3;                #Ratio of feedback resistor and 3rd input resistor\n",
-    "\n",
-    "#(i) First input combination\n",
-    "V1=10;                                   #Input voltage to 1st input resistor, V\n",
-    "V2=0;                                    #Input voltage to 2nd input resistor, V\n",
-    "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-    "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-    "print(\"(i) The output voltage=%.1fV\"%Vout);\n",
-    "\n",
-    "#(i) First input combination\n",
-    "V1=0;                                   #Input voltage to 1st input resistor, V\n",
-    "V2=10;                                    #Input voltage to 2nd input resistor, V\n",
-    "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-    "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-    "print(\"(ii) The output voltage=%.1fV\"%Vout);\n",
-    "\n",
-    "\n",
-    "#(i) First input combination\n",
-    "V1=10;                                   #Input voltage to 1st input resistor, V\n",
-    "V2=10;                                    #Input voltage to 2nd input resistor, V\n",
-    "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-    "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-    "print(\"(iii) The output voltage=%.1fV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.49 : Page number 711"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 50,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vout=-[0.5sin(1000t)+0.33sin(3000t)]V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=330;                 #Feedback resistor, kΩ\n",
-    "R1=33.0;                  #Input resistor 1, kΩ\n",
-    "R2=10.0;                  #Input resistor 2, kΩ\n",
-    "V1_m=50;                #Peak voltage of 1st input, mV\n",
-    "V2_m=10;                #Peak voltage of 2nd input, mV\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Vout=-((Rf/R1)*V1 + (Rf/R2)*V2)\n",
-    "print(\"Vout=-[%.1fsin(1000t)+%.2fsin(3000t)]V\"%((V1_m/1000.0)*(Rf/R1),(V2_m/1000.0)*(Rf/R2)));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.50 : Page number 715"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 51,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vo=-1*(1/RC)∫vi dt.\n",
-      "=>Vo=-1*(1/1)∫vi dt\n",
-      "=>Vo=∫vi dt\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=100;                      #Input resistor, kΩ\n",
-    "C=10;                       #Feedback capacitor, μF\n",
-    "\n",
-    "#Calculation\n",
-    "RC=R*10**3*C*10**-6;        #product of input resistance and feedback capacitance, s\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vo=-1*(1/RC)∫vi dt.\");\n",
-    "print(\"=>Vo=-1*(1/%d)∫vi dt\"%RC);\n",
-    "print(\"=>Vo=∫vi dt\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.51 : Page number 715-716"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 52,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The critical frequency=159Hz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "Rf=100;                      #Feedback resistor, kΩ\n",
-    "C=0.01;                      #Feedback capacitor, μF\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "fc=1/(2*pi*Rf*1000*C*10**-6);               #Crictical frequency, Hz\n",
-    "\n",
-    "#Result\n",
-    "print(\"The critical frequency=%dHz.\"%fc);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.52 : Page number 716"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 53,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Vout=-1*(1/RC)∫vi dt.\n",
-      "    ΔVout/dt = -vin/RC = -50mV/μs.\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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/w4w8f/Q0SQ0NhCDpTElzJc2W9FNJ6zWyP7NaJk2C73/fd2ibDUfdYzFJ2hmY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\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f8b042c6b70>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "R=10.0;                      #Input resistor, kΩ\n",
-    "C=0.01;                    #Feedback capacitor, μF\n",
-    "vin=5;                     #Input voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "Vout_change_rate=-vin/(R*C);            #Rate of change of output voltage, V/μs   \n",
-    "print(\"(i) Vout=-1*(1/RC)∫vi dt.\");\n",
-    "print(\"    ΔVout/dt = -vin/RC = %dmV/μs.\"%Vout_change_rate);\n",
-    "\n",
-    "#(ii) Plotting the output waveform\n",
-    "vin_plot=[];                    #Plotting variable for input waveform, V\n",
-    "dt=100;                          #time between edges, μs\n",
-    "for i in range(0,3*dt+1):\n",
-    "    if i<dt or i>2*dt :\n",
-    "        vin_plot.append(0);\n",
-    "    else:\n",
-    "        vin_plot.append(5); \n",
-    "\n",
-    "plt.subplot(211);\n",
-    "plt.plot(vin_plot);\n",
-    "plt.xlim([0,300])\n",
-    "plt.ylim([-5,10])\n",
-    "plt.xlabel(\"t(microsecond)\");\n",
-    "plt.ylabel(\"Vin(V)\");\n",
-    "plt.title(\"Input waveform\");\n",
-    "\n",
-    "        \n",
-    "vout_plot=[];                  #Plotting variable for output waveform, V\n",
-    "t=[i for i in range(0,301)];  #Time scale, μs\n",
-    "for i in t[:] :\n",
-    "    if i<dt:\n",
-    "        vout_plot.append(0);\n",
-    "    elif i>2*dt:\n",
-    "        vout_plot.append((Vout_change_rate/1000.0)*dt);\n",
-    "    else :\n",
-    "        vout_plot.append((-vin_plot[i]/(R*C))/1000*(i-dt));\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout_plot);\n",
-    "plt.xlim([0,300])\n",
-    "plt.ylim([-5,5]);\n",
-    "plt.xlabel('t(microsecond)');\n",
-    "plt.ylabel(\"Vout(V)\");\n",
-    "plt.title(\"output waveform\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.53 : Page number 716-717"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 54,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vout=-1*(1/RC)∫vi dt.\n",
-      "Vout=-5*t volts\n",
-      "Time required=2.6seconds.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_supply=15;              #Supply voltage, V\n",
-    "R=10;                     #Input resistor, kΩ\n",
-    "C=0.2;                    #Feedback capacitor, μF\n",
-    "vin=10;                   #Input voltage, mV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Vs=-V_supply+2;         #Saturation voltage, V\n",
-    "print(\"Vout=-1*(1/RC)∫vi dt.\");\n",
-    "print(\"Vout=%d*t volts\"%(-vin/(R*C)));\n",
-    "t=Vs/(-vin/(R*C));                      #Time required, seconds\n",
-    "print(\"Time required=%.1fseconds.\"%t);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.54 : Page number 717-718"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 55,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vo=-5V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=1;                     #Feedback resistor, kΩ\n",
-    "C=0.1;                   #Input capacitor, μF\n",
-    "Vin_change=5;            #Change in input voltage, V\n",
-    "t=0.1;                   #Time taken for change in input voltage, ms\n",
-    "\n",
-    "#Calcualtion\n",
-    "dvi_dt=Vin_change/(t/1000);        #Rate of change of input voltage, V/s\n",
-    "RC=R*1000*C*10**-6;                #Product of feedback resistance and input capacitance, s\n",
-    "#Since, Vo=-R*C*(dvi/dt);\n",
-    "Vo=-RC*dvi_dt;                      #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vo=%dV.\"%Vo);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.55 : Page number 718"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 56,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vo=-0.55V.\n",
-      "The output voltage stays constant at -0.55V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=10;                     #Feedback resistor, kΩ\n",
-    "C=2.2;                   #Input capacitor, μF\n",
-    "Vin_change=10;            #Change in input voltage, V\n",
-    "t=0.4;                   #Time taken for change in input voltage, s\n",
-    "\n",
-    "#Calcualtin\n",
-    "dvi_dt=Vin_change/t;        #Rate of change of input voltage, V/s\n",
-    "RC=R*1000*C*10**-6;         #Product of feedback resistance and input capacitance, s\n",
-    "#Since, Vo=-R*C*(dvi/dt);\n",
-    "Vo=-RC*dvi_dt;                      #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vo=%.2fV.\"%Vo);\n",
-    "print(\"The output voltage stays constant at %.2fV.\"%Vo);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.56 : Page number 718-719"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 57,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "vo=-1*(dvi/dt).\n",
-      "vo=-5V.\n",
-      "Therefore, between 0 to 0.2s, the output voltage is constant at -5V.\n",
-      "For t>0.2s, the input is constant so that output voltage is zero.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=100;                     #Feedback resistor, kΩ\n",
-    "C=10;                   #Input capacitor, μF\n",
-    "Vin_change=1;            #Change in input voltage, V\n",
-    "t=0.2;                   #Time taken for change in input voltage, s\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "RC=R*1000*C*10**-6;             #Product of feedback resistor and input capacitance, s\n",
-    "#(i)\n",
-    "print(\"vo=-%d*(dvi/dt).\"%RC);\n",
-    "\n",
-    "#(ii)\n",
-    "dvi_dt=Vin_change/t;                #Rate of change of input voltage, V\n",
-    "vo=-dvi_dt;                         #Output voltage, V\n",
-    "print(\"vo=%dV.\"%vo);\n",
-    "\n",
-    "print(\"Therefore, between 0 to 0.2s, the output voltage is constant at %dV.\"%vo);\n",
-    "print(\"For t>0.2s, the input is constant so that output voltage is zero.\");\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_5.ipynb
deleted file mode 100755
index 92d6f1ec..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_5.ipynb
+++ /dev/null
@@ -1,2522 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# CHAPTER 25 : OPERATIONAL AMPLIFIERS"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": [
-    "%matplotlib inline"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.1: Page number 664"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output voltage of the differential amplifier = 10V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A=100.0;              #Open-circuit voltage gain of differential amplifier\n",
-    "V1=3.25;              #Input voltage to terminal 1 in V\n",
-    "V2=3.15;              #Input voltage to terminal 2 in V\n",
-    "\n",
-    "#Calculations\n",
-    "V0=A*(V1-V2);         #Output voltage in V\n",
-    "\n",
-    "#Results\n",
-    "print(\"The output voltage of the differential amplifier = %dV\"%V0);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.2: Page number 672"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The common mode rejection ratio = 10000.\n",
-      "The common mode rejection ratio in decibels= 80dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "A_DM=2000.0;              #Differential mode voltage gain\n",
-    "A_CM=0.2;                 #Common mode voltage gain\n",
-    "\n",
-    "#Calculations\n",
-    "CMRR=A_DM/A_CM;           #Common mode rejection ratio\n",
-    "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"The common mode rejection ratio = %d.\"%CMRR);\n",
-    "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.3: Page number 672"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The common mode rejection ratio in decibels= 46dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "VD_in=10.0;               #Differential mode input in mV\n",
-    "VD_out=1.0;               #Output for differential mode input in V\n",
-    "VC_in=10.0;                 #Common mode input in mV\n",
-    "VC_out=5.0;                 #Output for common mode input in mV\\\n",
-    "\n",
-    "#Calculations\n",
-    "A_DM=(VD_out*1000)/VD_in;       #Differntial mode voltage gain\n",
-    "A_CM=VC_out/VC_in;       #Common mode voltage gain\n",
-    "CMRR=A_DM/A_CM;           #Common mode rejection ratio\n",
-    "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-    "\n",
-    "#Results\n",
-    "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.4: Page number 672"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage =7.5V\n",
-      "Noise on output = 4.7x10^-6V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_DM=150.0;           #Differential mode voltage gain\n",
-    "CMRR_dB=90.0;         #Common mode rejection ratio\n",
-    "V1=100.0;             #Input voltage for terminal 1 in mV\n",
-    "V2=50.0;              #Input voltage for terminal 2 in mV\n",
-    "V_noise=1.0;          #Voltage of noise signal in mV\n",
-    "\n",
-    "#Calculation\n",
-    "\n",
-    "#Case(i)\n",
-    "V_out=A_DM*(V1-V2)/1000.0;            #Output voltage for differntial mode input, in V\n",
-    "\n",
-    "#Since CMRR_dB=20*log10(differential mode gain/common mode gain),\n",
-    "A_CM=A_DM/pow(10,(CMRR_dB/20));         #Common mode gain\n",
-    "V_OUT_noise=A_CM*(V_noise/1000);        #Noise on output in V\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"Output voltage =%.1fV\"%V_out);\n",
-    "print(\"Noise on output = %.1fx10^-6V\"%(V_OUT_noise*pow(10,6)));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.5 : Page number 672-673"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Common mode gain =0.083\n",
-      "Common mode rejection ratio in decibels=89.5dB\n",
-      "r.m.s output signal =1.25V\n",
-      "r.m.s interfernce output voltage = 83mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "A_DM=2500.0;                    #Differential mode voltage gain\n",
-    "CMRR=30000.0;                   #Common mode rejection ratio\n",
-    "Input_signal=500.0;             #Single ended input r.m.s signal in microvolts\n",
-    "Interference=1.0;               #Interference signal, in V\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#(i)\n",
-    "A_CM=A_DM/CMRR;                 #Common mode gain\n",
-    "\n",
-    "#(ii)\n",
-    "CMRR_dB=20*log10(CMRR);         #Common mode rejection ratio in decibels\n",
-    "\n",
-    "#(iii)\n",
-    "V_out=A_DM*(Input_signal/pow(10,6)-0);        #r.m.s output signal  in V\n",
-    "\n",
-    "#(iv)\n",
-    "Interference_out=A_CM*Interference;          #r.m.s interference output in V\n",
-    "Interference_out=Interference_out*1000;      #r.m.s interference output in mV\n",
-    "\n",
-    "\n",
-    "#Results\n",
-    "print(\"Common mode gain =%.3f\"%A_CM);\n",
-    "print(\"Common mode rejection ratio in decibels=%.1fdB\"%CMRR_dB);\n",
-    "print(\"r.m.s output signal =%.2fV\"%V_out);\n",
-    "print(\"r.m.s interfernce output voltage = %dmV\"%Interference_out);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.6 : Page number 674-675"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VE=-0.7V\n",
-      "IE=0.452mA\n",
-      "IE1=0.226mA\n",
-      "IE2=0.226mA\n",
-      "IC1=0.226mA\n",
-      "IC2=0.226mA\n",
-      "IB1=2.26μA\n",
-      "IB2=2.26μA\n",
-      "VC1=12V\n",
-      "VC2=9.7V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RB=10;                  #Base resistor, kΩ\n",
-    "RC2=10;                 #Collector resistor, kΩ\n",
-    "RE=25;                  #Emitter resistor, kΩ\n",
-    "VBE=0.7;               #Base-emitter voltage, V\n",
-    "beta=100;               #Base amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-    "IE=(VEE-VBE)/RE;            #Tail current, mA\n",
-    "IE1=IE/2;                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-    "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-    "IC2=IC1;                    #Collector current of 2nd transistor, mA\n",
-    "IB1=(IC1/beta)*1000;        #Base current of 1st transistor, μA\n",
-    "IB2=IB1;                    #Base current of 2nd transistor, μA\n",
-    "VC1=VCC;                    #Collector voltage of 1st transistor, V\n",
-    "VC2=VCC-IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"VE=%.1fV\"%VE);\n",
-    "print(\"IE=%.3fmA\"%IE);\n",
-    "print(\"IE1=%.3fmA\"%IE1);\n",
-    "print(\"IE2=%.3fmA\"%IE2);\n",
-    "print(\"IC1=%.3fmA\"%IC1);\n",
-    "print(\"IC2=%.3fmA\"%IC2);\n",
-    "print(\"IB1=%.2fμA\"%IB1);\n",
-    "print(\"IB2=%.2fμA\"%IB2);\n",
-    "print(\"VC1=%dV\"%VC1);\n",
-    "print(\"VC2=%.1fV\"%VC2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.7 : Page number 675"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output voltage=7.85V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15;                 #Collector supply voltage, V\n",
-    "VEE=15;                 #Emitter supply voltage, V\n",
-    "RB=33;                  #Base resistor, kΩ\n",
-    "RC=15;                  #Collector resistor, kΩ\n",
-    "RE=15;                  #Emitter resistor, kΩ\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "IE_tail=(VEE-VBE)/RE;            #Tail current, mA\n",
-    "IE=round(IE_tail/2,3);                    #Emitter current in each transistor, mA\n",
-    "IC=IE;                           #Collector current(=emitter current), mA\n",
-    "Vout=VCC-IC*RC;                  #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output voltage=%.2fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.8 : Page number 675"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) IB1=5.56μA\n",
-      "    IB2=4.55μA\n",
-      "(ii) VB1=-0.183V\n",
-      "     VB2=-0.15V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15.0;                 #Collector supply voltage, V\n",
-    "VEE=15.0;                 #Emitter supply voltage, V\n",
-    "RB=33.0;                  #Base resistor, kΩ\n",
-    "RC=15.0;                  #Collector resistor, kΩ\n",
-    "RE=15.0;                  #Emitter resistor, kΩ\n",
-    "VBE=0;                  #Base-emitter voltage, V\n",
-    "beta_dc_l=90.0;           #base current amplification factor for left transistor\n",
-    "beta_dc_r=110.0;          #base current amplification factor for right transistor\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IE_tail=(VEE-VBE)/RE;             #Tail current, mA\n",
-    "IE=IE_tail/2;                      #Emitter current in each transistor, mA\n",
-    "IB1=(IE/beta_dc_l)*1000;          #Base current of 1st transistor, μA\n",
-    "IB2=(IE/beta_dc_r)*1000;          #Base current of 2nd transistor, μA\n",
-    "\n",
-    "#(ii)\n",
-    "VB1=-IB1/1000*RB;                    #Base voltage of 1st transistor, V\n",
-    "VB2=-IB2/1000*RB;                    #Base voltage of 1st transistor, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) IB1=%.2fμA\"%IB1);\n",
-    "print(\"    IB2=%.2fμA\"%IB2);\n",
-    "print(\"(ii) VB1=%.3fV\"%VB1);\n",
-    "print(\"     VB2=%.2fV\"%VB2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.9 : Page number 675-676"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VE=-0.7V\n",
-      "Emitter current in each transistor=0.5mA.\n",
-      "IC1~IE1=0.5mA and IC2~IE2=0.5mA\n",
-      "VC1=VC2=10V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15.0;                 #Collector supply voltage, V\n",
-    "VEE=15.0;                 #Emitter supply voltage, V\n",
-    "RB=10.0;                  #Base resistor, kΩ\n",
-    "RC1=10.0;                 #Collector resistor of 1st transistor, kΩ\n",
-    "RC2=10.0;                 #Collector resistor of 2nd transistor, kΩ\n",
-    "IE=1.0;                   #Tail current, mA\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-    "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-    "IC2=IE2;                    #Collector current of 2nd transistor, mA\n",
-    "VC1=VCC-IC1*RC1;            #Collector voltage of 1st transistor, V\n",
-    "VC2=VCC-IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"VE=%.1fV\"%VE);\n",
-    "print(\"Emitter current in each transistor=%.1fmA.\"%(IE/2.0));\n",
-    "print(\"IC1~IE1=%.1fmA and IC2~IE2=%.1fmA\"%(IE1,IE2));\n",
-    "print(\"VC1=VC2=%dV.\"%VC2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.10 : Page number 676-677"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "VE=0.7V\n",
-      "Tail current=0.452mA.\n",
-      "Emitter current in each transistor=0.226mA.\n",
-      "IC1~IE1=0.226mA and IC2~IE2=0.226mA\n",
-      "VC1=-12V\n",
-      "VC2=-9.74V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12.0;                 #Collector supply voltage, V\n",
-    "VEE=12.0;                 #Emitter supply voltage, V\n",
-    "RC2=10.0;                 #Collector resistor of 2nd transistor, kΩ\n",
-    "RE=25.0;                  #Emitter current, kΩ\n",
-    "VBE=-0.7;               #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "VE=-VBE;                    #Emitter voltage, V (Ignoring the base current)\n",
-    "IE=(VCC-VE)/RE;             #Tail current, mA\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, mA\n",
-    "IC1=IE1;                    #Collector current(= emitter current) of 1st transistor, mA\n",
-    "IC2=IE2;                    #Collector current of 2nd transistor, mA\n",
-    "VC1=-VEE;                   #Collector voltage of 1st transistor, V\n",
-    "VC2=-VEE+IC2*RC2;            #Collector voltage of 2nd transistor, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"VE=%.1fV\"%VE);\n",
-    "print(\"Tail current=%.3fmA.\"%IE);\n",
-    "print(\"Emitter current in each transistor=%.3fmA.\"%(IE/2.0));\n",
-    "print(\"IC1~IE1=%.3fmA and IC2~IE2=%.3fmA\"%(IC1,IC2));\n",
-    "print(\"VC1=%dV\"%VC1);\n",
-    "print(\"VC2=%.2fV\"%VC2);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.11 : Page number 679"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The input offset current=15.1nA\n",
-      "(ii) The input bias current=75.8nA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15;                 #Collector supply voltage, V\n",
-    "VEE=15;                 #Emitter supply voltage, V\n",
-    "RB=1;                   #Base resistor, MΩ\n",
-    "RC2=1;                  #Collector resistor, MΩ\n",
-    "RE=1;                   #Emitter resistor, MΩ\n",
-    "VBE=0;                  #Base-emitter voltage, V (Neglected)\n",
-    "beta_dc_l=90.0;           #base current amplification factor for left transistor\n",
-    "beta_dc_r=110.0;          #base current amplification factor for right transistor\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "IE=(VEE-VBE)/RE;                    #Tail current, μA\n",
-    "IE1=IE/2.0;                           #Emitter current of 1st transistor, μA\n",
-    "IE2=IE1;                            #Emitter current of 2nd transistor, μA\n",
-    "IB1=round((IE1/beta_dc_l)*1000,1);           #Base current of 1st transistor, nA\n",
-    "IB2=round((IE2/beta_dc_r)*1000,1);           #Base current of 2nd transistor, nA\n",
-    "I_in_offset=IB1-IB2;                 #Input offset current, nA\n",
-    "\n",
-    "#(ii)\n",
-    "I_in_bias=(IB1+IB2)/2;                  #Input bias current, nA\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  The input offset current=%.1fnA\"%I_in_offset);\n",
-    "print(\"(ii) The input bias current=%.1fnA\"%I_in_bias);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.12 : Page number 679"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The two base currents are: IB1=90nA and IB2=70nA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_in_offset=20;                 #Input offset current, nA\n",
-    "I_in_bias=80;                   #Input bias current, nA\n",
-    "\n",
-    "#Calculation\n",
-    "IB1=I_in_bias+I_in_offset/2;            #Base current in 1st transistor, nA\n",
-    "IB2=I_in_bias-I_in_offset/2;            #Base current in 2nd transistor, nA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The two base currents are: IB1=%dnA and IB2=%dnA.\"%(IB1,IB2));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.13 : Page number 679-680"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input offset voltage=2mV.\n",
-      "The output offset voltage=0.3V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declration\n",
-    "I_in_offset=20;                 #Input offset current, nA\n",
-    "I_in_bias=80;                   #Input bias current, nA\n",
-    "A=150;                          #Voltage gain\n",
-    "RB=100;                         #Base resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_io=(I_in_offset*10**-9*RB*1000)*1000;       #Input offset voltage, mV\n",
-    "V_out_offset=(A*V_io)/1000;                     #Output offset voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input offset voltage=%dmV.\"%V_io);\n",
-    "print(\"The output offset voltage=%.1fV.\"%V_out_offset);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.14 : Page number 682"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Output voltage=0.15V.\n",
-      "(ii) Output voltage=-0.15V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=15;                 #Collector supply voltage, V\n",
-    "VEE=15;                 #Emitter supply voltage, V\n",
-    "RE=1;                   #Emitter resistor, MΩ\n",
-    "RC=1;                   #Collector resistor, MΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "IE=VEE/RE;                  #Tail current, μA\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, μA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, μA\n",
-    "re=25/IE1;                  #a.c emitter resistance, kΩ\n",
-    "A_DM=RC/(2.0*re);        #Differential voltage gain,\n",
-    "\n",
-    "#(i)\n",
-    "vin=1;                      #Input voltage, V\n",
-    "Vout=A_DM*vin;              #Output voltage, V\n",
-    "\n",
-    "print(\"(i) Output voltage=%.2fV.\"%Vout);\n",
-    "\n",
-    "#(ii)\n",
-    "vin=-1;                      #Input voltage, V\n",
-    "Vout=A_DM*vin;              #Output voltage, V;\n",
-    "print(\"(ii) Output voltage=%.2fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.15 : Page number 682-683"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The input impedance=194kΩ.\n",
-      "(ii) The differential voltage gain=136.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RE=100;                 #Emitter resistor, kΩ\n",
-    "RC1=120;                #Collector resistor of 1st transistor, kΩ\n",
-    "RC2=120;                #Collector resistor of 2nd transistor, kΩ\n",
-    "beta=220;               #Base amplification factor\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calcualtion\n",
-    "IE=((VEE-VBE)/RE)*1000;     #Tail current, μA\n",
-    "IE1=IE/2.0;                   #Emitter current of 1st transistor, μA\n",
-    "IE2=IE1;                    #Emitter current of 2nd transistor, μA\n",
-    "re=(25/IE1)*1000;           #a.c emitter resistance, Ω\n",
-    "Zin=2*beta*re/1000;         #Input impedance, kΩ\n",
-    "A_DM=RC1*1000/(2.0*re);             #Differential voltage gain,\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) The input impedance=%dkΩ.\"%Zin);\n",
-    "print(\"(ii) The differential voltage gain=%.0f.\"%A_DM);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.16: Page number 683-684"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Differential voltage gain=56.6.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RE=200;                 #Emitter resistor, kΩ\n",
-    "RC=100;                 #Collector resistor, kΩ\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "IE=round((VEE-VBE)/RE,4);            #Tail current, mA\n",
-    "IE1=round(IE/2,4);                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                             #Emitter current of 2nd transistor, mA\n",
-    "re=round(25/IE1,1);                  #a.c emitter resistance, Ω\n",
-    "A_DM=RC*1000/(2*re);                 #Differential voltage gain,\n",
-    "\n",
-    "#Result\n",
-    "print(\"Differential voltage gain=%.1f.\"%A_DM);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.17 : Page number 685-686"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Common mode rejection ratio=666.7.\n",
-      "Common mode rejection ratio in decibel=56.48dB\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "v1=0.5;               #Voltage in terminal 1, mV\n",
-    "v2=-0.5;               #Voltage in terminal 2, mV\n",
-    "vo=8.0;                   #Output voltage, V\n",
-    "vo_cm=12.0;               #Common mode output, mV\n",
-    "\n",
-    "#Calculation\n",
-    "vin=v1-v2;                  #Differential input, mV\n",
-    "A_DM=vo/(vin/1000.0);         #Differential mode gain,\n",
-    "vin_cm=1;                   #Common mode input, mV\n",
-    "A_CM=vo_cm/vin_cm;          #Common mode gain\n",
-    "CMRR=A_DM/A_CM;             #Common mode rejection ratio\n",
-    "CMRR_dB=20*log10(CMRR);     #Common mode rejection ratio in dB\n",
-    "\n",
-    "#Result\n",
-    "print(\"Common mode rejection ratio=%.1f.\"%CMRR)\n",
-    "print(\"Common mode rejection ratio in decibel=%.2fdB\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.18 : Page number 686"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 19,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Common mode voltage gain=6.32.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_DM=200000;                #Differential mode gain\n",
-    "CMRR_dB=90;                 #Common mode rejection ratio, dB\n",
-    "\n",
-    "#Calculation\n",
-    "CMRR=10**(CMRR_dB/20.0);              #Common mode rejection ratio\n",
-    "A_CM=A_DM/CMRR;                     #Common mode gain\n",
-    "\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Common mode voltage gain=%.2f.\"%A_CM);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.19 : Page number 686"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 20,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  The Common mode gain=0.0081\n",
-      "(ii) The common mode rejection ratio=81.8dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "vin_cm=3.2;                     #Common input voltage, V\n",
-    "vout=26;                        #Output voltage, V\n",
-    "A_DM=100;                       #Open-circuit voltage gain\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CM=vout*10**-3/vin_cm;                #Common mode gain\n",
-    "\n",
-    "#(ii)\n",
-    "CMRR_dB=20*log10(A_DM/A_CM);            #Common mode rejection ratio, dB\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  The Common mode gain=%.4f\"%A_CM);\n",
-    "print(\"(ii) The common mode rejection ratio=%.1fdB.\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.20 : Page number 686-687"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 21,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Common mode gain=0.25\n",
-      "(ii)Common mode rejection ratio=47.09dB\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "from math import floor\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=12;                 #Collector supply voltage, V\n",
-    "VEE=12;                 #Emitter supply voltage, V\n",
-    "RE=200.0;                 #Emitter resistor, kΩ\n",
-    "RC=100.0;                 #Collector resistor, kΩ\n",
-    "VBE=0.7;                #Base-emitter voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CM=round(RC/(2*RE),2);             #Common mode voltage gain\n",
-    "\n",
-    "#(ii)\n",
-    "IE=round((VEE-VBE)/RE,4);            #Tail current, mA\n",
-    "IE1=round(IE/2,4);                   #Emitter current of 1st transistor, mA\n",
-    "IE2=IE1;                             #Emitter current of 2nd transistor, mA\n",
-    "re=round(25/IE1,1);                  #a.c emitter resistance, Ω\n",
-    "A_DM=RC*1000/(2*re);                 #Differential voltage gain,\n",
-    "CMRR_dB=floor(20*log10(A_DM/A_CM)*100)/100;         #Common mode rejection ratio, dB\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) Common mode gain=%.2f\"%A_CM);\n",
-    "print(\"(ii)Common mode rejection ratio=%.2fdB\"%CMRR_dB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.21 : Page number 691"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "f2=30kHz\n",
-      "ACL=75 or 37.5dB.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log10\n",
-    "\n",
-    "#Variable declaration\n",
-    "ACL=500;                    #closed loop gain\n",
-    "f_unity=15;                 #frequency with cloased-loop unity gain, MHz\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "f2=f_unity*1000/500                 #Upper frequency of bandwidth,kHz\n",
-    "BW=f2-0;                            #Bandwidth, kHz\n",
-    "A_CL=f_unity*1000/200;                    #Maximum value of A_CL when f2=200kHz\n",
-    "A_CL_dB=20*log10(A_CL);             #Maximum value of A_CL in decibel\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"f2=%dkHz\"%f2);\n",
-    "print(\"ACL=%d or %.1fdB.\"%(A_CL,A_CL_dB));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.22 : Page number 691-692"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Operating Bandwidth=1.5MHz.\n",
-      "(ii) Operating Bandwidth=150kHz.\n",
-      "(iii) Operating Bandwidth=15kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "GBW=1.5;                    #Gain-bandwidth, MHz\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) For A_CL=1;\n",
-    "A_CL=1;                         #Closed loop gain\n",
-    "BW=GBW/A_CL;                    #Bandwidth, MHz\n",
-    "\n",
-    "print(\"(i) Operating Bandwidth=%.1fMHz.\"%BW);\n",
-    "\n",
-    "#(ii) For A_CL=10;\n",
-    "A_CL=10;                         #Closed loop gain\n",
-    "BW=(GBW/A_CL)*1000;              #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(ii) Operating Bandwidth=%dkHz.\"%BW);\n",
-    "\n",
-    "#(iii) For A_CL=100;\n",
-    "A_CL=100;                         #Closed loop gain\n",
-    "BW=(GBW/A_CL)*1000;               #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(iii) Operating Bandwidth=%dkHz.\"%BW);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.23 : Page number 692"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Maximum operating frequency=9.95kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "slew_rate=0.5;                  #Slew rate, V/μs\n",
-    "V_supply=10;                     #Supply voltage, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_sat=V_supply-2;                           #Saturation voltage, V\n",
-    "V_pk=V_sat;                                 #Maximum peak-output voltage, V\n",
-    "f_max=((slew_rate*10**6)/(2*pi*V_pk))/1000;       #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Maximum operating frequency=%.2fkHz.\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.24 : Page number 692"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Maximum operating frequency=796kHz\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "slew_rate=0.5;                  #Slew rate, V/μs\n",
-    "V_pk=100.0;                       #Peak-output voltage, mV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "V_pk=V_pk/1000.0;                                #Peak-output voltage, V\n",
-    "f_max=(slew_rate*10**6/(2*pi*V_pk))/1000.0;      #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"Maximum operating frequency=%.0fkHz\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.25 : Page number 695-696"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Feedback resistor=220kΩ\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_CL=-100;              #Closed-loop voltage gain\n",
-    "Ri=2.2;                 #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, A_CL=-(Rf/Ri)\n",
-    "Rf=-A_CL*Ri;                #Feedback resistor, kΩ\n",
-    "\n",
-    "#Result\n",
-    "print(\"Feedback resistor=%dkΩ\"%Rf);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.26 : Page number 696"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-0.25V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "vin=2.5;                #Input voltage, mV\n",
-    "Rf=200;                 #Feedback resistor, kΩ\n",
-    "Ri=2;                   #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "vout=A_CL*vin/1000;                  #Output voltage,V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.2fV\"%vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.27 : Page number 696"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 28,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Closed-loop voltage gain=-1\n",
-      "Therefore, output will have same amplitude but 180° phase inversion.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Varaiable declaration\n",
-    "Rf=1.0;                   #Feedback resistor, kΩ\n",
-    "Ri=1.0;                   #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"Closed-loop voltage gain=%d\"%A_CL);\n",
-    "print(\"Therefore, output will have same amplitude but 180° phase inversion.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.28 : Page number 696-697"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 29,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Closed-loop voltage gain=-40\n",
-      "Supply voltage=±15V, saturation voltage=±13V. Since gain=-40, op-Amp will be driven to saturation.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=40;                   #Feedback resistor, kΩ\n",
-    "Ri=1;                   #Input resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Closed-loop voltage gain=%d\"%A_CL);\n",
-    "print(\"Supply voltage=±15V, saturation voltage=±13V. Since gain=-40, op-Amp will be driven to saturation.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.29 : Page number 697"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 30,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  A_CL=-10.\n",
-      "(ii) Zi=10kΩ\n",
-      "(iii) Maximum operating frequency=15.9kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "Rf=100;                   #Feedback resistor, kΩ\n",
-    "Ri=10;                    #Input resistor, kΩ\n",
-    "Vpp=1;                    #Input peak-peak voltage, V\n",
-    "slew_rate=0.5;            #Slew rate, V//μs\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CL=-(Rf/Ri);                  #Closed-loop voltage gain\n",
-    "\n",
-    "#(ii)\n",
-    "Zi=Ri;                  #Input impedance(~ Input resistor), kΩ\n",
-    "\n",
-    "#(iii)\n",
-    "Vout=A_CL*Vpp;                           #Peak-to-peak voltage, V\n",
-    "Vpk=Vout/2;                              #Peak output voltage, V\n",
-    "f_max=(slew_rate*10**6/(2*pi*abs(Vpk)))/1000;        #Maximum operating frequency, kHz\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  A_CL=%d.\"%A_CL);\n",
-    "print(\"(ii) Zi=%dkΩ\"%Zi);\n",
-    "print(\"(iii) Maximum operating frequency=%.1fkHz.\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.30 : Page number 697"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Rf=20kΩ and Ri=5kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A_CL=-4;                    #Closed loop voltage gain\n",
-    "R=[1.0,5.0,10.0,20.0];              #List of available resistors, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "for i in R[:]:\n",
-    "    for j in R[:]:\n",
-    "        if -(i/j)==A_CL :\n",
-    "            print(\"Rf=%dkΩ and Ri=%dkΩ.\"%(i,j));\n",
-    "            break;\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.31 : Page  number 697-698"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 32,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Closed loop voltage gain=-100.\n",
-      "(ii) Closed loop voltage gain=-50.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=100;                   #Feedback resistor, kΩ\n",
-    "Ri=1;                    #Input resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "R_source=0;               #Source resistor, kΩ\n",
-    "A_CL=-Rf/(R_source+Ri);    #Closed-loop voltage gain\n",
-    "\n",
-    "print(\"(i) Closed loop voltage gain=%d.\"%A_CL);\n",
-    "\n",
-    "#(ii)\n",
-    "R_source=1;               #Source resistor, kΩ\n",
-    "A_CL=-Rf/(R_source+Ri);    #Closed-loop voltage gain\n",
-    "\n",
-    "print(\"(ii) Closed loop voltage gain=%d.\"%A_CL);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.32 : Page number 699-700"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 33,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=12.12mV\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=240;                   #Feedback resistor, kΩ\n",
-    "Ri=2.4;                    #Input resistor, kΩ\n",
-    "Vin=120;                    #Input voltage, μV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=1+(Rf/Ri);             #Closed loop voltage gain\n",
-    "Vout=(A_CL*Vin)/1000;       #Output voltage, mV\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.2fmV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.33 : Page number 700"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Output voltage=11V\n",
-      "(ii) Output voltage=-11V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=10;                   #Feedback resistor, kΩ\n",
-    "Ri=1;                    #Input resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "A_CL=1+(Rf/Ri);             #Closed loop voltage gain\n",
-    "#(i)\n",
-    "Vin=1;                      #Input voltage, V\n",
-    "Vout=A_CL*Vin;              #Output voltage, V\n",
-    "\n",
-    "print(\"(i)  Output voltage=%dV\"%Vout);\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "Vin=-1;                      #Input voltage, V\n",
-    "Vout=A_CL*Vin;               #Output voltage, V\n",
-    "\n",
-    "print(\"(ii) Output voltage=%dV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.34 : Page number 700"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 35,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Peak to peak output voltage=12V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=5;                       #Feedback resistor, kΩ\n",
-    "Ri=1;                       #Input resistor, kΩ\n",
-    "Vin_max=1;                  #Maximum input voltage, V\n",
-    "Vin_min=-1;                 #Minimum input voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "V_inpp=Vin_max-Vin_min;               #Peak-peak input voltage, V\n",
-    "A_CL=1+(Rf/Ri);                     #Closed loop voltage gain\n",
-    "Vout_pp=A_CL*V_inpp;                #Peak-peak output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Peak to peak output voltage=%dV\"%Vout_pp);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.35 : Page number 700-701"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 36,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Closed-loop voltage gain=11\n",
-      "(ii) Maximum operating frequency=14.47kHz\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "Rf=100;                       #Feedback resistor, kΩ\n",
-    "Ri=10;                       #Input resistor, kΩ\n",
-    "Vpp=1;                    #Input peak-peak voltage, V\n",
-    "slew_rate=0.5;              #Slew rate, V/μs\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "A_CL=1+(Rf/Ri);                                      #Closed loop voltage gain\n",
-    "\n",
-    "#(ii)\n",
-    "Vout_pp=A_CL*Vpp;                                    #Peak-peak output voltage, V\n",
-    "Vpk=Vout_pp/2.0;                                          #Peak output voltage, V\n",
-    "f_max=((slew_rate*10**6)/(2*pi*Vpk))/1000.0;           #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i)  Closed-loop voltage gain=%d\"%A_CL);\n",
-    "\n",
-    "print(\"(ii) Maximum operating frequency=%.2fkHz\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.36 : Page number 701"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 37,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)  Bandwidth=44.3kHz.\n",
-      "(ii)  Bandwidth=63.8kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=220;                       #Feedback resistor, kΩ\n",
-    "Ri=3.3;                       #Input resistor, kΩ\n",
-    "unity_gain_BW=3;              #Unity gain bandwidth, MHz\n",
-    "\n",
-    "#Calculation\n",
-    "#(i) For non-inverting amplifier\n",
-    "A_CL=1+(Rf/Ri);                    #Closed loop voltage gain\n",
-    "BW=unity_gain_BW*1000.0/A_CL;        #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(i)  Bandwidth=%.1fkHz.\"%BW);\n",
-    "\n",
-    "#(ii) For inverting amplifier\n",
-    "Rf=47;                          #Feedback resistor, kΩ\n",
-    "Ri=1;                           #Input resistor, kΩ\n",
-    "A_CL=-(Rf/Ri);                    #Closed loop voltage gain\n",
-    "BW=unity_gain_BW*1000.0/abs(A_CL);        #Bandwidth, kHz\n",
-    "\n",
-    "print(\"(ii)  Bandwidth=%.1fkHz.\"%BW);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.37 : Page number 701-702"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 38,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) For voltage follower A_CL=1.\n",
-      "(ii) The maximum output frequency=26.53kHz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#(i)\n",
-    "A_CL=1;                     #Closed loop voltage gain for voltage follower\n",
-    "print(\"(i) For voltage follower A_CL=1.\");\n",
-    "\n",
-    "\n",
-    "#(ii)\n",
-    "slew_rate=0.5;              #Slew rate, V/μs\n",
-    "V_inpp=6;                   #peak-peak input voltage, V\n",
-    "Vout=A_CL*V_inpp;           #Peak-peak output voltage, V\n",
-    "Vpk=Vout/2;                 #Peak output voltage, V\n",
-    "\n",
-    "f_max=(slew_rate*10**6/(2*pi*Vpk))/1000;       #Maximum operating frequency, kHz\n",
-    "\n",
-    "#Result\n",
-    "print(\"(ii) The maximum output frequency=%.2fkHz.\"%f_max);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.38 : Page number 702"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 39,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=1.78V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=470.0;                 #Feedback resistor, kΩ\n",
-    "R1=4.3;                 #Input resistor of 1st op-Amp, kΩ\n",
-    "R2=33.0;                 #Input resistor of 2nd op-Amp, kΩ\n",
-    "R3=33.0;                 #Input resistor of 3rd op-Amp, kΩ\n",
-    "Vin=80.0;                 #Input voltage, μV.\n",
-    "\n",
-    "#Calculation\n",
-    "A1=1+Rf/R1;                 #Gain of first op-Amp\n",
-    "A2=-round(Rf/R2,1);         #Gain of second op-Amp\n",
-    "A3=-round(Rf/R3,1);         #Gain of third op-Amp\n",
-    "A=A1*A2*A3;                 #Overall gain\n",
-    "Vout=A*Vin*10**-6;          #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.2fV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.39 : Page number 702-703"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "R1=30kΩ, R2=15kΩ and R3=10kΩ.\n",
-      "Output voltage=0.729V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "A1=10;                  #Voltage gain of 1st op-Amp\n",
-    "A2=-18;                 #Voltage gain of 2nd op-Amp\n",
-    "A3=-27;                 #Voltage gain of 3rd op-Amp\n",
-    "Rf=270;                 #Feedback resistor, kΩ\n",
-    "Vin=150;                #Input voltage, μV  \n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "R1=Rf/(A1-1);                   #Input resistor of 1st op-Amp, kΩ\n",
-    "R2=-Rf/A2;                      #Input resistr of 2nd op-Amp, kΩ\n",
-    "R3=-Rf/A3;                      #Input resistor of 3rd op-Amp, kΩ\n",
-    "\n",
-    "A=A1*A2*A3;                     #overall gain,\n",
-    "Vout=Vin*10**-6*A;              #Output voltage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"R1=%dkΩ, R2=%dkΩ and R3=%dkΩ.\"%(R1,R2,R3));\n",
-    "print(\"Output voltage=%.3fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.40 : Page number 703-704"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 41,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "R1=50kΩ, R2=25kΩ and R3=10kΩ.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=500;                 #Feedback resistor, kΩ\n",
-    "A1=-10;                 #Gain of 1st op-Amp\n",
-    "A2=-20;                 #Gain of 2nd op-Amp\n",
-    "A3=-50;                 #Gain of 3rd op-Amp\n",
-    "\n",
-    "#Calculation\n",
-    "R1=-Rf/A1;                      #Input resistor of 1st op-Amp, kΩ\n",
-    "R2=-Rf/A2;                      #Input resistor of 2nd op-Amp, kΩ\n",
-    "R3=-Rf/A3;                      #Input resistor of 3rd op-Amp, kΩ\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"R1=%dkΩ, R2=%dkΩ and R3=%dkΩ.\"%(R1,R2,R3));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.41 : Page number 705"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 42,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The input impedance=17202MΩ and output impedance=8.7e-03Ω.\n",
-      "(ii) The closed loop voltage gain=23.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Zin=2.0;              #Input impedance of op-Amp, MΩ\n",
-    "Zout=75.0;            #Output impedance of op-Amp, Ω\n",
-    "A_OL=200000.0;        #Open-loop voltage gain\n",
-    "Rf=220.0;             #Feedback resistor, kΩ\n",
-    "Ri=10.0;              #Input resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "mv=round(Ri/(Ri+Rf),3);       #Feedback fraction\n",
-    "Zin_NI=Zin*(1+(A_OL*mv));     #Input impedance, MΩ\n",
-    "Zout_NI=Zout/(1+A_OL*mv);     #Output impedance, Ω\n",
-    "\n",
-    "#(ii)\n",
-    "A_CL=1+Rf/Ri;               #Closed loop voltage gain\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) The input impedance=%dMΩ and output impedance=%.1eΩ.\"%(Zin_NI,Zout_NI));\n",
-    "print(\"(ii) The closed loop voltage gain=%d.\"%A_CL);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.42 : Page number 705-706"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 43,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance=400002MΩ and output impedance=0.38e-03Ω.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "#For voltage follower,\n",
-    "mv=1.0;               #Feedback fraction\n",
-    "A_OL=200000.0;        #Open-loop voltage gain\n",
-    "Zin=2.0;              #Input impedance of op-Amp, MΩ\n",
-    "Zout=75.0;            #Output impedance of op-Amp, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Zin_VF=Zin*(1+(A_OL*mv));     #Input impedance, MΩ\n",
-    "Zout_VF=round(round(Zout/(1+A_OL*mv),6),5);     #Output impedance, Ω\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance=%dMΩ and output impedance=%.2fe-03Ω.\"%(Zin_VF,Zout_VF*1000));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.43 : Page number 706"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 44,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The input impedance=1kΩ and output impedance=50Ω.\n",
-      "Closed-loop voltage gain=-100\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=100;             #Feedback resistor, kΩ\n",
-    "Ri=1.0;               #Input resistor, kΩ\n",
-    "Zin=4;              #Input impedance of op-Amp, MΩ\n",
-    "Zout=50;            #Output impedance of op-Amp, Ω\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Zin_I=Ri;                   #Input impedance, kΩ\n",
-    "Zout_I=Zout;                #Output impedance, Ω\n",
-    "A_CL=-(Rf/Ri);              #Closed loop voltage gain\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The input impedance=%dkΩ and output impedance=%dΩ.\"%(Zin_I,Zout_I));\n",
-    "print(\"Closed-loop voltage gain=%d\"%A_CL);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.44 : Page number 709"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-12V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=10;               #Feedback resistor, kΩ\n",
-    "Ri=10;               #Input resistor, kΩ\n",
-    "V1=3;                #Input voltage 1st, V\n",
-    "V2=1;                #Input voltage 2nd, V\n",
-    "V3=8;                #Input voltage 3rd, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Rf=Ri, Vout=-(Rf/Ri)*(V1+V2+V3)= -(V1+V2+V3);\n",
-    "Vout=-(V1+V2+V3);           #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.45 : Page number 709"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 46,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-7V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=10;               #Feedback resistor, kΩ\n",
-    "R1=1;                #Input resistor for input 1, kΩ\n",
-    "R2=1;                #Input resistor for input 2, kΩ\n",
-    "V1=0.2;              #Input voltage 1st, V\n",
-    "V2=0.5;              #Input voltage 2nd, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "R=R1;                       #Input resistor(=R1 or R2), kΩ\n",
-    "Vout=-(Rf/R)*(V1+V2);       #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%dV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.46 : Page number 709-710"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 47,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=-2.5V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=1;                #Feedback resistor, kΩ\n",
-    "Ri=10.0;               #Input resistor, kΩ\n",
-    "V1=10;                #Input voltage 1st, V\n",
-    "V2=8.0;                #Input voltage 2nd, V\n",
-    "V3=7.0;                #Input voltage 3rd, V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Vout=-(Rf/Ri)*(V1+V2+V3);\n",
-    "Vout=-(Rf/Ri)*(V1+V2+V3);           #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.1fV.\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.47 : Page number 710"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 48,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output voltage=2.5V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V1=0.6;                 #Input voltage to 1st input resistor, V\n",
-    "V2=-1.4;                #Input voltage to 2nd input resistor, V\n",
-    "Rf=200;                 #Feedback resistor, kΩ\n",
-    "R1=400;                 #Input resistor 1, kΩ\n",
-    "R2=100.0;                 #Input resistor 2, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "Vout=-Rf*(V1/R1 +V2/R2);                 #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Output voltage=%.1fV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.48 : Page number 710-711"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 49,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) The output voltage=-12.5V\n",
-      "(ii) The output voltage=-7.5V\n",
-      "(iii) The output voltage=-17.5V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=1.0;                 #Feedback resistor, kΩ\n",
-    "R1=1.0;                 #Input resistor 1, kΩ\n",
-    "R2=2.0;                 #Input resistor 2, kΩ\n",
-    "R3=4.0;                 #Input resistor 3, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Rf_R1=Rf/R1;                #Ratio of feedback resistor and 1st input resistor\n",
-    "Rf_R2=Rf/R2;                #Ratio of feedback resistor and 2nd input resistor\n",
-    "Rf_R3=Rf/R3;                #Ratio of feedback resistor and 3rd input resistor\n",
-    "\n",
-    "#(i) First input combination\n",
-    "V1=10;                                   #Input voltage to 1st input resistor, V\n",
-    "V2=0;                                    #Input voltage to 2nd input resistor, V\n",
-    "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-    "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-    "print(\"(i) The output voltage=%.1fV\"%Vout);\n",
-    "\n",
-    "#(i) First input combination\n",
-    "V1=0;                                   #Input voltage to 1st input resistor, V\n",
-    "V2=10;                                    #Input voltage to 2nd input resistor, V\n",
-    "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-    "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-    "print(\"(ii) The output voltage=%.1fV\"%Vout);\n",
-    "\n",
-    "\n",
-    "#(i) First input combination\n",
-    "V1=10;                                   #Input voltage to 1st input resistor, V\n",
-    "V2=10;                                    #Input voltage to 2nd input resistor, V\n",
-    "V3=10;                                   #Input voltage to 3rd input resistor, V\n",
-    "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3);          #Output voltage, V\n",
-    "print(\"(iii) The output voltage=%.1fV\"%Vout);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.49 : Page number 711"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 50,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vout=-[0.5sin(1000t)+0.33sin(3000t)]V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Rf=330;                 #Feedback resistor, kΩ\n",
-    "R1=33.0;                  #Input resistor 1, kΩ\n",
-    "R2=10.0;                  #Input resistor 2, kΩ\n",
-    "V1_m=50;                #Peak voltage of 1st input, mV\n",
-    "V2_m=10;                #Peak voltage of 2nd input, mV\n",
-    "\n",
-    "#Calculation\n",
-    "#Since, Vout=-((Rf/R1)*V1 + (Rf/R2)*V2)\n",
-    "print(\"Vout=-[%.1fsin(1000t)+%.2fsin(3000t)]V\"%((V1_m/1000.0)*(Rf/R1),(V2_m/1000.0)*(Rf/R2)));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.50 : Page number 715"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 51,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vo=-1*(1/RC)∫vi dt.\n",
-      "=>Vo=-1*(1/1)∫vi dt\n",
-      "=>Vo=∫vi dt\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=100;                      #Input resistor, kΩ\n",
-    "C=10;                       #Feedback capacitor, μF\n",
-    "\n",
-    "#Calculation\n",
-    "RC=R*10**3*C*10**-6;        #product of input resistance and feedback capacitance, s\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vo=-1*(1/RC)∫vi dt.\");\n",
-    "print(\"=>Vo=-1*(1/%d)∫vi dt\"%RC);\n",
-    "print(\"=>Vo=∫vi dt\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.51 : Page number 715-716"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 52,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The critical frequency=159Hz.\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import pi\n",
-    "\n",
-    "#Variable declaration\n",
-    "Rf=100;                      #Feedback resistor, kΩ\n",
-    "C=0.01;                      #Feedback capacitor, μF\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "fc=1/(2*pi*Rf*1000*C*10**-6);               #Crictical frequency, Hz\n",
-    "\n",
-    "#Result\n",
-    "print(\"The critical frequency=%dHz.\"%fc);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.52 : Page number 716"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 53,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Vout=-1*(1/RC)∫vi dt.\n",
-      "    ΔVout/dt = -vin/RC = -50mV/μs.\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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/w4w8f/Q0SQ0NhCDpTElzJc2W9FNJ6zWyP7NaJk2C73/fd2ibDUfdYzFJ2hmY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\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f8b042c6b70>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "R=10.0;                      #Input resistor, kΩ\n",
-    "C=0.01;                    #Feedback capacitor, μF\n",
-    "vin=5;                     #Input voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "Vout_change_rate=-vin/(R*C);            #Rate of change of output voltage, V/μs   \n",
-    "print(\"(i) Vout=-1*(1/RC)∫vi dt.\");\n",
-    "print(\"    ΔVout/dt = -vin/RC = %dmV/μs.\"%Vout_change_rate);\n",
-    "\n",
-    "#(ii) Plotting the output waveform\n",
-    "vin_plot=[];                    #Plotting variable for input waveform, V\n",
-    "dt=100;                          #time between edges, μs\n",
-    "for i in range(0,3*dt+1):\n",
-    "    if i<dt or i>2*dt :\n",
-    "        vin_plot.append(0);\n",
-    "    else:\n",
-    "        vin_plot.append(5); \n",
-    "\n",
-    "plt.subplot(211);\n",
-    "plt.plot(vin_plot);\n",
-    "plt.xlim([0,300])\n",
-    "plt.ylim([-5,10])\n",
-    "plt.xlabel(\"t(microsecond)\");\n",
-    "plt.ylabel(\"Vin(V)\");\n",
-    "plt.title(\"Input waveform\");\n",
-    "\n",
-    "        \n",
-    "vout_plot=[];                  #Plotting variable for output waveform, V\n",
-    "t=[i for i in range(0,301)];  #Time scale, μs\n",
-    "for i in t[:] :\n",
-    "    if i<dt:\n",
-    "        vout_plot.append(0);\n",
-    "    elif i>2*dt:\n",
-    "        vout_plot.append((Vout_change_rate/1000.0)*dt);\n",
-    "    else :\n",
-    "        vout_plot.append((-vin_plot[i]/(R*C))/1000*(i-dt));\n",
-    "\n",
-    "plt.subplot(212)\n",
-    "plt.plot(vout_plot);\n",
-    "plt.xlim([0,300])\n",
-    "plt.ylim([-5,5]);\n",
-    "plt.xlabel('t(microsecond)');\n",
-    "plt.ylabel(\"Vout(V)\");\n",
-    "plt.title(\"output waveform\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.53 : Page number 716-717"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 54,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vout=-1*(1/RC)∫vi dt.\n",
-      "Vout=-5*t volts\n",
-      "Time required=2.6seconds.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_supply=15;              #Supply voltage, V\n",
-    "R=10;                     #Input resistor, kΩ\n",
-    "C=0.2;                    #Feedback capacitor, μF\n",
-    "vin=10;                   #Input voltage, mV\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "Vs=-V_supply+2;         #Saturation voltage, V\n",
-    "print(\"Vout=-1*(1/RC)∫vi dt.\");\n",
-    "print(\"Vout=%d*t volts\"%(-vin/(R*C)));\n",
-    "t=Vs/(-vin/(R*C));                      #Time required, seconds\n",
-    "print(\"Time required=%.1fseconds.\"%t);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.54 : Page number 717-718"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 55,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vo=-5V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=1;                     #Feedback resistor, kΩ\n",
-    "C=0.1;                   #Input capacitor, μF\n",
-    "Vin_change=5;            #Change in input voltage, V\n",
-    "t=0.1;                   #Time taken for change in input voltage, ms\n",
-    "\n",
-    "#Calcualtion\n",
-    "dvi_dt=Vin_change/(t/1000);        #Rate of change of input voltage, V/s\n",
-    "RC=R*1000*C*10**-6;                #Product of feedback resistance and input capacitance, s\n",
-    "#Since, Vo=-R*C*(dvi/dt);\n",
-    "Vo=-RC*dvi_dt;                      #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vo=%dV.\"%Vo);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.55 : Page number 718"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 56,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vo=-0.55V.\n",
-      "The output voltage stays constant at -0.55V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=10;                     #Feedback resistor, kΩ\n",
-    "C=2.2;                   #Input capacitor, μF\n",
-    "Vin_change=10;            #Change in input voltage, V\n",
-    "t=0.4;                   #Time taken for change in input voltage, s\n",
-    "\n",
-    "#Calcualtin\n",
-    "dvi_dt=Vin_change/t;        #Rate of change of input voltage, V/s\n",
-    "RC=R*1000*C*10**-6;         #Product of feedback resistance and input capacitance, s\n",
-    "#Since, Vo=-R*C*(dvi/dt);\n",
-    "Vo=-RC*dvi_dt;                      #Output voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vo=%.2fV.\"%Vo);\n",
-    "print(\"The output voltage stays constant at %.2fV.\"%Vo);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 25.56 : Page number 718-719"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 57,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "vo=-1*(dvi/dt).\n",
-      "vo=-5V.\n",
-      "Therefore, between 0 to 0.2s, the output voltage is constant at -5V.\n",
-      "For t>0.2s, the input is constant so that output voltage is zero.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R=100;                     #Feedback resistor, kΩ\n",
-    "C=10;                   #Input capacitor, μF\n",
-    "Vin_change=1;            #Change in input voltage, V\n",
-    "t=0.2;                   #Time taken for change in input voltage, s\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "RC=R*1000*C*10**-6;             #Product of feedback resistor and input capacitance, s\n",
-    "#(i)\n",
-    "print(\"vo=-%d*(dvi/dt).\"%RC);\n",
-    "\n",
-    "#(ii)\n",
-    "dvi_dt=Vin_change/t;                #Rate of change of input voltage, V\n",
-    "vo=-dvi_dt;                         #Output voltage, V\n",
-    "print(\"vo=%dV.\"%vo);\n",
-    "\n",
-    "print(\"Therefore, between 0 to 0.2s, the output voltage is constant at %dV.\"%vo);\n",
-    "print(\"For t>0.2s, the input is constant so that output voltage is zero.\");\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26.ipynb
deleted file mode 100755
index 670a61b0..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26.ipynb
+++ /dev/null
@@ -1,472 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a1845801144904256bc26f3ca2e0294eb55dcabb139a523d403624121bc6876a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 26 : DIGITAL ELECTRONICS"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.1 : Page 732"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "d=37;         #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "b=int(bin(d)[2:]);                #Equivalent Octal number \n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent binary number=100101.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.2 : Page number 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "d=23;      #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "b=int(bin(d)[2:]);                   #Equivalent Octal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent binary number=%d.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent binary number=10111.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.3 : Page number 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "b=0b110001;               #Given binary number\n",
-      "\n",
-      "#Calculation\n",
-      "d=int(b);                   #Equivalent decimal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent decimal number=%d.\"%d);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent decimal number=49.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.4 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d1=76;                #Given decimal number\n",
-      "d2=255;               #Given decimal number\n",
-      "d3=372;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o1=int(oct(d1)[1:]);                   #Equivalent octal number\n",
-      "o2=int(oct(d2)[1:]);                   #Equivalent octal number\n",
-      "o3=int(oct(d3)[1:]);                   #Equivalent octal number\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Equivalent octal number=%d.\"%o1);\n",
-      "print(\"(ii)  Equivalent octal number=%d.\"%o2);\n",
-      "print(\"(iii) Equivalent octal number=%d.\"%o3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Equivalent octal number=114.\n",
-        "(ii)  Equivalent octal number=377.\n",
-        "(iii) Equivalent octal number=564.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.5 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "o=24.6;               #Given octal number\n",
-      "\n",
-      "#Calculation\n",
-      "o_f=o%1;                             #Floating part of octal number\n",
-      "o_i=(int)(o-(o%1));                  #Integer part of octal number\n",
-      "d=int(str(o_i),8);                   #Equivalent decimal number\n",
-      "\n",
-      "s=str(o_f);                         #String value of floating  part \n",
-      "i=2\n",
-      "while(i<len(s)):\n",
-      "    d=d+int(s[i])*8**-(i-1);\n",
-      "    i+=1;\n",
-      "#Result\n",
-      "print(\"Equivalent decimal number=%.2f.\"%d);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent decimal number=20.75.\n"
-       ]
-      }
-     ],
-     "prompt_number": 64
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.6 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d=177;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o=oct(d)[1:];                   #Equivalent octal number\n",
-      "\n",
-      "b=\"\";\n",
-      "for i in o:\n",
-      "    bo=bin(int(i))[2:];                     #Binary of individual octal digit\n",
-      "    b=b+((\"0\" if len(bo)==2 else (\"00\" if len(bo)==1 else\"\")) +bo);                   #Equivalent binary number\n",
-      "    \n",
-      "#Result\n",
-      "print(\"Equivalent octal number=%s.\"%o);\n",
-      "print(\"Equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent octal number=261.\n",
-        "Equivalent binary number=010110001.\n"
-       ]
-      }
-     ],
-     "prompt_number": 63
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.7 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d=541;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "h=hex(d)[2:];                   #Equivalent hexadecimal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent hexadecimal number=%s.\"%h);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent hexadecimal number=21d.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.8 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "hex_to_dec={'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9,'a':10,'b':11,'c':12,'d':13,'e':14,'f':15};\n",
-      "        \n",
-      "#Given \n",
-      "d=378;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "h=hex(d)[2:];                   #Equivalent Hexadecimal number\n",
-      "\n",
-      "\n",
-      "b=\"\";\n",
-      "for i in h:\n",
-      "    bh=bin(hex_to_dec[i])[2:];                 #Binary of individual hexadecimaldigit\n",
-      "    b=b+((\"0\" if len(bh)==3 else (\"00\" if len(bh)==2 else (\"000\" if len(bh)==1 else \"\")))+bh);    #Equivalent binary number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent hexadeciaml number=%s.\"%h);\n",
-      "print(\"Equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent hexadeciaml number=17a.\n",
-        "Equivalent binary number=000101111010.\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 26.9 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "h=0xB2F;        #Given hexadecimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o=oct(h)[1:];       #Equivalent octal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent octal number=%s.\"%o);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent octal number=5457.\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.10 : Page number 738"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "BCD=\"0100 0000 0010\"             #Given BCD string\n",
-      "BCD_split=BCD.split(\" \");        #Splitting th binary string into individual BCD \n",
-      "d=0;\n",
-      "for i in range(len(BCD_split),0,-1):\n",
-      "    d+=int(BCD_split[len(BCD_split)-i],2)*10**(i-1);\n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent decimal =%d.\"%d);\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent decimal =402.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.11 : Page number 745"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print(\"Boolean Expression obtained from the circuit: \\n Y'=A+B \\n Y=((A+B).A)\");\n",
-      "print(\"Truth Table:\");\n",
-      "print(\"a\\tb\\tY'=A+B\\t  Y=Y'.A\");\n",
-      "for b in range(0,2):\n",
-      "    for a in range(0,2):\n",
-      "        Y_dash=1 if a or b else 0;\n",
-      "        Y=1 if Y_dash and a else 0;\n",
-      "        print(\"%d\\t%d\\t%d\\t  %d\"%(a,b,Y_dash,Y));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boolean Expression obtained from the circuit: \n",
-        " Y'=A+B \n",
-        " Y=((A+B).A)\n",
-        "Truth Table:\n",
-        "a\tb\tY'=A+B\t  Y=Y'.A\n",
-        "0\t0\t0\t  0\n",
-        "1\t0\t1\t  1\n",
-        "0\t1\t1\t  0\n",
-        "1\t1\t1\t  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.12 : Page number 745-746"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print(\"Boolean Expression obtained from the circuit: \\n Y'=A'.B \\n Y=Y'+B'\");\n",
-      "print(\"Truth Table:\");\n",
-      "print(\"A\\tB\\tA'\\tY'=A'.B\\t  B'\\tY=Y'+B'\");\n",
-      "for b in range(0,2):\n",
-      "    for a in range(0,2):\n",
-      "        a_dash=1 if not a else 0;\n",
-      "        b_dash=1 if not b else 0;\n",
-      "        Y_dash=1 if a_dash and b else 0;\n",
-      "        Y=1 if Y_dash or b_dash else 0;\n",
-      "        print(\"%d\\t%d\\t%d\\t%d\\t  %d\\t%d\"%(a,b,a_dash,Y_dash,b_dash,Y));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boolean Expression obtained from the circuit: \n",
-        " Y'=A'.B \n",
-        " Y=Y'+B'\n",
-        "Truth Table:\n",
-        "A\tB\tA'\tY'=A'.B\t  B'\tY=Y'+B'\n",
-        "0\t0\t1\t0\t  1\t1\n",
-        "1\t0\t0\t0\t  1\t1\n",
-        "0\t1\t1\t1\t  0\t1\n",
-        "1\t1\t0\t0\t  0\t0\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_1.ipynb
deleted file mode 100755
index 670a61b0..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_1.ipynb
+++ /dev/null
@@ -1,472 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a1845801144904256bc26f3ca2e0294eb55dcabb139a523d403624121bc6876a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 26 : DIGITAL ELECTRONICS"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.1 : Page 732"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "d=37;         #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "b=int(bin(d)[2:]);                #Equivalent Octal number \n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent binary number=100101.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.2 : Page number 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "d=23;      #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "b=int(bin(d)[2:]);                   #Equivalent Octal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent binary number=%d.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent binary number=10111.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.3 : Page number 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "b=0b110001;               #Given binary number\n",
-      "\n",
-      "#Calculation\n",
-      "d=int(b);                   #Equivalent decimal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent decimal number=%d.\"%d);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent decimal number=49.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.4 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d1=76;                #Given decimal number\n",
-      "d2=255;               #Given decimal number\n",
-      "d3=372;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o1=int(oct(d1)[1:]);                   #Equivalent octal number\n",
-      "o2=int(oct(d2)[1:]);                   #Equivalent octal number\n",
-      "o3=int(oct(d3)[1:]);                   #Equivalent octal number\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Equivalent octal number=%d.\"%o1);\n",
-      "print(\"(ii)  Equivalent octal number=%d.\"%o2);\n",
-      "print(\"(iii) Equivalent octal number=%d.\"%o3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Equivalent octal number=114.\n",
-        "(ii)  Equivalent octal number=377.\n",
-        "(iii) Equivalent octal number=564.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.5 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "o=24.6;               #Given octal number\n",
-      "\n",
-      "#Calculation\n",
-      "o_f=o%1;                             #Floating part of octal number\n",
-      "o_i=(int)(o-(o%1));                  #Integer part of octal number\n",
-      "d=int(str(o_i),8);                   #Equivalent decimal number\n",
-      "\n",
-      "s=str(o_f);                         #String value of floating  part \n",
-      "i=2\n",
-      "while(i<len(s)):\n",
-      "    d=d+int(s[i])*8**-(i-1);\n",
-      "    i+=1;\n",
-      "#Result\n",
-      "print(\"Equivalent decimal number=%.2f.\"%d);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent decimal number=20.75.\n"
-       ]
-      }
-     ],
-     "prompt_number": 64
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.6 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d=177;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o=oct(d)[1:];                   #Equivalent octal number\n",
-      "\n",
-      "b=\"\";\n",
-      "for i in o:\n",
-      "    bo=bin(int(i))[2:];                     #Binary of individual octal digit\n",
-      "    b=b+((\"0\" if len(bo)==2 else (\"00\" if len(bo)==1 else\"\")) +bo);                   #Equivalent binary number\n",
-      "    \n",
-      "#Result\n",
-      "print(\"Equivalent octal number=%s.\"%o);\n",
-      "print(\"Equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent octal number=261.\n",
-        "Equivalent binary number=010110001.\n"
-       ]
-      }
-     ],
-     "prompt_number": 63
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.7 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d=541;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "h=hex(d)[2:];                   #Equivalent hexadecimal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent hexadecimal number=%s.\"%h);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent hexadecimal number=21d.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.8 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "hex_to_dec={'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9,'a':10,'b':11,'c':12,'d':13,'e':14,'f':15};\n",
-      "        \n",
-      "#Given \n",
-      "d=378;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "h=hex(d)[2:];                   #Equivalent Hexadecimal number\n",
-      "\n",
-      "\n",
-      "b=\"\";\n",
-      "for i in h:\n",
-      "    bh=bin(hex_to_dec[i])[2:];                 #Binary of individual hexadecimaldigit\n",
-      "    b=b+((\"0\" if len(bh)==3 else (\"00\" if len(bh)==2 else (\"000\" if len(bh)==1 else \"\")))+bh);    #Equivalent binary number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent hexadeciaml number=%s.\"%h);\n",
-      "print(\"Equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent hexadeciaml number=17a.\n",
-        "Equivalent binary number=000101111010.\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 26.9 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "h=0xB2F;        #Given hexadecimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o=oct(h)[1:];       #Equivalent octal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent octal number=%s.\"%o);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent octal number=5457.\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.10 : Page number 738"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "BCD=\"0100 0000 0010\"             #Given BCD string\n",
-      "BCD_split=BCD.split(\" \");        #Splitting th binary string into individual BCD \n",
-      "d=0;\n",
-      "for i in range(len(BCD_split),0,-1):\n",
-      "    d+=int(BCD_split[len(BCD_split)-i],2)*10**(i-1);\n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent decimal =%d.\"%d);\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent decimal =402.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.11 : Page number 745"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print(\"Boolean Expression obtained from the circuit: \\n Y'=A+B \\n Y=((A+B).A)\");\n",
-      "print(\"Truth Table:\");\n",
-      "print(\"a\\tb\\tY'=A+B\\t  Y=Y'.A\");\n",
-      "for b in range(0,2):\n",
-      "    for a in range(0,2):\n",
-      "        Y_dash=1 if a or b else 0;\n",
-      "        Y=1 if Y_dash and a else 0;\n",
-      "        print(\"%d\\t%d\\t%d\\t  %d\"%(a,b,Y_dash,Y));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boolean Expression obtained from the circuit: \n",
-        " Y'=A+B \n",
-        " Y=((A+B).A)\n",
-        "Truth Table:\n",
-        "a\tb\tY'=A+B\t  Y=Y'.A\n",
-        "0\t0\t0\t  0\n",
-        "1\t0\t1\t  1\n",
-        "0\t1\t1\t  0\n",
-        "1\t1\t1\t  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.12 : Page number 745-746"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print(\"Boolean Expression obtained from the circuit: \\n Y'=A'.B \\n Y=Y'+B'\");\n",
-      "print(\"Truth Table:\");\n",
-      "print(\"A\\tB\\tA'\\tY'=A'.B\\t  B'\\tY=Y'+B'\");\n",
-      "for b in range(0,2):\n",
-      "    for a in range(0,2):\n",
-      "        a_dash=1 if not a else 0;\n",
-      "        b_dash=1 if not b else 0;\n",
-      "        Y_dash=1 if a_dash and b else 0;\n",
-      "        Y=1 if Y_dash or b_dash else 0;\n",
-      "        print(\"%d\\t%d\\t%d\\t%d\\t  %d\\t%d\"%(a,b,a_dash,Y_dash,b_dash,Y));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boolean Expression obtained from the circuit: \n",
-        " Y'=A'.B \n",
-        " Y=Y'+B'\n",
-        "Truth Table:\n",
-        "A\tB\tA'\tY'=A'.B\t  B'\tY=Y'+B'\n",
-        "0\t0\t1\t0\t  1\t1\n",
-        "1\t0\t0\t0\t  1\t1\n",
-        "0\t1\t1\t1\t  0\t1\n",
-        "1\t1\t0\t0\t  0\t0\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_2.ipynb
deleted file mode 100755
index 670a61b0..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_2.ipynb
+++ /dev/null
@@ -1,472 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a1845801144904256bc26f3ca2e0294eb55dcabb139a523d403624121bc6876a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 26 : DIGITAL ELECTRONICS"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.1 : Page 732"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "d=37;         #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "b=int(bin(d)[2:]);                #Equivalent Octal number \n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent binary number=100101.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.2 : Page number 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "d=23;      #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "b=int(bin(d)[2:]);                   #Equivalent Octal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent binary number=%d.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent binary number=10111.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.3 : Page number 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "b=0b110001;               #Given binary number\n",
-      "\n",
-      "#Calculation\n",
-      "d=int(b);                   #Equivalent decimal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent decimal number=%d.\"%d);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent decimal number=49.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.4 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d1=76;                #Given decimal number\n",
-      "d2=255;               #Given decimal number\n",
-      "d3=372;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o1=int(oct(d1)[1:]);                   #Equivalent octal number\n",
-      "o2=int(oct(d2)[1:]);                   #Equivalent octal number\n",
-      "o3=int(oct(d3)[1:]);                   #Equivalent octal number\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Equivalent octal number=%d.\"%o1);\n",
-      "print(\"(ii)  Equivalent octal number=%d.\"%o2);\n",
-      "print(\"(iii) Equivalent octal number=%d.\"%o3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Equivalent octal number=114.\n",
-        "(ii)  Equivalent octal number=377.\n",
-        "(iii) Equivalent octal number=564.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.5 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "o=24.6;               #Given octal number\n",
-      "\n",
-      "#Calculation\n",
-      "o_f=o%1;                             #Floating part of octal number\n",
-      "o_i=(int)(o-(o%1));                  #Integer part of octal number\n",
-      "d=int(str(o_i),8);                   #Equivalent decimal number\n",
-      "\n",
-      "s=str(o_f);                         #String value of floating  part \n",
-      "i=2\n",
-      "while(i<len(s)):\n",
-      "    d=d+int(s[i])*8**-(i-1);\n",
-      "    i+=1;\n",
-      "#Result\n",
-      "print(\"Equivalent decimal number=%.2f.\"%d);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent decimal number=20.75.\n"
-       ]
-      }
-     ],
-     "prompt_number": 64
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.6 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d=177;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o=oct(d)[1:];                   #Equivalent octal number\n",
-      "\n",
-      "b=\"\";\n",
-      "for i in o:\n",
-      "    bo=bin(int(i))[2:];                     #Binary of individual octal digit\n",
-      "    b=b+((\"0\" if len(bo)==2 else (\"00\" if len(bo)==1 else\"\")) +bo);                   #Equivalent binary number\n",
-      "    \n",
-      "#Result\n",
-      "print(\"Equivalent octal number=%s.\"%o);\n",
-      "print(\"Equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent octal number=261.\n",
-        "Equivalent binary number=010110001.\n"
-       ]
-      }
-     ],
-     "prompt_number": 63
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.7 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d=541;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "h=hex(d)[2:];                   #Equivalent hexadecimal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent hexadecimal number=%s.\"%h);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent hexadecimal number=21d.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.8 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "hex_to_dec={'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9,'a':10,'b':11,'c':12,'d':13,'e':14,'f':15};\n",
-      "        \n",
-      "#Given \n",
-      "d=378;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "h=hex(d)[2:];                   #Equivalent Hexadecimal number\n",
-      "\n",
-      "\n",
-      "b=\"\";\n",
-      "for i in h:\n",
-      "    bh=bin(hex_to_dec[i])[2:];                 #Binary of individual hexadecimaldigit\n",
-      "    b=b+((\"0\" if len(bh)==3 else (\"00\" if len(bh)==2 else (\"000\" if len(bh)==1 else \"\")))+bh);    #Equivalent binary number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent hexadeciaml number=%s.\"%h);\n",
-      "print(\"Equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent hexadeciaml number=17a.\n",
-        "Equivalent binary number=000101111010.\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 26.9 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "h=0xB2F;        #Given hexadecimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o=oct(h)[1:];       #Equivalent octal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent octal number=%s.\"%o);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent octal number=5457.\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.10 : Page number 738"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "BCD=\"0100 0000 0010\"             #Given BCD string\n",
-      "BCD_split=BCD.split(\" \");        #Splitting th binary string into individual BCD \n",
-      "d=0;\n",
-      "for i in range(len(BCD_split),0,-1):\n",
-      "    d+=int(BCD_split[len(BCD_split)-i],2)*10**(i-1);\n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent decimal =%d.\"%d);\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent decimal =402.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.11 : Page number 745"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print(\"Boolean Expression obtained from the circuit: \\n Y'=A+B \\n Y=((A+B).A)\");\n",
-      "print(\"Truth Table:\");\n",
-      "print(\"a\\tb\\tY'=A+B\\t  Y=Y'.A\");\n",
-      "for b in range(0,2):\n",
-      "    for a in range(0,2):\n",
-      "        Y_dash=1 if a or b else 0;\n",
-      "        Y=1 if Y_dash and a else 0;\n",
-      "        print(\"%d\\t%d\\t%d\\t  %d\"%(a,b,Y_dash,Y));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boolean Expression obtained from the circuit: \n",
-        " Y'=A+B \n",
-        " Y=((A+B).A)\n",
-        "Truth Table:\n",
-        "a\tb\tY'=A+B\t  Y=Y'.A\n",
-        "0\t0\t0\t  0\n",
-        "1\t0\t1\t  1\n",
-        "0\t1\t1\t  0\n",
-        "1\t1\t1\t  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.12 : Page number 745-746"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print(\"Boolean Expression obtained from the circuit: \\n Y'=A'.B \\n Y=Y'+B'\");\n",
-      "print(\"Truth Table:\");\n",
-      "print(\"A\\tB\\tA'\\tY'=A'.B\\t  B'\\tY=Y'+B'\");\n",
-      "for b in range(0,2):\n",
-      "    for a in range(0,2):\n",
-      "        a_dash=1 if not a else 0;\n",
-      "        b_dash=1 if not b else 0;\n",
-      "        Y_dash=1 if a_dash and b else 0;\n",
-      "        Y=1 if Y_dash or b_dash else 0;\n",
-      "        print(\"%d\\t%d\\t%d\\t%d\\t  %d\\t%d\"%(a,b,a_dash,Y_dash,b_dash,Y));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boolean Expression obtained from the circuit: \n",
-        " Y'=A'.B \n",
-        " Y=Y'+B'\n",
-        "Truth Table:\n",
-        "A\tB\tA'\tY'=A'.B\t  B'\tY=Y'+B'\n",
-        "0\t0\t1\t0\t  1\t1\n",
-        "1\t0\t0\t0\t  1\t1\n",
-        "0\t1\t1\t1\t  0\t1\n",
-        "1\t1\t0\t0\t  0\t0\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_3.ipynb
deleted file mode 100755
index 670a61b0..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_3.ipynb
+++ /dev/null
@@ -1,472 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a1845801144904256bc26f3ca2e0294eb55dcabb139a523d403624121bc6876a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 26 : DIGITAL ELECTRONICS"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.1 : Page 732"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "d=37;         #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "b=int(bin(d)[2:]);                #Equivalent Octal number \n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent binary number=100101.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.2 : Page number 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "d=23;      #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "b=int(bin(d)[2:]);                   #Equivalent Octal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent binary number=%d.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent binary number=10111.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.3 : Page number 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "b=0b110001;               #Given binary number\n",
-      "\n",
-      "#Calculation\n",
-      "d=int(b);                   #Equivalent decimal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent decimal number=%d.\"%d);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent decimal number=49.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.4 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d1=76;                #Given decimal number\n",
-      "d2=255;               #Given decimal number\n",
-      "d3=372;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o1=int(oct(d1)[1:]);                   #Equivalent octal number\n",
-      "o2=int(oct(d2)[1:]);                   #Equivalent octal number\n",
-      "o3=int(oct(d3)[1:]);                   #Equivalent octal number\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Equivalent octal number=%d.\"%o1);\n",
-      "print(\"(ii)  Equivalent octal number=%d.\"%o2);\n",
-      "print(\"(iii) Equivalent octal number=%d.\"%o3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Equivalent octal number=114.\n",
-        "(ii)  Equivalent octal number=377.\n",
-        "(iii) Equivalent octal number=564.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.5 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "o=24.6;               #Given octal number\n",
-      "\n",
-      "#Calculation\n",
-      "o_f=o%1;                             #Floating part of octal number\n",
-      "o_i=(int)(o-(o%1));                  #Integer part of octal number\n",
-      "d=int(str(o_i),8);                   #Equivalent decimal number\n",
-      "\n",
-      "s=str(o_f);                         #String value of floating  part \n",
-      "i=2\n",
-      "while(i<len(s)):\n",
-      "    d=d+int(s[i])*8**-(i-1);\n",
-      "    i+=1;\n",
-      "#Result\n",
-      "print(\"Equivalent decimal number=%.2f.\"%d);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent decimal number=20.75.\n"
-       ]
-      }
-     ],
-     "prompt_number": 64
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.6 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d=177;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o=oct(d)[1:];                   #Equivalent octal number\n",
-      "\n",
-      "b=\"\";\n",
-      "for i in o:\n",
-      "    bo=bin(int(i))[2:];                     #Binary of individual octal digit\n",
-      "    b=b+((\"0\" if len(bo)==2 else (\"00\" if len(bo)==1 else\"\")) +bo);                   #Equivalent binary number\n",
-      "    \n",
-      "#Result\n",
-      "print(\"Equivalent octal number=%s.\"%o);\n",
-      "print(\"Equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent octal number=261.\n",
-        "Equivalent binary number=010110001.\n"
-       ]
-      }
-     ],
-     "prompt_number": 63
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.7 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d=541;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "h=hex(d)[2:];                   #Equivalent hexadecimal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent hexadecimal number=%s.\"%h);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent hexadecimal number=21d.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.8 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "hex_to_dec={'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9,'a':10,'b':11,'c':12,'d':13,'e':14,'f':15};\n",
-      "        \n",
-      "#Given \n",
-      "d=378;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "h=hex(d)[2:];                   #Equivalent Hexadecimal number\n",
-      "\n",
-      "\n",
-      "b=\"\";\n",
-      "for i in h:\n",
-      "    bh=bin(hex_to_dec[i])[2:];                 #Binary of individual hexadecimaldigit\n",
-      "    b=b+((\"0\" if len(bh)==3 else (\"00\" if len(bh)==2 else (\"000\" if len(bh)==1 else \"\")))+bh);    #Equivalent binary number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent hexadeciaml number=%s.\"%h);\n",
-      "print(\"Equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent hexadeciaml number=17a.\n",
-        "Equivalent binary number=000101111010.\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 26.9 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "h=0xB2F;        #Given hexadecimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o=oct(h)[1:];       #Equivalent octal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent octal number=%s.\"%o);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent octal number=5457.\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.10 : Page number 738"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "BCD=\"0100 0000 0010\"             #Given BCD string\n",
-      "BCD_split=BCD.split(\" \");        #Splitting th binary string into individual BCD \n",
-      "d=0;\n",
-      "for i in range(len(BCD_split),0,-1):\n",
-      "    d+=int(BCD_split[len(BCD_split)-i],2)*10**(i-1);\n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent decimal =%d.\"%d);\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent decimal =402.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.11 : Page number 745"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print(\"Boolean Expression obtained from the circuit: \\n Y'=A+B \\n Y=((A+B).A)\");\n",
-      "print(\"Truth Table:\");\n",
-      "print(\"a\\tb\\tY'=A+B\\t  Y=Y'.A\");\n",
-      "for b in range(0,2):\n",
-      "    for a in range(0,2):\n",
-      "        Y_dash=1 if a or b else 0;\n",
-      "        Y=1 if Y_dash and a else 0;\n",
-      "        print(\"%d\\t%d\\t%d\\t  %d\"%(a,b,Y_dash,Y));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boolean Expression obtained from the circuit: \n",
-        " Y'=A+B \n",
-        " Y=((A+B).A)\n",
-        "Truth Table:\n",
-        "a\tb\tY'=A+B\t  Y=Y'.A\n",
-        "0\t0\t0\t  0\n",
-        "1\t0\t1\t  1\n",
-        "0\t1\t1\t  0\n",
-        "1\t1\t1\t  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.12 : Page number 745-746"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print(\"Boolean Expression obtained from the circuit: \\n Y'=A'.B \\n Y=Y'+B'\");\n",
-      "print(\"Truth Table:\");\n",
-      "print(\"A\\tB\\tA'\\tY'=A'.B\\t  B'\\tY=Y'+B'\");\n",
-      "for b in range(0,2):\n",
-      "    for a in range(0,2):\n",
-      "        a_dash=1 if not a else 0;\n",
-      "        b_dash=1 if not b else 0;\n",
-      "        Y_dash=1 if a_dash and b else 0;\n",
-      "        Y=1 if Y_dash or b_dash else 0;\n",
-      "        print(\"%d\\t%d\\t%d\\t%d\\t  %d\\t%d\"%(a,b,a_dash,Y_dash,b_dash,Y));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boolean Expression obtained from the circuit: \n",
-        " Y'=A'.B \n",
-        " Y=Y'+B'\n",
-        "Truth Table:\n",
-        "A\tB\tA'\tY'=A'.B\t  B'\tY=Y'+B'\n",
-        "0\t0\t1\t0\t  1\t1\n",
-        "1\t0\t0\t0\t  1\t1\n",
-        "0\t1\t1\t1\t  0\t1\n",
-        "1\t1\t0\t0\t  0\t0\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_4.ipynb
deleted file mode 100755
index 670a61b0..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_4.ipynb
+++ /dev/null
@@ -1,472 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a1845801144904256bc26f3ca2e0294eb55dcabb139a523d403624121bc6876a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 26 : DIGITAL ELECTRONICS"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.1 : Page 732"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "d=37;         #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "b=int(bin(d)[2:]);                #Equivalent Octal number \n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent binary number=100101.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.2 : Page number 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "d=23;      #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "b=int(bin(d)[2:]);                   #Equivalent Octal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent binary number=%d.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent binary number=10111.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.3 : Page number 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "b=0b110001;               #Given binary number\n",
-      "\n",
-      "#Calculation\n",
-      "d=int(b);                   #Equivalent decimal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent decimal number=%d.\"%d);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent decimal number=49.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.4 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d1=76;                #Given decimal number\n",
-      "d2=255;               #Given decimal number\n",
-      "d3=372;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o1=int(oct(d1)[1:]);                   #Equivalent octal number\n",
-      "o2=int(oct(d2)[1:]);                   #Equivalent octal number\n",
-      "o3=int(oct(d3)[1:]);                   #Equivalent octal number\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Equivalent octal number=%d.\"%o1);\n",
-      "print(\"(ii)  Equivalent octal number=%d.\"%o2);\n",
-      "print(\"(iii) Equivalent octal number=%d.\"%o3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Equivalent octal number=114.\n",
-        "(ii)  Equivalent octal number=377.\n",
-        "(iii) Equivalent octal number=564.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.5 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "o=24.6;               #Given octal number\n",
-      "\n",
-      "#Calculation\n",
-      "o_f=o%1;                             #Floating part of octal number\n",
-      "o_i=(int)(o-(o%1));                  #Integer part of octal number\n",
-      "d=int(str(o_i),8);                   #Equivalent decimal number\n",
-      "\n",
-      "s=str(o_f);                         #String value of floating  part \n",
-      "i=2\n",
-      "while(i<len(s)):\n",
-      "    d=d+int(s[i])*8**-(i-1);\n",
-      "    i+=1;\n",
-      "#Result\n",
-      "print(\"Equivalent decimal number=%.2f.\"%d);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent decimal number=20.75.\n"
-       ]
-      }
-     ],
-     "prompt_number": 64
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.6 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d=177;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o=oct(d)[1:];                   #Equivalent octal number\n",
-      "\n",
-      "b=\"\";\n",
-      "for i in o:\n",
-      "    bo=bin(int(i))[2:];                     #Binary of individual octal digit\n",
-      "    b=b+((\"0\" if len(bo)==2 else (\"00\" if len(bo)==1 else\"\")) +bo);                   #Equivalent binary number\n",
-      "    \n",
-      "#Result\n",
-      "print(\"Equivalent octal number=%s.\"%o);\n",
-      "print(\"Equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent octal number=261.\n",
-        "Equivalent binary number=010110001.\n"
-       ]
-      }
-     ],
-     "prompt_number": 63
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.7 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d=541;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "h=hex(d)[2:];                   #Equivalent hexadecimal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent hexadecimal number=%s.\"%h);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent hexadecimal number=21d.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.8 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "hex_to_dec={'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9,'a':10,'b':11,'c':12,'d':13,'e':14,'f':15};\n",
-      "        \n",
-      "#Given \n",
-      "d=378;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "h=hex(d)[2:];                   #Equivalent Hexadecimal number\n",
-      "\n",
-      "\n",
-      "b=\"\";\n",
-      "for i in h:\n",
-      "    bh=bin(hex_to_dec[i])[2:];                 #Binary of individual hexadecimaldigit\n",
-      "    b=b+((\"0\" if len(bh)==3 else (\"00\" if len(bh)==2 else (\"000\" if len(bh)==1 else \"\")))+bh);    #Equivalent binary number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent hexadeciaml number=%s.\"%h);\n",
-      "print(\"Equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent hexadeciaml number=17a.\n",
-        "Equivalent binary number=000101111010.\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 26.9 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "h=0xB2F;        #Given hexadecimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o=oct(h)[1:];       #Equivalent octal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent octal number=%s.\"%o);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent octal number=5457.\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.10 : Page number 738"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "BCD=\"0100 0000 0010\"             #Given BCD string\n",
-      "BCD_split=BCD.split(\" \");        #Splitting th binary string into individual BCD \n",
-      "d=0;\n",
-      "for i in range(len(BCD_split),0,-1):\n",
-      "    d+=int(BCD_split[len(BCD_split)-i],2)*10**(i-1);\n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent decimal =%d.\"%d);\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent decimal =402.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.11 : Page number 745"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print(\"Boolean Expression obtained from the circuit: \\n Y'=A+B \\n Y=((A+B).A)\");\n",
-      "print(\"Truth Table:\");\n",
-      "print(\"a\\tb\\tY'=A+B\\t  Y=Y'.A\");\n",
-      "for b in range(0,2):\n",
-      "    for a in range(0,2):\n",
-      "        Y_dash=1 if a or b else 0;\n",
-      "        Y=1 if Y_dash and a else 0;\n",
-      "        print(\"%d\\t%d\\t%d\\t  %d\"%(a,b,Y_dash,Y));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boolean Expression obtained from the circuit: \n",
-        " Y'=A+B \n",
-        " Y=((A+B).A)\n",
-        "Truth Table:\n",
-        "a\tb\tY'=A+B\t  Y=Y'.A\n",
-        "0\t0\t0\t  0\n",
-        "1\t0\t1\t  1\n",
-        "0\t1\t1\t  0\n",
-        "1\t1\t1\t  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.12 : Page number 745-746"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print(\"Boolean Expression obtained from the circuit: \\n Y'=A'.B \\n Y=Y'+B'\");\n",
-      "print(\"Truth Table:\");\n",
-      "print(\"A\\tB\\tA'\\tY'=A'.B\\t  B'\\tY=Y'+B'\");\n",
-      "for b in range(0,2):\n",
-      "    for a in range(0,2):\n",
-      "        a_dash=1 if not a else 0;\n",
-      "        b_dash=1 if not b else 0;\n",
-      "        Y_dash=1 if a_dash and b else 0;\n",
-      "        Y=1 if Y_dash or b_dash else 0;\n",
-      "        print(\"%d\\t%d\\t%d\\t%d\\t  %d\\t%d\"%(a,b,a_dash,Y_dash,b_dash,Y));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boolean Expression obtained from the circuit: \n",
-        " Y'=A'.B \n",
-        " Y=Y'+B'\n",
-        "Truth Table:\n",
-        "A\tB\tA'\tY'=A'.B\t  B'\tY=Y'+B'\n",
-        "0\t0\t1\t0\t  1\t1\n",
-        "1\t0\t0\t0\t  1\t1\n",
-        "0\t1\t1\t1\t  0\t1\n",
-        "1\t1\t0\t0\t  0\t0\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_5.ipynb
deleted file mode 100755
index 670a61b0..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_5.ipynb
+++ /dev/null
@@ -1,472 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a1845801144904256bc26f3ca2e0294eb55dcabb139a523d403624121bc6876a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "#CHAPTER 26 : DIGITAL ELECTRONICS"
-     ]
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.1 : Page 732"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "d=37;         #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "b=int(bin(d)[2:]);                #Equivalent Octal number \n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent binary number=100101.\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.2 : Page number 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "d=23;      #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "b=int(bin(d)[2:]);                   #Equivalent Octal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent binary number=%d.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent binary number=10111.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.3 : Page number 733"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "b=0b110001;               #Given binary number\n",
-      "\n",
-      "#Calculation\n",
-      "d=int(b);                   #Equivalent decimal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent decimal number=%d.\"%d);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent decimal number=49.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.4 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d1=76;                #Given decimal number\n",
-      "d2=255;               #Given decimal number\n",
-      "d3=372;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o1=int(oct(d1)[1:]);                   #Equivalent octal number\n",
-      "o2=int(oct(d2)[1:]);                   #Equivalent octal number\n",
-      "o3=int(oct(d3)[1:]);                   #Equivalent octal number\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i)   Equivalent octal number=%d.\"%o1);\n",
-      "print(\"(ii)  Equivalent octal number=%d.\"%o2);\n",
-      "print(\"(iii) Equivalent octal number=%d.\"%o3);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)   Equivalent octal number=114.\n",
-        "(ii)  Equivalent octal number=377.\n",
-        "(iii) Equivalent octal number=564.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.5 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "o=24.6;               #Given octal number\n",
-      "\n",
-      "#Calculation\n",
-      "o_f=o%1;                             #Floating part of octal number\n",
-      "o_i=(int)(o-(o%1));                  #Integer part of octal number\n",
-      "d=int(str(o_i),8);                   #Equivalent decimal number\n",
-      "\n",
-      "s=str(o_f);                         #String value of floating  part \n",
-      "i=2\n",
-      "while(i<len(s)):\n",
-      "    d=d+int(s[i])*8**-(i-1);\n",
-      "    i+=1;\n",
-      "#Result\n",
-      "print(\"Equivalent decimal number=%.2f.\"%d);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent decimal number=20.75.\n"
-       ]
-      }
-     ],
-     "prompt_number": 64
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.6 : Page number 735"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d=177;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o=oct(d)[1:];                   #Equivalent octal number\n",
-      "\n",
-      "b=\"\";\n",
-      "for i in o:\n",
-      "    bo=bin(int(i))[2:];                     #Binary of individual octal digit\n",
-      "    b=b+((\"0\" if len(bo)==2 else (\"00\" if len(bo)==1 else\"\")) +bo);                   #Equivalent binary number\n",
-      "    \n",
-      "#Result\n",
-      "print(\"Equivalent octal number=%s.\"%o);\n",
-      "print(\"Equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent octal number=261.\n",
-        "Equivalent binary number=010110001.\n"
-       ]
-      }
-     ],
-     "prompt_number": 63
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.7 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given \n",
-      "d=541;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "h=hex(d)[2:];                   #Equivalent hexadecimal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent hexadecimal number=%s.\"%h);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent hexadecimal number=21d.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.8 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "hex_to_dec={'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9,'a':10,'b':11,'c':12,'d':13,'e':14,'f':15};\n",
-      "        \n",
-      "#Given \n",
-      "d=378;               #Given decimal number\n",
-      "\n",
-      "#Calculation\n",
-      "h=hex(d)[2:];                   #Equivalent Hexadecimal number\n",
-      "\n",
-      "\n",
-      "b=\"\";\n",
-      "for i in h:\n",
-      "    bh=bin(hex_to_dec[i])[2:];                 #Binary of individual hexadecimaldigit\n",
-      "    b=b+((\"0\" if len(bh)==3 else (\"00\" if len(bh)==2 else (\"000\" if len(bh)==1 else \"\")))+bh);    #Equivalent binary number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent hexadeciaml number=%s.\"%h);\n",
-      "print(\"Equivalent binary number=%s.\"%b);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent hexadeciaml number=17a.\n",
-        "Equivalent binary number=000101111010.\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 26.9 : Page number 737"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "h=0xB2F;        #Given hexadecimal number\n",
-      "\n",
-      "#Calculation\n",
-      "o=oct(h)[1:];       #Equivalent octal number\n",
-      "\n",
-      "#Result\n",
-      "print(\"Equivalent octal number=%s.\"%o);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Equivalent octal number=5457.\n"
-       ]
-      }
-     ],
-     "prompt_number": 58
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.10 : Page number 738"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given\n",
-      "BCD=\"0100 0000 0010\"             #Given BCD string\n",
-      "BCD_split=BCD.split(\" \");        #Splitting th binary string into individual BCD \n",
-      "d=0;\n",
-      "for i in range(len(BCD_split),0,-1):\n",
-      "    d+=int(BCD_split[len(BCD_split)-i],2)*10**(i-1);\n",
-      "\n",
-      "#Result\n",
-      "print(\"The equivalent decimal =%d.\"%d);\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The equivalent decimal =402.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.11 : Page number 745"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print(\"Boolean Expression obtained from the circuit: \\n Y'=A+B \\n Y=((A+B).A)\");\n",
-      "print(\"Truth Table:\");\n",
-      "print(\"a\\tb\\tY'=A+B\\t  Y=Y'.A\");\n",
-      "for b in range(0,2):\n",
-      "    for a in range(0,2):\n",
-      "        Y_dash=1 if a or b else 0;\n",
-      "        Y=1 if Y_dash and a else 0;\n",
-      "        print(\"%d\\t%d\\t%d\\t  %d\"%(a,b,Y_dash,Y));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boolean Expression obtained from the circuit: \n",
-        " Y'=A+B \n",
-        " Y=((A+B).A)\n",
-        "Truth Table:\n",
-        "a\tb\tY'=A+B\t  Y=Y'.A\n",
-        "0\t0\t0\t  0\n",
-        "1\t0\t1\t  1\n",
-        "0\t1\t1\t  0\n",
-        "1\t1\t1\t  1\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 26.12 : Page number 745-746"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "print(\"Boolean Expression obtained from the circuit: \\n Y'=A'.B \\n Y=Y'+B'\");\n",
-      "print(\"Truth Table:\");\n",
-      "print(\"A\\tB\\tA'\\tY'=A'.B\\t  B'\\tY=Y'+B'\");\n",
-      "for b in range(0,2):\n",
-      "    for a in range(0,2):\n",
-      "        a_dash=1 if not a else 0;\n",
-      "        b_dash=1 if not b else 0;\n",
-      "        Y_dash=1 if a_dash and b else 0;\n",
-      "        Y=1 if Y_dash or b_dash else 0;\n",
-      "        print(\"%d\\t%d\\t%d\\t%d\\t  %d\\t%d\"%(a,b,a_dash,Y_dash,b_dash,Y));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Boolean Expression obtained from the circuit: \n",
-        " Y'=A'.B \n",
-        " Y=Y'+B'\n",
-        "Truth Table:\n",
-        "A\tB\tA'\tY'=A'.B\t  B'\tY=Y'+B'\n",
-        "0\t0\t1\t0\t  1\t1\n",
-        "1\t0\t0\t0\t  1\t1\n",
-        "0\t1\t1\t1\t  0\t1\n",
-        "1\t1\t0\t0\t  0\t0\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_1.ipynb
deleted file mode 100755
index 06a555a6..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_1.ipynb
+++ /dev/null
@@ -1,125 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c0ff4f67576afe73a11c06eedd0a50709b7f5831737f83db1cd640098e3e9740"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 2 : ELECTRONIC EMISSION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1: Page number 31\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "\n",
-      "from math import exp\n",
-      "from math import pi\n",
-      "\n",
-      "l=5.0;                        #length of tungsten filament in cm\n",
-      "d=0.01;                       #diameter of the filament in cm\n",
-      "T=2500.0;                     #operating temperature in K\n",
-      "A=60.2*pow(10,4);             #constant, depending upon the type of thermionic emitter, in amp/m\u00b2/K\u00b2\n",
-      "phi=4.517;                    #work function of emitter in eV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "b=round(11600*phi,-1);                       #constant for a metal, in K\n",
-      "Js=round(A*T*T*exp(-b/T),-2);                #Emission current density in amp/m\u00b2\n",
-      "a=pi*(d/100)*(l/100);                        #Surface area of the cathode in m\u00b2\n",
-      "E_I=Js*a;                                     #Emission current in A\n",
-      "\n",
-      "#Result\n",
-      "print(\"emission current =%.3f A\"%E_I);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "emission current =0.047 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2:Page number 31\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "Js=0.1;                   #Emission current density in amp/cm\u00b2\n",
-      "A=60.2;                   #Constant depending upon the type of thermionic emitter, in amp/cm\u00b2/K\u00b2\n",
-      "T=1900.0;                 #Absolute temperature in K\n",
-      "\n",
-      "\n",
-      "#calculations\n",
-      "#Calculating b according to the formula Js=A*T\u00b2*exp(-b/T) for emission current density\n",
-      "b=-T*(log(Js/(A*T*T)));          #constant for emitter, in K\n",
-      "phi= round(b/11600,2);           # work function in eV\n",
-      "\n",
-      "print (\"Work function of the tungsten wire = %.2f eV\"%phi);\n",
-      "\n",
-      "if(phi==4.52):\n",
-      "\tprint(\"Given sample is pure Tungsten\");\n",
-      "elif(phi!=4.52 and phi>=2.63 and phi<=4.52):\n",
-      "\tprint (\"The sample is not pure Tungsten\");\n",
-      "    \n",
-      "#Note : In the text book, the work function has been approximated to 3.56eV, but in the code it calculates as  3.52eV\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Work function of the tungsten wire = 3.52 eV\n",
-        "The sample is not pure Tungsten\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_2.ipynb
deleted file mode 100755
index 06a555a6..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_2.ipynb
+++ /dev/null
@@ -1,125 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c0ff4f67576afe73a11c06eedd0a50709b7f5831737f83db1cd640098e3e9740"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 2 : ELECTRONIC EMISSION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1: Page number 31\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "\n",
-      "from math import exp\n",
-      "from math import pi\n",
-      "\n",
-      "l=5.0;                        #length of tungsten filament in cm\n",
-      "d=0.01;                       #diameter of the filament in cm\n",
-      "T=2500.0;                     #operating temperature in K\n",
-      "A=60.2*pow(10,4);             #constant, depending upon the type of thermionic emitter, in amp/m\u00b2/K\u00b2\n",
-      "phi=4.517;                    #work function of emitter in eV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "b=round(11600*phi,-1);                       #constant for a metal, in K\n",
-      "Js=round(A*T*T*exp(-b/T),-2);                #Emission current density in amp/m\u00b2\n",
-      "a=pi*(d/100)*(l/100);                        #Surface area of the cathode in m\u00b2\n",
-      "E_I=Js*a;                                     #Emission current in A\n",
-      "\n",
-      "#Result\n",
-      "print(\"emission current =%.3f A\"%E_I);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "emission current =0.047 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2:Page number 31\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "Js=0.1;                   #Emission current density in amp/cm\u00b2\n",
-      "A=60.2;                   #Constant depending upon the type of thermionic emitter, in amp/cm\u00b2/K\u00b2\n",
-      "T=1900.0;                 #Absolute temperature in K\n",
-      "\n",
-      "\n",
-      "#calculations\n",
-      "#Calculating b according to the formula Js=A*T\u00b2*exp(-b/T) for emission current density\n",
-      "b=-T*(log(Js/(A*T*T)));          #constant for emitter, in K\n",
-      "phi= round(b/11600,2);           # work function in eV\n",
-      "\n",
-      "print (\"Work function of the tungsten wire = %.2f eV\"%phi);\n",
-      "\n",
-      "if(phi==4.52):\n",
-      "\tprint(\"Given sample is pure Tungsten\");\n",
-      "elif(phi!=4.52 and phi>=2.63 and phi<=4.52):\n",
-      "\tprint (\"The sample is not pure Tungsten\");\n",
-      "    \n",
-      "#Note : In the text book, the work function has been approximated to 3.56eV, but in the code it calculates as  3.52eV\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Work function of the tungsten wire = 3.52 eV\n",
-        "The sample is not pure Tungsten\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_3.ipynb
deleted file mode 100755
index 06a555a6..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_3.ipynb
+++ /dev/null
@@ -1,125 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c0ff4f67576afe73a11c06eedd0a50709b7f5831737f83db1cd640098e3e9740"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 2 : ELECTRONIC EMISSION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1: Page number 31\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "\n",
-      "from math import exp\n",
-      "from math import pi\n",
-      "\n",
-      "l=5.0;                        #length of tungsten filament in cm\n",
-      "d=0.01;                       #diameter of the filament in cm\n",
-      "T=2500.0;                     #operating temperature in K\n",
-      "A=60.2*pow(10,4);             #constant, depending upon the type of thermionic emitter, in amp/m\u00b2/K\u00b2\n",
-      "phi=4.517;                    #work function of emitter in eV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "b=round(11600*phi,-1);                       #constant for a metal, in K\n",
-      "Js=round(A*T*T*exp(-b/T),-2);                #Emission current density in amp/m\u00b2\n",
-      "a=pi*(d/100)*(l/100);                        #Surface area of the cathode in m\u00b2\n",
-      "E_I=Js*a;                                     #Emission current in A\n",
-      "\n",
-      "#Result\n",
-      "print(\"emission current =%.3f A\"%E_I);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "emission current =0.047 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2:Page number 31\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "Js=0.1;                   #Emission current density in amp/cm\u00b2\n",
-      "A=60.2;                   #Constant depending upon the type of thermionic emitter, in amp/cm\u00b2/K\u00b2\n",
-      "T=1900.0;                 #Absolute temperature in K\n",
-      "\n",
-      "\n",
-      "#calculations\n",
-      "#Calculating b according to the formula Js=A*T\u00b2*exp(-b/T) for emission current density\n",
-      "b=-T*(log(Js/(A*T*T)));          #constant for emitter, in K\n",
-      "phi= round(b/11600,2);           # work function in eV\n",
-      "\n",
-      "print (\"Work function of the tungsten wire = %.2f eV\"%phi);\n",
-      "\n",
-      "if(phi==4.52):\n",
-      "\tprint(\"Given sample is pure Tungsten\");\n",
-      "elif(phi!=4.52 and phi>=2.63 and phi<=4.52):\n",
-      "\tprint (\"The sample is not pure Tungsten\");\n",
-      "    \n",
-      "#Note : In the text book, the work function has been approximated to 3.56eV, but in the code it calculates as  3.52eV\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Work function of the tungsten wire = 3.52 eV\n",
-        "The sample is not pure Tungsten\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_4.ipynb
deleted file mode 100755
index 06a555a6..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_4.ipynb
+++ /dev/null
@@ -1,125 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c0ff4f67576afe73a11c06eedd0a50709b7f5831737f83db1cd640098e3e9740"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 2 : ELECTRONIC EMISSION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1: Page number 31\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "\n",
-      "from math import exp\n",
-      "from math import pi\n",
-      "\n",
-      "l=5.0;                        #length of tungsten filament in cm\n",
-      "d=0.01;                       #diameter of the filament in cm\n",
-      "T=2500.0;                     #operating temperature in K\n",
-      "A=60.2*pow(10,4);             #constant, depending upon the type of thermionic emitter, in amp/m\u00b2/K\u00b2\n",
-      "phi=4.517;                    #work function of emitter in eV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "b=round(11600*phi,-1);                       #constant for a metal, in K\n",
-      "Js=round(A*T*T*exp(-b/T),-2);                #Emission current density in amp/m\u00b2\n",
-      "a=pi*(d/100)*(l/100);                        #Surface area of the cathode in m\u00b2\n",
-      "E_I=Js*a;                                     #Emission current in A\n",
-      "\n",
-      "#Result\n",
-      "print(\"emission current =%.3f A\"%E_I);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "emission current =0.047 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2:Page number 31\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "Js=0.1;                   #Emission current density in amp/cm\u00b2\n",
-      "A=60.2;                   #Constant depending upon the type of thermionic emitter, in amp/cm\u00b2/K\u00b2\n",
-      "T=1900.0;                 #Absolute temperature in K\n",
-      "\n",
-      "\n",
-      "#calculations\n",
-      "#Calculating b according to the formula Js=A*T\u00b2*exp(-b/T) for emission current density\n",
-      "b=-T*(log(Js/(A*T*T)));          #constant for emitter, in K\n",
-      "phi= round(b/11600,2);           # work function in eV\n",
-      "\n",
-      "print (\"Work function of the tungsten wire = %.2f eV\"%phi);\n",
-      "\n",
-      "if(phi==4.52):\n",
-      "\tprint(\"Given sample is pure Tungsten\");\n",
-      "elif(phi!=4.52 and phi>=2.63 and phi<=4.52):\n",
-      "\tprint (\"The sample is not pure Tungsten\");\n",
-      "    \n",
-      "#Note : In the text book, the work function has been approximated to 3.56eV, but in the code it calculates as  3.52eV\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Work function of the tungsten wire = 3.52 eV\n",
-        "The sample is not pure Tungsten\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_5.ipynb
deleted file mode 100755
index 06a555a6..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_5.ipynb
+++ /dev/null
@@ -1,125 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c0ff4f67576afe73a11c06eedd0a50709b7f5831737f83db1cd640098e3e9740"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 2 : ELECTRONIC EMISSION"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.1: Page number 31\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "\n",
-      "from math import exp\n",
-      "from math import pi\n",
-      "\n",
-      "l=5.0;                        #length of tungsten filament in cm\n",
-      "d=0.01;                       #diameter of the filament in cm\n",
-      "T=2500.0;                     #operating temperature in K\n",
-      "A=60.2*pow(10,4);             #constant, depending upon the type of thermionic emitter, in amp/m\u00b2/K\u00b2\n",
-      "phi=4.517;                    #work function of emitter in eV\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "b=round(11600*phi,-1);                       #constant for a metal, in K\n",
-      "Js=round(A*T*T*exp(-b/T),-2);                #Emission current density in amp/m\u00b2\n",
-      "a=pi*(d/100)*(l/100);                        #Surface area of the cathode in m\u00b2\n",
-      "E_I=Js*a;                                     #Emission current in A\n",
-      "\n",
-      "#Result\n",
-      "print(\"emission current =%.3f A\"%E_I);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "emission current =0.047 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.2:Page number 31\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import log\n",
-      "\n",
-      "#Variable declaration\n",
-      "Js=0.1;                   #Emission current density in amp/cm\u00b2\n",
-      "A=60.2;                   #Constant depending upon the type of thermionic emitter, in amp/cm\u00b2/K\u00b2\n",
-      "T=1900.0;                 #Absolute temperature in K\n",
-      "\n",
-      "\n",
-      "#calculations\n",
-      "#Calculating b according to the formula Js=A*T\u00b2*exp(-b/T) for emission current density\n",
-      "b=-T*(log(Js/(A*T*T)));          #constant for emitter, in K\n",
-      "phi= round(b/11600,2);           # work function in eV\n",
-      "\n",
-      "print (\"Work function of the tungsten wire = %.2f eV\"%phi);\n",
-      "\n",
-      "if(phi==4.52):\n",
-      "\tprint(\"Given sample is pure Tungsten\");\n",
-      "elif(phi!=4.52 and phi>=2.63 and phi<=4.52):\n",
-      "\tprint (\"The sample is not pure Tungsten\");\n",
-      "    \n",
-      "#Note : In the text book, the work function has been approximated to 3.56eV, but in the code it calculates as  3.52eV\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Work function of the tungsten wire = 3.52 eV\n",
-        "The sample is not pure Tungsten\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6.ipynb
deleted file mode 100755
index a6008ee1..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6.ipynb
+++ /dev/null
@@ -1,1624 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3452607f2168b562d941493f83083042eaa5a2d316715f9d9f089ff03d73fdb8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 6: SEMICONDUCTOR DIODE"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2, Page number 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration \n",
-      "Vf =20;                #Peak Input Voltage in V\n",
-      "rf=10;                  #Forward Resistance in ohms\n",
-      "RL=500.0;                 #Load Resistance in ohms\n",
-      "V0=0.7;                  #Potential Barrier Voltage of the diodes in V\n",
-      "\n",
-      "#Calculation\n",
-      "#(1)\n",
-      "If_peak=(Vf-V0)/(rf+RL);                  #Peak current through the diode in A\n",
-      "If_peak=If_peak*1000;                    #Peak current through the diode in mA\n",
-      "#(2)\n",
-      "V_out_peak =If_peak * RL/1000 ;               #Peak output voltage in V\n",
-      "\n",
-      "#For an Ideal diode\n",
-      "If_peak_ideal=Vf/RL;                   #Peak current through the ideal diode in A\n",
-      "If_peak_ideal=If_peak_ideal*1000;      #Peak current through the ideal diode in mA\n",
-      "\n",
-      "V_out_peak_ideal=If_peak_ideal * RL/1000;     # Peak output voltage in case of the ideal diode in V\n",
-      "\n",
-      "#Result\n",
-      "print '(i) Peak current through the diode = %.1f mA '%If_peak;\n",
-      "print '(ii) Peak output voltage = %.1f V'%V_out_peak;\n",
-      "print '(iii) Peak current through the ideal diode = %d mA '%If_peak_ideal;\n",
-      "print '(iv) Peak output voltage in case of the ideal diode = %d V'%V_out_peak_ideal;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Peak current through the diode = 37.8 mA \n",
-        "(ii) Peak output voltage = 18.9 V\n",
-        "(iii) Peak current through the ideal diode = 40 mA \n",
-        "(iv) Peak output voltage in case of the ideal diode = 20 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3, Page number 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V =10.0;                #Battery voltage in V\n",
-      "R1=50.0;                  #Resistor 1's resistance in ohms\n",
-      "R2=5.0;                 #Resistor 2's resistance in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "#Using Thevenin's Theorem to find current in the diode\n",
-      "E0=(R2/(R1+R2))*V;          #Thevenin's Voltage in V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's Resistance in ohms\n",
-      "\n",
-      "I0=E0/R0;                   #Current through the diode in A\n",
-      "I0=I0*1000;                 #Current through the diode in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Current through the diode = %d mA '%Io;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through the diode = 200 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4, Page number 82-83 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V =10.0;                #Battery voltage in V\n",
-      "R0=48.0;                  #Resistance of the resistor in ohms\n",
-      "Rd=1.0;                 #Forward resistance of the diodes in ohms\n",
-      "Vd=0.7;                 #Potential barrier of the diodes in V\n",
-      "#Calculation\n",
-      "V_net=V-Vd-Vd;        #Net voltage in the circuit in V\n",
-      "R_net=R0+Rd+Rd        #Net resistance of the circuit in ohms\n",
-      "I_net=V_net/R_net;    #Net current in the circuit in A\n",
-      "I_net=I_net*1000;     #Net current in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Net current in the circuit = %d mA '%I_net;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Net current in the circuit = 172 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page number 83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E1=24;           #Voltage of first source in V\n",
-      "E2=4;            #Voltage of second source in V\n",
-      "V0=0.7;          #Potential barrier of diodes in V\n",
-      "R=2000;          #Resistance of the given resistor in ohms\n",
-      "Rd=0;            #Forward resistance of the diodes in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I=(E1-E2-V0)/(R+Rd);  #Current in the circuit in A\n",
-      "I=I*1000;             #Current in the circuit in mA \n",
-      "\n",
-      "#Result\n",
-      "print 'Current in the circuit = %.2f mA '%I;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in the circuit = 9.65 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page number 83-84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=20;           #Voltage of source in V\n",
-      "V0=0.3;         #Potential barrier of Germanium diode in V\n",
-      "V0_Si=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "\n",
-      "#Calculation\n",
-      "#As only Ge diode is turned on due to less potential barrier,\n",
-      "VA=V-V0;        #Voltage VA acroos resistor of 3k ohms\n",
-      "\n",
-      "#Result\n",
-      "print 'Voltage VA = %.1f mA '%VA;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage VA = 19.7 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7, Page number 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=10;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "# Resistance of all resistors in ohms\n",
-      "R1=2000;\n",
-      "R2=2000;\n",
-      "R3=2000;\n",
-      "\n",
-      "#Calculation\n",
-      "Id=(V-V0)/(R2+2*R3);     #Current through the diodes in A\n",
-      "VQ=2*Id*R3;      #Voltage VQ across the grounded 2k ohm resistor in V\n",
-      "Id=Id*1000;      #Current through the diodes in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Voltage VQ = %.1f V '%VQ;\n",
-      "print 'Current through the diodes, Id = %.2f mA '%Id;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage VQ = 6.2 V \n",
-        "Current through the diodes, Id = 1.55 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8, Page number 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=15;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "R=500       # Resistance of all resistors in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I1=(V-V0)/R;      #total current in the circuit in A\n",
-      "Id1=I1/2;         #current in first diode in A\n",
-      "Id1=Id1*1000;     #current in first diode in mA\n",
-      "Id2=Id1           #current in second diode in mA\n",
-      "\n",
-      "#Result\n",
-      "print ('Current in first diode = %.1f mA'%Id1);\n",
-      "print ('Current in second diode = %.1f mA'%Id2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in first diode = 14.3 mA\n",
-        "Current in second diode = 14.3 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9, Page number 85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E=20;           #Voltage of source in V\n",
-      "V0_d1=0.7;      #Potetial barrier of first Silicon diode in V\n",
-      "V0_d2=0.7;      #Potetial barrier of second Silicon diode in V\n",
-      "R1=5600;       # Resistance of first resistor in ohms\n",
-      "R2=3300;       #  Resistance of second resistor in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I2=V0_d2/R2;                  #Current I2 through resistor R2 in A\n",
-      "I2=round((I2*1000),3);                   #Current I2 through resistor R2 in mA\n",
-      "I1=(E-V0_d1-V0_d2)/R1;        #Current I1 through resistor R1 in  A\n",
-      "I1=round((I1*1000),2);                   #Current I1 through resistor R1 in  mA\n",
-      "I3=I1-I2;                     #Current I3 through diode D2 in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Current I1= %.2f mA'%I1;\n",
-      "print 'Current I1= %.3f mA'%I2;\n",
-      "print 'Current I1= %.3f mA'%I3;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current I1= 3.32 mA\n",
-        "Current I1= 0.212 mA\n",
-        "Current I1= 3.108 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10, Page number 85-86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E=10.0;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V\n",
-      "R1=2000;       # Resistance of first resistor in ohms\n",
-      "R2=8000;       #  Resistance of second resistor in ohms\n",
-      "R3=4000;       #Resistance of third resistor in ohms\n",
-      "R4=6000;       #Resistance of fourth resistor in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "#Assuming the given diode to be reverse bised and calculating voltage across it's terminals\n",
-      "V1=(E/(R1+R2))*R2;      #voltage at the P side of the diode, i.e, voltage across R2 resistor,according to voltage divider rule, in V\n",
-      "V2=(E/(R3+R4))*R4;      #voltage at the N side of the diode, i.e, voltage across R4 resistor,according to voltage divider rule, in V\n",
-      "\n",
-      "#Result\n",
-      "if((V1-V2)>=V0):\n",
-      "    print 'Our assumption was wrong and, the diode is forward biased';\n",
-      "else:\n",
-      "    print 'The diode is reverse biased';\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Our assumption was wrong and, the diode is forward biased\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.11, Page number 86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=2;                              #Supply voltage in V\n",
-      "V0=0.7;                           #Potential barrier voltage of the diode in V \n",
-      "R1=4000.0;                        #Resistance of first resistor in \u03a9\n",
-      "R2=1000.0;                        ##Resistance of second resistor in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Assuming the diode to be in ON state\n",
-      "I1=((V-V0)/R1)*1000;                  #Current through resistor R1, in mA\n",
-      "I2=(V0/R2)*1000;                    #Current through resistor R2, in mA\n",
-      "ID=I1-I2;                           #Diode current, in  mA\n",
-      "\n",
-      "if(ID<0):\n",
-      "    #Since the diode current is negative, the diode must be OFF \n",
-      "    ID=0;                     #True value of diode current, mA\n",
-      "    \n",
-      "#As the diode is in OFF state it can be replaced by an open ciruit equivalent \n",
-      "VD=V*R2/(R1 +R2);                #Voltage across the diode, in V\n",
-      "\n",
-      "#Result\n",
-      "print 'ID =%d mA'%ID;\n",
-      "print 'VD =%.1f V'%VD;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ID =0 mA\n",
-        "VD =0.4 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12, Page number 89-90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "AC_Input_Power=100.0;          #Input AC Power in watts\n",
-      "AC_Output_Power=40.0;          #Output AC Power in watts\n",
-      "Accepted_Power=50.0;           #Power accepted by the half-wave rectifier in watt\n",
-      "\n",
-      "#Calculation\n",
-      "R_eff=(AC_Output_Power/AC_Input_Power)*100;             #Rectification efficiency of the half-wave rectifier\n",
-      "Unused_power=AC_Input_Power-Accepted_Power;       #Power not used by the half_wave rectifier due to open circuited condition of the diode in watt\n",
-      "Power_dissipated=Accepted_Power-AC_Output_Power;  #Power dissipated by the diode watt\n",
-      "\n",
-      "#Result\n",
-      "print 'The rectification efficiency of the half-wave rectifier= %d%% '%R_eff;\n",
-      "\n",
-      "print 'Rest 60%% of the power is the unused power and power dissipated by the diode = %d watts and %d watts' %(Unused_power ,Power_dissipated);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rectification efficiency of the half-wave rectifier= 40% \n",
-        "Rest 60% of the power is the unused power and power dissipated by the diode = 50 watts and 10 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.13, Page number 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "Vrms=230.0;                #AC supply RMS voltage in V\n",
-      "Turns_Ratio=10/1;        #turn ratio of the transformer \n",
-      "\n",
-      "#Calculation\n",
-      "Vpm=sqrt(2)*Vrms;         #Maximum primary voltage in V\n",
-      "Vsm=Vpm/Turns_Ratio;     #Maximum secondary voltage in V\n",
-      "#Case 1\n",
-      "Vdc=Vsm/(round(pi,2));              #Output D.C voltage, which is the average voltage in V\n",
-      "Vdc=round(Vdc,2);\n",
-      "#Case 2\n",
-      "PIV=Vsm;                 #Peak Inverse Voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print 'The output d.c voltage= %.2f V'%Vdc;\n",
-      "print 'The peak inverse voltage= %.2f V'%PIV;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output d.c voltage= 10.36 V\n",
-        "The peak inverse voltage= 32.53 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.14, Page number 90-91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "rf=20.0;           #Internal resistance of the crystal diode in ohms\n",
-      "Vm=50.0;           #Maximum applied voltage in V\n",
-      "RL=800.0;          #Load Resistance in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "# 1\n",
-      "Im=Vm/(rf+RL);         #Maximum current in A\n",
-      "Im=Im*1000;            #Maximum current in \n",
-      "Im=round(Im,0);\n",
-      "Idc=Im/pi;             #Average voltage in mA\n",
-      "Idc=round(Idc,1);\n",
-      "Irms=Im/2;             #RMS value of the current in mA\n",
-      "Irms=round(Irms,1)\n",
-      "\n",
-      "# 2\n",
-      "AC_Input_Power=pow(Irms/1000,2)*(rf+RL);     #Input a.c power in watt\n",
-      "\n",
-      "DC_Output_Power=pow(Idc/1000,2)*RL;          #Output d.c power in watt\n",
-      "\n",
-      "# 3\n",
-      "DC_Output_Voltage=(Idc/1000)*RL;               #Output d.c voltage in V\n",
-      "\n",
-      "# 4\n",
-      "Rectifier_efficiency=(DC_Output_Power/AC_Input_Power)*100;   # Efficiency of rectification of the half-wave rectifier\n",
-      "\n",
-      "#Result\n",
-      "print ' i:';\n",
-      "print '   Im   = %d mA'%Im;\n",
-      "print '   Idc  = %.1f mA'%Idc;\n",
-      "print '   Irms = %.1f mA'%Irms;\n",
-      "print ' ii: ';\n",
-      "print '   a.c input power= %.3f watt'%AC_Input_Power;\n",
-      "print '   d.c output power= %.3f watt'%DC_Output_Power;\n",
-      "print ' iii: ';\n",
-      "print '   d.c output voltage = %.2f volts'%DC_Output_Voltage;\n",
-      "print ' iv: '\n",
-      "print '   Efficiency of rectification = %.1f%%'%Rectifier_efficiency;\n",
-      "\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " i:\n",
-        "   Im   = 61 mA\n",
-        "   Idc  = 19.4 mA\n",
-        "   Irms = 30.5 mA\n",
-        " ii: \n",
-        "   a.c input power= 0.763 watt\n",
-        "   d.c output power= 0.301 watt\n",
-        " iii: \n",
-        "   d.c output voltage = 15.52 volts\n",
-        " iv: \n",
-        "   Efficiency of rectification = 39.5%\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.15, Page number 91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "Vdc=50.0;           #Output d.c voltage in V\n",
-      "rf=25;            #Diode resistance in ohm\n",
-      "RL=800;           #Load resistance in ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vm=(pi*(rf+RL)*Vdc)/RL;   #[ Vdc=Vm*RL/(pi*(rf+RL)) ]Maximum value of a.c voltage required to get a volatge  of Vdc from the half-wave rectifier, in V\n",
-      "Vm=round(Vm,0);  \n",
-      "#Result\n",
-      "print 'The a.c voltage required should have maximum value of = %d V' %Vm;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c voltage required should have maximum value of = 162 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.16, Page number 95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "rf=20;              #Internal resistance of the diodes in ohm\n",
-      "Vrms=50;            #RMS value of transformer's secondary voltage from centre tap to each end of secondary\n",
-      "RL=980;             #Load resistance in ohm\n",
-      "\n",
-      "#Calculation\n",
-      "Vm=Vrms*sqrt(2);      #Maximum a.c voltage in V\n",
-      "Im=Vm/(rf+RL);        #Maximum load current in A\n",
-      "Im=Im*1000;           #Maximum load current in mA\n",
-      " \n",
-      "# 1:\n",
-      "Idc=2*Im/pi;          #Mean load current\n",
-      "\n",
-      "# 2:\n",
-      "Irms=Im/sqrt(2);       #RMS value of load current in A\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print'   The mean load current= %d mA'%Idc;\n",
-      "print 'ii:';\n",
-      "print '  The r.m.s value of the load current = %d mA'%Irms; "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The mean load current= 45 mA\n",
-        "ii:\n",
-        "  The r.m.s value of the load current = 50 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.17, Page number 95-96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "#Variable declaration\n",
-      "RL=100;                          #Load resistance in ohm       \n",
-      "rf=0;                            #Internal resistance of the diodes in ohm\n",
-      "Turns_ratio=5/1;                 #Primary to secondary turns ratio of transformer \n",
-      "P_Vrms=230;                      #R.M.S value of voltage in primary winding in V\n",
-      "S_Vrms=P_Vrms/Turns_ratio;       #R.M.S value of voltage in secondary winding in V\n",
-      "S_Vm=S_Vrms*sqrt(2);        #Maximum voltage across secondary winding in V\n",
-      "Vm=S_Vm/2;                  #Maximum voltage across half seconfdary winding in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Idc=2*Vm/(pi*RL);            #Average current in A\n",
-      "Vdc=Idc*RL;                  #d.c output voltage in V\n",
-      "\n",
-      "# 2:\n",
-      "PIV=S_Vm;                    #Peak Invers Voltage(= Maximum secondary voltage) in V\n",
-      "\n",
-      "# 3:\n",
-      "Pac=pow(Vm/(RL*sqrt(2)),2)*(rf+RL);         #a.c input power in watt\n",
-      "Pdc=(pow(Idc,2)*RL);                        #d.c output power in watt\n",
-      "R_eff=(Pdc/Pac)*100;      #Rectification efficiency\n",
-      "R_eff=round(R_eff,1);\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   The d.c output voltage= %.1f V'%Vdc;\n",
-      "print 'ii:';\n",
-      "print '   The peak inverse voltage= %d V'%PIV;\n",
-      "print 'iii:';\n",
-      "print '   Rectification efficiency= %.1f%%'%R_eff;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The d.c output voltage= 20.7 V\n",
-        "ii:\n",
-        "   The peak inverse voltage= 65 V\n",
-        "iii:\n",
-        "   Rectification efficiency= 81.1%\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The value of rectification efficiency is calculated as 81.2% in the textbook using the formula 0.812/(1 + (rf/RL)), but by calculating using the correct values in the formula we get 81.1%."
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.18, Page number 96-97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "#Variable declaration\n",
-      "fin=50;                       #frequency of input ac source in Hz\n",
-      "RL=200;                       #Load resistance in ohm\n",
-      "Turns_ratio=4/1;              #Transformers turns ratio, primary to secondary.\n",
-      "P_Vrms=230.0;                 #R.M.S value of voltage in primary winding in V\n",
-      "S_Vrms=P_Vrms/Turns_ratio     #R.M.S value of voltage in secondary winding in V\n",
-      "Vm=S_Vrms*sqrt(2);            #Maximum voltage across secondary winding in V\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Idc=2*Vm/(pi*RL);           # Average current in A\n",
-      "Vdc=Idc*RL;                 #Output d.c voltage in V\n",
-      "Vdc=round(Vdc,0);\n",
-      "# 2:\n",
-      "PIV= Vm;                    #Peak Inverse Voltage(= Maximum volutage across secondary winding) in V\n",
-      "\n",
-      "# 3:\n",
-      "fout=2*fin;                 #Output frequency in Hz\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   The d.c output voltage = %d V' %Vdc;\n",
-      "print 'ii:';\n",
-      "print '   The peak inverse voltage = %.1f V'%PIV;\n",
-      "print 'iii:';\n",
-      "print '   The output frequency = %d Hz'%fout;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The d.c output voltage = 52 V\n",
-        "ii:\n",
-        "   The peak inverse voltage = 81.3 V\n",
-        "iii:\n",
-        "   The output frequency = 100 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.19, Page number 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=100.0;                        #Load Resistance in ohm\n",
-      "Turns_ratio=5/1;                 #Primary to secondary turns ratio of the transformer\n",
-      "Vin=230.0;                       #R.M.S value of input voltage in V\n",
-      "fin=50;                          #Input frequency in Hz\n",
-      "\n",
-      "#Calculation\n",
-      "Vs_rms=Vin/Turns_ratio;          #R.M.S value of the voltage in secondary winding, in v\n",
-      "Vs_max=Vs_rms*sqrt(2);           #Maximum voltage across secondary, in V\n",
-      "\n",
-      "# (i)\n",
-      "#Case i: Centre-tap circuit\n",
-      "Vm=Vs_max/2;                     #Maximum voltage across half secondary winding, in V \n",
-      "Vdc=2*Vm*RL/(pi*RL);             #DC output voltage, in V \n",
-      "print 'The d.c output voltage for the centre-tap circuit = %.1f V'%Vdc;\n",
-      "\n",
-      "#Case ii:\n",
-      "Vm=Vs_max;                      #Maximum voltage across secondary, in V\n",
-      "Vdc=2*Vm*RL/(pi*RL);            #DC output voltage, in V \n",
-      "print 'The d.c output voltage for the bridge circuit = %.1f V'%Vdc; \n",
-      "\n",
-      "# ii:\n",
-      "#Case i: Centre-tap circuit\n",
-      "Turns_ratio=5/1;                      #Turns ratio of the transformer\n",
-      "Vs_rms=Vin/Turns_ratio;               #R.M.S value of the secondary voltage in V\n",
-      "Vs_max=Vs_rms*sqrt(2);                #Maximum voltage across the secondary in V\n",
-      "Vm=Vs_max/2;                          #Maximum voltage across half of the secondary in V\n",
-      "PIV=2*Vm;                             #Peak Inverse Voltage in V\n",
-      "print 'PIV in case of centre-tap circuit = %d V'%PIV;\n",
-      "\n",
-      "#Case ii: Bridge circuit\n",
-      "Turns_ratio=10/1;                             #Turns ratio of the transformer\n",
-      "Vs_rms=Vin/Turns_ratio;                       #R.M.S value of the secondary voltage in V\n",
-      "Vs_max=Vs_rms*sqrt(2);                        #Maximum voltage across the secondary in V\n",
-      "PIV=Vm;                                        #Peak Inverse Voltage in V\n",
-      "print 'PIV in case of bridge circuit = %.1f V'%PIV;\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The d.c output voltage for the centre-tap circuit = 20.7 V\n",
-        "The d.c output voltage for the bridge circuit = 41.4 V\n",
-        "PIV in case of centre-tap circuit = 65 V\n",
-        "PIV in case of bridge circuit = 32.5 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.20, Page number 98"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "rf=1;                    #forward resistance of diodes of the rectifier in ohm\n",
-      "RL=480;                  #Load resistance in ohm\n",
-      "Vrms=240.0;                #a.c supply voltage in V\n",
-      "Vm=Vrms*sqrt(2);         #Maximum a.c voltage in V \n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Rt=2*rf+RL;             #Total circuit resistance at any instance in ohm\n",
-      "Im=Vm/Rt;               #Maximum load current in A\n",
-      "Idc=2*Im/pi;            #Mean load current in A\n",
-      "\n",
-      "# 2:\n",
-      "Irms=Im/2;              #R.M.S value of current  in A\n",
-      "P=pow(Irms,2)*rf;       #Power dissipated in each diode in watt\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   Mean load current = %.2f A'%Idc;\n",
-      "print 'ii:';\n",
-      "print '   Power dissipated in each diode= %.3f W'%P;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   Mean load current = 0.45 A\n",
-        "ii:\n",
-        "   Power dissipated in each diode= 0.124 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The value of power dissipated is approximately 0.124 W , but in the textbook it is approximated as 0.123W."
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.21, Page number 98-99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt,pi\n",
-      "#Variable declaration\n",
-      "RL=12000;                #Load resistance in ohm\n",
-      "V0=0.7;                   #Potential barrier voltage of diodes in V\n",
-      "Vrms=12;                  #R.M.S value of input a.c voltage in V\n",
-      "Vs_pk=Vrms*sqrt(2);       #Peak secondary voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Vout_pk=Vs_pk-(2*V0);      #Peak output voltage in  V\n",
-      "Vav=2*Vout_pk/pi;          #Average output voltage in V\n",
-      "Vav=round(Vav,2);\n",
-      "\n",
-      "# 2:\n",
-      "Iav=Vav/RL;                #Average output current in A\n",
-      "Iav=Iav*pow(10,6);         #Average output current in \u03bcA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   Average output voltage=%.2f V'%Vav;\n",
-      "print 'ii:';\n",
-      "print '   Average output current=%.1f \u03bcA'%Iav;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   Average output voltage=9.91 V\n",
-        "ii:\n",
-        "   Average output current=825.8 \u03bcA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.22, Page number 102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vdc_A=10;                      #Supply voltage of A in V\n",
-      "Vdc_B=25;                      #Supply voltage of B in V\n",
-      "Vac_rms_a=0.5;                 #Ripples in power supply A in V\n",
-      "Vac_rms_b=0.001;               #Ripples in power supply B in V\n",
-      "\n",
-      "#Calculation\n",
-      "#For power supply A\n",
-      "ripple_factor_A=Vac_rms_a/Vdc_A;      #Ripple factor of power supply A\n",
-      "\n",
-      "#For power supply B\n",
-      "ripple_factor_B=Vac_rms_b/Vdc_B;      #Ripple factor of power supply B\n",
-      "\n",
-      "#Result\n",
-      "if(ripple_factor_A<ripple_factor_B):\n",
-      "    print 'Power supply A is better';\n",
-      "else :\n",
-      "    print 'Power supply B is better';"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supply B is better\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.23, Page number 105-106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "RL=2200;                    #Load resistance in ohm\n",
-      "C=50*pow(10,-6);            #Capacitance of the capacitor used in filter circuit in F\n",
-      "V0=0.7;                     #Potential barrier voltage of the diodes of the rectifier in  V\n",
-      "Vrms=115.0;                   #R.M.S value of input a.c voltage in V \n",
-      "fin=60;                     #Frequency of input a.c voltage in Hz\n",
-      "Turns_ratio=10/1;           #Primary to secondary, turns ratio of the transformer \n",
-      "\n",
-      "#Calculation\n",
-      "Vp_prim=Vrms*sqrt(2);               #Peak primary voltage in V\n",
-      "Vp_sec=Vp_prim/Turns_ratio;         #Peak secondary voltage in V\n",
-      "Vp_in= Vp_sec - 2*V0;               #Peak full wave rectified voltage at the filter input in V\n",
-      "f=2*fin;                            #Output frequency in Hz\n",
-      "Vdc=Vp_in*(1-(1/(2*f*RL*C)));       #Output d.c voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print 'The output d.c voltage  is = %.1f V'%Vdc;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output d.c voltage  is = 14.3 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.24, Page number 106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "R=25;                        #d.c resistance of the choke in ohm\n",
-      "RL=750;                      #Load resistance in ohm\n",
-      "Vm=25.7;                     #Maximum value of the pulsating output from the rectifier in V\n",
-      "\n",
-      "#Calculation\n",
-      "V_dc=2*Vm/pi;                #d.c component of the pulsating output in V\n",
-      "V_dc=round(V_dc,1);\n",
-      "V_dc_out=(V_dc*RL)/(R+RL);   #Output d.c voltage in V\n",
-      "V_dc_out=round(V_dc_out,1);\n",
-      "\n",
-      "#Result\n",
-      "print ' The output d.c voltage accross the load resistance is = %.1f V'%V_dc_out;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The output d.c voltage accross the load resistance is = 15.9 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.25, Page number 113-114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=120.0;                   #Input Voltage in V\n",
-      "Vz=50.0;                    #Zener Voltage in V\n",
-      "R=5000.0;                   #Resistance of the series resistor in ohm\n",
-      "RL=10000.0;                 #Load resistance in ohm\n",
-      "\n",
-      "#Calculation\n",
-      "V=Ei*RL/(R+RL);             #Voltage across the open circuit if the zener diode is removed\n",
-      "if(V>Vz):\n",
-      "    #Zener diode is in ON state\n",
-      "    # i:\n",
-      "    Output_voltage=Vz;             #Voltage across load resistance, in V\n",
-      "    #ii:\n",
-      "    Voltage_R=Ei-Vz;               #Voltage across the series resistance R, in V\n",
-      "    #iii:\n",
-      "    IL=Vz/RL;                      #Load current through RL in A\n",
-      "    IL=IL*1000;                    #Load current through RL in mA\n",
-      "    I=Voltage_R/R;                 #Current through the series resistance in A\n",
-      "    I=I*1000;                      #Current through the series resistance in mA\n",
-      "    Iz=I-IL;                       #Applying Kirchhoff's first law, Zener current in mA\n",
-      "    \n",
-      "    #Result\n",
-      "    print 'i)   The output voltage across the load resistance RL = %d V'%Output_voltage;\n",
-      "    print 'ii)  The voltage drop across the series resistance  R = %d V'%Voltage_R;\n",
-      "    print 'iii) The current through the zener diode = %d mA'%Iz;\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)   The output voltage across the load resistance RL = 50 V\n",
-        "ii)  The voltage drop across the series resistance  R = 70 V\n",
-        "iii) The current through the zener diode = 9 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.26, Page number 114-115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Max_V=120.0;                  #Maximum input voltage in V\n",
-      "Min_V=80.0;                   #Minimum input voltage in V\n",
-      "R=5000.0;                     #Series resistance in ohm\n",
-      "RL=10000.0;                   #Load resistance in ohm\n",
-      "Vz=50.0;                      #Zener voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i: Maximum zener current\n",
-      "#Zener current will be maximum when the input voltage is maximum\n",
-      "V_R_max=Max_V-Vz;                #Voltage across series resistance R, in  V\n",
-      "I_max=V_R_max/R;                 #Current through series resistance R, in A\n",
-      "I_max=I_max*1000;                #Current through series resistance R, in mA\n",
-      "IL_max=Vz/RL;                        #Load current in A\n",
-      "IL_max=IL_max*1000;                      #Load current in mA\n",
-      "Iz_max=I_max-IL_max;                     #Applying Kirchhoff's first law, Zener current in mA;\n",
-      "\n",
-      "#Case ii: Minimum zener current\n",
-      "#The zener will conduct minimum current when the input voltage is minimum\n",
-      "V_R_min=Min_V-Vz;                #Voltage across series resistance R, in V\n",
-      "I_min=V_R_min/R;                 #Current through series resistance R, in A\n",
-      "I_min=I_min*1000;                #Current through series resistance R, in mA\n",
-      "IL_min=Vz/RL;                        #Load current in A\n",
-      "IL_min=IL_min*1000;                      #Load current in mA\n",
-      "Iz_min=I_min-IL_min;                     #Applying Kirchhoff's first law, Zener current in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Case i: ';\n",
-      "print 'Maximum zener current = %d mA'%Iz_max;\n",
-      "print 'Case ii: ';\n",
-      "print 'Minimum zener current = %d mA'%Iz_min;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case i: \n",
-        "Maximum zener current = 9 mA\n",
-        "Case ii: \n",
-        "Minimum zener current = 1 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.27, Page number 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=12;                      #Input voltage in V\n",
-      "Vz=7.2;                     #Zener voltage in V\n",
-      "E0=Vz;                      #Voltage to be maintained across the load in  V\n",
-      "IL_max=0.1;                 #Maximum load current in A\n",
-      "IL_min=0.012;                  #Minimum load current in A\n",
-      "Iz_min=0.01;                  #Minimum zener current in A\n",
-      "\n",
-      "#Calculation\n",
-      "#When the load current is maximum at minimum value of RL, the zener current is minimum and, as the load current decreases due to increase in value of RL\n",
-      "R=(Ei-E0)/(Iz_min+IL_max);        #The value of series resistance R to maintain a voltage=E0 across load, in ohm\n",
-      "\n",
-      "#Result\n",
-      "print 'The minimum value of series resistance R to maintain a constant value of 7.2 V is = %.1f \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum value of series resistance R to maintain a constant value of 7.2 V is = 43.6 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The actual value of R is 43.636363 (recurring) but, in the textbook the value of R is wrongly approximated 43.5 \u03a9"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.28, Page number 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei_min=22;                #Minimum input voltage in V\n",
-      "Ei_max=28;                #Maximum input voltage in V\n",
-      "Vz=18;                   #Zener voltage in V\n",
-      "E0=Vz;                    #Constant voltage maintained across the load resistance in V\n",
-      "Iz_min=0.2;               #Minimum zener current in A\n",
-      "Iz_max=2;                 #Maximum zener current in A\n",
-      "RL=18;                    #Load resistance in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "IL=Vz/RL;                      #Constant value of load current in A\n",
-      "#When the input voltage is minimum, the zener current will be minimum\n",
-      "R=(Ei_min-E0)/(Iz_min+IL)      #The value of series resistance so that the voltage E0 across RL remains constant\n",
-      "\n",
-      "print 'The value of series resistance R, to maintain constant voltage E0 across RL = %.2f \u03a9.'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistance R, to maintain constant voltage E0 across RL = 3.33 \u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.29, Page number 116 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10                                #Zener voltage in V\n",
-      "Ei_min=13;                           #Minimum input voltage in V\n",
-      "Ei_max=16;                           #Maximum input voltage in V\n",
-      "Iz_min=0.015;                        #Minimum zener current in A\n",
-      "IL_min=0.01;                         #Minimum load current in A \n",
-      "IL_max=0.085;                        #Maximum load curremt in A\n",
-      "E0=Vz;                               #Constant voltage to be maintained in V \n",
-      "\n",
-      "#Calculation\n",
-      "#The zener current will be minimum when the input voltage will be minimum  and at that time the load current will be maximum\n",
-      "R=(Ei_min-E0)/(Iz_min+IL_max);       #The value of series resistance R to maintain a constant voltage across load\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'The value of series resistance to maintain a constant voltage across the load resistance is = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistance to maintain a constant voltage across the load resistance is = 30 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.30, Page number 116"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Iz=0.2;               #Current rating of each zener in A\n",
-      "Vz=15;                #Voltage rating of each zener in V\n",
-      "Ei=45;                #Input voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "# i: Regulated output voltage across the two zener diodes \n",
-      "E0=2*Vz;                     # V\n",
-      "\n",
-      "# ii: Value of series resistance \n",
-      "R=(Ei-E0)/Iz;                  # \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print 'i) The regulated output voltage = %d V'%E0;\n",
-      "print 'ii) The value of the series resistance = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) The regulated output voltage = 30 V\n",
-        "ii) The value of the series resistance = 75 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.31, Page number 116-117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10;                   #Voltage rating of each zener in V\n",
-      "Iz=1;                    #Current rating of each zener in A\n",
-      "Ei=45;                   #Input unregulated voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "#Regulated output voltage across the three zener diodes\n",
-      "E0=3*Vz;               # V\n",
-      "\n",
-      "#Value of series resistance to obtain a 30V regulated output voltage\n",
-      "R=(Ei-E0)/Iz;          # \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print 'Value of series resistance to obtain a 30V regulated output voltage = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of series resistance to obtain a 30V regulated output voltage = 15 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.32, Page number 117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "RL=2000.0;                   #Load resistance in \u03a9\n",
-      "R=200.0;                     #Series resistance in \u03a9\n",
-      "Iz=0.025;                  #Zener current rating in A\n",
-      "E0=30.0;                     #Output regulated voltage in V \n",
-      "\n",
-      "#Calculation\n",
-      "#Minimum input voltage will be required when Iz=0 A, and at  this condition\n",
-      "IL=E0/RL;                   #Load current during Iz=0, in A\n",
-      "I=IL;                       #According to Kirchhoff's law, total current, in A\n",
-      "Ei_min=E0+(I*R);            #Minimum input voltage in V\n",
-      "\n",
-      "#The maximum input voltage required will be when Iz=0.025 A, and at that condition \n",
-      "I=IL+Iz;                    #According to Kirchhoff's law, total current, in A\n",
-      "Ei_max=E0+(I*R);            #maximum input voltage in V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'The required range of input voltage is from %d V to %d V'%(Ei_min,Ei_max);  \n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required range of input voltage is from 33 V to 38 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.33, Page number 117-118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=16;                   #Unregulated input voltage in V\n",
-      "E0=12;                   #Output regulated voltage in V\n",
-      "IL_min=0;                #Minimum load current in A\n",
-      "IL_max=0.2;              #Maximum load current in A\n",
-      "Iz_min=0;                #Minimum zener current in A\n",
-      "Iz_max=0.2;              #Maximum zener current in A\n",
-      "\n",
-      "#Calculation\n",
-      "#As the regulated voltage required across the load is 12V\n",
-      "Vz=E0;                   #Voltage rating of zener diode in V\n",
-      "V_R=Ei-E0;               #Constant Voltage that should remain across series resistance \n",
-      "#The minimum zener current will occur when the curent in the load in maximum\n",
-      "R=V_R/(Iz_min+IL_max);   #Series resistance in \u03a9\n",
-      "\n",
-      "Max_power_rating=Vz*Iz_max;     #Maximum power rating of zener diode in W\n",
-      "\n",
-      "#Result\n",
-      "print 'The regulator is designed using a Seris resistance of %d \u03a9 and a zener diode of zener voltage %d V'%(R,Vz);\n",
-      "print 'The maximum power rating of the zener diode is = %.1f W '%Max_power_rating;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulator is designed using a Seris resistance of 20 \u03a9 and a zener diode of zener voltage 12 V\n",
-        "The maximum power rating of the zener diode is = 2.4 W \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.34, Page number 118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=12;                       #Source voltage in V\n",
-      "R=1000;                     #Series resistance in \u03a9\n",
-      "RL=5000;                    #Load resistance in \u03a9\n",
-      "Vz=6;                       #Voltage rating of zener in V\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i: zener is working properly\n",
-      "#The output voltage across the load will be equal to the zener voltage.\n",
-      "V0=Vz;                        # V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case i: Output voltage when zener is working properly is %d V'%V0;\n",
-      "\n",
-      "#Case ii: zener is shorted\n",
-      "#As the zener is shorted, the potential difference across the load will be zero\n",
-      "V0=0;                        #V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case ii: Output voltage when zener is short circuited is %d V'%V0;\n",
-      "     \n",
-      "#Case iii: zener is open circuited\n",
-      "#If the zener is open circuited, the total voltage will drop across R and RL according to the voltage divider rule\n",
-      "V0=V*RL/(R+RL);                      #V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case iii: Output voltage when zener is open circuited is %d V'%V0;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case i: Output voltage when zener is working properly is 6 V\n",
-        "Case ii: Output voltage when zener is short circuited is 0 V\n",
-        "Case iii: Output voltage when zener is open circuited is 10 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_1.ipynb
deleted file mode 100755
index a6008ee1..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_1.ipynb
+++ /dev/null
@@ -1,1624 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3452607f2168b562d941493f83083042eaa5a2d316715f9d9f089ff03d73fdb8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 6: SEMICONDUCTOR DIODE"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2, Page number 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration \n",
-      "Vf =20;                #Peak Input Voltage in V\n",
-      "rf=10;                  #Forward Resistance in ohms\n",
-      "RL=500.0;                 #Load Resistance in ohms\n",
-      "V0=0.7;                  #Potential Barrier Voltage of the diodes in V\n",
-      "\n",
-      "#Calculation\n",
-      "#(1)\n",
-      "If_peak=(Vf-V0)/(rf+RL);                  #Peak current through the diode in A\n",
-      "If_peak=If_peak*1000;                    #Peak current through the diode in mA\n",
-      "#(2)\n",
-      "V_out_peak =If_peak * RL/1000 ;               #Peak output voltage in V\n",
-      "\n",
-      "#For an Ideal diode\n",
-      "If_peak_ideal=Vf/RL;                   #Peak current through the ideal diode in A\n",
-      "If_peak_ideal=If_peak_ideal*1000;      #Peak current through the ideal diode in mA\n",
-      "\n",
-      "V_out_peak_ideal=If_peak_ideal * RL/1000;     # Peak output voltage in case of the ideal diode in V\n",
-      "\n",
-      "#Result\n",
-      "print '(i) Peak current through the diode = %.1f mA '%If_peak;\n",
-      "print '(ii) Peak output voltage = %.1f V'%V_out_peak;\n",
-      "print '(iii) Peak current through the ideal diode = %d mA '%If_peak_ideal;\n",
-      "print '(iv) Peak output voltage in case of the ideal diode = %d V'%V_out_peak_ideal;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Peak current through the diode = 37.8 mA \n",
-        "(ii) Peak output voltage = 18.9 V\n",
-        "(iii) Peak current through the ideal diode = 40 mA \n",
-        "(iv) Peak output voltage in case of the ideal diode = 20 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3, Page number 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V =10.0;                #Battery voltage in V\n",
-      "R1=50.0;                  #Resistor 1's resistance in ohms\n",
-      "R2=5.0;                 #Resistor 2's resistance in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "#Using Thevenin's Theorem to find current in the diode\n",
-      "E0=(R2/(R1+R2))*V;          #Thevenin's Voltage in V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's Resistance in ohms\n",
-      "\n",
-      "I0=E0/R0;                   #Current through the diode in A\n",
-      "I0=I0*1000;                 #Current through the diode in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Current through the diode = %d mA '%Io;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through the diode = 200 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4, Page number 82-83 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V =10.0;                #Battery voltage in V\n",
-      "R0=48.0;                  #Resistance of the resistor in ohms\n",
-      "Rd=1.0;                 #Forward resistance of the diodes in ohms\n",
-      "Vd=0.7;                 #Potential barrier of the diodes in V\n",
-      "#Calculation\n",
-      "V_net=V-Vd-Vd;        #Net voltage in the circuit in V\n",
-      "R_net=R0+Rd+Rd        #Net resistance of the circuit in ohms\n",
-      "I_net=V_net/R_net;    #Net current in the circuit in A\n",
-      "I_net=I_net*1000;     #Net current in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Net current in the circuit = %d mA '%I_net;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Net current in the circuit = 172 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page number 83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E1=24;           #Voltage of first source in V\n",
-      "E2=4;            #Voltage of second source in V\n",
-      "V0=0.7;          #Potential barrier of diodes in V\n",
-      "R=2000;          #Resistance of the given resistor in ohms\n",
-      "Rd=0;            #Forward resistance of the diodes in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I=(E1-E2-V0)/(R+Rd);  #Current in the circuit in A\n",
-      "I=I*1000;             #Current in the circuit in mA \n",
-      "\n",
-      "#Result\n",
-      "print 'Current in the circuit = %.2f mA '%I;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in the circuit = 9.65 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page number 83-84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=20;           #Voltage of source in V\n",
-      "V0=0.3;         #Potential barrier of Germanium diode in V\n",
-      "V0_Si=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "\n",
-      "#Calculation\n",
-      "#As only Ge diode is turned on due to less potential barrier,\n",
-      "VA=V-V0;        #Voltage VA acroos resistor of 3k ohms\n",
-      "\n",
-      "#Result\n",
-      "print 'Voltage VA = %.1f mA '%VA;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage VA = 19.7 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7, Page number 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=10;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "# Resistance of all resistors in ohms\n",
-      "R1=2000;\n",
-      "R2=2000;\n",
-      "R3=2000;\n",
-      "\n",
-      "#Calculation\n",
-      "Id=(V-V0)/(R2+2*R3);     #Current through the diodes in A\n",
-      "VQ=2*Id*R3;      #Voltage VQ across the grounded 2k ohm resistor in V\n",
-      "Id=Id*1000;      #Current through the diodes in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Voltage VQ = %.1f V '%VQ;\n",
-      "print 'Current through the diodes, Id = %.2f mA '%Id;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage VQ = 6.2 V \n",
-        "Current through the diodes, Id = 1.55 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8, Page number 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=15;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "R=500       # Resistance of all resistors in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I1=(V-V0)/R;      #total current in the circuit in A\n",
-      "Id1=I1/2;         #current in first diode in A\n",
-      "Id1=Id1*1000;     #current in first diode in mA\n",
-      "Id2=Id1           #current in second diode in mA\n",
-      "\n",
-      "#Result\n",
-      "print ('Current in first diode = %.1f mA'%Id1);\n",
-      "print ('Current in second diode = %.1f mA'%Id2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in first diode = 14.3 mA\n",
-        "Current in second diode = 14.3 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9, Page number 85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E=20;           #Voltage of source in V\n",
-      "V0_d1=0.7;      #Potetial barrier of first Silicon diode in V\n",
-      "V0_d2=0.7;      #Potetial barrier of second Silicon diode in V\n",
-      "R1=5600;       # Resistance of first resistor in ohms\n",
-      "R2=3300;       #  Resistance of second resistor in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I2=V0_d2/R2;                  #Current I2 through resistor R2 in A\n",
-      "I2=round((I2*1000),3);                   #Current I2 through resistor R2 in mA\n",
-      "I1=(E-V0_d1-V0_d2)/R1;        #Current I1 through resistor R1 in  A\n",
-      "I1=round((I1*1000),2);                   #Current I1 through resistor R1 in  mA\n",
-      "I3=I1-I2;                     #Current I3 through diode D2 in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Current I1= %.2f mA'%I1;\n",
-      "print 'Current I1= %.3f mA'%I2;\n",
-      "print 'Current I1= %.3f mA'%I3;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current I1= 3.32 mA\n",
-        "Current I1= 0.212 mA\n",
-        "Current I1= 3.108 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10, Page number 85-86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E=10.0;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V\n",
-      "R1=2000;       # Resistance of first resistor in ohms\n",
-      "R2=8000;       #  Resistance of second resistor in ohms\n",
-      "R3=4000;       #Resistance of third resistor in ohms\n",
-      "R4=6000;       #Resistance of fourth resistor in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "#Assuming the given diode to be reverse bised and calculating voltage across it's terminals\n",
-      "V1=(E/(R1+R2))*R2;      #voltage at the P side of the diode, i.e, voltage across R2 resistor,according to voltage divider rule, in V\n",
-      "V2=(E/(R3+R4))*R4;      #voltage at the N side of the diode, i.e, voltage across R4 resistor,according to voltage divider rule, in V\n",
-      "\n",
-      "#Result\n",
-      "if((V1-V2)>=V0):\n",
-      "    print 'Our assumption was wrong and, the diode is forward biased';\n",
-      "else:\n",
-      "    print 'The diode is reverse biased';\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Our assumption was wrong and, the diode is forward biased\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.11, Page number 86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=2;                              #Supply voltage in V\n",
-      "V0=0.7;                           #Potential barrier voltage of the diode in V \n",
-      "R1=4000.0;                        #Resistance of first resistor in \u03a9\n",
-      "R2=1000.0;                        ##Resistance of second resistor in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Assuming the diode to be in ON state\n",
-      "I1=((V-V0)/R1)*1000;                  #Current through resistor R1, in mA\n",
-      "I2=(V0/R2)*1000;                    #Current through resistor R2, in mA\n",
-      "ID=I1-I2;                           #Diode current, in  mA\n",
-      "\n",
-      "if(ID<0):\n",
-      "    #Since the diode current is negative, the diode must be OFF \n",
-      "    ID=0;                     #True value of diode current, mA\n",
-      "    \n",
-      "#As the diode is in OFF state it can be replaced by an open ciruit equivalent \n",
-      "VD=V*R2/(R1 +R2);                #Voltage across the diode, in V\n",
-      "\n",
-      "#Result\n",
-      "print 'ID =%d mA'%ID;\n",
-      "print 'VD =%.1f V'%VD;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ID =0 mA\n",
-        "VD =0.4 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12, Page number 89-90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "AC_Input_Power=100.0;          #Input AC Power in watts\n",
-      "AC_Output_Power=40.0;          #Output AC Power in watts\n",
-      "Accepted_Power=50.0;           #Power accepted by the half-wave rectifier in watt\n",
-      "\n",
-      "#Calculation\n",
-      "R_eff=(AC_Output_Power/AC_Input_Power)*100;             #Rectification efficiency of the half-wave rectifier\n",
-      "Unused_power=AC_Input_Power-Accepted_Power;       #Power not used by the half_wave rectifier due to open circuited condition of the diode in watt\n",
-      "Power_dissipated=Accepted_Power-AC_Output_Power;  #Power dissipated by the diode watt\n",
-      "\n",
-      "#Result\n",
-      "print 'The rectification efficiency of the half-wave rectifier= %d%% '%R_eff;\n",
-      "\n",
-      "print 'Rest 60%% of the power is the unused power and power dissipated by the diode = %d watts and %d watts' %(Unused_power ,Power_dissipated);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rectification efficiency of the half-wave rectifier= 40% \n",
-        "Rest 60% of the power is the unused power and power dissipated by the diode = 50 watts and 10 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.13, Page number 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "Vrms=230.0;                #AC supply RMS voltage in V\n",
-      "Turns_Ratio=10/1;        #turn ratio of the transformer \n",
-      "\n",
-      "#Calculation\n",
-      "Vpm=sqrt(2)*Vrms;         #Maximum primary voltage in V\n",
-      "Vsm=Vpm/Turns_Ratio;     #Maximum secondary voltage in V\n",
-      "#Case 1\n",
-      "Vdc=Vsm/(round(pi,2));              #Output D.C voltage, which is the average voltage in V\n",
-      "Vdc=round(Vdc,2);\n",
-      "#Case 2\n",
-      "PIV=Vsm;                 #Peak Inverse Voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print 'The output d.c voltage= %.2f V'%Vdc;\n",
-      "print 'The peak inverse voltage= %.2f V'%PIV;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output d.c voltage= 10.36 V\n",
-        "The peak inverse voltage= 32.53 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.14, Page number 90-91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "rf=20.0;           #Internal resistance of the crystal diode in ohms\n",
-      "Vm=50.0;           #Maximum applied voltage in V\n",
-      "RL=800.0;          #Load Resistance in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "# 1\n",
-      "Im=Vm/(rf+RL);         #Maximum current in A\n",
-      "Im=Im*1000;            #Maximum current in \n",
-      "Im=round(Im,0);\n",
-      "Idc=Im/pi;             #Average voltage in mA\n",
-      "Idc=round(Idc,1);\n",
-      "Irms=Im/2;             #RMS value of the current in mA\n",
-      "Irms=round(Irms,1)\n",
-      "\n",
-      "# 2\n",
-      "AC_Input_Power=pow(Irms/1000,2)*(rf+RL);     #Input a.c power in watt\n",
-      "\n",
-      "DC_Output_Power=pow(Idc/1000,2)*RL;          #Output d.c power in watt\n",
-      "\n",
-      "# 3\n",
-      "DC_Output_Voltage=(Idc/1000)*RL;               #Output d.c voltage in V\n",
-      "\n",
-      "# 4\n",
-      "Rectifier_efficiency=(DC_Output_Power/AC_Input_Power)*100;   # Efficiency of rectification of the half-wave rectifier\n",
-      "\n",
-      "#Result\n",
-      "print ' i:';\n",
-      "print '   Im   = %d mA'%Im;\n",
-      "print '   Idc  = %.1f mA'%Idc;\n",
-      "print '   Irms = %.1f mA'%Irms;\n",
-      "print ' ii: ';\n",
-      "print '   a.c input power= %.3f watt'%AC_Input_Power;\n",
-      "print '   d.c output power= %.3f watt'%DC_Output_Power;\n",
-      "print ' iii: ';\n",
-      "print '   d.c output voltage = %.2f volts'%DC_Output_Voltage;\n",
-      "print ' iv: '\n",
-      "print '   Efficiency of rectification = %.1f%%'%Rectifier_efficiency;\n",
-      "\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " i:\n",
-        "   Im   = 61 mA\n",
-        "   Idc  = 19.4 mA\n",
-        "   Irms = 30.5 mA\n",
-        " ii: \n",
-        "   a.c input power= 0.763 watt\n",
-        "   d.c output power= 0.301 watt\n",
-        " iii: \n",
-        "   d.c output voltage = 15.52 volts\n",
-        " iv: \n",
-        "   Efficiency of rectification = 39.5%\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.15, Page number 91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "Vdc=50.0;           #Output d.c voltage in V\n",
-      "rf=25;            #Diode resistance in ohm\n",
-      "RL=800;           #Load resistance in ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vm=(pi*(rf+RL)*Vdc)/RL;   #[ Vdc=Vm*RL/(pi*(rf+RL)) ]Maximum value of a.c voltage required to get a volatge  of Vdc from the half-wave rectifier, in V\n",
-      "Vm=round(Vm,0);  \n",
-      "#Result\n",
-      "print 'The a.c voltage required should have maximum value of = %d V' %Vm;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c voltage required should have maximum value of = 162 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.16, Page number 95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "rf=20;              #Internal resistance of the diodes in ohm\n",
-      "Vrms=50;            #RMS value of transformer's secondary voltage from centre tap to each end of secondary\n",
-      "RL=980;             #Load resistance in ohm\n",
-      "\n",
-      "#Calculation\n",
-      "Vm=Vrms*sqrt(2);      #Maximum a.c voltage in V\n",
-      "Im=Vm/(rf+RL);        #Maximum load current in A\n",
-      "Im=Im*1000;           #Maximum load current in mA\n",
-      " \n",
-      "# 1:\n",
-      "Idc=2*Im/pi;          #Mean load current\n",
-      "\n",
-      "# 2:\n",
-      "Irms=Im/sqrt(2);       #RMS value of load current in A\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print'   The mean load current= %d mA'%Idc;\n",
-      "print 'ii:';\n",
-      "print '  The r.m.s value of the load current = %d mA'%Irms; "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The mean load current= 45 mA\n",
-        "ii:\n",
-        "  The r.m.s value of the load current = 50 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.17, Page number 95-96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "#Variable declaration\n",
-      "RL=100;                          #Load resistance in ohm       \n",
-      "rf=0;                            #Internal resistance of the diodes in ohm\n",
-      "Turns_ratio=5/1;                 #Primary to secondary turns ratio of transformer \n",
-      "P_Vrms=230;                      #R.M.S value of voltage in primary winding in V\n",
-      "S_Vrms=P_Vrms/Turns_ratio;       #R.M.S value of voltage in secondary winding in V\n",
-      "S_Vm=S_Vrms*sqrt(2);        #Maximum voltage across secondary winding in V\n",
-      "Vm=S_Vm/2;                  #Maximum voltage across half seconfdary winding in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Idc=2*Vm/(pi*RL);            #Average current in A\n",
-      "Vdc=Idc*RL;                  #d.c output voltage in V\n",
-      "\n",
-      "# 2:\n",
-      "PIV=S_Vm;                    #Peak Invers Voltage(= Maximum secondary voltage) in V\n",
-      "\n",
-      "# 3:\n",
-      "Pac=pow(Vm/(RL*sqrt(2)),2)*(rf+RL);         #a.c input power in watt\n",
-      "Pdc=(pow(Idc,2)*RL);                        #d.c output power in watt\n",
-      "R_eff=(Pdc/Pac)*100;      #Rectification efficiency\n",
-      "R_eff=round(R_eff,1);\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   The d.c output voltage= %.1f V'%Vdc;\n",
-      "print 'ii:';\n",
-      "print '   The peak inverse voltage= %d V'%PIV;\n",
-      "print 'iii:';\n",
-      "print '   Rectification efficiency= %.1f%%'%R_eff;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The d.c output voltage= 20.7 V\n",
-        "ii:\n",
-        "   The peak inverse voltage= 65 V\n",
-        "iii:\n",
-        "   Rectification efficiency= 81.1%\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The value of rectification efficiency is calculated as 81.2% in the textbook using the formula 0.812/(1 + (rf/RL)), but by calculating using the correct values in the formula we get 81.1%."
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.18, Page number 96-97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "#Variable declaration\n",
-      "fin=50;                       #frequency of input ac source in Hz\n",
-      "RL=200;                       #Load resistance in ohm\n",
-      "Turns_ratio=4/1;              #Transformers turns ratio, primary to secondary.\n",
-      "P_Vrms=230.0;                 #R.M.S value of voltage in primary winding in V\n",
-      "S_Vrms=P_Vrms/Turns_ratio     #R.M.S value of voltage in secondary winding in V\n",
-      "Vm=S_Vrms*sqrt(2);            #Maximum voltage across secondary winding in V\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Idc=2*Vm/(pi*RL);           # Average current in A\n",
-      "Vdc=Idc*RL;                 #Output d.c voltage in V\n",
-      "Vdc=round(Vdc,0);\n",
-      "# 2:\n",
-      "PIV= Vm;                    #Peak Inverse Voltage(= Maximum volutage across secondary winding) in V\n",
-      "\n",
-      "# 3:\n",
-      "fout=2*fin;                 #Output frequency in Hz\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   The d.c output voltage = %d V' %Vdc;\n",
-      "print 'ii:';\n",
-      "print '   The peak inverse voltage = %.1f V'%PIV;\n",
-      "print 'iii:';\n",
-      "print '   The output frequency = %d Hz'%fout;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The d.c output voltage = 52 V\n",
-        "ii:\n",
-        "   The peak inverse voltage = 81.3 V\n",
-        "iii:\n",
-        "   The output frequency = 100 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.19, Page number 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=100.0;                        #Load Resistance in ohm\n",
-      "Turns_ratio=5/1;                 #Primary to secondary turns ratio of the transformer\n",
-      "Vin=230.0;                       #R.M.S value of input voltage in V\n",
-      "fin=50;                          #Input frequency in Hz\n",
-      "\n",
-      "#Calculation\n",
-      "Vs_rms=Vin/Turns_ratio;          #R.M.S value of the voltage in secondary winding, in v\n",
-      "Vs_max=Vs_rms*sqrt(2);           #Maximum voltage across secondary, in V\n",
-      "\n",
-      "# (i)\n",
-      "#Case i: Centre-tap circuit\n",
-      "Vm=Vs_max/2;                     #Maximum voltage across half secondary winding, in V \n",
-      "Vdc=2*Vm*RL/(pi*RL);             #DC output voltage, in V \n",
-      "print 'The d.c output voltage for the centre-tap circuit = %.1f V'%Vdc;\n",
-      "\n",
-      "#Case ii:\n",
-      "Vm=Vs_max;                      #Maximum voltage across secondary, in V\n",
-      "Vdc=2*Vm*RL/(pi*RL);            #DC output voltage, in V \n",
-      "print 'The d.c output voltage for the bridge circuit = %.1f V'%Vdc; \n",
-      "\n",
-      "# ii:\n",
-      "#Case i: Centre-tap circuit\n",
-      "Turns_ratio=5/1;                      #Turns ratio of the transformer\n",
-      "Vs_rms=Vin/Turns_ratio;               #R.M.S value of the secondary voltage in V\n",
-      "Vs_max=Vs_rms*sqrt(2);                #Maximum voltage across the secondary in V\n",
-      "Vm=Vs_max/2;                          #Maximum voltage across half of the secondary in V\n",
-      "PIV=2*Vm;                             #Peak Inverse Voltage in V\n",
-      "print 'PIV in case of centre-tap circuit = %d V'%PIV;\n",
-      "\n",
-      "#Case ii: Bridge circuit\n",
-      "Turns_ratio=10/1;                             #Turns ratio of the transformer\n",
-      "Vs_rms=Vin/Turns_ratio;                       #R.M.S value of the secondary voltage in V\n",
-      "Vs_max=Vs_rms*sqrt(2);                        #Maximum voltage across the secondary in V\n",
-      "PIV=Vm;                                        #Peak Inverse Voltage in V\n",
-      "print 'PIV in case of bridge circuit = %.1f V'%PIV;\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The d.c output voltage for the centre-tap circuit = 20.7 V\n",
-        "The d.c output voltage for the bridge circuit = 41.4 V\n",
-        "PIV in case of centre-tap circuit = 65 V\n",
-        "PIV in case of bridge circuit = 32.5 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.20, Page number 98"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "rf=1;                    #forward resistance of diodes of the rectifier in ohm\n",
-      "RL=480;                  #Load resistance in ohm\n",
-      "Vrms=240.0;                #a.c supply voltage in V\n",
-      "Vm=Vrms*sqrt(2);         #Maximum a.c voltage in V \n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Rt=2*rf+RL;             #Total circuit resistance at any instance in ohm\n",
-      "Im=Vm/Rt;               #Maximum load current in A\n",
-      "Idc=2*Im/pi;            #Mean load current in A\n",
-      "\n",
-      "# 2:\n",
-      "Irms=Im/2;              #R.M.S value of current  in A\n",
-      "P=pow(Irms,2)*rf;       #Power dissipated in each diode in watt\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   Mean load current = %.2f A'%Idc;\n",
-      "print 'ii:';\n",
-      "print '   Power dissipated in each diode= %.3f W'%P;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   Mean load current = 0.45 A\n",
-        "ii:\n",
-        "   Power dissipated in each diode= 0.124 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The value of power dissipated is approximately 0.124 W , but in the textbook it is approximated as 0.123W."
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.21, Page number 98-99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt,pi\n",
-      "#Variable declaration\n",
-      "RL=12000;                #Load resistance in ohm\n",
-      "V0=0.7;                   #Potential barrier voltage of diodes in V\n",
-      "Vrms=12;                  #R.M.S value of input a.c voltage in V\n",
-      "Vs_pk=Vrms*sqrt(2);       #Peak secondary voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Vout_pk=Vs_pk-(2*V0);      #Peak output voltage in  V\n",
-      "Vav=2*Vout_pk/pi;          #Average output voltage in V\n",
-      "Vav=round(Vav,2);\n",
-      "\n",
-      "# 2:\n",
-      "Iav=Vav/RL;                #Average output current in A\n",
-      "Iav=Iav*pow(10,6);         #Average output current in \u03bcA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   Average output voltage=%.2f V'%Vav;\n",
-      "print 'ii:';\n",
-      "print '   Average output current=%.1f \u03bcA'%Iav;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   Average output voltage=9.91 V\n",
-        "ii:\n",
-        "   Average output current=825.8 \u03bcA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.22, Page number 102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vdc_A=10;                      #Supply voltage of A in V\n",
-      "Vdc_B=25;                      #Supply voltage of B in V\n",
-      "Vac_rms_a=0.5;                 #Ripples in power supply A in V\n",
-      "Vac_rms_b=0.001;               #Ripples in power supply B in V\n",
-      "\n",
-      "#Calculation\n",
-      "#For power supply A\n",
-      "ripple_factor_A=Vac_rms_a/Vdc_A;      #Ripple factor of power supply A\n",
-      "\n",
-      "#For power supply B\n",
-      "ripple_factor_B=Vac_rms_b/Vdc_B;      #Ripple factor of power supply B\n",
-      "\n",
-      "#Result\n",
-      "if(ripple_factor_A<ripple_factor_B):\n",
-      "    print 'Power supply A is better';\n",
-      "else :\n",
-      "    print 'Power supply B is better';"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supply B is better\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.23, Page number 105-106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "RL=2200;                    #Load resistance in ohm\n",
-      "C=50*pow(10,-6);            #Capacitance of the capacitor used in filter circuit in F\n",
-      "V0=0.7;                     #Potential barrier voltage of the diodes of the rectifier in  V\n",
-      "Vrms=115.0;                   #R.M.S value of input a.c voltage in V \n",
-      "fin=60;                     #Frequency of input a.c voltage in Hz\n",
-      "Turns_ratio=10/1;           #Primary to secondary, turns ratio of the transformer \n",
-      "\n",
-      "#Calculation\n",
-      "Vp_prim=Vrms*sqrt(2);               #Peak primary voltage in V\n",
-      "Vp_sec=Vp_prim/Turns_ratio;         #Peak secondary voltage in V\n",
-      "Vp_in= Vp_sec - 2*V0;               #Peak full wave rectified voltage at the filter input in V\n",
-      "f=2*fin;                            #Output frequency in Hz\n",
-      "Vdc=Vp_in*(1-(1/(2*f*RL*C)));       #Output d.c voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print 'The output d.c voltage  is = %.1f V'%Vdc;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output d.c voltage  is = 14.3 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.24, Page number 106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "R=25;                        #d.c resistance of the choke in ohm\n",
-      "RL=750;                      #Load resistance in ohm\n",
-      "Vm=25.7;                     #Maximum value of the pulsating output from the rectifier in V\n",
-      "\n",
-      "#Calculation\n",
-      "V_dc=2*Vm/pi;                #d.c component of the pulsating output in V\n",
-      "V_dc=round(V_dc,1);\n",
-      "V_dc_out=(V_dc*RL)/(R+RL);   #Output d.c voltage in V\n",
-      "V_dc_out=round(V_dc_out,1);\n",
-      "\n",
-      "#Result\n",
-      "print ' The output d.c voltage accross the load resistance is = %.1f V'%V_dc_out;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The output d.c voltage accross the load resistance is = 15.9 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.25, Page number 113-114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=120.0;                   #Input Voltage in V\n",
-      "Vz=50.0;                    #Zener Voltage in V\n",
-      "R=5000.0;                   #Resistance of the series resistor in ohm\n",
-      "RL=10000.0;                 #Load resistance in ohm\n",
-      "\n",
-      "#Calculation\n",
-      "V=Ei*RL/(R+RL);             #Voltage across the open circuit if the zener diode is removed\n",
-      "if(V>Vz):\n",
-      "    #Zener diode is in ON state\n",
-      "    # i:\n",
-      "    Output_voltage=Vz;             #Voltage across load resistance, in V\n",
-      "    #ii:\n",
-      "    Voltage_R=Ei-Vz;               #Voltage across the series resistance R, in V\n",
-      "    #iii:\n",
-      "    IL=Vz/RL;                      #Load current through RL in A\n",
-      "    IL=IL*1000;                    #Load current through RL in mA\n",
-      "    I=Voltage_R/R;                 #Current through the series resistance in A\n",
-      "    I=I*1000;                      #Current through the series resistance in mA\n",
-      "    Iz=I-IL;                       #Applying Kirchhoff's first law, Zener current in mA\n",
-      "    \n",
-      "    #Result\n",
-      "    print 'i)   The output voltage across the load resistance RL = %d V'%Output_voltage;\n",
-      "    print 'ii)  The voltage drop across the series resistance  R = %d V'%Voltage_R;\n",
-      "    print 'iii) The current through the zener diode = %d mA'%Iz;\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)   The output voltage across the load resistance RL = 50 V\n",
-        "ii)  The voltage drop across the series resistance  R = 70 V\n",
-        "iii) The current through the zener diode = 9 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.26, Page number 114-115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Max_V=120.0;                  #Maximum input voltage in V\n",
-      "Min_V=80.0;                   #Minimum input voltage in V\n",
-      "R=5000.0;                     #Series resistance in ohm\n",
-      "RL=10000.0;                   #Load resistance in ohm\n",
-      "Vz=50.0;                      #Zener voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i: Maximum zener current\n",
-      "#Zener current will be maximum when the input voltage is maximum\n",
-      "V_R_max=Max_V-Vz;                #Voltage across series resistance R, in  V\n",
-      "I_max=V_R_max/R;                 #Current through series resistance R, in A\n",
-      "I_max=I_max*1000;                #Current through series resistance R, in mA\n",
-      "IL_max=Vz/RL;                        #Load current in A\n",
-      "IL_max=IL_max*1000;                      #Load current in mA\n",
-      "Iz_max=I_max-IL_max;                     #Applying Kirchhoff's first law, Zener current in mA;\n",
-      "\n",
-      "#Case ii: Minimum zener current\n",
-      "#The zener will conduct minimum current when the input voltage is minimum\n",
-      "V_R_min=Min_V-Vz;                #Voltage across series resistance R, in V\n",
-      "I_min=V_R_min/R;                 #Current through series resistance R, in A\n",
-      "I_min=I_min*1000;                #Current through series resistance R, in mA\n",
-      "IL_min=Vz/RL;                        #Load current in A\n",
-      "IL_min=IL_min*1000;                      #Load current in mA\n",
-      "Iz_min=I_min-IL_min;                     #Applying Kirchhoff's first law, Zener current in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Case i: ';\n",
-      "print 'Maximum zener current = %d mA'%Iz_max;\n",
-      "print 'Case ii: ';\n",
-      "print 'Minimum zener current = %d mA'%Iz_min;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case i: \n",
-        "Maximum zener current = 9 mA\n",
-        "Case ii: \n",
-        "Minimum zener current = 1 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.27, Page number 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=12;                      #Input voltage in V\n",
-      "Vz=7.2;                     #Zener voltage in V\n",
-      "E0=Vz;                      #Voltage to be maintained across the load in  V\n",
-      "IL_max=0.1;                 #Maximum load current in A\n",
-      "IL_min=0.012;                  #Minimum load current in A\n",
-      "Iz_min=0.01;                  #Minimum zener current in A\n",
-      "\n",
-      "#Calculation\n",
-      "#When the load current is maximum at minimum value of RL, the zener current is minimum and, as the load current decreases due to increase in value of RL\n",
-      "R=(Ei-E0)/(Iz_min+IL_max);        #The value of series resistance R to maintain a voltage=E0 across load, in ohm\n",
-      "\n",
-      "#Result\n",
-      "print 'The minimum value of series resistance R to maintain a constant value of 7.2 V is = %.1f \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum value of series resistance R to maintain a constant value of 7.2 V is = 43.6 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The actual value of R is 43.636363 (recurring) but, in the textbook the value of R is wrongly approximated 43.5 \u03a9"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.28, Page number 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei_min=22;                #Minimum input voltage in V\n",
-      "Ei_max=28;                #Maximum input voltage in V\n",
-      "Vz=18;                   #Zener voltage in V\n",
-      "E0=Vz;                    #Constant voltage maintained across the load resistance in V\n",
-      "Iz_min=0.2;               #Minimum zener current in A\n",
-      "Iz_max=2;                 #Maximum zener current in A\n",
-      "RL=18;                    #Load resistance in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "IL=Vz/RL;                      #Constant value of load current in A\n",
-      "#When the input voltage is minimum, the zener current will be minimum\n",
-      "R=(Ei_min-E0)/(Iz_min+IL)      #The value of series resistance so that the voltage E0 across RL remains constant\n",
-      "\n",
-      "print 'The value of series resistance R, to maintain constant voltage E0 across RL = %.2f \u03a9.'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistance R, to maintain constant voltage E0 across RL = 3.33 \u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.29, Page number 116 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10                                #Zener voltage in V\n",
-      "Ei_min=13;                           #Minimum input voltage in V\n",
-      "Ei_max=16;                           #Maximum input voltage in V\n",
-      "Iz_min=0.015;                        #Minimum zener current in A\n",
-      "IL_min=0.01;                         #Minimum load current in A \n",
-      "IL_max=0.085;                        #Maximum load curremt in A\n",
-      "E0=Vz;                               #Constant voltage to be maintained in V \n",
-      "\n",
-      "#Calculation\n",
-      "#The zener current will be minimum when the input voltage will be minimum  and at that time the load current will be maximum\n",
-      "R=(Ei_min-E0)/(Iz_min+IL_max);       #The value of series resistance R to maintain a constant voltage across load\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'The value of series resistance to maintain a constant voltage across the load resistance is = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistance to maintain a constant voltage across the load resistance is = 30 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.30, Page number 116"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Iz=0.2;               #Current rating of each zener in A\n",
-      "Vz=15;                #Voltage rating of each zener in V\n",
-      "Ei=45;                #Input voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "# i: Regulated output voltage across the two zener diodes \n",
-      "E0=2*Vz;                     # V\n",
-      "\n",
-      "# ii: Value of series resistance \n",
-      "R=(Ei-E0)/Iz;                  # \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print 'i) The regulated output voltage = %d V'%E0;\n",
-      "print 'ii) The value of the series resistance = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) The regulated output voltage = 30 V\n",
-        "ii) The value of the series resistance = 75 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.31, Page number 116-117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10;                   #Voltage rating of each zener in V\n",
-      "Iz=1;                    #Current rating of each zener in A\n",
-      "Ei=45;                   #Input unregulated voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "#Regulated output voltage across the three zener diodes\n",
-      "E0=3*Vz;               # V\n",
-      "\n",
-      "#Value of series resistance to obtain a 30V regulated output voltage\n",
-      "R=(Ei-E0)/Iz;          # \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print 'Value of series resistance to obtain a 30V regulated output voltage = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of series resistance to obtain a 30V regulated output voltage = 15 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.32, Page number 117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "RL=2000.0;                   #Load resistance in \u03a9\n",
-      "R=200.0;                     #Series resistance in \u03a9\n",
-      "Iz=0.025;                  #Zener current rating in A\n",
-      "E0=30.0;                     #Output regulated voltage in V \n",
-      "\n",
-      "#Calculation\n",
-      "#Minimum input voltage will be required when Iz=0 A, and at  this condition\n",
-      "IL=E0/RL;                   #Load current during Iz=0, in A\n",
-      "I=IL;                       #According to Kirchhoff's law, total current, in A\n",
-      "Ei_min=E0+(I*R);            #Minimum input voltage in V\n",
-      "\n",
-      "#The maximum input voltage required will be when Iz=0.025 A, and at that condition \n",
-      "I=IL+Iz;                    #According to Kirchhoff's law, total current, in A\n",
-      "Ei_max=E0+(I*R);            #maximum input voltage in V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'The required range of input voltage is from %d V to %d V'%(Ei_min,Ei_max);  \n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required range of input voltage is from 33 V to 38 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.33, Page number 117-118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=16;                   #Unregulated input voltage in V\n",
-      "E0=12;                   #Output regulated voltage in V\n",
-      "IL_min=0;                #Minimum load current in A\n",
-      "IL_max=0.2;              #Maximum load current in A\n",
-      "Iz_min=0;                #Minimum zener current in A\n",
-      "Iz_max=0.2;              #Maximum zener current in A\n",
-      "\n",
-      "#Calculation\n",
-      "#As the regulated voltage required across the load is 12V\n",
-      "Vz=E0;                   #Voltage rating of zener diode in V\n",
-      "V_R=Ei-E0;               #Constant Voltage that should remain across series resistance \n",
-      "#The minimum zener current will occur when the curent in the load in maximum\n",
-      "R=V_R/(Iz_min+IL_max);   #Series resistance in \u03a9\n",
-      "\n",
-      "Max_power_rating=Vz*Iz_max;     #Maximum power rating of zener diode in W\n",
-      "\n",
-      "#Result\n",
-      "print 'The regulator is designed using a Seris resistance of %d \u03a9 and a zener diode of zener voltage %d V'%(R,Vz);\n",
-      "print 'The maximum power rating of the zener diode is = %.1f W '%Max_power_rating;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulator is designed using a Seris resistance of 20 \u03a9 and a zener diode of zener voltage 12 V\n",
-        "The maximum power rating of the zener diode is = 2.4 W \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.34, Page number 118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=12;                       #Source voltage in V\n",
-      "R=1000;                     #Series resistance in \u03a9\n",
-      "RL=5000;                    #Load resistance in \u03a9\n",
-      "Vz=6;                       #Voltage rating of zener in V\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i: zener is working properly\n",
-      "#The output voltage across the load will be equal to the zener voltage.\n",
-      "V0=Vz;                        # V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case i: Output voltage when zener is working properly is %d V'%V0;\n",
-      "\n",
-      "#Case ii: zener is shorted\n",
-      "#As the zener is shorted, the potential difference across the load will be zero\n",
-      "V0=0;                        #V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case ii: Output voltage when zener is short circuited is %d V'%V0;\n",
-      "     \n",
-      "#Case iii: zener is open circuited\n",
-      "#If the zener is open circuited, the total voltage will drop across R and RL according to the voltage divider rule\n",
-      "V0=V*RL/(R+RL);                      #V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case iii: Output voltage when zener is open circuited is %d V'%V0;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case i: Output voltage when zener is working properly is 6 V\n",
-        "Case ii: Output voltage when zener is short circuited is 0 V\n",
-        "Case iii: Output voltage when zener is open circuited is 10 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_2.ipynb
deleted file mode 100755
index a6008ee1..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_2.ipynb
+++ /dev/null
@@ -1,1624 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3452607f2168b562d941493f83083042eaa5a2d316715f9d9f089ff03d73fdb8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 6: SEMICONDUCTOR DIODE"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2, Page number 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration \n",
-      "Vf =20;                #Peak Input Voltage in V\n",
-      "rf=10;                  #Forward Resistance in ohms\n",
-      "RL=500.0;                 #Load Resistance in ohms\n",
-      "V0=0.7;                  #Potential Barrier Voltage of the diodes in V\n",
-      "\n",
-      "#Calculation\n",
-      "#(1)\n",
-      "If_peak=(Vf-V0)/(rf+RL);                  #Peak current through the diode in A\n",
-      "If_peak=If_peak*1000;                    #Peak current through the diode in mA\n",
-      "#(2)\n",
-      "V_out_peak =If_peak * RL/1000 ;               #Peak output voltage in V\n",
-      "\n",
-      "#For an Ideal diode\n",
-      "If_peak_ideal=Vf/RL;                   #Peak current through the ideal diode in A\n",
-      "If_peak_ideal=If_peak_ideal*1000;      #Peak current through the ideal diode in mA\n",
-      "\n",
-      "V_out_peak_ideal=If_peak_ideal * RL/1000;     # Peak output voltage in case of the ideal diode in V\n",
-      "\n",
-      "#Result\n",
-      "print '(i) Peak current through the diode = %.1f mA '%If_peak;\n",
-      "print '(ii) Peak output voltage = %.1f V'%V_out_peak;\n",
-      "print '(iii) Peak current through the ideal diode = %d mA '%If_peak_ideal;\n",
-      "print '(iv) Peak output voltage in case of the ideal diode = %d V'%V_out_peak_ideal;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Peak current through the diode = 37.8 mA \n",
-        "(ii) Peak output voltage = 18.9 V\n",
-        "(iii) Peak current through the ideal diode = 40 mA \n",
-        "(iv) Peak output voltage in case of the ideal diode = 20 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3, Page number 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V =10.0;                #Battery voltage in V\n",
-      "R1=50.0;                  #Resistor 1's resistance in ohms\n",
-      "R2=5.0;                 #Resistor 2's resistance in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "#Using Thevenin's Theorem to find current in the diode\n",
-      "E0=(R2/(R1+R2))*V;          #Thevenin's Voltage in V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's Resistance in ohms\n",
-      "\n",
-      "I0=E0/R0;                   #Current through the diode in A\n",
-      "I0=I0*1000;                 #Current through the diode in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Current through the diode = %d mA '%Io;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through the diode = 200 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4, Page number 82-83 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V =10.0;                #Battery voltage in V\n",
-      "R0=48.0;                  #Resistance of the resistor in ohms\n",
-      "Rd=1.0;                 #Forward resistance of the diodes in ohms\n",
-      "Vd=0.7;                 #Potential barrier of the diodes in V\n",
-      "#Calculation\n",
-      "V_net=V-Vd-Vd;        #Net voltage in the circuit in V\n",
-      "R_net=R0+Rd+Rd        #Net resistance of the circuit in ohms\n",
-      "I_net=V_net/R_net;    #Net current in the circuit in A\n",
-      "I_net=I_net*1000;     #Net current in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Net current in the circuit = %d mA '%I_net;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Net current in the circuit = 172 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page number 83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E1=24;           #Voltage of first source in V\n",
-      "E2=4;            #Voltage of second source in V\n",
-      "V0=0.7;          #Potential barrier of diodes in V\n",
-      "R=2000;          #Resistance of the given resistor in ohms\n",
-      "Rd=0;            #Forward resistance of the diodes in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I=(E1-E2-V0)/(R+Rd);  #Current in the circuit in A\n",
-      "I=I*1000;             #Current in the circuit in mA \n",
-      "\n",
-      "#Result\n",
-      "print 'Current in the circuit = %.2f mA '%I;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in the circuit = 9.65 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page number 83-84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=20;           #Voltage of source in V\n",
-      "V0=0.3;         #Potential barrier of Germanium diode in V\n",
-      "V0_Si=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "\n",
-      "#Calculation\n",
-      "#As only Ge diode is turned on due to less potential barrier,\n",
-      "VA=V-V0;        #Voltage VA acroos resistor of 3k ohms\n",
-      "\n",
-      "#Result\n",
-      "print 'Voltage VA = %.1f mA '%VA;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage VA = 19.7 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7, Page number 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=10;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "# Resistance of all resistors in ohms\n",
-      "R1=2000;\n",
-      "R2=2000;\n",
-      "R3=2000;\n",
-      "\n",
-      "#Calculation\n",
-      "Id=(V-V0)/(R2+2*R3);     #Current through the diodes in A\n",
-      "VQ=2*Id*R3;      #Voltage VQ across the grounded 2k ohm resistor in V\n",
-      "Id=Id*1000;      #Current through the diodes in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Voltage VQ = %.1f V '%VQ;\n",
-      "print 'Current through the diodes, Id = %.2f mA '%Id;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage VQ = 6.2 V \n",
-        "Current through the diodes, Id = 1.55 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8, Page number 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=15;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "R=500       # Resistance of all resistors in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I1=(V-V0)/R;      #total current in the circuit in A\n",
-      "Id1=I1/2;         #current in first diode in A\n",
-      "Id1=Id1*1000;     #current in first diode in mA\n",
-      "Id2=Id1           #current in second diode in mA\n",
-      "\n",
-      "#Result\n",
-      "print ('Current in first diode = %.1f mA'%Id1);\n",
-      "print ('Current in second diode = %.1f mA'%Id2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in first diode = 14.3 mA\n",
-        "Current in second diode = 14.3 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9, Page number 85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E=20;           #Voltage of source in V\n",
-      "V0_d1=0.7;      #Potetial barrier of first Silicon diode in V\n",
-      "V0_d2=0.7;      #Potetial barrier of second Silicon diode in V\n",
-      "R1=5600;       # Resistance of first resistor in ohms\n",
-      "R2=3300;       #  Resistance of second resistor in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I2=V0_d2/R2;                  #Current I2 through resistor R2 in A\n",
-      "I2=round((I2*1000),3);                   #Current I2 through resistor R2 in mA\n",
-      "I1=(E-V0_d1-V0_d2)/R1;        #Current I1 through resistor R1 in  A\n",
-      "I1=round((I1*1000),2);                   #Current I1 through resistor R1 in  mA\n",
-      "I3=I1-I2;                     #Current I3 through diode D2 in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Current I1= %.2f mA'%I1;\n",
-      "print 'Current I1= %.3f mA'%I2;\n",
-      "print 'Current I1= %.3f mA'%I3;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current I1= 3.32 mA\n",
-        "Current I1= 0.212 mA\n",
-        "Current I1= 3.108 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10, Page number 85-86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E=10.0;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V\n",
-      "R1=2000;       # Resistance of first resistor in ohms\n",
-      "R2=8000;       #  Resistance of second resistor in ohms\n",
-      "R3=4000;       #Resistance of third resistor in ohms\n",
-      "R4=6000;       #Resistance of fourth resistor in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "#Assuming the given diode to be reverse bised and calculating voltage across it's terminals\n",
-      "V1=(E/(R1+R2))*R2;      #voltage at the P side of the diode, i.e, voltage across R2 resistor,according to voltage divider rule, in V\n",
-      "V2=(E/(R3+R4))*R4;      #voltage at the N side of the diode, i.e, voltage across R4 resistor,according to voltage divider rule, in V\n",
-      "\n",
-      "#Result\n",
-      "if((V1-V2)>=V0):\n",
-      "    print 'Our assumption was wrong and, the diode is forward biased';\n",
-      "else:\n",
-      "    print 'The diode is reverse biased';\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Our assumption was wrong and, the diode is forward biased\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.11, Page number 86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=2;                              #Supply voltage in V\n",
-      "V0=0.7;                           #Potential barrier voltage of the diode in V \n",
-      "R1=4000.0;                        #Resistance of first resistor in \u03a9\n",
-      "R2=1000.0;                        ##Resistance of second resistor in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Assuming the diode to be in ON state\n",
-      "I1=((V-V0)/R1)*1000;                  #Current through resistor R1, in mA\n",
-      "I2=(V0/R2)*1000;                    #Current through resistor R2, in mA\n",
-      "ID=I1-I2;                           #Diode current, in  mA\n",
-      "\n",
-      "if(ID<0):\n",
-      "    #Since the diode current is negative, the diode must be OFF \n",
-      "    ID=0;                     #True value of diode current, mA\n",
-      "    \n",
-      "#As the diode is in OFF state it can be replaced by an open ciruit equivalent \n",
-      "VD=V*R2/(R1 +R2);                #Voltage across the diode, in V\n",
-      "\n",
-      "#Result\n",
-      "print 'ID =%d mA'%ID;\n",
-      "print 'VD =%.1f V'%VD;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ID =0 mA\n",
-        "VD =0.4 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12, Page number 89-90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "AC_Input_Power=100.0;          #Input AC Power in watts\n",
-      "AC_Output_Power=40.0;          #Output AC Power in watts\n",
-      "Accepted_Power=50.0;           #Power accepted by the half-wave rectifier in watt\n",
-      "\n",
-      "#Calculation\n",
-      "R_eff=(AC_Output_Power/AC_Input_Power)*100;             #Rectification efficiency of the half-wave rectifier\n",
-      "Unused_power=AC_Input_Power-Accepted_Power;       #Power not used by the half_wave rectifier due to open circuited condition of the diode in watt\n",
-      "Power_dissipated=Accepted_Power-AC_Output_Power;  #Power dissipated by the diode watt\n",
-      "\n",
-      "#Result\n",
-      "print 'The rectification efficiency of the half-wave rectifier= %d%% '%R_eff;\n",
-      "\n",
-      "print 'Rest 60%% of the power is the unused power and power dissipated by the diode = %d watts and %d watts' %(Unused_power ,Power_dissipated);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rectification efficiency of the half-wave rectifier= 40% \n",
-        "Rest 60% of the power is the unused power and power dissipated by the diode = 50 watts and 10 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.13, Page number 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "Vrms=230.0;                #AC supply RMS voltage in V\n",
-      "Turns_Ratio=10/1;        #turn ratio of the transformer \n",
-      "\n",
-      "#Calculation\n",
-      "Vpm=sqrt(2)*Vrms;         #Maximum primary voltage in V\n",
-      "Vsm=Vpm/Turns_Ratio;     #Maximum secondary voltage in V\n",
-      "#Case 1\n",
-      "Vdc=Vsm/(round(pi,2));              #Output D.C voltage, which is the average voltage in V\n",
-      "Vdc=round(Vdc,2);\n",
-      "#Case 2\n",
-      "PIV=Vsm;                 #Peak Inverse Voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print 'The output d.c voltage= %.2f V'%Vdc;\n",
-      "print 'The peak inverse voltage= %.2f V'%PIV;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output d.c voltage= 10.36 V\n",
-        "The peak inverse voltage= 32.53 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.14, Page number 90-91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "rf=20.0;           #Internal resistance of the crystal diode in ohms\n",
-      "Vm=50.0;           #Maximum applied voltage in V\n",
-      "RL=800.0;          #Load Resistance in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "# 1\n",
-      "Im=Vm/(rf+RL);         #Maximum current in A\n",
-      "Im=Im*1000;            #Maximum current in \n",
-      "Im=round(Im,0);\n",
-      "Idc=Im/pi;             #Average voltage in mA\n",
-      "Idc=round(Idc,1);\n",
-      "Irms=Im/2;             #RMS value of the current in mA\n",
-      "Irms=round(Irms,1)\n",
-      "\n",
-      "# 2\n",
-      "AC_Input_Power=pow(Irms/1000,2)*(rf+RL);     #Input a.c power in watt\n",
-      "\n",
-      "DC_Output_Power=pow(Idc/1000,2)*RL;          #Output d.c power in watt\n",
-      "\n",
-      "# 3\n",
-      "DC_Output_Voltage=(Idc/1000)*RL;               #Output d.c voltage in V\n",
-      "\n",
-      "# 4\n",
-      "Rectifier_efficiency=(DC_Output_Power/AC_Input_Power)*100;   # Efficiency of rectification of the half-wave rectifier\n",
-      "\n",
-      "#Result\n",
-      "print ' i:';\n",
-      "print '   Im   = %d mA'%Im;\n",
-      "print '   Idc  = %.1f mA'%Idc;\n",
-      "print '   Irms = %.1f mA'%Irms;\n",
-      "print ' ii: ';\n",
-      "print '   a.c input power= %.3f watt'%AC_Input_Power;\n",
-      "print '   d.c output power= %.3f watt'%DC_Output_Power;\n",
-      "print ' iii: ';\n",
-      "print '   d.c output voltage = %.2f volts'%DC_Output_Voltage;\n",
-      "print ' iv: '\n",
-      "print '   Efficiency of rectification = %.1f%%'%Rectifier_efficiency;\n",
-      "\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " i:\n",
-        "   Im   = 61 mA\n",
-        "   Idc  = 19.4 mA\n",
-        "   Irms = 30.5 mA\n",
-        " ii: \n",
-        "   a.c input power= 0.763 watt\n",
-        "   d.c output power= 0.301 watt\n",
-        " iii: \n",
-        "   d.c output voltage = 15.52 volts\n",
-        " iv: \n",
-        "   Efficiency of rectification = 39.5%\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.15, Page number 91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "Vdc=50.0;           #Output d.c voltage in V\n",
-      "rf=25;            #Diode resistance in ohm\n",
-      "RL=800;           #Load resistance in ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vm=(pi*(rf+RL)*Vdc)/RL;   #[ Vdc=Vm*RL/(pi*(rf+RL)) ]Maximum value of a.c voltage required to get a volatge  of Vdc from the half-wave rectifier, in V\n",
-      "Vm=round(Vm,0);  \n",
-      "#Result\n",
-      "print 'The a.c voltage required should have maximum value of = %d V' %Vm;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c voltage required should have maximum value of = 162 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.16, Page number 95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "rf=20;              #Internal resistance of the diodes in ohm\n",
-      "Vrms=50;            #RMS value of transformer's secondary voltage from centre tap to each end of secondary\n",
-      "RL=980;             #Load resistance in ohm\n",
-      "\n",
-      "#Calculation\n",
-      "Vm=Vrms*sqrt(2);      #Maximum a.c voltage in V\n",
-      "Im=Vm/(rf+RL);        #Maximum load current in A\n",
-      "Im=Im*1000;           #Maximum load current in mA\n",
-      " \n",
-      "# 1:\n",
-      "Idc=2*Im/pi;          #Mean load current\n",
-      "\n",
-      "# 2:\n",
-      "Irms=Im/sqrt(2);       #RMS value of load current in A\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print'   The mean load current= %d mA'%Idc;\n",
-      "print 'ii:';\n",
-      "print '  The r.m.s value of the load current = %d mA'%Irms; "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The mean load current= 45 mA\n",
-        "ii:\n",
-        "  The r.m.s value of the load current = 50 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.17, Page number 95-96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "#Variable declaration\n",
-      "RL=100;                          #Load resistance in ohm       \n",
-      "rf=0;                            #Internal resistance of the diodes in ohm\n",
-      "Turns_ratio=5/1;                 #Primary to secondary turns ratio of transformer \n",
-      "P_Vrms=230;                      #R.M.S value of voltage in primary winding in V\n",
-      "S_Vrms=P_Vrms/Turns_ratio;       #R.M.S value of voltage in secondary winding in V\n",
-      "S_Vm=S_Vrms*sqrt(2);        #Maximum voltage across secondary winding in V\n",
-      "Vm=S_Vm/2;                  #Maximum voltage across half seconfdary winding in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Idc=2*Vm/(pi*RL);            #Average current in A\n",
-      "Vdc=Idc*RL;                  #d.c output voltage in V\n",
-      "\n",
-      "# 2:\n",
-      "PIV=S_Vm;                    #Peak Invers Voltage(= Maximum secondary voltage) in V\n",
-      "\n",
-      "# 3:\n",
-      "Pac=pow(Vm/(RL*sqrt(2)),2)*(rf+RL);         #a.c input power in watt\n",
-      "Pdc=(pow(Idc,2)*RL);                        #d.c output power in watt\n",
-      "R_eff=(Pdc/Pac)*100;      #Rectification efficiency\n",
-      "R_eff=round(R_eff,1);\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   The d.c output voltage= %.1f V'%Vdc;\n",
-      "print 'ii:';\n",
-      "print '   The peak inverse voltage= %d V'%PIV;\n",
-      "print 'iii:';\n",
-      "print '   Rectification efficiency= %.1f%%'%R_eff;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The d.c output voltage= 20.7 V\n",
-        "ii:\n",
-        "   The peak inverse voltage= 65 V\n",
-        "iii:\n",
-        "   Rectification efficiency= 81.1%\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The value of rectification efficiency is calculated as 81.2% in the textbook using the formula 0.812/(1 + (rf/RL)), but by calculating using the correct values in the formula we get 81.1%."
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.18, Page number 96-97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "#Variable declaration\n",
-      "fin=50;                       #frequency of input ac source in Hz\n",
-      "RL=200;                       #Load resistance in ohm\n",
-      "Turns_ratio=4/1;              #Transformers turns ratio, primary to secondary.\n",
-      "P_Vrms=230.0;                 #R.M.S value of voltage in primary winding in V\n",
-      "S_Vrms=P_Vrms/Turns_ratio     #R.M.S value of voltage in secondary winding in V\n",
-      "Vm=S_Vrms*sqrt(2);            #Maximum voltage across secondary winding in V\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Idc=2*Vm/(pi*RL);           # Average current in A\n",
-      "Vdc=Idc*RL;                 #Output d.c voltage in V\n",
-      "Vdc=round(Vdc,0);\n",
-      "# 2:\n",
-      "PIV= Vm;                    #Peak Inverse Voltage(= Maximum volutage across secondary winding) in V\n",
-      "\n",
-      "# 3:\n",
-      "fout=2*fin;                 #Output frequency in Hz\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   The d.c output voltage = %d V' %Vdc;\n",
-      "print 'ii:';\n",
-      "print '   The peak inverse voltage = %.1f V'%PIV;\n",
-      "print 'iii:';\n",
-      "print '   The output frequency = %d Hz'%fout;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The d.c output voltage = 52 V\n",
-        "ii:\n",
-        "   The peak inverse voltage = 81.3 V\n",
-        "iii:\n",
-        "   The output frequency = 100 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.19, Page number 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=100.0;                        #Load Resistance in ohm\n",
-      "Turns_ratio=5/1;                 #Primary to secondary turns ratio of the transformer\n",
-      "Vin=230.0;                       #R.M.S value of input voltage in V\n",
-      "fin=50;                          #Input frequency in Hz\n",
-      "\n",
-      "#Calculation\n",
-      "Vs_rms=Vin/Turns_ratio;          #R.M.S value of the voltage in secondary winding, in v\n",
-      "Vs_max=Vs_rms*sqrt(2);           #Maximum voltage across secondary, in V\n",
-      "\n",
-      "# (i)\n",
-      "#Case i: Centre-tap circuit\n",
-      "Vm=Vs_max/2;                     #Maximum voltage across half secondary winding, in V \n",
-      "Vdc=2*Vm*RL/(pi*RL);             #DC output voltage, in V \n",
-      "print 'The d.c output voltage for the centre-tap circuit = %.1f V'%Vdc;\n",
-      "\n",
-      "#Case ii:\n",
-      "Vm=Vs_max;                      #Maximum voltage across secondary, in V\n",
-      "Vdc=2*Vm*RL/(pi*RL);            #DC output voltage, in V \n",
-      "print 'The d.c output voltage for the bridge circuit = %.1f V'%Vdc; \n",
-      "\n",
-      "# ii:\n",
-      "#Case i: Centre-tap circuit\n",
-      "Turns_ratio=5/1;                      #Turns ratio of the transformer\n",
-      "Vs_rms=Vin/Turns_ratio;               #R.M.S value of the secondary voltage in V\n",
-      "Vs_max=Vs_rms*sqrt(2);                #Maximum voltage across the secondary in V\n",
-      "Vm=Vs_max/2;                          #Maximum voltage across half of the secondary in V\n",
-      "PIV=2*Vm;                             #Peak Inverse Voltage in V\n",
-      "print 'PIV in case of centre-tap circuit = %d V'%PIV;\n",
-      "\n",
-      "#Case ii: Bridge circuit\n",
-      "Turns_ratio=10/1;                             #Turns ratio of the transformer\n",
-      "Vs_rms=Vin/Turns_ratio;                       #R.M.S value of the secondary voltage in V\n",
-      "Vs_max=Vs_rms*sqrt(2);                        #Maximum voltage across the secondary in V\n",
-      "PIV=Vm;                                        #Peak Inverse Voltage in V\n",
-      "print 'PIV in case of bridge circuit = %.1f V'%PIV;\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The d.c output voltage for the centre-tap circuit = 20.7 V\n",
-        "The d.c output voltage for the bridge circuit = 41.4 V\n",
-        "PIV in case of centre-tap circuit = 65 V\n",
-        "PIV in case of bridge circuit = 32.5 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.20, Page number 98"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "rf=1;                    #forward resistance of diodes of the rectifier in ohm\n",
-      "RL=480;                  #Load resistance in ohm\n",
-      "Vrms=240.0;                #a.c supply voltage in V\n",
-      "Vm=Vrms*sqrt(2);         #Maximum a.c voltage in V \n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Rt=2*rf+RL;             #Total circuit resistance at any instance in ohm\n",
-      "Im=Vm/Rt;               #Maximum load current in A\n",
-      "Idc=2*Im/pi;            #Mean load current in A\n",
-      "\n",
-      "# 2:\n",
-      "Irms=Im/2;              #R.M.S value of current  in A\n",
-      "P=pow(Irms,2)*rf;       #Power dissipated in each diode in watt\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   Mean load current = %.2f A'%Idc;\n",
-      "print 'ii:';\n",
-      "print '   Power dissipated in each diode= %.3f W'%P;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   Mean load current = 0.45 A\n",
-        "ii:\n",
-        "   Power dissipated in each diode= 0.124 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The value of power dissipated is approximately 0.124 W , but in the textbook it is approximated as 0.123W."
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.21, Page number 98-99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt,pi\n",
-      "#Variable declaration\n",
-      "RL=12000;                #Load resistance in ohm\n",
-      "V0=0.7;                   #Potential barrier voltage of diodes in V\n",
-      "Vrms=12;                  #R.M.S value of input a.c voltage in V\n",
-      "Vs_pk=Vrms*sqrt(2);       #Peak secondary voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Vout_pk=Vs_pk-(2*V0);      #Peak output voltage in  V\n",
-      "Vav=2*Vout_pk/pi;          #Average output voltage in V\n",
-      "Vav=round(Vav,2);\n",
-      "\n",
-      "# 2:\n",
-      "Iav=Vav/RL;                #Average output current in A\n",
-      "Iav=Iav*pow(10,6);         #Average output current in \u03bcA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   Average output voltage=%.2f V'%Vav;\n",
-      "print 'ii:';\n",
-      "print '   Average output current=%.1f \u03bcA'%Iav;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   Average output voltage=9.91 V\n",
-        "ii:\n",
-        "   Average output current=825.8 \u03bcA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.22, Page number 102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vdc_A=10;                      #Supply voltage of A in V\n",
-      "Vdc_B=25;                      #Supply voltage of B in V\n",
-      "Vac_rms_a=0.5;                 #Ripples in power supply A in V\n",
-      "Vac_rms_b=0.001;               #Ripples in power supply B in V\n",
-      "\n",
-      "#Calculation\n",
-      "#For power supply A\n",
-      "ripple_factor_A=Vac_rms_a/Vdc_A;      #Ripple factor of power supply A\n",
-      "\n",
-      "#For power supply B\n",
-      "ripple_factor_B=Vac_rms_b/Vdc_B;      #Ripple factor of power supply B\n",
-      "\n",
-      "#Result\n",
-      "if(ripple_factor_A<ripple_factor_B):\n",
-      "    print 'Power supply A is better';\n",
-      "else :\n",
-      "    print 'Power supply B is better';"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supply B is better\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.23, Page number 105-106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "RL=2200;                    #Load resistance in ohm\n",
-      "C=50*pow(10,-6);            #Capacitance of the capacitor used in filter circuit in F\n",
-      "V0=0.7;                     #Potential barrier voltage of the diodes of the rectifier in  V\n",
-      "Vrms=115.0;                   #R.M.S value of input a.c voltage in V \n",
-      "fin=60;                     #Frequency of input a.c voltage in Hz\n",
-      "Turns_ratio=10/1;           #Primary to secondary, turns ratio of the transformer \n",
-      "\n",
-      "#Calculation\n",
-      "Vp_prim=Vrms*sqrt(2);               #Peak primary voltage in V\n",
-      "Vp_sec=Vp_prim/Turns_ratio;         #Peak secondary voltage in V\n",
-      "Vp_in= Vp_sec - 2*V0;               #Peak full wave rectified voltage at the filter input in V\n",
-      "f=2*fin;                            #Output frequency in Hz\n",
-      "Vdc=Vp_in*(1-(1/(2*f*RL*C)));       #Output d.c voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print 'The output d.c voltage  is = %.1f V'%Vdc;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output d.c voltage  is = 14.3 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.24, Page number 106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "R=25;                        #d.c resistance of the choke in ohm\n",
-      "RL=750;                      #Load resistance in ohm\n",
-      "Vm=25.7;                     #Maximum value of the pulsating output from the rectifier in V\n",
-      "\n",
-      "#Calculation\n",
-      "V_dc=2*Vm/pi;                #d.c component of the pulsating output in V\n",
-      "V_dc=round(V_dc,1);\n",
-      "V_dc_out=(V_dc*RL)/(R+RL);   #Output d.c voltage in V\n",
-      "V_dc_out=round(V_dc_out,1);\n",
-      "\n",
-      "#Result\n",
-      "print ' The output d.c voltage accross the load resistance is = %.1f V'%V_dc_out;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The output d.c voltage accross the load resistance is = 15.9 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.25, Page number 113-114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=120.0;                   #Input Voltage in V\n",
-      "Vz=50.0;                    #Zener Voltage in V\n",
-      "R=5000.0;                   #Resistance of the series resistor in ohm\n",
-      "RL=10000.0;                 #Load resistance in ohm\n",
-      "\n",
-      "#Calculation\n",
-      "V=Ei*RL/(R+RL);             #Voltage across the open circuit if the zener diode is removed\n",
-      "if(V>Vz):\n",
-      "    #Zener diode is in ON state\n",
-      "    # i:\n",
-      "    Output_voltage=Vz;             #Voltage across load resistance, in V\n",
-      "    #ii:\n",
-      "    Voltage_R=Ei-Vz;               #Voltage across the series resistance R, in V\n",
-      "    #iii:\n",
-      "    IL=Vz/RL;                      #Load current through RL in A\n",
-      "    IL=IL*1000;                    #Load current through RL in mA\n",
-      "    I=Voltage_R/R;                 #Current through the series resistance in A\n",
-      "    I=I*1000;                      #Current through the series resistance in mA\n",
-      "    Iz=I-IL;                       #Applying Kirchhoff's first law, Zener current in mA\n",
-      "    \n",
-      "    #Result\n",
-      "    print 'i)   The output voltage across the load resistance RL = %d V'%Output_voltage;\n",
-      "    print 'ii)  The voltage drop across the series resistance  R = %d V'%Voltage_R;\n",
-      "    print 'iii) The current through the zener diode = %d mA'%Iz;\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)   The output voltage across the load resistance RL = 50 V\n",
-        "ii)  The voltage drop across the series resistance  R = 70 V\n",
-        "iii) The current through the zener diode = 9 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.26, Page number 114-115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Max_V=120.0;                  #Maximum input voltage in V\n",
-      "Min_V=80.0;                   #Minimum input voltage in V\n",
-      "R=5000.0;                     #Series resistance in ohm\n",
-      "RL=10000.0;                   #Load resistance in ohm\n",
-      "Vz=50.0;                      #Zener voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i: Maximum zener current\n",
-      "#Zener current will be maximum when the input voltage is maximum\n",
-      "V_R_max=Max_V-Vz;                #Voltage across series resistance R, in  V\n",
-      "I_max=V_R_max/R;                 #Current through series resistance R, in A\n",
-      "I_max=I_max*1000;                #Current through series resistance R, in mA\n",
-      "IL_max=Vz/RL;                        #Load current in A\n",
-      "IL_max=IL_max*1000;                      #Load current in mA\n",
-      "Iz_max=I_max-IL_max;                     #Applying Kirchhoff's first law, Zener current in mA;\n",
-      "\n",
-      "#Case ii: Minimum zener current\n",
-      "#The zener will conduct minimum current when the input voltage is minimum\n",
-      "V_R_min=Min_V-Vz;                #Voltage across series resistance R, in V\n",
-      "I_min=V_R_min/R;                 #Current through series resistance R, in A\n",
-      "I_min=I_min*1000;                #Current through series resistance R, in mA\n",
-      "IL_min=Vz/RL;                        #Load current in A\n",
-      "IL_min=IL_min*1000;                      #Load current in mA\n",
-      "Iz_min=I_min-IL_min;                     #Applying Kirchhoff's first law, Zener current in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Case i: ';\n",
-      "print 'Maximum zener current = %d mA'%Iz_max;\n",
-      "print 'Case ii: ';\n",
-      "print 'Minimum zener current = %d mA'%Iz_min;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case i: \n",
-        "Maximum zener current = 9 mA\n",
-        "Case ii: \n",
-        "Minimum zener current = 1 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.27, Page number 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=12;                      #Input voltage in V\n",
-      "Vz=7.2;                     #Zener voltage in V\n",
-      "E0=Vz;                      #Voltage to be maintained across the load in  V\n",
-      "IL_max=0.1;                 #Maximum load current in A\n",
-      "IL_min=0.012;                  #Minimum load current in A\n",
-      "Iz_min=0.01;                  #Minimum zener current in A\n",
-      "\n",
-      "#Calculation\n",
-      "#When the load current is maximum at minimum value of RL, the zener current is minimum and, as the load current decreases due to increase in value of RL\n",
-      "R=(Ei-E0)/(Iz_min+IL_max);        #The value of series resistance R to maintain a voltage=E0 across load, in ohm\n",
-      "\n",
-      "#Result\n",
-      "print 'The minimum value of series resistance R to maintain a constant value of 7.2 V is = %.1f \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum value of series resistance R to maintain a constant value of 7.2 V is = 43.6 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The actual value of R is 43.636363 (recurring) but, in the textbook the value of R is wrongly approximated 43.5 \u03a9"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.28, Page number 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei_min=22;                #Minimum input voltage in V\n",
-      "Ei_max=28;                #Maximum input voltage in V\n",
-      "Vz=18;                   #Zener voltage in V\n",
-      "E0=Vz;                    #Constant voltage maintained across the load resistance in V\n",
-      "Iz_min=0.2;               #Minimum zener current in A\n",
-      "Iz_max=2;                 #Maximum zener current in A\n",
-      "RL=18;                    #Load resistance in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "IL=Vz/RL;                      #Constant value of load current in A\n",
-      "#When the input voltage is minimum, the zener current will be minimum\n",
-      "R=(Ei_min-E0)/(Iz_min+IL)      #The value of series resistance so that the voltage E0 across RL remains constant\n",
-      "\n",
-      "print 'The value of series resistance R, to maintain constant voltage E0 across RL = %.2f \u03a9.'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistance R, to maintain constant voltage E0 across RL = 3.33 \u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.29, Page number 116 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10                                #Zener voltage in V\n",
-      "Ei_min=13;                           #Minimum input voltage in V\n",
-      "Ei_max=16;                           #Maximum input voltage in V\n",
-      "Iz_min=0.015;                        #Minimum zener current in A\n",
-      "IL_min=0.01;                         #Minimum load current in A \n",
-      "IL_max=0.085;                        #Maximum load curremt in A\n",
-      "E0=Vz;                               #Constant voltage to be maintained in V \n",
-      "\n",
-      "#Calculation\n",
-      "#The zener current will be minimum when the input voltage will be minimum  and at that time the load current will be maximum\n",
-      "R=(Ei_min-E0)/(Iz_min+IL_max);       #The value of series resistance R to maintain a constant voltage across load\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'The value of series resistance to maintain a constant voltage across the load resistance is = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistance to maintain a constant voltage across the load resistance is = 30 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.30, Page number 116"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Iz=0.2;               #Current rating of each zener in A\n",
-      "Vz=15;                #Voltage rating of each zener in V\n",
-      "Ei=45;                #Input voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "# i: Regulated output voltage across the two zener diodes \n",
-      "E0=2*Vz;                     # V\n",
-      "\n",
-      "# ii: Value of series resistance \n",
-      "R=(Ei-E0)/Iz;                  # \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print 'i) The regulated output voltage = %d V'%E0;\n",
-      "print 'ii) The value of the series resistance = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) The regulated output voltage = 30 V\n",
-        "ii) The value of the series resistance = 75 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.31, Page number 116-117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10;                   #Voltage rating of each zener in V\n",
-      "Iz=1;                    #Current rating of each zener in A\n",
-      "Ei=45;                   #Input unregulated voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "#Regulated output voltage across the three zener diodes\n",
-      "E0=3*Vz;               # V\n",
-      "\n",
-      "#Value of series resistance to obtain a 30V regulated output voltage\n",
-      "R=(Ei-E0)/Iz;          # \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print 'Value of series resistance to obtain a 30V regulated output voltage = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of series resistance to obtain a 30V regulated output voltage = 15 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.32, Page number 117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "RL=2000.0;                   #Load resistance in \u03a9\n",
-      "R=200.0;                     #Series resistance in \u03a9\n",
-      "Iz=0.025;                  #Zener current rating in A\n",
-      "E0=30.0;                     #Output regulated voltage in V \n",
-      "\n",
-      "#Calculation\n",
-      "#Minimum input voltage will be required when Iz=0 A, and at  this condition\n",
-      "IL=E0/RL;                   #Load current during Iz=0, in A\n",
-      "I=IL;                       #According to Kirchhoff's law, total current, in A\n",
-      "Ei_min=E0+(I*R);            #Minimum input voltage in V\n",
-      "\n",
-      "#The maximum input voltage required will be when Iz=0.025 A, and at that condition \n",
-      "I=IL+Iz;                    #According to Kirchhoff's law, total current, in A\n",
-      "Ei_max=E0+(I*R);            #maximum input voltage in V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'The required range of input voltage is from %d V to %d V'%(Ei_min,Ei_max);  \n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required range of input voltage is from 33 V to 38 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.33, Page number 117-118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=16;                   #Unregulated input voltage in V\n",
-      "E0=12;                   #Output regulated voltage in V\n",
-      "IL_min=0;                #Minimum load current in A\n",
-      "IL_max=0.2;              #Maximum load current in A\n",
-      "Iz_min=0;                #Minimum zener current in A\n",
-      "Iz_max=0.2;              #Maximum zener current in A\n",
-      "\n",
-      "#Calculation\n",
-      "#As the regulated voltage required across the load is 12V\n",
-      "Vz=E0;                   #Voltage rating of zener diode in V\n",
-      "V_R=Ei-E0;               #Constant Voltage that should remain across series resistance \n",
-      "#The minimum zener current will occur when the curent in the load in maximum\n",
-      "R=V_R/(Iz_min+IL_max);   #Series resistance in \u03a9\n",
-      "\n",
-      "Max_power_rating=Vz*Iz_max;     #Maximum power rating of zener diode in W\n",
-      "\n",
-      "#Result\n",
-      "print 'The regulator is designed using a Seris resistance of %d \u03a9 and a zener diode of zener voltage %d V'%(R,Vz);\n",
-      "print 'The maximum power rating of the zener diode is = %.1f W '%Max_power_rating;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulator is designed using a Seris resistance of 20 \u03a9 and a zener diode of zener voltage 12 V\n",
-        "The maximum power rating of the zener diode is = 2.4 W \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.34, Page number 118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=12;                       #Source voltage in V\n",
-      "R=1000;                     #Series resistance in \u03a9\n",
-      "RL=5000;                    #Load resistance in \u03a9\n",
-      "Vz=6;                       #Voltage rating of zener in V\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i: zener is working properly\n",
-      "#The output voltage across the load will be equal to the zener voltage.\n",
-      "V0=Vz;                        # V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case i: Output voltage when zener is working properly is %d V'%V0;\n",
-      "\n",
-      "#Case ii: zener is shorted\n",
-      "#As the zener is shorted, the potential difference across the load will be zero\n",
-      "V0=0;                        #V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case ii: Output voltage when zener is short circuited is %d V'%V0;\n",
-      "     \n",
-      "#Case iii: zener is open circuited\n",
-      "#If the zener is open circuited, the total voltage will drop across R and RL according to the voltage divider rule\n",
-      "V0=V*RL/(R+RL);                      #V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case iii: Output voltage when zener is open circuited is %d V'%V0;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case i: Output voltage when zener is working properly is 6 V\n",
-        "Case ii: Output voltage when zener is short circuited is 0 V\n",
-        "Case iii: Output voltage when zener is open circuited is 10 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_3.ipynb
deleted file mode 100755
index a6008ee1..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_3.ipynb
+++ /dev/null
@@ -1,1624 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3452607f2168b562d941493f83083042eaa5a2d316715f9d9f089ff03d73fdb8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 6: SEMICONDUCTOR DIODE"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2, Page number 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration \n",
-      "Vf =20;                #Peak Input Voltage in V\n",
-      "rf=10;                  #Forward Resistance in ohms\n",
-      "RL=500.0;                 #Load Resistance in ohms\n",
-      "V0=0.7;                  #Potential Barrier Voltage of the diodes in V\n",
-      "\n",
-      "#Calculation\n",
-      "#(1)\n",
-      "If_peak=(Vf-V0)/(rf+RL);                  #Peak current through the diode in A\n",
-      "If_peak=If_peak*1000;                    #Peak current through the diode in mA\n",
-      "#(2)\n",
-      "V_out_peak =If_peak * RL/1000 ;               #Peak output voltage in V\n",
-      "\n",
-      "#For an Ideal diode\n",
-      "If_peak_ideal=Vf/RL;                   #Peak current through the ideal diode in A\n",
-      "If_peak_ideal=If_peak_ideal*1000;      #Peak current through the ideal diode in mA\n",
-      "\n",
-      "V_out_peak_ideal=If_peak_ideal * RL/1000;     # Peak output voltage in case of the ideal diode in V\n",
-      "\n",
-      "#Result\n",
-      "print '(i) Peak current through the diode = %.1f mA '%If_peak;\n",
-      "print '(ii) Peak output voltage = %.1f V'%V_out_peak;\n",
-      "print '(iii) Peak current through the ideal diode = %d mA '%If_peak_ideal;\n",
-      "print '(iv) Peak output voltage in case of the ideal diode = %d V'%V_out_peak_ideal;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Peak current through the diode = 37.8 mA \n",
-        "(ii) Peak output voltage = 18.9 V\n",
-        "(iii) Peak current through the ideal diode = 40 mA \n",
-        "(iv) Peak output voltage in case of the ideal diode = 20 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3, Page number 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V =10.0;                #Battery voltage in V\n",
-      "R1=50.0;                  #Resistor 1's resistance in ohms\n",
-      "R2=5.0;                 #Resistor 2's resistance in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "#Using Thevenin's Theorem to find current in the diode\n",
-      "E0=(R2/(R1+R2))*V;          #Thevenin's Voltage in V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's Resistance in ohms\n",
-      "\n",
-      "I0=E0/R0;                   #Current through the diode in A\n",
-      "I0=I0*1000;                 #Current through the diode in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Current through the diode = %d mA '%Io;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through the diode = 200 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4, Page number 82-83 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V =10.0;                #Battery voltage in V\n",
-      "R0=48.0;                  #Resistance of the resistor in ohms\n",
-      "Rd=1.0;                 #Forward resistance of the diodes in ohms\n",
-      "Vd=0.7;                 #Potential barrier of the diodes in V\n",
-      "#Calculation\n",
-      "V_net=V-Vd-Vd;        #Net voltage in the circuit in V\n",
-      "R_net=R0+Rd+Rd        #Net resistance of the circuit in ohms\n",
-      "I_net=V_net/R_net;    #Net current in the circuit in A\n",
-      "I_net=I_net*1000;     #Net current in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Net current in the circuit = %d mA '%I_net;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Net current in the circuit = 172 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page number 83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E1=24;           #Voltage of first source in V\n",
-      "E2=4;            #Voltage of second source in V\n",
-      "V0=0.7;          #Potential barrier of diodes in V\n",
-      "R=2000;          #Resistance of the given resistor in ohms\n",
-      "Rd=0;            #Forward resistance of the diodes in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I=(E1-E2-V0)/(R+Rd);  #Current in the circuit in A\n",
-      "I=I*1000;             #Current in the circuit in mA \n",
-      "\n",
-      "#Result\n",
-      "print 'Current in the circuit = %.2f mA '%I;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in the circuit = 9.65 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page number 83-84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=20;           #Voltage of source in V\n",
-      "V0=0.3;         #Potential barrier of Germanium diode in V\n",
-      "V0_Si=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "\n",
-      "#Calculation\n",
-      "#As only Ge diode is turned on due to less potential barrier,\n",
-      "VA=V-V0;        #Voltage VA acroos resistor of 3k ohms\n",
-      "\n",
-      "#Result\n",
-      "print 'Voltage VA = %.1f mA '%VA;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage VA = 19.7 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7, Page number 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=10;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "# Resistance of all resistors in ohms\n",
-      "R1=2000;\n",
-      "R2=2000;\n",
-      "R3=2000;\n",
-      "\n",
-      "#Calculation\n",
-      "Id=(V-V0)/(R2+2*R3);     #Current through the diodes in A\n",
-      "VQ=2*Id*R3;      #Voltage VQ across the grounded 2k ohm resistor in V\n",
-      "Id=Id*1000;      #Current through the diodes in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Voltage VQ = %.1f V '%VQ;\n",
-      "print 'Current through the diodes, Id = %.2f mA '%Id;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage VQ = 6.2 V \n",
-        "Current through the diodes, Id = 1.55 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8, Page number 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=15;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "R=500       # Resistance of all resistors in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I1=(V-V0)/R;      #total current in the circuit in A\n",
-      "Id1=I1/2;         #current in first diode in A\n",
-      "Id1=Id1*1000;     #current in first diode in mA\n",
-      "Id2=Id1           #current in second diode in mA\n",
-      "\n",
-      "#Result\n",
-      "print ('Current in first diode = %.1f mA'%Id1);\n",
-      "print ('Current in second diode = %.1f mA'%Id2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in first diode = 14.3 mA\n",
-        "Current in second diode = 14.3 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9, Page number 85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E=20;           #Voltage of source in V\n",
-      "V0_d1=0.7;      #Potetial barrier of first Silicon diode in V\n",
-      "V0_d2=0.7;      #Potetial barrier of second Silicon diode in V\n",
-      "R1=5600;       # Resistance of first resistor in ohms\n",
-      "R2=3300;       #  Resistance of second resistor in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I2=V0_d2/R2;                  #Current I2 through resistor R2 in A\n",
-      "I2=round((I2*1000),3);                   #Current I2 through resistor R2 in mA\n",
-      "I1=(E-V0_d1-V0_d2)/R1;        #Current I1 through resistor R1 in  A\n",
-      "I1=round((I1*1000),2);                   #Current I1 through resistor R1 in  mA\n",
-      "I3=I1-I2;                     #Current I3 through diode D2 in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Current I1= %.2f mA'%I1;\n",
-      "print 'Current I1= %.3f mA'%I2;\n",
-      "print 'Current I1= %.3f mA'%I3;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current I1= 3.32 mA\n",
-        "Current I1= 0.212 mA\n",
-        "Current I1= 3.108 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10, Page number 85-86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E=10.0;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V\n",
-      "R1=2000;       # Resistance of first resistor in ohms\n",
-      "R2=8000;       #  Resistance of second resistor in ohms\n",
-      "R3=4000;       #Resistance of third resistor in ohms\n",
-      "R4=6000;       #Resistance of fourth resistor in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "#Assuming the given diode to be reverse bised and calculating voltage across it's terminals\n",
-      "V1=(E/(R1+R2))*R2;      #voltage at the P side of the diode, i.e, voltage across R2 resistor,according to voltage divider rule, in V\n",
-      "V2=(E/(R3+R4))*R4;      #voltage at the N side of the diode, i.e, voltage across R4 resistor,according to voltage divider rule, in V\n",
-      "\n",
-      "#Result\n",
-      "if((V1-V2)>=V0):\n",
-      "    print 'Our assumption was wrong and, the diode is forward biased';\n",
-      "else:\n",
-      "    print 'The diode is reverse biased';\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Our assumption was wrong and, the diode is forward biased\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.11, Page number 86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=2;                              #Supply voltage in V\n",
-      "V0=0.7;                           #Potential barrier voltage of the diode in V \n",
-      "R1=4000.0;                        #Resistance of first resistor in \u03a9\n",
-      "R2=1000.0;                        ##Resistance of second resistor in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Assuming the diode to be in ON state\n",
-      "I1=((V-V0)/R1)*1000;                  #Current through resistor R1, in mA\n",
-      "I2=(V0/R2)*1000;                    #Current through resistor R2, in mA\n",
-      "ID=I1-I2;                           #Diode current, in  mA\n",
-      "\n",
-      "if(ID<0):\n",
-      "    #Since the diode current is negative, the diode must be OFF \n",
-      "    ID=0;                     #True value of diode current, mA\n",
-      "    \n",
-      "#As the diode is in OFF state it can be replaced by an open ciruit equivalent \n",
-      "VD=V*R2/(R1 +R2);                #Voltage across the diode, in V\n",
-      "\n",
-      "#Result\n",
-      "print 'ID =%d mA'%ID;\n",
-      "print 'VD =%.1f V'%VD;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ID =0 mA\n",
-        "VD =0.4 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12, Page number 89-90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "AC_Input_Power=100.0;          #Input AC Power in watts\n",
-      "AC_Output_Power=40.0;          #Output AC Power in watts\n",
-      "Accepted_Power=50.0;           #Power accepted by the half-wave rectifier in watt\n",
-      "\n",
-      "#Calculation\n",
-      "R_eff=(AC_Output_Power/AC_Input_Power)*100;             #Rectification efficiency of the half-wave rectifier\n",
-      "Unused_power=AC_Input_Power-Accepted_Power;       #Power not used by the half_wave rectifier due to open circuited condition of the diode in watt\n",
-      "Power_dissipated=Accepted_Power-AC_Output_Power;  #Power dissipated by the diode watt\n",
-      "\n",
-      "#Result\n",
-      "print 'The rectification efficiency of the half-wave rectifier= %d%% '%R_eff;\n",
-      "\n",
-      "print 'Rest 60%% of the power is the unused power and power dissipated by the diode = %d watts and %d watts' %(Unused_power ,Power_dissipated);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rectification efficiency of the half-wave rectifier= 40% \n",
-        "Rest 60% of the power is the unused power and power dissipated by the diode = 50 watts and 10 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.13, Page number 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "Vrms=230.0;                #AC supply RMS voltage in V\n",
-      "Turns_Ratio=10/1;        #turn ratio of the transformer \n",
-      "\n",
-      "#Calculation\n",
-      "Vpm=sqrt(2)*Vrms;         #Maximum primary voltage in V\n",
-      "Vsm=Vpm/Turns_Ratio;     #Maximum secondary voltage in V\n",
-      "#Case 1\n",
-      "Vdc=Vsm/(round(pi,2));              #Output D.C voltage, which is the average voltage in V\n",
-      "Vdc=round(Vdc,2);\n",
-      "#Case 2\n",
-      "PIV=Vsm;                 #Peak Inverse Voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print 'The output d.c voltage= %.2f V'%Vdc;\n",
-      "print 'The peak inverse voltage= %.2f V'%PIV;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output d.c voltage= 10.36 V\n",
-        "The peak inverse voltage= 32.53 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.14, Page number 90-91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "rf=20.0;           #Internal resistance of the crystal diode in ohms\n",
-      "Vm=50.0;           #Maximum applied voltage in V\n",
-      "RL=800.0;          #Load Resistance in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "# 1\n",
-      "Im=Vm/(rf+RL);         #Maximum current in A\n",
-      "Im=Im*1000;            #Maximum current in \n",
-      "Im=round(Im,0);\n",
-      "Idc=Im/pi;             #Average voltage in mA\n",
-      "Idc=round(Idc,1);\n",
-      "Irms=Im/2;             #RMS value of the current in mA\n",
-      "Irms=round(Irms,1)\n",
-      "\n",
-      "# 2\n",
-      "AC_Input_Power=pow(Irms/1000,2)*(rf+RL);     #Input a.c power in watt\n",
-      "\n",
-      "DC_Output_Power=pow(Idc/1000,2)*RL;          #Output d.c power in watt\n",
-      "\n",
-      "# 3\n",
-      "DC_Output_Voltage=(Idc/1000)*RL;               #Output d.c voltage in V\n",
-      "\n",
-      "# 4\n",
-      "Rectifier_efficiency=(DC_Output_Power/AC_Input_Power)*100;   # Efficiency of rectification of the half-wave rectifier\n",
-      "\n",
-      "#Result\n",
-      "print ' i:';\n",
-      "print '   Im   = %d mA'%Im;\n",
-      "print '   Idc  = %.1f mA'%Idc;\n",
-      "print '   Irms = %.1f mA'%Irms;\n",
-      "print ' ii: ';\n",
-      "print '   a.c input power= %.3f watt'%AC_Input_Power;\n",
-      "print '   d.c output power= %.3f watt'%DC_Output_Power;\n",
-      "print ' iii: ';\n",
-      "print '   d.c output voltage = %.2f volts'%DC_Output_Voltage;\n",
-      "print ' iv: '\n",
-      "print '   Efficiency of rectification = %.1f%%'%Rectifier_efficiency;\n",
-      "\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " i:\n",
-        "   Im   = 61 mA\n",
-        "   Idc  = 19.4 mA\n",
-        "   Irms = 30.5 mA\n",
-        " ii: \n",
-        "   a.c input power= 0.763 watt\n",
-        "   d.c output power= 0.301 watt\n",
-        " iii: \n",
-        "   d.c output voltage = 15.52 volts\n",
-        " iv: \n",
-        "   Efficiency of rectification = 39.5%\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.15, Page number 91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "Vdc=50.0;           #Output d.c voltage in V\n",
-      "rf=25;            #Diode resistance in ohm\n",
-      "RL=800;           #Load resistance in ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vm=(pi*(rf+RL)*Vdc)/RL;   #[ Vdc=Vm*RL/(pi*(rf+RL)) ]Maximum value of a.c voltage required to get a volatge  of Vdc from the half-wave rectifier, in V\n",
-      "Vm=round(Vm,0);  \n",
-      "#Result\n",
-      "print 'The a.c voltage required should have maximum value of = %d V' %Vm;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c voltage required should have maximum value of = 162 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.16, Page number 95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "rf=20;              #Internal resistance of the diodes in ohm\n",
-      "Vrms=50;            #RMS value of transformer's secondary voltage from centre tap to each end of secondary\n",
-      "RL=980;             #Load resistance in ohm\n",
-      "\n",
-      "#Calculation\n",
-      "Vm=Vrms*sqrt(2);      #Maximum a.c voltage in V\n",
-      "Im=Vm/(rf+RL);        #Maximum load current in A\n",
-      "Im=Im*1000;           #Maximum load current in mA\n",
-      " \n",
-      "# 1:\n",
-      "Idc=2*Im/pi;          #Mean load current\n",
-      "\n",
-      "# 2:\n",
-      "Irms=Im/sqrt(2);       #RMS value of load current in A\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print'   The mean load current= %d mA'%Idc;\n",
-      "print 'ii:';\n",
-      "print '  The r.m.s value of the load current = %d mA'%Irms; "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The mean load current= 45 mA\n",
-        "ii:\n",
-        "  The r.m.s value of the load current = 50 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.17, Page number 95-96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "#Variable declaration\n",
-      "RL=100;                          #Load resistance in ohm       \n",
-      "rf=0;                            #Internal resistance of the diodes in ohm\n",
-      "Turns_ratio=5/1;                 #Primary to secondary turns ratio of transformer \n",
-      "P_Vrms=230;                      #R.M.S value of voltage in primary winding in V\n",
-      "S_Vrms=P_Vrms/Turns_ratio;       #R.M.S value of voltage in secondary winding in V\n",
-      "S_Vm=S_Vrms*sqrt(2);        #Maximum voltage across secondary winding in V\n",
-      "Vm=S_Vm/2;                  #Maximum voltage across half seconfdary winding in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Idc=2*Vm/(pi*RL);            #Average current in A\n",
-      "Vdc=Idc*RL;                  #d.c output voltage in V\n",
-      "\n",
-      "# 2:\n",
-      "PIV=S_Vm;                    #Peak Invers Voltage(= Maximum secondary voltage) in V\n",
-      "\n",
-      "# 3:\n",
-      "Pac=pow(Vm/(RL*sqrt(2)),2)*(rf+RL);         #a.c input power in watt\n",
-      "Pdc=(pow(Idc,2)*RL);                        #d.c output power in watt\n",
-      "R_eff=(Pdc/Pac)*100;      #Rectification efficiency\n",
-      "R_eff=round(R_eff,1);\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   The d.c output voltage= %.1f V'%Vdc;\n",
-      "print 'ii:';\n",
-      "print '   The peak inverse voltage= %d V'%PIV;\n",
-      "print 'iii:';\n",
-      "print '   Rectification efficiency= %.1f%%'%R_eff;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The d.c output voltage= 20.7 V\n",
-        "ii:\n",
-        "   The peak inverse voltage= 65 V\n",
-        "iii:\n",
-        "   Rectification efficiency= 81.1%\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The value of rectification efficiency is calculated as 81.2% in the textbook using the formula 0.812/(1 + (rf/RL)), but by calculating using the correct values in the formula we get 81.1%."
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.18, Page number 96-97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "#Variable declaration\n",
-      "fin=50;                       #frequency of input ac source in Hz\n",
-      "RL=200;                       #Load resistance in ohm\n",
-      "Turns_ratio=4/1;              #Transformers turns ratio, primary to secondary.\n",
-      "P_Vrms=230.0;                 #R.M.S value of voltage in primary winding in V\n",
-      "S_Vrms=P_Vrms/Turns_ratio     #R.M.S value of voltage in secondary winding in V\n",
-      "Vm=S_Vrms*sqrt(2);            #Maximum voltage across secondary winding in V\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Idc=2*Vm/(pi*RL);           # Average current in A\n",
-      "Vdc=Idc*RL;                 #Output d.c voltage in V\n",
-      "Vdc=round(Vdc,0);\n",
-      "# 2:\n",
-      "PIV= Vm;                    #Peak Inverse Voltage(= Maximum volutage across secondary winding) in V\n",
-      "\n",
-      "# 3:\n",
-      "fout=2*fin;                 #Output frequency in Hz\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   The d.c output voltage = %d V' %Vdc;\n",
-      "print 'ii:';\n",
-      "print '   The peak inverse voltage = %.1f V'%PIV;\n",
-      "print 'iii:';\n",
-      "print '   The output frequency = %d Hz'%fout;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The d.c output voltage = 52 V\n",
-        "ii:\n",
-        "   The peak inverse voltage = 81.3 V\n",
-        "iii:\n",
-        "   The output frequency = 100 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.19, Page number 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=100.0;                        #Load Resistance in ohm\n",
-      "Turns_ratio=5/1;                 #Primary to secondary turns ratio of the transformer\n",
-      "Vin=230.0;                       #R.M.S value of input voltage in V\n",
-      "fin=50;                          #Input frequency in Hz\n",
-      "\n",
-      "#Calculation\n",
-      "Vs_rms=Vin/Turns_ratio;          #R.M.S value of the voltage in secondary winding, in v\n",
-      "Vs_max=Vs_rms*sqrt(2);           #Maximum voltage across secondary, in V\n",
-      "\n",
-      "# (i)\n",
-      "#Case i: Centre-tap circuit\n",
-      "Vm=Vs_max/2;                     #Maximum voltage across half secondary winding, in V \n",
-      "Vdc=2*Vm*RL/(pi*RL);             #DC output voltage, in V \n",
-      "print 'The d.c output voltage for the centre-tap circuit = %.1f V'%Vdc;\n",
-      "\n",
-      "#Case ii:\n",
-      "Vm=Vs_max;                      #Maximum voltage across secondary, in V\n",
-      "Vdc=2*Vm*RL/(pi*RL);            #DC output voltage, in V \n",
-      "print 'The d.c output voltage for the bridge circuit = %.1f V'%Vdc; \n",
-      "\n",
-      "# ii:\n",
-      "#Case i: Centre-tap circuit\n",
-      "Turns_ratio=5/1;                      #Turns ratio of the transformer\n",
-      "Vs_rms=Vin/Turns_ratio;               #R.M.S value of the secondary voltage in V\n",
-      "Vs_max=Vs_rms*sqrt(2);                #Maximum voltage across the secondary in V\n",
-      "Vm=Vs_max/2;                          #Maximum voltage across half of the secondary in V\n",
-      "PIV=2*Vm;                             #Peak Inverse Voltage in V\n",
-      "print 'PIV in case of centre-tap circuit = %d V'%PIV;\n",
-      "\n",
-      "#Case ii: Bridge circuit\n",
-      "Turns_ratio=10/1;                             #Turns ratio of the transformer\n",
-      "Vs_rms=Vin/Turns_ratio;                       #R.M.S value of the secondary voltage in V\n",
-      "Vs_max=Vs_rms*sqrt(2);                        #Maximum voltage across the secondary in V\n",
-      "PIV=Vm;                                        #Peak Inverse Voltage in V\n",
-      "print 'PIV in case of bridge circuit = %.1f V'%PIV;\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The d.c output voltage for the centre-tap circuit = 20.7 V\n",
-        "The d.c output voltage for the bridge circuit = 41.4 V\n",
-        "PIV in case of centre-tap circuit = 65 V\n",
-        "PIV in case of bridge circuit = 32.5 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.20, Page number 98"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "rf=1;                    #forward resistance of diodes of the rectifier in ohm\n",
-      "RL=480;                  #Load resistance in ohm\n",
-      "Vrms=240.0;                #a.c supply voltage in V\n",
-      "Vm=Vrms*sqrt(2);         #Maximum a.c voltage in V \n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Rt=2*rf+RL;             #Total circuit resistance at any instance in ohm\n",
-      "Im=Vm/Rt;               #Maximum load current in A\n",
-      "Idc=2*Im/pi;            #Mean load current in A\n",
-      "\n",
-      "# 2:\n",
-      "Irms=Im/2;              #R.M.S value of current  in A\n",
-      "P=pow(Irms,2)*rf;       #Power dissipated in each diode in watt\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   Mean load current = %.2f A'%Idc;\n",
-      "print 'ii:';\n",
-      "print '   Power dissipated in each diode= %.3f W'%P;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   Mean load current = 0.45 A\n",
-        "ii:\n",
-        "   Power dissipated in each diode= 0.124 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The value of power dissipated is approximately 0.124 W , but in the textbook it is approximated as 0.123W."
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.21, Page number 98-99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt,pi\n",
-      "#Variable declaration\n",
-      "RL=12000;                #Load resistance in ohm\n",
-      "V0=0.7;                   #Potential barrier voltage of diodes in V\n",
-      "Vrms=12;                  #R.M.S value of input a.c voltage in V\n",
-      "Vs_pk=Vrms*sqrt(2);       #Peak secondary voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Vout_pk=Vs_pk-(2*V0);      #Peak output voltage in  V\n",
-      "Vav=2*Vout_pk/pi;          #Average output voltage in V\n",
-      "Vav=round(Vav,2);\n",
-      "\n",
-      "# 2:\n",
-      "Iav=Vav/RL;                #Average output current in A\n",
-      "Iav=Iav*pow(10,6);         #Average output current in \u03bcA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   Average output voltage=%.2f V'%Vav;\n",
-      "print 'ii:';\n",
-      "print '   Average output current=%.1f \u03bcA'%Iav;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   Average output voltage=9.91 V\n",
-        "ii:\n",
-        "   Average output current=825.8 \u03bcA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.22, Page number 102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vdc_A=10;                      #Supply voltage of A in V\n",
-      "Vdc_B=25;                      #Supply voltage of B in V\n",
-      "Vac_rms_a=0.5;                 #Ripples in power supply A in V\n",
-      "Vac_rms_b=0.001;               #Ripples in power supply B in V\n",
-      "\n",
-      "#Calculation\n",
-      "#For power supply A\n",
-      "ripple_factor_A=Vac_rms_a/Vdc_A;      #Ripple factor of power supply A\n",
-      "\n",
-      "#For power supply B\n",
-      "ripple_factor_B=Vac_rms_b/Vdc_B;      #Ripple factor of power supply B\n",
-      "\n",
-      "#Result\n",
-      "if(ripple_factor_A<ripple_factor_B):\n",
-      "    print 'Power supply A is better';\n",
-      "else :\n",
-      "    print 'Power supply B is better';"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supply B is better\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.23, Page number 105-106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "RL=2200;                    #Load resistance in ohm\n",
-      "C=50*pow(10,-6);            #Capacitance of the capacitor used in filter circuit in F\n",
-      "V0=0.7;                     #Potential barrier voltage of the diodes of the rectifier in  V\n",
-      "Vrms=115.0;                   #R.M.S value of input a.c voltage in V \n",
-      "fin=60;                     #Frequency of input a.c voltage in Hz\n",
-      "Turns_ratio=10/1;           #Primary to secondary, turns ratio of the transformer \n",
-      "\n",
-      "#Calculation\n",
-      "Vp_prim=Vrms*sqrt(2);               #Peak primary voltage in V\n",
-      "Vp_sec=Vp_prim/Turns_ratio;         #Peak secondary voltage in V\n",
-      "Vp_in= Vp_sec - 2*V0;               #Peak full wave rectified voltage at the filter input in V\n",
-      "f=2*fin;                            #Output frequency in Hz\n",
-      "Vdc=Vp_in*(1-(1/(2*f*RL*C)));       #Output d.c voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print 'The output d.c voltage  is = %.1f V'%Vdc;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output d.c voltage  is = 14.3 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.24, Page number 106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "R=25;                        #d.c resistance of the choke in ohm\n",
-      "RL=750;                      #Load resistance in ohm\n",
-      "Vm=25.7;                     #Maximum value of the pulsating output from the rectifier in V\n",
-      "\n",
-      "#Calculation\n",
-      "V_dc=2*Vm/pi;                #d.c component of the pulsating output in V\n",
-      "V_dc=round(V_dc,1);\n",
-      "V_dc_out=(V_dc*RL)/(R+RL);   #Output d.c voltage in V\n",
-      "V_dc_out=round(V_dc_out,1);\n",
-      "\n",
-      "#Result\n",
-      "print ' The output d.c voltage accross the load resistance is = %.1f V'%V_dc_out;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The output d.c voltage accross the load resistance is = 15.9 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.25, Page number 113-114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=120.0;                   #Input Voltage in V\n",
-      "Vz=50.0;                    #Zener Voltage in V\n",
-      "R=5000.0;                   #Resistance of the series resistor in ohm\n",
-      "RL=10000.0;                 #Load resistance in ohm\n",
-      "\n",
-      "#Calculation\n",
-      "V=Ei*RL/(R+RL);             #Voltage across the open circuit if the zener diode is removed\n",
-      "if(V>Vz):\n",
-      "    #Zener diode is in ON state\n",
-      "    # i:\n",
-      "    Output_voltage=Vz;             #Voltage across load resistance, in V\n",
-      "    #ii:\n",
-      "    Voltage_R=Ei-Vz;               #Voltage across the series resistance R, in V\n",
-      "    #iii:\n",
-      "    IL=Vz/RL;                      #Load current through RL in A\n",
-      "    IL=IL*1000;                    #Load current through RL in mA\n",
-      "    I=Voltage_R/R;                 #Current through the series resistance in A\n",
-      "    I=I*1000;                      #Current through the series resistance in mA\n",
-      "    Iz=I-IL;                       #Applying Kirchhoff's first law, Zener current in mA\n",
-      "    \n",
-      "    #Result\n",
-      "    print 'i)   The output voltage across the load resistance RL = %d V'%Output_voltage;\n",
-      "    print 'ii)  The voltage drop across the series resistance  R = %d V'%Voltage_R;\n",
-      "    print 'iii) The current through the zener diode = %d mA'%Iz;\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)   The output voltage across the load resistance RL = 50 V\n",
-        "ii)  The voltage drop across the series resistance  R = 70 V\n",
-        "iii) The current through the zener diode = 9 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.26, Page number 114-115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Max_V=120.0;                  #Maximum input voltage in V\n",
-      "Min_V=80.0;                   #Minimum input voltage in V\n",
-      "R=5000.0;                     #Series resistance in ohm\n",
-      "RL=10000.0;                   #Load resistance in ohm\n",
-      "Vz=50.0;                      #Zener voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i: Maximum zener current\n",
-      "#Zener current will be maximum when the input voltage is maximum\n",
-      "V_R_max=Max_V-Vz;                #Voltage across series resistance R, in  V\n",
-      "I_max=V_R_max/R;                 #Current through series resistance R, in A\n",
-      "I_max=I_max*1000;                #Current through series resistance R, in mA\n",
-      "IL_max=Vz/RL;                        #Load current in A\n",
-      "IL_max=IL_max*1000;                      #Load current in mA\n",
-      "Iz_max=I_max-IL_max;                     #Applying Kirchhoff's first law, Zener current in mA;\n",
-      "\n",
-      "#Case ii: Minimum zener current\n",
-      "#The zener will conduct minimum current when the input voltage is minimum\n",
-      "V_R_min=Min_V-Vz;                #Voltage across series resistance R, in V\n",
-      "I_min=V_R_min/R;                 #Current through series resistance R, in A\n",
-      "I_min=I_min*1000;                #Current through series resistance R, in mA\n",
-      "IL_min=Vz/RL;                        #Load current in A\n",
-      "IL_min=IL_min*1000;                      #Load current in mA\n",
-      "Iz_min=I_min-IL_min;                     #Applying Kirchhoff's first law, Zener current in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Case i: ';\n",
-      "print 'Maximum zener current = %d mA'%Iz_max;\n",
-      "print 'Case ii: ';\n",
-      "print 'Minimum zener current = %d mA'%Iz_min;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case i: \n",
-        "Maximum zener current = 9 mA\n",
-        "Case ii: \n",
-        "Minimum zener current = 1 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.27, Page number 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=12;                      #Input voltage in V\n",
-      "Vz=7.2;                     #Zener voltage in V\n",
-      "E0=Vz;                      #Voltage to be maintained across the load in  V\n",
-      "IL_max=0.1;                 #Maximum load current in A\n",
-      "IL_min=0.012;                  #Minimum load current in A\n",
-      "Iz_min=0.01;                  #Minimum zener current in A\n",
-      "\n",
-      "#Calculation\n",
-      "#When the load current is maximum at minimum value of RL, the zener current is minimum and, as the load current decreases due to increase in value of RL\n",
-      "R=(Ei-E0)/(Iz_min+IL_max);        #The value of series resistance R to maintain a voltage=E0 across load, in ohm\n",
-      "\n",
-      "#Result\n",
-      "print 'The minimum value of series resistance R to maintain a constant value of 7.2 V is = %.1f \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum value of series resistance R to maintain a constant value of 7.2 V is = 43.6 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The actual value of R is 43.636363 (recurring) but, in the textbook the value of R is wrongly approximated 43.5 \u03a9"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.28, Page number 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei_min=22;                #Minimum input voltage in V\n",
-      "Ei_max=28;                #Maximum input voltage in V\n",
-      "Vz=18;                   #Zener voltage in V\n",
-      "E0=Vz;                    #Constant voltage maintained across the load resistance in V\n",
-      "Iz_min=0.2;               #Minimum zener current in A\n",
-      "Iz_max=2;                 #Maximum zener current in A\n",
-      "RL=18;                    #Load resistance in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "IL=Vz/RL;                      #Constant value of load current in A\n",
-      "#When the input voltage is minimum, the zener current will be minimum\n",
-      "R=(Ei_min-E0)/(Iz_min+IL)      #The value of series resistance so that the voltage E0 across RL remains constant\n",
-      "\n",
-      "print 'The value of series resistance R, to maintain constant voltage E0 across RL = %.2f \u03a9.'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistance R, to maintain constant voltage E0 across RL = 3.33 \u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.29, Page number 116 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10                                #Zener voltage in V\n",
-      "Ei_min=13;                           #Minimum input voltage in V\n",
-      "Ei_max=16;                           #Maximum input voltage in V\n",
-      "Iz_min=0.015;                        #Minimum zener current in A\n",
-      "IL_min=0.01;                         #Minimum load current in A \n",
-      "IL_max=0.085;                        #Maximum load curremt in A\n",
-      "E0=Vz;                               #Constant voltage to be maintained in V \n",
-      "\n",
-      "#Calculation\n",
-      "#The zener current will be minimum when the input voltage will be minimum  and at that time the load current will be maximum\n",
-      "R=(Ei_min-E0)/(Iz_min+IL_max);       #The value of series resistance R to maintain a constant voltage across load\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'The value of series resistance to maintain a constant voltage across the load resistance is = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistance to maintain a constant voltage across the load resistance is = 30 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.30, Page number 116"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Iz=0.2;               #Current rating of each zener in A\n",
-      "Vz=15;                #Voltage rating of each zener in V\n",
-      "Ei=45;                #Input voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "# i: Regulated output voltage across the two zener diodes \n",
-      "E0=2*Vz;                     # V\n",
-      "\n",
-      "# ii: Value of series resistance \n",
-      "R=(Ei-E0)/Iz;                  # \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print 'i) The regulated output voltage = %d V'%E0;\n",
-      "print 'ii) The value of the series resistance = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) The regulated output voltage = 30 V\n",
-        "ii) The value of the series resistance = 75 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.31, Page number 116-117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10;                   #Voltage rating of each zener in V\n",
-      "Iz=1;                    #Current rating of each zener in A\n",
-      "Ei=45;                   #Input unregulated voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "#Regulated output voltage across the three zener diodes\n",
-      "E0=3*Vz;               # V\n",
-      "\n",
-      "#Value of series resistance to obtain a 30V regulated output voltage\n",
-      "R=(Ei-E0)/Iz;          # \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print 'Value of series resistance to obtain a 30V regulated output voltage = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of series resistance to obtain a 30V regulated output voltage = 15 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.32, Page number 117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "RL=2000.0;                   #Load resistance in \u03a9\n",
-      "R=200.0;                     #Series resistance in \u03a9\n",
-      "Iz=0.025;                  #Zener current rating in A\n",
-      "E0=30.0;                     #Output regulated voltage in V \n",
-      "\n",
-      "#Calculation\n",
-      "#Minimum input voltage will be required when Iz=0 A, and at  this condition\n",
-      "IL=E0/RL;                   #Load current during Iz=0, in A\n",
-      "I=IL;                       #According to Kirchhoff's law, total current, in A\n",
-      "Ei_min=E0+(I*R);            #Minimum input voltage in V\n",
-      "\n",
-      "#The maximum input voltage required will be when Iz=0.025 A, and at that condition \n",
-      "I=IL+Iz;                    #According to Kirchhoff's law, total current, in A\n",
-      "Ei_max=E0+(I*R);            #maximum input voltage in V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'The required range of input voltage is from %d V to %d V'%(Ei_min,Ei_max);  \n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required range of input voltage is from 33 V to 38 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.33, Page number 117-118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=16;                   #Unregulated input voltage in V\n",
-      "E0=12;                   #Output regulated voltage in V\n",
-      "IL_min=0;                #Minimum load current in A\n",
-      "IL_max=0.2;              #Maximum load current in A\n",
-      "Iz_min=0;                #Minimum zener current in A\n",
-      "Iz_max=0.2;              #Maximum zener current in A\n",
-      "\n",
-      "#Calculation\n",
-      "#As the regulated voltage required across the load is 12V\n",
-      "Vz=E0;                   #Voltage rating of zener diode in V\n",
-      "V_R=Ei-E0;               #Constant Voltage that should remain across series resistance \n",
-      "#The minimum zener current will occur when the curent in the load in maximum\n",
-      "R=V_R/(Iz_min+IL_max);   #Series resistance in \u03a9\n",
-      "\n",
-      "Max_power_rating=Vz*Iz_max;     #Maximum power rating of zener diode in W\n",
-      "\n",
-      "#Result\n",
-      "print 'The regulator is designed using a Seris resistance of %d \u03a9 and a zener diode of zener voltage %d V'%(R,Vz);\n",
-      "print 'The maximum power rating of the zener diode is = %.1f W '%Max_power_rating;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulator is designed using a Seris resistance of 20 \u03a9 and a zener diode of zener voltage 12 V\n",
-        "The maximum power rating of the zener diode is = 2.4 W \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.34, Page number 118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=12;                       #Source voltage in V\n",
-      "R=1000;                     #Series resistance in \u03a9\n",
-      "RL=5000;                    #Load resistance in \u03a9\n",
-      "Vz=6;                       #Voltage rating of zener in V\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i: zener is working properly\n",
-      "#The output voltage across the load will be equal to the zener voltage.\n",
-      "V0=Vz;                        # V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case i: Output voltage when zener is working properly is %d V'%V0;\n",
-      "\n",
-      "#Case ii: zener is shorted\n",
-      "#As the zener is shorted, the potential difference across the load will be zero\n",
-      "V0=0;                        #V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case ii: Output voltage when zener is short circuited is %d V'%V0;\n",
-      "     \n",
-      "#Case iii: zener is open circuited\n",
-      "#If the zener is open circuited, the total voltage will drop across R and RL according to the voltage divider rule\n",
-      "V0=V*RL/(R+RL);                      #V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case iii: Output voltage when zener is open circuited is %d V'%V0;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case i: Output voltage when zener is working properly is 6 V\n",
-        "Case ii: Output voltage when zener is short circuited is 0 V\n",
-        "Case iii: Output voltage when zener is open circuited is 10 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_4.ipynb
deleted file mode 100755
index a6008ee1..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_4.ipynb
+++ /dev/null
@@ -1,1624 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3452607f2168b562d941493f83083042eaa5a2d316715f9d9f089ff03d73fdb8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 6: SEMICONDUCTOR DIODE"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2, Page number 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration \n",
-      "Vf =20;                #Peak Input Voltage in V\n",
-      "rf=10;                  #Forward Resistance in ohms\n",
-      "RL=500.0;                 #Load Resistance in ohms\n",
-      "V0=0.7;                  #Potential Barrier Voltage of the diodes in V\n",
-      "\n",
-      "#Calculation\n",
-      "#(1)\n",
-      "If_peak=(Vf-V0)/(rf+RL);                  #Peak current through the diode in A\n",
-      "If_peak=If_peak*1000;                    #Peak current through the diode in mA\n",
-      "#(2)\n",
-      "V_out_peak =If_peak * RL/1000 ;               #Peak output voltage in V\n",
-      "\n",
-      "#For an Ideal diode\n",
-      "If_peak_ideal=Vf/RL;                   #Peak current through the ideal diode in A\n",
-      "If_peak_ideal=If_peak_ideal*1000;      #Peak current through the ideal diode in mA\n",
-      "\n",
-      "V_out_peak_ideal=If_peak_ideal * RL/1000;     # Peak output voltage in case of the ideal diode in V\n",
-      "\n",
-      "#Result\n",
-      "print '(i) Peak current through the diode = %.1f mA '%If_peak;\n",
-      "print '(ii) Peak output voltage = %.1f V'%V_out_peak;\n",
-      "print '(iii) Peak current through the ideal diode = %d mA '%If_peak_ideal;\n",
-      "print '(iv) Peak output voltage in case of the ideal diode = %d V'%V_out_peak_ideal;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Peak current through the diode = 37.8 mA \n",
-        "(ii) Peak output voltage = 18.9 V\n",
-        "(iii) Peak current through the ideal diode = 40 mA \n",
-        "(iv) Peak output voltage in case of the ideal diode = 20 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3, Page number 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V =10.0;                #Battery voltage in V\n",
-      "R1=50.0;                  #Resistor 1's resistance in ohms\n",
-      "R2=5.0;                 #Resistor 2's resistance in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "#Using Thevenin's Theorem to find current in the diode\n",
-      "E0=(R2/(R1+R2))*V;          #Thevenin's Voltage in V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's Resistance in ohms\n",
-      "\n",
-      "I0=E0/R0;                   #Current through the diode in A\n",
-      "I0=I0*1000;                 #Current through the diode in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Current through the diode = %d mA '%Io;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through the diode = 200 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4, Page number 82-83 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V =10.0;                #Battery voltage in V\n",
-      "R0=48.0;                  #Resistance of the resistor in ohms\n",
-      "Rd=1.0;                 #Forward resistance of the diodes in ohms\n",
-      "Vd=0.7;                 #Potential barrier of the diodes in V\n",
-      "#Calculation\n",
-      "V_net=V-Vd-Vd;        #Net voltage in the circuit in V\n",
-      "R_net=R0+Rd+Rd        #Net resistance of the circuit in ohms\n",
-      "I_net=V_net/R_net;    #Net current in the circuit in A\n",
-      "I_net=I_net*1000;     #Net current in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Net current in the circuit = %d mA '%I_net;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Net current in the circuit = 172 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page number 83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E1=24;           #Voltage of first source in V\n",
-      "E2=4;            #Voltage of second source in V\n",
-      "V0=0.7;          #Potential barrier of diodes in V\n",
-      "R=2000;          #Resistance of the given resistor in ohms\n",
-      "Rd=0;            #Forward resistance of the diodes in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I=(E1-E2-V0)/(R+Rd);  #Current in the circuit in A\n",
-      "I=I*1000;             #Current in the circuit in mA \n",
-      "\n",
-      "#Result\n",
-      "print 'Current in the circuit = %.2f mA '%I;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in the circuit = 9.65 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page number 83-84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=20;           #Voltage of source in V\n",
-      "V0=0.3;         #Potential barrier of Germanium diode in V\n",
-      "V0_Si=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "\n",
-      "#Calculation\n",
-      "#As only Ge diode is turned on due to less potential barrier,\n",
-      "VA=V-V0;        #Voltage VA acroos resistor of 3k ohms\n",
-      "\n",
-      "#Result\n",
-      "print 'Voltage VA = %.1f mA '%VA;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage VA = 19.7 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7, Page number 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=10;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "# Resistance of all resistors in ohms\n",
-      "R1=2000;\n",
-      "R2=2000;\n",
-      "R3=2000;\n",
-      "\n",
-      "#Calculation\n",
-      "Id=(V-V0)/(R2+2*R3);     #Current through the diodes in A\n",
-      "VQ=2*Id*R3;      #Voltage VQ across the grounded 2k ohm resistor in V\n",
-      "Id=Id*1000;      #Current through the diodes in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Voltage VQ = %.1f V '%VQ;\n",
-      "print 'Current through the diodes, Id = %.2f mA '%Id;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage VQ = 6.2 V \n",
-        "Current through the diodes, Id = 1.55 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8, Page number 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=15;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "R=500       # Resistance of all resistors in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I1=(V-V0)/R;      #total current in the circuit in A\n",
-      "Id1=I1/2;         #current in first diode in A\n",
-      "Id1=Id1*1000;     #current in first diode in mA\n",
-      "Id2=Id1           #current in second diode in mA\n",
-      "\n",
-      "#Result\n",
-      "print ('Current in first diode = %.1f mA'%Id1);\n",
-      "print ('Current in second diode = %.1f mA'%Id2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in first diode = 14.3 mA\n",
-        "Current in second diode = 14.3 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9, Page number 85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E=20;           #Voltage of source in V\n",
-      "V0_d1=0.7;      #Potetial barrier of first Silicon diode in V\n",
-      "V0_d2=0.7;      #Potetial barrier of second Silicon diode in V\n",
-      "R1=5600;       # Resistance of first resistor in ohms\n",
-      "R2=3300;       #  Resistance of second resistor in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I2=V0_d2/R2;                  #Current I2 through resistor R2 in A\n",
-      "I2=round((I2*1000),3);                   #Current I2 through resistor R2 in mA\n",
-      "I1=(E-V0_d1-V0_d2)/R1;        #Current I1 through resistor R1 in  A\n",
-      "I1=round((I1*1000),2);                   #Current I1 through resistor R1 in  mA\n",
-      "I3=I1-I2;                     #Current I3 through diode D2 in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Current I1= %.2f mA'%I1;\n",
-      "print 'Current I1= %.3f mA'%I2;\n",
-      "print 'Current I1= %.3f mA'%I3;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current I1= 3.32 mA\n",
-        "Current I1= 0.212 mA\n",
-        "Current I1= 3.108 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10, Page number 85-86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E=10.0;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V\n",
-      "R1=2000;       # Resistance of first resistor in ohms\n",
-      "R2=8000;       #  Resistance of second resistor in ohms\n",
-      "R3=4000;       #Resistance of third resistor in ohms\n",
-      "R4=6000;       #Resistance of fourth resistor in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "#Assuming the given diode to be reverse bised and calculating voltage across it's terminals\n",
-      "V1=(E/(R1+R2))*R2;      #voltage at the P side of the diode, i.e, voltage across R2 resistor,according to voltage divider rule, in V\n",
-      "V2=(E/(R3+R4))*R4;      #voltage at the N side of the diode, i.e, voltage across R4 resistor,according to voltage divider rule, in V\n",
-      "\n",
-      "#Result\n",
-      "if((V1-V2)>=V0):\n",
-      "    print 'Our assumption was wrong and, the diode is forward biased';\n",
-      "else:\n",
-      "    print 'The diode is reverse biased';\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Our assumption was wrong and, the diode is forward biased\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.11, Page number 86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=2;                              #Supply voltage in V\n",
-      "V0=0.7;                           #Potential barrier voltage of the diode in V \n",
-      "R1=4000.0;                        #Resistance of first resistor in \u03a9\n",
-      "R2=1000.0;                        ##Resistance of second resistor in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Assuming the diode to be in ON state\n",
-      "I1=((V-V0)/R1)*1000;                  #Current through resistor R1, in mA\n",
-      "I2=(V0/R2)*1000;                    #Current through resistor R2, in mA\n",
-      "ID=I1-I2;                           #Diode current, in  mA\n",
-      "\n",
-      "if(ID<0):\n",
-      "    #Since the diode current is negative, the diode must be OFF \n",
-      "    ID=0;                     #True value of diode current, mA\n",
-      "    \n",
-      "#As the diode is in OFF state it can be replaced by an open ciruit equivalent \n",
-      "VD=V*R2/(R1 +R2);                #Voltage across the diode, in V\n",
-      "\n",
-      "#Result\n",
-      "print 'ID =%d mA'%ID;\n",
-      "print 'VD =%.1f V'%VD;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ID =0 mA\n",
-        "VD =0.4 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12, Page number 89-90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "AC_Input_Power=100.0;          #Input AC Power in watts\n",
-      "AC_Output_Power=40.0;          #Output AC Power in watts\n",
-      "Accepted_Power=50.0;           #Power accepted by the half-wave rectifier in watt\n",
-      "\n",
-      "#Calculation\n",
-      "R_eff=(AC_Output_Power/AC_Input_Power)*100;             #Rectification efficiency of the half-wave rectifier\n",
-      "Unused_power=AC_Input_Power-Accepted_Power;       #Power not used by the half_wave rectifier due to open circuited condition of the diode in watt\n",
-      "Power_dissipated=Accepted_Power-AC_Output_Power;  #Power dissipated by the diode watt\n",
-      "\n",
-      "#Result\n",
-      "print 'The rectification efficiency of the half-wave rectifier= %d%% '%R_eff;\n",
-      "\n",
-      "print 'Rest 60%% of the power is the unused power and power dissipated by the diode = %d watts and %d watts' %(Unused_power ,Power_dissipated);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rectification efficiency of the half-wave rectifier= 40% \n",
-        "Rest 60% of the power is the unused power and power dissipated by the diode = 50 watts and 10 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.13, Page number 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "Vrms=230.0;                #AC supply RMS voltage in V\n",
-      "Turns_Ratio=10/1;        #turn ratio of the transformer \n",
-      "\n",
-      "#Calculation\n",
-      "Vpm=sqrt(2)*Vrms;         #Maximum primary voltage in V\n",
-      "Vsm=Vpm/Turns_Ratio;     #Maximum secondary voltage in V\n",
-      "#Case 1\n",
-      "Vdc=Vsm/(round(pi,2));              #Output D.C voltage, which is the average voltage in V\n",
-      "Vdc=round(Vdc,2);\n",
-      "#Case 2\n",
-      "PIV=Vsm;                 #Peak Inverse Voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print 'The output d.c voltage= %.2f V'%Vdc;\n",
-      "print 'The peak inverse voltage= %.2f V'%PIV;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output d.c voltage= 10.36 V\n",
-        "The peak inverse voltage= 32.53 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.14, Page number 90-91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "rf=20.0;           #Internal resistance of the crystal diode in ohms\n",
-      "Vm=50.0;           #Maximum applied voltage in V\n",
-      "RL=800.0;          #Load Resistance in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "# 1\n",
-      "Im=Vm/(rf+RL);         #Maximum current in A\n",
-      "Im=Im*1000;            #Maximum current in \n",
-      "Im=round(Im,0);\n",
-      "Idc=Im/pi;             #Average voltage in mA\n",
-      "Idc=round(Idc,1);\n",
-      "Irms=Im/2;             #RMS value of the current in mA\n",
-      "Irms=round(Irms,1)\n",
-      "\n",
-      "# 2\n",
-      "AC_Input_Power=pow(Irms/1000,2)*(rf+RL);     #Input a.c power in watt\n",
-      "\n",
-      "DC_Output_Power=pow(Idc/1000,2)*RL;          #Output d.c power in watt\n",
-      "\n",
-      "# 3\n",
-      "DC_Output_Voltage=(Idc/1000)*RL;               #Output d.c voltage in V\n",
-      "\n",
-      "# 4\n",
-      "Rectifier_efficiency=(DC_Output_Power/AC_Input_Power)*100;   # Efficiency of rectification of the half-wave rectifier\n",
-      "\n",
-      "#Result\n",
-      "print ' i:';\n",
-      "print '   Im   = %d mA'%Im;\n",
-      "print '   Idc  = %.1f mA'%Idc;\n",
-      "print '   Irms = %.1f mA'%Irms;\n",
-      "print ' ii: ';\n",
-      "print '   a.c input power= %.3f watt'%AC_Input_Power;\n",
-      "print '   d.c output power= %.3f watt'%DC_Output_Power;\n",
-      "print ' iii: ';\n",
-      "print '   d.c output voltage = %.2f volts'%DC_Output_Voltage;\n",
-      "print ' iv: '\n",
-      "print '   Efficiency of rectification = %.1f%%'%Rectifier_efficiency;\n",
-      "\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " i:\n",
-        "   Im   = 61 mA\n",
-        "   Idc  = 19.4 mA\n",
-        "   Irms = 30.5 mA\n",
-        " ii: \n",
-        "   a.c input power= 0.763 watt\n",
-        "   d.c output power= 0.301 watt\n",
-        " iii: \n",
-        "   d.c output voltage = 15.52 volts\n",
-        " iv: \n",
-        "   Efficiency of rectification = 39.5%\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.15, Page number 91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "Vdc=50.0;           #Output d.c voltage in V\n",
-      "rf=25;            #Diode resistance in ohm\n",
-      "RL=800;           #Load resistance in ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vm=(pi*(rf+RL)*Vdc)/RL;   #[ Vdc=Vm*RL/(pi*(rf+RL)) ]Maximum value of a.c voltage required to get a volatge  of Vdc from the half-wave rectifier, in V\n",
-      "Vm=round(Vm,0);  \n",
-      "#Result\n",
-      "print 'The a.c voltage required should have maximum value of = %d V' %Vm;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c voltage required should have maximum value of = 162 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.16, Page number 95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "rf=20;              #Internal resistance of the diodes in ohm\n",
-      "Vrms=50;            #RMS value of transformer's secondary voltage from centre tap to each end of secondary\n",
-      "RL=980;             #Load resistance in ohm\n",
-      "\n",
-      "#Calculation\n",
-      "Vm=Vrms*sqrt(2);      #Maximum a.c voltage in V\n",
-      "Im=Vm/(rf+RL);        #Maximum load current in A\n",
-      "Im=Im*1000;           #Maximum load current in mA\n",
-      " \n",
-      "# 1:\n",
-      "Idc=2*Im/pi;          #Mean load current\n",
-      "\n",
-      "# 2:\n",
-      "Irms=Im/sqrt(2);       #RMS value of load current in A\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print'   The mean load current= %d mA'%Idc;\n",
-      "print 'ii:';\n",
-      "print '  The r.m.s value of the load current = %d mA'%Irms; "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The mean load current= 45 mA\n",
-        "ii:\n",
-        "  The r.m.s value of the load current = 50 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.17, Page number 95-96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "#Variable declaration\n",
-      "RL=100;                          #Load resistance in ohm       \n",
-      "rf=0;                            #Internal resistance of the diodes in ohm\n",
-      "Turns_ratio=5/1;                 #Primary to secondary turns ratio of transformer \n",
-      "P_Vrms=230;                      #R.M.S value of voltage in primary winding in V\n",
-      "S_Vrms=P_Vrms/Turns_ratio;       #R.M.S value of voltage in secondary winding in V\n",
-      "S_Vm=S_Vrms*sqrt(2);        #Maximum voltage across secondary winding in V\n",
-      "Vm=S_Vm/2;                  #Maximum voltage across half seconfdary winding in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Idc=2*Vm/(pi*RL);            #Average current in A\n",
-      "Vdc=Idc*RL;                  #d.c output voltage in V\n",
-      "\n",
-      "# 2:\n",
-      "PIV=S_Vm;                    #Peak Invers Voltage(= Maximum secondary voltage) in V\n",
-      "\n",
-      "# 3:\n",
-      "Pac=pow(Vm/(RL*sqrt(2)),2)*(rf+RL);         #a.c input power in watt\n",
-      "Pdc=(pow(Idc,2)*RL);                        #d.c output power in watt\n",
-      "R_eff=(Pdc/Pac)*100;      #Rectification efficiency\n",
-      "R_eff=round(R_eff,1);\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   The d.c output voltage= %.1f V'%Vdc;\n",
-      "print 'ii:';\n",
-      "print '   The peak inverse voltage= %d V'%PIV;\n",
-      "print 'iii:';\n",
-      "print '   Rectification efficiency= %.1f%%'%R_eff;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The d.c output voltage= 20.7 V\n",
-        "ii:\n",
-        "   The peak inverse voltage= 65 V\n",
-        "iii:\n",
-        "   Rectification efficiency= 81.1%\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The value of rectification efficiency is calculated as 81.2% in the textbook using the formula 0.812/(1 + (rf/RL)), but by calculating using the correct values in the formula we get 81.1%."
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.18, Page number 96-97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "#Variable declaration\n",
-      "fin=50;                       #frequency of input ac source in Hz\n",
-      "RL=200;                       #Load resistance in ohm\n",
-      "Turns_ratio=4/1;              #Transformers turns ratio, primary to secondary.\n",
-      "P_Vrms=230.0;                 #R.M.S value of voltage in primary winding in V\n",
-      "S_Vrms=P_Vrms/Turns_ratio     #R.M.S value of voltage in secondary winding in V\n",
-      "Vm=S_Vrms*sqrt(2);            #Maximum voltage across secondary winding in V\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Idc=2*Vm/(pi*RL);           # Average current in A\n",
-      "Vdc=Idc*RL;                 #Output d.c voltage in V\n",
-      "Vdc=round(Vdc,0);\n",
-      "# 2:\n",
-      "PIV= Vm;                    #Peak Inverse Voltage(= Maximum volutage across secondary winding) in V\n",
-      "\n",
-      "# 3:\n",
-      "fout=2*fin;                 #Output frequency in Hz\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   The d.c output voltage = %d V' %Vdc;\n",
-      "print 'ii:';\n",
-      "print '   The peak inverse voltage = %.1f V'%PIV;\n",
-      "print 'iii:';\n",
-      "print '   The output frequency = %d Hz'%fout;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The d.c output voltage = 52 V\n",
-        "ii:\n",
-        "   The peak inverse voltage = 81.3 V\n",
-        "iii:\n",
-        "   The output frequency = 100 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.19, Page number 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=100.0;                        #Load Resistance in ohm\n",
-      "Turns_ratio=5/1;                 #Primary to secondary turns ratio of the transformer\n",
-      "Vin=230.0;                       #R.M.S value of input voltage in V\n",
-      "fin=50;                          #Input frequency in Hz\n",
-      "\n",
-      "#Calculation\n",
-      "Vs_rms=Vin/Turns_ratio;          #R.M.S value of the voltage in secondary winding, in v\n",
-      "Vs_max=Vs_rms*sqrt(2);           #Maximum voltage across secondary, in V\n",
-      "\n",
-      "# (i)\n",
-      "#Case i: Centre-tap circuit\n",
-      "Vm=Vs_max/2;                     #Maximum voltage across half secondary winding, in V \n",
-      "Vdc=2*Vm*RL/(pi*RL);             #DC output voltage, in V \n",
-      "print 'The d.c output voltage for the centre-tap circuit = %.1f V'%Vdc;\n",
-      "\n",
-      "#Case ii:\n",
-      "Vm=Vs_max;                      #Maximum voltage across secondary, in V\n",
-      "Vdc=2*Vm*RL/(pi*RL);            #DC output voltage, in V \n",
-      "print 'The d.c output voltage for the bridge circuit = %.1f V'%Vdc; \n",
-      "\n",
-      "# ii:\n",
-      "#Case i: Centre-tap circuit\n",
-      "Turns_ratio=5/1;                      #Turns ratio of the transformer\n",
-      "Vs_rms=Vin/Turns_ratio;               #R.M.S value of the secondary voltage in V\n",
-      "Vs_max=Vs_rms*sqrt(2);                #Maximum voltage across the secondary in V\n",
-      "Vm=Vs_max/2;                          #Maximum voltage across half of the secondary in V\n",
-      "PIV=2*Vm;                             #Peak Inverse Voltage in V\n",
-      "print 'PIV in case of centre-tap circuit = %d V'%PIV;\n",
-      "\n",
-      "#Case ii: Bridge circuit\n",
-      "Turns_ratio=10/1;                             #Turns ratio of the transformer\n",
-      "Vs_rms=Vin/Turns_ratio;                       #R.M.S value of the secondary voltage in V\n",
-      "Vs_max=Vs_rms*sqrt(2);                        #Maximum voltage across the secondary in V\n",
-      "PIV=Vm;                                        #Peak Inverse Voltage in V\n",
-      "print 'PIV in case of bridge circuit = %.1f V'%PIV;\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The d.c output voltage for the centre-tap circuit = 20.7 V\n",
-        "The d.c output voltage for the bridge circuit = 41.4 V\n",
-        "PIV in case of centre-tap circuit = 65 V\n",
-        "PIV in case of bridge circuit = 32.5 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.20, Page number 98"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "rf=1;                    #forward resistance of diodes of the rectifier in ohm\n",
-      "RL=480;                  #Load resistance in ohm\n",
-      "Vrms=240.0;                #a.c supply voltage in V\n",
-      "Vm=Vrms*sqrt(2);         #Maximum a.c voltage in V \n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Rt=2*rf+RL;             #Total circuit resistance at any instance in ohm\n",
-      "Im=Vm/Rt;               #Maximum load current in A\n",
-      "Idc=2*Im/pi;            #Mean load current in A\n",
-      "\n",
-      "# 2:\n",
-      "Irms=Im/2;              #R.M.S value of current  in A\n",
-      "P=pow(Irms,2)*rf;       #Power dissipated in each diode in watt\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   Mean load current = %.2f A'%Idc;\n",
-      "print 'ii:';\n",
-      "print '   Power dissipated in each diode= %.3f W'%P;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   Mean load current = 0.45 A\n",
-        "ii:\n",
-        "   Power dissipated in each diode= 0.124 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The value of power dissipated is approximately 0.124 W , but in the textbook it is approximated as 0.123W."
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.21, Page number 98-99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt,pi\n",
-      "#Variable declaration\n",
-      "RL=12000;                #Load resistance in ohm\n",
-      "V0=0.7;                   #Potential barrier voltage of diodes in V\n",
-      "Vrms=12;                  #R.M.S value of input a.c voltage in V\n",
-      "Vs_pk=Vrms*sqrt(2);       #Peak secondary voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Vout_pk=Vs_pk-(2*V0);      #Peak output voltage in  V\n",
-      "Vav=2*Vout_pk/pi;          #Average output voltage in V\n",
-      "Vav=round(Vav,2);\n",
-      "\n",
-      "# 2:\n",
-      "Iav=Vav/RL;                #Average output current in A\n",
-      "Iav=Iav*pow(10,6);         #Average output current in \u03bcA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   Average output voltage=%.2f V'%Vav;\n",
-      "print 'ii:';\n",
-      "print '   Average output current=%.1f \u03bcA'%Iav;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   Average output voltage=9.91 V\n",
-        "ii:\n",
-        "   Average output current=825.8 \u03bcA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.22, Page number 102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vdc_A=10;                      #Supply voltage of A in V\n",
-      "Vdc_B=25;                      #Supply voltage of B in V\n",
-      "Vac_rms_a=0.5;                 #Ripples in power supply A in V\n",
-      "Vac_rms_b=0.001;               #Ripples in power supply B in V\n",
-      "\n",
-      "#Calculation\n",
-      "#For power supply A\n",
-      "ripple_factor_A=Vac_rms_a/Vdc_A;      #Ripple factor of power supply A\n",
-      "\n",
-      "#For power supply B\n",
-      "ripple_factor_B=Vac_rms_b/Vdc_B;      #Ripple factor of power supply B\n",
-      "\n",
-      "#Result\n",
-      "if(ripple_factor_A<ripple_factor_B):\n",
-      "    print 'Power supply A is better';\n",
-      "else :\n",
-      "    print 'Power supply B is better';"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supply B is better\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.23, Page number 105-106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "RL=2200;                    #Load resistance in ohm\n",
-      "C=50*pow(10,-6);            #Capacitance of the capacitor used in filter circuit in F\n",
-      "V0=0.7;                     #Potential barrier voltage of the diodes of the rectifier in  V\n",
-      "Vrms=115.0;                   #R.M.S value of input a.c voltage in V \n",
-      "fin=60;                     #Frequency of input a.c voltage in Hz\n",
-      "Turns_ratio=10/1;           #Primary to secondary, turns ratio of the transformer \n",
-      "\n",
-      "#Calculation\n",
-      "Vp_prim=Vrms*sqrt(2);               #Peak primary voltage in V\n",
-      "Vp_sec=Vp_prim/Turns_ratio;         #Peak secondary voltage in V\n",
-      "Vp_in= Vp_sec - 2*V0;               #Peak full wave rectified voltage at the filter input in V\n",
-      "f=2*fin;                            #Output frequency in Hz\n",
-      "Vdc=Vp_in*(1-(1/(2*f*RL*C)));       #Output d.c voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print 'The output d.c voltage  is = %.1f V'%Vdc;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output d.c voltage  is = 14.3 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.24, Page number 106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "R=25;                        #d.c resistance of the choke in ohm\n",
-      "RL=750;                      #Load resistance in ohm\n",
-      "Vm=25.7;                     #Maximum value of the pulsating output from the rectifier in V\n",
-      "\n",
-      "#Calculation\n",
-      "V_dc=2*Vm/pi;                #d.c component of the pulsating output in V\n",
-      "V_dc=round(V_dc,1);\n",
-      "V_dc_out=(V_dc*RL)/(R+RL);   #Output d.c voltage in V\n",
-      "V_dc_out=round(V_dc_out,1);\n",
-      "\n",
-      "#Result\n",
-      "print ' The output d.c voltage accross the load resistance is = %.1f V'%V_dc_out;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The output d.c voltage accross the load resistance is = 15.9 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.25, Page number 113-114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=120.0;                   #Input Voltage in V\n",
-      "Vz=50.0;                    #Zener Voltage in V\n",
-      "R=5000.0;                   #Resistance of the series resistor in ohm\n",
-      "RL=10000.0;                 #Load resistance in ohm\n",
-      "\n",
-      "#Calculation\n",
-      "V=Ei*RL/(R+RL);             #Voltage across the open circuit if the zener diode is removed\n",
-      "if(V>Vz):\n",
-      "    #Zener diode is in ON state\n",
-      "    # i:\n",
-      "    Output_voltage=Vz;             #Voltage across load resistance, in V\n",
-      "    #ii:\n",
-      "    Voltage_R=Ei-Vz;               #Voltage across the series resistance R, in V\n",
-      "    #iii:\n",
-      "    IL=Vz/RL;                      #Load current through RL in A\n",
-      "    IL=IL*1000;                    #Load current through RL in mA\n",
-      "    I=Voltage_R/R;                 #Current through the series resistance in A\n",
-      "    I=I*1000;                      #Current through the series resistance in mA\n",
-      "    Iz=I-IL;                       #Applying Kirchhoff's first law, Zener current in mA\n",
-      "    \n",
-      "    #Result\n",
-      "    print 'i)   The output voltage across the load resistance RL = %d V'%Output_voltage;\n",
-      "    print 'ii)  The voltage drop across the series resistance  R = %d V'%Voltage_R;\n",
-      "    print 'iii) The current through the zener diode = %d mA'%Iz;\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)   The output voltage across the load resistance RL = 50 V\n",
-        "ii)  The voltage drop across the series resistance  R = 70 V\n",
-        "iii) The current through the zener diode = 9 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.26, Page number 114-115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Max_V=120.0;                  #Maximum input voltage in V\n",
-      "Min_V=80.0;                   #Minimum input voltage in V\n",
-      "R=5000.0;                     #Series resistance in ohm\n",
-      "RL=10000.0;                   #Load resistance in ohm\n",
-      "Vz=50.0;                      #Zener voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i: Maximum zener current\n",
-      "#Zener current will be maximum when the input voltage is maximum\n",
-      "V_R_max=Max_V-Vz;                #Voltage across series resistance R, in  V\n",
-      "I_max=V_R_max/R;                 #Current through series resistance R, in A\n",
-      "I_max=I_max*1000;                #Current through series resistance R, in mA\n",
-      "IL_max=Vz/RL;                        #Load current in A\n",
-      "IL_max=IL_max*1000;                      #Load current in mA\n",
-      "Iz_max=I_max-IL_max;                     #Applying Kirchhoff's first law, Zener current in mA;\n",
-      "\n",
-      "#Case ii: Minimum zener current\n",
-      "#The zener will conduct minimum current when the input voltage is minimum\n",
-      "V_R_min=Min_V-Vz;                #Voltage across series resistance R, in V\n",
-      "I_min=V_R_min/R;                 #Current through series resistance R, in A\n",
-      "I_min=I_min*1000;                #Current through series resistance R, in mA\n",
-      "IL_min=Vz/RL;                        #Load current in A\n",
-      "IL_min=IL_min*1000;                      #Load current in mA\n",
-      "Iz_min=I_min-IL_min;                     #Applying Kirchhoff's first law, Zener current in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Case i: ';\n",
-      "print 'Maximum zener current = %d mA'%Iz_max;\n",
-      "print 'Case ii: ';\n",
-      "print 'Minimum zener current = %d mA'%Iz_min;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case i: \n",
-        "Maximum zener current = 9 mA\n",
-        "Case ii: \n",
-        "Minimum zener current = 1 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.27, Page number 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=12;                      #Input voltage in V\n",
-      "Vz=7.2;                     #Zener voltage in V\n",
-      "E0=Vz;                      #Voltage to be maintained across the load in  V\n",
-      "IL_max=0.1;                 #Maximum load current in A\n",
-      "IL_min=0.012;                  #Minimum load current in A\n",
-      "Iz_min=0.01;                  #Minimum zener current in A\n",
-      "\n",
-      "#Calculation\n",
-      "#When the load current is maximum at minimum value of RL, the zener current is minimum and, as the load current decreases due to increase in value of RL\n",
-      "R=(Ei-E0)/(Iz_min+IL_max);        #The value of series resistance R to maintain a voltage=E0 across load, in ohm\n",
-      "\n",
-      "#Result\n",
-      "print 'The minimum value of series resistance R to maintain a constant value of 7.2 V is = %.1f \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum value of series resistance R to maintain a constant value of 7.2 V is = 43.6 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The actual value of R is 43.636363 (recurring) but, in the textbook the value of R is wrongly approximated 43.5 \u03a9"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.28, Page number 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei_min=22;                #Minimum input voltage in V\n",
-      "Ei_max=28;                #Maximum input voltage in V\n",
-      "Vz=18;                   #Zener voltage in V\n",
-      "E0=Vz;                    #Constant voltage maintained across the load resistance in V\n",
-      "Iz_min=0.2;               #Minimum zener current in A\n",
-      "Iz_max=2;                 #Maximum zener current in A\n",
-      "RL=18;                    #Load resistance in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "IL=Vz/RL;                      #Constant value of load current in A\n",
-      "#When the input voltage is minimum, the zener current will be minimum\n",
-      "R=(Ei_min-E0)/(Iz_min+IL)      #The value of series resistance so that the voltage E0 across RL remains constant\n",
-      "\n",
-      "print 'The value of series resistance R, to maintain constant voltage E0 across RL = %.2f \u03a9.'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistance R, to maintain constant voltage E0 across RL = 3.33 \u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.29, Page number 116 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10                                #Zener voltage in V\n",
-      "Ei_min=13;                           #Minimum input voltage in V\n",
-      "Ei_max=16;                           #Maximum input voltage in V\n",
-      "Iz_min=0.015;                        #Minimum zener current in A\n",
-      "IL_min=0.01;                         #Minimum load current in A \n",
-      "IL_max=0.085;                        #Maximum load curremt in A\n",
-      "E0=Vz;                               #Constant voltage to be maintained in V \n",
-      "\n",
-      "#Calculation\n",
-      "#The zener current will be minimum when the input voltage will be minimum  and at that time the load current will be maximum\n",
-      "R=(Ei_min-E0)/(Iz_min+IL_max);       #The value of series resistance R to maintain a constant voltage across load\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'The value of series resistance to maintain a constant voltage across the load resistance is = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistance to maintain a constant voltage across the load resistance is = 30 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.30, Page number 116"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Iz=0.2;               #Current rating of each zener in A\n",
-      "Vz=15;                #Voltage rating of each zener in V\n",
-      "Ei=45;                #Input voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "# i: Regulated output voltage across the two zener diodes \n",
-      "E0=2*Vz;                     # V\n",
-      "\n",
-      "# ii: Value of series resistance \n",
-      "R=(Ei-E0)/Iz;                  # \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print 'i) The regulated output voltage = %d V'%E0;\n",
-      "print 'ii) The value of the series resistance = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) The regulated output voltage = 30 V\n",
-        "ii) The value of the series resistance = 75 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.31, Page number 116-117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10;                   #Voltage rating of each zener in V\n",
-      "Iz=1;                    #Current rating of each zener in A\n",
-      "Ei=45;                   #Input unregulated voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "#Regulated output voltage across the three zener diodes\n",
-      "E0=3*Vz;               # V\n",
-      "\n",
-      "#Value of series resistance to obtain a 30V regulated output voltage\n",
-      "R=(Ei-E0)/Iz;          # \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print 'Value of series resistance to obtain a 30V regulated output voltage = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of series resistance to obtain a 30V regulated output voltage = 15 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.32, Page number 117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "RL=2000.0;                   #Load resistance in \u03a9\n",
-      "R=200.0;                     #Series resistance in \u03a9\n",
-      "Iz=0.025;                  #Zener current rating in A\n",
-      "E0=30.0;                     #Output regulated voltage in V \n",
-      "\n",
-      "#Calculation\n",
-      "#Minimum input voltage will be required when Iz=0 A, and at  this condition\n",
-      "IL=E0/RL;                   #Load current during Iz=0, in A\n",
-      "I=IL;                       #According to Kirchhoff's law, total current, in A\n",
-      "Ei_min=E0+(I*R);            #Minimum input voltage in V\n",
-      "\n",
-      "#The maximum input voltage required will be when Iz=0.025 A, and at that condition \n",
-      "I=IL+Iz;                    #According to Kirchhoff's law, total current, in A\n",
-      "Ei_max=E0+(I*R);            #maximum input voltage in V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'The required range of input voltage is from %d V to %d V'%(Ei_min,Ei_max);  \n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required range of input voltage is from 33 V to 38 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.33, Page number 117-118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=16;                   #Unregulated input voltage in V\n",
-      "E0=12;                   #Output regulated voltage in V\n",
-      "IL_min=0;                #Minimum load current in A\n",
-      "IL_max=0.2;              #Maximum load current in A\n",
-      "Iz_min=0;                #Minimum zener current in A\n",
-      "Iz_max=0.2;              #Maximum zener current in A\n",
-      "\n",
-      "#Calculation\n",
-      "#As the regulated voltage required across the load is 12V\n",
-      "Vz=E0;                   #Voltage rating of zener diode in V\n",
-      "V_R=Ei-E0;               #Constant Voltage that should remain across series resistance \n",
-      "#The minimum zener current will occur when the curent in the load in maximum\n",
-      "R=V_R/(Iz_min+IL_max);   #Series resistance in \u03a9\n",
-      "\n",
-      "Max_power_rating=Vz*Iz_max;     #Maximum power rating of zener diode in W\n",
-      "\n",
-      "#Result\n",
-      "print 'The regulator is designed using a Seris resistance of %d \u03a9 and a zener diode of zener voltage %d V'%(R,Vz);\n",
-      "print 'The maximum power rating of the zener diode is = %.1f W '%Max_power_rating;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulator is designed using a Seris resistance of 20 \u03a9 and a zener diode of zener voltage 12 V\n",
-        "The maximum power rating of the zener diode is = 2.4 W \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.34, Page number 118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=12;                       #Source voltage in V\n",
-      "R=1000;                     #Series resistance in \u03a9\n",
-      "RL=5000;                    #Load resistance in \u03a9\n",
-      "Vz=6;                       #Voltage rating of zener in V\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i: zener is working properly\n",
-      "#The output voltage across the load will be equal to the zener voltage.\n",
-      "V0=Vz;                        # V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case i: Output voltage when zener is working properly is %d V'%V0;\n",
-      "\n",
-      "#Case ii: zener is shorted\n",
-      "#As the zener is shorted, the potential difference across the load will be zero\n",
-      "V0=0;                        #V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case ii: Output voltage when zener is short circuited is %d V'%V0;\n",
-      "     \n",
-      "#Case iii: zener is open circuited\n",
-      "#If the zener is open circuited, the total voltage will drop across R and RL according to the voltage divider rule\n",
-      "V0=V*RL/(R+RL);                      #V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case iii: Output voltage when zener is open circuited is %d V'%V0;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case i: Output voltage when zener is working properly is 6 V\n",
-        "Case ii: Output voltage when zener is short circuited is 0 V\n",
-        "Case iii: Output voltage when zener is open circuited is 10 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_5.ipynb
deleted file mode 100755
index a6008ee1..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_5.ipynb
+++ /dev/null
@@ -1,1624 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:3452607f2168b562d941493f83083042eaa5a2d316715f9d9f089ff03d73fdb8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 6: SEMICONDUCTOR DIODE"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.2, Page number 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration \n",
-      "Vf =20;                #Peak Input Voltage in V\n",
-      "rf=10;                  #Forward Resistance in ohms\n",
-      "RL=500.0;                 #Load Resistance in ohms\n",
-      "V0=0.7;                  #Potential Barrier Voltage of the diodes in V\n",
-      "\n",
-      "#Calculation\n",
-      "#(1)\n",
-      "If_peak=(Vf-V0)/(rf+RL);                  #Peak current through the diode in A\n",
-      "If_peak=If_peak*1000;                    #Peak current through the diode in mA\n",
-      "#(2)\n",
-      "V_out_peak =If_peak * RL/1000 ;               #Peak output voltage in V\n",
-      "\n",
-      "#For an Ideal diode\n",
-      "If_peak_ideal=Vf/RL;                   #Peak current through the ideal diode in A\n",
-      "If_peak_ideal=If_peak_ideal*1000;      #Peak current through the ideal diode in mA\n",
-      "\n",
-      "V_out_peak_ideal=If_peak_ideal * RL/1000;     # Peak output voltage in case of the ideal diode in V\n",
-      "\n",
-      "#Result\n",
-      "print '(i) Peak current through the diode = %.1f mA '%If_peak;\n",
-      "print '(ii) Peak output voltage = %.1f V'%V_out_peak;\n",
-      "print '(iii) Peak current through the ideal diode = %d mA '%If_peak_ideal;\n",
-      "print '(iv) Peak output voltage in case of the ideal diode = %d V'%V_out_peak_ideal;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Peak current through the diode = 37.8 mA \n",
-        "(ii) Peak output voltage = 18.9 V\n",
-        "(iii) Peak current through the ideal diode = 40 mA \n",
-        "(iv) Peak output voltage in case of the ideal diode = 20 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.3, Page number 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V =10.0;                #Battery voltage in V\n",
-      "R1=50.0;                  #Resistor 1's resistance in ohms\n",
-      "R2=5.0;                 #Resistor 2's resistance in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "#Using Thevenin's Theorem to find current in the diode\n",
-      "E0=(R2/(R1+R2))*V;          #Thevenin's Voltage in V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's Resistance in ohms\n",
-      "\n",
-      "I0=E0/R0;                   #Current through the diode in A\n",
-      "I0=I0*1000;                 #Current through the diode in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Current through the diode = %d mA '%Io;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current through the diode = 200 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.4, Page number 82-83 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V =10.0;                #Battery voltage in V\n",
-      "R0=48.0;                  #Resistance of the resistor in ohms\n",
-      "Rd=1.0;                 #Forward resistance of the diodes in ohms\n",
-      "Vd=0.7;                 #Potential barrier of the diodes in V\n",
-      "#Calculation\n",
-      "V_net=V-Vd-Vd;        #Net voltage in the circuit in V\n",
-      "R_net=R0+Rd+Rd        #Net resistance of the circuit in ohms\n",
-      "I_net=V_net/R_net;    #Net current in the circuit in A\n",
-      "I_net=I_net*1000;     #Net current in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Net current in the circuit = %d mA '%I_net;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Net current in the circuit = 172 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page number 83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E1=24;           #Voltage of first source in V\n",
-      "E2=4;            #Voltage of second source in V\n",
-      "V0=0.7;          #Potential barrier of diodes in V\n",
-      "R=2000;          #Resistance of the given resistor in ohms\n",
-      "Rd=0;            #Forward resistance of the diodes in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I=(E1-E2-V0)/(R+Rd);  #Current in the circuit in A\n",
-      "I=I*1000;             #Current in the circuit in mA \n",
-      "\n",
-      "#Result\n",
-      "print 'Current in the circuit = %.2f mA '%I;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in the circuit = 9.65 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page number 83-84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=20;           #Voltage of source in V\n",
-      "V0=0.3;         #Potential barrier of Germanium diode in V\n",
-      "V0_Si=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "\n",
-      "#Calculation\n",
-      "#As only Ge diode is turned on due to less potential barrier,\n",
-      "VA=V-V0;        #Voltage VA acroos resistor of 3k ohms\n",
-      "\n",
-      "#Result\n",
-      "print 'Voltage VA = %.1f mA '%VA;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage VA = 19.7 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.7, Page number 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=10;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "# Resistance of all resistors in ohms\n",
-      "R1=2000;\n",
-      "R2=2000;\n",
-      "R3=2000;\n",
-      "\n",
-      "#Calculation\n",
-      "Id=(V-V0)/(R2+2*R3);     #Current through the diodes in A\n",
-      "VQ=2*Id*R3;      #Voltage VQ across the grounded 2k ohm resistor in V\n",
-      "Id=Id*1000;      #Current through the diodes in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Voltage VQ = %.1f V '%VQ;\n",
-      "print 'Current through the diodes, Id = %.2f mA '%Id;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage VQ = 6.2 V \n",
-        "Current through the diodes, Id = 1.55 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.8, Page number 84"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "V=15;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V \n",
-      "R=500       # Resistance of all resistors in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I1=(V-V0)/R;      #total current in the circuit in A\n",
-      "Id1=I1/2;         #current in first diode in A\n",
-      "Id1=Id1*1000;     #current in first diode in mA\n",
-      "Id2=Id1           #current in second diode in mA\n",
-      "\n",
-      "#Result\n",
-      "print ('Current in first diode = %.1f mA'%Id1);\n",
-      "print ('Current in second diode = %.1f mA'%Id2);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in first diode = 14.3 mA\n",
-        "Current in second diode = 14.3 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.9, Page number 85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E=20;           #Voltage of source in V\n",
-      "V0_d1=0.7;      #Potetial barrier of first Silicon diode in V\n",
-      "V0_d2=0.7;      #Potetial barrier of second Silicon diode in V\n",
-      "R1=5600;       # Resistance of first resistor in ohms\n",
-      "R2=3300;       #  Resistance of second resistor in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "I2=V0_d2/R2;                  #Current I2 through resistor R2 in A\n",
-      "I2=round((I2*1000),3);                   #Current I2 through resistor R2 in mA\n",
-      "I1=(E-V0_d1-V0_d2)/R1;        #Current I1 through resistor R1 in  A\n",
-      "I1=round((I1*1000),2);                   #Current I1 through resistor R1 in  mA\n",
-      "I3=I1-I2;                     #Current I3 through diode D2 in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Current I1= %.2f mA'%I1;\n",
-      "print 'Current I1= %.3f mA'%I2;\n",
-      "print 'Current I1= %.3f mA'%I3;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current I1= 3.32 mA\n",
-        "Current I1= 0.212 mA\n",
-        "Current I1= 3.108 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.10, Page number 85-86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration\n",
-      "E=10.0;           #Voltage of source in V\n",
-      "V0=0.7;      #Potetial barrier of Silicon diode in V\n",
-      "R1=2000;       # Resistance of first resistor in ohms\n",
-      "R2=8000;       #  Resistance of second resistor in ohms\n",
-      "R3=4000;       #Resistance of third resistor in ohms\n",
-      "R4=6000;       #Resistance of fourth resistor in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "#Assuming the given diode to be reverse bised and calculating voltage across it's terminals\n",
-      "V1=(E/(R1+R2))*R2;      #voltage at the P side of the diode, i.e, voltage across R2 resistor,according to voltage divider rule, in V\n",
-      "V2=(E/(R3+R4))*R4;      #voltage at the N side of the diode, i.e, voltage across R4 resistor,according to voltage divider rule, in V\n",
-      "\n",
-      "#Result\n",
-      "if((V1-V2)>=V0):\n",
-      "    print 'Our assumption was wrong and, the diode is forward biased';\n",
-      "else:\n",
-      "    print 'The diode is reverse biased';\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Our assumption was wrong and, the diode is forward biased\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.11, Page number 86"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=2;                              #Supply voltage in V\n",
-      "V0=0.7;                           #Potential barrier voltage of the diode in V \n",
-      "R1=4000.0;                        #Resistance of first resistor in \u03a9\n",
-      "R2=1000.0;                        ##Resistance of second resistor in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#Assuming the diode to be in ON state\n",
-      "I1=((V-V0)/R1)*1000;                  #Current through resistor R1, in mA\n",
-      "I2=(V0/R2)*1000;                    #Current through resistor R2, in mA\n",
-      "ID=I1-I2;                           #Diode current, in  mA\n",
-      "\n",
-      "if(ID<0):\n",
-      "    #Since the diode current is negative, the diode must be OFF \n",
-      "    ID=0;                     #True value of diode current, mA\n",
-      "    \n",
-      "#As the diode is in OFF state it can be replaced by an open ciruit equivalent \n",
-      "VD=V*R2/(R1 +R2);                #Voltage across the diode, in V\n",
-      "\n",
-      "#Result\n",
-      "print 'ID =%d mA'%ID;\n",
-      "print 'VD =%.1f V'%VD;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "ID =0 mA\n",
-        "VD =0.4 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.12, Page number 89-90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "AC_Input_Power=100.0;          #Input AC Power in watts\n",
-      "AC_Output_Power=40.0;          #Output AC Power in watts\n",
-      "Accepted_Power=50.0;           #Power accepted by the half-wave rectifier in watt\n",
-      "\n",
-      "#Calculation\n",
-      "R_eff=(AC_Output_Power/AC_Input_Power)*100;             #Rectification efficiency of the half-wave rectifier\n",
-      "Unused_power=AC_Input_Power-Accepted_Power;       #Power not used by the half_wave rectifier due to open circuited condition of the diode in watt\n",
-      "Power_dissipated=Accepted_Power-AC_Output_Power;  #Power dissipated by the diode watt\n",
-      "\n",
-      "#Result\n",
-      "print 'The rectification efficiency of the half-wave rectifier= %d%% '%R_eff;\n",
-      "\n",
-      "print 'Rest 60%% of the power is the unused power and power dissipated by the diode = %d watts and %d watts' %(Unused_power ,Power_dissipated);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The rectification efficiency of the half-wave rectifier= 40% \n",
-        "Rest 60% of the power is the unused power and power dissipated by the diode = 50 watts and 10 watts\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.13, Page number 90"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "Vrms=230.0;                #AC supply RMS voltage in V\n",
-      "Turns_Ratio=10/1;        #turn ratio of the transformer \n",
-      "\n",
-      "#Calculation\n",
-      "Vpm=sqrt(2)*Vrms;         #Maximum primary voltage in V\n",
-      "Vsm=Vpm/Turns_Ratio;     #Maximum secondary voltage in V\n",
-      "#Case 1\n",
-      "Vdc=Vsm/(round(pi,2));              #Output D.C voltage, which is the average voltage in V\n",
-      "Vdc=round(Vdc,2);\n",
-      "#Case 2\n",
-      "PIV=Vsm;                 #Peak Inverse Voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print 'The output d.c voltage= %.2f V'%Vdc;\n",
-      "print 'The peak inverse voltage= %.2f V'%PIV;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output d.c voltage= 10.36 V\n",
-        "The peak inverse voltage= 32.53 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.14, Page number 90-91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "rf=20.0;           #Internal resistance of the crystal diode in ohms\n",
-      "Vm=50.0;           #Maximum applied voltage in V\n",
-      "RL=800.0;          #Load Resistance in ohms\n",
-      "\n",
-      "#Calculation\n",
-      "# 1\n",
-      "Im=Vm/(rf+RL);         #Maximum current in A\n",
-      "Im=Im*1000;            #Maximum current in \n",
-      "Im=round(Im,0);\n",
-      "Idc=Im/pi;             #Average voltage in mA\n",
-      "Idc=round(Idc,1);\n",
-      "Irms=Im/2;             #RMS value of the current in mA\n",
-      "Irms=round(Irms,1)\n",
-      "\n",
-      "# 2\n",
-      "AC_Input_Power=pow(Irms/1000,2)*(rf+RL);     #Input a.c power in watt\n",
-      "\n",
-      "DC_Output_Power=pow(Idc/1000,2)*RL;          #Output d.c power in watt\n",
-      "\n",
-      "# 3\n",
-      "DC_Output_Voltage=(Idc/1000)*RL;               #Output d.c voltage in V\n",
-      "\n",
-      "# 4\n",
-      "Rectifier_efficiency=(DC_Output_Power/AC_Input_Power)*100;   # Efficiency of rectification of the half-wave rectifier\n",
-      "\n",
-      "#Result\n",
-      "print ' i:';\n",
-      "print '   Im   = %d mA'%Im;\n",
-      "print '   Idc  = %.1f mA'%Idc;\n",
-      "print '   Irms = %.1f mA'%Irms;\n",
-      "print ' ii: ';\n",
-      "print '   a.c input power= %.3f watt'%AC_Input_Power;\n",
-      "print '   d.c output power= %.3f watt'%DC_Output_Power;\n",
-      "print ' iii: ';\n",
-      "print '   d.c output voltage = %.2f volts'%DC_Output_Voltage;\n",
-      "print ' iv: '\n",
-      "print '   Efficiency of rectification = %.1f%%'%Rectifier_efficiency;\n",
-      "\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " i:\n",
-        "   Im   = 61 mA\n",
-        "   Idc  = 19.4 mA\n",
-        "   Irms = 30.5 mA\n",
-        " ii: \n",
-        "   a.c input power= 0.763 watt\n",
-        "   d.c output power= 0.301 watt\n",
-        " iii: \n",
-        "   d.c output voltage = 15.52 volts\n",
-        " iv: \n",
-        "   Efficiency of rectification = 39.5%\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.15, Page number 91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "Vdc=50.0;           #Output d.c voltage in V\n",
-      "rf=25;            #Diode resistance in ohm\n",
-      "RL=800;           #Load resistance in ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "Vm=(pi*(rf+RL)*Vdc)/RL;   #[ Vdc=Vm*RL/(pi*(rf+RL)) ]Maximum value of a.c voltage required to get a volatge  of Vdc from the half-wave rectifier, in V\n",
-      "Vm=round(Vm,0);  \n",
-      "#Result\n",
-      "print 'The a.c voltage required should have maximum value of = %d V' %Vm;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The a.c voltage required should have maximum value of = 162 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.16, Page number 95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "rf=20;              #Internal resistance of the diodes in ohm\n",
-      "Vrms=50;            #RMS value of transformer's secondary voltage from centre tap to each end of secondary\n",
-      "RL=980;             #Load resistance in ohm\n",
-      "\n",
-      "#Calculation\n",
-      "Vm=Vrms*sqrt(2);      #Maximum a.c voltage in V\n",
-      "Im=Vm/(rf+RL);        #Maximum load current in A\n",
-      "Im=Im*1000;           #Maximum load current in mA\n",
-      " \n",
-      "# 1:\n",
-      "Idc=2*Im/pi;          #Mean load current\n",
-      "\n",
-      "# 2:\n",
-      "Irms=Im/sqrt(2);       #RMS value of load current in A\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print'   The mean load current= %d mA'%Idc;\n",
-      "print 'ii:';\n",
-      "print '  The r.m.s value of the load current = %d mA'%Irms; "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The mean load current= 45 mA\n",
-        "ii:\n",
-        "  The r.m.s value of the load current = 50 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.17, Page number 95-96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "#Variable declaration\n",
-      "RL=100;                          #Load resistance in ohm       \n",
-      "rf=0;                            #Internal resistance of the diodes in ohm\n",
-      "Turns_ratio=5/1;                 #Primary to secondary turns ratio of transformer \n",
-      "P_Vrms=230;                      #R.M.S value of voltage in primary winding in V\n",
-      "S_Vrms=P_Vrms/Turns_ratio;       #R.M.S value of voltage in secondary winding in V\n",
-      "S_Vm=S_Vrms*sqrt(2);        #Maximum voltage across secondary winding in V\n",
-      "Vm=S_Vm/2;                  #Maximum voltage across half seconfdary winding in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Idc=2*Vm/(pi*RL);            #Average current in A\n",
-      "Vdc=Idc*RL;                  #d.c output voltage in V\n",
-      "\n",
-      "# 2:\n",
-      "PIV=S_Vm;                    #Peak Invers Voltage(= Maximum secondary voltage) in V\n",
-      "\n",
-      "# 3:\n",
-      "Pac=pow(Vm/(RL*sqrt(2)),2)*(rf+RL);         #a.c input power in watt\n",
-      "Pdc=(pow(Idc,2)*RL);                        #d.c output power in watt\n",
-      "R_eff=(Pdc/Pac)*100;      #Rectification efficiency\n",
-      "R_eff=round(R_eff,1);\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   The d.c output voltage= %.1f V'%Vdc;\n",
-      "print 'ii:';\n",
-      "print '   The peak inverse voltage= %d V'%PIV;\n",
-      "print 'iii:';\n",
-      "print '   Rectification efficiency= %.1f%%'%R_eff;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The d.c output voltage= 20.7 V\n",
-        "ii:\n",
-        "   The peak inverse voltage= 65 V\n",
-        "iii:\n",
-        "   Rectification efficiency= 81.1%\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The value of rectification efficiency is calculated as 81.2% in the textbook using the formula 0.812/(1 + (rf/RL)), but by calculating using the correct values in the formula we get 81.1%."
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.18, Page number 96-97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt \n",
-      "#Variable declaration\n",
-      "fin=50;                       #frequency of input ac source in Hz\n",
-      "RL=200;                       #Load resistance in ohm\n",
-      "Turns_ratio=4/1;              #Transformers turns ratio, primary to secondary.\n",
-      "P_Vrms=230.0;                 #R.M.S value of voltage in primary winding in V\n",
-      "S_Vrms=P_Vrms/Turns_ratio     #R.M.S value of voltage in secondary winding in V\n",
-      "Vm=S_Vrms*sqrt(2);            #Maximum voltage across secondary winding in V\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Idc=2*Vm/(pi*RL);           # Average current in A\n",
-      "Vdc=Idc*RL;                 #Output d.c voltage in V\n",
-      "Vdc=round(Vdc,0);\n",
-      "# 2:\n",
-      "PIV= Vm;                    #Peak Inverse Voltage(= Maximum volutage across secondary winding) in V\n",
-      "\n",
-      "# 3:\n",
-      "fout=2*fin;                 #Output frequency in Hz\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   The d.c output voltage = %d V' %Vdc;\n",
-      "print 'ii:';\n",
-      "print '   The peak inverse voltage = %.1f V'%PIV;\n",
-      "print 'iii:';\n",
-      "print '   The output frequency = %d Hz'%fout;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   The d.c output voltage = 52 V\n",
-        "ii:\n",
-        "   The peak inverse voltage = 81.3 V\n",
-        "iii:\n",
-        "   The output frequency = 100 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.19, Page number 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "RL=100.0;                        #Load Resistance in ohm\n",
-      "Turns_ratio=5/1;                 #Primary to secondary turns ratio of the transformer\n",
-      "Vin=230.0;                       #R.M.S value of input voltage in V\n",
-      "fin=50;                          #Input frequency in Hz\n",
-      "\n",
-      "#Calculation\n",
-      "Vs_rms=Vin/Turns_ratio;          #R.M.S value of the voltage in secondary winding, in v\n",
-      "Vs_max=Vs_rms*sqrt(2);           #Maximum voltage across secondary, in V\n",
-      "\n",
-      "# (i)\n",
-      "#Case i: Centre-tap circuit\n",
-      "Vm=Vs_max/2;                     #Maximum voltage across half secondary winding, in V \n",
-      "Vdc=2*Vm*RL/(pi*RL);             #DC output voltage, in V \n",
-      "print 'The d.c output voltage for the centre-tap circuit = %.1f V'%Vdc;\n",
-      "\n",
-      "#Case ii:\n",
-      "Vm=Vs_max;                      #Maximum voltage across secondary, in V\n",
-      "Vdc=2*Vm*RL/(pi*RL);            #DC output voltage, in V \n",
-      "print 'The d.c output voltage for the bridge circuit = %.1f V'%Vdc; \n",
-      "\n",
-      "# ii:\n",
-      "#Case i: Centre-tap circuit\n",
-      "Turns_ratio=5/1;                      #Turns ratio of the transformer\n",
-      "Vs_rms=Vin/Turns_ratio;               #R.M.S value of the secondary voltage in V\n",
-      "Vs_max=Vs_rms*sqrt(2);                #Maximum voltage across the secondary in V\n",
-      "Vm=Vs_max/2;                          #Maximum voltage across half of the secondary in V\n",
-      "PIV=2*Vm;                             #Peak Inverse Voltage in V\n",
-      "print 'PIV in case of centre-tap circuit = %d V'%PIV;\n",
-      "\n",
-      "#Case ii: Bridge circuit\n",
-      "Turns_ratio=10/1;                             #Turns ratio of the transformer\n",
-      "Vs_rms=Vin/Turns_ratio;                       #R.M.S value of the secondary voltage in V\n",
-      "Vs_max=Vs_rms*sqrt(2);                        #Maximum voltage across the secondary in V\n",
-      "PIV=Vm;                                        #Peak Inverse Voltage in V\n",
-      "print 'PIV in case of bridge circuit = %.1f V'%PIV;\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The d.c output voltage for the centre-tap circuit = 20.7 V\n",
-        "The d.c output voltage for the bridge circuit = 41.4 V\n",
-        "PIV in case of centre-tap circuit = 65 V\n",
-        "PIV in case of bridge circuit = 32.5 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.20, Page number 98"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "rf=1;                    #forward resistance of diodes of the rectifier in ohm\n",
-      "RL=480;                  #Load resistance in ohm\n",
-      "Vrms=240.0;                #a.c supply voltage in V\n",
-      "Vm=Vrms*sqrt(2);         #Maximum a.c voltage in V \n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Rt=2*rf+RL;             #Total circuit resistance at any instance in ohm\n",
-      "Im=Vm/Rt;               #Maximum load current in A\n",
-      "Idc=2*Im/pi;            #Mean load current in A\n",
-      "\n",
-      "# 2:\n",
-      "Irms=Im/2;              #R.M.S value of current  in A\n",
-      "P=pow(Irms,2)*rf;       #Power dissipated in each diode in watt\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   Mean load current = %.2f A'%Idc;\n",
-      "print 'ii:';\n",
-      "print '   Power dissipated in each diode= %.3f W'%P;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   Mean load current = 0.45 A\n",
-        "ii:\n",
-        "   Power dissipated in each diode= 0.124 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The value of power dissipated is approximately 0.124 W , but in the textbook it is approximated as 0.123W."
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.21, Page number 98-99"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt,pi\n",
-      "#Variable declaration\n",
-      "RL=12000;                #Load resistance in ohm\n",
-      "V0=0.7;                   #Potential barrier voltage of diodes in V\n",
-      "Vrms=12;                  #R.M.S value of input a.c voltage in V\n",
-      "Vs_pk=Vrms*sqrt(2);       #Peak secondary voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "# 1:\n",
-      "Vout_pk=Vs_pk-(2*V0);      #Peak output voltage in  V\n",
-      "Vav=2*Vout_pk/pi;          #Average output voltage in V\n",
-      "Vav=round(Vav,2);\n",
-      "\n",
-      "# 2:\n",
-      "Iav=Vav/RL;                #Average output current in A\n",
-      "Iav=Iav*pow(10,6);         #Average output current in \u03bcA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'i:';\n",
-      "print '   Average output voltage=%.2f V'%Vav;\n",
-      "print 'ii:';\n",
-      "print '   Average output current=%.1f \u03bcA'%Iav;\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i:\n",
-        "   Average output voltage=9.91 V\n",
-        "ii:\n",
-        "   Average output current=825.8 \u03bcA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.22, Page number 102"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vdc_A=10;                      #Supply voltage of A in V\n",
-      "Vdc_B=25;                      #Supply voltage of B in V\n",
-      "Vac_rms_a=0.5;                 #Ripples in power supply A in V\n",
-      "Vac_rms_b=0.001;               #Ripples in power supply B in V\n",
-      "\n",
-      "#Calculation\n",
-      "#For power supply A\n",
-      "ripple_factor_A=Vac_rms_a/Vdc_A;      #Ripple factor of power supply A\n",
-      "\n",
-      "#For power supply B\n",
-      "ripple_factor_B=Vac_rms_b/Vdc_B;      #Ripple factor of power supply B\n",
-      "\n",
-      "#Result\n",
-      "if(ripple_factor_A<ripple_factor_B):\n",
-      "    print 'Power supply A is better';\n",
-      "else :\n",
-      "    print 'Power supply B is better';"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supply B is better\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.23, Page number 105-106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "#Variable declaration\n",
-      "RL=2200;                    #Load resistance in ohm\n",
-      "C=50*pow(10,-6);            #Capacitance of the capacitor used in filter circuit in F\n",
-      "V0=0.7;                     #Potential barrier voltage of the diodes of the rectifier in  V\n",
-      "Vrms=115.0;                   #R.M.S value of input a.c voltage in V \n",
-      "fin=60;                     #Frequency of input a.c voltage in Hz\n",
-      "Turns_ratio=10/1;           #Primary to secondary, turns ratio of the transformer \n",
-      "\n",
-      "#Calculation\n",
-      "Vp_prim=Vrms*sqrt(2);               #Peak primary voltage in V\n",
-      "Vp_sec=Vp_prim/Turns_ratio;         #Peak secondary voltage in V\n",
-      "Vp_in= Vp_sec - 2*V0;               #Peak full wave rectified voltage at the filter input in V\n",
-      "f=2*fin;                            #Output frequency in Hz\n",
-      "Vdc=Vp_in*(1-(1/(2*f*RL*C)));       #Output d.c voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print 'The output d.c voltage  is = %.1f V'%Vdc;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output d.c voltage  is = 14.3 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.24, Page number 106"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "#Variable declaration\n",
-      "R=25;                        #d.c resistance of the choke in ohm\n",
-      "RL=750;                      #Load resistance in ohm\n",
-      "Vm=25.7;                     #Maximum value of the pulsating output from the rectifier in V\n",
-      "\n",
-      "#Calculation\n",
-      "V_dc=2*Vm/pi;                #d.c component of the pulsating output in V\n",
-      "V_dc=round(V_dc,1);\n",
-      "V_dc_out=(V_dc*RL)/(R+RL);   #Output d.c voltage in V\n",
-      "V_dc_out=round(V_dc_out,1);\n",
-      "\n",
-      "#Result\n",
-      "print ' The output d.c voltage accross the load resistance is = %.1f V'%V_dc_out;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The output d.c voltage accross the load resistance is = 15.9 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.25, Page number 113-114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=120.0;                   #Input Voltage in V\n",
-      "Vz=50.0;                    #Zener Voltage in V\n",
-      "R=5000.0;                   #Resistance of the series resistor in ohm\n",
-      "RL=10000.0;                 #Load resistance in ohm\n",
-      "\n",
-      "#Calculation\n",
-      "V=Ei*RL/(R+RL);             #Voltage across the open circuit if the zener diode is removed\n",
-      "if(V>Vz):\n",
-      "    #Zener diode is in ON state\n",
-      "    # i:\n",
-      "    Output_voltage=Vz;             #Voltage across load resistance, in V\n",
-      "    #ii:\n",
-      "    Voltage_R=Ei-Vz;               #Voltage across the series resistance R, in V\n",
-      "    #iii:\n",
-      "    IL=Vz/RL;                      #Load current through RL in A\n",
-      "    IL=IL*1000;                    #Load current through RL in mA\n",
-      "    I=Voltage_R/R;                 #Current through the series resistance in A\n",
-      "    I=I*1000;                      #Current through the series resistance in mA\n",
-      "    Iz=I-IL;                       #Applying Kirchhoff's first law, Zener current in mA\n",
-      "    \n",
-      "    #Result\n",
-      "    print 'i)   The output voltage across the load resistance RL = %d V'%Output_voltage;\n",
-      "    print 'ii)  The voltage drop across the series resistance  R = %d V'%Voltage_R;\n",
-      "    print 'iii) The current through the zener diode = %d mA'%Iz;\n",
-      "    "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i)   The output voltage across the load resistance RL = 50 V\n",
-        "ii)  The voltage drop across the series resistance  R = 70 V\n",
-        "iii) The current through the zener diode = 9 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.26, Page number 114-115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Max_V=120.0;                  #Maximum input voltage in V\n",
-      "Min_V=80.0;                   #Minimum input voltage in V\n",
-      "R=5000.0;                     #Series resistance in ohm\n",
-      "RL=10000.0;                   #Load resistance in ohm\n",
-      "Vz=50.0;                      #Zener voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i: Maximum zener current\n",
-      "#Zener current will be maximum when the input voltage is maximum\n",
-      "V_R_max=Max_V-Vz;                #Voltage across series resistance R, in  V\n",
-      "I_max=V_R_max/R;                 #Current through series resistance R, in A\n",
-      "I_max=I_max*1000;                #Current through series resistance R, in mA\n",
-      "IL_max=Vz/RL;                        #Load current in A\n",
-      "IL_max=IL_max*1000;                      #Load current in mA\n",
-      "Iz_max=I_max-IL_max;                     #Applying Kirchhoff's first law, Zener current in mA;\n",
-      "\n",
-      "#Case ii: Minimum zener current\n",
-      "#The zener will conduct minimum current when the input voltage is minimum\n",
-      "V_R_min=Min_V-Vz;                #Voltage across series resistance R, in V\n",
-      "I_min=V_R_min/R;                 #Current through series resistance R, in A\n",
-      "I_min=I_min*1000;                #Current through series resistance R, in mA\n",
-      "IL_min=Vz/RL;                        #Load current in A\n",
-      "IL_min=IL_min*1000;                      #Load current in mA\n",
-      "Iz_min=I_min-IL_min;                     #Applying Kirchhoff's first law, Zener current in mA\n",
-      "\n",
-      "#Result\n",
-      "print 'Case i: ';\n",
-      "print 'Maximum zener current = %d mA'%Iz_max;\n",
-      "print 'Case ii: ';\n",
-      "print 'Minimum zener current = %d mA'%Iz_min;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case i: \n",
-        "Maximum zener current = 9 mA\n",
-        "Case ii: \n",
-        "Minimum zener current = 1 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.27, Page number 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=12;                      #Input voltage in V\n",
-      "Vz=7.2;                     #Zener voltage in V\n",
-      "E0=Vz;                      #Voltage to be maintained across the load in  V\n",
-      "IL_max=0.1;                 #Maximum load current in A\n",
-      "IL_min=0.012;                  #Minimum load current in A\n",
-      "Iz_min=0.01;                  #Minimum zener current in A\n",
-      "\n",
-      "#Calculation\n",
-      "#When the load current is maximum at minimum value of RL, the zener current is minimum and, as the load current decreases due to increase in value of RL\n",
-      "R=(Ei-E0)/(Iz_min+IL_max);        #The value of series resistance R to maintain a voltage=E0 across load, in ohm\n",
-      "\n",
-      "#Result\n",
-      "print 'The minimum value of series resistance R to maintain a constant value of 7.2 V is = %.1f \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum value of series resistance R to maintain a constant value of 7.2 V is = 43.6 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "NOTE: The actual value of R is 43.636363 (recurring) but, in the textbook the value of R is wrongly approximated 43.5 \u03a9"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.28, Page number 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei_min=22;                #Minimum input voltage in V\n",
-      "Ei_max=28;                #Maximum input voltage in V\n",
-      "Vz=18;                   #Zener voltage in V\n",
-      "E0=Vz;                    #Constant voltage maintained across the load resistance in V\n",
-      "Iz_min=0.2;               #Minimum zener current in A\n",
-      "Iz_max=2;                 #Maximum zener current in A\n",
-      "RL=18;                    #Load resistance in \u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "IL=Vz/RL;                      #Constant value of load current in A\n",
-      "#When the input voltage is minimum, the zener current will be minimum\n",
-      "R=(Ei_min-E0)/(Iz_min+IL)      #The value of series resistance so that the voltage E0 across RL remains constant\n",
-      "\n",
-      "print 'The value of series resistance R, to maintain constant voltage E0 across RL = %.2f \u03a9.'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistance R, to maintain constant voltage E0 across RL = 3.33 \u03a9.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.29, Page number 116 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10                                #Zener voltage in V\n",
-      "Ei_min=13;                           #Minimum input voltage in V\n",
-      "Ei_max=16;                           #Maximum input voltage in V\n",
-      "Iz_min=0.015;                        #Minimum zener current in A\n",
-      "IL_min=0.01;                         #Minimum load current in A \n",
-      "IL_max=0.085;                        #Maximum load curremt in A\n",
-      "E0=Vz;                               #Constant voltage to be maintained in V \n",
-      "\n",
-      "#Calculation\n",
-      "#The zener current will be minimum when the input voltage will be minimum  and at that time the load current will be maximum\n",
-      "R=(Ei_min-E0)/(Iz_min+IL_max);       #The value of series resistance R to maintain a constant voltage across load\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'The value of series resistance to maintain a constant voltage across the load resistance is = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistance to maintain a constant voltage across the load resistance is = 30 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.30, Page number 116"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Iz=0.2;               #Current rating of each zener in A\n",
-      "Vz=15;                #Voltage rating of each zener in V\n",
-      "Ei=45;                #Input voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "# i: Regulated output voltage across the two zener diodes \n",
-      "E0=2*Vz;                     # V\n",
-      "\n",
-      "# ii: Value of series resistance \n",
-      "R=(Ei-E0)/Iz;                  # \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print 'i) The regulated output voltage = %d V'%E0;\n",
-      "print 'ii) The value of the series resistance = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "i) The regulated output voltage = 30 V\n",
-        "ii) The value of the series resistance = 75 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.31, Page number 116-117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Vz=10;                   #Voltage rating of each zener in V\n",
-      "Iz=1;                    #Current rating of each zener in A\n",
-      "Ei=45;                   #Input unregulated voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "#Regulated output voltage across the three zener diodes\n",
-      "E0=3*Vz;               # V\n",
-      "\n",
-      "#Value of series resistance to obtain a 30V regulated output voltage\n",
-      "R=(Ei-E0)/Iz;          # \u03a9\n",
-      "\n",
-      "#Result\n",
-      "print 'Value of series resistance to obtain a 30V regulated output voltage = %d \u03a9'%R;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of series resistance to obtain a 30V regulated output voltage = 15 \u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.32, Page number 117"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "RL=2000.0;                   #Load resistance in \u03a9\n",
-      "R=200.0;                     #Series resistance in \u03a9\n",
-      "Iz=0.025;                  #Zener current rating in A\n",
-      "E0=30.0;                     #Output regulated voltage in V \n",
-      "\n",
-      "#Calculation\n",
-      "#Minimum input voltage will be required when Iz=0 A, and at  this condition\n",
-      "IL=E0/RL;                   #Load current during Iz=0, in A\n",
-      "I=IL;                       #According to Kirchhoff's law, total current, in A\n",
-      "Ei_min=E0+(I*R);            #Minimum input voltage in V\n",
-      "\n",
-      "#The maximum input voltage required will be when Iz=0.025 A, and at that condition \n",
-      "I=IL+Iz;                    #According to Kirchhoff's law, total current, in A\n",
-      "Ei_max=E0+(I*R);            #maximum input voltage in V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print 'The required range of input voltage is from %d V to %d V'%(Ei_min,Ei_max);  \n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required range of input voltage is from 33 V to 38 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.33, Page number 117-118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ei=16;                   #Unregulated input voltage in V\n",
-      "E0=12;                   #Output regulated voltage in V\n",
-      "IL_min=0;                #Minimum load current in A\n",
-      "IL_max=0.2;              #Maximum load current in A\n",
-      "Iz_min=0;                #Minimum zener current in A\n",
-      "Iz_max=0.2;              #Maximum zener current in A\n",
-      "\n",
-      "#Calculation\n",
-      "#As the regulated voltage required across the load is 12V\n",
-      "Vz=E0;                   #Voltage rating of zener diode in V\n",
-      "V_R=Ei-E0;               #Constant Voltage that should remain across series resistance \n",
-      "#The minimum zener current will occur when the curent in the load in maximum\n",
-      "R=V_R/(Iz_min+IL_max);   #Series resistance in \u03a9\n",
-      "\n",
-      "Max_power_rating=Vz*Iz_max;     #Maximum power rating of zener diode in W\n",
-      "\n",
-      "#Result\n",
-      "print 'The regulator is designed using a Seris resistance of %d \u03a9 and a zener diode of zener voltage %d V'%(R,Vz);\n",
-      "print 'The maximum power rating of the zener diode is = %.1f W '%Max_power_rating;"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The regulator is designed using a Seris resistance of 20 \u03a9 and a zener diode of zener voltage 12 V\n",
-        "The maximum power rating of the zener diode is = 2.4 W \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 6.34, Page number 118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V=12;                       #Source voltage in V\n",
-      "R=1000;                     #Series resistance in \u03a9\n",
-      "RL=5000;                    #Load resistance in \u03a9\n",
-      "Vz=6;                       #Voltage rating of zener in V\n",
-      "\n",
-      "#Calculation\n",
-      "#Case i: zener is working properly\n",
-      "#The output voltage across the load will be equal to the zener voltage.\n",
-      "V0=Vz;                        # V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case i: Output voltage when zener is working properly is %d V'%V0;\n",
-      "\n",
-      "#Case ii: zener is shorted\n",
-      "#As the zener is shorted, the potential difference across the load will be zero\n",
-      "V0=0;                        #V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case ii: Output voltage when zener is short circuited is %d V'%V0;\n",
-      "     \n",
-      "#Case iii: zener is open circuited\n",
-      "#If the zener is open circuited, the total voltage will drop across R and RL according to the voltage divider rule\n",
-      "V0=V*RL/(R+RL);                      #V\n",
-      "\n",
-      "#Result\n",
-      "print 'Case iii: Output voltage when zener is open circuited is %d V'%V0;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Case i: Output voltage when zener is working properly is 6 V\n",
-        "Case ii: Output voltage when zener is short circuited is 0 V\n",
-        "Case iii: Output voltage when zener is open circuited is 10 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7.ipynb
deleted file mode 100755
index 537a179e..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7.ipynb
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e210474f5c4fc6668f4c7b5af2adf833a1c7f62577017a980ab8d11cd8ce2886"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 7 : SPECIAL-PURPOSE DIODES"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1 : Page number 127-128\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_S=10.0;                        #Supply voltage in V\n",
-      "V_D=1.6;                         #Forward voltage drop of LED, in V\n",
-      "I_F=20.0;                          #Required limited current through LED, in mA\n",
-      "\n",
-      "#Calculations\n",
-      "R_S=(V_S-V_D)/(I_F/1000);               #Series resistor required to limit the current through the LED, in \u2126\n",
-      "\n",
-      "#Result \n",
-      "print(\"The value of series resistor required to limit the current through the LED = %d \u2126.\"%R_S);\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistor required to limit the current through the LED = 420 \u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2: Page number 128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_S=15.0;                        #Supply voltage in V\n",
-      "V_D=2.0;                         #Forward voltage drop of LED, in V\n",
-      "R_S=2200.0;                         #Series resistor required to limit the current through the LED, in \u2126\n",
-      "\n",
-      "#Calculations\n",
-      "I_F=((V_S-V_D)/R_S)*1000;                          #Required limited current through LED, in mA\n",
-      "\n",
-      "#Result \n",
-      "print(\"The current through the LED in the circuit = %.2f mA\"%I_F);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the LED in the circuit = 5.91 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3: Page number 132-133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ir=50.0;                                 #Dark current as observed from the current Illumination curve, in mA \n",
-      "V_R=10.0;                                #Reverse voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "R_R=V_R/(Ir/pow(10,6));                    #Dark Resistance in \u2126\n",
-      "\n",
-      "#Result\n",
-      "print(\"The dark resistance is=%d k\u2126\"%(R_R/1000));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dark resistance is=200 k\u2126\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4: Page number 133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "E=2.5;                      #Illumination in mW/cm\u00b2\n",
-      "m=37.4;                     #sensitivity of the photodiode in \ud835\udf07A/mW/cm\u00b2\n",
-      "\n",
-      "#Calculations\n",
-      "I_R=m*E;                     #Reverse current in \ud835\udf07A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The reverese current in the photodiode = %.1f  \ud835\udf07A\"%I_R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The reverese current in the photodiode = 93.5  \ud835\udf07A\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5: Page number 137"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\t\n",
-      "#Variable declaration\n",
-      "L=1.0;                        #Inductance of the inductor in mH\n",
-      "C=100.0;                      #Capacitance of the varactor in pF\n",
-      "\n",
-      "#Result\n",
-      "f_r=1/(2*pi*sqrt(L*pow(10,-3)*C*pow(10,-12)));        #Resonant frequency of the circuit in Hz\n",
-      "f_r=f_r/1000;                                         #Resonant frequency of the circuit in kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The resonant frequency of the circuit = %.1f kHz\"%f_r);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resonant frequency of the circuit = 503.3 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_1.ipynb
deleted file mode 100755
index 537a179e..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_1.ipynb
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e210474f5c4fc6668f4c7b5af2adf833a1c7f62577017a980ab8d11cd8ce2886"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 7 : SPECIAL-PURPOSE DIODES"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1 : Page number 127-128\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_S=10.0;                        #Supply voltage in V\n",
-      "V_D=1.6;                         #Forward voltage drop of LED, in V\n",
-      "I_F=20.0;                          #Required limited current through LED, in mA\n",
-      "\n",
-      "#Calculations\n",
-      "R_S=(V_S-V_D)/(I_F/1000);               #Series resistor required to limit the current through the LED, in \u2126\n",
-      "\n",
-      "#Result \n",
-      "print(\"The value of series resistor required to limit the current through the LED = %d \u2126.\"%R_S);\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistor required to limit the current through the LED = 420 \u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2: Page number 128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_S=15.0;                        #Supply voltage in V\n",
-      "V_D=2.0;                         #Forward voltage drop of LED, in V\n",
-      "R_S=2200.0;                         #Series resistor required to limit the current through the LED, in \u2126\n",
-      "\n",
-      "#Calculations\n",
-      "I_F=((V_S-V_D)/R_S)*1000;                          #Required limited current through LED, in mA\n",
-      "\n",
-      "#Result \n",
-      "print(\"The current through the LED in the circuit = %.2f mA\"%I_F);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the LED in the circuit = 5.91 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3: Page number 132-133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ir=50.0;                                 #Dark current as observed from the current Illumination curve, in mA \n",
-      "V_R=10.0;                                #Reverse voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "R_R=V_R/(Ir/pow(10,6));                    #Dark Resistance in \u2126\n",
-      "\n",
-      "#Result\n",
-      "print(\"The dark resistance is=%d k\u2126\"%(R_R/1000));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dark resistance is=200 k\u2126\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4: Page number 133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "E=2.5;                      #Illumination in mW/cm\u00b2\n",
-      "m=37.4;                     #sensitivity of the photodiode in \ud835\udf07A/mW/cm\u00b2\n",
-      "\n",
-      "#Calculations\n",
-      "I_R=m*E;                     #Reverse current in \ud835\udf07A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The reverese current in the photodiode = %.1f  \ud835\udf07A\"%I_R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The reverese current in the photodiode = 93.5  \ud835\udf07A\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5: Page number 137"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\t\n",
-      "#Variable declaration\n",
-      "L=1.0;                        #Inductance of the inductor in mH\n",
-      "C=100.0;                      #Capacitance of the varactor in pF\n",
-      "\n",
-      "#Result\n",
-      "f_r=1/(2*pi*sqrt(L*pow(10,-3)*C*pow(10,-12)));        #Resonant frequency of the circuit in Hz\n",
-      "f_r=f_r/1000;                                         #Resonant frequency of the circuit in kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The resonant frequency of the circuit = %.1f kHz\"%f_r);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resonant frequency of the circuit = 503.3 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_2.ipynb
deleted file mode 100755
index 537a179e..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_2.ipynb
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e210474f5c4fc6668f4c7b5af2adf833a1c7f62577017a980ab8d11cd8ce2886"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 7 : SPECIAL-PURPOSE DIODES"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1 : Page number 127-128\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_S=10.0;                        #Supply voltage in V\n",
-      "V_D=1.6;                         #Forward voltage drop of LED, in V\n",
-      "I_F=20.0;                          #Required limited current through LED, in mA\n",
-      "\n",
-      "#Calculations\n",
-      "R_S=(V_S-V_D)/(I_F/1000);               #Series resistor required to limit the current through the LED, in \u2126\n",
-      "\n",
-      "#Result \n",
-      "print(\"The value of series resistor required to limit the current through the LED = %d \u2126.\"%R_S);\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistor required to limit the current through the LED = 420 \u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2: Page number 128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_S=15.0;                        #Supply voltage in V\n",
-      "V_D=2.0;                         #Forward voltage drop of LED, in V\n",
-      "R_S=2200.0;                         #Series resistor required to limit the current through the LED, in \u2126\n",
-      "\n",
-      "#Calculations\n",
-      "I_F=((V_S-V_D)/R_S)*1000;                          #Required limited current through LED, in mA\n",
-      "\n",
-      "#Result \n",
-      "print(\"The current through the LED in the circuit = %.2f mA\"%I_F);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the LED in the circuit = 5.91 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3: Page number 132-133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ir=50.0;                                 #Dark current as observed from the current Illumination curve, in mA \n",
-      "V_R=10.0;                                #Reverse voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "R_R=V_R/(Ir/pow(10,6));                    #Dark Resistance in \u2126\n",
-      "\n",
-      "#Result\n",
-      "print(\"The dark resistance is=%d k\u2126\"%(R_R/1000));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dark resistance is=200 k\u2126\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4: Page number 133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "E=2.5;                      #Illumination in mW/cm\u00b2\n",
-      "m=37.4;                     #sensitivity of the photodiode in \ud835\udf07A/mW/cm\u00b2\n",
-      "\n",
-      "#Calculations\n",
-      "I_R=m*E;                     #Reverse current in \ud835\udf07A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The reverese current in the photodiode = %.1f  \ud835\udf07A\"%I_R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The reverese current in the photodiode = 93.5  \ud835\udf07A\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5: Page number 137"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\t\n",
-      "#Variable declaration\n",
-      "L=1.0;                        #Inductance of the inductor in mH\n",
-      "C=100.0;                      #Capacitance of the varactor in pF\n",
-      "\n",
-      "#Result\n",
-      "f_r=1/(2*pi*sqrt(L*pow(10,-3)*C*pow(10,-12)));        #Resonant frequency of the circuit in Hz\n",
-      "f_r=f_r/1000;                                         #Resonant frequency of the circuit in kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The resonant frequency of the circuit = %.1f kHz\"%f_r);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resonant frequency of the circuit = 503.3 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_3.ipynb
deleted file mode 100755
index 537a179e..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_3.ipynb
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e210474f5c4fc6668f4c7b5af2adf833a1c7f62577017a980ab8d11cd8ce2886"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 7 : SPECIAL-PURPOSE DIODES"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1 : Page number 127-128\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_S=10.0;                        #Supply voltage in V\n",
-      "V_D=1.6;                         #Forward voltage drop of LED, in V\n",
-      "I_F=20.0;                          #Required limited current through LED, in mA\n",
-      "\n",
-      "#Calculations\n",
-      "R_S=(V_S-V_D)/(I_F/1000);               #Series resistor required to limit the current through the LED, in \u2126\n",
-      "\n",
-      "#Result \n",
-      "print(\"The value of series resistor required to limit the current through the LED = %d \u2126.\"%R_S);\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistor required to limit the current through the LED = 420 \u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2: Page number 128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_S=15.0;                        #Supply voltage in V\n",
-      "V_D=2.0;                         #Forward voltage drop of LED, in V\n",
-      "R_S=2200.0;                         #Series resistor required to limit the current through the LED, in \u2126\n",
-      "\n",
-      "#Calculations\n",
-      "I_F=((V_S-V_D)/R_S)*1000;                          #Required limited current through LED, in mA\n",
-      "\n",
-      "#Result \n",
-      "print(\"The current through the LED in the circuit = %.2f mA\"%I_F);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the LED in the circuit = 5.91 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3: Page number 132-133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ir=50.0;                                 #Dark current as observed from the current Illumination curve, in mA \n",
-      "V_R=10.0;                                #Reverse voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "R_R=V_R/(Ir/pow(10,6));                    #Dark Resistance in \u2126\n",
-      "\n",
-      "#Result\n",
-      "print(\"The dark resistance is=%d k\u2126\"%(R_R/1000));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dark resistance is=200 k\u2126\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4: Page number 133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "E=2.5;                      #Illumination in mW/cm\u00b2\n",
-      "m=37.4;                     #sensitivity of the photodiode in \ud835\udf07A/mW/cm\u00b2\n",
-      "\n",
-      "#Calculations\n",
-      "I_R=m*E;                     #Reverse current in \ud835\udf07A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The reverese current in the photodiode = %.1f  \ud835\udf07A\"%I_R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The reverese current in the photodiode = 93.5  \ud835\udf07A\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5: Page number 137"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\t\n",
-      "#Variable declaration\n",
-      "L=1.0;                        #Inductance of the inductor in mH\n",
-      "C=100.0;                      #Capacitance of the varactor in pF\n",
-      "\n",
-      "#Result\n",
-      "f_r=1/(2*pi*sqrt(L*pow(10,-3)*C*pow(10,-12)));        #Resonant frequency of the circuit in Hz\n",
-      "f_r=f_r/1000;                                         #Resonant frequency of the circuit in kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The resonant frequency of the circuit = %.1f kHz\"%f_r);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resonant frequency of the circuit = 503.3 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_4.ipynb
deleted file mode 100755
index 537a179e..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_4.ipynb
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e210474f5c4fc6668f4c7b5af2adf833a1c7f62577017a980ab8d11cd8ce2886"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 7 : SPECIAL-PURPOSE DIODES"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1 : Page number 127-128\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_S=10.0;                        #Supply voltage in V\n",
-      "V_D=1.6;                         #Forward voltage drop of LED, in V\n",
-      "I_F=20.0;                          #Required limited current through LED, in mA\n",
-      "\n",
-      "#Calculations\n",
-      "R_S=(V_S-V_D)/(I_F/1000);               #Series resistor required to limit the current through the LED, in \u2126\n",
-      "\n",
-      "#Result \n",
-      "print(\"The value of series resistor required to limit the current through the LED = %d \u2126.\"%R_S);\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistor required to limit the current through the LED = 420 \u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2: Page number 128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_S=15.0;                        #Supply voltage in V\n",
-      "V_D=2.0;                         #Forward voltage drop of LED, in V\n",
-      "R_S=2200.0;                         #Series resistor required to limit the current through the LED, in \u2126\n",
-      "\n",
-      "#Calculations\n",
-      "I_F=((V_S-V_D)/R_S)*1000;                          #Required limited current through LED, in mA\n",
-      "\n",
-      "#Result \n",
-      "print(\"The current through the LED in the circuit = %.2f mA\"%I_F);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the LED in the circuit = 5.91 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3: Page number 132-133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ir=50.0;                                 #Dark current as observed from the current Illumination curve, in mA \n",
-      "V_R=10.0;                                #Reverse voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "R_R=V_R/(Ir/pow(10,6));                    #Dark Resistance in \u2126\n",
-      "\n",
-      "#Result\n",
-      "print(\"The dark resistance is=%d k\u2126\"%(R_R/1000));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dark resistance is=200 k\u2126\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4: Page number 133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "E=2.5;                      #Illumination in mW/cm\u00b2\n",
-      "m=37.4;                     #sensitivity of the photodiode in \ud835\udf07A/mW/cm\u00b2\n",
-      "\n",
-      "#Calculations\n",
-      "I_R=m*E;                     #Reverse current in \ud835\udf07A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The reverese current in the photodiode = %.1f  \ud835\udf07A\"%I_R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The reverese current in the photodiode = 93.5  \ud835\udf07A\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5: Page number 137"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\t\n",
-      "#Variable declaration\n",
-      "L=1.0;                        #Inductance of the inductor in mH\n",
-      "C=100.0;                      #Capacitance of the varactor in pF\n",
-      "\n",
-      "#Result\n",
-      "f_r=1/(2*pi*sqrt(L*pow(10,-3)*C*pow(10,-12)));        #Resonant frequency of the circuit in Hz\n",
-      "f_r=f_r/1000;                                         #Resonant frequency of the circuit in kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The resonant frequency of the circuit = %.1f kHz\"%f_r);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resonant frequency of the circuit = 503.3 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_5.ipynb
deleted file mode 100755
index 537a179e..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_5.ipynb
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:e210474f5c4fc6668f4c7b5af2adf833a1c7f62577017a980ab8d11cd8ce2886"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 7 : SPECIAL-PURPOSE DIODES"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.1 : Page number 127-128\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_S=10.0;                        #Supply voltage in V\n",
-      "V_D=1.6;                         #Forward voltage drop of LED, in V\n",
-      "I_F=20.0;                          #Required limited current through LED, in mA\n",
-      "\n",
-      "#Calculations\n",
-      "R_S=(V_S-V_D)/(I_F/1000);               #Series resistor required to limit the current through the LED, in \u2126\n",
-      "\n",
-      "#Result \n",
-      "print(\"The value of series resistor required to limit the current through the LED = %d \u2126.\"%R_S);\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of series resistor required to limit the current through the LED = 420 \u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.2: Page number 128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_S=15.0;                        #Supply voltage in V\n",
-      "V_D=2.0;                         #Forward voltage drop of LED, in V\n",
-      "R_S=2200.0;                         #Series resistor required to limit the current through the LED, in \u2126\n",
-      "\n",
-      "#Calculations\n",
-      "I_F=((V_S-V_D)/R_S)*1000;                          #Required limited current through LED, in mA\n",
-      "\n",
-      "#Result \n",
-      "print(\"The current through the LED in the circuit = %.2f mA\"%I_F);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current through the LED in the circuit = 5.91 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.3: Page number 132-133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Ir=50.0;                                 #Dark current as observed from the current Illumination curve, in mA \n",
-      "V_R=10.0;                                #Reverse voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "R_R=V_R/(Ir/pow(10,6));                    #Dark Resistance in \u2126\n",
-      "\n",
-      "#Result\n",
-      "print(\"The dark resistance is=%d k\u2126\"%(R_R/1000));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The dark resistance is=200 k\u2126\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.4: Page number 133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "E=2.5;                      #Illumination in mW/cm\u00b2\n",
-      "m=37.4;                     #sensitivity of the photodiode in \ud835\udf07A/mW/cm\u00b2\n",
-      "\n",
-      "#Calculations\n",
-      "I_R=m*E;                     #Reverse current in \ud835\udf07A\n",
-      "\n",
-      "#Result\n",
-      "print(\"The reverese current in the photodiode = %.1f  \ud835\udf07A\"%I_R);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The reverese current in the photodiode = 93.5  \ud835\udf07A\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 7.5: Page number 137"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import pi\n",
-      "from math import sqrt\t\n",
-      "#Variable declaration\n",
-      "L=1.0;                        #Inductance of the inductor in mH\n",
-      "C=100.0;                      #Capacitance of the varactor in pF\n",
-      "\n",
-      "#Result\n",
-      "f_r=1/(2*pi*sqrt(L*pow(10,-3)*C*pow(10,-12)));        #Resonant frequency of the circuit in Hz\n",
-      "f_r=f_r/1000;                                         #Resonant frequency of the circuit in kHz\n",
-      "\n",
-      "#Result\n",
-      "print(\"The resonant frequency of the circuit = %.1f kHz\"%f_r);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resonant frequency of the circuit = 503.3 kHz\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8.ipynb
deleted file mode 100755
index 4afe0858..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8.ipynb
+++ /dev/null
@@ -1,1851 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:9c13bdd66a3dbb3eae04903205b69bc52bf35e6dadf8b1b3ade1bab68394ae3b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1: Page number 147-148\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Signal=500.0;                    #Signal voltage in V\n",
-      "Rin=20.0;                        #Input resistance in \u03a9 \n",
-      "Rout=100.0;                      #Output resistance in \u03a9\n",
-      "R_C=1000.0;                      #Collector load in \u03a9\n",
-      "alpha_ac=1.0;                    #current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_E=(Signal/1000)/Rin;        \t#Input current in mA\n",
-      "I_C=I_E*alpha_ac;               #Output current in mA\n",
-      "Vout=I_C*R_C;                   #Output voltage in V \n",
-      "Av=Vout/(Signal/1000);          #Voltage amplification \n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage amplification = %d. \"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage amplification = 50. \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.2: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_E=1;                            #Emitter curent in mA\n",
-      "I_C=0.95;                         #Collector current in mA\n",
-      "\n",
-      "#Calculation\n",
-      "I_B=I_E-I_C;                      #Base current in mA\n",
-      "\n",
-      "#Result \n",
-      "print(\"The base current = %.2f mA \"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current = 0.05 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 8.3: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "alpha=0.9;                      #Current amplification factor\n",
-      "I_E=1;                          #Emitter current in mA\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=alpha*I_E;                  #Collector current in mA\n",
-      "I_B=I_E-I_C;                    #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The base current =%.1f mA\"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current =0.1 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_C=0.95;\t\t\t#Collector current in mA\n",
-      "I_B=0.05;\t\t\t#Base current in mA\n",
-      "\n",
-      "#Calculation\n",
-      "I_E=I_B+I_C;                    #Emitter current in mA\n",
-      "alpha=I_C/I_E;                  #Current amplification factor \n",
-      "\n",
-      "#Result\n",
-      "print(\"The current amplification factor = %.2f .\"%alpha);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current amplification factor = 0.95 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.5: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_E=1;                  #Emitter current in mA\n",
-      "I_CBO=50.0;               #Collector current with emitter circuit open, in microAmp\n",
-      "alpha=0.92;             #Current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=alpha*I_E + (I_CBO/1000);           #Total collector current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total collector current = %.2f mA.\"%I_C);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total collector current = 0.97 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.6: Page number 150-151\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "alpha=0.95;               #Current amplification factor\n",
-      "Rc=2.0;                   #Resistor connected to the collector, in kilo ohm\n",
-      "V_Rc=2.0;                 #Voltage drop across the resistor connected to the collector in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=V_Rc/Rc;              #Collector current in mA\n",
-      "I_E=I_C/alpha;            #Emitter current in mA\n",
-      "I_B=I_E-I_C;              #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The base current = %.2f mA\"%I_B); \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current = 0.05 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.7: Page number 151\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_EE=8.0;                 #Supply voltage at the emitter in V\n",
-      "V_CC=18.0;                #Supply voltage at the collector in V\n",
-      "V_BE=0.7;                 #Base to emitter voltage in V\n",
-      "R_E=1.5;                  #Emitter resistance in \u03a9\n",
-      "R_C=1.2;                  #Collector resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "I_E=(V_EE-V_BE)/R_E;              #Emitter current in mA\n",
-      "I_C=I_E;                          #Collector current in mA (approximately equal to emitter current)\n",
-      "V_CB=V_CC-(I_C*R_C);              #Collector to base voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The collector current =%.2f mA\"%I_C);\n",
-      "print(\"The collector to base voltage = %.2f V\"%V_CB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The collector current =4.87 mA\n",
-        "The collector to base voltage = 12.16 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8:Page number 155\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating beta from alpha\n",
-      "def calc_beta(a):                   #a is the value of alpha\n",
-      "\treturn(a/(1-a));\n",
-      "\n",
-      "#Case (i)\n",
-      "alpha=0.9;                      #current amplification factor\n",
-      "beta=calc_beta(alpha);\t\t#Base current amplification factor    \n",
-      "print(\"(i) Value of beta =%d\"%beta );\t\t\t\t\t\t\t\t\t\n",
-      "\n",
-      "#Case (ii)\n",
-      "alpha=0.98;                     #current amplification factor\n",
-      "beta=calc_beta(alpha);          #Base current amplification factor\n",
-      "print(\"(ii) Value of beta =%.0f\"%beta );\n",
-      "\n",
-      "\n",
-      "#Case (iii)\n",
-      "alpha=0.99;                     #current amplification factor\n",
-      "beta=calc_beta(alpha);          #Base current amplification factor                      \n",
-      "print(\"(iii) Value of beta =%.0f\"%beta );\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Value of beta =9\n",
-        "(ii) Value of beta =49\n",
-        "(iii) Value of beta =99\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.9: Page number 155\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=50.0;                   #Base current amplification factor\n",
-      "I_B=20.0;                    #Base current in microAmp\n",
-      "\n",
-      "#Calculation\n",
-      "I_B=I_B/1000;               #Base current in mA\n",
-      "I_C=beta*I_B;               #Collector current in mA\n",
-      "I_E=I_B+I_C;                #Emitter current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The emitter curent = %.2f mA\"%I_E);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The emitter curent = 1.02 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.10: Page number 155\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_B=240.0;                    #Base current in microAmp\n",
-      "I_E=12;                       #Emitter current in mA\n",
-      "beta=49.0;                      #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "alpha=beta/(1+beta);              #current amplification factor \n",
-      "I_C_alpha=alpha*I_E;              #Collector current in mA calculated using alpha\n",
-      "I_C_beta=beta*(I_B/1000);         #Collector current in mA calculated using beta\n",
-      "\n",
-      "#Results\n",
-      "print(\"alpha=%.2f.\"%alpha);\n",
-      "print(\"Collector current determined using alpha =%.2f mA\"%I_C_alpha);\n",
-      "print(\"Collector current determined using beta =%.2f mA\"%I_C_beta);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "alpha=0.98.\n",
-        "Collector current determined using alpha =11.76 mA\n",
-        "Collector current determined using beta =11.76 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.11: Page number 156\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=45.0;                       #Base current amplification factor\n",
-      "R_C=1.0;                         #Resistance of the collector resistance in k\u03a9\n",
-      "V_R_C=1.0;                       #Voltage drop across the collector resistance in V\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=V_R_C/R_C;                   #Collector current in mA\n",
-      "I_B=I_C/beta;                           #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The base current =%.3f mA\"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current =0.022 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.12: Page number 156\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=8.0;                  #Collector supply voltage in V\n",
-      "R_C=800.0;                  #Resistance of the collector resistance in \u03a9\n",
-      "V_R_C=0.5;                 #Voltage drop across collector resistance in V\n",
-      "alpha=0.96;                #current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "V_CE=V_CC-V_R_C;           #Collector to emitter voltage in V\n",
-      "I_C=V_R_C/R_C;             #Collector current in A\n",
-      "I_C=I_C*1000;              #Collector current in mA\n",
-      "beta=alpha/(1-alpha);      #Base current amplification factor\n",
-      "I_B=I_C/beta;              #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Collector to emitter voltage = %.1f V\"%V_CE);\n",
-      "print(\"Base current= %.3f mA\"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector to emitter voltage = 7.5 V\n",
-        "Base current= 0.026 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.13: Page number 156-157\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=5;                  \t#Collector supply voltage in V\n",
-      "I_CBO=0.2;               \t#Leakage current at collector base junction with emitter open, in  \u03bcA\n",
-      "I_CEO=20.0;              \t#Leakage current with base open, in  \u03bcA\n",
-      "I_C=1.0;                        #Collector current in mA\n",
-      "I_C=I_C*1000;                  \t#Collector current in  \u03bcA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "alpha=1-(I_CBO/I_CEO);\t\t#current amplification factor\n",
-      "I_E=(I_C-I_CBO)/alpha;          #Emitter current in  \u03bcA\n",
-      "I_E=round(I_E,-1);\n",
-      "I_B=I_E-I_C;                    #Base current in  \u03bcA\n",
-      "I_B=round(I_B,-1);\n",
-      "\n",
-      "#Result\n",
-      "print(\"Current amplification factor = %.2f \"%alpha);\n",
-      "print(\"The emitter curent =%d  \u03bcA \"%I_E);\n",
-      "print(\"The base curent =%d  \u03bcA \"%I_B);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current amplification factor = 0.99 \n",
-        "The emitter curent =1010  \u03bcA \n",
-        "The base curent =10  \u03bcA \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.14: Page number 157\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_CEO=300.0;              #Leakage current in common emitter configuration, in  \u03bcA\n",
-      "beta=120.0;               #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "alpha=beta/(1+beta);               #Current amplification factor\n",
-      "alpha=round(alpha,3);\n",
-      "I_CBO=(1-alpha)*I_CEO;             #Leakage current in common base configuration, in  \u03bcA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Vale of I_CBO= %.1f \u03bcA\"%I_CBO);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vale of I_CBO= 2.4 \u03bcA\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.15: Page number 157\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_B=20.0;                       #Base current in \u03bcA\n",
-      "I_C=2.0;                        #Collector current in mA\n",
-      "beta=80.0;                      #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_CEO=I_C-(beta*I_B/1000);            #Leakage current with base open, in mA \n",
-      "alpha=beta/(beta+1);                  #Current amplification factor\n",
-      "alpha=round(alpha,3);\n",
-      "I_CBO=(1-alpha)*I_CEO;                #Leakage current with emitter open, in mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Value of I_CBO=%.4f mA\"%I_CBO);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of I_CBO=0.0048 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.17: Page number 158\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=150.0;               \t#Base current amplification factor\n",
-      "R_B=10.0;                   \t#Base resistance in kilo ohm\n",
-      "R_C=100.0;                   \t#Collector resistance in kilo ohm\n",
-      "V_CC=10.0;                      #Collector supply voltage in V\n",
-      "V_BB=5.0;                       #Base supply voltage in V\n",
-      "V_BE=0.7;                       #Base to emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_B=(V_BB-V_BE)/R_B;              #Base current in mA\n",
-      "I_C=beta*I_B;                     #Collector current in mA\n",
-      "V_CE=V_CC - (I_C/1000)*R_C;       #Collector to emitter voltage in V\n",
-      "V_CB=V_CE-V_BE;                   #Collector to base voltage in V\n",
-      "\n",
-      "\n",
-      "#Result \n",
-      "print(\"Collector to base voltage, V_CB= %.2f V\"%V_CB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector to base voltage, V_CB= 2.85 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.18: Page number158-159\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_B=68.0;              #Base current in \u03bcA\n",
-      "I_E=30.0;              #Emitter current in mA\n",
-      "beta=440.0;\t     #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "alpha=beta/(beta + 1);          #current amplification factor\n",
-      "I_C_alpha=alpha*I_E;\t\t#Collector current using alpha rating, in mA\n",
-      "I_C_beta=beta*(I_B/1000.0);       #Collector current using beta rating, in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Collector current determined using alpha rating =%.2f mA\"%I_C_alpha);\n",
-      "print(\"Collector current determined using beta rating =%.2f mA\"%I_C_beta);\n",
-      "\n",
-      "#Note: In the textbook, the collector current obtained from beta rating is approximated to 29.93 mA\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current determined using alpha rating =29.93 mA\n",
-        "Collector current determined using beta rating =29.92 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.19: Page number 159\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_C_max=500.0;                   #Maximum collector current in mA\n",
-      "beta_max=300.0;                  #Maximum base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_B_max=I_C_max/beta_max;           #Maximum base current in mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable value of base current = %.2f mA\"%I_B_max);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable value of base current = 1.67 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.22 : Page number 167-168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.5;                #Collector supply voltage, V\n",
-      "RC=2.5;                  #Collector resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC-VCE)/RC;           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,15])\n",
-      "limit.set_ylim([0,6])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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tROJz+mmhwcHBePbZZ3H48GHs2LEDHTt2lDOXZum919NC1673c6wVomUWLS+gwevUm9XX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\nrkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f2eadbe6710>"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.23 : Page number 168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                #Collector supply voltage, V\n",
-      "RC=6.0;                  #Collector resistor, k\u03a9\n",
-      "IB=20.0;                     #Zero signal base current,  \u03bcA\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC-VCE)/RC;           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,15])\n",
-      "limit.set_ylim([0,5])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n",
-      "#Calculating Q-point\n",
-      "IC=beta*(IB/1000);                                  #Collector current, mA\n",
-      "VCE=VCC-IC*RC;                               #Collector emitter voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Operating point: IC=%dmA and VCE=%dV.\"%(IC,VCE));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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EVut2p75u3TrcunXrkf3V1dVYv369fdIREZHdmB3qpaWlmDZt2iP7p06dim+++Ua2UFrG\nXk9+ouUFmFkJouUF1MlsdqjX1tZ2+lhzc7PdwxARkW3MdupxcXFYu3Ytnn766Xb78/LysGvXLhw5\ncsS2xdmpExFZrdufUVpSUoI5c+YgJiYG48aNgyRJOHfuHM6cOYNDhw7ZfAUMhzoRkfW6/UJpcHAw\nioqKMHXqVJSVleHKlSuYNm0aioqKnPaSRvZ68hMtL8DMShAtL6DB69QBoFevXkhNTVUiCxER2chs\n/dKvXz/odLqOv9EOH5LB+oWIyHrd7tRtZTAYsGzZMty4cQM6nQ6rVq3CT3/6U4uCERFRx2z+kIzu\ncnNzw/bt21FcXIzCwkK89957uHz5spxLyo69nvxEywswsxJEywto8Dp1Ww0ZMgSRkZEA7lc5ISEh\nqKqqknNJIiKnptj91MvLyzFt2jQUFxejX79+9xdn/UJEZDXV6pc2dXV1SExMREZGhmmgt0lIAPbt\nA+rqlEhCROTYLPo4O1s0NzcjISEBS5Yswfz58x95/Nq1FXjttUAsWwY88YQH5s+PxEsv6dGv3w99\nlF6vB6CN7YsXLyItLU0zeSzZbtunlTyOlvfBrFrJY8n2jh07EBkZqZk8jpa3wI7zoqCgAFlZWQCA\nwMBAmCXJqLW1VVq6dKmUlpbW4eMPLl9dLUmZmZI0e7YkubtLUny8JOXkSFJtrZwJrZefn692BKuJ\nllm0vJLEzEoQLa8kyZfZ3OiWtVP/6quvMHXqVIwePdp0vfumTZswe/ZsAJ33QrdvA59/DuzfD5w5\nAzz5JLBoEfD008BD7Q0RkdNR7Tr1rljyQikHPBFRe6q/UGqLgQOB1FTgyBGgrOz+MM/KAoYOVedF\n1ge7U1GIllm0vAAzK0G0vIADXqdub1ob8EREWqP5+sUSrGiIyJkI3albiwOeiByd0J26teSuaNjr\nyU+0vAAzK0G0vAA7dbtjB09Ezsbh6hdLsKIhIpE5VaduLQ54IhKNU3Xq1rK2omGvJz/R8gLMrATR\n8gLs1FVnyYD//nu1UxIRdc7p6xdLsKIhIi1hp25HHPBEpDZ26nZUVFQg3GWSonWRouUFmFkJouUF\n2KkLh9fBE5HWsH6RASsaIpITO3UVccATkb2xU7cjazsyLVQ0onWRouUFmFkJouUF2Kk7PC0MeCJy\nbKxfNIAVDRFZg526QDjgiagr7NTtSO6OTI6KRrQuUrS8ADMrQbS8ADt1egg7eCKyFusXAbGiIXJu\n7NQdGAc8kfNhp25HWuv1LKlojhwpUDumVbR2jC3BzPITLS/ATp1s1NmAT0xkB0/kLFi/OAFWNESO\nhZ06mXDAE4mPnbodid7riXCZpOjHWBSiZRYtL8BOnRQmwoAnIuuwfqFHsKIh0jZ26tRtHPBE2sNO\n3Y6crddTo6JxtmOsFtEyi5YXYKdOGscOnkj7WL+QzVjRECmLnTophgOeSH6qdeqpqanw9vZGRESE\nnMsoir2eefaoaHiMlSFaZtHyAg7YqaekpODo0aNyLkEaxg6eSHmy1y/l5eWYO3cuLl269OjirF+c\nEisaItvwkkbSFJ7BE8nHVe0AK1asQGBgIADAw8MDkZGR0Ov1AH7oo7S0ffHiRaSlpWkmjyXbbfu0\nkufh7dRUPVJTgYMHC/DVV8C2bcCqVXqMGVMAvR546SU9+vXTTt6Oth8+1mrnsWR7x44dmv99Ezmv\nPedFQUEBsrKyAMA0LzslyaysrEwKDw/v8DEFlre7/Px8tSNYTbTM+fn5UnW1JGVmStLs2ZLk7i5J\n8fGSlJMjSbW1aqfrmGjHWJLEyyxaXkmSL7O52clOnTSPHTxRe6p16snJyYiJiUFJSQn8/f3x4Ycf\nyrkcOSh28ESWk3WoZ2dno6qqCo2NjTAYDEhJSZFzOUU82J2KQrTM5vJqdcCLdowB8TKLlhdwwOvU\nieSk1QFPpCbeJoAcDjt4cnS89ws5LQ54ckR885EdsdeTnz3zKlXRiHaMAfEyi5YXYKdOJCt28OQM\nWL+Q02NFQ6Jhp05kIQ54EgE7dTtiryc/NfN2t6IR7RgD4mUWLS/ATp1IU9jBk4hYvxBZiRUNqY2d\nOpFMOOBJDezU7Yi9nvxEyttW0bz8coFwFY1IxxkQLy/ATp1IaOzgSQtYvxDJjBUN2Rs7dSKN4IAn\ne2Cnbkfs9eQnWl7A8sxaqmhEO86i5QXYqRM5FS0NeHIcrF+INIYVDXWFnTqRoDjgqSPs1O2IvZ78\nRMsLyJdZzopGtOMsWl6AnToRmcEOnizB+oVIcKxonA87dSInwQHvHNip2xF7PfmJlhfQTmZrKhqt\nZLaUaHkBdupEZEddDfgvv2QH74hYvxA5GVY04mOnTkQd4oAXEzt1O2KvJz/R8gLiZhbpMklRj7HS\nONSJCACvg3cUrF+IyCxWNNrDTp2I7IIDXhvYqdsRez35iZYXcJ7MalY0znKMbcWhTkTdwg5em1i/\nEJFdsaKRHzt1IlIFB7w8VOvUjx49ilGjRmHkyJHYvHmznEsphr2e/ETLCzBzZ+xZ0fAYW0a2oW40\nGvH888/j6NGj+Oc//4ns7GxcvnxZruUUc/HiRbUjWE20zKLlBZjZErYOeB5jy8g21M+ePYsRI0Yg\nMDAQbm5uSEpKQm5urlzLKebOnTtqR7CaaJlFywsws7W6M+B5jC0j21CvrKyEv7+/advPzw+VlZVy\nLUdEguJVNPYl21DX6XRyPbWqysvL1Y5gNdEyi5YXYGZ7MTfgP/mkXO14VlPjGMt29UthYSE2btyI\no0ePAgA2bdoEFxcXvPzyyz8s7qCDn4hIbopf0tjS0oInnngCJ0+ehK+vLyZMmIDs7GyEhITIsRwR\nEQFwle2JXV3x61//Gk899RSMRiOeffZZDnQiIpmp+uYjIiKyL9Xu/SLaG5MMBgNiY2MRFhaG8PBw\n7Ny5U+1IFjEajYiKisLcuXPVjmKRO3fuIDExESEhIQgNDUVhYaHakbq0fft2hIeHIyIiAosXL0Zj\nY6PakdpJTU2Ft7c3IiIiTPtu376NmTNnIjg4GLNmzdLc5YIdZf7FL36BkJAQjBkzBvHx8bh7966K\nCdvrKG+bbdu2wcXFBbdv31YkiypDXcQ3Jrm5uWH79u0oLi5GYWEh3nvvPc1nBoCMjAyEhoYK86L0\n+vXrERcXh8uXL6OoqEjzlV1lZSV27dqFc+fO4dKlSzAajcjJyVE7VjspKSmmCxbavP3225g5cyZK\nSkowY8YMvP322yql61hHmWfNmoXi4mJ88803CA4OxqZNm1RK96iO8gL3TwaPHz+OYcOGKZZFlaEu\n4huThgwZgsjISABAv379EBISgqqqKpVTmVdRUYG8vDysXLlSiHvs3L17F6dOnUJqaiqA+6/L9O/f\nX+VUXWtpaUF9fb3pf4cOHap2pHamTJmCAQMGtNt38OBBLF++HACwfPlyfP7552pE61RHmWfOnAkX\nl/sja+LEiaioqFAjWoc6ygsAL774It555x1Fs6gy1EV/Y1J5eTkuXLiAiRMnqh3FrBdeeAFbtmwx\n/SJoXVlZGTw9PZGSkoKxY8fiueeeQ319vdqxzBo6dCh+9rOfISAgAL6+vvDw8MCTTz6pdqwuXb9+\nHd7e3gAAb29vXL9+XeVE1tm9ezfi4uLUjmFWbm4u/Pz8MHr0aEXXVeW3XZQqoCN1dXVITExERkYG\n+mn4NnOHDh2Cl5cXoqKihDhLB+6f8Z4/fx5r1qzB+fPn0bdvX83VAg+rqanBwYMHUV5ejqqqKtTV\n1eGTTz5RO5ZVdDqdUL+Tb731Fnr27InFixerHaVT9fX1SE9Px69+9SvTPqV+D1UZ6kOHDoXBYDBt\nGwwG+Pn5qRHFKs3NzUhISMCSJUswf/58teOYdebMGRw8eBBBQUFITk7Gl19+iWXLlqkdyyw/Pz/4\n+flh/PjxAIDExEScP39e5VTmnThxAkFBQRg0aBBcXV0RHx+PM2fOqB2rS97e3rh27RoA4OrVq/Dy\n8lI5kWWysrKQl5en+T+c//nPf1BeXo4xY8YgKCgIFRUVGDduHG7cuCH72qoM9ejoaHz77bcoLy9H\nU1MT9u3bh3nz5qkRxWKSJOHZZ59FaGgo0tLS1I7TpfT0dBgMBpSVlSEnJwfTp0/Hnj171I5l1pAh\nQ+Dv74+SkhIA9wdmWFiYyqnMGzZsGAoLC/H9999DkiScOHECoaGhasfq0rx58/DRRx8BAD766CPN\nn6QA96+Y27JlC3Jzc9GrVy+145gVERGB69evo6ysDGVlZfDz88P58+eV+eMpqSQvL08KDg6Whg8f\nLqWnp6sVw2KnTp2SdDqdNGbMGCkyMlKKjIyUjhw5onYsixQUFEhz585VO4ZFLl68KEVHR0ujR4+W\nFixYIN25c0ftSF3asGGDNGrUKCk8PFxatmyZ1NTUpHakdpKSkiQfHx/Jzc1N8vPzk3bv3i1VV1dL\nM2bMkEaOHCnNnDlTqqmpUTtmOw9nzszMlEaMGCEFBASYfv9Wr16tdkyTtrw9e/Y0HeMHBQUFSdXV\n1Ypk4ZuPiIgciBiXRRARkUU41ImIHAiHOhGRA+FQJyJyIBzqREQOhEOdiMiBcKgTETkQDnVySNOn\nT8exY8fa7duxYwfWrFmDkpISxMXFITg4GOPGjcMzzzyDGzduoKCgAP3790dUVJTp38mTJwEA33//\nPfR6PVpbW/H444+b3vXaJi0tDe+88w7+8Y9/ICUlRbGfk+hhHOrkkJKTkx+5r/m+ffuQnJyMOXPm\nYO3atSgpKcG5c+ewZs0a3Lx5EzqdDlOnTsWFCxdM/2bMmAHg/l0BExIS4OLi8shzt7a24rPPPkNy\ncjLCw8NRUVHR7t5GREriUCeHlJCQgMOHD6OlpQUATHdR/PbbbxETE4Onn37a9LXTpk1DWFiY2bvo\n7d27Fz/+8Y8B3P+DsW/fPtNjf/3rXzFs2DDT7aTnzp2ruQ/KIOfBoU4OaeDAgZgwYQLy8vIAADk5\nOVi0aBGKi4sxduzYTr/v1KlT7eqXsrIyNDU14b///S8CAgIAAOHh4XBxcUFRUZHpuR+8DWx0dDRO\nnTol409H1DkOdXJYD9Yk+/bts+j+21OmTGlXvwQFBeHWrVvw8PDo8LmNRiNyc3OxcOFC02Oenp6a\n/1Qsclwc6uSw5s2bh5MnT+LChQuor69HVFQUwsLCcO7cOauep3fv3mhoaGi3LykpCZ9++ilOnDiB\n0aNHw9PT0/RYQ0MDevfubZefgchaHOrksPr164fY2FikpKSYztIXL16MM2fOmGoZ4H4nXlxc3Onz\nDBgwAEajEU1NTaZ9jz/+OAYPHoxXXnnlkf8CKCkpQXh4uJ1/GiLLcKiTQ0tOTsalS5eQnJwMAOjV\nqxcOHTqEXbt2ITg4GGFhYfjtb38LT09P6HS6Rzr1AwcOALj/SfYP9+TJycn497//jfj4+Hb78/Pz\nMWfOHGV+QKKH8H7qRBa4cOECtm/f3uWnRzU2NkKv1+P06dPCfOA3ORb+v47IAlFRUYiNjUVra6vZ\nrzMYDNi8eTMHOqmGZ+pERA6EpxNERA6EQ52IyIFwqBMRORAOdSIiB8KhTkTkQP4PTEiMVfQcO/gA\nAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8947e0f0d0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: IC=1mA and VCE=6V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.24 : Page number 168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=4.0;               #Collector load, k\u03a9\n",
-      "IC_Q=1.0;             #Quiescent current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "VCC=10;                  #Collector supply voltage, V\n",
-      "VCE=VCC-IC*RC;            #Collector emitter voltage, V\n",
-      "\n",
-      "print(\"(i) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n",
-      "\n",
-      "#(ii)\n",
-      "RC=5.0;               #Collector load, k\u03a9\n",
-      "VCE=VCC-IC*RC;            #Collector emitter voltage, V\n",
-      "print(\"(ii) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Operating point: VCE=6V and IC=1mA.\n",
-        "(ii) Operating point: VCE=5V and IC=1mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 8.25 : Page number 168-169"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                #Collector supply voltage, V\n",
-      "VBB=10.0;                  #Base supply voltage, V\n",
-      "RC=330.0;                  #Collector resistor, \u03a9\n",
-      "RB=47.0;                #Base resistoe, k\u03a9\n",
-      "beta=200.0;                 #Base current amplification factor\n",
-      "VBE=0.7;                      #Base -emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#VBB-IB*RB-VBE=0\n",
-      "IB=round(((VBB-VBE)/RB)*1000,0);                     #Base current,  \u03bcA\n",
-      "IC=beta*IB/1000;                    #Collector current, mA\n",
-      "VCE=VCC-IC*(RC/1000);                       #Collector-emitter voltage, V\n",
-      "\n",
-      "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n",
-      "\n",
-      "#For d.c load line\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC-VCE)/(RC/1000.0);           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,25])\n",
-      "limit.set_ylim([0,65])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: IC=39.6mA and VCE=6.93V.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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Tw549e0oGpjEB8VInT8Jdd8GuXbB0KbRqZXZEYiUeOyaQmZnJ9u3b6dq1Kzk5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\nIRiBRQAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8947baebd0>"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.26 : Page number 169-170"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                #Collector supply voltage, V\n",
-      "VEE=10.0;                  #Emitter supply voltage, V\n",
-      "RC=1.0;                  #Collector resistor, k\u03a9\n",
-      "RE=4.7;                  #Collector resistor, k\u03a9\n",
-      "RB=47.0;                #Base resistoe, k\u03a9\n",
-      "beta=100.0;                 #Base current amplification factor\n",
-      "VBE=0.7;                      #Base -emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#-IB*RB-VBE-IE*RE+VEE=0\n",
-      "#AS, IC=beta*IB and IC~IE\n",
-      "IE=round((VEE-VBE)/(RE+(RB/beta)),1);      #Emitter current,  mA\n",
-      "IC=IE;                                     #Collector current, mA\n",
-      "\n",
-      "#VCC-IC*RC-VCE-IE*RE+VEE=0\n",
-      "#IC~IE\n",
-      "VCE=VCC+VEE-IC*(RC+RE);                       #Collector-emitter voltage, V\n",
-      "\n",
-      "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n",
-      "\n",
-      "\n",
-      "#For d.c load line\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC+VEE-IC*(RC+RE);               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC+VEE-VCE)/(RC+RE);           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,25])\n",
-      "limit.set_ylim([0,5])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: IC=1.8mA and VCE=9.74V.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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CpCTzBRCodFx4R8VeJMh17mzm4w8ZAj17wt13Q16e3VGJv6lnLxJCDhyA8eNN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\nQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8947c78950>"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.27 : Page number 170-171"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VEE=10.0;                    #Emitter supply voltage, V\n",
-      "IE=1.8;                      #Emitter current, mA\n",
-      "RE=4.7;                      #Emitter resistor, k\u03a9\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "VCC=10.0;                     #Collector supply voltage, V\n",
-      "IC=1.8;                       #Collector current, mA\n",
-      "RC=1.0;                       #Collector resistor, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "VE=-VEE+IE*RE;               #Emitter voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "VB=VEE+VBE;                   #Base voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "VC=VCC-IC*RC;                 #Collector voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) Emitter voltage=%.2fV.\"%VE);\n",
-      "print(\"(i) Base voltage=%.1fV.\"%VB);\n",
-      "print(\"(i) Collector voltage=%.1fV.\"%VC);\n",
-      "\n",
-      "#Note: In the textbook, VB=VE+VBE has been written, which is worng. It should be VB=VEE+VBE. "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Emitter voltage=-1.54V.\n",
-        "(i) Base voltage=10.7V.\n",
-        "(i) Collector voltage=8.2V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "\n",
-      "Example 8.28: Page number 173-174\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BE_change=200.0;                #Change in base-emitter voltage in mV\n",
-      "I_B_change=100.0;                  #Change in base current in  \u03bcA\n",
-      "\n",
-      "#Calculations\n",
-      "Ri=V_BE_change/I_B_change;              #Input resistance in k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"Input resistance =%d k\u03a9\"%Ri);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input resistance =2 k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.29; Page number 174\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CE_final=10.0;\t\t\t#Final value of collector-emitter voltage in V\n",
-      "V_CE_initial=2.0;                 #Initial value of collector-emitter voltage in V\n",
-      "I_C_final=3.0;                    #Final value of collector current in mA\n",
-      "I_C_initial=2.0;                  #Initial value of collector current in mA\n",
-      "\n",
-      "#Calculations\n",
-      "V_CE_change=V_CE_final-V_CE_initial;\t\t#Change in collector to emitter voltage in V\n",
-      "I_C_change=I_C_final-I_C_initial;               #Change in collector current in mA\n",
-      "R0=V_CE_change/I_C_change;                      #Output resistance in k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output resistance =%dk\u03a9\"%R0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output resistance =8k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.30: Page number 174\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R_C=2.0;\t\t#Collector load in kilo ohm\n",
-      "R_i=1.0;\t\t#Input resistance in kilo ohm\n",
-      "R_AC=R_C;               #Effective collector load for single stage in kilo ohm(appoximately equal to collector load for single stage)\n",
-      "beta=50.0;              #Current gain\n",
-      "\n",
-      "#Calculations\n",
-      "A_v=beta*(R_AC/R_i);\t\t#Voltage gain of the amplifier\n",
-      "\n",
-      "#Result \n",
-      "print(\"The voltage gain of the amplifier =%d \"%A_v);\t\t\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain of the amplifier =100 \n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.31: Page number 175-176\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=20;\t\t#Collector supply voltage in V\n",
-      "R_C=1;                  #Collector resistance in kilo ohm\n",
-      "V_knee_Si=1;\t\t#Knee voltage of V_CE for Si in V \n",
-      "V_knee_Ge=0.5;\t\t#Knee voltage of V_CE for Ge in V\n",
-      "\n",
-      "#Calculations\n",
-      "I_C_sat_Si=(V_CC-V_knee_Si)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Si transistor)\n",
-      "I_C_sat_Ge=(V_CC-V_knee_Ge)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Ge transistor)\n",
-      "I_C_sat=(V_CC)/R_C;\t\t\t\t#Saturation (maximum) value of collector current in mA (neglecting knee voltage)\n",
-      "V_CE_cut_off=V_CC;                              #Collector to emitter voltage in cutoff when base current=0, in V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Collector current during saturation = %d mA\"%I_C_sat);\n",
-      "print(\"Collector emitter voltage during cutoff = %d V.\"%V_CE_cut_off);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current during saturation = 20 mA\n",
-        "Collector emitter voltage during cutoff = 20 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.32: Page number 176-177\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=12.0;\t\t#Collector supply voltage in V\n",
-      "V_EE=12.0;\t\t#Emitter supply voltage in V\n",
-      "R_C=750.0;\t\t#Collector resistance in ohm\n",
-      "R_E=1.5;\t\t#Emitter resistance in kilo ohm\n",
-      "R_B=100.0;\t\t#Base resistance in ohm\n",
-      "beta=200;\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to the collector side of the circuit\n",
-      "#using the equation: Vcc -IcRc-Vce -IeRe+Vee=0\n",
-      "#we get Vce=Vcc+Vee-Ic(Rc+Re), [Ie=Ic, approximately]\n",
-      "#We get Vce(off), when Ic=0;\n",
-      "\n",
-      "I_C_Vce_off=0;\t\t\t\t\t#Collector current for Vce(off) in mA\n",
-      "V_CE_off=V_CC+V_EE -(I_C_Vce_off * (R_C +R_E));\t#Collector to emitter voltage in V, during transistor in off state\n",
-      "\n",
-      "#We get Ic(sat), when Vce=0\n",
-      "V_CE_Ic_sat=0;\t\t\t\t\t\t#Collector to emitter voltage for saturation current of collector in V\n",
-      "I_C_sat=(V_CC+V_EE-V_CE_Ic_sat)/(R_C+(R_E*1000));\t#Saturated collector current in A \n",
-      "I_C_sat=I_C_sat*1000;\t\t\t\t\t#Saturated collector current in mA\n",
-      "#Result\n",
-      "print(\"Vce(off)= %dV\"%V_CE_off);\n",
-      "print(\"Ic(sat) = %.2f mA\"%I_C_sat);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vce(off)= 24V\n",
-        "Ic(sat) = 10.67 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.33 : Page number 177\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_knee=0.2;\t\t\t\t#Knee voltage of collector-emitter voltage in V\n",
-      "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BB=3.0;\t\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V \t\n",
-      "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=1.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=50.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#applying Kirchhoff's voltage law along the collector side of the circuit,\n",
-      "#We get Vcc-Ic(sat)*Rc-V_knee=0\n",
-      "#From the above equation, we get:\n",
-      "I_C_sat=(V_CC-V_knee)/R_C;\t\t#Saturated collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along base emitter side,\n",
-      "#We get VBB-IB*RB-VBE=0;\n",
-      "#From the above equation, we get:\n",
-      "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-      "\n",
-      "\n",
-      "I_C=beta*I_B\t\t\t\t#Collector current in mA\n",
-      "\n",
-      "#Result\n",
-      "if(I_C>I_C_sat):\n",
-      "\tprint(\"The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\");\n",
-      "else:\n",
-      "\tprint(\"The base current is not large enough to produce Ic greater than Ic(sat), therefore the transistor isn't saturated. \");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.34: Page number 177-178\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BE=0.95;\t\t\t\t#Base-emitter voltage in V \t\n",
-      "I_B=100.0;\t\t\t\t#Base current in microAmp\n",
-      "R_C=970.0;\t\t\t\t#Collector resistor's resistance in ohm\n",
-      "beta=100.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "I_C=(I_B/1000)*beta;\t\t\t\t#Collector current in mA \n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along collector side\n",
-      "#We get Vcc-IcRc-Vce=0\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "V_CE=V_CC-((I_C/1000)*R_C);\t\t\t\t#Collector-emitter voltage in V\n",
-      "\n",
-      "#From the equation, V_CE=V_CB+V_BE,\n",
-      "V_CB=V_CE-V_BE;\t\t\t\t\t\t#Collector-base voltage in V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "if(V_CB<0 and V_BE >0):\n",
-      "\tprint(\"As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \");\n",
-      "else:\n",
-      "\tprint(\"No. The transistor isn't operating in the saturation region.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.35: Page number 178\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=10.0;\t\t\t\t#Collector supplu voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-      "R_B=50.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=2.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=200.0;\t\t\t\t#Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along the collector side,\n",
-      "#We get, Vcc-Ic(sat)*Rc-Vce=0;\n",
-      "#From the above equation, we get:\n",
-      "#I_C_sat=(V_CC-V_CE)/R_C, but as transistor goes into saturation, Vce=0;\n",
-      "\n",
-      "V_CE=0;\t\t\t\t\t\t#Collector-emiter voltage in V, for transistor in saturation \n",
-      "I_C_sat=(V_CC-V_CE)/R_C;\t\t\t#Saturated collector current in mA\n",
-      "\n",
-      "I_B=I_C_sat/beta;\t\t\t\t#Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to the base circuit,\n",
-      "#We get, VBB - IB*RB - VBE=0\n",
-      "#From the above equation. we get:\n",
-      "V_BB=V_BE+ I_B*R_B;\t\t\t\t#Base supply voltage to put transistor in saturation, in V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Therefore, for putting transistor in saturation, VBB >= %.2f V\"%V_BB);\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Therefore, for putting transistor in saturation, VBB >= 1.95 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.36: Page number 178-179\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=10.0;\t\t\t#Collector supply voltage in V\n",
-      "V_BB=2.7;\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n",
-      "beta=100.0;\t\t\t#Base current amplification factor\n",
-      "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calcultaion\t\n",
-      "V_B=V_BB;\t\t\t#Base voltage in V\n",
-      "V_E=V_B-V_BE;\t\t\t#Emitter voltage in V\n",
-      "I_E=V_E/R_E;\t\t\t#Emitter current in mA\n",
-      "I_C=I_E;\t\t\t#Collector current (approximately equal to emitter current) in mA\n",
-      "I_B=I_C/beta;\t\t\t#Base current in mA\n",
-      "\n",
-      "#Case (i):\n",
-      "R_C=2;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "#We get,Vcc-IcRc=Vc,\n",
-      "\n",
-      "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-      "\n",
-      "if(V_C>V_E):\n",
-      "\tprint(\"(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\");\n",
-      "elif(V_C<V_E):\n",
-      "\tprint(\"(i)Our assumption was wrong, the transistor is in saturation for Rc=2 kilo ohm.\");\n",
-      "elif(V_C==V_E):\n",
-      "\tprint(\"(i)The transistor is at the edge of saturation for Rc=2 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n",
-      "\n",
-      "#Case (ii):\n",
-      "R_C=4;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "#We get,Vcc-IcRc=Vc,\n",
-      "\n",
-      "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-      "if(V_C>V_E):\n",
-      "\tprint(\"(ii)Our assumption was correct, the transistor is in active state for Rc=4 kilo ohm.\");\n",
-      "elif(V_C==V_E):\n",
-      "\tprint(\"(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n",
-      "elif(V_C<V_E):\n",
-      "\tprint(\"(ii)Our assumption was wrong, the transistor is in saturation for Rc=4 kilo ohm.\");\n",
-      "\n",
-      "\n",
-      "#Case (iii):\n",
-      "R_C=8;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "#We get,Vcc-IcRc=Vc,\n",
-      "\n",
-      "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-      "if(V_C>V_E):\n",
-      "\tprint(\"(iii)Our assumption was correct, the transistor is in active state for Rc=8 kilo ohm.\");\n",
-      "elif(V_C<V_E):\n",
-      "\tprint(\"(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\");\n",
-      "elif(V_C==V_E):\n",
-      "\tprint(\"(iii)The transistor is at the edge of saturation for Rc=8 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\n",
-        "(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\n",
-        "(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.37 : Page number 179-180"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=15.0;\t\t\t#Collector supply voltage in V\n",
-      "R_C=10.0;\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n",
-      "beta=100.0;\t\t\t#Base current amplification factor\n",
-      "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\t\n",
-      "\n",
-      "#Case (i):\n",
-      "V_BB=0.5;\t\t\t#Base supply voltage in V\n",
-      "VB=V_BB;            #Base voltage, V\n",
-      "print(\"(i) Base voltage =%.1fV is less than VBE=%.1fV, therefore, transistor is cut-off.\"%(VB,V_BE));\n",
-      "\n",
-      "\n",
-      "#Case (ii):\n",
-      "V_BB=1.5;\t\t\t#Base supply voltage in V\n",
-      "VB=V_BB;            #Base voltage, V\n",
-      "VE=VB-V_BE;          #Emitter voltage, V\n",
-      "IE=round(VE/R_E,1);  #Emitter current, mA\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "IC=IE;              #Collector current, mA\n",
-      "IB=IC/beta;         #Base  current, mA\n",
-      "VC=V_CC-IC*R_C;      #Collector voltage, V\n",
-      "print(VE,IE,VC);\n",
-      "print(\"(ii) VC=%dV > VE=%.1fV, therefore the transistor is active. Our assumption was correct.\"%(VC,VE));\n",
-      "\n",
-      "#Case (iii):\n",
-      "V_BB=3; \t\t\t#Base supply voltage in V\n",
-      "VB=V_BB;            #Base voltage, V\n",
-      "VE=VB-V_BE;          #Emitter voltage, V\n",
-      "IE=round(VE/R_E,1);  #Emitter current, mA\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "IC=IE;              #Collector current, mA\n",
-      "IB=IC/beta;         #Base  current, mA\n",
-      "VC=V_CC-IC*R_C;      #Collector voltage, V\n",
-      "\n",
-      "print(\"(iii) VC=%dV < VE=%.1fV, therefore the transistor is saturated. Our assumption was wrong.\"%(VC,VE));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Base voltage =0.5V is less than VBE=0.7V, therefore, transistor is cut-off.\n",
-        "(0.8, 0.8, 7.0)\n",
-        "(ii) VC=7V > VE=0.8V, therefore the transistor is active. Our assumption was correct.\n",
-        "(iii) VC=-8V < VE=2.3V, therefore the transistor is saturated. Our assumption was wrong.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.38: Page number 181\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_D_max=100.0;\t\t\t#Maximum power dissipation of a transistor in mW\n",
-      "V_CE=20.0;\t\t\t#Collector emitter voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "#As power=curent*voltage\n",
-      "#P_D_max=I_C_max*V_CE\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "I_C_max=P_D_max/V_CE;\t\t#Maximum collector current that can be allowed without destruction of the transistor, in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum collector current that can be allowed without destruction of the transistor = %d mA.\"%I_C_max); \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum collector current that can be allowed without destruction of the transistor = 5 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.39: Page number 181\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BB=5.0;\t\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-      "R_B=1.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=200.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along base circuit<\n",
-      "#We get, VBB- IB*RB - VBE=0.\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-      "\n",
-      "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along collector circuit:\n",
-      "\n",
-      "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n",
-      "\n",
-      "#As power=curent*voltage\n",
-      "#P_D=I_C*V_CE\n",
-      "#From the above equation, we get:\n",
-      "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n",
-      "P_D=P_D/1000;\t\t\t\t#Power dissipated in W\n",
-      "\n",
-      "#Result\n",
-      "print(\"Power dissipated = %.1fW\"%P_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power dissipated = 4.3W\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.40: Page number 181-182\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BB=1.0;\t\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-      "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=1.0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=100.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along base circuit<\n",
-      "#We get, VBB- IB*RB - VBE=0.\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-      "\n",
-      "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along collector circuit:\n",
-      "\n",
-      "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n",
-      "\n",
-      "#As power=curent*voltage\n",
-      "#P_D=I_C*V_CE\n",
-      "#From the above equation, we get:\n",
-      "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Power dissipated = %.0fmW\"%P_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power dissipated = 6mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.41 : Page number 182"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VBB=5.0;                     #Base supply voltage, V\n",
-      "RB=22.0;                  #Base resistor, kilo ohm\n",
-      "RC=1.0;                   #Collector resistor, kilo ohm\n",
-      "beta=100.0;               #Base current amplification factor\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "PD_max=800.0;             #Maximum power dissipation, mW\n",
-      "VCE_max=15.0;             #Maximum collector-emitter voltage, V\n",
-      "IC_max=100.0;             #Maximum collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "IB=((VBB-VBE)/RB)*1000;               #Base current,  \u03bcA\n",
-      "IC=beta*IB/1000;                      #Collector current, mA\n",
-      "\n",
-      "print(\"IC=%.1fmA is much less than IC_max=%dmA. Therefore, will not change with VCC and current rating is not exceeded.\"%(IC,IC_max));\n",
-      "\n",
-      "#VCC=VCE+IC*RC\n",
-      "VCC_max=VCE_max+IC*RC;           #Maximum value of Collector supply voltage, V\n",
-      "PD=VCE_max*IC;                  #Power dissipation, mW\n",
-      "\n",
-      "print(\"PD=%dmW  is less than PD_max=%dmW. Therefore, power rating is not exceeded.\"%(PD,PD_max));\n",
-      "\n",
-      "print(\"If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\");"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IC=19.5mA is much less than IC_max=100mA. Therefore, will not change with VCC and current rating is not exceeded.\n",
-        "PD=293mW  is less than PD_max=800mW. Therefore, power rating is not exceeded.\n",
-        "If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_1.ipynb
deleted file mode 100755
index 4afe0858..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_1.ipynb
+++ /dev/null
@@ -1,1851 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:9c13bdd66a3dbb3eae04903205b69bc52bf35e6dadf8b1b3ade1bab68394ae3b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1: Page number 147-148\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Signal=500.0;                    #Signal voltage in V\n",
-      "Rin=20.0;                        #Input resistance in \u03a9 \n",
-      "Rout=100.0;                      #Output resistance in \u03a9\n",
-      "R_C=1000.0;                      #Collector load in \u03a9\n",
-      "alpha_ac=1.0;                    #current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_E=(Signal/1000)/Rin;        \t#Input current in mA\n",
-      "I_C=I_E*alpha_ac;               #Output current in mA\n",
-      "Vout=I_C*R_C;                   #Output voltage in V \n",
-      "Av=Vout/(Signal/1000);          #Voltage amplification \n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage amplification = %d. \"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage amplification = 50. \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.2: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_E=1;                            #Emitter curent in mA\n",
-      "I_C=0.95;                         #Collector current in mA\n",
-      "\n",
-      "#Calculation\n",
-      "I_B=I_E-I_C;                      #Base current in mA\n",
-      "\n",
-      "#Result \n",
-      "print(\"The base current = %.2f mA \"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current = 0.05 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 8.3: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "alpha=0.9;                      #Current amplification factor\n",
-      "I_E=1;                          #Emitter current in mA\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=alpha*I_E;                  #Collector current in mA\n",
-      "I_B=I_E-I_C;                    #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The base current =%.1f mA\"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current =0.1 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_C=0.95;\t\t\t#Collector current in mA\n",
-      "I_B=0.05;\t\t\t#Base current in mA\n",
-      "\n",
-      "#Calculation\n",
-      "I_E=I_B+I_C;                    #Emitter current in mA\n",
-      "alpha=I_C/I_E;                  #Current amplification factor \n",
-      "\n",
-      "#Result\n",
-      "print(\"The current amplification factor = %.2f .\"%alpha);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current amplification factor = 0.95 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.5: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_E=1;                  #Emitter current in mA\n",
-      "I_CBO=50.0;               #Collector current with emitter circuit open, in microAmp\n",
-      "alpha=0.92;             #Current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=alpha*I_E + (I_CBO/1000);           #Total collector current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total collector current = %.2f mA.\"%I_C);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total collector current = 0.97 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.6: Page number 150-151\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "alpha=0.95;               #Current amplification factor\n",
-      "Rc=2.0;                   #Resistor connected to the collector, in kilo ohm\n",
-      "V_Rc=2.0;                 #Voltage drop across the resistor connected to the collector in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=V_Rc/Rc;              #Collector current in mA\n",
-      "I_E=I_C/alpha;            #Emitter current in mA\n",
-      "I_B=I_E-I_C;              #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The base current = %.2f mA\"%I_B); \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current = 0.05 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.7: Page number 151\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_EE=8.0;                 #Supply voltage at the emitter in V\n",
-      "V_CC=18.0;                #Supply voltage at the collector in V\n",
-      "V_BE=0.7;                 #Base to emitter voltage in V\n",
-      "R_E=1.5;                  #Emitter resistance in \u03a9\n",
-      "R_C=1.2;                  #Collector resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "I_E=(V_EE-V_BE)/R_E;              #Emitter current in mA\n",
-      "I_C=I_E;                          #Collector current in mA (approximately equal to emitter current)\n",
-      "V_CB=V_CC-(I_C*R_C);              #Collector to base voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The collector current =%.2f mA\"%I_C);\n",
-      "print(\"The collector to base voltage = %.2f V\"%V_CB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The collector current =4.87 mA\n",
-        "The collector to base voltage = 12.16 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8:Page number 155\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating beta from alpha\n",
-      "def calc_beta(a):                   #a is the value of alpha\n",
-      "\treturn(a/(1-a));\n",
-      "\n",
-      "#Case (i)\n",
-      "alpha=0.9;                      #current amplification factor\n",
-      "beta=calc_beta(alpha);\t\t#Base current amplification factor    \n",
-      "print(\"(i) Value of beta =%d\"%beta );\t\t\t\t\t\t\t\t\t\n",
-      "\n",
-      "#Case (ii)\n",
-      "alpha=0.98;                     #current amplification factor\n",
-      "beta=calc_beta(alpha);          #Base current amplification factor\n",
-      "print(\"(ii) Value of beta =%.0f\"%beta );\n",
-      "\n",
-      "\n",
-      "#Case (iii)\n",
-      "alpha=0.99;                     #current amplification factor\n",
-      "beta=calc_beta(alpha);          #Base current amplification factor                      \n",
-      "print(\"(iii) Value of beta =%.0f\"%beta );\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Value of beta =9\n",
-        "(ii) Value of beta =49\n",
-        "(iii) Value of beta =99\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.9: Page number 155\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=50.0;                   #Base current amplification factor\n",
-      "I_B=20.0;                    #Base current in microAmp\n",
-      "\n",
-      "#Calculation\n",
-      "I_B=I_B/1000;               #Base current in mA\n",
-      "I_C=beta*I_B;               #Collector current in mA\n",
-      "I_E=I_B+I_C;                #Emitter current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The emitter curent = %.2f mA\"%I_E);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The emitter curent = 1.02 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.10: Page number 155\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_B=240.0;                    #Base current in microAmp\n",
-      "I_E=12;                       #Emitter current in mA\n",
-      "beta=49.0;                      #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "alpha=beta/(1+beta);              #current amplification factor \n",
-      "I_C_alpha=alpha*I_E;              #Collector current in mA calculated using alpha\n",
-      "I_C_beta=beta*(I_B/1000);         #Collector current in mA calculated using beta\n",
-      "\n",
-      "#Results\n",
-      "print(\"alpha=%.2f.\"%alpha);\n",
-      "print(\"Collector current determined using alpha =%.2f mA\"%I_C_alpha);\n",
-      "print(\"Collector current determined using beta =%.2f mA\"%I_C_beta);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "alpha=0.98.\n",
-        "Collector current determined using alpha =11.76 mA\n",
-        "Collector current determined using beta =11.76 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.11: Page number 156\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=45.0;                       #Base current amplification factor\n",
-      "R_C=1.0;                         #Resistance of the collector resistance in k\u03a9\n",
-      "V_R_C=1.0;                       #Voltage drop across the collector resistance in V\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=V_R_C/R_C;                   #Collector current in mA\n",
-      "I_B=I_C/beta;                           #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The base current =%.3f mA\"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current =0.022 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.12: Page number 156\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=8.0;                  #Collector supply voltage in V\n",
-      "R_C=800.0;                  #Resistance of the collector resistance in \u03a9\n",
-      "V_R_C=0.5;                 #Voltage drop across collector resistance in V\n",
-      "alpha=0.96;                #current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "V_CE=V_CC-V_R_C;           #Collector to emitter voltage in V\n",
-      "I_C=V_R_C/R_C;             #Collector current in A\n",
-      "I_C=I_C*1000;              #Collector current in mA\n",
-      "beta=alpha/(1-alpha);      #Base current amplification factor\n",
-      "I_B=I_C/beta;              #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Collector to emitter voltage = %.1f V\"%V_CE);\n",
-      "print(\"Base current= %.3f mA\"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector to emitter voltage = 7.5 V\n",
-        "Base current= 0.026 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.13: Page number 156-157\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=5;                  \t#Collector supply voltage in V\n",
-      "I_CBO=0.2;               \t#Leakage current at collector base junction with emitter open, in  \u03bcA\n",
-      "I_CEO=20.0;              \t#Leakage current with base open, in  \u03bcA\n",
-      "I_C=1.0;                        #Collector current in mA\n",
-      "I_C=I_C*1000;                  \t#Collector current in  \u03bcA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "alpha=1-(I_CBO/I_CEO);\t\t#current amplification factor\n",
-      "I_E=(I_C-I_CBO)/alpha;          #Emitter current in  \u03bcA\n",
-      "I_E=round(I_E,-1);\n",
-      "I_B=I_E-I_C;                    #Base current in  \u03bcA\n",
-      "I_B=round(I_B,-1);\n",
-      "\n",
-      "#Result\n",
-      "print(\"Current amplification factor = %.2f \"%alpha);\n",
-      "print(\"The emitter curent =%d  \u03bcA \"%I_E);\n",
-      "print(\"The base curent =%d  \u03bcA \"%I_B);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current amplification factor = 0.99 \n",
-        "The emitter curent =1010  \u03bcA \n",
-        "The base curent =10  \u03bcA \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.14: Page number 157\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_CEO=300.0;              #Leakage current in common emitter configuration, in  \u03bcA\n",
-      "beta=120.0;               #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "alpha=beta/(1+beta);               #Current amplification factor\n",
-      "alpha=round(alpha,3);\n",
-      "I_CBO=(1-alpha)*I_CEO;             #Leakage current in common base configuration, in  \u03bcA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Vale of I_CBO= %.1f \u03bcA\"%I_CBO);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vale of I_CBO= 2.4 \u03bcA\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.15: Page number 157\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_B=20.0;                       #Base current in \u03bcA\n",
-      "I_C=2.0;                        #Collector current in mA\n",
-      "beta=80.0;                      #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_CEO=I_C-(beta*I_B/1000);            #Leakage current with base open, in mA \n",
-      "alpha=beta/(beta+1);                  #Current amplification factor\n",
-      "alpha=round(alpha,3);\n",
-      "I_CBO=(1-alpha)*I_CEO;                #Leakage current with emitter open, in mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Value of I_CBO=%.4f mA\"%I_CBO);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of I_CBO=0.0048 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.17: Page number 158\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=150.0;               \t#Base current amplification factor\n",
-      "R_B=10.0;                   \t#Base resistance in kilo ohm\n",
-      "R_C=100.0;                   \t#Collector resistance in kilo ohm\n",
-      "V_CC=10.0;                      #Collector supply voltage in V\n",
-      "V_BB=5.0;                       #Base supply voltage in V\n",
-      "V_BE=0.7;                       #Base to emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_B=(V_BB-V_BE)/R_B;              #Base current in mA\n",
-      "I_C=beta*I_B;                     #Collector current in mA\n",
-      "V_CE=V_CC - (I_C/1000)*R_C;       #Collector to emitter voltage in V\n",
-      "V_CB=V_CE-V_BE;                   #Collector to base voltage in V\n",
-      "\n",
-      "\n",
-      "#Result \n",
-      "print(\"Collector to base voltage, V_CB= %.2f V\"%V_CB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector to base voltage, V_CB= 2.85 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.18: Page number158-159\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_B=68.0;              #Base current in \u03bcA\n",
-      "I_E=30.0;              #Emitter current in mA\n",
-      "beta=440.0;\t     #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "alpha=beta/(beta + 1);          #current amplification factor\n",
-      "I_C_alpha=alpha*I_E;\t\t#Collector current using alpha rating, in mA\n",
-      "I_C_beta=beta*(I_B/1000.0);       #Collector current using beta rating, in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Collector current determined using alpha rating =%.2f mA\"%I_C_alpha);\n",
-      "print(\"Collector current determined using beta rating =%.2f mA\"%I_C_beta);\n",
-      "\n",
-      "#Note: In the textbook, the collector current obtained from beta rating is approximated to 29.93 mA\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current determined using alpha rating =29.93 mA\n",
-        "Collector current determined using beta rating =29.92 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.19: Page number 159\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_C_max=500.0;                   #Maximum collector current in mA\n",
-      "beta_max=300.0;                  #Maximum base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_B_max=I_C_max/beta_max;           #Maximum base current in mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable value of base current = %.2f mA\"%I_B_max);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable value of base current = 1.67 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.22 : Page number 167-168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.5;                #Collector supply voltage, V\n",
-      "RC=2.5;                  #Collector resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC-VCE)/RC;           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,15])\n",
-      "limit.set_ylim([0,6])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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tROJz+mmhwcHBePbZZ3H48GHs2LEDHTt2lDOXZum919NC1673c6wVomUWLS+gwevUm9XX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\nrkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f2eadbe6710>"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.23 : Page number 168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                #Collector supply voltage, V\n",
-      "RC=6.0;                  #Collector resistor, k\u03a9\n",
-      "IB=20.0;                     #Zero signal base current,  \u03bcA\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC-VCE)/RC;           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,15])\n",
-      "limit.set_ylim([0,5])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n",
-      "#Calculating Q-point\n",
-      "IC=beta*(IB/1000);                                  #Collector current, mA\n",
-      "VCE=VCC-IC*RC;                               #Collector emitter voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Operating point: IC=%dmA and VCE=%dV.\"%(IC,VCE));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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EVut2p75u3TrcunXrkf3V1dVYv369fdIREZHdmB3qpaWlmDZt2iP7p06dim+++Ua2UFrG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\nAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8947e0f0d0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: IC=1mA and VCE=6V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.24 : Page number 168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=4.0;               #Collector load, k\u03a9\n",
-      "IC_Q=1.0;             #Quiescent current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "VCC=10;                  #Collector supply voltage, V\n",
-      "VCE=VCC-IC*RC;            #Collector emitter voltage, V\n",
-      "\n",
-      "print(\"(i) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n",
-      "\n",
-      "#(ii)\n",
-      "RC=5.0;               #Collector load, k\u03a9\n",
-      "VCE=VCC-IC*RC;            #Collector emitter voltage, V\n",
-      "print(\"(ii) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Operating point: VCE=6V and IC=1mA.\n",
-        "(ii) Operating point: VCE=5V and IC=1mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 8.25 : Page number 168-169"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                #Collector supply voltage, V\n",
-      "VBB=10.0;                  #Base supply voltage, V\n",
-      "RC=330.0;                  #Collector resistor, \u03a9\n",
-      "RB=47.0;                #Base resistoe, k\u03a9\n",
-      "beta=200.0;                 #Base current amplification factor\n",
-      "VBE=0.7;                      #Base -emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#VBB-IB*RB-VBE=0\n",
-      "IB=round(((VBB-VBE)/RB)*1000,0);                     #Base current,  \u03bcA\n",
-      "IC=beta*IB/1000;                    #Collector current, mA\n",
-      "VCE=VCC-IC*(RC/1000);                       #Collector-emitter voltage, V\n",
-      "\n",
-      "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n",
-      "\n",
-      "#For d.c load line\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC-VCE)/(RC/1000.0);           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,25])\n",
-      "limit.set_ylim([0,65])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: IC=39.6mA and VCE=6.93V.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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Tw549e0oGpjEB8VInT8Jdd8GuXbB0KbRqZXZEYiUeOyaQmZnJ9u3b6dq1Kzk5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\nIRiBRQAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8947baebd0>"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.26 : Page number 169-170"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                #Collector supply voltage, V\n",
-      "VEE=10.0;                  #Emitter supply voltage, V\n",
-      "RC=1.0;                  #Collector resistor, k\u03a9\n",
-      "RE=4.7;                  #Collector resistor, k\u03a9\n",
-      "RB=47.0;                #Base resistoe, k\u03a9\n",
-      "beta=100.0;                 #Base current amplification factor\n",
-      "VBE=0.7;                      #Base -emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#-IB*RB-VBE-IE*RE+VEE=0\n",
-      "#AS, IC=beta*IB and IC~IE\n",
-      "IE=round((VEE-VBE)/(RE+(RB/beta)),1);      #Emitter current,  mA\n",
-      "IC=IE;                                     #Collector current, mA\n",
-      "\n",
-      "#VCC-IC*RC-VCE-IE*RE+VEE=0\n",
-      "#IC~IE\n",
-      "VCE=VCC+VEE-IC*(RC+RE);                       #Collector-emitter voltage, V\n",
-      "\n",
-      "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n",
-      "\n",
-      "\n",
-      "#For d.c load line\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC+VEE-IC*(RC+RE);               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC+VEE-VCE)/(RC+RE);           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,25])\n",
-      "limit.set_ylim([0,5])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: IC=1.8mA and VCE=9.74V.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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CpCTzBRCodFx4R8VeJMh17mzm4w8ZAj17wt13Q16e3VGJv6lnLxJCDhyA8eNN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\nQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8947c78950>"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.27 : Page number 170-171"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VEE=10.0;                    #Emitter supply voltage, V\n",
-      "IE=1.8;                      #Emitter current, mA\n",
-      "RE=4.7;                      #Emitter resistor, k\u03a9\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "VCC=10.0;                     #Collector supply voltage, V\n",
-      "IC=1.8;                       #Collector current, mA\n",
-      "RC=1.0;                       #Collector resistor, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "VE=-VEE+IE*RE;               #Emitter voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "VB=VEE+VBE;                   #Base voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "VC=VCC-IC*RC;                 #Collector voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) Emitter voltage=%.2fV.\"%VE);\n",
-      "print(\"(i) Base voltage=%.1fV.\"%VB);\n",
-      "print(\"(i) Collector voltage=%.1fV.\"%VC);\n",
-      "\n",
-      "#Note: In the textbook, VB=VE+VBE has been written, which is worng. It should be VB=VEE+VBE. "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Emitter voltage=-1.54V.\n",
-        "(i) Base voltage=10.7V.\n",
-        "(i) Collector voltage=8.2V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "\n",
-      "Example 8.28: Page number 173-174\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BE_change=200.0;                #Change in base-emitter voltage in mV\n",
-      "I_B_change=100.0;                  #Change in base current in  \u03bcA\n",
-      "\n",
-      "#Calculations\n",
-      "Ri=V_BE_change/I_B_change;              #Input resistance in k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"Input resistance =%d k\u03a9\"%Ri);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input resistance =2 k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.29; Page number 174\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CE_final=10.0;\t\t\t#Final value of collector-emitter voltage in V\n",
-      "V_CE_initial=2.0;                 #Initial value of collector-emitter voltage in V\n",
-      "I_C_final=3.0;                    #Final value of collector current in mA\n",
-      "I_C_initial=2.0;                  #Initial value of collector current in mA\n",
-      "\n",
-      "#Calculations\n",
-      "V_CE_change=V_CE_final-V_CE_initial;\t\t#Change in collector to emitter voltage in V\n",
-      "I_C_change=I_C_final-I_C_initial;               #Change in collector current in mA\n",
-      "R0=V_CE_change/I_C_change;                      #Output resistance in k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output resistance =%dk\u03a9\"%R0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output resistance =8k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.30: Page number 174\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R_C=2.0;\t\t#Collector load in kilo ohm\n",
-      "R_i=1.0;\t\t#Input resistance in kilo ohm\n",
-      "R_AC=R_C;               #Effective collector load for single stage in kilo ohm(appoximately equal to collector load for single stage)\n",
-      "beta=50.0;              #Current gain\n",
-      "\n",
-      "#Calculations\n",
-      "A_v=beta*(R_AC/R_i);\t\t#Voltage gain of the amplifier\n",
-      "\n",
-      "#Result \n",
-      "print(\"The voltage gain of the amplifier =%d \"%A_v);\t\t\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain of the amplifier =100 \n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.31: Page number 175-176\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=20;\t\t#Collector supply voltage in V\n",
-      "R_C=1;                  #Collector resistance in kilo ohm\n",
-      "V_knee_Si=1;\t\t#Knee voltage of V_CE for Si in V \n",
-      "V_knee_Ge=0.5;\t\t#Knee voltage of V_CE for Ge in V\n",
-      "\n",
-      "#Calculations\n",
-      "I_C_sat_Si=(V_CC-V_knee_Si)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Si transistor)\n",
-      "I_C_sat_Ge=(V_CC-V_knee_Ge)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Ge transistor)\n",
-      "I_C_sat=(V_CC)/R_C;\t\t\t\t#Saturation (maximum) value of collector current in mA (neglecting knee voltage)\n",
-      "V_CE_cut_off=V_CC;                              #Collector to emitter voltage in cutoff when base current=0, in V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Collector current during saturation = %d mA\"%I_C_sat);\n",
-      "print(\"Collector emitter voltage during cutoff = %d V.\"%V_CE_cut_off);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current during saturation = 20 mA\n",
-        "Collector emitter voltage during cutoff = 20 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.32: Page number 176-177\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=12.0;\t\t#Collector supply voltage in V\n",
-      "V_EE=12.0;\t\t#Emitter supply voltage in V\n",
-      "R_C=750.0;\t\t#Collector resistance in ohm\n",
-      "R_E=1.5;\t\t#Emitter resistance in kilo ohm\n",
-      "R_B=100.0;\t\t#Base resistance in ohm\n",
-      "beta=200;\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to the collector side of the circuit\n",
-      "#using the equation: Vcc -IcRc-Vce -IeRe+Vee=0\n",
-      "#we get Vce=Vcc+Vee-Ic(Rc+Re), [Ie=Ic, approximately]\n",
-      "#We get Vce(off), when Ic=0;\n",
-      "\n",
-      "I_C_Vce_off=0;\t\t\t\t\t#Collector current for Vce(off) in mA\n",
-      "V_CE_off=V_CC+V_EE -(I_C_Vce_off * (R_C +R_E));\t#Collector to emitter voltage in V, during transistor in off state\n",
-      "\n",
-      "#We get Ic(sat), when Vce=0\n",
-      "V_CE_Ic_sat=0;\t\t\t\t\t\t#Collector to emitter voltage for saturation current of collector in V\n",
-      "I_C_sat=(V_CC+V_EE-V_CE_Ic_sat)/(R_C+(R_E*1000));\t#Saturated collector current in A \n",
-      "I_C_sat=I_C_sat*1000;\t\t\t\t\t#Saturated collector current in mA\n",
-      "#Result\n",
-      "print(\"Vce(off)= %dV\"%V_CE_off);\n",
-      "print(\"Ic(sat) = %.2f mA\"%I_C_sat);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vce(off)= 24V\n",
-        "Ic(sat) = 10.67 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.33 : Page number 177\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_knee=0.2;\t\t\t\t#Knee voltage of collector-emitter voltage in V\n",
-      "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BB=3.0;\t\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V \t\n",
-      "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=1.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=50.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#applying Kirchhoff's voltage law along the collector side of the circuit,\n",
-      "#We get Vcc-Ic(sat)*Rc-V_knee=0\n",
-      "#From the above equation, we get:\n",
-      "I_C_sat=(V_CC-V_knee)/R_C;\t\t#Saturated collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along base emitter side,\n",
-      "#We get VBB-IB*RB-VBE=0;\n",
-      "#From the above equation, we get:\n",
-      "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-      "\n",
-      "\n",
-      "I_C=beta*I_B\t\t\t\t#Collector current in mA\n",
-      "\n",
-      "#Result\n",
-      "if(I_C>I_C_sat):\n",
-      "\tprint(\"The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\");\n",
-      "else:\n",
-      "\tprint(\"The base current is not large enough to produce Ic greater than Ic(sat), therefore the transistor isn't saturated. \");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.34: Page number 177-178\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BE=0.95;\t\t\t\t#Base-emitter voltage in V \t\n",
-      "I_B=100.0;\t\t\t\t#Base current in microAmp\n",
-      "R_C=970.0;\t\t\t\t#Collector resistor's resistance in ohm\n",
-      "beta=100.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "I_C=(I_B/1000)*beta;\t\t\t\t#Collector current in mA \n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along collector side\n",
-      "#We get Vcc-IcRc-Vce=0\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "V_CE=V_CC-((I_C/1000)*R_C);\t\t\t\t#Collector-emitter voltage in V\n",
-      "\n",
-      "#From the equation, V_CE=V_CB+V_BE,\n",
-      "V_CB=V_CE-V_BE;\t\t\t\t\t\t#Collector-base voltage in V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "if(V_CB<0 and V_BE >0):\n",
-      "\tprint(\"As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \");\n",
-      "else:\n",
-      "\tprint(\"No. The transistor isn't operating in the saturation region.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.35: Page number 178\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=10.0;\t\t\t\t#Collector supplu voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-      "R_B=50.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=2.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=200.0;\t\t\t\t#Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along the collector side,\n",
-      "#We get, Vcc-Ic(sat)*Rc-Vce=0;\n",
-      "#From the above equation, we get:\n",
-      "#I_C_sat=(V_CC-V_CE)/R_C, but as transistor goes into saturation, Vce=0;\n",
-      "\n",
-      "V_CE=0;\t\t\t\t\t\t#Collector-emiter voltage in V, for transistor in saturation \n",
-      "I_C_sat=(V_CC-V_CE)/R_C;\t\t\t#Saturated collector current in mA\n",
-      "\n",
-      "I_B=I_C_sat/beta;\t\t\t\t#Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to the base circuit,\n",
-      "#We get, VBB - IB*RB - VBE=0\n",
-      "#From the above equation. we get:\n",
-      "V_BB=V_BE+ I_B*R_B;\t\t\t\t#Base supply voltage to put transistor in saturation, in V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Therefore, for putting transistor in saturation, VBB >= %.2f V\"%V_BB);\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Therefore, for putting transistor in saturation, VBB >= 1.95 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.36: Page number 178-179\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=10.0;\t\t\t#Collector supply voltage in V\n",
-      "V_BB=2.7;\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n",
-      "beta=100.0;\t\t\t#Base current amplification factor\n",
-      "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calcultaion\t\n",
-      "V_B=V_BB;\t\t\t#Base voltage in V\n",
-      "V_E=V_B-V_BE;\t\t\t#Emitter voltage in V\n",
-      "I_E=V_E/R_E;\t\t\t#Emitter current in mA\n",
-      "I_C=I_E;\t\t\t#Collector current (approximately equal to emitter current) in mA\n",
-      "I_B=I_C/beta;\t\t\t#Base current in mA\n",
-      "\n",
-      "#Case (i):\n",
-      "R_C=2;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "#We get,Vcc-IcRc=Vc,\n",
-      "\n",
-      "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-      "\n",
-      "if(V_C>V_E):\n",
-      "\tprint(\"(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\");\n",
-      "elif(V_C<V_E):\n",
-      "\tprint(\"(i)Our assumption was wrong, the transistor is in saturation for Rc=2 kilo ohm.\");\n",
-      "elif(V_C==V_E):\n",
-      "\tprint(\"(i)The transistor is at the edge of saturation for Rc=2 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n",
-      "\n",
-      "#Case (ii):\n",
-      "R_C=4;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "#We get,Vcc-IcRc=Vc,\n",
-      "\n",
-      "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-      "if(V_C>V_E):\n",
-      "\tprint(\"(ii)Our assumption was correct, the transistor is in active state for Rc=4 kilo ohm.\");\n",
-      "elif(V_C==V_E):\n",
-      "\tprint(\"(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n",
-      "elif(V_C<V_E):\n",
-      "\tprint(\"(ii)Our assumption was wrong, the transistor is in saturation for Rc=4 kilo ohm.\");\n",
-      "\n",
-      "\n",
-      "#Case (iii):\n",
-      "R_C=8;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "#We get,Vcc-IcRc=Vc,\n",
-      "\n",
-      "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-      "if(V_C>V_E):\n",
-      "\tprint(\"(iii)Our assumption was correct, the transistor is in active state for Rc=8 kilo ohm.\");\n",
-      "elif(V_C<V_E):\n",
-      "\tprint(\"(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\");\n",
-      "elif(V_C==V_E):\n",
-      "\tprint(\"(iii)The transistor is at the edge of saturation for Rc=8 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\n",
-        "(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\n",
-        "(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.37 : Page number 179-180"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=15.0;\t\t\t#Collector supply voltage in V\n",
-      "R_C=10.0;\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n",
-      "beta=100.0;\t\t\t#Base current amplification factor\n",
-      "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\t\n",
-      "\n",
-      "#Case (i):\n",
-      "V_BB=0.5;\t\t\t#Base supply voltage in V\n",
-      "VB=V_BB;            #Base voltage, V\n",
-      "print(\"(i) Base voltage =%.1fV is less than VBE=%.1fV, therefore, transistor is cut-off.\"%(VB,V_BE));\n",
-      "\n",
-      "\n",
-      "#Case (ii):\n",
-      "V_BB=1.5;\t\t\t#Base supply voltage in V\n",
-      "VB=V_BB;            #Base voltage, V\n",
-      "VE=VB-V_BE;          #Emitter voltage, V\n",
-      "IE=round(VE/R_E,1);  #Emitter current, mA\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "IC=IE;              #Collector current, mA\n",
-      "IB=IC/beta;         #Base  current, mA\n",
-      "VC=V_CC-IC*R_C;      #Collector voltage, V\n",
-      "print(VE,IE,VC);\n",
-      "print(\"(ii) VC=%dV > VE=%.1fV, therefore the transistor is active. Our assumption was correct.\"%(VC,VE));\n",
-      "\n",
-      "#Case (iii):\n",
-      "V_BB=3; \t\t\t#Base supply voltage in V\n",
-      "VB=V_BB;            #Base voltage, V\n",
-      "VE=VB-V_BE;          #Emitter voltage, V\n",
-      "IE=round(VE/R_E,1);  #Emitter current, mA\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "IC=IE;              #Collector current, mA\n",
-      "IB=IC/beta;         #Base  current, mA\n",
-      "VC=V_CC-IC*R_C;      #Collector voltage, V\n",
-      "\n",
-      "print(\"(iii) VC=%dV < VE=%.1fV, therefore the transistor is saturated. Our assumption was wrong.\"%(VC,VE));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Base voltage =0.5V is less than VBE=0.7V, therefore, transistor is cut-off.\n",
-        "(0.8, 0.8, 7.0)\n",
-        "(ii) VC=7V > VE=0.8V, therefore the transistor is active. Our assumption was correct.\n",
-        "(iii) VC=-8V < VE=2.3V, therefore the transistor is saturated. Our assumption was wrong.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.38: Page number 181\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_D_max=100.0;\t\t\t#Maximum power dissipation of a transistor in mW\n",
-      "V_CE=20.0;\t\t\t#Collector emitter voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "#As power=curent*voltage\n",
-      "#P_D_max=I_C_max*V_CE\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "I_C_max=P_D_max/V_CE;\t\t#Maximum collector current that can be allowed without destruction of the transistor, in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum collector current that can be allowed without destruction of the transistor = %d mA.\"%I_C_max); \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum collector current that can be allowed without destruction of the transistor = 5 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.39: Page number 181\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BB=5.0;\t\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-      "R_B=1.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=200.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along base circuit<\n",
-      "#We get, VBB- IB*RB - VBE=0.\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-      "\n",
-      "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along collector circuit:\n",
-      "\n",
-      "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n",
-      "\n",
-      "#As power=curent*voltage\n",
-      "#P_D=I_C*V_CE\n",
-      "#From the above equation, we get:\n",
-      "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n",
-      "P_D=P_D/1000;\t\t\t\t#Power dissipated in W\n",
-      "\n",
-      "#Result\n",
-      "print(\"Power dissipated = %.1fW\"%P_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power dissipated = 4.3W\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.40: Page number 181-182\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BB=1.0;\t\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-      "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=1.0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=100.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along base circuit<\n",
-      "#We get, VBB- IB*RB - VBE=0.\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-      "\n",
-      "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along collector circuit:\n",
-      "\n",
-      "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n",
-      "\n",
-      "#As power=curent*voltage\n",
-      "#P_D=I_C*V_CE\n",
-      "#From the above equation, we get:\n",
-      "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Power dissipated = %.0fmW\"%P_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power dissipated = 6mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.41 : Page number 182"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VBB=5.0;                     #Base supply voltage, V\n",
-      "RB=22.0;                  #Base resistor, kilo ohm\n",
-      "RC=1.0;                   #Collector resistor, kilo ohm\n",
-      "beta=100.0;               #Base current amplification factor\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "PD_max=800.0;             #Maximum power dissipation, mW\n",
-      "VCE_max=15.0;             #Maximum collector-emitter voltage, V\n",
-      "IC_max=100.0;             #Maximum collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "IB=((VBB-VBE)/RB)*1000;               #Base current,  \u03bcA\n",
-      "IC=beta*IB/1000;                      #Collector current, mA\n",
-      "\n",
-      "print(\"IC=%.1fmA is much less than IC_max=%dmA. Therefore, will not change with VCC and current rating is not exceeded.\"%(IC,IC_max));\n",
-      "\n",
-      "#VCC=VCE+IC*RC\n",
-      "VCC_max=VCE_max+IC*RC;           #Maximum value of Collector supply voltage, V\n",
-      "PD=VCE_max*IC;                  #Power dissipation, mW\n",
-      "\n",
-      "print(\"PD=%dmW  is less than PD_max=%dmW. Therefore, power rating is not exceeded.\"%(PD,PD_max));\n",
-      "\n",
-      "print(\"If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\");"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IC=19.5mA is much less than IC_max=100mA. Therefore, will not change with VCC and current rating is not exceeded.\n",
-        "PD=293mW  is less than PD_max=800mW. Therefore, power rating is not exceeded.\n",
-        "If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_2.ipynb
deleted file mode 100755
index 4afe0858..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_2.ipynb
+++ /dev/null
@@ -1,1851 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:9c13bdd66a3dbb3eae04903205b69bc52bf35e6dadf8b1b3ade1bab68394ae3b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1: Page number 147-148\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Signal=500.0;                    #Signal voltage in V\n",
-      "Rin=20.0;                        #Input resistance in \u03a9 \n",
-      "Rout=100.0;                      #Output resistance in \u03a9\n",
-      "R_C=1000.0;                      #Collector load in \u03a9\n",
-      "alpha_ac=1.0;                    #current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_E=(Signal/1000)/Rin;        \t#Input current in mA\n",
-      "I_C=I_E*alpha_ac;               #Output current in mA\n",
-      "Vout=I_C*R_C;                   #Output voltage in V \n",
-      "Av=Vout/(Signal/1000);          #Voltage amplification \n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage amplification = %d. \"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage amplification = 50. \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.2: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_E=1;                            #Emitter curent in mA\n",
-      "I_C=0.95;                         #Collector current in mA\n",
-      "\n",
-      "#Calculation\n",
-      "I_B=I_E-I_C;                      #Base current in mA\n",
-      "\n",
-      "#Result \n",
-      "print(\"The base current = %.2f mA \"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current = 0.05 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 8.3: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "alpha=0.9;                      #Current amplification factor\n",
-      "I_E=1;                          #Emitter current in mA\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=alpha*I_E;                  #Collector current in mA\n",
-      "I_B=I_E-I_C;                    #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The base current =%.1f mA\"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current =0.1 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_C=0.95;\t\t\t#Collector current in mA\n",
-      "I_B=0.05;\t\t\t#Base current in mA\n",
-      "\n",
-      "#Calculation\n",
-      "I_E=I_B+I_C;                    #Emitter current in mA\n",
-      "alpha=I_C/I_E;                  #Current amplification factor \n",
-      "\n",
-      "#Result\n",
-      "print(\"The current amplification factor = %.2f .\"%alpha);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current amplification factor = 0.95 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.5: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_E=1;                  #Emitter current in mA\n",
-      "I_CBO=50.0;               #Collector current with emitter circuit open, in microAmp\n",
-      "alpha=0.92;             #Current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=alpha*I_E + (I_CBO/1000);           #Total collector current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total collector current = %.2f mA.\"%I_C);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total collector current = 0.97 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.6: Page number 150-151\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "alpha=0.95;               #Current amplification factor\n",
-      "Rc=2.0;                   #Resistor connected to the collector, in kilo ohm\n",
-      "V_Rc=2.0;                 #Voltage drop across the resistor connected to the collector in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=V_Rc/Rc;              #Collector current in mA\n",
-      "I_E=I_C/alpha;            #Emitter current in mA\n",
-      "I_B=I_E-I_C;              #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The base current = %.2f mA\"%I_B); \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current = 0.05 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.7: Page number 151\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_EE=8.0;                 #Supply voltage at the emitter in V\n",
-      "V_CC=18.0;                #Supply voltage at the collector in V\n",
-      "V_BE=0.7;                 #Base to emitter voltage in V\n",
-      "R_E=1.5;                  #Emitter resistance in \u03a9\n",
-      "R_C=1.2;                  #Collector resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "I_E=(V_EE-V_BE)/R_E;              #Emitter current in mA\n",
-      "I_C=I_E;                          #Collector current in mA (approximately equal to emitter current)\n",
-      "V_CB=V_CC-(I_C*R_C);              #Collector to base voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The collector current =%.2f mA\"%I_C);\n",
-      "print(\"The collector to base voltage = %.2f V\"%V_CB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The collector current =4.87 mA\n",
-        "The collector to base voltage = 12.16 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8:Page number 155\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating beta from alpha\n",
-      "def calc_beta(a):                   #a is the value of alpha\n",
-      "\treturn(a/(1-a));\n",
-      "\n",
-      "#Case (i)\n",
-      "alpha=0.9;                      #current amplification factor\n",
-      "beta=calc_beta(alpha);\t\t#Base current amplification factor    \n",
-      "print(\"(i) Value of beta =%d\"%beta );\t\t\t\t\t\t\t\t\t\n",
-      "\n",
-      "#Case (ii)\n",
-      "alpha=0.98;                     #current amplification factor\n",
-      "beta=calc_beta(alpha);          #Base current amplification factor\n",
-      "print(\"(ii) Value of beta =%.0f\"%beta );\n",
-      "\n",
-      "\n",
-      "#Case (iii)\n",
-      "alpha=0.99;                     #current amplification factor\n",
-      "beta=calc_beta(alpha);          #Base current amplification factor                      \n",
-      "print(\"(iii) Value of beta =%.0f\"%beta );\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Value of beta =9\n",
-        "(ii) Value of beta =49\n",
-        "(iii) Value of beta =99\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.9: Page number 155\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=50.0;                   #Base current amplification factor\n",
-      "I_B=20.0;                    #Base current in microAmp\n",
-      "\n",
-      "#Calculation\n",
-      "I_B=I_B/1000;               #Base current in mA\n",
-      "I_C=beta*I_B;               #Collector current in mA\n",
-      "I_E=I_B+I_C;                #Emitter current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The emitter curent = %.2f mA\"%I_E);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The emitter curent = 1.02 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.10: Page number 155\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_B=240.0;                    #Base current in microAmp\n",
-      "I_E=12;                       #Emitter current in mA\n",
-      "beta=49.0;                      #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "alpha=beta/(1+beta);              #current amplification factor \n",
-      "I_C_alpha=alpha*I_E;              #Collector current in mA calculated using alpha\n",
-      "I_C_beta=beta*(I_B/1000);         #Collector current in mA calculated using beta\n",
-      "\n",
-      "#Results\n",
-      "print(\"alpha=%.2f.\"%alpha);\n",
-      "print(\"Collector current determined using alpha =%.2f mA\"%I_C_alpha);\n",
-      "print(\"Collector current determined using beta =%.2f mA\"%I_C_beta);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "alpha=0.98.\n",
-        "Collector current determined using alpha =11.76 mA\n",
-        "Collector current determined using beta =11.76 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.11: Page number 156\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=45.0;                       #Base current amplification factor\n",
-      "R_C=1.0;                         #Resistance of the collector resistance in k\u03a9\n",
-      "V_R_C=1.0;                       #Voltage drop across the collector resistance in V\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=V_R_C/R_C;                   #Collector current in mA\n",
-      "I_B=I_C/beta;                           #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The base current =%.3f mA\"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current =0.022 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.12: Page number 156\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=8.0;                  #Collector supply voltage in V\n",
-      "R_C=800.0;                  #Resistance of the collector resistance in \u03a9\n",
-      "V_R_C=0.5;                 #Voltage drop across collector resistance in V\n",
-      "alpha=0.96;                #current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "V_CE=V_CC-V_R_C;           #Collector to emitter voltage in V\n",
-      "I_C=V_R_C/R_C;             #Collector current in A\n",
-      "I_C=I_C*1000;              #Collector current in mA\n",
-      "beta=alpha/(1-alpha);      #Base current amplification factor\n",
-      "I_B=I_C/beta;              #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Collector to emitter voltage = %.1f V\"%V_CE);\n",
-      "print(\"Base current= %.3f mA\"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector to emitter voltage = 7.5 V\n",
-        "Base current= 0.026 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.13: Page number 156-157\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=5;                  \t#Collector supply voltage in V\n",
-      "I_CBO=0.2;               \t#Leakage current at collector base junction with emitter open, in  \u03bcA\n",
-      "I_CEO=20.0;              \t#Leakage current with base open, in  \u03bcA\n",
-      "I_C=1.0;                        #Collector current in mA\n",
-      "I_C=I_C*1000;                  \t#Collector current in  \u03bcA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "alpha=1-(I_CBO/I_CEO);\t\t#current amplification factor\n",
-      "I_E=(I_C-I_CBO)/alpha;          #Emitter current in  \u03bcA\n",
-      "I_E=round(I_E,-1);\n",
-      "I_B=I_E-I_C;                    #Base current in  \u03bcA\n",
-      "I_B=round(I_B,-1);\n",
-      "\n",
-      "#Result\n",
-      "print(\"Current amplification factor = %.2f \"%alpha);\n",
-      "print(\"The emitter curent =%d  \u03bcA \"%I_E);\n",
-      "print(\"The base curent =%d  \u03bcA \"%I_B);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current amplification factor = 0.99 \n",
-        "The emitter curent =1010  \u03bcA \n",
-        "The base curent =10  \u03bcA \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.14: Page number 157\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_CEO=300.0;              #Leakage current in common emitter configuration, in  \u03bcA\n",
-      "beta=120.0;               #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "alpha=beta/(1+beta);               #Current amplification factor\n",
-      "alpha=round(alpha,3);\n",
-      "I_CBO=(1-alpha)*I_CEO;             #Leakage current in common base configuration, in  \u03bcA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Vale of I_CBO= %.1f \u03bcA\"%I_CBO);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vale of I_CBO= 2.4 \u03bcA\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.15: Page number 157\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_B=20.0;                       #Base current in \u03bcA\n",
-      "I_C=2.0;                        #Collector current in mA\n",
-      "beta=80.0;                      #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_CEO=I_C-(beta*I_B/1000);            #Leakage current with base open, in mA \n",
-      "alpha=beta/(beta+1);                  #Current amplification factor\n",
-      "alpha=round(alpha,3);\n",
-      "I_CBO=(1-alpha)*I_CEO;                #Leakage current with emitter open, in mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Value of I_CBO=%.4f mA\"%I_CBO);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of I_CBO=0.0048 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.17: Page number 158\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=150.0;               \t#Base current amplification factor\n",
-      "R_B=10.0;                   \t#Base resistance in kilo ohm\n",
-      "R_C=100.0;                   \t#Collector resistance in kilo ohm\n",
-      "V_CC=10.0;                      #Collector supply voltage in V\n",
-      "V_BB=5.0;                       #Base supply voltage in V\n",
-      "V_BE=0.7;                       #Base to emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_B=(V_BB-V_BE)/R_B;              #Base current in mA\n",
-      "I_C=beta*I_B;                     #Collector current in mA\n",
-      "V_CE=V_CC - (I_C/1000)*R_C;       #Collector to emitter voltage in V\n",
-      "V_CB=V_CE-V_BE;                   #Collector to base voltage in V\n",
-      "\n",
-      "\n",
-      "#Result \n",
-      "print(\"Collector to base voltage, V_CB= %.2f V\"%V_CB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector to base voltage, V_CB= 2.85 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.18: Page number158-159\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_B=68.0;              #Base current in \u03bcA\n",
-      "I_E=30.0;              #Emitter current in mA\n",
-      "beta=440.0;\t     #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "alpha=beta/(beta + 1);          #current amplification factor\n",
-      "I_C_alpha=alpha*I_E;\t\t#Collector current using alpha rating, in mA\n",
-      "I_C_beta=beta*(I_B/1000.0);       #Collector current using beta rating, in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Collector current determined using alpha rating =%.2f mA\"%I_C_alpha);\n",
-      "print(\"Collector current determined using beta rating =%.2f mA\"%I_C_beta);\n",
-      "\n",
-      "#Note: In the textbook, the collector current obtained from beta rating is approximated to 29.93 mA\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current determined using alpha rating =29.93 mA\n",
-        "Collector current determined using beta rating =29.92 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.19: Page number 159\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_C_max=500.0;                   #Maximum collector current in mA\n",
-      "beta_max=300.0;                  #Maximum base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_B_max=I_C_max/beta_max;           #Maximum base current in mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable value of base current = %.2f mA\"%I_B_max);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable value of base current = 1.67 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.22 : Page number 167-168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.5;                #Collector supply voltage, V\n",
-      "RC=2.5;                  #Collector resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC-VCE)/RC;           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,15])\n",
-      "limit.set_ylim([0,6])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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tROJz+mmhwcHBePbZZ3H48GHs2LEDHTt2lDOXZum919NC1673c6wVomUWLS+gwevUm9XX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\nrkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f2eadbe6710>"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.23 : Page number 168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                #Collector supply voltage, V\n",
-      "RC=6.0;                  #Collector resistor, k\u03a9\n",
-      "IB=20.0;                     #Zero signal base current,  \u03bcA\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC-VCE)/RC;           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,15])\n",
-      "limit.set_ylim([0,5])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n",
-      "#Calculating Q-point\n",
-      "IC=beta*(IB/1000);                                  #Collector current, mA\n",
-      "VCE=VCC-IC*RC;                               #Collector emitter voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Operating point: IC=%dmA and VCE=%dV.\"%(IC,VCE));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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EVut2p75u3TrcunXrkf3V1dVYv369fdIREZHdmB3qpaWlmDZt2iP7p06dim+++Ua2UFrG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\nAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8947e0f0d0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: IC=1mA and VCE=6V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.24 : Page number 168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=4.0;               #Collector load, k\u03a9\n",
-      "IC_Q=1.0;             #Quiescent current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "VCC=10;                  #Collector supply voltage, V\n",
-      "VCE=VCC-IC*RC;            #Collector emitter voltage, V\n",
-      "\n",
-      "print(\"(i) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n",
-      "\n",
-      "#(ii)\n",
-      "RC=5.0;               #Collector load, k\u03a9\n",
-      "VCE=VCC-IC*RC;            #Collector emitter voltage, V\n",
-      "print(\"(ii) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Operating point: VCE=6V and IC=1mA.\n",
-        "(ii) Operating point: VCE=5V and IC=1mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 8.25 : Page number 168-169"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                #Collector supply voltage, V\n",
-      "VBB=10.0;                  #Base supply voltage, V\n",
-      "RC=330.0;                  #Collector resistor, \u03a9\n",
-      "RB=47.0;                #Base resistoe, k\u03a9\n",
-      "beta=200.0;                 #Base current amplification factor\n",
-      "VBE=0.7;                      #Base -emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#VBB-IB*RB-VBE=0\n",
-      "IB=round(((VBB-VBE)/RB)*1000,0);                     #Base current,  \u03bcA\n",
-      "IC=beta*IB/1000;                    #Collector current, mA\n",
-      "VCE=VCC-IC*(RC/1000);                       #Collector-emitter voltage, V\n",
-      "\n",
-      "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n",
-      "\n",
-      "#For d.c load line\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC-VCE)/(RC/1000.0);           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,25])\n",
-      "limit.set_ylim([0,65])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: IC=39.6mA and VCE=6.93V.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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Tw549e0oGpjEB8VInT8Jdd8GuXbB0KbRqZXZEYiUeOyaQmZnJ9u3b6dq1Kzk5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\nIRiBRQAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8947baebd0>"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.26 : Page number 169-170"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                #Collector supply voltage, V\n",
-      "VEE=10.0;                  #Emitter supply voltage, V\n",
-      "RC=1.0;                  #Collector resistor, k\u03a9\n",
-      "RE=4.7;                  #Collector resistor, k\u03a9\n",
-      "RB=47.0;                #Base resistoe, k\u03a9\n",
-      "beta=100.0;                 #Base current amplification factor\n",
-      "VBE=0.7;                      #Base -emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#-IB*RB-VBE-IE*RE+VEE=0\n",
-      "#AS, IC=beta*IB and IC~IE\n",
-      "IE=round((VEE-VBE)/(RE+(RB/beta)),1);      #Emitter current,  mA\n",
-      "IC=IE;                                     #Collector current, mA\n",
-      "\n",
-      "#VCC-IC*RC-VCE-IE*RE+VEE=0\n",
-      "#IC~IE\n",
-      "VCE=VCC+VEE-IC*(RC+RE);                       #Collector-emitter voltage, V\n",
-      "\n",
-      "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n",
-      "\n",
-      "\n",
-      "#For d.c load line\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC+VEE-IC*(RC+RE);               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC+VEE-VCE)/(RC+RE);           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,25])\n",
-      "limit.set_ylim([0,5])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: IC=1.8mA and VCE=9.74V.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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CpCTzBRCodFx4R8VeJMh17mzm4w8ZAj17wt13Q16e3VGJv6lnLxJCDhyA8eNN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\nQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8947c78950>"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.27 : Page number 170-171"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VEE=10.0;                    #Emitter supply voltage, V\n",
-      "IE=1.8;                      #Emitter current, mA\n",
-      "RE=4.7;                      #Emitter resistor, k\u03a9\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "VCC=10.0;                     #Collector supply voltage, V\n",
-      "IC=1.8;                       #Collector current, mA\n",
-      "RC=1.0;                       #Collector resistor, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "VE=-VEE+IE*RE;               #Emitter voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "VB=VEE+VBE;                   #Base voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "VC=VCC-IC*RC;                 #Collector voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) Emitter voltage=%.2fV.\"%VE);\n",
-      "print(\"(i) Base voltage=%.1fV.\"%VB);\n",
-      "print(\"(i) Collector voltage=%.1fV.\"%VC);\n",
-      "\n",
-      "#Note: In the textbook, VB=VE+VBE has been written, which is worng. It should be VB=VEE+VBE. "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Emitter voltage=-1.54V.\n",
-        "(i) Base voltage=10.7V.\n",
-        "(i) Collector voltage=8.2V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "\n",
-      "Example 8.28: Page number 173-174\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BE_change=200.0;                #Change in base-emitter voltage in mV\n",
-      "I_B_change=100.0;                  #Change in base current in  \u03bcA\n",
-      "\n",
-      "#Calculations\n",
-      "Ri=V_BE_change/I_B_change;              #Input resistance in k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"Input resistance =%d k\u03a9\"%Ri);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input resistance =2 k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.29; Page number 174\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CE_final=10.0;\t\t\t#Final value of collector-emitter voltage in V\n",
-      "V_CE_initial=2.0;                 #Initial value of collector-emitter voltage in V\n",
-      "I_C_final=3.0;                    #Final value of collector current in mA\n",
-      "I_C_initial=2.0;                  #Initial value of collector current in mA\n",
-      "\n",
-      "#Calculations\n",
-      "V_CE_change=V_CE_final-V_CE_initial;\t\t#Change in collector to emitter voltage in V\n",
-      "I_C_change=I_C_final-I_C_initial;               #Change in collector current in mA\n",
-      "R0=V_CE_change/I_C_change;                      #Output resistance in k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output resistance =%dk\u03a9\"%R0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output resistance =8k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.30: Page number 174\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R_C=2.0;\t\t#Collector load in kilo ohm\n",
-      "R_i=1.0;\t\t#Input resistance in kilo ohm\n",
-      "R_AC=R_C;               #Effective collector load for single stage in kilo ohm(appoximately equal to collector load for single stage)\n",
-      "beta=50.0;              #Current gain\n",
-      "\n",
-      "#Calculations\n",
-      "A_v=beta*(R_AC/R_i);\t\t#Voltage gain of the amplifier\n",
-      "\n",
-      "#Result \n",
-      "print(\"The voltage gain of the amplifier =%d \"%A_v);\t\t\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain of the amplifier =100 \n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.31: Page number 175-176\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=20;\t\t#Collector supply voltage in V\n",
-      "R_C=1;                  #Collector resistance in kilo ohm\n",
-      "V_knee_Si=1;\t\t#Knee voltage of V_CE for Si in V \n",
-      "V_knee_Ge=0.5;\t\t#Knee voltage of V_CE for Ge in V\n",
-      "\n",
-      "#Calculations\n",
-      "I_C_sat_Si=(V_CC-V_knee_Si)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Si transistor)\n",
-      "I_C_sat_Ge=(V_CC-V_knee_Ge)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Ge transistor)\n",
-      "I_C_sat=(V_CC)/R_C;\t\t\t\t#Saturation (maximum) value of collector current in mA (neglecting knee voltage)\n",
-      "V_CE_cut_off=V_CC;                              #Collector to emitter voltage in cutoff when base current=0, in V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Collector current during saturation = %d mA\"%I_C_sat);\n",
-      "print(\"Collector emitter voltage during cutoff = %d V.\"%V_CE_cut_off);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current during saturation = 20 mA\n",
-        "Collector emitter voltage during cutoff = 20 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.32: Page number 176-177\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=12.0;\t\t#Collector supply voltage in V\n",
-      "V_EE=12.0;\t\t#Emitter supply voltage in V\n",
-      "R_C=750.0;\t\t#Collector resistance in ohm\n",
-      "R_E=1.5;\t\t#Emitter resistance in kilo ohm\n",
-      "R_B=100.0;\t\t#Base resistance in ohm\n",
-      "beta=200;\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to the collector side of the circuit\n",
-      "#using the equation: Vcc -IcRc-Vce -IeRe+Vee=0\n",
-      "#we get Vce=Vcc+Vee-Ic(Rc+Re), [Ie=Ic, approximately]\n",
-      "#We get Vce(off), when Ic=0;\n",
-      "\n",
-      "I_C_Vce_off=0;\t\t\t\t\t#Collector current for Vce(off) in mA\n",
-      "V_CE_off=V_CC+V_EE -(I_C_Vce_off * (R_C +R_E));\t#Collector to emitter voltage in V, during transistor in off state\n",
-      "\n",
-      "#We get Ic(sat), when Vce=0\n",
-      "V_CE_Ic_sat=0;\t\t\t\t\t\t#Collector to emitter voltage for saturation current of collector in V\n",
-      "I_C_sat=(V_CC+V_EE-V_CE_Ic_sat)/(R_C+(R_E*1000));\t#Saturated collector current in A \n",
-      "I_C_sat=I_C_sat*1000;\t\t\t\t\t#Saturated collector current in mA\n",
-      "#Result\n",
-      "print(\"Vce(off)= %dV\"%V_CE_off);\n",
-      "print(\"Ic(sat) = %.2f mA\"%I_C_sat);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vce(off)= 24V\n",
-        "Ic(sat) = 10.67 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.33 : Page number 177\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_knee=0.2;\t\t\t\t#Knee voltage of collector-emitter voltage in V\n",
-      "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BB=3.0;\t\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V \t\n",
-      "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=1.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=50.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#applying Kirchhoff's voltage law along the collector side of the circuit,\n",
-      "#We get Vcc-Ic(sat)*Rc-V_knee=0\n",
-      "#From the above equation, we get:\n",
-      "I_C_sat=(V_CC-V_knee)/R_C;\t\t#Saturated collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along base emitter side,\n",
-      "#We get VBB-IB*RB-VBE=0;\n",
-      "#From the above equation, we get:\n",
-      "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-      "\n",
-      "\n",
-      "I_C=beta*I_B\t\t\t\t#Collector current in mA\n",
-      "\n",
-      "#Result\n",
-      "if(I_C>I_C_sat):\n",
-      "\tprint(\"The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\");\n",
-      "else:\n",
-      "\tprint(\"The base current is not large enough to produce Ic greater than Ic(sat), therefore the transistor isn't saturated. \");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.34: Page number 177-178\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BE=0.95;\t\t\t\t#Base-emitter voltage in V \t\n",
-      "I_B=100.0;\t\t\t\t#Base current in microAmp\n",
-      "R_C=970.0;\t\t\t\t#Collector resistor's resistance in ohm\n",
-      "beta=100.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "I_C=(I_B/1000)*beta;\t\t\t\t#Collector current in mA \n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along collector side\n",
-      "#We get Vcc-IcRc-Vce=0\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "V_CE=V_CC-((I_C/1000)*R_C);\t\t\t\t#Collector-emitter voltage in V\n",
-      "\n",
-      "#From the equation, V_CE=V_CB+V_BE,\n",
-      "V_CB=V_CE-V_BE;\t\t\t\t\t\t#Collector-base voltage in V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "if(V_CB<0 and V_BE >0):\n",
-      "\tprint(\"As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \");\n",
-      "else:\n",
-      "\tprint(\"No. The transistor isn't operating in the saturation region.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.35: Page number 178\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=10.0;\t\t\t\t#Collector supplu voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-      "R_B=50.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=2.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=200.0;\t\t\t\t#Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along the collector side,\n",
-      "#We get, Vcc-Ic(sat)*Rc-Vce=0;\n",
-      "#From the above equation, we get:\n",
-      "#I_C_sat=(V_CC-V_CE)/R_C, but as transistor goes into saturation, Vce=0;\n",
-      "\n",
-      "V_CE=0;\t\t\t\t\t\t#Collector-emiter voltage in V, for transistor in saturation \n",
-      "I_C_sat=(V_CC-V_CE)/R_C;\t\t\t#Saturated collector current in mA\n",
-      "\n",
-      "I_B=I_C_sat/beta;\t\t\t\t#Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to the base circuit,\n",
-      "#We get, VBB - IB*RB - VBE=0\n",
-      "#From the above equation. we get:\n",
-      "V_BB=V_BE+ I_B*R_B;\t\t\t\t#Base supply voltage to put transistor in saturation, in V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Therefore, for putting transistor in saturation, VBB >= %.2f V\"%V_BB);\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Therefore, for putting transistor in saturation, VBB >= 1.95 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.36: Page number 178-179\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=10.0;\t\t\t#Collector supply voltage in V\n",
-      "V_BB=2.7;\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n",
-      "beta=100.0;\t\t\t#Base current amplification factor\n",
-      "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calcultaion\t\n",
-      "V_B=V_BB;\t\t\t#Base voltage in V\n",
-      "V_E=V_B-V_BE;\t\t\t#Emitter voltage in V\n",
-      "I_E=V_E/R_E;\t\t\t#Emitter current in mA\n",
-      "I_C=I_E;\t\t\t#Collector current (approximately equal to emitter current) in mA\n",
-      "I_B=I_C/beta;\t\t\t#Base current in mA\n",
-      "\n",
-      "#Case (i):\n",
-      "R_C=2;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "#We get,Vcc-IcRc=Vc,\n",
-      "\n",
-      "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-      "\n",
-      "if(V_C>V_E):\n",
-      "\tprint(\"(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\");\n",
-      "elif(V_C<V_E):\n",
-      "\tprint(\"(i)Our assumption was wrong, the transistor is in saturation for Rc=2 kilo ohm.\");\n",
-      "elif(V_C==V_E):\n",
-      "\tprint(\"(i)The transistor is at the edge of saturation for Rc=2 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n",
-      "\n",
-      "#Case (ii):\n",
-      "R_C=4;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "#We get,Vcc-IcRc=Vc,\n",
-      "\n",
-      "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-      "if(V_C>V_E):\n",
-      "\tprint(\"(ii)Our assumption was correct, the transistor is in active state for Rc=4 kilo ohm.\");\n",
-      "elif(V_C==V_E):\n",
-      "\tprint(\"(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n",
-      "elif(V_C<V_E):\n",
-      "\tprint(\"(ii)Our assumption was wrong, the transistor is in saturation for Rc=4 kilo ohm.\");\n",
-      "\n",
-      "\n",
-      "#Case (iii):\n",
-      "R_C=8;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "#We get,Vcc-IcRc=Vc,\n",
-      "\n",
-      "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-      "if(V_C>V_E):\n",
-      "\tprint(\"(iii)Our assumption was correct, the transistor is in active state for Rc=8 kilo ohm.\");\n",
-      "elif(V_C<V_E):\n",
-      "\tprint(\"(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\");\n",
-      "elif(V_C==V_E):\n",
-      "\tprint(\"(iii)The transistor is at the edge of saturation for Rc=8 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\n",
-        "(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\n",
-        "(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.37 : Page number 179-180"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=15.0;\t\t\t#Collector supply voltage in V\n",
-      "R_C=10.0;\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n",
-      "beta=100.0;\t\t\t#Base current amplification factor\n",
-      "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\t\n",
-      "\n",
-      "#Case (i):\n",
-      "V_BB=0.5;\t\t\t#Base supply voltage in V\n",
-      "VB=V_BB;            #Base voltage, V\n",
-      "print(\"(i) Base voltage =%.1fV is less than VBE=%.1fV, therefore, transistor is cut-off.\"%(VB,V_BE));\n",
-      "\n",
-      "\n",
-      "#Case (ii):\n",
-      "V_BB=1.5;\t\t\t#Base supply voltage in V\n",
-      "VB=V_BB;            #Base voltage, V\n",
-      "VE=VB-V_BE;          #Emitter voltage, V\n",
-      "IE=round(VE/R_E,1);  #Emitter current, mA\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "IC=IE;              #Collector current, mA\n",
-      "IB=IC/beta;         #Base  current, mA\n",
-      "VC=V_CC-IC*R_C;      #Collector voltage, V\n",
-      "print(VE,IE,VC);\n",
-      "print(\"(ii) VC=%dV > VE=%.1fV, therefore the transistor is active. Our assumption was correct.\"%(VC,VE));\n",
-      "\n",
-      "#Case (iii):\n",
-      "V_BB=3; \t\t\t#Base supply voltage in V\n",
-      "VB=V_BB;            #Base voltage, V\n",
-      "VE=VB-V_BE;          #Emitter voltage, V\n",
-      "IE=round(VE/R_E,1);  #Emitter current, mA\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "IC=IE;              #Collector current, mA\n",
-      "IB=IC/beta;         #Base  current, mA\n",
-      "VC=V_CC-IC*R_C;      #Collector voltage, V\n",
-      "\n",
-      "print(\"(iii) VC=%dV < VE=%.1fV, therefore the transistor is saturated. Our assumption was wrong.\"%(VC,VE));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Base voltage =0.5V is less than VBE=0.7V, therefore, transistor is cut-off.\n",
-        "(0.8, 0.8, 7.0)\n",
-        "(ii) VC=7V > VE=0.8V, therefore the transistor is active. Our assumption was correct.\n",
-        "(iii) VC=-8V < VE=2.3V, therefore the transistor is saturated. Our assumption was wrong.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.38: Page number 181\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_D_max=100.0;\t\t\t#Maximum power dissipation of a transistor in mW\n",
-      "V_CE=20.0;\t\t\t#Collector emitter voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "#As power=curent*voltage\n",
-      "#P_D_max=I_C_max*V_CE\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "I_C_max=P_D_max/V_CE;\t\t#Maximum collector current that can be allowed without destruction of the transistor, in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum collector current that can be allowed without destruction of the transistor = %d mA.\"%I_C_max); \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum collector current that can be allowed without destruction of the transistor = 5 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.39: Page number 181\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BB=5.0;\t\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-      "R_B=1.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=200.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along base circuit<\n",
-      "#We get, VBB- IB*RB - VBE=0.\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-      "\n",
-      "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along collector circuit:\n",
-      "\n",
-      "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n",
-      "\n",
-      "#As power=curent*voltage\n",
-      "#P_D=I_C*V_CE\n",
-      "#From the above equation, we get:\n",
-      "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n",
-      "P_D=P_D/1000;\t\t\t\t#Power dissipated in W\n",
-      "\n",
-      "#Result\n",
-      "print(\"Power dissipated = %.1fW\"%P_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power dissipated = 4.3W\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.40: Page number 181-182\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BB=1.0;\t\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-      "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=1.0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=100.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along base circuit<\n",
-      "#We get, VBB- IB*RB - VBE=0.\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-      "\n",
-      "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along collector circuit:\n",
-      "\n",
-      "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n",
-      "\n",
-      "#As power=curent*voltage\n",
-      "#P_D=I_C*V_CE\n",
-      "#From the above equation, we get:\n",
-      "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Power dissipated = %.0fmW\"%P_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power dissipated = 6mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.41 : Page number 182"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VBB=5.0;                     #Base supply voltage, V\n",
-      "RB=22.0;                  #Base resistor, kilo ohm\n",
-      "RC=1.0;                   #Collector resistor, kilo ohm\n",
-      "beta=100.0;               #Base current amplification factor\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "PD_max=800.0;             #Maximum power dissipation, mW\n",
-      "VCE_max=15.0;             #Maximum collector-emitter voltage, V\n",
-      "IC_max=100.0;             #Maximum collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "IB=((VBB-VBE)/RB)*1000;               #Base current,  \u03bcA\n",
-      "IC=beta*IB/1000;                      #Collector current, mA\n",
-      "\n",
-      "print(\"IC=%.1fmA is much less than IC_max=%dmA. Therefore, will not change with VCC and current rating is not exceeded.\"%(IC,IC_max));\n",
-      "\n",
-      "#VCC=VCE+IC*RC\n",
-      "VCC_max=VCE_max+IC*RC;           #Maximum value of Collector supply voltage, V\n",
-      "PD=VCE_max*IC;                  #Power dissipation, mW\n",
-      "\n",
-      "print(\"PD=%dmW  is less than PD_max=%dmW. Therefore, power rating is not exceeded.\"%(PD,PD_max));\n",
-      "\n",
-      "print(\"If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\");"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IC=19.5mA is much less than IC_max=100mA. Therefore, will not change with VCC and current rating is not exceeded.\n",
-        "PD=293mW  is less than PD_max=800mW. Therefore, power rating is not exceeded.\n",
-        "If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_3.ipynb
deleted file mode 100755
index 4afe0858..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_3.ipynb
+++ /dev/null
@@ -1,1851 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:9c13bdd66a3dbb3eae04903205b69bc52bf35e6dadf8b1b3ade1bab68394ae3b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.1: Page number 147-148\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "Signal=500.0;                    #Signal voltage in V\n",
-      "Rin=20.0;                        #Input resistance in \u03a9 \n",
-      "Rout=100.0;                      #Output resistance in \u03a9\n",
-      "R_C=1000.0;                      #Collector load in \u03a9\n",
-      "alpha_ac=1.0;                    #current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_E=(Signal/1000)/Rin;        \t#Input current in mA\n",
-      "I_C=I_E*alpha_ac;               #Output current in mA\n",
-      "Vout=I_C*R_C;                   #Output voltage in V \n",
-      "Av=Vout/(Signal/1000);          #Voltage amplification \n",
-      "\n",
-      "#Result\n",
-      "print(\"The voltage amplification = %d. \"%Av);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage amplification = 50. \n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.2: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_E=1;                            #Emitter curent in mA\n",
-      "I_C=0.95;                         #Collector current in mA\n",
-      "\n",
-      "#Calculation\n",
-      "I_B=I_E-I_C;                      #Base current in mA\n",
-      "\n",
-      "#Result \n",
-      "print(\"The base current = %.2f mA \"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current = 0.05 mA \n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 8.3: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration\n",
-      "alpha=0.9;                      #Current amplification factor\n",
-      "I_E=1;                          #Emitter current in mA\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=alpha*I_E;                  #Collector current in mA\n",
-      "I_B=I_E-I_C;                    #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The base current =%.1f mA\"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current =0.1 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.4: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_C=0.95;\t\t\t#Collector current in mA\n",
-      "I_B=0.05;\t\t\t#Base current in mA\n",
-      "\n",
-      "#Calculation\n",
-      "I_E=I_B+I_C;                    #Emitter current in mA\n",
-      "alpha=I_C/I_E;                  #Current amplification factor \n",
-      "\n",
-      "#Result\n",
-      "print(\"The current amplification factor = %.2f .\"%alpha);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The current amplification factor = 0.95 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.5: Page number 150\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_E=1;                  #Emitter current in mA\n",
-      "I_CBO=50.0;               #Collector current with emitter circuit open, in microAmp\n",
-      "alpha=0.92;             #Current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=alpha*I_E + (I_CBO/1000);           #Total collector current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The total collector current = %.2f mA.\"%I_C);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total collector current = 0.97 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.6: Page number 150-151\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "alpha=0.95;               #Current amplification factor\n",
-      "Rc=2.0;                   #Resistor connected to the collector, in kilo ohm\n",
-      "V_Rc=2.0;                 #Voltage drop across the resistor connected to the collector in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=V_Rc/Rc;              #Collector current in mA\n",
-      "I_E=I_C/alpha;            #Emitter current in mA\n",
-      "I_B=I_E-I_C;              #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The base current = %.2f mA\"%I_B); \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current = 0.05 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.7: Page number 151\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_EE=8.0;                 #Supply voltage at the emitter in V\n",
-      "V_CC=18.0;                #Supply voltage at the collector in V\n",
-      "V_BE=0.7;                 #Base to emitter voltage in V\n",
-      "R_E=1.5;                  #Emitter resistance in \u03a9\n",
-      "R_C=1.2;                  #Collector resistance in \u03a9\n",
-      "\n",
-      "#Calculations\n",
-      "I_E=(V_EE-V_BE)/R_E;              #Emitter current in mA\n",
-      "I_C=I_E;                          #Collector current in mA (approximately equal to emitter current)\n",
-      "V_CB=V_CC-(I_C*R_C);              #Collector to base voltage in V\n",
-      "\n",
-      "#Result\n",
-      "print(\"The collector current =%.2f mA\"%I_C);\n",
-      "print(\"The collector to base voltage = %.2f V\"%V_CB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The collector current =4.87 mA\n",
-        "The collector to base voltage = 12.16 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.8:Page number 155\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Function for calculating beta from alpha\n",
-      "def calc_beta(a):                   #a is the value of alpha\n",
-      "\treturn(a/(1-a));\n",
-      "\n",
-      "#Case (i)\n",
-      "alpha=0.9;                      #current amplification factor\n",
-      "beta=calc_beta(alpha);\t\t#Base current amplification factor    \n",
-      "print(\"(i) Value of beta =%d\"%beta );\t\t\t\t\t\t\t\t\t\n",
-      "\n",
-      "#Case (ii)\n",
-      "alpha=0.98;                     #current amplification factor\n",
-      "beta=calc_beta(alpha);          #Base current amplification factor\n",
-      "print(\"(ii) Value of beta =%.0f\"%beta );\n",
-      "\n",
-      "\n",
-      "#Case (iii)\n",
-      "alpha=0.99;                     #current amplification factor\n",
-      "beta=calc_beta(alpha);          #Base current amplification factor                      \n",
-      "print(\"(iii) Value of beta =%.0f\"%beta );\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Value of beta =9\n",
-        "(ii) Value of beta =49\n",
-        "(iii) Value of beta =99\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.9: Page number 155\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=50.0;                   #Base current amplification factor\n",
-      "I_B=20.0;                    #Base current in microAmp\n",
-      "\n",
-      "#Calculation\n",
-      "I_B=I_B/1000;               #Base current in mA\n",
-      "I_C=beta*I_B;               #Collector current in mA\n",
-      "I_E=I_B+I_C;                #Emitter current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The emitter curent = %.2f mA\"%I_E);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The emitter curent = 1.02 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.10: Page number 155\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_B=240.0;                    #Base current in microAmp\n",
-      "I_E=12;                       #Emitter current in mA\n",
-      "beta=49.0;                      #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "alpha=beta/(1+beta);              #current amplification factor \n",
-      "I_C_alpha=alpha*I_E;              #Collector current in mA calculated using alpha\n",
-      "I_C_beta=beta*(I_B/1000);         #Collector current in mA calculated using beta\n",
-      "\n",
-      "#Results\n",
-      "print(\"alpha=%.2f.\"%alpha);\n",
-      "print(\"Collector current determined using alpha =%.2f mA\"%I_C_alpha);\n",
-      "print(\"Collector current determined using beta =%.2f mA\"%I_C_beta);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "alpha=0.98.\n",
-        "Collector current determined using alpha =11.76 mA\n",
-        "Collector current determined using beta =11.76 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.11: Page number 156\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=45.0;                       #Base current amplification factor\n",
-      "R_C=1.0;                         #Resistance of the collector resistance in k\u03a9\n",
-      "V_R_C=1.0;                       #Voltage drop across the collector resistance in V\n",
-      "\n",
-      "#Calculation\n",
-      "I_C=V_R_C/R_C;                   #Collector current in mA\n",
-      "I_B=I_C/beta;                           #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"The base current =%.3f mA\"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current =0.022 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.12: Page number 156\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=8.0;                  #Collector supply voltage in V\n",
-      "R_C=800.0;                  #Resistance of the collector resistance in \u03a9\n",
-      "V_R_C=0.5;                 #Voltage drop across collector resistance in V\n",
-      "alpha=0.96;                #current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "V_CE=V_CC-V_R_C;           #Collector to emitter voltage in V\n",
-      "I_C=V_R_C/R_C;             #Collector current in A\n",
-      "I_C=I_C*1000;              #Collector current in mA\n",
-      "beta=alpha/(1-alpha);      #Base current amplification factor\n",
-      "I_B=I_C/beta;              #Base current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Collector to emitter voltage = %.1f V\"%V_CE);\n",
-      "print(\"Base current= %.3f mA\"%I_B);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector to emitter voltage = 7.5 V\n",
-        "Base current= 0.026 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.13: Page number 156-157\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=5;                  \t#Collector supply voltage in V\n",
-      "I_CBO=0.2;               \t#Leakage current at collector base junction with emitter open, in  \u03bcA\n",
-      "I_CEO=20.0;              \t#Leakage current with base open, in  \u03bcA\n",
-      "I_C=1.0;                        #Collector current in mA\n",
-      "I_C=I_C*1000;                  \t#Collector current in  \u03bcA\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "alpha=1-(I_CBO/I_CEO);\t\t#current amplification factor\n",
-      "I_E=(I_C-I_CBO)/alpha;          #Emitter current in  \u03bcA\n",
-      "I_E=round(I_E,-1);\n",
-      "I_B=I_E-I_C;                    #Base current in  \u03bcA\n",
-      "I_B=round(I_B,-1);\n",
-      "\n",
-      "#Result\n",
-      "print(\"Current amplification factor = %.2f \"%alpha);\n",
-      "print(\"The emitter curent =%d  \u03bcA \"%I_E);\n",
-      "print(\"The base curent =%d  \u03bcA \"%I_B);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current amplification factor = 0.99 \n",
-        "The emitter curent =1010  \u03bcA \n",
-        "The base curent =10  \u03bcA \n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.14: Page number 157\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_CEO=300.0;              #Leakage current in common emitter configuration, in  \u03bcA\n",
-      "beta=120.0;               #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "alpha=beta/(1+beta);               #Current amplification factor\n",
-      "alpha=round(alpha,3);\n",
-      "I_CBO=(1-alpha)*I_CEO;             #Leakage current in common base configuration, in  \u03bcA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Vale of I_CBO= %.1f \u03bcA\"%I_CBO);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vale of I_CBO= 2.4 \u03bcA\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.15: Page number 157\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_B=20.0;                       #Base current in \u03bcA\n",
-      "I_C=2.0;                        #Collector current in mA\n",
-      "beta=80.0;                      #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_CEO=I_C-(beta*I_B/1000);            #Leakage current with base open, in mA \n",
-      "alpha=beta/(beta+1);                  #Current amplification factor\n",
-      "alpha=round(alpha,3);\n",
-      "I_CBO=(1-alpha)*I_CEO;                #Leakage current with emitter open, in mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Value of I_CBO=%.4f mA\"%I_CBO);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of I_CBO=0.0048 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.17: Page number 158\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "beta=150.0;               \t#Base current amplification factor\n",
-      "R_B=10.0;                   \t#Base resistance in kilo ohm\n",
-      "R_C=100.0;                   \t#Collector resistance in kilo ohm\n",
-      "V_CC=10.0;                      #Collector supply voltage in V\n",
-      "V_BB=5.0;                       #Base supply voltage in V\n",
-      "V_BE=0.7;                       #Base to emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "I_B=(V_BB-V_BE)/R_B;              #Base current in mA\n",
-      "I_C=beta*I_B;                     #Collector current in mA\n",
-      "V_CE=V_CC - (I_C/1000)*R_C;       #Collector to emitter voltage in V\n",
-      "V_CB=V_CE-V_BE;                   #Collector to base voltage in V\n",
-      "\n",
-      "\n",
-      "#Result \n",
-      "print(\"Collector to base voltage, V_CB= %.2f V\"%V_CB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector to base voltage, V_CB= 2.85 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.18: Page number158-159\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_B=68.0;              #Base current in \u03bcA\n",
-      "I_E=30.0;              #Emitter current in mA\n",
-      "beta=440.0;\t     #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "alpha=beta/(beta + 1);          #current amplification factor\n",
-      "I_C_alpha=alpha*I_E;\t\t#Collector current using alpha rating, in mA\n",
-      "I_C_beta=beta*(I_B/1000.0);       #Collector current using beta rating, in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Collector current determined using alpha rating =%.2f mA\"%I_C_alpha);\n",
-      "print(\"Collector current determined using beta rating =%.2f mA\"%I_C_beta);\n",
-      "\n",
-      "#Note: In the textbook, the collector current obtained from beta rating is approximated to 29.93 mA\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current determined using alpha rating =29.93 mA\n",
-        "Collector current determined using beta rating =29.92 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.19: Page number 159\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "I_C_max=500.0;                   #Maximum collector current in mA\n",
-      "beta_max=300.0;                  #Maximum base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "I_B_max=I_C_max/beta_max;           #Maximum base current in mA\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"The maximum allowable value of base current = %.2f mA\"%I_B_max);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowable value of base current = 1.67 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.22 : Page number 167-168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.5;                #Collector supply voltage, V\n",
-      "RC=2.5;                  #Collector resistor, k\u03a9\n",
-      "\n",
-      "#Calculation\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC-VCE)/RC;           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,15])\n",
-      "limit.set_ylim([0,6])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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tROJz+mmhwcHBePbZZ3H48GHs2LEDHTt2lDOXZum919NC1673c6wVomUWLS+gwevUm9XX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\nrkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f2eadbe6710>"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.23 : Page number 168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                #Collector supply voltage, V\n",
-      "RC=6.0;                  #Collector resistor, k\u03a9\n",
-      "IB=20.0;                     #Zero signal base current,  \u03bcA\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC-VCE)/RC;           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,15])\n",
-      "limit.set_ylim([0,5])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n",
-      "#Calculating Q-point\n",
-      "IC=beta*(IB/1000);                                  #Collector current, mA\n",
-      "VCE=VCC-IC*RC;                               #Collector emitter voltage, V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Operating point: IC=%dmA and VCE=%dV.\"%(IC,VCE));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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EVut2p75u3TrcunXrkf3V1dVYv369fdIREZHdmB3qpaWlmDZt2iP7p06dim+++Ua2UFrG\nXk9+ouUFmFkJouUF1MlsdqjX1tZ2+lhzc7PdwxARkW3MdupxcXFYu3Ytnn766Xb78/LysGvXLhw5\ncsS2xdmpExFZrdufUVpSUoI5c+YgJiYG48aNgyRJOHfuHM6cOYNDhw7ZfAUMhzoRkfW6/UJpcHAw\nioqKMHXqVJSVleHKlSuYNm0aioqKnPaSRvZ68hMtL8DMShAtL6DB69QBoFevXkhNTVUiCxER2chs\n/dKvXz/odLqOv9EOH5LB+oWIyHrd7tRtZTAYsGzZMty4cQM6nQ6rVq3CT3/6U4uCERFRx2z+kIzu\ncnNzw/bt21FcXIzCwkK89957uHz5spxLyo69nvxEywswsxJEywto8Dp1Ww0ZMgSRkZEA7lc5ISEh\nqKqqknNJIiKnptj91MvLyzFt2jQUFxejX79+9xdn/UJEZDXV6pc2dXV1SExMREZGhmmgt0lIAPbt\nA+rqlEhCROTYLPo4O1s0NzcjISEBS5Yswfz58x95/Nq1FXjttUAsWwY88YQH5s+PxEsv6dGv3w99\nlF6vB6CN7YsXLyItLU0zeSzZbtunlTyOlvfBrFrJY8n2jh07EBkZqZk8jpa3wI7zoqCgAFlZWQCA\nwMBAmCXJqLW1VVq6dKmUlpbW4eMPLl9dLUmZmZI0e7YkubtLUny8JOXkSFJtrZwJrZefn692BKuJ\nllm0vJLEzEoQLa8kyZfZ3OiWtVP/6quvMHXqVIwePdp0vfumTZswe/ZsAJ33QrdvA59/DuzfD5w5\nAzz5JLBoEfD008BD7Q0RkdNR7Tr1rljyQikHPBFRe6q/UGqLgQOB1FTgyBGgrOz+MM/KAoYOVedF\n1ge7U1GIllm0vAAzK0G0vIADXqdub1ob8EREWqP5+sUSrGiIyJkI3albiwOeiByd0J26teSuaNjr\nyU+0vAAzK0G0vAA7dbtjB09Ezsbh6hdLsKIhIpE5VaduLQ54IhKNU3Xq1rK2omGvJz/R8gLMrATR\n8gLs1FVnyYD//nu1UxIRdc7p6xdLsKIhIi1hp25HHPBEpDZ26nZUVFQg3GWSonWRouUFmFkJouUF\n2KkLh9fBE5HWsH6RASsaIpITO3UVccATkb2xU7cjazsyLVQ0onWRouUFmFkJouUF2Kk7PC0MeCJy\nbKxfNIAVDRFZg526QDjgiagr7NTtSO6OTI6KRrQuUrS8ADMrQbS8ADt1egg7eCKyFusXAbGiIXJu\n7NQdGAc8kfNhp25HWuv1LKlojhwpUDumVbR2jC3BzPITLS/ATp1s1NmAT0xkB0/kLFi/OAFWNESO\nhZ06mXDAE4mPnbodid7riXCZpOjHWBSiZRYtL8BOnRQmwoAnIuuwfqFHsKIh0jZ26tRtHPBE2sNO\n3Y6crddTo6JxtmOsFtEyi5YXYKdOGscOnkj7WL+QzVjRECmLnTophgOeSH6qdeqpqanw9vZGRESE\nnMsoir2eefaoaHiMlSFaZtHyAg7YqaekpODo0aNyLkEaxg6eSHmy1y/l5eWYO3cuLl269OjirF+c\nEisaItvwkkbSFJ7BE8nHVe0AK1asQGBgIADAw8MDkZGR0Ov1AH7oo7S0ffHiRaSlpWkmjyXbbfu0\nkufh7dRUPVJTgYMHC/DVV8C2bcCqVXqMGVMAvR546SU9+vXTTt6Oth8+1mrnsWR7x44dmv99Ezmv\nPedFQUEBsrKyAMA0LzslyaysrEwKDw/v8DEFlre7/Px8tSNYTbTM+fn5UnW1JGVmStLs2ZLk7i5J\n8fGSlJMjSbW1aqfrmGjHWJLEyyxaXkmSL7O52clOnTSPHTxRe6p16snJyYiJiUFJSQn8/f3x4Ycf\nyrkcOSh28ESWk3WoZ2dno6qqCo2NjTAYDEhJSZFzOUU82J2KQrTM5vJqdcCLdowB8TKLlhdwwOvU\nieSk1QFPpCbeJoAcDjt4cnS89ws5LQ54ckR885EdsdeTnz3zKlXRiHaMAfEyi5YXYKdOJCt28OQM\nWL+Q02NFQ6Jhp05kIQ54EgE7dTtiryc/NfN2t6IR7RgD4mUWLS/ATp1IU9jBk4hYvxBZiRUNqY2d\nOpFMOOBJDezU7Yi9nvxEyttW0bz8coFwFY1IxxkQLy/ATp1IaOzgSQtYvxDJjBUN2Rs7dSKN4IAn\ne2Cnbkfs9eQnWl7A8sxaqmhEO86i5QXYqRM5FS0NeHIcrF+INIYVDXWFnTqRoDjgqSPs1O2IvZ78\nRMsLyJdZzopGtOMsWl6AnToRmcEOnizB+oVIcKxonA87dSInwQHvHNip2xF7PfmJlhfQTmZrKhqt\nZLaUaHkBdupEZEddDfgvv2QH74hYvxA5GVY04mOnTkQd4oAXEzt1O2KvJz/R8gLiZhbpMklRj7HS\nONSJCACvg3cUrF+IyCxWNNrDTp2I7IIDXhvYqdsRez35iZYXcJ7MalY0znKMbcWhTkTdwg5em1i/\nEJFdsaKRHzt1IlIFB7w8VOvUjx49ilGjRmHkyJHYvHmznEsphr2e/ETLCzBzZ+xZ0fAYW0a2oW40\nGvH888/j6NGj+Oc//4ns7GxcvnxZruUUc/HiRbUjWE20zKLlBZjZErYOeB5jy8g21M+ePYsRI0Yg\nMDAQbm5uSEpKQm5urlzLKebOnTtqR7CaaJlFywsws7W6M+B5jC0j21CvrKyEv7+/advPzw+VlZVy\nLUdEguJVNPYl21DX6XRyPbWqysvL1Y5gNdEyi5YXYGZ7MTfgP/mkXO14VlPjGMt29UthYSE2btyI\no0ePAgA2bdoEFxcXvPzyyz8s7qCDn4hIbopf0tjS0oInnngCJ0+ehK+vLyZMmIDs7GyEhITIsRwR\nEQFwle2JXV3x61//Gk899RSMRiOeffZZDnQiIpmp+uYjIiKyL9Xu/SLaG5MMBgNiY2MRFhaG8PBw\n7Ny5U+1IFjEajYiKisLcuXPVjmKRO3fuIDExESEhIQgNDUVhYaHakbq0fft2hIeHIyIiAosXL0Zj\nY6PakdpJTU2Ft7c3IiIiTPtu376NmTNnIjg4GLNmzdLc5YIdZf7FL36BkJAQjBkzBvHx8bh7966K\nCdvrKG+bbdu2wcXFBbdv31YkiypDXcQ3Jrm5uWH79u0oLi5GYWEh3nvvPc1nBoCMjAyEhoYK86L0\n+vXrERcXh8uXL6OoqEjzlV1lZSV27dqFc+fO4dKlSzAajcjJyVE7VjspKSmmCxbavP3225g5cyZK\nSkowY8YMvP322yql61hHmWfNmoXi4mJ88803CA4OxqZNm1RK96iO8gL3TwaPHz+OYcOGKZZFlaEu\n4huThgwZgsjISABAv379EBISgqqqKpVTmVdRUYG8vDysXLlSiHvs3L17F6dOnUJqaiqA+6/L9O/f\nX+VUXWtpaUF9fb3pf4cOHap2pHamTJmCAQMGtNt38OBBLF++HACwfPlyfP7552pE61RHmWfOnAkX\nl/sja+LEiaioqFAjWoc6ygsAL774It555x1Fs6gy1EV/Y1J5eTkuXLiAiRMnqh3FrBdeeAFbtmwx\n/SJoXVlZGTw9PZGSkoKxY8fiueeeQ319vdqxzBo6dCh+9rOfISAgAL6+vvDw8MCTTz6pdqwuXb9+\nHd7e3gAAb29vXL9+XeVE1tm9ezfi4uLUjmFWbm4u/Pz8MHr0aEXXVeW3XZQqoCN1dXVITExERkYG\n+mn4NnOHDh2Cl5cXoqKihDhLB+6f8Z4/fx5r1qzB+fPn0bdvX83VAg+rqanBwYMHUV5ejqqqKtTV\n1eGTTz5RO5ZVdDqdUL+Tb731Fnr27InFixerHaVT9fX1SE9Px69+9SvTPqV+D1UZ6kOHDoXBYDBt\nGwwG+Pn5qRHFKs3NzUhISMCSJUswf/58teOYdebMGRw8eBBBQUFITk7Gl19+iWXLlqkdyyw/Pz/4\n+flh/PjxAIDExEScP39e5VTmnThxAkFBQRg0aBBcXV0RHx+PM2fOqB2rS97e3rh27RoA4OrVq/Dy\n8lI5kWWysrKQl5en+T+c//nPf1BeXo4xY8YgKCgIFRUVGDduHG7cuCH72qoM9ejoaHz77bcoLy9H\nU1MT9u3bh3nz5qkRxWKSJOHZZ59FaGgo0tLS1I7TpfT0dBgMBpSVlSEnJwfTp0/Hnj171I5l1pAh\nQ+Dv74+SkhIA9wdmWFiYyqnMGzZsGAoLC/H9999DkiScOHECoaGhasfq0rx58/DRRx8BAD766CPN\nn6QA96+Y27JlC3Jzc9GrVy+145gVERGB69evo6ysDGVlZfDz88P58+eV+eMpqSQvL08KDg6Whg8f\nLqWnp6sVw2KnTp2SdDqdNGbMGCkyMlKKjIyUjhw5onYsixQUFEhz585VO4ZFLl68KEVHR0ujR4+W\nFixYIN25c0ftSF3asGGDNGrUKCk8PFxatmyZ1NTUpHakdpKSkiQfHx/Jzc1N8vPzk3bv3i1VV1dL\nM2bMkEaOHCnNnDlTqqmpUTtmOw9nzszMlEaMGCEFBASYfv9Wr16tdkyTtrw9e/Y0HeMHBQUFSdXV\n1Ypk4ZuPiIgciBiXRRARkUU41ImIHAiHOhGRA+FQJyJyIBzqREQOhEOdiMiBcKgTETkQDnVySNOn\nT8exY8fa7duxYwfWrFmDkpISxMXFITg4GOPGjcMzzzyDGzduoKCgAP3790dUVJTp38mTJwEA33//\nPfR6PVpbW/H444+b3vXaJi0tDe+88w7+8Y9/ICUlRbGfk+hhHOrkkJKTkx+5r/m+ffuQnJyMOXPm\nYO3atSgpKcG5c+ewZs0a3Lx5EzqdDlOnTsWFCxdM/2bMmAHg/l0BExIS4OLi8shzt7a24rPPPkNy\ncjLCw8NRUVHR7t5GREriUCeHlJCQgMOHD6OlpQUATHdR/PbbbxETE4Onn37a9LXTpk1DWFiY2bvo\n7d27Fz/+8Y8B3P+DsW/fPtNjf/3rXzFs2DDT7aTnzp2ruQ/KIOfBoU4OaeDAgZgwYQLy8vIAADk5\nOVi0aBGKi4sxduzYTr/v1KlT7eqXsrIyNDU14b///S8CAgIAAOHh4XBxcUFRUZHpuR+8DWx0dDRO\nnTol409H1DkOdXJYD9Yk+/bts+j+21OmTGlXvwQFBeHWrVvw8PDo8LmNRiNyc3OxcOFC02Oenp6a\n/1Qsclwc6uSw5s2bh5MnT+LChQuor69HVFQUwsLCcO7cOauep3fv3mhoaGi3LykpCZ9++ilOnDiB\n0aNHw9PT0/RYQ0MDevfubZefgchaHOrksPr164fY2FikpKSYztIXL16MM2fOmGoZ4H4nXlxc3Onz\nDBgwAEajEU1NTaZ9jz/+OAYPHoxXXnnlkf8CKCkpQXh4uJ1/GiLLcKiTQ0tOTsalS5eQnJwMAOjV\nqxcOHTqEXbt2ITg4GGFhYfjtb38LT09P6HS6Rzr1AwcOALj/SfYP9+TJycn497//jfj4+Hb78/Pz\nMWfOHGV+QKKH8H7qRBa4cOECtm/f3uWnRzU2NkKv1+P06dPCfOA3ORb+v47IAlFRUYiNjUVra6vZ\nrzMYDNi8eTMHOqmGZ+pERA6EpxNERA6EQ52IyIFwqBMRORAOdSIiB8KhTkTkQP4PTEiMVfQcO/gA\nAAAASUVORK5CYII=\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8947e0f0d0>"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: IC=1mA and VCE=6V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.24 : Page number 168"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "RC=4.0;               #Collector load, k\u03a9\n",
-      "IC_Q=1.0;             #Quiescent current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "VCC=10;                  #Collector supply voltage, V\n",
-      "VCE=VCC-IC*RC;            #Collector emitter voltage, V\n",
-      "\n",
-      "print(\"(i) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n",
-      "\n",
-      "#(ii)\n",
-      "RC=5.0;               #Collector load, k\u03a9\n",
-      "VCE=VCC-IC*RC;            #Collector emitter voltage, V\n",
-      "print(\"(ii) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Operating point: VCE=6V and IC=1mA.\n",
-        "(ii) Operating point: VCE=5V and IC=1mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "markdown",
-     "metadata": {},
-     "source": [
-      "##Example 8.25 : Page number 168-169"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                #Collector supply voltage, V\n",
-      "VBB=10.0;                  #Base supply voltage, V\n",
-      "RC=330.0;                  #Collector resistor, \u03a9\n",
-      "RB=47.0;                #Base resistoe, k\u03a9\n",
-      "beta=200.0;                 #Base current amplification factor\n",
-      "VBE=0.7;                      #Base -emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#VBB-IB*RB-VBE=0\n",
-      "IB=round(((VBB-VBE)/RB)*1000,0);                     #Base current,  \u03bcA\n",
-      "IC=beta*IB/1000;                    #Collector current, mA\n",
-      "VCE=VCC-IC*(RC/1000);                       #Collector-emitter voltage, V\n",
-      "\n",
-      "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n",
-      "\n",
-      "#For d.c load line\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC-VCE)/(RC/1000.0);           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,25])\n",
-      "limit.set_ylim([0,65])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: IC=39.6mA and VCE=6.93V.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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Tw549e0oGpjEB8VInT8Jdd8GuXbB0KbRqZXZEYiUeOyaQmZnJ9u3b6dq1Kzk5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\nIRiBRQAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8947baebd0>"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.26 : Page number 169-170"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                #Collector supply voltage, V\n",
-      "VEE=10.0;                  #Emitter supply voltage, V\n",
-      "RC=1.0;                  #Collector resistor, k\u03a9\n",
-      "RE=4.7;                  #Collector resistor, k\u03a9\n",
-      "RB=47.0;                #Base resistoe, k\u03a9\n",
-      "beta=100.0;                 #Base current amplification factor\n",
-      "VBE=0.7;                      #Base -emitter voltage, V\n",
-      "\n",
-      "#Calculation\n",
-      "#-IB*RB-VBE-IE*RE+VEE=0\n",
-      "#AS, IC=beta*IB and IC~IE\n",
-      "IE=round((VEE-VBE)/(RE+(RB/beta)),1);      #Emitter current,  mA\n",
-      "IC=IE;                                     #Collector current, mA\n",
-      "\n",
-      "#VCC-IC*RC-VCE-IE*RE+VEE=0\n",
-      "#IC~IE\n",
-      "VCE=VCC+VEE-IC*(RC+RE);                       #Collector-emitter voltage, V\n",
-      "\n",
-      "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n",
-      "\n",
-      "\n",
-      "#For d.c load line\n",
-      "#VCE=VCC-IC*RC\n",
-      "#For calculating VCE, IC=0\n",
-      "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-      "VCE_max=VCC+VEE-IC*(RC+RE);               #Maximum collector-emitter voltage, V\n",
-      "\n",
-      "#For calculating VCE, IC=0\n",
-      "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-      "IC_max=(VCC+VEE-VCE)/(RC+RE);           #Maximum collector current, mA\n",
-      "\n",
-      "\n",
-      "#Plotting of d.c load line\n",
-      "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-      "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-      "p=plot(VCE_plot,IC_plot);\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,25])\n",
-      "limit.set_ylim([0,5])\n",
-      "xlabel('VCE(V)');\n",
-      "ylabel('IC(mA)');\n",
-      "title('d.c load line');\n",
-      "plt.grid();\n",
-      "show(p);\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: IC=1.8mA and VCE=9.74V.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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CpCTzBRCodFx4R8VeJMh17mzm4w8ZAj17wt13Q16e3VGJv6lnLxJCDhyA8eNN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\nQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8947c78950>"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.27 : Page number 170-171"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VEE=10.0;                    #Emitter supply voltage, V\n",
-      "IE=1.8;                      #Emitter current, mA\n",
-      "RE=4.7;                      #Emitter resistor, k\u03a9\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "VCC=10.0;                     #Collector supply voltage, V\n",
-      "IC=1.8;                       #Collector current, mA\n",
-      "RC=1.0;                       #Collector resistor, k\u03a9\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
-      "#(i)\n",
-      "VE=-VEE+IE*RE;               #Emitter voltage, V\n",
-      "\n",
-      "#(ii)\n",
-      "VB=VEE+VBE;                   #Base voltage, V\n",
-      "\n",
-      "#(iii)\n",
-      "VC=VCC-IC*RC;                 #Collector voltage, V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) Emitter voltage=%.2fV.\"%VE);\n",
-      "print(\"(i) Base voltage=%.1fV.\"%VB);\n",
-      "print(\"(i) Collector voltage=%.1fV.\"%VC);\n",
-      "\n",
-      "#Note: In the textbook, VB=VE+VBE has been written, which is worng. It should be VB=VEE+VBE. "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Emitter voltage=-1.54V.\n",
-        "(i) Base voltage=10.7V.\n",
-        "(i) Collector voltage=8.2V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "\n",
-      "Example 8.28: Page number 173-174\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_BE_change=200.0;                #Change in base-emitter voltage in mV\n",
-      "I_B_change=100.0;                  #Change in base current in  \u03bcA\n",
-      "\n",
-      "#Calculations\n",
-      "Ri=V_BE_change/I_B_change;              #Input resistance in k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"Input resistance =%d k\u03a9\"%Ri);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Input resistance =2 k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.29; Page number 174\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CE_final=10.0;\t\t\t#Final value of collector-emitter voltage in V\n",
-      "V_CE_initial=2.0;                 #Initial value of collector-emitter voltage in V\n",
-      "I_C_final=3.0;                    #Final value of collector current in mA\n",
-      "I_C_initial=2.0;                  #Initial value of collector current in mA\n",
-      "\n",
-      "#Calculations\n",
-      "V_CE_change=V_CE_final-V_CE_initial;\t\t#Change in collector to emitter voltage in V\n",
-      "I_C_change=I_C_final-I_C_initial;               #Change in collector current in mA\n",
-      "R0=V_CE_change/I_C_change;                      #Output resistance in k\u03a9\n",
-      "\n",
-      "#Result\n",
-      "print(\"The output resistance =%dk\u03a9\"%R0);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The output resistance =8k\u03a9\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.30: Page number 174\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "R_C=2.0;\t\t#Collector load in kilo ohm\n",
-      "R_i=1.0;\t\t#Input resistance in kilo ohm\n",
-      "R_AC=R_C;               #Effective collector load for single stage in kilo ohm(appoximately equal to collector load for single stage)\n",
-      "beta=50.0;              #Current gain\n",
-      "\n",
-      "#Calculations\n",
-      "A_v=beta*(R_AC/R_i);\t\t#Voltage gain of the amplifier\n",
-      "\n",
-      "#Result \n",
-      "print(\"The voltage gain of the amplifier =%d \"%A_v);\t\t\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage gain of the amplifier =100 \n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.31: Page number 175-176\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=20;\t\t#Collector supply voltage in V\n",
-      "R_C=1;                  #Collector resistance in kilo ohm\n",
-      "V_knee_Si=1;\t\t#Knee voltage of V_CE for Si in V \n",
-      "V_knee_Ge=0.5;\t\t#Knee voltage of V_CE for Ge in V\n",
-      "\n",
-      "#Calculations\n",
-      "I_C_sat_Si=(V_CC-V_knee_Si)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Si transistor)\n",
-      "I_C_sat_Ge=(V_CC-V_knee_Ge)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Ge transistor)\n",
-      "I_C_sat=(V_CC)/R_C;\t\t\t\t#Saturation (maximum) value of collector current in mA (neglecting knee voltage)\n",
-      "V_CE_cut_off=V_CC;                              #Collector to emitter voltage in cutoff when base current=0, in V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Collector current during saturation = %d mA\"%I_C_sat);\n",
-      "print(\"Collector emitter voltage during cutoff = %d V.\"%V_CE_cut_off);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current during saturation = 20 mA\n",
-        "Collector emitter voltage during cutoff = 20 V.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.32: Page number 176-177\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=12.0;\t\t#Collector supply voltage in V\n",
-      "V_EE=12.0;\t\t#Emitter supply voltage in V\n",
-      "R_C=750.0;\t\t#Collector resistance in ohm\n",
-      "R_E=1.5;\t\t#Emitter resistance in kilo ohm\n",
-      "R_B=100.0;\t\t#Base resistance in ohm\n",
-      "beta=200;\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to the collector side of the circuit\n",
-      "#using the equation: Vcc -IcRc-Vce -IeRe+Vee=0\n",
-      "#we get Vce=Vcc+Vee-Ic(Rc+Re), [Ie=Ic, approximately]\n",
-      "#We get Vce(off), when Ic=0;\n",
-      "\n",
-      "I_C_Vce_off=0;\t\t\t\t\t#Collector current for Vce(off) in mA\n",
-      "V_CE_off=V_CC+V_EE -(I_C_Vce_off * (R_C +R_E));\t#Collector to emitter voltage in V, during transistor in off state\n",
-      "\n",
-      "#We get Ic(sat), when Vce=0\n",
-      "V_CE_Ic_sat=0;\t\t\t\t\t\t#Collector to emitter voltage for saturation current of collector in V\n",
-      "I_C_sat=(V_CC+V_EE-V_CE_Ic_sat)/(R_C+(R_E*1000));\t#Saturated collector current in A \n",
-      "I_C_sat=I_C_sat*1000;\t\t\t\t\t#Saturated collector current in mA\n",
-      "#Result\n",
-      "print(\"Vce(off)= %dV\"%V_CE_off);\n",
-      "print(\"Ic(sat) = %.2f mA\"%I_C_sat);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Vce(off)= 24V\n",
-        "Ic(sat) = 10.67 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.33 : Page number 177\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_knee=0.2;\t\t\t\t#Knee voltage of collector-emitter voltage in V\n",
-      "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BB=3.0;\t\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V \t\n",
-      "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=1.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=50.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#applying Kirchhoff's voltage law along the collector side of the circuit,\n",
-      "#We get Vcc-Ic(sat)*Rc-V_knee=0\n",
-      "#From the above equation, we get:\n",
-      "I_C_sat=(V_CC-V_knee)/R_C;\t\t#Saturated collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along base emitter side,\n",
-      "#We get VBB-IB*RB-VBE=0;\n",
-      "#From the above equation, we get:\n",
-      "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-      "\n",
-      "\n",
-      "I_C=beta*I_B\t\t\t\t#Collector current in mA\n",
-      "\n",
-      "#Result\n",
-      "if(I_C>I_C_sat):\n",
-      "\tprint(\"The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\");\n",
-      "else:\n",
-      "\tprint(\"The base current is not large enough to produce Ic greater than Ic(sat), therefore the transistor isn't saturated. \");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.34: Page number 177-178\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BE=0.95;\t\t\t\t#Base-emitter voltage in V \t\n",
-      "I_B=100.0;\t\t\t\t#Base current in microAmp\n",
-      "R_C=970.0;\t\t\t\t#Collector resistor's resistance in ohm\n",
-      "beta=100.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "I_C=(I_B/1000)*beta;\t\t\t\t#Collector current in mA \n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along collector side\n",
-      "#We get Vcc-IcRc-Vce=0\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "V_CE=V_CC-((I_C/1000)*R_C);\t\t\t\t#Collector-emitter voltage in V\n",
-      "\n",
-      "#From the equation, V_CE=V_CB+V_BE,\n",
-      "V_CB=V_CE-V_BE;\t\t\t\t\t\t#Collector-base voltage in V\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "if(V_CB<0 and V_BE >0):\n",
-      "\tprint(\"As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \");\n",
-      "else:\n",
-      "\tprint(\"No. The transistor isn't operating in the saturation region.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.35: Page number 178\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=10.0;\t\t\t\t#Collector supplu voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-      "R_B=50.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=2.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=200.0;\t\t\t\t#Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along the collector side,\n",
-      "#We get, Vcc-Ic(sat)*Rc-Vce=0;\n",
-      "#From the above equation, we get:\n",
-      "#I_C_sat=(V_CC-V_CE)/R_C, but as transistor goes into saturation, Vce=0;\n",
-      "\n",
-      "V_CE=0;\t\t\t\t\t\t#Collector-emiter voltage in V, for transistor in saturation \n",
-      "I_C_sat=(V_CC-V_CE)/R_C;\t\t\t#Saturated collector current in mA\n",
-      "\n",
-      "I_B=I_C_sat/beta;\t\t\t\t#Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law to the base circuit,\n",
-      "#We get, VBB - IB*RB - VBE=0\n",
-      "#From the above equation. we get:\n",
-      "V_BB=V_BE+ I_B*R_B;\t\t\t\t#Base supply voltage to put transistor in saturation, in V\n",
-      "\n",
-      "#Result\n",
-      "print(\"Therefore, for putting transistor in saturation, VBB >= %.2f V\"%V_BB);\n",
-      " \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Therefore, for putting transistor in saturation, VBB >= 1.95 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.36: Page number 178-179\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=10.0;\t\t\t#Collector supply voltage in V\n",
-      "V_BB=2.7;\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n",
-      "beta=100.0;\t\t\t#Base current amplification factor\n",
-      "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calcultaion\t\n",
-      "V_B=V_BB;\t\t\t#Base voltage in V\n",
-      "V_E=V_B-V_BE;\t\t\t#Emitter voltage in V\n",
-      "I_E=V_E/R_E;\t\t\t#Emitter current in mA\n",
-      "I_C=I_E;\t\t\t#Collector current (approximately equal to emitter current) in mA\n",
-      "I_B=I_C/beta;\t\t\t#Base current in mA\n",
-      "\n",
-      "#Case (i):\n",
-      "R_C=2;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "#We get,Vcc-IcRc=Vc,\n",
-      "\n",
-      "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-      "\n",
-      "if(V_C>V_E):\n",
-      "\tprint(\"(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\");\n",
-      "elif(V_C<V_E):\n",
-      "\tprint(\"(i)Our assumption was wrong, the transistor is in saturation for Rc=2 kilo ohm.\");\n",
-      "elif(V_C==V_E):\n",
-      "\tprint(\"(i)The transistor is at the edge of saturation for Rc=2 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n",
-      "\n",
-      "#Case (ii):\n",
-      "R_C=4;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "#We get,Vcc-IcRc=Vc,\n",
-      "\n",
-      "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-      "if(V_C>V_E):\n",
-      "\tprint(\"(ii)Our assumption was correct, the transistor is in active state for Rc=4 kilo ohm.\");\n",
-      "elif(V_C==V_E):\n",
-      "\tprint(\"(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n",
-      "elif(V_C<V_E):\n",
-      "\tprint(\"(ii)Our assumption was wrong, the transistor is in saturation for Rc=4 kilo ohm.\");\n",
-      "\n",
-      "\n",
-      "#Case (iii):\n",
-      "R_C=8;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "#We get,Vcc-IcRc=Vc,\n",
-      "\n",
-      "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-      "if(V_C>V_E):\n",
-      "\tprint(\"(iii)Our assumption was correct, the transistor is in active state for Rc=8 kilo ohm.\");\n",
-      "elif(V_C<V_E):\n",
-      "\tprint(\"(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\");\n",
-      "elif(V_C==V_E):\n",
-      "\tprint(\"(iii)The transistor is at the edge of saturation for Rc=8 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\n",
-        "(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\n",
-        "(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.37 : Page number 179-180"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=15.0;\t\t\t#Collector supply voltage in V\n",
-      "R_C=10.0;\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n",
-      "beta=100.0;\t\t\t#Base current amplification factor\n",
-      "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n",
-      "\n",
-      "\n",
-      "#Calculation\t\n",
-      "\n",
-      "#Case (i):\n",
-      "V_BB=0.5;\t\t\t#Base supply voltage in V\n",
-      "VB=V_BB;            #Base voltage, V\n",
-      "print(\"(i) Base voltage =%.1fV is less than VBE=%.1fV, therefore, transistor is cut-off.\"%(VB,V_BE));\n",
-      "\n",
-      "\n",
-      "#Case (ii):\n",
-      "V_BB=1.5;\t\t\t#Base supply voltage in V\n",
-      "VB=V_BB;            #Base voltage, V\n",
-      "VE=VB-V_BE;          #Emitter voltage, V\n",
-      "IE=round(VE/R_E,1);  #Emitter current, mA\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "IC=IE;              #Collector current, mA\n",
-      "IB=IC/beta;         #Base  current, mA\n",
-      "VC=V_CC-IC*R_C;      #Collector voltage, V\n",
-      "print(VE,IE,VC);\n",
-      "print(\"(ii) VC=%dV > VE=%.1fV, therefore the transistor is active. Our assumption was correct.\"%(VC,VE));\n",
-      "\n",
-      "#Case (iii):\n",
-      "V_BB=3; \t\t\t#Base supply voltage in V\n",
-      "VB=V_BB;            #Base voltage, V\n",
-      "VE=VB-V_BE;          #Emitter voltage, V\n",
-      "IE=round(VE/R_E,1);  #Emitter current, mA\n",
-      "#Assuming transistor to be in active state\n",
-      "#Applying Kirchhoff's voltage law along collector side,\n",
-      "IC=IE;              #Collector current, mA\n",
-      "IB=IC/beta;         #Base  current, mA\n",
-      "VC=V_CC-IC*R_C;      #Collector voltage, V\n",
-      "\n",
-      "print(\"(iii) VC=%dV < VE=%.1fV, therefore the transistor is saturated. Our assumption was wrong.\"%(VC,VE));"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Base voltage =0.5V is less than VBE=0.7V, therefore, transistor is cut-off.\n",
-        "(0.8, 0.8, 7.0)\n",
-        "(ii) VC=7V > VE=0.8V, therefore the transistor is active. Our assumption was correct.\n",
-        "(iii) VC=-8V < VE=2.3V, therefore the transistor is saturated. Our assumption was wrong.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.38: Page number 181\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "P_D_max=100.0;\t\t\t#Maximum power dissipation of a transistor in mW\n",
-      "V_CE=20.0;\t\t\t#Collector emitter voltage in V\n",
-      "\n",
-      "#Calculation\n",
-      "#As power=curent*voltage\n",
-      "#P_D_max=I_C_max*V_CE\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "I_C_max=P_D_max/V_CE;\t\t#Maximum collector current that can be allowed without destruction of the transistor, in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Maximum collector current that can be allowed without destruction of the transistor = %d mA.\"%I_C_max); \n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum collector current that can be allowed without destruction of the transistor = 5 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.39: Page number 181\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BB=5.0;\t\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-      "R_B=1.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=200.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along base circuit<\n",
-      "#We get, VBB- IB*RB - VBE=0.\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-      "\n",
-      "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along collector circuit:\n",
-      "\n",
-      "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n",
-      "\n",
-      "#As power=curent*voltage\n",
-      "#P_D=I_C*V_CE\n",
-      "#From the above equation, we get:\n",
-      "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n",
-      "P_D=P_D/1000;\t\t\t\t#Power dissipated in W\n",
-      "\n",
-      "#Result\n",
-      "print(\"Power dissipated = %.1fW\"%P_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power dissipated = 4.3W\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.40: Page number 181-182\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n",
-      "V_BB=1.0;\t\t\t\t#Base supply voltage in V\n",
-      "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-      "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-      "R_C=1.0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-      "beta=100.0;\t\t\t\t#base current amplification factor\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along base circuit<\n",
-      "#We get, VBB- IB*RB - VBE=0.\n",
-      "#From the above equation, we get:\n",
-      "\n",
-      "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-      "\n",
-      "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along collector circuit:\n",
-      "\n",
-      "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n",
-      "\n",
-      "#As power=curent*voltage\n",
-      "#P_D=I_C*V_CE\n",
-      "#From the above equation, we get:\n",
-      "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"Power dissipated = %.0fmW\"%P_D);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power dissipated = 6mW\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 8.41 : Page number 182"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VBB=5.0;                     #Base supply voltage, V\n",
-      "RB=22.0;                  #Base resistor, kilo ohm\n",
-      "RC=1.0;                   #Collector resistor, kilo ohm\n",
-      "beta=100.0;               #Base current amplification factor\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "PD_max=800.0;             #Maximum power dissipation, mW\n",
-      "VCE_max=15.0;             #Maximum collector-emitter voltage, V\n",
-      "IC_max=100.0;             #Maximum collector current, mA\n",
-      "\n",
-      "#Calculation\n",
-      "IB=((VBB-VBE)/RB)*1000;               #Base current,  \u03bcA\n",
-      "IC=beta*IB/1000;                      #Collector current, mA\n",
-      "\n",
-      "print(\"IC=%.1fmA is much less than IC_max=%dmA. Therefore, will not change with VCC and current rating is not exceeded.\"%(IC,IC_max));\n",
-      "\n",
-      "#VCC=VCE+IC*RC\n",
-      "VCC_max=VCE_max+IC*RC;           #Maximum value of Collector supply voltage, V\n",
-      "PD=VCE_max*IC;                  #Power dissipation, mW\n",
-      "\n",
-      "print(\"PD=%dmW  is less than PD_max=%dmW. Therefore, power rating is not exceeded.\"%(PD,PD_max));\n",
-      "\n",
-      "print(\"If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\");"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IC=19.5mA is much less than IC_max=100mA. Therefore, will not change with VCC and current rating is not exceeded.\n",
-        "PD=293mW  is less than PD_max=800mW. Therefore, power rating is not exceeded.\n",
-        "If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_4.ipynb
deleted file mode 100755
index c13922ee..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_4.ipynb
+++ /dev/null
@@ -1,1845 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.1: Page number 147-148"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage amplification = 50. \n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Signal=500.0;                    #Signal voltage in V\n",
-    "Rin=20.0;                        #Input resistance in Ω \n",
-    "Rout=100.0;                      #Output resistance in Ω\n",
-    "R_C=1000.0;                      #Collector load in Ω\n",
-    "alpha_ac=1.0;                    #current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "I_E=(Signal/1000)/Rin;        \t#Input current in mA\n",
-    "I_C=I_E*alpha_ac;               #Output current in mA\n",
-    "Vout=I_C*R_C;                   #Output voltage in V \n",
-    "Av=Vout/(Signal/1000);          #Voltage amplification \n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage amplification = %d. \"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.2: Page number 150"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The base current = 0.05 mA \n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_E=1;                            #Emitter curent in mA\n",
-    "I_C=0.95;                         #Collector current in mA\n",
-    "\n",
-    "#Calculation\n",
-    "I_B=I_E-I_C;                      #Base current in mA\n",
-    "\n",
-    "#Result \n",
-    "print(\"The base current = %.2f mA \"%I_B);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# Example 8.3: Page number 150\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The base current =0.1 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#variable declaration\n",
-    "alpha=0.9;                      #Current amplification factor\n",
-    "I_E=1;                          #Emitter current in mA\n",
-    "\n",
-    "#Calculation\n",
-    "I_C=alpha*I_E;                  #Collector current in mA\n",
-    "I_B=I_E-I_C;                    #Base current in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The base current =%.1f mA\"%I_B);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.4: Page number 150"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The current amplification factor = 0.95 .\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_C=0.95;\t\t\t#Collector current in mA\n",
-    "I_B=0.05;\t\t\t#Base current in mA\n",
-    "\n",
-    "#Calculation\n",
-    "I_E=I_B+I_C;                    #Emitter current in mA\n",
-    "alpha=I_C/I_E;                  #Current amplification factor \n",
-    "\n",
-    "#Result\n",
-    "print(\"The current amplification factor = %.2f .\"%alpha);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.5: Page number 150"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The total collector current = 0.97 mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_E=1;                  #Emitter current in mA\n",
-    "I_CBO=50.0;               #Collector current with emitter circuit open, in microAmp\n",
-    "alpha=0.92;             #Current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "I_C=alpha*I_E + (I_CBO/1000);           #Total collector current in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The total collector current = %.2f mA.\"%I_C);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.6: Page number 150-151"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The base current = 0.05 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "alpha=0.95;               #Current amplification factor\n",
-    "Rc=2.0;                   #Resistor connected to the collector, in kilo ohm\n",
-    "V_Rc=2.0;                 #Voltage drop across the resistor connected to the collector in V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_C=V_Rc/Rc;              #Collector current in mA\n",
-    "I_E=I_C/alpha;            #Emitter current in mA\n",
-    "I_B=I_E-I_C;              #Base current in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The base current = %.2f mA\"%I_B); \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.7: Page number 151"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The collector current =4.87 mA\n",
-      "The collector to base voltage = 12.16 V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_EE=8.0;                 #Supply voltage at the emitter in V\n",
-    "V_CC=18.0;                #Supply voltage at the collector in V\n",
-    "V_BE=0.7;                 #Base to emitter voltage in V\n",
-    "R_E=1.5;                  #Emitter resistance in Ω\n",
-    "R_C=1.2;                  #Collector resistance in Ω\n",
-    "\n",
-    "#Calculations\n",
-    "I_E=(V_EE-V_BE)/R_E;              #Emitter current in mA\n",
-    "I_C=I_E;                          #Collector current in mA (approximately equal to emitter current)\n",
-    "V_CB=V_CC-(I_C*R_C);              #Collector to base voltage in V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The collector current =%.2f mA\"%I_C);\n",
-    "print(\"The collector to base voltage = %.2f V\"%V_CB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.8:Page number 155"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Value of beta =9\n",
-      "(ii) Value of beta =49\n",
-      "(iii) Value of beta =99\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Function for calculating beta from alpha\n",
-    "def calc_beta(a):                   #a is the value of alpha\n",
-    "\treturn(a/(1-a));\n",
-    "\n",
-    "#Case (i)\n",
-    "alpha=0.9;                      #current amplification factor\n",
-    "beta=calc_beta(alpha);\t\t#Base current amplification factor    \n",
-    "print(\"(i) Value of beta =%d\"%beta );\t\t\t\t\t\t\t\t\t\n",
-    "\n",
-    "#Case (ii)\n",
-    "alpha=0.98;                     #current amplification factor\n",
-    "beta=calc_beta(alpha);          #Base current amplification factor\n",
-    "print(\"(ii) Value of beta =%.0f\"%beta );\n",
-    "\n",
-    "\n",
-    "#Case (iii)\n",
-    "alpha=0.99;                     #current amplification factor\n",
-    "beta=calc_beta(alpha);          #Base current amplification factor                      \n",
-    "print(\"(iii) Value of beta =%.0f\"%beta );\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.9: Page number 155"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The emitter curent = 1.02 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "beta=50.0;                   #Base current amplification factor\n",
-    "I_B=20.0;                    #Base current in microAmp\n",
-    "\n",
-    "#Calculation\n",
-    "I_B=I_B/1000;               #Base current in mA\n",
-    "I_C=beta*I_B;               #Collector current in mA\n",
-    "I_E=I_B+I_C;                #Emitter current in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The emitter curent = %.2f mA\"%I_E);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.10: Page number 155"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "alpha=0.98.\n",
-      "Collector current determined using alpha =11.76 mA\n",
-      "Collector current determined using beta =11.76 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_B=240.0;                    #Base current in microAmp\n",
-    "I_E=12;                       #Emitter current in mA\n",
-    "beta=49.0;                      #Base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "alpha=beta/(1+beta);              #current amplification factor \n",
-    "I_C_alpha=alpha*I_E;              #Collector current in mA calculated using alpha\n",
-    "I_C_beta=beta*(I_B/1000);         #Collector current in mA calculated using beta\n",
-    "\n",
-    "#Results\n",
-    "print(\"alpha=%.2f.\"%alpha);\n",
-    "print(\"Collector current determined using alpha =%.2f mA\"%I_C_alpha);\n",
-    "print(\"Collector current determined using beta =%.2f mA\"%I_C_beta);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.11: Page number 156"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The base current =0.022 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "beta=45.0;                       #Base current amplification factor\n",
-    "R_C=1.0;                         #Resistance of the collector resistance in kΩ\n",
-    "V_R_C=1.0;                       #Voltage drop across the collector resistance in V\n",
-    "\n",
-    "#Calculation\n",
-    "I_C=V_R_C/R_C;                   #Collector current in mA\n",
-    "I_B=I_C/beta;                           #Base current in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The base current =%.3f mA\"%I_B);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.12: Page number 156"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Collector to emitter voltage = 7.5 V\n",
-      "Base current= 0.026 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=8.0;                  #Collector supply voltage in V\n",
-    "R_C=800.0;                  #Resistance of the collector resistance in Ω\n",
-    "V_R_C=0.5;                 #Voltage drop across collector resistance in V\n",
-    "alpha=0.96;                #current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "V_CE=V_CC-V_R_C;           #Collector to emitter voltage in V\n",
-    "I_C=V_R_C/R_C;             #Collector current in A\n",
-    "I_C=I_C*1000;              #Collector current in mA\n",
-    "beta=alpha/(1-alpha);      #Base current amplification factor\n",
-    "I_B=I_C/beta;              #Base current in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Collector to emitter voltage = %.1f V\"%V_CE);\n",
-    "print(\"Base current= %.3f mA\"%I_B);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.13: Page number 156-157"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Current amplification factor = 0.99 \n",
-      "The emitter curent =1010  μA \n",
-      "The base curent =10  μA \n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=5;                  \t#Collector supply voltage in V\n",
-    "I_CBO=0.2;               \t#Leakage current at collector base junction with emitter open, in  μA\n",
-    "I_CEO=20.0;              \t#Leakage current with base open, in  μA\n",
-    "I_C=1.0;                        #Collector current in mA\n",
-    "I_C=I_C*1000;                  \t#Collector current in  μA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "alpha=1-(I_CBO/I_CEO);\t\t#current amplification factor\n",
-    "I_E=(I_C-I_CBO)/alpha;          #Emitter current in  μA\n",
-    "I_E=round(I_E,-1);\n",
-    "I_B=I_E-I_C;                    #Base current in  μA\n",
-    "I_B=round(I_B,-1);\n",
-    "\n",
-    "#Result\n",
-    "print(\"Current amplification factor = %.2f \"%alpha);\n",
-    "print(\"The emitter curent =%d  μA \"%I_E);\n",
-    "print(\"The base curent =%d  μA \"%I_B);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.14: Page number 157"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vale of I_CBO= 2.4 μA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_CEO=300.0;              #Leakage current in common emitter configuration, in  μA\n",
-    "beta=120.0;               #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "alpha=beta/(1+beta);               #Current amplification factor\n",
-    "alpha=round(alpha,3);\n",
-    "I_CBO=(1-alpha)*I_CEO;             #Leakage current in common base configuration, in  μA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vale of I_CBO= %.1f μA\"%I_CBO);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.15: Page number 157"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Value of I_CBO=0.0048 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_B=20.0;                       #Base current in μA\n",
-    "I_C=2.0;                        #Collector current in mA\n",
-    "beta=80.0;                      #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "I_CEO=I_C-(beta*I_B/1000);            #Leakage current with base open, in mA \n",
-    "alpha=beta/(beta+1);                  #Current amplification factor\n",
-    "alpha=round(alpha,3);\n",
-    "I_CBO=(1-alpha)*I_CEO;                #Leakage current with emitter open, in mA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Value of I_CBO=%.4f mA\"%I_CBO);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.17: Page number 158"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Collector to base voltage, V_CB= 2.85 V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "beta=150.0;               \t#Base current amplification factor\n",
-    "R_B=10.0;                   \t#Base resistance in kilo ohm\n",
-    "R_C=100.0;                   \t#Collector resistance in kilo ohm\n",
-    "V_CC=10.0;                      #Collector supply voltage in V\n",
-    "V_BB=5.0;                       #Base supply voltage in V\n",
-    "V_BE=0.7;                       #Base to emitter voltage in V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_B=(V_BB-V_BE)/R_B;              #Base current in mA\n",
-    "I_C=beta*I_B;                     #Collector current in mA\n",
-    "V_CE=V_CC - (I_C/1000)*R_C;       #Collector to emitter voltage in V\n",
-    "V_CB=V_CE-V_BE;                   #Collector to base voltage in V\n",
-    "\n",
-    "\n",
-    "#Result \n",
-    "print(\"Collector to base voltage, V_CB= %.2f V\"%V_CB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.18: Page number158-159"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Collector current determined using alpha rating =29.93 mA\n",
-      "Collector current determined using beta rating =29.92 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_B=68.0;              #Base current in μA\n",
-    "I_E=30.0;              #Emitter current in mA\n",
-    "beta=440.0;\t     #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "alpha=beta/(beta + 1);          #current amplification factor\n",
-    "I_C_alpha=alpha*I_E;\t\t#Collector current using alpha rating, in mA\n",
-    "I_C_beta=beta*(I_B/1000.0);       #Collector current using beta rating, in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Collector current determined using alpha rating =%.2f mA\"%I_C_alpha);\n",
-    "print(\"Collector current determined using beta rating =%.2f mA\"%I_C_beta);\n",
-    "\n",
-    "#Note: In the textbook, the collector current obtained from beta rating is approximated to 29.93 mA\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.19: Page number 159"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The maximum allowable value of base current = 1.67 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_C_max=500.0;                   #Maximum collector current in mA\n",
-    "beta_max=300.0;                  #Maximum base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "I_B_max=I_C_max/beta_max;           #Maximum base current in mA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The maximum allowable value of base current = %.2f mA\"%I_B_max);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.22 : Page number 167-168"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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5R9IrOjzo79B4NarZ0Lx6NxudXgb7auASSZ+UdG728Ska/fqqfONVk3u+fLR27xs2zJXy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\neLCbmdWMB7uZWc38f6b7SUAyF1dTAAAAAElFTkSuQmCC\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f9cdbfbafd0>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=12.5;                #Collector supply voltage, V\n",
-    "RC=2.5;                  #Collector resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "#VCE=VCC-IC*RC\n",
-    "#For calculating VCE, IC=0\n",
-    "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-    "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-    "\n",
-    "#For calculating VCE, IC=0\n",
-    "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-    "IC_max=(VCC-VCE)/RC;           #Maximum collector current, mA\n",
-    "\n",
-    "\n",
-    "#Plotting of d.c load line\n",
-    "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-    "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-    "p=plt.plot(VCE_plot,IC_plot);\n",
-    "limit = plt.gca()\n",
-    "limit.set_xlim([0,15])\n",
-    "limit.set_ylim([0,6])\n",
-    "plt.xlabel('VCE(V)');\n",
-    "plt.ylabel('IC(mA)');\n",
-    "plt.title('d.c load line');\n",
-    "plt.grid();\n",
-    "plt.show(p);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.23 : Page number 168"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f9cec03e9b0>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Operating point: IC=1mA and VCE=6V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=12.0;                #Collector supply voltage, V\n",
-    "RC=6.0;                  #Collector resistor, kΩ\n",
-    "IB=20.0;                     #Zero signal base current,  μA\n",
-    "beta=50.0;                  #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "#VCE=VCC-IC*RC\n",
-    "#For calculating VCE, IC=0\n",
-    "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-    "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-    "\n",
-    "#For calculating VCE, IC=0\n",
-    "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-    "IC_max=(VCC-VCE)/RC;           #Maximum collector current, mA\n",
-    "\n",
-    "\n",
-    "#Plotting of d.c load line\n",
-    "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-    "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-    "p=plt.plot(VCE_plot,IC_plot);\n",
-    "limit = plt.gca()\n",
-    "limit.set_xlim([0,15])\n",
-    "limit.set_ylim([0,5])\n",
-    "plt.xlabel('VCE(V)');\n",
-    "plt.ylabel('IC(mA)');\n",
-    "plt.title('d.c load line');\n",
-    "plt.grid();\n",
-    "plt.show(p);\n",
-    "\n",
-    "#Calculating Q-point\n",
-    "IC=beta*(IB/1000);                                  #Collector current, mA\n",
-    "VCE=VCC-IC*RC;                               #Collector emitter voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Operating point: IC=%dmA and VCE=%dV.\"%(IC,VCE));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.24 : Page number 168"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Operating point: VCE=6V and IC=1mA.\n",
-      "(ii) Operating point: VCE=5V and IC=1mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=4.0;               #Collector load, kΩ\n",
-    "IC_Q=1.0;             #Quiescent current, mA\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "VCC=10;                  #Collector supply voltage, V\n",
-    "VCE=VCC-IC*RC;            #Collector emitter voltage, V\n",
-    "\n",
-    "print(\"(i) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n",
-    "\n",
-    "#(ii)\n",
-    "RC=5.0;               #Collector load, kΩ\n",
-    "VCE=VCC-IC*RC;            #Collector emitter voltage, V\n",
-    "print(\"(ii) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# Example 8.25 : Page number 168-169"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Operating point: IC=39.6mA and VCE=6.93V.\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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+faI2kUhu3XVXuCZhzJjQPpLikMY20Thg/2r7RgDT3X1LYCbwqxzHICK1OOYY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\nSUVORK5CYII=\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f9cec03e978>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "from matplotlib import pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=20.0;                #Collector supply voltage, V\n",
-    "VBB=10.0;                  #Base supply voltage, V\n",
-    "RC=330.0;                  #Collector resistor, Ω\n",
-    "RB=47.0;                #Base resistoe, kΩ\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "VBE=0.7;                      #Base -emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#VBB-IB*RB-VBE=0\n",
-    "IB=round(((VBB-VBE)/RB)*1000,0);                     #Base current,  μA\n",
-    "IC=beta*IB/1000;                    #Collector current, mA\n",
-    "VCE=VCC-IC*(RC/1000);                       #Collector-emitter voltage, V\n",
-    "\n",
-    "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n",
-    "\n",
-    "#For d.c load line\n",
-    "#VCE=VCC-IC*RC\n",
-    "#For calculating VCE, IC=0\n",
-    "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-    "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-    "\n",
-    "#For calculating VCE, IC=0\n",
-    "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-    "IC_max=(VCC-VCE)/(RC/1000.0);           #Maximum collector current, mA\n",
-    "\n",
-    "\n",
-    "#Plotting of d.c load line\n",
-    "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-    "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-    "p=plt.plot(VCE_plot,IC_plot);\n",
-    "limit = plt.gca()\n",
-    "limit.set_xlim([0,25])\n",
-    "limit.set_ylim([0,65])\n",
-    "plt.xlabel('VCE(V)');\n",
-    "plt.ylabel('IC(mA)');\n",
-    "plt.title('d.c load line');\n",
-    "plt.grid();\n",
-    "plt.show(p);\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.26 : Page number 169-170"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 29,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Operating point: IC=1.8mA and VCE=9.74V.\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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qq6vjDiERlIdImnPRvTtMngx//zuMHAmffFLNhReGeXske3kdlmlmOwMXufuQ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\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f9cdbd4bdd8>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "from matplotlib import pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=10.0;                #Collector supply voltage, V\n",
-    "VEE=10.0;                  #Emitter supply voltage, V\n",
-    "RC=1.0;                  #Collector resistor, kΩ\n",
-    "RE=4.7;                  #Collector resistor, kΩ\n",
-    "RB=47.0;                #Base resistoe, kΩ\n",
-    "beta=100.0;                 #Base current amplification factor\n",
-    "VBE=0.7;                      #Base -emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#-IB*RB-VBE-IE*RE+VEE=0\n",
-    "#AS, IC=beta*IB and IC~IE\n",
-    "IE=round((VEE-VBE)/(RE+(RB/beta)),1);      #Emitter current,  mA\n",
-    "IC=IE;                                     #Collector current, mA\n",
-    "\n",
-    "#VCC-IC*RC-VCE-IE*RE+VEE=0\n",
-    "#IC~IE\n",
-    "VCE=VCC+VEE-IC*(RC+RE);                       #Collector-emitter voltage, V\n",
-    "\n",
-    "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n",
-    "\n",
-    "\n",
-    "#For d.c load line\n",
-    "#VCE=VCC-IC*RC\n",
-    "#For calculating VCE, IC=0\n",
-    "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-    "VCE_max=VCC+VEE-IC*(RC+RE);               #Maximum collector-emitter voltage, V\n",
-    "\n",
-    "#For calculating VCE, IC=0\n",
-    "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-    "IC_max=(VCC+VEE-VCE)/(RC+RE);           #Maximum collector current, mA\n",
-    "\n",
-    "\n",
-    "#Plotting of d.c load line\n",
-    "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-    "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-    "p=plt.plot(VCE_plot,IC_plot);\n",
-    "limit = plt.gca()\n",
-    "limit.set_xlim([0,25])\n",
-    "limit.set_ylim([0,5])\n",
-    "plt.xlabel('VCE(V)');\n",
-    "plt.ylabel('IC(mA)');\n",
-    "plt.title('d.c load line');\n",
-    "plt.grid();\n",
-    "plt.show(p);\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.27 : Page number 170-171"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 30,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Emitter voltage=-1.54V.\n",
-      "(i) Base voltage=10.7V.\n",
-      "(i) Collector voltage=8.2V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VEE=10.0;                    #Emitter supply voltage, V\n",
-    "IE=1.8;                      #Emitter current, mA\n",
-    "RE=4.7;                      #Emitter resistor, kΩ\n",
-    "VBE=0.7;                     #Base-emitter voltage, V\n",
-    "VCC=10.0;                     #Collector supply voltage, V\n",
-    "IC=1.8;                       #Collector current, mA\n",
-    "RC=1.0;                       #Collector resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "VE=-VEE+IE*RE;               #Emitter voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "VB=VEE+VBE;                   #Base voltage, V\n",
-    "\n",
-    "#(iii)\n",
-    "VC=VCC-IC*RC;                 #Collector voltage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) Emitter voltage=%.2fV.\"%VE);\n",
-    "print(\"(i) Base voltage=%.1fV.\"%VB);\n",
-    "print(\"(i) Collector voltage=%.1fV.\"%VC);\n",
-    "\n",
-    "#Note: In the textbook, VB=VE+VBE has been written, which is worng. It should be VB=VEE+VBE. "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##  Example 8.28: Page number 173-174"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Input resistance =2 kΩ\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_BE_change=200.0;                #Change in base-emitter voltage in mV\n",
-    "I_B_change=100.0;                  #Change in base current in  μA\n",
-    "\n",
-    "#Calculations\n",
-    "Ri=V_BE_change/I_B_change;              #Input resistance in kΩ\n",
-    "\n",
-    "#Result\n",
-    "print(\"Input resistance =%d kΩ\"%Ri);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.29; Page number 174"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 32,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output resistance =8kΩ\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CE_final=10.0;\t\t\t#Final value of collector-emitter voltage in V\n",
-    "V_CE_initial=2.0;                 #Initial value of collector-emitter voltage in V\n",
-    "I_C_final=3.0;                    #Final value of collector current in mA\n",
-    "I_C_initial=2.0;                  #Initial value of collector current in mA\n",
-    "\n",
-    "#Calculations\n",
-    "V_CE_change=V_CE_final-V_CE_initial;\t\t#Change in collector to emitter voltage in V\n",
-    "I_C_change=I_C_final-I_C_initial;               #Change in collector current in mA\n",
-    "R0=V_CE_change/I_C_change;                      #Output resistance in kΩ\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output resistance =%dkΩ\"%R0);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.30: Page number 174"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 33,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the amplifier =100 \n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R_C=2.0;\t\t#Collector load in kilo ohm\n",
-    "R_i=1.0;\t\t#Input resistance in kilo ohm\n",
-    "R_AC=R_C;               #Effective collector load for single stage in kilo ohm(appoximately equal to collector load for single stage)\n",
-    "beta=50.0;              #Current gain\n",
-    "\n",
-    "#Calculations\n",
-    "A_v=beta*(R_AC/R_i);\t\t#Voltage gain of the amplifier\n",
-    "\n",
-    "#Result \n",
-    "print(\"The voltage gain of the amplifier =%d \"%A_v);\t\t\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.31: Page number 175-176"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Collector current during saturation = 20 mA\n",
-      "Collector emitter voltage during cutoff = 20 V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=20;\t\t#Collector supply voltage in V\n",
-    "R_C=1;                  #Collector resistance in kilo ohm\n",
-    "V_knee_Si=1;\t\t#Knee voltage of V_CE for Si in V \n",
-    "V_knee_Ge=0.5;\t\t#Knee voltage of V_CE for Ge in V\n",
-    "\n",
-    "#Calculations\n",
-    "I_C_sat_Si=(V_CC-V_knee_Si)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Si transistor)\n",
-    "I_C_sat_Ge=(V_CC-V_knee_Ge)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Ge transistor)\n",
-    "I_C_sat=(V_CC)/R_C;\t\t\t\t#Saturation (maximum) value of collector current in mA (neglecting knee voltage)\n",
-    "V_CE_cut_off=V_CC;                              #Collector to emitter voltage in cutoff when base current=0, in V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Collector current during saturation = %d mA\"%I_C_sat);\n",
-    "print(\"Collector emitter voltage during cutoff = %d V.\"%V_CE_cut_off);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.32: Page number 176-177"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 35,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vce(off)= 24V\n",
-      "Ic(sat) = 10.67 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=12.0;\t\t#Collector supply voltage in V\n",
-    "V_EE=12.0;\t\t#Emitter supply voltage in V\n",
-    "R_C=750.0;\t\t#Collector resistance in ohm\n",
-    "R_E=1.5;\t\t#Emitter resistance in kilo ohm\n",
-    "R_B=100.0;\t\t#Base resistance in ohm\n",
-    "beta=200;\t\t#base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law to the collector side of the circuit\n",
-    "#using the equation: Vcc -IcRc-Vce -IeRe+Vee=0\n",
-    "#we get Vce=Vcc+Vee-Ic(Rc+Re), [Ie=Ic, approximately]\n",
-    "#We get Vce(off), when Ic=0;\n",
-    "\n",
-    "I_C_Vce_off=0;\t\t\t\t\t#Collector current for Vce(off) in mA\n",
-    "V_CE_off=V_CC+V_EE -(I_C_Vce_off * (R_C +R_E));\t#Collector to emitter voltage in V, during transistor in off state\n",
-    "\n",
-    "#We get Ic(sat), when Vce=0\n",
-    "V_CE_Ic_sat=0;\t\t\t\t\t\t#Collector to emitter voltage for saturation current of collector in V\n",
-    "I_C_sat=(V_CC+V_EE-V_CE_Ic_sat)/(R_C+(R_E*1000));\t#Saturated collector current in A \n",
-    "I_C_sat=I_C_sat*1000;\t\t\t\t\t#Saturated collector current in mA\n",
-    "#Result\n",
-    "print(\"Vce(off)= %dV\"%V_CE_off);\n",
-    "print(\"Ic(sat) = %.2f mA\"%I_C_sat);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.33 : Page number 177"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 36,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_knee=0.2;\t\t\t\t#Knee voltage of collector-emitter voltage in V\n",
-    "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n",
-    "V_BB=3.0;\t\t\t\t#Base supply voltage in V\n",
-    "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V \t\n",
-    "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-    "R_C=1.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "beta=50.0;\t\t\t\t#base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#applying Kirchhoff's voltage law along the collector side of the circuit,\n",
-    "#We get Vcc-Ic(sat)*Rc-V_knee=0\n",
-    "#From the above equation, we get:\n",
-    "I_C_sat=(V_CC-V_knee)/R_C;\t\t#Saturated collector current in mA\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along base emitter side,\n",
-    "#We get VBB-IB*RB-VBE=0;\n",
-    "#From the above equation, we get:\n",
-    "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-    "\n",
-    "\n",
-    "I_C=beta*I_B\t\t\t\t#Collector current in mA\n",
-    "\n",
-    "#Result\n",
-    "if(I_C>I_C_sat):\n",
-    "\tprint(\"The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\");\n",
-    "else:\n",
-    "\tprint(\"The base current is not large enough to produce Ic greater than Ic(sat), therefore the transistor isn't saturated. \");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.34: Page number 177-178"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 37,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n",
-    "V_BE=0.95;\t\t\t\t#Base-emitter voltage in V \t\n",
-    "I_B=100.0;\t\t\t\t#Base current in microAmp\n",
-    "R_C=970.0;\t\t\t\t#Collector resistor's resistance in ohm\n",
-    "beta=100.0;\t\t\t\t#base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "I_C=(I_B/1000)*beta;\t\t\t\t#Collector current in mA \n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along collector side\n",
-    "#We get Vcc-IcRc-Vce=0\n",
-    "#From the above equation, we get:\n",
-    "\n",
-    "V_CE=V_CC-((I_C/1000)*R_C);\t\t\t\t#Collector-emitter voltage in V\n",
-    "\n",
-    "#From the equation, V_CE=V_CB+V_BE,\n",
-    "V_CB=V_CE-V_BE;\t\t\t\t\t\t#Collector-base voltage in V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "if(V_CB<0 and V_BE >0):\n",
-    "\tprint(\"As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \");\n",
-    "else:\n",
-    "\tprint(\"No. The transistor isn't operating in the saturation region.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.35: Page number 178"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 38,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Therefore, for putting transistor in saturation, VBB >= 1.95 V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=10.0;\t\t\t\t#Collector supplu voltage in V\n",
-    "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-    "R_B=50.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-    "R_C=2.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "beta=200.0;\t\t\t\t#Base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along the collector side,\n",
-    "#We get, Vcc-Ic(sat)*Rc-Vce=0;\n",
-    "#From the above equation, we get:\n",
-    "#I_C_sat=(V_CC-V_CE)/R_C, but as transistor goes into saturation, Vce=0;\n",
-    "\n",
-    "V_CE=0;\t\t\t\t\t\t#Collector-emiter voltage in V, for transistor in saturation \n",
-    "I_C_sat=(V_CC-V_CE)/R_C;\t\t\t#Saturated collector current in mA\n",
-    "\n",
-    "I_B=I_C_sat/beta;\t\t\t\t#Base current in mA\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law to the base circuit,\n",
-    "#We get, VBB - IB*RB - VBE=0\n",
-    "#From the above equation. we get:\n",
-    "V_BB=V_BE+ I_B*R_B;\t\t\t\t#Base supply voltage to put transistor in saturation, in V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Therefore, for putting transistor in saturation, VBB >= %.2f V\"%V_BB);\n",
-    " \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.36: Page number 178-179"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 39,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\n",
-      "(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\n",
-      "(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=10.0;\t\t\t#Collector supply voltage in V\n",
-    "V_BB=2.7;\t\t\t#Base supply voltage in V\n",
-    "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n",
-    "beta=100.0;\t\t\t#Base current amplification factor\n",
-    "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n",
-    "\n",
-    "\n",
-    "#Calcultaion\t\n",
-    "V_B=V_BB;\t\t\t#Base voltage in V\n",
-    "V_E=V_B-V_BE;\t\t\t#Emitter voltage in V\n",
-    "I_E=V_E/R_E;\t\t\t#Emitter current in mA\n",
-    "I_C=I_E;\t\t\t#Collector current (approximately equal to emitter current) in mA\n",
-    "I_B=I_C/beta;\t\t\t#Base current in mA\n",
-    "\n",
-    "#Case (i):\n",
-    "R_C=2;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "\n",
-    "#Assuming transistor to be in active state\n",
-    "#Applying Kirchhoff's voltage law along collector side,\n",
-    "#We get,Vcc-IcRc=Vc,\n",
-    "\n",
-    "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-    "\n",
-    "if(V_C>V_E):\n",
-    "\tprint(\"(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\");\n",
-    "elif(V_C<V_E):\n",
-    "\tprint(\"(i)Our assumption was wrong, the transistor is in saturation for Rc=2 kilo ohm.\");\n",
-    "elif(V_C==V_E):\n",
-    "\tprint(\"(i)The transistor is at the edge of saturation for Rc=2 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n",
-    "\n",
-    "#Case (ii):\n",
-    "R_C=4;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "\n",
-    "#Assuming transistor to be in active state\n",
-    "#Applying Kirchhoff's voltage law along collector side,\n",
-    "#We get,Vcc-IcRc=Vc,\n",
-    "\n",
-    "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-    "if(V_C>V_E):\n",
-    "\tprint(\"(ii)Our assumption was correct, the transistor is in active state for Rc=4 kilo ohm.\");\n",
-    "elif(V_C==V_E):\n",
-    "\tprint(\"(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n",
-    "elif(V_C<V_E):\n",
-    "\tprint(\"(ii)Our assumption was wrong, the transistor is in saturation for Rc=4 kilo ohm.\");\n",
-    "\n",
-    "\n",
-    "#Case (iii):\n",
-    "R_C=8;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "\n",
-    "#Assuming transistor to be in active state\n",
-    "#Applying Kirchhoff's voltage law along collector side,\n",
-    "#We get,Vcc-IcRc=Vc,\n",
-    "\n",
-    "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-    "if(V_C>V_E):\n",
-    "\tprint(\"(iii)Our assumption was correct, the transistor is in active state for Rc=8 kilo ohm.\");\n",
-    "elif(V_C<V_E):\n",
-    "\tprint(\"(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\");\n",
-    "elif(V_C==V_E):\n",
-    "\tprint(\"(iii)The transistor is at the edge of saturation for Rc=8 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.37 : Page number 179-180"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Base voltage =0.5V is less than VBE=0.7V, therefore, transistor is cut-off.\n",
-      "0.8 0.8 7.0\n",
-      "(ii) VC=7V > VE=0.8V, therefore the transistor is active. Our assumption was correct.\n",
-      "(iii) VC=-8V < VE=2.3V, therefore the transistor is saturated. Our assumption was wrong.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=15.0;\t\t\t#Collector supply voltage in V\n",
-    "R_C=10.0;\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n",
-    "beta=100.0;\t\t\t#Base current amplification factor\n",
-    "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n",
-    "\n",
-    "\n",
-    "#Calculation\t\n",
-    "\n",
-    "#Case (i):\n",
-    "V_BB=0.5;\t\t\t#Base supply voltage in V\n",
-    "VB=V_BB;            #Base voltage, V\n",
-    "print(\"(i) Base voltage =%.1fV is less than VBE=%.1fV, therefore, transistor is cut-off.\"%(VB,V_BE));\n",
-    "\n",
-    "\n",
-    "#Case (ii):\n",
-    "V_BB=1.5;\t\t\t#Base supply voltage in V\n",
-    "VB=V_BB;            #Base voltage, V\n",
-    "VE=VB-V_BE;          #Emitter voltage, V\n",
-    "IE=round(VE/R_E,1);  #Emitter current, mA\n",
-    "#Assuming transistor to be in active state\n",
-    "#Applying Kirchhoff's voltage law along collector side,\n",
-    "IC=IE;              #Collector current, mA\n",
-    "IB=IC/beta;         #Base  current, mA\n",
-    "VC=V_CC-IC*R_C;      #Collector voltage, V\n",
-    "print(VE,IE,VC);\n",
-    "print(\"(ii) VC=%dV > VE=%.1fV, therefore the transistor is active. Our assumption was correct.\"%(VC,VE));\n",
-    "\n",
-    "#Case (iii):\n",
-    "V_BB=3; \t\t\t#Base supply voltage in V\n",
-    "VB=V_BB;            #Base voltage, V\n",
-    "VE=VB-V_BE;          #Emitter voltage, V\n",
-    "IE=round(VE/R_E,1);  #Emitter current, mA\n",
-    "#Assuming transistor to be in active state\n",
-    "#Applying Kirchhoff's voltage law along collector side,\n",
-    "IC=IE;              #Collector current, mA\n",
-    "IB=IC/beta;         #Base  current, mA\n",
-    "VC=V_CC-IC*R_C;      #Collector voltage, V\n",
-    "\n",
-    "print(\"(iii) VC=%dV < VE=%.1fV, therefore the transistor is saturated. Our assumption was wrong.\"%(VC,VE));"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.38: Page number 181"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 41,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Maximum collector current that can be allowed without destruction of the transistor = 5 mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "P_D_max=100.0;\t\t\t#Maximum power dissipation of a transistor in mW\n",
-    "V_CE=20.0;\t\t\t#Collector emitter voltage in V\n",
-    "\n",
-    "#Calculation\n",
-    "#As power=curent*voltage\n",
-    "#P_D_max=I_C_max*V_CE\n",
-    "#From the above equation, we get:\n",
-    "\n",
-    "I_C_max=P_D_max/V_CE;\t\t#Maximum collector current that can be allowed without destruction of the transistor, in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Maximum collector current that can be allowed without destruction of the transistor = %d mA.\"%I_C_max); \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.39: Page number 181"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 42,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Power dissipated = 4.3W\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n",
-    "V_BB=5.0;\t\t\t\t#Base supply voltage in V\n",
-    "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-    "R_B=1.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-    "R_C=0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "beta=200.0;\t\t\t\t#base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along base circuit<\n",
-    "#We get, VBB- IB*RB - VBE=0.\n",
-    "#From the above equation, we get:\n",
-    "\n",
-    "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-    "\n",
-    "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along collector circuit:\n",
-    "\n",
-    "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n",
-    "\n",
-    "#As power=curent*voltage\n",
-    "#P_D=I_C*V_CE\n",
-    "#From the above equation, we get:\n",
-    "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n",
-    "P_D=P_D/1000;\t\t\t\t#Power dissipated in W\n",
-    "\n",
-    "#Result\n",
-    "print(\"Power dissipated = %.1fW\"%P_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.40: Page number 181-182"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 43,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Power dissipated = 6mW\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n",
-    "V_BB=1.0;\t\t\t\t#Base supply voltage in V\n",
-    "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-    "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-    "R_C=1.0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "beta=100.0;\t\t\t\t#base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along base circuit<\n",
-    "#We get, VBB- IB*RB - VBE=0.\n",
-    "#From the above equation, we get:\n",
-    "\n",
-    "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-    "\n",
-    "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along collector circuit:\n",
-    "\n",
-    "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n",
-    "\n",
-    "#As power=curent*voltage\n",
-    "#P_D=I_C*V_CE\n",
-    "#From the above equation, we get:\n",
-    "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Power dissipated = %.0fmW\"%P_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.41 : Page number 182"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 44,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "IC=19.5mA is much less than IC_max=100mA. Therefore, will not change with VCC and current rating is not exceeded.\n",
-      "PD=293mW  is less than PD_max=800mW. Therefore, power rating is not exceeded.\n",
-      "If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VBB=5.0;                     #Base supply voltage, V\n",
-    "RB=22.0;                  #Base resistor, kilo ohm\n",
-    "RC=1.0;                   #Collector resistor, kilo ohm\n",
-    "beta=100.0;               #Base current amplification factor\n",
-    "VBE=0.7;                  #Base-emitter voltage, V\n",
-    "PD_max=800.0;             #Maximum power dissipation, mW\n",
-    "VCE_max=15.0;             #Maximum collector-emitter voltage, V\n",
-    "IC_max=100.0;             #Maximum collector current, mA\n",
-    "\n",
-    "#Calculation\n",
-    "IB=((VBB-VBE)/RB)*1000;               #Base current,  μA\n",
-    "IC=beta*IB/1000;                      #Collector current, mA\n",
-    "\n",
-    "print(\"IC=%.1fmA is much less than IC_max=%dmA. Therefore, will not change with VCC and current rating is not exceeded.\"%(IC,IC_max));\n",
-    "\n",
-    "#VCC=VCE+IC*RC\n",
-    "VCC_max=VCE_max+IC*RC;           #Maximum value of Collector supply voltage, V\n",
-    "PD=VCE_max*IC;                  #Power dissipation, mW\n",
-    "\n",
-    "print(\"PD=%dmW  is less than PD_max=%dmW. Therefore, power rating is not exceeded.\"%(PD,PD_max));\n",
-    "\n",
-    "print(\"If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\");"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_5.ipynb
deleted file mode 100755
index c13922ee..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_5.ipynb
+++ /dev/null
@@ -1,1845 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.1: Page number 147-148"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage amplification = 50. \n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "Signal=500.0;                    #Signal voltage in V\n",
-    "Rin=20.0;                        #Input resistance in Ω \n",
-    "Rout=100.0;                      #Output resistance in Ω\n",
-    "R_C=1000.0;                      #Collector load in Ω\n",
-    "alpha_ac=1.0;                    #current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "I_E=(Signal/1000)/Rin;        \t#Input current in mA\n",
-    "I_C=I_E*alpha_ac;               #Output current in mA\n",
-    "Vout=I_C*R_C;                   #Output voltage in V \n",
-    "Av=Vout/(Signal/1000);          #Voltage amplification \n",
-    "\n",
-    "#Result\n",
-    "print(\"The voltage amplification = %d. \"%Av);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.2: Page number 150"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The base current = 0.05 mA \n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_E=1;                            #Emitter curent in mA\n",
-    "I_C=0.95;                         #Collector current in mA\n",
-    "\n",
-    "#Calculation\n",
-    "I_B=I_E-I_C;                      #Base current in mA\n",
-    "\n",
-    "#Result \n",
-    "print(\"The base current = %.2f mA \"%I_B);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# Example 8.3: Page number 150\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The base current =0.1 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#variable declaration\n",
-    "alpha=0.9;                      #Current amplification factor\n",
-    "I_E=1;                          #Emitter current in mA\n",
-    "\n",
-    "#Calculation\n",
-    "I_C=alpha*I_E;                  #Collector current in mA\n",
-    "I_B=I_E-I_C;                    #Base current in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The base current =%.1f mA\"%I_B);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.4: Page number 150"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The current amplification factor = 0.95 .\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_C=0.95;\t\t\t#Collector current in mA\n",
-    "I_B=0.05;\t\t\t#Base current in mA\n",
-    "\n",
-    "#Calculation\n",
-    "I_E=I_B+I_C;                    #Emitter current in mA\n",
-    "alpha=I_C/I_E;                  #Current amplification factor \n",
-    "\n",
-    "#Result\n",
-    "print(\"The current amplification factor = %.2f .\"%alpha);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.5: Page number 150"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The total collector current = 0.97 mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_E=1;                  #Emitter current in mA\n",
-    "I_CBO=50.0;               #Collector current with emitter circuit open, in microAmp\n",
-    "alpha=0.92;             #Current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "I_C=alpha*I_E + (I_CBO/1000);           #Total collector current in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The total collector current = %.2f mA.\"%I_C);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.6: Page number 150-151"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 6,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The base current = 0.05 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "alpha=0.95;               #Current amplification factor\n",
-    "Rc=2.0;                   #Resistor connected to the collector, in kilo ohm\n",
-    "V_Rc=2.0;                 #Voltage drop across the resistor connected to the collector in V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_C=V_Rc/Rc;              #Collector current in mA\n",
-    "I_E=I_C/alpha;            #Emitter current in mA\n",
-    "I_B=I_E-I_C;              #Base current in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The base current = %.2f mA\"%I_B); \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.7: Page number 151"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The collector current =4.87 mA\n",
-      "The collector to base voltage = 12.16 V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_EE=8.0;                 #Supply voltage at the emitter in V\n",
-    "V_CC=18.0;                #Supply voltage at the collector in V\n",
-    "V_BE=0.7;                 #Base to emitter voltage in V\n",
-    "R_E=1.5;                  #Emitter resistance in Ω\n",
-    "R_C=1.2;                  #Collector resistance in Ω\n",
-    "\n",
-    "#Calculations\n",
-    "I_E=(V_EE-V_BE)/R_E;              #Emitter current in mA\n",
-    "I_C=I_E;                          #Collector current in mA (approximately equal to emitter current)\n",
-    "V_CB=V_CC-(I_C*R_C);              #Collector to base voltage in V\n",
-    "\n",
-    "#Result\n",
-    "print(\"The collector current =%.2f mA\"%I_C);\n",
-    "print(\"The collector to base voltage = %.2f V\"%V_CB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.8:Page number 155"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 8,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Value of beta =9\n",
-      "(ii) Value of beta =49\n",
-      "(iii) Value of beta =99\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Function for calculating beta from alpha\n",
-    "def calc_beta(a):                   #a is the value of alpha\n",
-    "\treturn(a/(1-a));\n",
-    "\n",
-    "#Case (i)\n",
-    "alpha=0.9;                      #current amplification factor\n",
-    "beta=calc_beta(alpha);\t\t#Base current amplification factor    \n",
-    "print(\"(i) Value of beta =%d\"%beta );\t\t\t\t\t\t\t\t\t\n",
-    "\n",
-    "#Case (ii)\n",
-    "alpha=0.98;                     #current amplification factor\n",
-    "beta=calc_beta(alpha);          #Base current amplification factor\n",
-    "print(\"(ii) Value of beta =%.0f\"%beta );\n",
-    "\n",
-    "\n",
-    "#Case (iii)\n",
-    "alpha=0.99;                     #current amplification factor\n",
-    "beta=calc_beta(alpha);          #Base current amplification factor                      \n",
-    "print(\"(iii) Value of beta =%.0f\"%beta );\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.9: Page number 155"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The emitter curent = 1.02 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "beta=50.0;                   #Base current amplification factor\n",
-    "I_B=20.0;                    #Base current in microAmp\n",
-    "\n",
-    "#Calculation\n",
-    "I_B=I_B/1000;               #Base current in mA\n",
-    "I_C=beta*I_B;               #Collector current in mA\n",
-    "I_E=I_B+I_C;                #Emitter current in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The emitter curent = %.2f mA\"%I_E);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.10: Page number 155"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "alpha=0.98.\n",
-      "Collector current determined using alpha =11.76 mA\n",
-      "Collector current determined using beta =11.76 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_B=240.0;                    #Base current in microAmp\n",
-    "I_E=12;                       #Emitter current in mA\n",
-    "beta=49.0;                      #Base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "alpha=beta/(1+beta);              #current amplification factor \n",
-    "I_C_alpha=alpha*I_E;              #Collector current in mA calculated using alpha\n",
-    "I_C_beta=beta*(I_B/1000);         #Collector current in mA calculated using beta\n",
-    "\n",
-    "#Results\n",
-    "print(\"alpha=%.2f.\"%alpha);\n",
-    "print(\"Collector current determined using alpha =%.2f mA\"%I_C_alpha);\n",
-    "print(\"Collector current determined using beta =%.2f mA\"%I_C_beta);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.11: Page number 156"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The base current =0.022 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "beta=45.0;                       #Base current amplification factor\n",
-    "R_C=1.0;                         #Resistance of the collector resistance in kΩ\n",
-    "V_R_C=1.0;                       #Voltage drop across the collector resistance in V\n",
-    "\n",
-    "#Calculation\n",
-    "I_C=V_R_C/R_C;                   #Collector current in mA\n",
-    "I_B=I_C/beta;                           #Base current in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"The base current =%.3f mA\"%I_B);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.12: Page number 156"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Collector to emitter voltage = 7.5 V\n",
-      "Base current= 0.026 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=8.0;                  #Collector supply voltage in V\n",
-    "R_C=800.0;                  #Resistance of the collector resistance in Ω\n",
-    "V_R_C=0.5;                 #Voltage drop across collector resistance in V\n",
-    "alpha=0.96;                #current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "V_CE=V_CC-V_R_C;           #Collector to emitter voltage in V\n",
-    "I_C=V_R_C/R_C;             #Collector current in A\n",
-    "I_C=I_C*1000;              #Collector current in mA\n",
-    "beta=alpha/(1-alpha);      #Base current amplification factor\n",
-    "I_B=I_C/beta;              #Base current in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Collector to emitter voltage = %.1f V\"%V_CE);\n",
-    "print(\"Base current= %.3f mA\"%I_B);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.13: Page number 156-157"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Current amplification factor = 0.99 \n",
-      "The emitter curent =1010  μA \n",
-      "The base curent =10  μA \n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=5;                  \t#Collector supply voltage in V\n",
-    "I_CBO=0.2;               \t#Leakage current at collector base junction with emitter open, in  μA\n",
-    "I_CEO=20.0;              \t#Leakage current with base open, in  μA\n",
-    "I_C=1.0;                        #Collector current in mA\n",
-    "I_C=I_C*1000;                  \t#Collector current in  μA\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "alpha=1-(I_CBO/I_CEO);\t\t#current amplification factor\n",
-    "I_E=(I_C-I_CBO)/alpha;          #Emitter current in  μA\n",
-    "I_E=round(I_E,-1);\n",
-    "I_B=I_E-I_C;                    #Base current in  μA\n",
-    "I_B=round(I_B,-1);\n",
-    "\n",
-    "#Result\n",
-    "print(\"Current amplification factor = %.2f \"%alpha);\n",
-    "print(\"The emitter curent =%d  μA \"%I_E);\n",
-    "print(\"The base curent =%d  μA \"%I_B);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.14: Page number 157"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vale of I_CBO= 2.4 μA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_CEO=300.0;              #Leakage current in common emitter configuration, in  μA\n",
-    "beta=120.0;               #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "alpha=beta/(1+beta);               #Current amplification factor\n",
-    "alpha=round(alpha,3);\n",
-    "I_CBO=(1-alpha)*I_CEO;             #Leakage current in common base configuration, in  μA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Vale of I_CBO= %.1f μA\"%I_CBO);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.15: Page number 157"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 15,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Value of I_CBO=0.0048 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_B=20.0;                       #Base current in μA\n",
-    "I_C=2.0;                        #Collector current in mA\n",
-    "beta=80.0;                      #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "I_CEO=I_C-(beta*I_B/1000);            #Leakage current with base open, in mA \n",
-    "alpha=beta/(beta+1);                  #Current amplification factor\n",
-    "alpha=round(alpha,3);\n",
-    "I_CBO=(1-alpha)*I_CEO;                #Leakage current with emitter open, in mA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Value of I_CBO=%.4f mA\"%I_CBO);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.17: Page number 158"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 16,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Collector to base voltage, V_CB= 2.85 V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "beta=150.0;               \t#Base current amplification factor\n",
-    "R_B=10.0;                   \t#Base resistance in kilo ohm\n",
-    "R_C=100.0;                   \t#Collector resistance in kilo ohm\n",
-    "V_CC=10.0;                      #Collector supply voltage in V\n",
-    "V_BB=5.0;                       #Base supply voltage in V\n",
-    "V_BE=0.7;                       #Base to emitter voltage in V\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "I_B=(V_BB-V_BE)/R_B;              #Base current in mA\n",
-    "I_C=beta*I_B;                     #Collector current in mA\n",
-    "V_CE=V_CC - (I_C/1000)*R_C;       #Collector to emitter voltage in V\n",
-    "V_CB=V_CE-V_BE;                   #Collector to base voltage in V\n",
-    "\n",
-    "\n",
-    "#Result \n",
-    "print(\"Collector to base voltage, V_CB= %.2f V\"%V_CB);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.18: Page number158-159"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 17,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Collector current determined using alpha rating =29.93 mA\n",
-      "Collector current determined using beta rating =29.92 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_B=68.0;              #Base current in μA\n",
-    "I_E=30.0;              #Emitter current in mA\n",
-    "beta=440.0;\t     #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "alpha=beta/(beta + 1);          #current amplification factor\n",
-    "I_C_alpha=alpha*I_E;\t\t#Collector current using alpha rating, in mA\n",
-    "I_C_beta=beta*(I_B/1000.0);       #Collector current using beta rating, in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Collector current determined using alpha rating =%.2f mA\"%I_C_alpha);\n",
-    "print(\"Collector current determined using beta rating =%.2f mA\"%I_C_beta);\n",
-    "\n",
-    "#Note: In the textbook, the collector current obtained from beta rating is approximated to 29.93 mA\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.19: Page number 159"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 18,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The maximum allowable value of base current = 1.67 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "I_C_max=500.0;                   #Maximum collector current in mA\n",
-    "beta_max=300.0;                  #Maximum base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "I_B_max=I_C_max/beta_max;           #Maximum base current in mA\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"The maximum allowable value of base current = %.2f mA\"%I_B_max);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.22 : Page number 167-168"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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5R9IrOjzo79B4NarZ0Lx6NxudXgb7auASSZ+UdG728Ska/fqqfONVk3u+fLR27xs2zJXy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\neLCbmdWMB7uZWc38f6b7SUAyF1dTAAAAAElFTkSuQmCC\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f9cdbfbafd0>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=12.5;                #Collector supply voltage, V\n",
-    "RC=2.5;                  #Collector resistor, kΩ\n",
-    "\n",
-    "#Calculation\n",
-    "#VCE=VCC-IC*RC\n",
-    "#For calculating VCE, IC=0\n",
-    "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-    "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-    "\n",
-    "#For calculating VCE, IC=0\n",
-    "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-    "IC_max=(VCC-VCE)/RC;           #Maximum collector current, mA\n",
-    "\n",
-    "\n",
-    "#Plotting of d.c load line\n",
-    "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-    "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-    "p=plt.plot(VCE_plot,IC_plot);\n",
-    "limit = plt.gca()\n",
-    "limit.set_xlim([0,15])\n",
-    "limit.set_ylim([0,6])\n",
-    "plt.xlabel('VCE(V)');\n",
-    "plt.ylabel('IC(mA)');\n",
-    "plt.title('d.c load line');\n",
-    "plt.grid();\n",
-    "plt.show(p);"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.23 : Page number 168"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f9cec03e9b0>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Operating point: IC=1mA and VCE=6V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "import matplotlib.pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=12.0;                #Collector supply voltage, V\n",
-    "RC=6.0;                  #Collector resistor, kΩ\n",
-    "IB=20.0;                     #Zero signal base current,  μA\n",
-    "beta=50.0;                  #Base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "#VCE=VCC-IC*RC\n",
-    "#For calculating VCE, IC=0\n",
-    "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-    "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-    "\n",
-    "#For calculating VCE, IC=0\n",
-    "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-    "IC_max=(VCC-VCE)/RC;           #Maximum collector current, mA\n",
-    "\n",
-    "\n",
-    "#Plotting of d.c load line\n",
-    "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-    "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-    "p=plt.plot(VCE_plot,IC_plot);\n",
-    "limit = plt.gca()\n",
-    "limit.set_xlim([0,15])\n",
-    "limit.set_ylim([0,5])\n",
-    "plt.xlabel('VCE(V)');\n",
-    "plt.ylabel('IC(mA)');\n",
-    "plt.title('d.c load line');\n",
-    "plt.grid();\n",
-    "plt.show(p);\n",
-    "\n",
-    "#Calculating Q-point\n",
-    "IC=beta*(IB/1000);                                  #Collector current, mA\n",
-    "VCE=VCC-IC*RC;                               #Collector emitter voltage, V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Operating point: IC=%dmA and VCE=%dV.\"%(IC,VCE));\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.24 : Page number 168"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Operating point: VCE=6V and IC=1mA.\n",
-      "(ii) Operating point: VCE=5V and IC=1mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "RC=4.0;               #Collector load, kΩ\n",
-    "IC_Q=1.0;             #Quiescent current, mA\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "VCC=10;                  #Collector supply voltage, V\n",
-    "VCE=VCC-IC*RC;            #Collector emitter voltage, V\n",
-    "\n",
-    "print(\"(i) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n",
-    "\n",
-    "#(ii)\n",
-    "RC=5.0;               #Collector load, kΩ\n",
-    "VCE=VCC-IC*RC;            #Collector emitter voltage, V\n",
-    "print(\"(ii) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# Example 8.25 : Page number 168-169"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 27,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Operating point: IC=39.6mA and VCE=6.93V.\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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+faI2kUhu3XVXuCZhzJjQPpLikMY20Thg/2r7RgDT3X1LYCbwqxzHICK1OOYY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\nSUVORK5CYII=\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f9cec03e978>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "from matplotlib import pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=20.0;                #Collector supply voltage, V\n",
-    "VBB=10.0;                  #Base supply voltage, V\n",
-    "RC=330.0;                  #Collector resistor, Ω\n",
-    "RB=47.0;                #Base resistoe, kΩ\n",
-    "beta=200.0;                 #Base current amplification factor\n",
-    "VBE=0.7;                      #Base -emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#VBB-IB*RB-VBE=0\n",
-    "IB=round(((VBB-VBE)/RB)*1000,0);                     #Base current,  μA\n",
-    "IC=beta*IB/1000;                    #Collector current, mA\n",
-    "VCE=VCC-IC*(RC/1000);                       #Collector-emitter voltage, V\n",
-    "\n",
-    "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n",
-    "\n",
-    "#For d.c load line\n",
-    "#VCE=VCC-IC*RC\n",
-    "#For calculating VCE, IC=0\n",
-    "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-    "VCE_max=VCC-IC*RC;               #Maximum collector-emitter voltage, V\n",
-    "\n",
-    "#For calculating VCE, IC=0\n",
-    "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-    "IC_max=(VCC-VCE)/(RC/1000.0);           #Maximum collector current, mA\n",
-    "\n",
-    "\n",
-    "#Plotting of d.c load line\n",
-    "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-    "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-    "p=plt.plot(VCE_plot,IC_plot);\n",
-    "limit = plt.gca()\n",
-    "limit.set_xlim([0,25])\n",
-    "limit.set_ylim([0,65])\n",
-    "plt.xlabel('VCE(V)');\n",
-    "plt.ylabel('IC(mA)');\n",
-    "plt.title('d.c load line');\n",
-    "plt.grid();\n",
-    "plt.show(p);\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.26 : Page number 169-170"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 29,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Operating point: IC=1.8mA and VCE=9.74V.\n"
-     ]
-    },
-    {
-     "data": {
-      "image/png": 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qq6vjDiERlIdImnPRvTtMngx//zuMHAmffFLNhReGeXske3kdlmlmOwMXufuQ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\n",
-      "text/plain": [
-       "<matplotlib.figure.Figure at 0x7f9cdbd4bdd8>"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%matplotlib inline\n",
-    "from matplotlib import pyplot as plt\n",
-    "\n",
-    "#Variable declaration\n",
-    "VCC=10.0;                #Collector supply voltage, V\n",
-    "VEE=10.0;                  #Emitter supply voltage, V\n",
-    "RC=1.0;                  #Collector resistor, kΩ\n",
-    "RE=4.7;                  #Collector resistor, kΩ\n",
-    "RB=47.0;                #Base resistoe, kΩ\n",
-    "beta=100.0;                 #Base current amplification factor\n",
-    "VBE=0.7;                      #Base -emitter voltage, V\n",
-    "\n",
-    "#Calculation\n",
-    "#-IB*RB-VBE-IE*RE+VEE=0\n",
-    "#AS, IC=beta*IB and IC~IE\n",
-    "IE=round((VEE-VBE)/(RE+(RB/beta)),1);      #Emitter current,  mA\n",
-    "IC=IE;                                     #Collector current, mA\n",
-    "\n",
-    "#VCC-IC*RC-VCE-IE*RE+VEE=0\n",
-    "#IC~IE\n",
-    "VCE=VCC+VEE-IC*(RC+RE);                       #Collector-emitter voltage, V\n",
-    "\n",
-    "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n",
-    "\n",
-    "\n",
-    "#For d.c load line\n",
-    "#VCE=VCC-IC*RC\n",
-    "#For calculating VCE, IC=0\n",
-    "IC=0;                           #Collector current for maximum Collector-emitter voltage, mA\n",
-    "VCE_max=VCC+VEE-IC*(RC+RE);               #Maximum collector-emitter voltage, V\n",
-    "\n",
-    "#For calculating VCE, IC=0\n",
-    "VCE=0;                         #Collector emitter voltage for maximum collector current, V\n",
-    "IC_max=(VCC+VEE-VCE)/(RC+RE);           #Maximum collector current, mA\n",
-    "\n",
-    "\n",
-    "#Plotting of d.c load line\n",
-    "VCE_plot=[0,VCE_max];                 #Plotting variable for VCE\n",
-    "IC_plot=[IC_max,0];                   #Plotting variable for IC\n",
-    "p=plt.plot(VCE_plot,IC_plot);\n",
-    "limit = plt.gca()\n",
-    "limit.set_xlim([0,25])\n",
-    "limit.set_ylim([0,5])\n",
-    "plt.xlabel('VCE(V)');\n",
-    "plt.ylabel('IC(mA)');\n",
-    "plt.title('d.c load line');\n",
-    "plt.grid();\n",
-    "plt.show(p);\n",
-    "\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.27 : Page number 170-171"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 30,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Emitter voltage=-1.54V.\n",
-      "(i) Base voltage=10.7V.\n",
-      "(i) Collector voltage=8.2V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VEE=10.0;                    #Emitter supply voltage, V\n",
-    "IE=1.8;                      #Emitter current, mA\n",
-    "RE=4.7;                      #Emitter resistor, kΩ\n",
-    "VBE=0.7;                     #Base-emitter voltage, V\n",
-    "VCC=10.0;                     #Collector supply voltage, V\n",
-    "IC=1.8;                       #Collector current, mA\n",
-    "RC=1.0;                       #Collector resistor, kΩ\n",
-    "\n",
-    "\n",
-    "#Calculation\n",
-    "#(i)\n",
-    "VE=-VEE+IE*RE;               #Emitter voltage, V\n",
-    "\n",
-    "#(ii)\n",
-    "VB=VEE+VBE;                   #Base voltage, V\n",
-    "\n",
-    "#(iii)\n",
-    "VC=VCC-IC*RC;                 #Collector voltage, V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"(i) Emitter voltage=%.2fV.\"%VE);\n",
-    "print(\"(i) Base voltage=%.1fV.\"%VB);\n",
-    "print(\"(i) Collector voltage=%.1fV.\"%VC);\n",
-    "\n",
-    "#Note: In the textbook, VB=VE+VBE has been written, which is worng. It should be VB=VEE+VBE. "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "##  Example 8.28: Page number 173-174"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Input resistance =2 kΩ\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_BE_change=200.0;                #Change in base-emitter voltage in mV\n",
-    "I_B_change=100.0;                  #Change in base current in  μA\n",
-    "\n",
-    "#Calculations\n",
-    "Ri=V_BE_change/I_B_change;              #Input resistance in kΩ\n",
-    "\n",
-    "#Result\n",
-    "print(\"Input resistance =%d kΩ\"%Ri);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.29; Page number 174"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 32,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The output resistance =8kΩ\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CE_final=10.0;\t\t\t#Final value of collector-emitter voltage in V\n",
-    "V_CE_initial=2.0;                 #Initial value of collector-emitter voltage in V\n",
-    "I_C_final=3.0;                    #Final value of collector current in mA\n",
-    "I_C_initial=2.0;                  #Initial value of collector current in mA\n",
-    "\n",
-    "#Calculations\n",
-    "V_CE_change=V_CE_final-V_CE_initial;\t\t#Change in collector to emitter voltage in V\n",
-    "I_C_change=I_C_final-I_C_initial;               #Change in collector current in mA\n",
-    "R0=V_CE_change/I_C_change;                      #Output resistance in kΩ\n",
-    "\n",
-    "#Result\n",
-    "print(\"The output resistance =%dkΩ\"%R0);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.30: Page number 174"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 33,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The voltage gain of the amplifier =100 \n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "R_C=2.0;\t\t#Collector load in kilo ohm\n",
-    "R_i=1.0;\t\t#Input resistance in kilo ohm\n",
-    "R_AC=R_C;               #Effective collector load for single stage in kilo ohm(appoximately equal to collector load for single stage)\n",
-    "beta=50.0;              #Current gain\n",
-    "\n",
-    "#Calculations\n",
-    "A_v=beta*(R_AC/R_i);\t\t#Voltage gain of the amplifier\n",
-    "\n",
-    "#Result \n",
-    "print(\"The voltage gain of the amplifier =%d \"%A_v);\t\t\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.31: Page number 175-176"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Collector current during saturation = 20 mA\n",
-      "Collector emitter voltage during cutoff = 20 V.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=20;\t\t#Collector supply voltage in V\n",
-    "R_C=1;                  #Collector resistance in kilo ohm\n",
-    "V_knee_Si=1;\t\t#Knee voltage of V_CE for Si in V \n",
-    "V_knee_Ge=0.5;\t\t#Knee voltage of V_CE for Ge in V\n",
-    "\n",
-    "#Calculations\n",
-    "I_C_sat_Si=(V_CC-V_knee_Si)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Si transistor)\n",
-    "I_C_sat_Ge=(V_CC-V_knee_Ge)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Ge transistor)\n",
-    "I_C_sat=(V_CC)/R_C;\t\t\t\t#Saturation (maximum) value of collector current in mA (neglecting knee voltage)\n",
-    "V_CE_cut_off=V_CC;                              #Collector to emitter voltage in cutoff when base current=0, in V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Collector current during saturation = %d mA\"%I_C_sat);\n",
-    "print(\"Collector emitter voltage during cutoff = %d V.\"%V_CE_cut_off);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.32: Page number 176-177"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 35,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Vce(off)= 24V\n",
-      "Ic(sat) = 10.67 mA\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=12.0;\t\t#Collector supply voltage in V\n",
-    "V_EE=12.0;\t\t#Emitter supply voltage in V\n",
-    "R_C=750.0;\t\t#Collector resistance in ohm\n",
-    "R_E=1.5;\t\t#Emitter resistance in kilo ohm\n",
-    "R_B=100.0;\t\t#Base resistance in ohm\n",
-    "beta=200;\t\t#base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law to the collector side of the circuit\n",
-    "#using the equation: Vcc -IcRc-Vce -IeRe+Vee=0\n",
-    "#we get Vce=Vcc+Vee-Ic(Rc+Re), [Ie=Ic, approximately]\n",
-    "#We get Vce(off), when Ic=0;\n",
-    "\n",
-    "I_C_Vce_off=0;\t\t\t\t\t#Collector current for Vce(off) in mA\n",
-    "V_CE_off=V_CC+V_EE -(I_C_Vce_off * (R_C +R_E));\t#Collector to emitter voltage in V, during transistor in off state\n",
-    "\n",
-    "#We get Ic(sat), when Vce=0\n",
-    "V_CE_Ic_sat=0;\t\t\t\t\t\t#Collector to emitter voltage for saturation current of collector in V\n",
-    "I_C_sat=(V_CC+V_EE-V_CE_Ic_sat)/(R_C+(R_E*1000));\t#Saturated collector current in A \n",
-    "I_C_sat=I_C_sat*1000;\t\t\t\t\t#Saturated collector current in mA\n",
-    "#Result\n",
-    "print(\"Vce(off)= %dV\"%V_CE_off);\n",
-    "print(\"Ic(sat) = %.2f mA\"%I_C_sat);\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.33 : Page number 177"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 36,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_knee=0.2;\t\t\t\t#Knee voltage of collector-emitter voltage in V\n",
-    "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n",
-    "V_BB=3.0;\t\t\t\t#Base supply voltage in V\n",
-    "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V \t\n",
-    "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-    "R_C=1.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "beta=50.0;\t\t\t\t#base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#applying Kirchhoff's voltage law along the collector side of the circuit,\n",
-    "#We get Vcc-Ic(sat)*Rc-V_knee=0\n",
-    "#From the above equation, we get:\n",
-    "I_C_sat=(V_CC-V_knee)/R_C;\t\t#Saturated collector current in mA\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along base emitter side,\n",
-    "#We get VBB-IB*RB-VBE=0;\n",
-    "#From the above equation, we get:\n",
-    "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-    "\n",
-    "\n",
-    "I_C=beta*I_B\t\t\t\t#Collector current in mA\n",
-    "\n",
-    "#Result\n",
-    "if(I_C>I_C_sat):\n",
-    "\tprint(\"The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\");\n",
-    "else:\n",
-    "\tprint(\"The base current is not large enough to produce Ic greater than Ic(sat), therefore the transistor isn't saturated. \");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.34: Page number 177-178"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 37,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \n"
-     ]
-    }
-   ],
-   "source": [
-    "\n",
-    "#Variable declaration\n",
-    "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n",
-    "V_BE=0.95;\t\t\t\t#Base-emitter voltage in V \t\n",
-    "I_B=100.0;\t\t\t\t#Base current in microAmp\n",
-    "R_C=970.0;\t\t\t\t#Collector resistor's resistance in ohm\n",
-    "beta=100.0;\t\t\t\t#base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "I_C=(I_B/1000)*beta;\t\t\t\t#Collector current in mA \n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along collector side\n",
-    "#We get Vcc-IcRc-Vce=0\n",
-    "#From the above equation, we get:\n",
-    "\n",
-    "V_CE=V_CC-((I_C/1000)*R_C);\t\t\t\t#Collector-emitter voltage in V\n",
-    "\n",
-    "#From the equation, V_CE=V_CB+V_BE,\n",
-    "V_CB=V_CE-V_BE;\t\t\t\t\t\t#Collector-base voltage in V\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "if(V_CB<0 and V_BE >0):\n",
-    "\tprint(\"As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \");\n",
-    "else:\n",
-    "\tprint(\"No. The transistor isn't operating in the saturation region.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.35: Page number 178"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 38,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Therefore, for putting transistor in saturation, VBB >= 1.95 V\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=10.0;\t\t\t\t#Collector supplu voltage in V\n",
-    "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-    "R_B=50.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-    "R_C=2.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "beta=200.0;\t\t\t\t#Base current amplification factor\n",
-    "\n",
-    "#Calculations\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along the collector side,\n",
-    "#We get, Vcc-Ic(sat)*Rc-Vce=0;\n",
-    "#From the above equation, we get:\n",
-    "#I_C_sat=(V_CC-V_CE)/R_C, but as transistor goes into saturation, Vce=0;\n",
-    "\n",
-    "V_CE=0;\t\t\t\t\t\t#Collector-emiter voltage in V, for transistor in saturation \n",
-    "I_C_sat=(V_CC-V_CE)/R_C;\t\t\t#Saturated collector current in mA\n",
-    "\n",
-    "I_B=I_C_sat/beta;\t\t\t\t#Base current in mA\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law to the base circuit,\n",
-    "#We get, VBB - IB*RB - VBE=0\n",
-    "#From the above equation. we get:\n",
-    "V_BB=V_BE+ I_B*R_B;\t\t\t\t#Base supply voltage to put transistor in saturation, in V\n",
-    "\n",
-    "#Result\n",
-    "print(\"Therefore, for putting transistor in saturation, VBB >= %.2f V\"%V_BB);\n",
-    " \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.36: Page number 178-179"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 39,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\n",
-      "(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\n",
-      "(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=10.0;\t\t\t#Collector supply voltage in V\n",
-    "V_BB=2.7;\t\t\t#Base supply voltage in V\n",
-    "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n",
-    "beta=100.0;\t\t\t#Base current amplification factor\n",
-    "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n",
-    "\n",
-    "\n",
-    "#Calcultaion\t\n",
-    "V_B=V_BB;\t\t\t#Base voltage in V\n",
-    "V_E=V_B-V_BE;\t\t\t#Emitter voltage in V\n",
-    "I_E=V_E/R_E;\t\t\t#Emitter current in mA\n",
-    "I_C=I_E;\t\t\t#Collector current (approximately equal to emitter current) in mA\n",
-    "I_B=I_C/beta;\t\t\t#Base current in mA\n",
-    "\n",
-    "#Case (i):\n",
-    "R_C=2;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "\n",
-    "#Assuming transistor to be in active state\n",
-    "#Applying Kirchhoff's voltage law along collector side,\n",
-    "#We get,Vcc-IcRc=Vc,\n",
-    "\n",
-    "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-    "\n",
-    "if(V_C>V_E):\n",
-    "\tprint(\"(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\");\n",
-    "elif(V_C<V_E):\n",
-    "\tprint(\"(i)Our assumption was wrong, the transistor is in saturation for Rc=2 kilo ohm.\");\n",
-    "elif(V_C==V_E):\n",
-    "\tprint(\"(i)The transistor is at the edge of saturation for Rc=2 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n",
-    "\n",
-    "#Case (ii):\n",
-    "R_C=4;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "\n",
-    "#Assuming transistor to be in active state\n",
-    "#Applying Kirchhoff's voltage law along collector side,\n",
-    "#We get,Vcc-IcRc=Vc,\n",
-    "\n",
-    "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-    "if(V_C>V_E):\n",
-    "\tprint(\"(ii)Our assumption was correct, the transistor is in active state for Rc=4 kilo ohm.\");\n",
-    "elif(V_C==V_E):\n",
-    "\tprint(\"(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n",
-    "elif(V_C<V_E):\n",
-    "\tprint(\"(ii)Our assumption was wrong, the transistor is in saturation for Rc=4 kilo ohm.\");\n",
-    "\n",
-    "\n",
-    "#Case (iii):\n",
-    "R_C=8;\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "\n",
-    "#Assuming transistor to be in active state\n",
-    "#Applying Kirchhoff's voltage law along collector side,\n",
-    "#We get,Vcc-IcRc=Vc,\n",
-    "\n",
-    "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n",
-    "if(V_C>V_E):\n",
-    "\tprint(\"(iii)Our assumption was correct, the transistor is in active state for Rc=8 kilo ohm.\");\n",
-    "elif(V_C<V_E):\n",
-    "\tprint(\"(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\");\n",
-    "elif(V_C==V_E):\n",
-    "\tprint(\"(iii)The transistor is at the edge of saturation for Rc=8 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.37 : Page number 179-180"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "(i) Base voltage =0.5V is less than VBE=0.7V, therefore, transistor is cut-off.\n",
-      "0.8 0.8 7.0\n",
-      "(ii) VC=7V > VE=0.8V, therefore the transistor is active. Our assumption was correct.\n",
-      "(iii) VC=-8V < VE=2.3V, therefore the transistor is saturated. Our assumption was wrong.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=15.0;\t\t\t#Collector supply voltage in V\n",
-    "R_C=10.0;\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n",
-    "beta=100.0;\t\t\t#Base current amplification factor\n",
-    "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n",
-    "\n",
-    "\n",
-    "#Calculation\t\n",
-    "\n",
-    "#Case (i):\n",
-    "V_BB=0.5;\t\t\t#Base supply voltage in V\n",
-    "VB=V_BB;            #Base voltage, V\n",
-    "print(\"(i) Base voltage =%.1fV is less than VBE=%.1fV, therefore, transistor is cut-off.\"%(VB,V_BE));\n",
-    "\n",
-    "\n",
-    "#Case (ii):\n",
-    "V_BB=1.5;\t\t\t#Base supply voltage in V\n",
-    "VB=V_BB;            #Base voltage, V\n",
-    "VE=VB-V_BE;          #Emitter voltage, V\n",
-    "IE=round(VE/R_E,1);  #Emitter current, mA\n",
-    "#Assuming transistor to be in active state\n",
-    "#Applying Kirchhoff's voltage law along collector side,\n",
-    "IC=IE;              #Collector current, mA\n",
-    "IB=IC/beta;         #Base  current, mA\n",
-    "VC=V_CC-IC*R_C;      #Collector voltage, V\n",
-    "print(VE,IE,VC);\n",
-    "print(\"(ii) VC=%dV > VE=%.1fV, therefore the transistor is active. Our assumption was correct.\"%(VC,VE));\n",
-    "\n",
-    "#Case (iii):\n",
-    "V_BB=3; \t\t\t#Base supply voltage in V\n",
-    "VB=V_BB;            #Base voltage, V\n",
-    "VE=VB-V_BE;          #Emitter voltage, V\n",
-    "IE=round(VE/R_E,1);  #Emitter current, mA\n",
-    "#Assuming transistor to be in active state\n",
-    "#Applying Kirchhoff's voltage law along collector side,\n",
-    "IC=IE;              #Collector current, mA\n",
-    "IB=IC/beta;         #Base  current, mA\n",
-    "VC=V_CC-IC*R_C;      #Collector voltage, V\n",
-    "\n",
-    "print(\"(iii) VC=%dV < VE=%.1fV, therefore the transistor is saturated. Our assumption was wrong.\"%(VC,VE));"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.38: Page number 181"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 41,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Maximum collector current that can be allowed without destruction of the transistor = 5 mA.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "P_D_max=100.0;\t\t\t#Maximum power dissipation of a transistor in mW\n",
-    "V_CE=20.0;\t\t\t#Collector emitter voltage in V\n",
-    "\n",
-    "#Calculation\n",
-    "#As power=curent*voltage\n",
-    "#P_D_max=I_C_max*V_CE\n",
-    "#From the above equation, we get:\n",
-    "\n",
-    "I_C_max=P_D_max/V_CE;\t\t#Maximum collector current that can be allowed without destruction of the transistor, in mA\n",
-    "\n",
-    "#Result\n",
-    "print(\"Maximum collector current that can be allowed without destruction of the transistor = %d mA.\"%I_C_max); \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.39: Page number 181"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 42,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Power dissipated = 4.3W\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n",
-    "V_BB=5.0;\t\t\t\t#Base supply voltage in V\n",
-    "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-    "R_B=1.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-    "R_C=0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "beta=200.0;\t\t\t\t#base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along base circuit<\n",
-    "#We get, VBB- IB*RB - VBE=0.\n",
-    "#From the above equation, we get:\n",
-    "\n",
-    "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-    "\n",
-    "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along collector circuit:\n",
-    "\n",
-    "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n",
-    "\n",
-    "#As power=curent*voltage\n",
-    "#P_D=I_C*V_CE\n",
-    "#From the above equation, we get:\n",
-    "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n",
-    "P_D=P_D/1000;\t\t\t\t#Power dissipated in W\n",
-    "\n",
-    "#Result\n",
-    "print(\"Power dissipated = %.1fW\"%P_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.40: Page number 181-182"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 43,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Power dissipated = 6mW\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n",
-    "V_BB=1.0;\t\t\t\t#Base supply voltage in V\n",
-    "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n",
-    "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n",
-    "R_C=1.0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n",
-    "beta=100.0;\t\t\t\t#base current amplification factor\n",
-    "\n",
-    "#Calculation\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along base circuit<\n",
-    "#We get, VBB- IB*RB - VBE=0.\n",
-    "#From the above equation, we get:\n",
-    "\n",
-    "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n",
-    "\n",
-    "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n",
-    "\n",
-    "#Applying Kirchhoff's voltage law along collector circuit:\n",
-    "\n",
-    "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n",
-    "\n",
-    "#As power=curent*voltage\n",
-    "#P_D=I_C*V_CE\n",
-    "#From the above equation, we get:\n",
-    "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n",
-    "\n",
-    "\n",
-    "#Result\n",
-    "print(\"Power dissipated = %.0fmW\"%P_D);\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example 8.41 : Page number 182"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 44,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "IC=19.5mA is much less than IC_max=100mA. Therefore, will not change with VCC and current rating is not exceeded.\n",
-      "PD=293mW  is less than PD_max=800mW. Therefore, power rating is not exceeded.\n",
-      "If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\n"
-     ]
-    }
-   ],
-   "source": [
-    "#Variable declaration\n",
-    "VBB=5.0;                     #Base supply voltage, V\n",
-    "RB=22.0;                  #Base resistor, kilo ohm\n",
-    "RC=1.0;                   #Collector resistor, kilo ohm\n",
-    "beta=100.0;               #Base current amplification factor\n",
-    "VBE=0.7;                  #Base-emitter voltage, V\n",
-    "PD_max=800.0;             #Maximum power dissipation, mW\n",
-    "VCE_max=15.0;             #Maximum collector-emitter voltage, V\n",
-    "IC_max=100.0;             #Maximum collector current, mA\n",
-    "\n",
-    "#Calculation\n",
-    "IB=((VBB-VBE)/RB)*1000;               #Base current,  μA\n",
-    "IC=beta*IB/1000;                      #Collector current, mA\n",
-    "\n",
-    "print(\"IC=%.1fmA is much less than IC_max=%dmA. Therefore, will not change with VCC and current rating is not exceeded.\"%(IC,IC_max));\n",
-    "\n",
-    "#VCC=VCE+IC*RC\n",
-    "VCC_max=VCE_max+IC*RC;           #Maximum value of Collector supply voltage, V\n",
-    "PD=VCE_max*IC;                  #Power dissipation, mW\n",
-    "\n",
-    "print(\"PD=%dmW  is less than PD_max=%dmW. Therefore, power rating is not exceeded.\"%(PD,PD_max));\n",
-    "\n",
-    "print(\"If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\");"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.5.1"
-  },
-  "widgets": {
-   "state": {},
-   "version": "1.1.2"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9.ipynb
deleted file mode 100755
index e8168ab8..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9.ipynb
+++ /dev/null
@@ -1,1907 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1e5463fc2e8c67a2f9099cf3f6c078bc1c9dccccd63589a75ad6ce5024fe6432"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 9 : TRANSISTOR BIASING"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1: Page number 195-196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=6.0;                     #Collector supply voltage\n",
-      "R_C=2.5;                      #Collector load in k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "#For faithful amplification Vce (collector-emitter voltage)> 1V for Si transistor.\n",
-      "V_CE_max=1;                                   #Maximum allowed collector-emitter voltage for faithful amplification, in V.\n",
-      "V_Rc_max=V_CC-V_CE_max;                       #maximum voltage drop across collector load in V.\n",
-      "I_C_max=V_Rc_max/R_C;                         #Maximum allowed collector current in mA\n",
-      "\n",
-      "#(ii)\n",
-      "IC_min_zero_signal=I_C_max/2;                            #Minimum zero signal collector current in mA\n",
-      "\n",
-      "#Results\n",
-      "print(\"The maximum allowed collector current during application of signal for faithful amplification = %d mA.\"%I_C_max);\n",
-      "print(\"The minimum zero signal collector current required = %d mA.\"%IC_min_zero_signal);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowed collector current during application of signal for faithful amplification = 2 mA.\n",
-        "The minimum zero signal collector current required = 1 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2: Page number 196\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=13.0;                     #Collector supply voltage in V\n",
-      "V_knee=1.0;                   #Knee voltage in V\n",
-      "R_C=4.0;                      #Collector load  in k\u2126\n",
-      "rate_IC_VBE=5.0;              #Rate of change of collector current IC with base-emitter voltage VBE in mA/V.\n",
-      "beta=100.0;                   #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V_Rc_max=VCC-V_knee;                    #Maximum allowed voltage across collector load in V\n",
-      "I_C_max=V_Rc_max/R_C;                   #Maximum allowed collector current in mA\n",
-      "I_B_max=I_C_max/beta;                   #Maximum base current in mA\n",
-      "I_B_max=I_B_max*1000;                   #Maximum base current in \ud835\udf07A\n",
-      "\n",
-      "V_B_max=I_C_max/rate_IC_VBE;            #Maximum base voltage signal in V\n",
-      "V_B_max=V_B_max*1000;                   #Maximum base voltage signal in mV\n",
-      "\n",
-      "#Results\n",
-      "print(\"Maximum base current =%d \ud835\udf07A.\"%I_B_max);\n",
-      "print(\"Maximum input signal voltage =%d mV.\"%V_B_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum base current =30 \ud835\udf07A.\n",
-        "Maximum input signal voltage =600 mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3: Page number 200-201"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=9.0;                  #Colector supply voltage in V\n",
-      "VBB=2.0;                  #Base supply voltage in V\n",
-      "R_B=100.0;                #Base resistor's resistance in k\u2126\n",
-      "R_C=2.0;                  #Collector load in k\u2126\n",
-      "beta=50.0;                #base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Case (i):\n",
-      "\n",
-      "#Applying Kirchhoff's law to the input circuit\n",
-      "#We get, IB*RB +VBE =VBB.\n",
-      "#Neglecting the small base-emitter voltage, we get:\n",
-      "I_B=VBB/R_B;                #Base current in mA\n",
-      "I_C=beta*I_B;               #Collector current in mA\n",
-      "\n",
-      "print(\"Collector current = %dmA\"%I_C);\n",
-      "\n",
-      "#Applying Kirchhoff's law to the output ciruit\n",
-      "#We get, IC*RC + VCE= VCC.\n",
-      "#From the above equation, we get:\n",
-      "V_CE=VCC-I_C*R_C;               #Collector emitter voltage in V\n",
-      "\n",
-      "print(\"Collector emitter voltage =%dV.\"%V_CE);\n",
-      "\n",
-      "\n",
-      "#Case (ii):\n",
-      "\n",
-      "R_B=50.0;\n",
-      "I_B=VBB/R_B;\n",
-      "I_C=beta*I_B;\n",
-      "V_CE=VCC  - I_C*R_C;\n",
-      "\n",
-      "print(\"The new operating point for base resistor RB=50 k\u2126 is, VCE=%dV and IC=%dmA.\"%(V_CE,I_C));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current = 1mA\n",
-        "Collector emitter voltage =7V.\n",
-        "The new operating point for base resistor RB=50 k\u2126 is, VCE=5V and IC=2mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4: Page number  201-202"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#variable declaration\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "VCC=6.0;                  #Collector suply voltagein V\n",
-      "VBE=0.7                   #Base emitter voltage in V\n",
-      "R_B=530.0;                #Base resistor's resistance in k\u2126 .\n",
-      "R_C=2.0;                  #Collector resistor's resistance in k\u2126 .\n",
-      "\n",
-      "#Calculation\n",
-      "#D.C load line equation : VCE=VCC-IC*RC;\n",
-      "#Calculating maximum VCE  ,by IC=0;\n",
-      "I_C_Vce_max=0;                      #Collector current for maximum collector-emitter voltage, in mA\n",
-      "VCE_max=VCC;-I_C_Vce_max*R_C;       #Maximum collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculating maximum collector current IC,by VCE=0;\n",
-      "V_CE_IC_max=0;             #Collector-emitter voltage for maximum collector current, in V         \n",
-      "I_C_max=(VCC-V_CE_IC_max)/R_C;        #Maximum collector current in mA\n",
-      "\n",
-      "\n",
-      "#Operating point:\n",
-      "#For input circuit, applying Kirchhoff's law, We get,\n",
-      "#VCC=IB*RB  + VBE.\n",
-      "#From the above equation,\n",
-      "IB=(VCC-VBE)/R_B;                #Base current in mA\n",
-      "IC=beta*IB;                    #Collector current\n",
-      "\n",
-      "#From the output circuit, applying Kirchhoff's law, we get:\n",
-      "VCE=VCC-IC*R_C;                 #Collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Stability factor\n",
-      "SF=beta+1;              \n",
-      "\n",
-      "#Result\n",
-      "print(\"Operating point: VCE= %dV and IC=%d mA\"%(VCE,IC));\n",
-      "print(\"Stability factor= %d.\"%SF);\n",
-      "\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,10])\n",
-      "limit.set_ylim([0,5])\n",
-      "VCE=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC=[((VCC-i)/(R_C)) for i in (VCE[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE,IC);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: VCE= 4V and IC=1 mA\n",
-        "Stability factor= 101.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8eea67df10>"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5: Page number 202"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage in V\n",
-      "beta=100.0;              #base current amplification factor\n",
-      "I_C_zero_signal=1.0;      #zero signal collector current in mA\n",
-      "VBE=0.3;                  #Base-emitter voltage of Ge transistor in V\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#Case(i)\n",
-      "I_B_zero_signal=I_C_zero_signal/beta;               #Zero signal base current in  mA\n",
-      "\n",
-      "#applying the Kirchhoff's law along input circuit:\n",
-      "#We get, VCC=IB*RB +VBE\n",
-      "#From the above equation we get,\n",
-      "R_B=(VCC-VBE)/I_B_zero_signal;              #Required base resistor's resistance in k\u2126\n",
-      "\n",
-      "print(\"Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = %d k\u2126\"%R_B);\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Case(ii)\n",
-      "beta=50;\n",
-      "I_B=(VCC-VBE)/R_B;                  #Base current of another transistor with beta=50, in mA\n",
-      "I_C_zero_signal=beta*I_B;           #Zero signal collector current for beta=50 , in mA\n",
-      "\n",
-      "print(\"The new value of zero signal collector current =%.1fmA\"%I_C_zero_signal);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = 1170 k\u2126\n",
-        "The new value of zero signal collector current =0.5mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6:Page number 202-203"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Varaible declaration\n",
-      "VCC=10.0;                     #Collector supply voltage in V\n",
-      "VBE=0;                        #Base emitter voltage in V(considering itas zero due to it's small value)\n",
-      "R_B=1.0;                      #Base resistor's resistance in M\u2126\n",
-      "R_C=2.0;                      #Collector resistor's resistance in k\u2126                  \n",
-      "R_E=1.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#using Kirchhoff's law in the input circuit, we get:\n",
-      "#VCC=IB*RB +VBE +IE*RE\n",
-      "#Since, IE=(beta +1)*I_B\n",
-      "#From the above equation we get:\n",
-      "I_B=round((VCC-VBE)/((beta + 1)*R_E + R_B*1000),4);           #Base current in mA\n",
-      "I_C=round(beta*I_B,2);                                   #Collector current in mA\n",
-      "I_E=I_B+I_C;                                    #Emitter current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Base current =%.4f mA\"%I_B);\n",
-      "print(\"Collector current =%.2f mA\"%I_C);\n",
-      "print(\"Emitter current =%.3f mA\"%I_E);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current =0.0091 mA\n",
-        "Collector current =0.91 mA\n",
-        "Emitter current =0.919 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.7: Page number 203-204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=8.0;                      #Collector-emitter voltage at operating point in V\n",
-      "IC=2.0;                       #Colector current at operating point in mA\n",
-      "VCC=15.0;                     #Collector supply voltagein V\n",
-      "beta=100.0;                   #base current amplification factor\n",
-      "VBE=0.6;                    #base emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC=VCE+IC*RC.\n",
-      "#So, from above equation we get:\n",
-      "RC=(VCC-VCE)/IC;                 #Collector resistor's resistance in k\u2126 .\n",
-      "IB=IC/beta;                      #Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the input circuit,\n",
-      "#we get, VCC=IB*RB + VBE\n",
-      "#So, from the above equation:\n",
-      "RB=(VCC-VBE)/IB;                #Base resistor's resistance in k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector load =%.1f k\u2126 .\"%RC);\n",
-      "print(\"Base resistor=%d k\u2126 .\"%RB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector load =3.5 k\u2126 .\n",
-        "Base resistor=720 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8: Page number 204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                #Collector supply voltage in V\n",
-      "VBE=0.7;                 #Base-emitter voltage in V\n",
-      "RB=100.0;                #Base resistor's resistance in k\u2126\n",
-      "RC=560.0;                #Collector resistor's resistance in \u2126\n",
-      "beta_25=100.0;           #base current amplification factor at 25 degree celsius\n",
-      "beta_75=150.0;           #base current amplification factor at 25 degree celsius\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit, we get\n",
-      "#VCC=IB*RB+VBE\n",
-      "IB=(VCC-VBE)/RB;                #Base current at 25 degree celsius, in mA\n",
-      "\n",
-      "\n",
-      "#For temperature 25 degree celsius\n",
-      "IC_25=beta_25*IB;                  #Collector current at 25 degree celsius, in mA\n",
-      "\n",
-      "\n",
-      "#Applying Kirchhoff's alw at the output circuit,\n",
-      "#we get: VCC=IC*RC + VCE\n",
-      "#From the above equation,\n",
-      "VCE_25=round(VCC-(IC_25/1000)*RC,2);                #Collector emitter voltage  at 25 degree celsius, in V\n",
-      "\n",
-      "\n",
-      "#For temperature 75 degree celsius\n",
-      "IC_75=round(beta_75*IB,0);                  #Collector current at 75 degree celsius, in mA\n",
-      "\n",
-      "#Applying Kirchhoff's alw at the output circuit,\n",
-      "#we get: VCC=IC*RC + VCE\n",
-      "#From the above equation,\n",
-      "VCE_75=round(VCC-(IC_75/1000)*RC,2);                #Collector emitter voltage at 75 degree celsius, in V\n",
-      "\n",
-      "\n",
-      "change_IC=(IC_75-IC_25)*100.0/IC_25;        #percentage change in collector current\n",
-      "change_VCE=(VCE_75-VCE_25)*100.0/VCE_25;    #Percentage change in collector-emitter voltage \n",
-      "\n",
-      "#Results\n",
-      "print(\"The percentage change in collector current =%d%%\"%change_IC);\n",
-      "print(\"The percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage change in collector current =50%\n",
-        "The percentage change in collector-emitter voltage =-56.3%\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.10: Page number 205"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE_max=20.0;             #Maximum collector-emitter voltage in V\n",
-      "VBE=0.7;                  #Base-emitter voltage in V\n",
-      "IC_max=8.0;               #Maximum collector current in mA\n",
-      "IB=40.0;                  #Base current in microampere\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#During cut off state the collector-emitter voltage is maximum and equal to collector supply voltage\n",
-      "VCC=VCE_max;                #Collector supply voltage in V\n",
-      "\n",
-      "#Maximum collector current IC_max=collector supply voltage(VCC)/collector load(RC)\n",
-      "#Collector load(RC)=VCC*IC_max\n",
-      "RC=VCC/IC_max;                  #Collector load in k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC=IB*RB +VBE.\n",
-      "#From the above equation, we get:\n",
-      "RB=(VCC-VBE)/(IB/1000);    #Base resistor's resistance in k\u2126 .\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector supply voltage = %dV\"%VCC);\n",
-      "print(\"Collector load=%.1f k\u2126 .\"%RC);\n",
-      "print(\"Base resistor's resistance=%.1f k\u2126 .\"%RB);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector supply voltage = 20V\n",
-        "Collector load=2.5 k\u2126 .\n",
-        "Base resistor's resistance=482.5 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.12: Page number 208"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage in V\n",
-      "VEE=-20.0;                     #Emitter supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=4.7;                       #Collector resistor's resistance in k\u2126\n",
-      "RE=10.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "VBE=0.7;                      #Base-emitter voltage in V\n",
-      "beta=85.0;                    #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n",
-      "#we get,IB*RB  +IE*RE +VBE -VEE=0.\n",
-      "#Since IB=IC/beta and IC~IE,\n",
-      "#(IE/beta)*RB + IE*RE + VBE + VEE =0.\n",
-      "IE=(-VEE-VBE)/(RE + RB/beta);                        #Emitter current in mA\n",
-      "IC=IE;                                              #Collector current (approximately equal to emitter current) in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law from VCC till collector terminal,\n",
-      "#we get, VCC - IC*RC =VC\n",
-      "VC=VCC-IC*RC;                               #voltage at collector terminal in V\n",
-      "\n",
-      "#Applying Kirchhoff's law from emitter terminal to VEE\n",
-      "#we get, VE -IE*RE =VEE\n",
-      "VE=VEE + IE*RE;                             #Voltage at emitter treminal in V\n",
-      "\n",
-      "VCE=VC-VE;                                  #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The collector current = %.2f mA\"%IC);\n",
-      "print(\"The emitter current = %.2f mA\"%IE);\n",
-      "print(\"The voltage at collector terminal = %.1f V\"%VC);\n",
-      "print(\"The collector-emitter voltage = %.1f V\"%VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The collector current = 1.73 mA\n",
-        "The emitter current = 1.73 mA\n",
-        "The voltage at collector terminal = 11.9 V\n",
-        "The collector-emitter voltage = 14.6 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.13: Page number 208-209\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage in V\n",
-      "VEE=-20.0;                     #Emitter supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=4.7;                       #Collector resistor's resistance in k\u2126\n",
-      "RE=10.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "beta1=85.0;                   #Base current amplification factor for case 1       \n",
-      "beta2=100.0;                  #Base current amplification factor for case 1\n",
-      "VBE_1=0.7;                    #Base emitter voltage for case 1 in V\n",
-      "VBE_2=0.6;                    #Base emitter voltage for case 2 in V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#For beta=85 and VBE=0.7,\n",
-      "#As calculated in the previous question,\n",
-      "IC_1=1.73;                          #Collector current in mA.\n",
-      "VCE_1=14.6;                           #Collector-emitter voltage in V.\n",
-      "\n",
-      "\n",
-      "#For case (ii)\n",
-      "#beta=100 and VBE=0.6\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n",
-      "#we get,IB*RB  +IE*RE +VBE -VEE=0.\n",
-      "#Since IB=IC/beta and IC~IE,\n",
-      "#(IE/beta)*RB + IE*RE + VBE +VEE =0.\n",
-      "IE_2=round((-VEE-VBE_2)/(RE + RB/beta2),2);                        #Emitter current in mA\n",
-      "IC_2=IE_2;                                                #Collector current (approximately equal to emitter current) in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law from VCC till collector terminal,\n",
-      "#we get, VCC - IC*RC =VC\n",
-      "VC=round(VCC-IC_2*RC,1);                               #voltage at collector terminal in V\n",
-      "\n",
-      "#Applying Kirchhoff's law from emitter terminal to VEE\n",
-      "#we get, VE -IE*RE =VEE\n",
-      "VE=round(VEE + IE_2*RE,1);                             #Voltage at emitter treminal in V\n",
-      "\n",
-      "VCE_2=VC-VE;                                #Collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "change_IC= (IC_2-IC_1)*100/IC_1;                    #%age change in collector current\n",
-      "\n",
-      "change_VCE=(VCE_2-VCE_1)*100/VCE_2;                 #%age change in collector-emitter voltage\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Percentage change in collector current =%.1f%%\"%change_IC);\n",
-      "print(\"Percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage change in collector current =1.7%\n",
-        "Percentage change in collector-emitter voltage =-3.5%\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.14: Page number 210\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                 #Collector supply voltage in V\n",
-      "VBE=0.7                   #Base-emitter voltage in V\n",
-      "RB=100.0;                 #Base resistor's resistance in k\u2126\n",
-      "RC=1.0;                   #Collector resistor's resistance in k\u2126\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC -IC*RC -IB*RB -VBE=0.\n",
-      "#since IC= beta*IB,\n",
-      "#We get,\n",
-      "IB=(VCC-VBE)/(RB + beta*RC);            #Base current in mA\n",
-      "IC=beta*IB;                             #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC=0.\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*RC;                          #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point : VCE=10.35V and IC=9.65mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.15: Page number 210-211\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                         #Collector supply voltage in V\n",
-      "VBE=0.3;                          #Base emitter voltage in V\n",
-      "IC=1.0;                           #Collector current in mA\n",
-      "VCE=8.0;                          #Collector emitter voltage in V\n",
-      "beta=100.0;                         #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Case(i)\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-IC*RC-VCE=0.\n",
-      "#from the above equation we get,\n",
-      "RC=(VCC-VCE)/IC;             #Collector load in kilo ohm\n",
-      "IB=IC/beta;                  #Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit\n",
-      "#we get, VCC-VBE-(beta*IB*RC)-IB*RB=0.\n",
-      "#From the above equation we get,\n",
-      "RB=round((VCC-VBE-beta*IB*RC)/IB,0);         #Base resistor's resistance in k\u2126\n",
-      "\n",
-      "#Results\n",
-      "print(\"The resistance value of base resistor=%d k\u2126 and collector load= %d k\u2126.\"%(RB,RC));\n",
-      "\n",
-      "#Case(ii)\n",
-      "\n",
-      "beta=50;\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC -IC*RC -IB*RB -VBE=0.\n",
-      "#since IC= beta*IB,\n",
-      "#We get,\n",
-      "IB=(VCC-VBE)/(RB + beta*RC);            #Base current in mA\n",
-      "IC=beta*IB;                             #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC=0.\n",
-      "#From the above equation,\n",
-      "VCE=round(VCC-IC*RC,1);                          #Collector emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.1fV and IC=%.1fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resistance value of base resistor=770 k\u2126 and collector load= 4 k\u2126.\n",
-        "The operating point : VCE=9.6V and IC=0.6mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.16 : Page number 211"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=2.0;                      #Collector-emitter voltage at operating point in V\n",
-      "VBE=0.7;                      #Base-emitter voltage in V            \n",
-      "IC=1.0;                       #Collector current at operating point in mA\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "IB=IC/beta;                     #Base current in mA\n",
-      "\n",
-      "#As, VCE=VCB +VBE\n",
-      "#we get,\n",
-      "VCB=VCE-VBE;                    #Collector-base voltage in V\n",
-      "RB=VCB/IB;                      #Base resistor's resistance in k\u2126\n",
-      "\n",
-      "#Results\n",
-      "print(\"Value of base resistor's resistance=%d k\u2126.\"%RB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of base resistor's resistance=130 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.17 : Page number 211-212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage in V\n",
-      "VBE=0.7                   #Base-emitter voltage in V\n",
-      "RB=400.0;                 #Base resistor's resistance in k\u2126\n",
-      "RC=4.0;                   #Collector resistor's resistance in k\u2126\n",
-      "RE=1.0;                   #Emitter resistor's resistance in k\u2126\n",
-      "beta=100.0;               #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along outut circuit,\n",
-      "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n",
-      "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n",
-      "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n",
-      "IC=(VCC-VBE)/(RB/beta + RC + RE);       #Collector current current in mA.\n",
-      "IE=IC;                                                #Emitter current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC -IE*RE=0. (IE~IC)\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*(RC+RE);                          #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.1fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point : VCE=5.7V and IC=1.26mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.18 : Page number 212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=10.0;                      #Collector resistor's resistance in k\u2126\n",
-      "RE=0;                         #Emitter resistor's  resistance in k\u2126\n",
-      "VBE=0.7;                      #Base-emitter voltage in V\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along outut circuit,\n",
-      "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n",
-      "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n",
-      "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n",
-      "IC=(VCC-VBE)/(RC +RB/beta + RE);                #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC =0. (IE~IC)\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*RC;                                  #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The d.c bias values are: VCE=%.2fV and IC=%.3fmA\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The d.c bias values are: VCE=1.55V and IC=0.845mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.19: Page number 214-215\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage in V\n",
-      "R1=10.0;                            #Resistor R1's resistance in k\u2126\n",
-      "R2=5.0;                             #Resistor R2's resistance in k\u2126\n",
-      "RC=1.0;                             #Collector resistor's resistance in k\u2126              \n",
-      "RE=2.0;                             #Emitter resistor's resistance in k\u2126\n",
-      "VBE=0.7;                            #Base-emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along output circuit\n",
-      "#VCE=VCC-IC*(RC+RE);\n",
-      "#IC=0, for VCE_max\n",
-      "VCE_max=VCC;                        #Maximum collector-emitter voltage in V\n",
-      "#VCE=0, for IC_max\n",
-      "IC_max=VCC/(RC+RE);                 #Maximum collector current in mA\n",
-      "\n",
-      "#Operating point\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage across R2 resistor V\n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "IC=IE;                              #Collector current(Approx. equal to emitter current) in mA\n",
-      "VCE=VCC-IC*(RC+RE);                 #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n",
-      "print(\"Collector current at operating point = %.2fmA\"%IC);\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,20])\n",
-      "limit.set_ylim([0,6])\n",
-      "VCE=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC=[((VCC-i)/(RC+RE)) for i in (VCE[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE,IC);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector-emitter voltage at operating point=8.55V\n",
-        "Collector current at operating point = 2.15mA\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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hKB07Mu+WiAStzuwdHBwwZswYrF69GiNHjgQADBgwAGfPnjXP4pzZi65xtPM//wPMny/s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\nAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8eeacf1ad0>"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.20: Page number 215-216\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage in V\n",
-      "R1=10.0;                            #Resistor R1's resistance in k\u2126 .\n",
-      "R2=5.0;                             #Resistor R2's resistance in k\u2126 .\n",
-      "RC=1.0;                             #Collector resistor's resistance in k\u2126 .              \n",
-      "RE=2.0;                             #Emitter resistor's resistance in k\u2126 .\n",
-      "VBE=0.7;                            #Base-emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's Theorem for replacing circuit consisting of VCC,R1,R2\n",
-      "E0=(VCC*R2)/(R1+R2);                #Thevenin's voltage in V\n",
-      "R0=(R1*R2)/(R1+R2);                 #Thevenin's equivalent resistance in k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "#IC=(E0-VBE)/(R0/beta +RE);\n",
-      "IC=(E0-VBE)/RE;                         #(Since R0/beta << RE) collector current in mA\n",
-      "VCE=VCC-IC*(RC+RE);                     #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n",
-      "print(\"Collector current at operating point = %.2fmA\"%IC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector-emitter voltage at operating point=8.55V\n",
-        "Collector current at operating point = 2.15mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.21: Page number 216-217\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                           #Collector  supply voltage in V\n",
-      "RE=1.0;                             #Emitter resistor, k\u2126 .\n",
-      "R1=50.0;                            #Resistor R1, k\u2126 .\n",
-      "R2=10.0;                            #Resistor R2, k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "VBE=0.1;                            #Base-emitter voltage in V\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V \n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "\n",
-      "print(\"(i)Emitter current= %.1fmA\"%IE);\n",
-      "\n",
-      "#(ii)\n",
-      "VBE=0.3;                            #Base-emitter voltage in V\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V\n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "\n",
-      "print(\"(ii)Emitter current= %.1fmA\"%IE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Emitter current= 1.9mA\n",
-        "(ii)Emitter current= 1.7mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.22: Page number 217\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                   #Collector supply voltage, V\n",
-      "R1=10.0;                    #Resistor R1, k\u2126\n",
-      "R2=10.0;                    #Resistor R2, k\u2126 .\n",
-      "RC=1.0;                     #Collector resistor, k\u2126 .\n",
-      "RE=5.0;                     #Emitter resistor, k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V\n",
-      "\n",
-      "#Applying kirchhoff's law from base terminal to emitter resistor\n",
-      "#V2=VBE+IE*RE\n",
-      "#VBE is neglected due to its small value\n",
-      "\n",
-      "IE=V2/RE;                     #Emitter current in mA\n",
-      "IC=IE;                              #Collector current (approx. equal to emitter current), mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along output circuit\n",
-      "VCE=VCC-IC*(RC+RE);                 #Collector-emitter voltage , V\n",
-      "VC=VCC-IC*RC;                       #Voltage at collector terminal,V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Emitter current =%dmA\"%IE);\n",
-      "print(\"Collector-emitter voltage=%dV\"%VCE);\n",
-      "print(\"Collector terminal's voltage=%dV\"%VC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Emitter current =2mA\n",
-        "Collector-emitter voltage=8V\n",
-        "Collector terminal's voltage=18V\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.23: Page number 219-220\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage, V\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "beta=50;                  #Base current amplification factor\n",
-      "R1=150;                   #Resistor R1, k\u2126 .\n",
-      "R2=100;                   #Resistor R2, k\u2126 .\n",
-      "RC=4.7;                   #Collector resistor, k\u2126 .\n",
-      "RE=2.2;                   #Emitter resistor, k\u2126  .\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "IB=round((E0-VBE)/(R0+beta*RE),3);              #Base current in mA\n",
-      "IC=round(beta*IB,1);                             #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-IC*(RC+RE);                     #Collector-emitter voltage, V\n",
-      "\n",
-      "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE);               #Stability factor\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n",
-      "print(\"Stability factor=%.1f\"%S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 3.72V and IC=1.2mA\n",
-        "Stability factor=18.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.24 : Page number 220\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                 #Collector supply voltage, V\n",
-      "VBE=0.7;                  #Base-emitter voltage , V\n",
-      "beta=100.0;                 #Base current amplification factor\n",
-      "R1=6.0;                     #Resistor R1, k\u2126 .\n",
-      "R2=3.0;                     #Resistor R2, k\u2126 .\n",
-      "RC=470.0;                   #Collector resistor, \u2126.\n",
-      "RE=1.0;                     #Emitter resistor, k\u2126 .\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "IB=round((E0-VBE)/(R0+beta*RE),3);              #Base current in mA\n",
-      "IC=round(beta*IB,1);                             #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-IC*(RC/1000+RE);                     #Collector-emitter voltage, V\n",
-      "\n",
-      "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE);               #Stability factor\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n",
-      "print(\"Stability factor=%.2f\"%S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 8.83V and IC=4.2mA\n",
-        "Stability factor=2.94\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.25 : Page number 221-222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "VCC=9;                    #Collector supply voltage, V\n",
-      "VCE=3;                    #Collector-emitter voltage, V\n",
-      "VBE=0.3;                    #Base-emitter voltage in V\n",
-      "RC=2.2;                     #Collector resistor , k\u2126 .\n",
-      "IC=2;                     #Collector current, mA\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "IB=IC/beta;                 #Base current in mA\n",
-      "\n",
-      "#According to given relation, I1=10*IB\n",
-      "I1=IB*10;                   #Current through the resistor R1, mA\n",
-      "\n",
-      "#I1=VCC/(R1+R2), .'s LAW\n",
-      "R1_R2_sum=VCC/I1;               #Sum of the resistor's R1 and R2, k\u2126 (OHM'S LAW).\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "#VCC=IC*RC+VCE+IE*RE\n",
-      "#IC~IE\n",
-      "RE=(VCC-IC*RC-VCE)/IC;      #Emitter resistor, k\u2126 .\n",
-      "RE=round(RE*1000,0);                 #Emittter resistor, \u2126 .\n",
-      "\n",
-      "IE=IC;                      #Emittter current(approximately equal to collector current), mA\n",
-      "VE=IE*(RE/1000);                   #Voltage at emitter terminal (OHM's LAW), V\n",
-      "V2=VBE+VE;                  #Voltage drop across resistor R2, V\n",
-      "\n",
-      "R2=V2/I1;                   #Resistor R2,(OHM's LAW), k\u2126 .\n",
-      "R1=R1_R2_sum-R2;            #Resistor R1, k\u2126 .\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"RE=%d \u2126., R1=%.2f k\u2126 . and R2=%.2f k\u2126 .\"%(RE,R1,R2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RE=800 \u2126., R1=17.75 k\u2126 . and R2=4.75 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.26 : Page number 222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=16.0;                     #Collector supply voltage, V\n",
-      "R2=20.0;                      #Resistor R2, k\u2126\n",
-      "RE=2.0;                       #Emitter resistor, k\u2126\n",
-      "VCE=6.0;                      #Collector-emitter voltage, V\n",
-      "IC=2.0;                       #Collector current , mA\n",
-      "VBE=0.3;                      #Base-emitter voltage,V\n",
-      "alpha=0.985;                  #Current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "beta=alpha/(1-alpha);               #Base current amplificatioon factor\n",
-      "IE=IC;                              #Emitter current, mA\n",
-      "IB=IC/beta;                         #Base current, mA\n",
-      "VE=IE*RE;                           #Emitter voltage,(OHM's LAW) V\n",
-      "V2=VBE+VE;                          #Voltage drop across resistor R2,(Kirchhoff's law) V\n",
-      "V_R1=VCC-V2;                        #Voltage drop across resistor R1, V\n",
-      "I1=V2/R2;                           #Current through resistor R2 an R1,(OHM's LAW) mA\n",
-      "R1=V_R1/I1;                         #Resistor R1,(OHM's LAW) k\u2126\n",
-      "\n",
-      "V_RC=(VCC-VCE-VE);                  #Voltage across collector resistor, V\n",
-      "RC=V_RC/IC;                         #Collector resistor,(OHM's LAW) k\u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"R1=%.1f k\u2126 and RC=%d k\u2126.\"%(R1,RC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R1=54.4 k\u2126 and RC=3 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.27 :Page number 222-223\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                     #Collector supply voltage, V\n",
-      "R1=10.0;                      #Resistor R1, k\u2126 \n",
-      "R2=5.0;                       #Resistor R2, k\u2126 \n",
-      "RC=1.0;                       #Collector resistor, k\u2126 \u007f\n",
-      "RE=2.0;                       #Emitter resistor, k\u2126 \n",
-      "VBE=0.7;                      #Base-emitter voltage, V\n",
-      "beta=100;                     #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 \n",
-      "\n",
-      "#Applying Kirchhoff' law along Thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IB=IE/beta,\n",
-      "IE=(E0-VBE)/(R0/beta + RE);                 #Emitter current , mA\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"The exact value of emitter current in the circuit = %.2fmA.\"%IE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The exact value of emitter current in the circuit = 2.11mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.28: Page number 223-224\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "IE=2.0;                     #Emitter current, mA\n",
-      "IB=50.0;                    #Base current, mA\n",
-      "VCC=10.0;                   #Collector supply voltage, V\n",
-      "VBE=0.2;                    #Base-emitter voltage, V\n",
-      "R2=10.0;                    #Resistor R2, k\u2126\n",
-      "RE=1.0;                     #Emitter resistance, k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law from the base to the emitter resistor,\n",
-      "V2=VBE+IE*RE;                   #Voltage at base terminal, V\n",
-      "I2=V2/R2;                       #Current through the resistor R2, mA\n",
-      "I1=I2+IB/1000;                  #Current through the resistor R2, mA\n",
-      "V1=VCC-V2;                      #Voltage drop across the resistor R2\n",
-      "R1=V1/I1;                       #Resistor R1, k\u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The value of the resistor R1=%.2f k\u2126.\"%R1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of the resistor R1=28.89 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.30 :Page number 225-226\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=8.0;                  #Collector supply voltage, V\n",
-      "RB=360.0;                 #Base resistor, k\u2126\n",
-      "RC=2.0;                   #Collector resistor, k\u2126\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "IC_max=VCC/RC;              #Maximum collector current, mA\n",
-      "VCE_max=VCC;                #Maximum collector voltage, V\n",
-      "\n",
-      "#Operating point\n",
-      "#Applying Kirchhoff's law along the input circuit\n",
-      "IB=(VCC-VBE)/RB;                    #Base current, mA\n",
-      "IC=beta*IB;                         #Collector current, mA\n",
-      "\n",
-      "#Kirchhoff' law along the output circuit\n",
-      "VCE=VCC-IC*RC;                      #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VCE=3.94V, is approximately half of VCC=8V \n",
-        " therefore it is mid-point biased.\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.31: page number 226\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage, V\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "beta=50.0;                     #Base current amplification factor\n",
-      "R1=12.0;                     #Resistor R1, k\u2126 \n",
-      "R2=2.7;                      #Resistor R2, k\u2126 \n",
-      "RC=620.0;                    #Collector resistor, \u2126 \n",
-      "RE=180.0;                    #Emitter resistor, \u2126\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Voltage divder rule across R1 and R2\n",
-      "V2=round((VCC*R2)/(R1+R2),2);            #Voltage drop across resistor R2, V\n",
-      "IE=round(((V2-VBE)/RE)*1000,2);                 #Emitter current, mA\n",
-      "IC=IE;                          #Collector current(Approximately equal to emitter current), mA\n",
-      "print(\"IC~IE=%.2fmA.\"%IC);\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-(IC/1000)*(RC+RE);                 #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IC~IE=6.33mA.\n",
-        "VCE=4.94V, is approximately half of VCC=10V \n",
-        " therefore it is mid-point biased.\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.32 : Page number 227\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=10.0;                    #Collector supply voltage, V\n",
-      "IC=10.0;                     #Collector  current, mA   \n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "R1=1.5;                      #Resistor R1, k\u2126 \n",
-      "R2=680.0;                    #Resistor R2, \u2126  \n",
-      "RC=260.0;                    #Collector resistor, \u2126 \n",
-      "RE=240.0;                    #Emitter resistor, \u2126 \n",
-      "beta_min=100;                #Minimum value of base current amplification factor\n",
-      "beta_max=400;                #Maximum value of base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Voltage divder rule across R1 and R2\n",
-      "V2=round((VCC*R2/1000)/(R1+R2/1000),2);               #Voltage drop across resistor R2, V\n",
-      "IE=round((V2-VBE)/(RE/1000),0);                         #OHM' LAW, Emitter current, mA\n",
-      "IC=IE;                                  #Collector current(approx. equal to emitter current),mA\n",
-      "beta_avg=sqrt(beta_min*beta_max);       #Average value of base current amplification factor\n",
-      "IB=IE/(beta_avg +1);                    #Base current, mA\n",
-      "IB=IB*1000;                             #Base current, \ud835\udf07A\n",
-      "\n",
-      "#Results\n",
-      "print(\"Base current= %.2f \ud835\udf07A\"%IB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current= 49.75 \ud835\udf07A\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.33 : Page number 227-228\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VEE=12.0;                    #Emitter supply voltage, V\n",
-      "RC=1.5;                     #Collector resistor, k\u2126\n",
-      "RB=120.0;                   #Base resistor k\u2126\n",
-      "RE=510.0;                   #Emitter resistor, \u2126                 \n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=60.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law,\n",
-      "#IB*RB - VBE - IE*RE +VEE=0\n",
-      "#Since IE~IC and IC=beta*IB,\n",
-      "IB=(VEE-VBE)/(RB + beta*RE/1000);           #Base current , mA\n",
-      "IC=round(beta*IB,1);                                 #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along output circuit,\n",
-      "VCE=VEE-IC*(RC + RE/1000);                        #Collector-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA.\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 2.96V and IC=4.5mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.34 : Page number 228-229\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "VEE=9.0;                    #Emitter supply voltage, V\n",
-      "RC=1.2;                     #Collector resistor, k\u2126\n",
-      "RB=100.0;                   #Base resistor ,k\u2126\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=45.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law,\n",
-      "#IB*RB + VBE=VEE\n",
-      "#Since IE~IC and IC=beta*IB,\n",
-      "IB=round((VEE-VBE)/RB,3);           #Base current , mA\n",
-      "IC=floor(beta*IB*100)/100;                                 #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along output circuit,\n",
-      "VCE=VEE-IC*RC;                        #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.2fmA.\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 4.52V and IC=3.73mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 68
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.35 : Page number 229\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=16.0;                   #Collector supply voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "IC=1.0;                     #Collector current, mA\n",
-      "VCE=6.0;                    #Collector-emitter voltage, V\n",
-      "beta=150.0;                 #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#For a good design, VE=VCC/10;\n",
-      "VE=VCC/10;                  #Emitter terminal's voltage, V\n",
-      "#OHM's Law\n",
-      "#And, taking IE~IC\n",
-      "RE=VE/IC;                   #Emitter resistor, k\u2126\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law alog output circuit:\n",
-      "#VCC=IC*RC + VCE + VE\n",
-      "RC=(VCC-VCE-VE)/IC;                       #Collector resistor, k\u2126\n",
-      "V2=VE+VBE;                                #Voltage drop across resistor R2,V\n",
-      "#From the relation I1=10*IB\n",
-      "R2=(beta*RE)/10;                          #Resistor R2, kilo ohm\n",
-      "\n",
-      "#From voltage divider rule across R1 and R2,\n",
-      "#V2=(VCC*R2)/(R1+R2)\n",
-      "R1=(VCC-V2)*R2/V2;                          #Resistor R1, k\u2126            \n",
-      "\n",
-      "#Results\n",
-      "print(\"RE=%.1f k\u2126 , RC=%.1f k\u2126, R1=%.0f k\u2126 and R2=%d k\u2126.\"%(RE,RC,R1,R2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RE=1.6 k\u2126 , RC=8.4 k\u2126, R1=143 k\u2126 and R2=24 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 69
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.36 : Page number 230-231\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ICBO=5.0;                          #Collector to base leakage current, microampere\n",
-      "beta=40.0;                         #Base current amplification factor\n",
-      "IC_zero_signal=2.0;                #Zero signal collector current, mA\n",
-      "op_temp=25.0;                      #operating temperature, degree celsius\n",
-      "temp_risen=55.0;                   #Temperature risen, degree celsius\n",
-      "temp_ICBO_doubles=10.0;            #Temperature after which ICBO doubles, degree celsius\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "ICEO=(beta+1)*ICBO;                 #Collector to emitter leakage current, microampere\n",
-      "\n",
-      "#(ii)\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;  #Number of times ICBO doubles\n",
-      "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled;                        #Final value of collector to base leakage current, microampere\n",
-      "ICEO_final=ICBO_final*(beta + 1);                                       #Final value of collector to emitter leakage current, microampere\n",
-      "\n",
-      "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal;                    #Zero signal collector current at 55 degree celius\n",
-      "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal;           #Percentage change in zero signal collector current\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The percentage change in the zero signal collector current=%.0f%%. \"%change)\n",
-      "\n",
-      "#(iii)\n",
-      "#For the silicon transistor\n",
-      "ICBO=0.1;                          #Collector to base leakage current, microampere\n",
-      "\n",
-      "ICEO=(beta+1)*ICBO;                 #Collector to emitter leakage current, microampere\n",
-      "\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;  #Number of times ICBO doubles\n",
-      "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled;                        #Final value of collector to base leakage current, microampere\n",
-      "ICEO_final=ICBO_final*(beta + 1);                                       #Final value of collector to emitter leakage current, microampere\n",
-      "\n",
-      "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal;                    #Zero signal collector current at 55 degree celius\n",
-      "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal;           #Percentage change in zero signal collector current\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) The percentage change in the zero signal collector current=%.1f%%. \"%change)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The percentage change in the zero signal collector current=82%. \n",
-        "(ii) The percentage change in the zero signal collector current=1.6%. \n"
-       ]
-      }
-     ],
-     "prompt_number": 70
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.37 : Page number 231\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ICBO=0.02                          #Collector to base leakage current, \ud835\udf07A\n",
-      "alpha=0.99;                        #Current amplification factor\n",
-      "IE=1.0;                            #Emitter current, mA\n",
-      "op_temp=27.0;                      #operating temperature, degree celsius\n",
-      "temp_risen=57.0;                   #Temperature risen, degree celsius\n",
-      "temp_ICBO_doubles=6.0;             #Temperature after which ICBO doubles, degree celsius\n",
-      "\n",
-      "#Calculations\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;       #Number of times ICBO doubles\n",
-      "ICBO_55=ICBO*2**Number_of_times_ICBO_doubled;                                #collector to base leakage current at 55 degree celsius, \ud835\udf07A\n",
-      "IC=alpha*IE + ICBO_55/1000;                                                     #Collector current, mA\n",
-      "IB=IE-IC;                                                                    #Base current, mA\n",
-      "IB=IB*1000;                                                                  #Base current,\ud835\udf07A\n",
-      "\n",
-      "#Results\n",
-      "print(\"Base current at 57 degree celsius=%.1f \ud835\udf07A \"%IB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current at 57 degree celsius=9.4 \ud835\udf07A \n"
-       ]
-      }
-     ],
-     "prompt_number": 71
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_1.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_1.ipynb
deleted file mode 100755
index e8168ab8..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_1.ipynb
+++ /dev/null
@@ -1,1907 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1e5463fc2e8c67a2f9099cf3f6c078bc1c9dccccd63589a75ad6ce5024fe6432"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 9 : TRANSISTOR BIASING"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1: Page number 195-196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=6.0;                     #Collector supply voltage\n",
-      "R_C=2.5;                      #Collector load in k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "#For faithful amplification Vce (collector-emitter voltage)> 1V for Si transistor.\n",
-      "V_CE_max=1;                                   #Maximum allowed collector-emitter voltage for faithful amplification, in V.\n",
-      "V_Rc_max=V_CC-V_CE_max;                       #maximum voltage drop across collector load in V.\n",
-      "I_C_max=V_Rc_max/R_C;                         #Maximum allowed collector current in mA\n",
-      "\n",
-      "#(ii)\n",
-      "IC_min_zero_signal=I_C_max/2;                            #Minimum zero signal collector current in mA\n",
-      "\n",
-      "#Results\n",
-      "print(\"The maximum allowed collector current during application of signal for faithful amplification = %d mA.\"%I_C_max);\n",
-      "print(\"The minimum zero signal collector current required = %d mA.\"%IC_min_zero_signal);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowed collector current during application of signal for faithful amplification = 2 mA.\n",
-        "The minimum zero signal collector current required = 1 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2: Page number 196\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=13.0;                     #Collector supply voltage in V\n",
-      "V_knee=1.0;                   #Knee voltage in V\n",
-      "R_C=4.0;                      #Collector load  in k\u2126\n",
-      "rate_IC_VBE=5.0;              #Rate of change of collector current IC with base-emitter voltage VBE in mA/V.\n",
-      "beta=100.0;                   #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V_Rc_max=VCC-V_knee;                    #Maximum allowed voltage across collector load in V\n",
-      "I_C_max=V_Rc_max/R_C;                   #Maximum allowed collector current in mA\n",
-      "I_B_max=I_C_max/beta;                   #Maximum base current in mA\n",
-      "I_B_max=I_B_max*1000;                   #Maximum base current in \ud835\udf07A\n",
-      "\n",
-      "V_B_max=I_C_max/rate_IC_VBE;            #Maximum base voltage signal in V\n",
-      "V_B_max=V_B_max*1000;                   #Maximum base voltage signal in mV\n",
-      "\n",
-      "#Results\n",
-      "print(\"Maximum base current =%d \ud835\udf07A.\"%I_B_max);\n",
-      "print(\"Maximum input signal voltage =%d mV.\"%V_B_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum base current =30 \ud835\udf07A.\n",
-        "Maximum input signal voltage =600 mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3: Page number 200-201"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=9.0;                  #Colector supply voltage in V\n",
-      "VBB=2.0;                  #Base supply voltage in V\n",
-      "R_B=100.0;                #Base resistor's resistance in k\u2126\n",
-      "R_C=2.0;                  #Collector load in k\u2126\n",
-      "beta=50.0;                #base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Case (i):\n",
-      "\n",
-      "#Applying Kirchhoff's law to the input circuit\n",
-      "#We get, IB*RB +VBE =VBB.\n",
-      "#Neglecting the small base-emitter voltage, we get:\n",
-      "I_B=VBB/R_B;                #Base current in mA\n",
-      "I_C=beta*I_B;               #Collector current in mA\n",
-      "\n",
-      "print(\"Collector current = %dmA\"%I_C);\n",
-      "\n",
-      "#Applying Kirchhoff's law to the output ciruit\n",
-      "#We get, IC*RC + VCE= VCC.\n",
-      "#From the above equation, we get:\n",
-      "V_CE=VCC-I_C*R_C;               #Collector emitter voltage in V\n",
-      "\n",
-      "print(\"Collector emitter voltage =%dV.\"%V_CE);\n",
-      "\n",
-      "\n",
-      "#Case (ii):\n",
-      "\n",
-      "R_B=50.0;\n",
-      "I_B=VBB/R_B;\n",
-      "I_C=beta*I_B;\n",
-      "V_CE=VCC  - I_C*R_C;\n",
-      "\n",
-      "print(\"The new operating point for base resistor RB=50 k\u2126 is, VCE=%dV and IC=%dmA.\"%(V_CE,I_C));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current = 1mA\n",
-        "Collector emitter voltage =7V.\n",
-        "The new operating point for base resistor RB=50 k\u2126 is, VCE=5V and IC=2mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4: Page number  201-202"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#variable declaration\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "VCC=6.0;                  #Collector suply voltagein V\n",
-      "VBE=0.7                   #Base emitter voltage in V\n",
-      "R_B=530.0;                #Base resistor's resistance in k\u2126 .\n",
-      "R_C=2.0;                  #Collector resistor's resistance in k\u2126 .\n",
-      "\n",
-      "#Calculation\n",
-      "#D.C load line equation : VCE=VCC-IC*RC;\n",
-      "#Calculating maximum VCE  ,by IC=0;\n",
-      "I_C_Vce_max=0;                      #Collector current for maximum collector-emitter voltage, in mA\n",
-      "VCE_max=VCC;-I_C_Vce_max*R_C;       #Maximum collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculating maximum collector current IC,by VCE=0;\n",
-      "V_CE_IC_max=0;             #Collector-emitter voltage for maximum collector current, in V         \n",
-      "I_C_max=(VCC-V_CE_IC_max)/R_C;        #Maximum collector current in mA\n",
-      "\n",
-      "\n",
-      "#Operating point:\n",
-      "#For input circuit, applying Kirchhoff's law, We get,\n",
-      "#VCC=IB*RB  + VBE.\n",
-      "#From the above equation,\n",
-      "IB=(VCC-VBE)/R_B;                #Base current in mA\n",
-      "IC=beta*IB;                    #Collector current\n",
-      "\n",
-      "#From the output circuit, applying Kirchhoff's law, we get:\n",
-      "VCE=VCC-IC*R_C;                 #Collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Stability factor\n",
-      "SF=beta+1;              \n",
-      "\n",
-      "#Result\n",
-      "print(\"Operating point: VCE= %dV and IC=%d mA\"%(VCE,IC));\n",
-      "print(\"Stability factor= %d.\"%SF);\n",
-      "\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,10])\n",
-      "limit.set_ylim([0,5])\n",
-      "VCE=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC=[((VCC-i)/(R_C)) for i in (VCE[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE,IC);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: VCE= 4V and IC=1 mA\n",
-        "Stability factor= 101.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8eea67df10>"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5: Page number 202"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage in V\n",
-      "beta=100.0;              #base current amplification factor\n",
-      "I_C_zero_signal=1.0;      #zero signal collector current in mA\n",
-      "VBE=0.3;                  #Base-emitter voltage of Ge transistor in V\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#Case(i)\n",
-      "I_B_zero_signal=I_C_zero_signal/beta;               #Zero signal base current in  mA\n",
-      "\n",
-      "#applying the Kirchhoff's law along input circuit:\n",
-      "#We get, VCC=IB*RB +VBE\n",
-      "#From the above equation we get,\n",
-      "R_B=(VCC-VBE)/I_B_zero_signal;              #Required base resistor's resistance in k\u2126\n",
-      "\n",
-      "print(\"Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = %d k\u2126\"%R_B);\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Case(ii)\n",
-      "beta=50;\n",
-      "I_B=(VCC-VBE)/R_B;                  #Base current of another transistor with beta=50, in mA\n",
-      "I_C_zero_signal=beta*I_B;           #Zero signal collector current for beta=50 , in mA\n",
-      "\n",
-      "print(\"The new value of zero signal collector current =%.1fmA\"%I_C_zero_signal);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = 1170 k\u2126\n",
-        "The new value of zero signal collector current =0.5mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6:Page number 202-203"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Varaible declaration\n",
-      "VCC=10.0;                     #Collector supply voltage in V\n",
-      "VBE=0;                        #Base emitter voltage in V(considering itas zero due to it's small value)\n",
-      "R_B=1.0;                      #Base resistor's resistance in M\u2126\n",
-      "R_C=2.0;                      #Collector resistor's resistance in k\u2126                  \n",
-      "R_E=1.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#using Kirchhoff's law in the input circuit, we get:\n",
-      "#VCC=IB*RB +VBE +IE*RE\n",
-      "#Since, IE=(beta +1)*I_B\n",
-      "#From the above equation we get:\n",
-      "I_B=round((VCC-VBE)/((beta + 1)*R_E + R_B*1000),4);           #Base current in mA\n",
-      "I_C=round(beta*I_B,2);                                   #Collector current in mA\n",
-      "I_E=I_B+I_C;                                    #Emitter current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Base current =%.4f mA\"%I_B);\n",
-      "print(\"Collector current =%.2f mA\"%I_C);\n",
-      "print(\"Emitter current =%.3f mA\"%I_E);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current =0.0091 mA\n",
-        "Collector current =0.91 mA\n",
-        "Emitter current =0.919 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.7: Page number 203-204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=8.0;                      #Collector-emitter voltage at operating point in V\n",
-      "IC=2.0;                       #Colector current at operating point in mA\n",
-      "VCC=15.0;                     #Collector supply voltagein V\n",
-      "beta=100.0;                   #base current amplification factor\n",
-      "VBE=0.6;                    #base emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC=VCE+IC*RC.\n",
-      "#So, from above equation we get:\n",
-      "RC=(VCC-VCE)/IC;                 #Collector resistor's resistance in k\u2126 .\n",
-      "IB=IC/beta;                      #Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the input circuit,\n",
-      "#we get, VCC=IB*RB + VBE\n",
-      "#So, from the above equation:\n",
-      "RB=(VCC-VBE)/IB;                #Base resistor's resistance in k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector load =%.1f k\u2126 .\"%RC);\n",
-      "print(\"Base resistor=%d k\u2126 .\"%RB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector load =3.5 k\u2126 .\n",
-        "Base resistor=720 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8: Page number 204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                #Collector supply voltage in V\n",
-      "VBE=0.7;                 #Base-emitter voltage in V\n",
-      "RB=100.0;                #Base resistor's resistance in k\u2126\n",
-      "RC=560.0;                #Collector resistor's resistance in \u2126\n",
-      "beta_25=100.0;           #base current amplification factor at 25 degree celsius\n",
-      "beta_75=150.0;           #base current amplification factor at 25 degree celsius\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit, we get\n",
-      "#VCC=IB*RB+VBE\n",
-      "IB=(VCC-VBE)/RB;                #Base current at 25 degree celsius, in mA\n",
-      "\n",
-      "\n",
-      "#For temperature 25 degree celsius\n",
-      "IC_25=beta_25*IB;                  #Collector current at 25 degree celsius, in mA\n",
-      "\n",
-      "\n",
-      "#Applying Kirchhoff's alw at the output circuit,\n",
-      "#we get: VCC=IC*RC + VCE\n",
-      "#From the above equation,\n",
-      "VCE_25=round(VCC-(IC_25/1000)*RC,2);                #Collector emitter voltage  at 25 degree celsius, in V\n",
-      "\n",
-      "\n",
-      "#For temperature 75 degree celsius\n",
-      "IC_75=round(beta_75*IB,0);                  #Collector current at 75 degree celsius, in mA\n",
-      "\n",
-      "#Applying Kirchhoff's alw at the output circuit,\n",
-      "#we get: VCC=IC*RC + VCE\n",
-      "#From the above equation,\n",
-      "VCE_75=round(VCC-(IC_75/1000)*RC,2);                #Collector emitter voltage at 75 degree celsius, in V\n",
-      "\n",
-      "\n",
-      "change_IC=(IC_75-IC_25)*100.0/IC_25;        #percentage change in collector current\n",
-      "change_VCE=(VCE_75-VCE_25)*100.0/VCE_25;    #Percentage change in collector-emitter voltage \n",
-      "\n",
-      "#Results\n",
-      "print(\"The percentage change in collector current =%d%%\"%change_IC);\n",
-      "print(\"The percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage change in collector current =50%\n",
-        "The percentage change in collector-emitter voltage =-56.3%\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.10: Page number 205"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE_max=20.0;             #Maximum collector-emitter voltage in V\n",
-      "VBE=0.7;                  #Base-emitter voltage in V\n",
-      "IC_max=8.0;               #Maximum collector current in mA\n",
-      "IB=40.0;                  #Base current in microampere\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#During cut off state the collector-emitter voltage is maximum and equal to collector supply voltage\n",
-      "VCC=VCE_max;                #Collector supply voltage in V\n",
-      "\n",
-      "#Maximum collector current IC_max=collector supply voltage(VCC)/collector load(RC)\n",
-      "#Collector load(RC)=VCC*IC_max\n",
-      "RC=VCC/IC_max;                  #Collector load in k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC=IB*RB +VBE.\n",
-      "#From the above equation, we get:\n",
-      "RB=(VCC-VBE)/(IB/1000);    #Base resistor's resistance in k\u2126 .\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector supply voltage = %dV\"%VCC);\n",
-      "print(\"Collector load=%.1f k\u2126 .\"%RC);\n",
-      "print(\"Base resistor's resistance=%.1f k\u2126 .\"%RB);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector supply voltage = 20V\n",
-        "Collector load=2.5 k\u2126 .\n",
-        "Base resistor's resistance=482.5 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.12: Page number 208"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage in V\n",
-      "VEE=-20.0;                     #Emitter supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=4.7;                       #Collector resistor's resistance in k\u2126\n",
-      "RE=10.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "VBE=0.7;                      #Base-emitter voltage in V\n",
-      "beta=85.0;                    #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n",
-      "#we get,IB*RB  +IE*RE +VBE -VEE=0.\n",
-      "#Since IB=IC/beta and IC~IE,\n",
-      "#(IE/beta)*RB + IE*RE + VBE + VEE =0.\n",
-      "IE=(-VEE-VBE)/(RE + RB/beta);                        #Emitter current in mA\n",
-      "IC=IE;                                              #Collector current (approximately equal to emitter current) in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law from VCC till collector terminal,\n",
-      "#we get, VCC - IC*RC =VC\n",
-      "VC=VCC-IC*RC;                               #voltage at collector terminal in V\n",
-      "\n",
-      "#Applying Kirchhoff's law from emitter terminal to VEE\n",
-      "#we get, VE -IE*RE =VEE\n",
-      "VE=VEE + IE*RE;                             #Voltage at emitter treminal in V\n",
-      "\n",
-      "VCE=VC-VE;                                  #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The collector current = %.2f mA\"%IC);\n",
-      "print(\"The emitter current = %.2f mA\"%IE);\n",
-      "print(\"The voltage at collector terminal = %.1f V\"%VC);\n",
-      "print(\"The collector-emitter voltage = %.1f V\"%VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The collector current = 1.73 mA\n",
-        "The emitter current = 1.73 mA\n",
-        "The voltage at collector terminal = 11.9 V\n",
-        "The collector-emitter voltage = 14.6 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.13: Page number 208-209\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage in V\n",
-      "VEE=-20.0;                     #Emitter supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=4.7;                       #Collector resistor's resistance in k\u2126\n",
-      "RE=10.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "beta1=85.0;                   #Base current amplification factor for case 1       \n",
-      "beta2=100.0;                  #Base current amplification factor for case 1\n",
-      "VBE_1=0.7;                    #Base emitter voltage for case 1 in V\n",
-      "VBE_2=0.6;                    #Base emitter voltage for case 2 in V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#For beta=85 and VBE=0.7,\n",
-      "#As calculated in the previous question,\n",
-      "IC_1=1.73;                          #Collector current in mA.\n",
-      "VCE_1=14.6;                           #Collector-emitter voltage in V.\n",
-      "\n",
-      "\n",
-      "#For case (ii)\n",
-      "#beta=100 and VBE=0.6\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n",
-      "#we get,IB*RB  +IE*RE +VBE -VEE=0.\n",
-      "#Since IB=IC/beta and IC~IE,\n",
-      "#(IE/beta)*RB + IE*RE + VBE +VEE =0.\n",
-      "IE_2=round((-VEE-VBE_2)/(RE + RB/beta2),2);                        #Emitter current in mA\n",
-      "IC_2=IE_2;                                                #Collector current (approximately equal to emitter current) in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law from VCC till collector terminal,\n",
-      "#we get, VCC - IC*RC =VC\n",
-      "VC=round(VCC-IC_2*RC,1);                               #voltage at collector terminal in V\n",
-      "\n",
-      "#Applying Kirchhoff's law from emitter terminal to VEE\n",
-      "#we get, VE -IE*RE =VEE\n",
-      "VE=round(VEE + IE_2*RE,1);                             #Voltage at emitter treminal in V\n",
-      "\n",
-      "VCE_2=VC-VE;                                #Collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "change_IC= (IC_2-IC_1)*100/IC_1;                    #%age change in collector current\n",
-      "\n",
-      "change_VCE=(VCE_2-VCE_1)*100/VCE_2;                 #%age change in collector-emitter voltage\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Percentage change in collector current =%.1f%%\"%change_IC);\n",
-      "print(\"Percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage change in collector current =1.7%\n",
-        "Percentage change in collector-emitter voltage =-3.5%\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.14: Page number 210\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                 #Collector supply voltage in V\n",
-      "VBE=0.7                   #Base-emitter voltage in V\n",
-      "RB=100.0;                 #Base resistor's resistance in k\u2126\n",
-      "RC=1.0;                   #Collector resistor's resistance in k\u2126\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC -IC*RC -IB*RB -VBE=0.\n",
-      "#since IC= beta*IB,\n",
-      "#We get,\n",
-      "IB=(VCC-VBE)/(RB + beta*RC);            #Base current in mA\n",
-      "IC=beta*IB;                             #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC=0.\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*RC;                          #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point : VCE=10.35V and IC=9.65mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.15: Page number 210-211\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                         #Collector supply voltage in V\n",
-      "VBE=0.3;                          #Base emitter voltage in V\n",
-      "IC=1.0;                           #Collector current in mA\n",
-      "VCE=8.0;                          #Collector emitter voltage in V\n",
-      "beta=100.0;                         #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Case(i)\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-IC*RC-VCE=0.\n",
-      "#from the above equation we get,\n",
-      "RC=(VCC-VCE)/IC;             #Collector load in kilo ohm\n",
-      "IB=IC/beta;                  #Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit\n",
-      "#we get, VCC-VBE-(beta*IB*RC)-IB*RB=0.\n",
-      "#From the above equation we get,\n",
-      "RB=round((VCC-VBE-beta*IB*RC)/IB,0);         #Base resistor's resistance in k\u2126\n",
-      "\n",
-      "#Results\n",
-      "print(\"The resistance value of base resistor=%d k\u2126 and collector load= %d k\u2126.\"%(RB,RC));\n",
-      "\n",
-      "#Case(ii)\n",
-      "\n",
-      "beta=50;\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC -IC*RC -IB*RB -VBE=0.\n",
-      "#since IC= beta*IB,\n",
-      "#We get,\n",
-      "IB=(VCC-VBE)/(RB + beta*RC);            #Base current in mA\n",
-      "IC=beta*IB;                             #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC=0.\n",
-      "#From the above equation,\n",
-      "VCE=round(VCC-IC*RC,1);                          #Collector emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.1fV and IC=%.1fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resistance value of base resistor=770 k\u2126 and collector load= 4 k\u2126.\n",
-        "The operating point : VCE=9.6V and IC=0.6mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.16 : Page number 211"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=2.0;                      #Collector-emitter voltage at operating point in V\n",
-      "VBE=0.7;                      #Base-emitter voltage in V            \n",
-      "IC=1.0;                       #Collector current at operating point in mA\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "IB=IC/beta;                     #Base current in mA\n",
-      "\n",
-      "#As, VCE=VCB +VBE\n",
-      "#we get,\n",
-      "VCB=VCE-VBE;                    #Collector-base voltage in V\n",
-      "RB=VCB/IB;                      #Base resistor's resistance in k\u2126\n",
-      "\n",
-      "#Results\n",
-      "print(\"Value of base resistor's resistance=%d k\u2126.\"%RB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of base resistor's resistance=130 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.17 : Page number 211-212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage in V\n",
-      "VBE=0.7                   #Base-emitter voltage in V\n",
-      "RB=400.0;                 #Base resistor's resistance in k\u2126\n",
-      "RC=4.0;                   #Collector resistor's resistance in k\u2126\n",
-      "RE=1.0;                   #Emitter resistor's resistance in k\u2126\n",
-      "beta=100.0;               #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along outut circuit,\n",
-      "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n",
-      "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n",
-      "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n",
-      "IC=(VCC-VBE)/(RB/beta + RC + RE);       #Collector current current in mA.\n",
-      "IE=IC;                                                #Emitter current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC -IE*RE=0. (IE~IC)\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*(RC+RE);                          #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.1fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point : VCE=5.7V and IC=1.26mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.18 : Page number 212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=10.0;                      #Collector resistor's resistance in k\u2126\n",
-      "RE=0;                         #Emitter resistor's  resistance in k\u2126\n",
-      "VBE=0.7;                      #Base-emitter voltage in V\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along outut circuit,\n",
-      "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n",
-      "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n",
-      "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n",
-      "IC=(VCC-VBE)/(RC +RB/beta + RE);                #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC =0. (IE~IC)\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*RC;                                  #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The d.c bias values are: VCE=%.2fV and IC=%.3fmA\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The d.c bias values are: VCE=1.55V and IC=0.845mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.19: Page number 214-215\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage in V\n",
-      "R1=10.0;                            #Resistor R1's resistance in k\u2126\n",
-      "R2=5.0;                             #Resistor R2's resistance in k\u2126\n",
-      "RC=1.0;                             #Collector resistor's resistance in k\u2126              \n",
-      "RE=2.0;                             #Emitter resistor's resistance in k\u2126\n",
-      "VBE=0.7;                            #Base-emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along output circuit\n",
-      "#VCE=VCC-IC*(RC+RE);\n",
-      "#IC=0, for VCE_max\n",
-      "VCE_max=VCC;                        #Maximum collector-emitter voltage in V\n",
-      "#VCE=0, for IC_max\n",
-      "IC_max=VCC/(RC+RE);                 #Maximum collector current in mA\n",
-      "\n",
-      "#Operating point\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage across R2 resistor V\n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "IC=IE;                              #Collector current(Approx. equal to emitter current) in mA\n",
-      "VCE=VCC-IC*(RC+RE);                 #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n",
-      "print(\"Collector current at operating point = %.2fmA\"%IC);\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,20])\n",
-      "limit.set_ylim([0,6])\n",
-      "VCE=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC=[((VCC-i)/(RC+RE)) for i in (VCE[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE,IC);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector-emitter voltage at operating point=8.55V\n",
-        "Collector current at operating point = 2.15mA\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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hKB07Mu+WiAStzuwdHBwwZswYrF69GiNHjgQADBgwAGfPnjXP4pzZi65xtPM//wPMny/s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\nAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8eeacf1ad0>"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.20: Page number 215-216\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage in V\n",
-      "R1=10.0;                            #Resistor R1's resistance in k\u2126 .\n",
-      "R2=5.0;                             #Resistor R2's resistance in k\u2126 .\n",
-      "RC=1.0;                             #Collector resistor's resistance in k\u2126 .              \n",
-      "RE=2.0;                             #Emitter resistor's resistance in k\u2126 .\n",
-      "VBE=0.7;                            #Base-emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's Theorem for replacing circuit consisting of VCC,R1,R2\n",
-      "E0=(VCC*R2)/(R1+R2);                #Thevenin's voltage in V\n",
-      "R0=(R1*R2)/(R1+R2);                 #Thevenin's equivalent resistance in k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "#IC=(E0-VBE)/(R0/beta +RE);\n",
-      "IC=(E0-VBE)/RE;                         #(Since R0/beta << RE) collector current in mA\n",
-      "VCE=VCC-IC*(RC+RE);                     #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n",
-      "print(\"Collector current at operating point = %.2fmA\"%IC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector-emitter voltage at operating point=8.55V\n",
-        "Collector current at operating point = 2.15mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.21: Page number 216-217\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                           #Collector  supply voltage in V\n",
-      "RE=1.0;                             #Emitter resistor, k\u2126 .\n",
-      "R1=50.0;                            #Resistor R1, k\u2126 .\n",
-      "R2=10.0;                            #Resistor R2, k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "VBE=0.1;                            #Base-emitter voltage in V\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V \n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "\n",
-      "print(\"(i)Emitter current= %.1fmA\"%IE);\n",
-      "\n",
-      "#(ii)\n",
-      "VBE=0.3;                            #Base-emitter voltage in V\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V\n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "\n",
-      "print(\"(ii)Emitter current= %.1fmA\"%IE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Emitter current= 1.9mA\n",
-        "(ii)Emitter current= 1.7mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.22: Page number 217\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                   #Collector supply voltage, V\n",
-      "R1=10.0;                    #Resistor R1, k\u2126\n",
-      "R2=10.0;                    #Resistor R2, k\u2126 .\n",
-      "RC=1.0;                     #Collector resistor, k\u2126 .\n",
-      "RE=5.0;                     #Emitter resistor, k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V\n",
-      "\n",
-      "#Applying kirchhoff's law from base terminal to emitter resistor\n",
-      "#V2=VBE+IE*RE\n",
-      "#VBE is neglected due to its small value\n",
-      "\n",
-      "IE=V2/RE;                     #Emitter current in mA\n",
-      "IC=IE;                              #Collector current (approx. equal to emitter current), mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along output circuit\n",
-      "VCE=VCC-IC*(RC+RE);                 #Collector-emitter voltage , V\n",
-      "VC=VCC-IC*RC;                       #Voltage at collector terminal,V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Emitter current =%dmA\"%IE);\n",
-      "print(\"Collector-emitter voltage=%dV\"%VCE);\n",
-      "print(\"Collector terminal's voltage=%dV\"%VC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Emitter current =2mA\n",
-        "Collector-emitter voltage=8V\n",
-        "Collector terminal's voltage=18V\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.23: Page number 219-220\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage, V\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "beta=50;                  #Base current amplification factor\n",
-      "R1=150;                   #Resistor R1, k\u2126 .\n",
-      "R2=100;                   #Resistor R2, k\u2126 .\n",
-      "RC=4.7;                   #Collector resistor, k\u2126 .\n",
-      "RE=2.2;                   #Emitter resistor, k\u2126  .\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "IB=round((E0-VBE)/(R0+beta*RE),3);              #Base current in mA\n",
-      "IC=round(beta*IB,1);                             #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-IC*(RC+RE);                     #Collector-emitter voltage, V\n",
-      "\n",
-      "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE);               #Stability factor\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n",
-      "print(\"Stability factor=%.1f\"%S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 3.72V and IC=1.2mA\n",
-        "Stability factor=18.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.24 : Page number 220\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                 #Collector supply voltage, V\n",
-      "VBE=0.7;                  #Base-emitter voltage , V\n",
-      "beta=100.0;                 #Base current amplification factor\n",
-      "R1=6.0;                     #Resistor R1, k\u2126 .\n",
-      "R2=3.0;                     #Resistor R2, k\u2126 .\n",
-      "RC=470.0;                   #Collector resistor, \u2126.\n",
-      "RE=1.0;                     #Emitter resistor, k\u2126 .\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "IB=round((E0-VBE)/(R0+beta*RE),3);              #Base current in mA\n",
-      "IC=round(beta*IB,1);                             #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-IC*(RC/1000+RE);                     #Collector-emitter voltage, V\n",
-      "\n",
-      "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE);               #Stability factor\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n",
-      "print(\"Stability factor=%.2f\"%S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 8.83V and IC=4.2mA\n",
-        "Stability factor=2.94\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.25 : Page number 221-222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "VCC=9;                    #Collector supply voltage, V\n",
-      "VCE=3;                    #Collector-emitter voltage, V\n",
-      "VBE=0.3;                    #Base-emitter voltage in V\n",
-      "RC=2.2;                     #Collector resistor , k\u2126 .\n",
-      "IC=2;                     #Collector current, mA\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "IB=IC/beta;                 #Base current in mA\n",
-      "\n",
-      "#According to given relation, I1=10*IB\n",
-      "I1=IB*10;                   #Current through the resistor R1, mA\n",
-      "\n",
-      "#I1=VCC/(R1+R2), .'s LAW\n",
-      "R1_R2_sum=VCC/I1;               #Sum of the resistor's R1 and R2, k\u2126 (OHM'S LAW).\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "#VCC=IC*RC+VCE+IE*RE\n",
-      "#IC~IE\n",
-      "RE=(VCC-IC*RC-VCE)/IC;      #Emitter resistor, k\u2126 .\n",
-      "RE=round(RE*1000,0);                 #Emittter resistor, \u2126 .\n",
-      "\n",
-      "IE=IC;                      #Emittter current(approximately equal to collector current), mA\n",
-      "VE=IE*(RE/1000);                   #Voltage at emitter terminal (OHM's LAW), V\n",
-      "V2=VBE+VE;                  #Voltage drop across resistor R2, V\n",
-      "\n",
-      "R2=V2/I1;                   #Resistor R2,(OHM's LAW), k\u2126 .\n",
-      "R1=R1_R2_sum-R2;            #Resistor R1, k\u2126 .\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"RE=%d \u2126., R1=%.2f k\u2126 . and R2=%.2f k\u2126 .\"%(RE,R1,R2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RE=800 \u2126., R1=17.75 k\u2126 . and R2=4.75 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.26 : Page number 222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=16.0;                     #Collector supply voltage, V\n",
-      "R2=20.0;                      #Resistor R2, k\u2126\n",
-      "RE=2.0;                       #Emitter resistor, k\u2126\n",
-      "VCE=6.0;                      #Collector-emitter voltage, V\n",
-      "IC=2.0;                       #Collector current , mA\n",
-      "VBE=0.3;                      #Base-emitter voltage,V\n",
-      "alpha=0.985;                  #Current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "beta=alpha/(1-alpha);               #Base current amplificatioon factor\n",
-      "IE=IC;                              #Emitter current, mA\n",
-      "IB=IC/beta;                         #Base current, mA\n",
-      "VE=IE*RE;                           #Emitter voltage,(OHM's LAW) V\n",
-      "V2=VBE+VE;                          #Voltage drop across resistor R2,(Kirchhoff's law) V\n",
-      "V_R1=VCC-V2;                        #Voltage drop across resistor R1, V\n",
-      "I1=V2/R2;                           #Current through resistor R2 an R1,(OHM's LAW) mA\n",
-      "R1=V_R1/I1;                         #Resistor R1,(OHM's LAW) k\u2126\n",
-      "\n",
-      "V_RC=(VCC-VCE-VE);                  #Voltage across collector resistor, V\n",
-      "RC=V_RC/IC;                         #Collector resistor,(OHM's LAW) k\u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"R1=%.1f k\u2126 and RC=%d k\u2126.\"%(R1,RC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R1=54.4 k\u2126 and RC=3 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.27 :Page number 222-223\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                     #Collector supply voltage, V\n",
-      "R1=10.0;                      #Resistor R1, k\u2126 \n",
-      "R2=5.0;                       #Resistor R2, k\u2126 \n",
-      "RC=1.0;                       #Collector resistor, k\u2126 \u007f\n",
-      "RE=2.0;                       #Emitter resistor, k\u2126 \n",
-      "VBE=0.7;                      #Base-emitter voltage, V\n",
-      "beta=100;                     #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 \n",
-      "\n",
-      "#Applying Kirchhoff' law along Thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IB=IE/beta,\n",
-      "IE=(E0-VBE)/(R0/beta + RE);                 #Emitter current , mA\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"The exact value of emitter current in the circuit = %.2fmA.\"%IE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The exact value of emitter current in the circuit = 2.11mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.28: Page number 223-224\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "IE=2.0;                     #Emitter current, mA\n",
-      "IB=50.0;                    #Base current, mA\n",
-      "VCC=10.0;                   #Collector supply voltage, V\n",
-      "VBE=0.2;                    #Base-emitter voltage, V\n",
-      "R2=10.0;                    #Resistor R2, k\u2126\n",
-      "RE=1.0;                     #Emitter resistance, k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law from the base to the emitter resistor,\n",
-      "V2=VBE+IE*RE;                   #Voltage at base terminal, V\n",
-      "I2=V2/R2;                       #Current through the resistor R2, mA\n",
-      "I1=I2+IB/1000;                  #Current through the resistor R2, mA\n",
-      "V1=VCC-V2;                      #Voltage drop across the resistor R2\n",
-      "R1=V1/I1;                       #Resistor R1, k\u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The value of the resistor R1=%.2f k\u2126.\"%R1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of the resistor R1=28.89 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.30 :Page number 225-226\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=8.0;                  #Collector supply voltage, V\n",
-      "RB=360.0;                 #Base resistor, k\u2126\n",
-      "RC=2.0;                   #Collector resistor, k\u2126\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "IC_max=VCC/RC;              #Maximum collector current, mA\n",
-      "VCE_max=VCC;                #Maximum collector voltage, V\n",
-      "\n",
-      "#Operating point\n",
-      "#Applying Kirchhoff's law along the input circuit\n",
-      "IB=(VCC-VBE)/RB;                    #Base current, mA\n",
-      "IC=beta*IB;                         #Collector current, mA\n",
-      "\n",
-      "#Kirchhoff' law along the output circuit\n",
-      "VCE=VCC-IC*RC;                      #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VCE=3.94V, is approximately half of VCC=8V \n",
-        " therefore it is mid-point biased.\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.31: page number 226\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage, V\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "beta=50.0;                     #Base current amplification factor\n",
-      "R1=12.0;                     #Resistor R1, k\u2126 \n",
-      "R2=2.7;                      #Resistor R2, k\u2126 \n",
-      "RC=620.0;                    #Collector resistor, \u2126 \n",
-      "RE=180.0;                    #Emitter resistor, \u2126\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Voltage divder rule across R1 and R2\n",
-      "V2=round((VCC*R2)/(R1+R2),2);            #Voltage drop across resistor R2, V\n",
-      "IE=round(((V2-VBE)/RE)*1000,2);                 #Emitter current, mA\n",
-      "IC=IE;                          #Collector current(Approximately equal to emitter current), mA\n",
-      "print(\"IC~IE=%.2fmA.\"%IC);\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-(IC/1000)*(RC+RE);                 #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IC~IE=6.33mA.\n",
-        "VCE=4.94V, is approximately half of VCC=10V \n",
-        " therefore it is mid-point biased.\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.32 : Page number 227\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=10.0;                    #Collector supply voltage, V\n",
-      "IC=10.0;                     #Collector  current, mA   \n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "R1=1.5;                      #Resistor R1, k\u2126 \n",
-      "R2=680.0;                    #Resistor R2, \u2126  \n",
-      "RC=260.0;                    #Collector resistor, \u2126 \n",
-      "RE=240.0;                    #Emitter resistor, \u2126 \n",
-      "beta_min=100;                #Minimum value of base current amplification factor\n",
-      "beta_max=400;                #Maximum value of base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Voltage divder rule across R1 and R2\n",
-      "V2=round((VCC*R2/1000)/(R1+R2/1000),2);               #Voltage drop across resistor R2, V\n",
-      "IE=round((V2-VBE)/(RE/1000),0);                         #OHM' LAW, Emitter current, mA\n",
-      "IC=IE;                                  #Collector current(approx. equal to emitter current),mA\n",
-      "beta_avg=sqrt(beta_min*beta_max);       #Average value of base current amplification factor\n",
-      "IB=IE/(beta_avg +1);                    #Base current, mA\n",
-      "IB=IB*1000;                             #Base current, \ud835\udf07A\n",
-      "\n",
-      "#Results\n",
-      "print(\"Base current= %.2f \ud835\udf07A\"%IB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current= 49.75 \ud835\udf07A\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.33 : Page number 227-228\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VEE=12.0;                    #Emitter supply voltage, V\n",
-      "RC=1.5;                     #Collector resistor, k\u2126\n",
-      "RB=120.0;                   #Base resistor k\u2126\n",
-      "RE=510.0;                   #Emitter resistor, \u2126                 \n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=60.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law,\n",
-      "#IB*RB - VBE - IE*RE +VEE=0\n",
-      "#Since IE~IC and IC=beta*IB,\n",
-      "IB=(VEE-VBE)/(RB + beta*RE/1000);           #Base current , mA\n",
-      "IC=round(beta*IB,1);                                 #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along output circuit,\n",
-      "VCE=VEE-IC*(RC + RE/1000);                        #Collector-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA.\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 2.96V and IC=4.5mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.34 : Page number 228-229\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "VEE=9.0;                    #Emitter supply voltage, V\n",
-      "RC=1.2;                     #Collector resistor, k\u2126\n",
-      "RB=100.0;                   #Base resistor ,k\u2126\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=45.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law,\n",
-      "#IB*RB + VBE=VEE\n",
-      "#Since IE~IC and IC=beta*IB,\n",
-      "IB=round((VEE-VBE)/RB,3);           #Base current , mA\n",
-      "IC=floor(beta*IB*100)/100;                                 #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along output circuit,\n",
-      "VCE=VEE-IC*RC;                        #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.2fmA.\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 4.52V and IC=3.73mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 68
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.35 : Page number 229\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=16.0;                   #Collector supply voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "IC=1.0;                     #Collector current, mA\n",
-      "VCE=6.0;                    #Collector-emitter voltage, V\n",
-      "beta=150.0;                 #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#For a good design, VE=VCC/10;\n",
-      "VE=VCC/10;                  #Emitter terminal's voltage, V\n",
-      "#OHM's Law\n",
-      "#And, taking IE~IC\n",
-      "RE=VE/IC;                   #Emitter resistor, k\u2126\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law alog output circuit:\n",
-      "#VCC=IC*RC + VCE + VE\n",
-      "RC=(VCC-VCE-VE)/IC;                       #Collector resistor, k\u2126\n",
-      "V2=VE+VBE;                                #Voltage drop across resistor R2,V\n",
-      "#From the relation I1=10*IB\n",
-      "R2=(beta*RE)/10;                          #Resistor R2, kilo ohm\n",
-      "\n",
-      "#From voltage divider rule across R1 and R2,\n",
-      "#V2=(VCC*R2)/(R1+R2)\n",
-      "R1=(VCC-V2)*R2/V2;                          #Resistor R1, k\u2126            \n",
-      "\n",
-      "#Results\n",
-      "print(\"RE=%.1f k\u2126 , RC=%.1f k\u2126, R1=%.0f k\u2126 and R2=%d k\u2126.\"%(RE,RC,R1,R2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RE=1.6 k\u2126 , RC=8.4 k\u2126, R1=143 k\u2126 and R2=24 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 69
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.36 : Page number 230-231\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ICBO=5.0;                          #Collector to base leakage current, microampere\n",
-      "beta=40.0;                         #Base current amplification factor\n",
-      "IC_zero_signal=2.0;                #Zero signal collector current, mA\n",
-      "op_temp=25.0;                      #operating temperature, degree celsius\n",
-      "temp_risen=55.0;                   #Temperature risen, degree celsius\n",
-      "temp_ICBO_doubles=10.0;            #Temperature after which ICBO doubles, degree celsius\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "ICEO=(beta+1)*ICBO;                 #Collector to emitter leakage current, microampere\n",
-      "\n",
-      "#(ii)\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;  #Number of times ICBO doubles\n",
-      "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled;                        #Final value of collector to base leakage current, microampere\n",
-      "ICEO_final=ICBO_final*(beta + 1);                                       #Final value of collector to emitter leakage current, microampere\n",
-      "\n",
-      "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal;                    #Zero signal collector current at 55 degree celius\n",
-      "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal;           #Percentage change in zero signal collector current\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The percentage change in the zero signal collector current=%.0f%%. \"%change)\n",
-      "\n",
-      "#(iii)\n",
-      "#For the silicon transistor\n",
-      "ICBO=0.1;                          #Collector to base leakage current, microampere\n",
-      "\n",
-      "ICEO=(beta+1)*ICBO;                 #Collector to emitter leakage current, microampere\n",
-      "\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;  #Number of times ICBO doubles\n",
-      "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled;                        #Final value of collector to base leakage current, microampere\n",
-      "ICEO_final=ICBO_final*(beta + 1);                                       #Final value of collector to emitter leakage current, microampere\n",
-      "\n",
-      "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal;                    #Zero signal collector current at 55 degree celius\n",
-      "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal;           #Percentage change in zero signal collector current\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) The percentage change in the zero signal collector current=%.1f%%. \"%change)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The percentage change in the zero signal collector current=82%. \n",
-        "(ii) The percentage change in the zero signal collector current=1.6%. \n"
-       ]
-      }
-     ],
-     "prompt_number": 70
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.37 : Page number 231\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ICBO=0.02                          #Collector to base leakage current, \ud835\udf07A\n",
-      "alpha=0.99;                        #Current amplification factor\n",
-      "IE=1.0;                            #Emitter current, mA\n",
-      "op_temp=27.0;                      #operating temperature, degree celsius\n",
-      "temp_risen=57.0;                   #Temperature risen, degree celsius\n",
-      "temp_ICBO_doubles=6.0;             #Temperature after which ICBO doubles, degree celsius\n",
-      "\n",
-      "#Calculations\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;       #Number of times ICBO doubles\n",
-      "ICBO_55=ICBO*2**Number_of_times_ICBO_doubled;                                #collector to base leakage current at 55 degree celsius, \ud835\udf07A\n",
-      "IC=alpha*IE + ICBO_55/1000;                                                     #Collector current, mA\n",
-      "IB=IE-IC;                                                                    #Base current, mA\n",
-      "IB=IB*1000;                                                                  #Base current,\ud835\udf07A\n",
-      "\n",
-      "#Results\n",
-      "print(\"Base current at 57 degree celsius=%.1f \ud835\udf07A \"%IB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current at 57 degree celsius=9.4 \ud835\udf07A \n"
-       ]
-      }
-     ],
-     "prompt_number": 71
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_2.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_2.ipynb
deleted file mode 100755
index e8168ab8..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_2.ipynb
+++ /dev/null
@@ -1,1907 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1e5463fc2e8c67a2f9099cf3f6c078bc1c9dccccd63589a75ad6ce5024fe6432"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 9 : TRANSISTOR BIASING"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1: Page number 195-196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=6.0;                     #Collector supply voltage\n",
-      "R_C=2.5;                      #Collector load in k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "#For faithful amplification Vce (collector-emitter voltage)> 1V for Si transistor.\n",
-      "V_CE_max=1;                                   #Maximum allowed collector-emitter voltage for faithful amplification, in V.\n",
-      "V_Rc_max=V_CC-V_CE_max;                       #maximum voltage drop across collector load in V.\n",
-      "I_C_max=V_Rc_max/R_C;                         #Maximum allowed collector current in mA\n",
-      "\n",
-      "#(ii)\n",
-      "IC_min_zero_signal=I_C_max/2;                            #Minimum zero signal collector current in mA\n",
-      "\n",
-      "#Results\n",
-      "print(\"The maximum allowed collector current during application of signal for faithful amplification = %d mA.\"%I_C_max);\n",
-      "print(\"The minimum zero signal collector current required = %d mA.\"%IC_min_zero_signal);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowed collector current during application of signal for faithful amplification = 2 mA.\n",
-        "The minimum zero signal collector current required = 1 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2: Page number 196\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=13.0;                     #Collector supply voltage in V\n",
-      "V_knee=1.0;                   #Knee voltage in V\n",
-      "R_C=4.0;                      #Collector load  in k\u2126\n",
-      "rate_IC_VBE=5.0;              #Rate of change of collector current IC with base-emitter voltage VBE in mA/V.\n",
-      "beta=100.0;                   #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V_Rc_max=VCC-V_knee;                    #Maximum allowed voltage across collector load in V\n",
-      "I_C_max=V_Rc_max/R_C;                   #Maximum allowed collector current in mA\n",
-      "I_B_max=I_C_max/beta;                   #Maximum base current in mA\n",
-      "I_B_max=I_B_max*1000;                   #Maximum base current in \ud835\udf07A\n",
-      "\n",
-      "V_B_max=I_C_max/rate_IC_VBE;            #Maximum base voltage signal in V\n",
-      "V_B_max=V_B_max*1000;                   #Maximum base voltage signal in mV\n",
-      "\n",
-      "#Results\n",
-      "print(\"Maximum base current =%d \ud835\udf07A.\"%I_B_max);\n",
-      "print(\"Maximum input signal voltage =%d mV.\"%V_B_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum base current =30 \ud835\udf07A.\n",
-        "Maximum input signal voltage =600 mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3: Page number 200-201"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=9.0;                  #Colector supply voltage in V\n",
-      "VBB=2.0;                  #Base supply voltage in V\n",
-      "R_B=100.0;                #Base resistor's resistance in k\u2126\n",
-      "R_C=2.0;                  #Collector load in k\u2126\n",
-      "beta=50.0;                #base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Case (i):\n",
-      "\n",
-      "#Applying Kirchhoff's law to the input circuit\n",
-      "#We get, IB*RB +VBE =VBB.\n",
-      "#Neglecting the small base-emitter voltage, we get:\n",
-      "I_B=VBB/R_B;                #Base current in mA\n",
-      "I_C=beta*I_B;               #Collector current in mA\n",
-      "\n",
-      "print(\"Collector current = %dmA\"%I_C);\n",
-      "\n",
-      "#Applying Kirchhoff's law to the output ciruit\n",
-      "#We get, IC*RC + VCE= VCC.\n",
-      "#From the above equation, we get:\n",
-      "V_CE=VCC-I_C*R_C;               #Collector emitter voltage in V\n",
-      "\n",
-      "print(\"Collector emitter voltage =%dV.\"%V_CE);\n",
-      "\n",
-      "\n",
-      "#Case (ii):\n",
-      "\n",
-      "R_B=50.0;\n",
-      "I_B=VBB/R_B;\n",
-      "I_C=beta*I_B;\n",
-      "V_CE=VCC  - I_C*R_C;\n",
-      "\n",
-      "print(\"The new operating point for base resistor RB=50 k\u2126 is, VCE=%dV and IC=%dmA.\"%(V_CE,I_C));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current = 1mA\n",
-        "Collector emitter voltage =7V.\n",
-        "The new operating point for base resistor RB=50 k\u2126 is, VCE=5V and IC=2mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4: Page number  201-202"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#variable declaration\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "VCC=6.0;                  #Collector suply voltagein V\n",
-      "VBE=0.7                   #Base emitter voltage in V\n",
-      "R_B=530.0;                #Base resistor's resistance in k\u2126 .\n",
-      "R_C=2.0;                  #Collector resistor's resistance in k\u2126 .\n",
-      "\n",
-      "#Calculation\n",
-      "#D.C load line equation : VCE=VCC-IC*RC;\n",
-      "#Calculating maximum VCE  ,by IC=0;\n",
-      "I_C_Vce_max=0;                      #Collector current for maximum collector-emitter voltage, in mA\n",
-      "VCE_max=VCC;-I_C_Vce_max*R_C;       #Maximum collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculating maximum collector current IC,by VCE=0;\n",
-      "V_CE_IC_max=0;             #Collector-emitter voltage for maximum collector current, in V         \n",
-      "I_C_max=(VCC-V_CE_IC_max)/R_C;        #Maximum collector current in mA\n",
-      "\n",
-      "\n",
-      "#Operating point:\n",
-      "#For input circuit, applying Kirchhoff's law, We get,\n",
-      "#VCC=IB*RB  + VBE.\n",
-      "#From the above equation,\n",
-      "IB=(VCC-VBE)/R_B;                #Base current in mA\n",
-      "IC=beta*IB;                    #Collector current\n",
-      "\n",
-      "#From the output circuit, applying Kirchhoff's law, we get:\n",
-      "VCE=VCC-IC*R_C;                 #Collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Stability factor\n",
-      "SF=beta+1;              \n",
-      "\n",
-      "#Result\n",
-      "print(\"Operating point: VCE= %dV and IC=%d mA\"%(VCE,IC));\n",
-      "print(\"Stability factor= %d.\"%SF);\n",
-      "\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,10])\n",
-      "limit.set_ylim([0,5])\n",
-      "VCE=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC=[((VCC-i)/(R_C)) for i in (VCE[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE,IC);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: VCE= 4V and IC=1 mA\n",
-        "Stability factor= 101.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8eea67df10>"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5: Page number 202"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage in V\n",
-      "beta=100.0;              #base current amplification factor\n",
-      "I_C_zero_signal=1.0;      #zero signal collector current in mA\n",
-      "VBE=0.3;                  #Base-emitter voltage of Ge transistor in V\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#Case(i)\n",
-      "I_B_zero_signal=I_C_zero_signal/beta;               #Zero signal base current in  mA\n",
-      "\n",
-      "#applying the Kirchhoff's law along input circuit:\n",
-      "#We get, VCC=IB*RB +VBE\n",
-      "#From the above equation we get,\n",
-      "R_B=(VCC-VBE)/I_B_zero_signal;              #Required base resistor's resistance in k\u2126\n",
-      "\n",
-      "print(\"Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = %d k\u2126\"%R_B);\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Case(ii)\n",
-      "beta=50;\n",
-      "I_B=(VCC-VBE)/R_B;                  #Base current of another transistor with beta=50, in mA\n",
-      "I_C_zero_signal=beta*I_B;           #Zero signal collector current for beta=50 , in mA\n",
-      "\n",
-      "print(\"The new value of zero signal collector current =%.1fmA\"%I_C_zero_signal);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = 1170 k\u2126\n",
-        "The new value of zero signal collector current =0.5mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6:Page number 202-203"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Varaible declaration\n",
-      "VCC=10.0;                     #Collector supply voltage in V\n",
-      "VBE=0;                        #Base emitter voltage in V(considering itas zero due to it's small value)\n",
-      "R_B=1.0;                      #Base resistor's resistance in M\u2126\n",
-      "R_C=2.0;                      #Collector resistor's resistance in k\u2126                  \n",
-      "R_E=1.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#using Kirchhoff's law in the input circuit, we get:\n",
-      "#VCC=IB*RB +VBE +IE*RE\n",
-      "#Since, IE=(beta +1)*I_B\n",
-      "#From the above equation we get:\n",
-      "I_B=round((VCC-VBE)/((beta + 1)*R_E + R_B*1000),4);           #Base current in mA\n",
-      "I_C=round(beta*I_B,2);                                   #Collector current in mA\n",
-      "I_E=I_B+I_C;                                    #Emitter current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Base current =%.4f mA\"%I_B);\n",
-      "print(\"Collector current =%.2f mA\"%I_C);\n",
-      "print(\"Emitter current =%.3f mA\"%I_E);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current =0.0091 mA\n",
-        "Collector current =0.91 mA\n",
-        "Emitter current =0.919 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.7: Page number 203-204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=8.0;                      #Collector-emitter voltage at operating point in V\n",
-      "IC=2.0;                       #Colector current at operating point in mA\n",
-      "VCC=15.0;                     #Collector supply voltagein V\n",
-      "beta=100.0;                   #base current amplification factor\n",
-      "VBE=0.6;                    #base emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC=VCE+IC*RC.\n",
-      "#So, from above equation we get:\n",
-      "RC=(VCC-VCE)/IC;                 #Collector resistor's resistance in k\u2126 .\n",
-      "IB=IC/beta;                      #Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the input circuit,\n",
-      "#we get, VCC=IB*RB + VBE\n",
-      "#So, from the above equation:\n",
-      "RB=(VCC-VBE)/IB;                #Base resistor's resistance in k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector load =%.1f k\u2126 .\"%RC);\n",
-      "print(\"Base resistor=%d k\u2126 .\"%RB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector load =3.5 k\u2126 .\n",
-        "Base resistor=720 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8: Page number 204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                #Collector supply voltage in V\n",
-      "VBE=0.7;                 #Base-emitter voltage in V\n",
-      "RB=100.0;                #Base resistor's resistance in k\u2126\n",
-      "RC=560.0;                #Collector resistor's resistance in \u2126\n",
-      "beta_25=100.0;           #base current amplification factor at 25 degree celsius\n",
-      "beta_75=150.0;           #base current amplification factor at 25 degree celsius\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit, we get\n",
-      "#VCC=IB*RB+VBE\n",
-      "IB=(VCC-VBE)/RB;                #Base current at 25 degree celsius, in mA\n",
-      "\n",
-      "\n",
-      "#For temperature 25 degree celsius\n",
-      "IC_25=beta_25*IB;                  #Collector current at 25 degree celsius, in mA\n",
-      "\n",
-      "\n",
-      "#Applying Kirchhoff's alw at the output circuit,\n",
-      "#we get: VCC=IC*RC + VCE\n",
-      "#From the above equation,\n",
-      "VCE_25=round(VCC-(IC_25/1000)*RC,2);                #Collector emitter voltage  at 25 degree celsius, in V\n",
-      "\n",
-      "\n",
-      "#For temperature 75 degree celsius\n",
-      "IC_75=round(beta_75*IB,0);                  #Collector current at 75 degree celsius, in mA\n",
-      "\n",
-      "#Applying Kirchhoff's alw at the output circuit,\n",
-      "#we get: VCC=IC*RC + VCE\n",
-      "#From the above equation,\n",
-      "VCE_75=round(VCC-(IC_75/1000)*RC,2);                #Collector emitter voltage at 75 degree celsius, in V\n",
-      "\n",
-      "\n",
-      "change_IC=(IC_75-IC_25)*100.0/IC_25;        #percentage change in collector current\n",
-      "change_VCE=(VCE_75-VCE_25)*100.0/VCE_25;    #Percentage change in collector-emitter voltage \n",
-      "\n",
-      "#Results\n",
-      "print(\"The percentage change in collector current =%d%%\"%change_IC);\n",
-      "print(\"The percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage change in collector current =50%\n",
-        "The percentage change in collector-emitter voltage =-56.3%\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.10: Page number 205"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE_max=20.0;             #Maximum collector-emitter voltage in V\n",
-      "VBE=0.7;                  #Base-emitter voltage in V\n",
-      "IC_max=8.0;               #Maximum collector current in mA\n",
-      "IB=40.0;                  #Base current in microampere\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#During cut off state the collector-emitter voltage is maximum and equal to collector supply voltage\n",
-      "VCC=VCE_max;                #Collector supply voltage in V\n",
-      "\n",
-      "#Maximum collector current IC_max=collector supply voltage(VCC)/collector load(RC)\n",
-      "#Collector load(RC)=VCC*IC_max\n",
-      "RC=VCC/IC_max;                  #Collector load in k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC=IB*RB +VBE.\n",
-      "#From the above equation, we get:\n",
-      "RB=(VCC-VBE)/(IB/1000);    #Base resistor's resistance in k\u2126 .\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector supply voltage = %dV\"%VCC);\n",
-      "print(\"Collector load=%.1f k\u2126 .\"%RC);\n",
-      "print(\"Base resistor's resistance=%.1f k\u2126 .\"%RB);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector supply voltage = 20V\n",
-        "Collector load=2.5 k\u2126 .\n",
-        "Base resistor's resistance=482.5 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.12: Page number 208"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage in V\n",
-      "VEE=-20.0;                     #Emitter supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=4.7;                       #Collector resistor's resistance in k\u2126\n",
-      "RE=10.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "VBE=0.7;                      #Base-emitter voltage in V\n",
-      "beta=85.0;                    #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n",
-      "#we get,IB*RB  +IE*RE +VBE -VEE=0.\n",
-      "#Since IB=IC/beta and IC~IE,\n",
-      "#(IE/beta)*RB + IE*RE + VBE + VEE =0.\n",
-      "IE=(-VEE-VBE)/(RE + RB/beta);                        #Emitter current in mA\n",
-      "IC=IE;                                              #Collector current (approximately equal to emitter current) in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law from VCC till collector terminal,\n",
-      "#we get, VCC - IC*RC =VC\n",
-      "VC=VCC-IC*RC;                               #voltage at collector terminal in V\n",
-      "\n",
-      "#Applying Kirchhoff's law from emitter terminal to VEE\n",
-      "#we get, VE -IE*RE =VEE\n",
-      "VE=VEE + IE*RE;                             #Voltage at emitter treminal in V\n",
-      "\n",
-      "VCE=VC-VE;                                  #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The collector current = %.2f mA\"%IC);\n",
-      "print(\"The emitter current = %.2f mA\"%IE);\n",
-      "print(\"The voltage at collector terminal = %.1f V\"%VC);\n",
-      "print(\"The collector-emitter voltage = %.1f V\"%VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The collector current = 1.73 mA\n",
-        "The emitter current = 1.73 mA\n",
-        "The voltage at collector terminal = 11.9 V\n",
-        "The collector-emitter voltage = 14.6 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.13: Page number 208-209\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage in V\n",
-      "VEE=-20.0;                     #Emitter supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=4.7;                       #Collector resistor's resistance in k\u2126\n",
-      "RE=10.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "beta1=85.0;                   #Base current amplification factor for case 1       \n",
-      "beta2=100.0;                  #Base current amplification factor for case 1\n",
-      "VBE_1=0.7;                    #Base emitter voltage for case 1 in V\n",
-      "VBE_2=0.6;                    #Base emitter voltage for case 2 in V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#For beta=85 and VBE=0.7,\n",
-      "#As calculated in the previous question,\n",
-      "IC_1=1.73;                          #Collector current in mA.\n",
-      "VCE_1=14.6;                           #Collector-emitter voltage in V.\n",
-      "\n",
-      "\n",
-      "#For case (ii)\n",
-      "#beta=100 and VBE=0.6\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n",
-      "#we get,IB*RB  +IE*RE +VBE -VEE=0.\n",
-      "#Since IB=IC/beta and IC~IE,\n",
-      "#(IE/beta)*RB + IE*RE + VBE +VEE =0.\n",
-      "IE_2=round((-VEE-VBE_2)/(RE + RB/beta2),2);                        #Emitter current in mA\n",
-      "IC_2=IE_2;                                                #Collector current (approximately equal to emitter current) in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law from VCC till collector terminal,\n",
-      "#we get, VCC - IC*RC =VC\n",
-      "VC=round(VCC-IC_2*RC,1);                               #voltage at collector terminal in V\n",
-      "\n",
-      "#Applying Kirchhoff's law from emitter terminal to VEE\n",
-      "#we get, VE -IE*RE =VEE\n",
-      "VE=round(VEE + IE_2*RE,1);                             #Voltage at emitter treminal in V\n",
-      "\n",
-      "VCE_2=VC-VE;                                #Collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "change_IC= (IC_2-IC_1)*100/IC_1;                    #%age change in collector current\n",
-      "\n",
-      "change_VCE=(VCE_2-VCE_1)*100/VCE_2;                 #%age change in collector-emitter voltage\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Percentage change in collector current =%.1f%%\"%change_IC);\n",
-      "print(\"Percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage change in collector current =1.7%\n",
-        "Percentage change in collector-emitter voltage =-3.5%\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.14: Page number 210\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                 #Collector supply voltage in V\n",
-      "VBE=0.7                   #Base-emitter voltage in V\n",
-      "RB=100.0;                 #Base resistor's resistance in k\u2126\n",
-      "RC=1.0;                   #Collector resistor's resistance in k\u2126\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC -IC*RC -IB*RB -VBE=0.\n",
-      "#since IC= beta*IB,\n",
-      "#We get,\n",
-      "IB=(VCC-VBE)/(RB + beta*RC);            #Base current in mA\n",
-      "IC=beta*IB;                             #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC=0.\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*RC;                          #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point : VCE=10.35V and IC=9.65mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.15: Page number 210-211\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                         #Collector supply voltage in V\n",
-      "VBE=0.3;                          #Base emitter voltage in V\n",
-      "IC=1.0;                           #Collector current in mA\n",
-      "VCE=8.0;                          #Collector emitter voltage in V\n",
-      "beta=100.0;                         #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Case(i)\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-IC*RC-VCE=0.\n",
-      "#from the above equation we get,\n",
-      "RC=(VCC-VCE)/IC;             #Collector load in kilo ohm\n",
-      "IB=IC/beta;                  #Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit\n",
-      "#we get, VCC-VBE-(beta*IB*RC)-IB*RB=0.\n",
-      "#From the above equation we get,\n",
-      "RB=round((VCC-VBE-beta*IB*RC)/IB,0);         #Base resistor's resistance in k\u2126\n",
-      "\n",
-      "#Results\n",
-      "print(\"The resistance value of base resistor=%d k\u2126 and collector load= %d k\u2126.\"%(RB,RC));\n",
-      "\n",
-      "#Case(ii)\n",
-      "\n",
-      "beta=50;\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC -IC*RC -IB*RB -VBE=0.\n",
-      "#since IC= beta*IB,\n",
-      "#We get,\n",
-      "IB=(VCC-VBE)/(RB + beta*RC);            #Base current in mA\n",
-      "IC=beta*IB;                             #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC=0.\n",
-      "#From the above equation,\n",
-      "VCE=round(VCC-IC*RC,1);                          #Collector emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.1fV and IC=%.1fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resistance value of base resistor=770 k\u2126 and collector load= 4 k\u2126.\n",
-        "The operating point : VCE=9.6V and IC=0.6mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.16 : Page number 211"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=2.0;                      #Collector-emitter voltage at operating point in V\n",
-      "VBE=0.7;                      #Base-emitter voltage in V            \n",
-      "IC=1.0;                       #Collector current at operating point in mA\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "IB=IC/beta;                     #Base current in mA\n",
-      "\n",
-      "#As, VCE=VCB +VBE\n",
-      "#we get,\n",
-      "VCB=VCE-VBE;                    #Collector-base voltage in V\n",
-      "RB=VCB/IB;                      #Base resistor's resistance in k\u2126\n",
-      "\n",
-      "#Results\n",
-      "print(\"Value of base resistor's resistance=%d k\u2126.\"%RB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of base resistor's resistance=130 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.17 : Page number 211-212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage in V\n",
-      "VBE=0.7                   #Base-emitter voltage in V\n",
-      "RB=400.0;                 #Base resistor's resistance in k\u2126\n",
-      "RC=4.0;                   #Collector resistor's resistance in k\u2126\n",
-      "RE=1.0;                   #Emitter resistor's resistance in k\u2126\n",
-      "beta=100.0;               #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along outut circuit,\n",
-      "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n",
-      "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n",
-      "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n",
-      "IC=(VCC-VBE)/(RB/beta + RC + RE);       #Collector current current in mA.\n",
-      "IE=IC;                                                #Emitter current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC -IE*RE=0. (IE~IC)\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*(RC+RE);                          #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.1fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point : VCE=5.7V and IC=1.26mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.18 : Page number 212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=10.0;                      #Collector resistor's resistance in k\u2126\n",
-      "RE=0;                         #Emitter resistor's  resistance in k\u2126\n",
-      "VBE=0.7;                      #Base-emitter voltage in V\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along outut circuit,\n",
-      "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n",
-      "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n",
-      "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n",
-      "IC=(VCC-VBE)/(RC +RB/beta + RE);                #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC =0. (IE~IC)\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*RC;                                  #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The d.c bias values are: VCE=%.2fV and IC=%.3fmA\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The d.c bias values are: VCE=1.55V and IC=0.845mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.19: Page number 214-215\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage in V\n",
-      "R1=10.0;                            #Resistor R1's resistance in k\u2126\n",
-      "R2=5.0;                             #Resistor R2's resistance in k\u2126\n",
-      "RC=1.0;                             #Collector resistor's resistance in k\u2126              \n",
-      "RE=2.0;                             #Emitter resistor's resistance in k\u2126\n",
-      "VBE=0.7;                            #Base-emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along output circuit\n",
-      "#VCE=VCC-IC*(RC+RE);\n",
-      "#IC=0, for VCE_max\n",
-      "VCE_max=VCC;                        #Maximum collector-emitter voltage in V\n",
-      "#VCE=0, for IC_max\n",
-      "IC_max=VCC/(RC+RE);                 #Maximum collector current in mA\n",
-      "\n",
-      "#Operating point\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage across R2 resistor V\n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "IC=IE;                              #Collector current(Approx. equal to emitter current) in mA\n",
-      "VCE=VCC-IC*(RC+RE);                 #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n",
-      "print(\"Collector current at operating point = %.2fmA\"%IC);\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,20])\n",
-      "limit.set_ylim([0,6])\n",
-      "VCE=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC=[((VCC-i)/(RC+RE)) for i in (VCE[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE,IC);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector-emitter voltage at operating point=8.55V\n",
-        "Collector current at operating point = 2.15mA\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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hKB07Mu+WiAStzuwdHBwwZswYrF69GiNHjgQADBgwAGfPnjXP4pzZi65xtPM//wPMny/s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\nAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8eeacf1ad0>"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.20: Page number 215-216\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage in V\n",
-      "R1=10.0;                            #Resistor R1's resistance in k\u2126 .\n",
-      "R2=5.0;                             #Resistor R2's resistance in k\u2126 .\n",
-      "RC=1.0;                             #Collector resistor's resistance in k\u2126 .              \n",
-      "RE=2.0;                             #Emitter resistor's resistance in k\u2126 .\n",
-      "VBE=0.7;                            #Base-emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's Theorem for replacing circuit consisting of VCC,R1,R2\n",
-      "E0=(VCC*R2)/(R1+R2);                #Thevenin's voltage in V\n",
-      "R0=(R1*R2)/(R1+R2);                 #Thevenin's equivalent resistance in k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "#IC=(E0-VBE)/(R0/beta +RE);\n",
-      "IC=(E0-VBE)/RE;                         #(Since R0/beta << RE) collector current in mA\n",
-      "VCE=VCC-IC*(RC+RE);                     #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n",
-      "print(\"Collector current at operating point = %.2fmA\"%IC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector-emitter voltage at operating point=8.55V\n",
-        "Collector current at operating point = 2.15mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.21: Page number 216-217\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                           #Collector  supply voltage in V\n",
-      "RE=1.0;                             #Emitter resistor, k\u2126 .\n",
-      "R1=50.0;                            #Resistor R1, k\u2126 .\n",
-      "R2=10.0;                            #Resistor R2, k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "VBE=0.1;                            #Base-emitter voltage in V\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V \n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "\n",
-      "print(\"(i)Emitter current= %.1fmA\"%IE);\n",
-      "\n",
-      "#(ii)\n",
-      "VBE=0.3;                            #Base-emitter voltage in V\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V\n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "\n",
-      "print(\"(ii)Emitter current= %.1fmA\"%IE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Emitter current= 1.9mA\n",
-        "(ii)Emitter current= 1.7mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.22: Page number 217\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                   #Collector supply voltage, V\n",
-      "R1=10.0;                    #Resistor R1, k\u2126\n",
-      "R2=10.0;                    #Resistor R2, k\u2126 .\n",
-      "RC=1.0;                     #Collector resistor, k\u2126 .\n",
-      "RE=5.0;                     #Emitter resistor, k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V\n",
-      "\n",
-      "#Applying kirchhoff's law from base terminal to emitter resistor\n",
-      "#V2=VBE+IE*RE\n",
-      "#VBE is neglected due to its small value\n",
-      "\n",
-      "IE=V2/RE;                     #Emitter current in mA\n",
-      "IC=IE;                              #Collector current (approx. equal to emitter current), mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along output circuit\n",
-      "VCE=VCC-IC*(RC+RE);                 #Collector-emitter voltage , V\n",
-      "VC=VCC-IC*RC;                       #Voltage at collector terminal,V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Emitter current =%dmA\"%IE);\n",
-      "print(\"Collector-emitter voltage=%dV\"%VCE);\n",
-      "print(\"Collector terminal's voltage=%dV\"%VC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Emitter current =2mA\n",
-        "Collector-emitter voltage=8V\n",
-        "Collector terminal's voltage=18V\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.23: Page number 219-220\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage, V\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "beta=50;                  #Base current amplification factor\n",
-      "R1=150;                   #Resistor R1, k\u2126 .\n",
-      "R2=100;                   #Resistor R2, k\u2126 .\n",
-      "RC=4.7;                   #Collector resistor, k\u2126 .\n",
-      "RE=2.2;                   #Emitter resistor, k\u2126  .\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "IB=round((E0-VBE)/(R0+beta*RE),3);              #Base current in mA\n",
-      "IC=round(beta*IB,1);                             #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-IC*(RC+RE);                     #Collector-emitter voltage, V\n",
-      "\n",
-      "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE);               #Stability factor\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n",
-      "print(\"Stability factor=%.1f\"%S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 3.72V and IC=1.2mA\n",
-        "Stability factor=18.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.24 : Page number 220\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                 #Collector supply voltage, V\n",
-      "VBE=0.7;                  #Base-emitter voltage , V\n",
-      "beta=100.0;                 #Base current amplification factor\n",
-      "R1=6.0;                     #Resistor R1, k\u2126 .\n",
-      "R2=3.0;                     #Resistor R2, k\u2126 .\n",
-      "RC=470.0;                   #Collector resistor, \u2126.\n",
-      "RE=1.0;                     #Emitter resistor, k\u2126 .\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "IB=round((E0-VBE)/(R0+beta*RE),3);              #Base current in mA\n",
-      "IC=round(beta*IB,1);                             #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-IC*(RC/1000+RE);                     #Collector-emitter voltage, V\n",
-      "\n",
-      "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE);               #Stability factor\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n",
-      "print(\"Stability factor=%.2f\"%S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 8.83V and IC=4.2mA\n",
-        "Stability factor=2.94\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.25 : Page number 221-222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "VCC=9;                    #Collector supply voltage, V\n",
-      "VCE=3;                    #Collector-emitter voltage, V\n",
-      "VBE=0.3;                    #Base-emitter voltage in V\n",
-      "RC=2.2;                     #Collector resistor , k\u2126 .\n",
-      "IC=2;                     #Collector current, mA\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "IB=IC/beta;                 #Base current in mA\n",
-      "\n",
-      "#According to given relation, I1=10*IB\n",
-      "I1=IB*10;                   #Current through the resistor R1, mA\n",
-      "\n",
-      "#I1=VCC/(R1+R2), .'s LAW\n",
-      "R1_R2_sum=VCC/I1;               #Sum of the resistor's R1 and R2, k\u2126 (OHM'S LAW).\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "#VCC=IC*RC+VCE+IE*RE\n",
-      "#IC~IE\n",
-      "RE=(VCC-IC*RC-VCE)/IC;      #Emitter resistor, k\u2126 .\n",
-      "RE=round(RE*1000,0);                 #Emittter resistor, \u2126 .\n",
-      "\n",
-      "IE=IC;                      #Emittter current(approximately equal to collector current), mA\n",
-      "VE=IE*(RE/1000);                   #Voltage at emitter terminal (OHM's LAW), V\n",
-      "V2=VBE+VE;                  #Voltage drop across resistor R2, V\n",
-      "\n",
-      "R2=V2/I1;                   #Resistor R2,(OHM's LAW), k\u2126 .\n",
-      "R1=R1_R2_sum-R2;            #Resistor R1, k\u2126 .\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"RE=%d \u2126., R1=%.2f k\u2126 . and R2=%.2f k\u2126 .\"%(RE,R1,R2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RE=800 \u2126., R1=17.75 k\u2126 . and R2=4.75 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.26 : Page number 222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=16.0;                     #Collector supply voltage, V\n",
-      "R2=20.0;                      #Resistor R2, k\u2126\n",
-      "RE=2.0;                       #Emitter resistor, k\u2126\n",
-      "VCE=6.0;                      #Collector-emitter voltage, V\n",
-      "IC=2.0;                       #Collector current , mA\n",
-      "VBE=0.3;                      #Base-emitter voltage,V\n",
-      "alpha=0.985;                  #Current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "beta=alpha/(1-alpha);               #Base current amplificatioon factor\n",
-      "IE=IC;                              #Emitter current, mA\n",
-      "IB=IC/beta;                         #Base current, mA\n",
-      "VE=IE*RE;                           #Emitter voltage,(OHM's LAW) V\n",
-      "V2=VBE+VE;                          #Voltage drop across resistor R2,(Kirchhoff's law) V\n",
-      "V_R1=VCC-V2;                        #Voltage drop across resistor R1, V\n",
-      "I1=V2/R2;                           #Current through resistor R2 an R1,(OHM's LAW) mA\n",
-      "R1=V_R1/I1;                         #Resistor R1,(OHM's LAW) k\u2126\n",
-      "\n",
-      "V_RC=(VCC-VCE-VE);                  #Voltage across collector resistor, V\n",
-      "RC=V_RC/IC;                         #Collector resistor,(OHM's LAW) k\u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"R1=%.1f k\u2126 and RC=%d k\u2126.\"%(R1,RC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R1=54.4 k\u2126 and RC=3 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.27 :Page number 222-223\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                     #Collector supply voltage, V\n",
-      "R1=10.0;                      #Resistor R1, k\u2126 \n",
-      "R2=5.0;                       #Resistor R2, k\u2126 \n",
-      "RC=1.0;                       #Collector resistor, k\u2126 \u007f\n",
-      "RE=2.0;                       #Emitter resistor, k\u2126 \n",
-      "VBE=0.7;                      #Base-emitter voltage, V\n",
-      "beta=100;                     #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 \n",
-      "\n",
-      "#Applying Kirchhoff' law along Thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IB=IE/beta,\n",
-      "IE=(E0-VBE)/(R0/beta + RE);                 #Emitter current , mA\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"The exact value of emitter current in the circuit = %.2fmA.\"%IE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The exact value of emitter current in the circuit = 2.11mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.28: Page number 223-224\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "IE=2.0;                     #Emitter current, mA\n",
-      "IB=50.0;                    #Base current, mA\n",
-      "VCC=10.0;                   #Collector supply voltage, V\n",
-      "VBE=0.2;                    #Base-emitter voltage, V\n",
-      "R2=10.0;                    #Resistor R2, k\u2126\n",
-      "RE=1.0;                     #Emitter resistance, k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law from the base to the emitter resistor,\n",
-      "V2=VBE+IE*RE;                   #Voltage at base terminal, V\n",
-      "I2=V2/R2;                       #Current through the resistor R2, mA\n",
-      "I1=I2+IB/1000;                  #Current through the resistor R2, mA\n",
-      "V1=VCC-V2;                      #Voltage drop across the resistor R2\n",
-      "R1=V1/I1;                       #Resistor R1, k\u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The value of the resistor R1=%.2f k\u2126.\"%R1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of the resistor R1=28.89 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.30 :Page number 225-226\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=8.0;                  #Collector supply voltage, V\n",
-      "RB=360.0;                 #Base resistor, k\u2126\n",
-      "RC=2.0;                   #Collector resistor, k\u2126\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "IC_max=VCC/RC;              #Maximum collector current, mA\n",
-      "VCE_max=VCC;                #Maximum collector voltage, V\n",
-      "\n",
-      "#Operating point\n",
-      "#Applying Kirchhoff's law along the input circuit\n",
-      "IB=(VCC-VBE)/RB;                    #Base current, mA\n",
-      "IC=beta*IB;                         #Collector current, mA\n",
-      "\n",
-      "#Kirchhoff' law along the output circuit\n",
-      "VCE=VCC-IC*RC;                      #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VCE=3.94V, is approximately half of VCC=8V \n",
-        " therefore it is mid-point biased.\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.31: page number 226\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage, V\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "beta=50.0;                     #Base current amplification factor\n",
-      "R1=12.0;                     #Resistor R1, k\u2126 \n",
-      "R2=2.7;                      #Resistor R2, k\u2126 \n",
-      "RC=620.0;                    #Collector resistor, \u2126 \n",
-      "RE=180.0;                    #Emitter resistor, \u2126\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Voltage divder rule across R1 and R2\n",
-      "V2=round((VCC*R2)/(R1+R2),2);            #Voltage drop across resistor R2, V\n",
-      "IE=round(((V2-VBE)/RE)*1000,2);                 #Emitter current, mA\n",
-      "IC=IE;                          #Collector current(Approximately equal to emitter current), mA\n",
-      "print(\"IC~IE=%.2fmA.\"%IC);\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-(IC/1000)*(RC+RE);                 #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IC~IE=6.33mA.\n",
-        "VCE=4.94V, is approximately half of VCC=10V \n",
-        " therefore it is mid-point biased.\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.32 : Page number 227\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=10.0;                    #Collector supply voltage, V\n",
-      "IC=10.0;                     #Collector  current, mA   \n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "R1=1.5;                      #Resistor R1, k\u2126 \n",
-      "R2=680.0;                    #Resistor R2, \u2126  \n",
-      "RC=260.0;                    #Collector resistor, \u2126 \n",
-      "RE=240.0;                    #Emitter resistor, \u2126 \n",
-      "beta_min=100;                #Minimum value of base current amplification factor\n",
-      "beta_max=400;                #Maximum value of base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Voltage divder rule across R1 and R2\n",
-      "V2=round((VCC*R2/1000)/(R1+R2/1000),2);               #Voltage drop across resistor R2, V\n",
-      "IE=round((V2-VBE)/(RE/1000),0);                         #OHM' LAW, Emitter current, mA\n",
-      "IC=IE;                                  #Collector current(approx. equal to emitter current),mA\n",
-      "beta_avg=sqrt(beta_min*beta_max);       #Average value of base current amplification factor\n",
-      "IB=IE/(beta_avg +1);                    #Base current, mA\n",
-      "IB=IB*1000;                             #Base current, \ud835\udf07A\n",
-      "\n",
-      "#Results\n",
-      "print(\"Base current= %.2f \ud835\udf07A\"%IB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current= 49.75 \ud835\udf07A\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.33 : Page number 227-228\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VEE=12.0;                    #Emitter supply voltage, V\n",
-      "RC=1.5;                     #Collector resistor, k\u2126\n",
-      "RB=120.0;                   #Base resistor k\u2126\n",
-      "RE=510.0;                   #Emitter resistor, \u2126                 \n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=60.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law,\n",
-      "#IB*RB - VBE - IE*RE +VEE=0\n",
-      "#Since IE~IC and IC=beta*IB,\n",
-      "IB=(VEE-VBE)/(RB + beta*RE/1000);           #Base current , mA\n",
-      "IC=round(beta*IB,1);                                 #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along output circuit,\n",
-      "VCE=VEE-IC*(RC + RE/1000);                        #Collector-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA.\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 2.96V and IC=4.5mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.34 : Page number 228-229\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "VEE=9.0;                    #Emitter supply voltage, V\n",
-      "RC=1.2;                     #Collector resistor, k\u2126\n",
-      "RB=100.0;                   #Base resistor ,k\u2126\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=45.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law,\n",
-      "#IB*RB + VBE=VEE\n",
-      "#Since IE~IC and IC=beta*IB,\n",
-      "IB=round((VEE-VBE)/RB,3);           #Base current , mA\n",
-      "IC=floor(beta*IB*100)/100;                                 #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along output circuit,\n",
-      "VCE=VEE-IC*RC;                        #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.2fmA.\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 4.52V and IC=3.73mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 68
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.35 : Page number 229\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=16.0;                   #Collector supply voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "IC=1.0;                     #Collector current, mA\n",
-      "VCE=6.0;                    #Collector-emitter voltage, V\n",
-      "beta=150.0;                 #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#For a good design, VE=VCC/10;\n",
-      "VE=VCC/10;                  #Emitter terminal's voltage, V\n",
-      "#OHM's Law\n",
-      "#And, taking IE~IC\n",
-      "RE=VE/IC;                   #Emitter resistor, k\u2126\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law alog output circuit:\n",
-      "#VCC=IC*RC + VCE + VE\n",
-      "RC=(VCC-VCE-VE)/IC;                       #Collector resistor, k\u2126\n",
-      "V2=VE+VBE;                                #Voltage drop across resistor R2,V\n",
-      "#From the relation I1=10*IB\n",
-      "R2=(beta*RE)/10;                          #Resistor R2, kilo ohm\n",
-      "\n",
-      "#From voltage divider rule across R1 and R2,\n",
-      "#V2=(VCC*R2)/(R1+R2)\n",
-      "R1=(VCC-V2)*R2/V2;                          #Resistor R1, k\u2126            \n",
-      "\n",
-      "#Results\n",
-      "print(\"RE=%.1f k\u2126 , RC=%.1f k\u2126, R1=%.0f k\u2126 and R2=%d k\u2126.\"%(RE,RC,R1,R2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RE=1.6 k\u2126 , RC=8.4 k\u2126, R1=143 k\u2126 and R2=24 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 69
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.36 : Page number 230-231\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ICBO=5.0;                          #Collector to base leakage current, microampere\n",
-      "beta=40.0;                         #Base current amplification factor\n",
-      "IC_zero_signal=2.0;                #Zero signal collector current, mA\n",
-      "op_temp=25.0;                      #operating temperature, degree celsius\n",
-      "temp_risen=55.0;                   #Temperature risen, degree celsius\n",
-      "temp_ICBO_doubles=10.0;            #Temperature after which ICBO doubles, degree celsius\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "ICEO=(beta+1)*ICBO;                 #Collector to emitter leakage current, microampere\n",
-      "\n",
-      "#(ii)\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;  #Number of times ICBO doubles\n",
-      "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled;                        #Final value of collector to base leakage current, microampere\n",
-      "ICEO_final=ICBO_final*(beta + 1);                                       #Final value of collector to emitter leakage current, microampere\n",
-      "\n",
-      "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal;                    #Zero signal collector current at 55 degree celius\n",
-      "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal;           #Percentage change in zero signal collector current\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The percentage change in the zero signal collector current=%.0f%%. \"%change)\n",
-      "\n",
-      "#(iii)\n",
-      "#For the silicon transistor\n",
-      "ICBO=0.1;                          #Collector to base leakage current, microampere\n",
-      "\n",
-      "ICEO=(beta+1)*ICBO;                 #Collector to emitter leakage current, microampere\n",
-      "\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;  #Number of times ICBO doubles\n",
-      "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled;                        #Final value of collector to base leakage current, microampere\n",
-      "ICEO_final=ICBO_final*(beta + 1);                                       #Final value of collector to emitter leakage current, microampere\n",
-      "\n",
-      "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal;                    #Zero signal collector current at 55 degree celius\n",
-      "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal;           #Percentage change in zero signal collector current\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) The percentage change in the zero signal collector current=%.1f%%. \"%change)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The percentage change in the zero signal collector current=82%. \n",
-        "(ii) The percentage change in the zero signal collector current=1.6%. \n"
-       ]
-      }
-     ],
-     "prompt_number": 70
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.37 : Page number 231\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ICBO=0.02                          #Collector to base leakage current, \ud835\udf07A\n",
-      "alpha=0.99;                        #Current amplification factor\n",
-      "IE=1.0;                            #Emitter current, mA\n",
-      "op_temp=27.0;                      #operating temperature, degree celsius\n",
-      "temp_risen=57.0;                   #Temperature risen, degree celsius\n",
-      "temp_ICBO_doubles=6.0;             #Temperature after which ICBO doubles, degree celsius\n",
-      "\n",
-      "#Calculations\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;       #Number of times ICBO doubles\n",
-      "ICBO_55=ICBO*2**Number_of_times_ICBO_doubled;                                #collector to base leakage current at 55 degree celsius, \ud835\udf07A\n",
-      "IC=alpha*IE + ICBO_55/1000;                                                     #Collector current, mA\n",
-      "IB=IE-IC;                                                                    #Base current, mA\n",
-      "IB=IB*1000;                                                                  #Base current,\ud835\udf07A\n",
-      "\n",
-      "#Results\n",
-      "print(\"Base current at 57 degree celsius=%.1f \ud835\udf07A \"%IB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current at 57 degree celsius=9.4 \ud835\udf07A \n"
-       ]
-      }
-     ],
-     "prompt_number": 71
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_3.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_3.ipynb
deleted file mode 100755
index e8168ab8..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_3.ipynb
+++ /dev/null
@@ -1,1907 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1e5463fc2e8c67a2f9099cf3f6c078bc1c9dccccd63589a75ad6ce5024fe6432"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 9 : TRANSISTOR BIASING"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1: Page number 195-196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=6.0;                     #Collector supply voltage\n",
-      "R_C=2.5;                      #Collector load in k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "#For faithful amplification Vce (collector-emitter voltage)> 1V for Si transistor.\n",
-      "V_CE_max=1;                                   #Maximum allowed collector-emitter voltage for faithful amplification, in V.\n",
-      "V_Rc_max=V_CC-V_CE_max;                       #maximum voltage drop across collector load in V.\n",
-      "I_C_max=V_Rc_max/R_C;                         #Maximum allowed collector current in mA\n",
-      "\n",
-      "#(ii)\n",
-      "IC_min_zero_signal=I_C_max/2;                            #Minimum zero signal collector current in mA\n",
-      "\n",
-      "#Results\n",
-      "print(\"The maximum allowed collector current during application of signal for faithful amplification = %d mA.\"%I_C_max);\n",
-      "print(\"The minimum zero signal collector current required = %d mA.\"%IC_min_zero_signal);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowed collector current during application of signal for faithful amplification = 2 mA.\n",
-        "The minimum zero signal collector current required = 1 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2: Page number 196\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=13.0;                     #Collector supply voltage in V\n",
-      "V_knee=1.0;                   #Knee voltage in V\n",
-      "R_C=4.0;                      #Collector load  in k\u2126\n",
-      "rate_IC_VBE=5.0;              #Rate of change of collector current IC with base-emitter voltage VBE in mA/V.\n",
-      "beta=100.0;                   #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V_Rc_max=VCC-V_knee;                    #Maximum allowed voltage across collector load in V\n",
-      "I_C_max=V_Rc_max/R_C;                   #Maximum allowed collector current in mA\n",
-      "I_B_max=I_C_max/beta;                   #Maximum base current in mA\n",
-      "I_B_max=I_B_max*1000;                   #Maximum base current in \ud835\udf07A\n",
-      "\n",
-      "V_B_max=I_C_max/rate_IC_VBE;            #Maximum base voltage signal in V\n",
-      "V_B_max=V_B_max*1000;                   #Maximum base voltage signal in mV\n",
-      "\n",
-      "#Results\n",
-      "print(\"Maximum base current =%d \ud835\udf07A.\"%I_B_max);\n",
-      "print(\"Maximum input signal voltage =%d mV.\"%V_B_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum base current =30 \ud835\udf07A.\n",
-        "Maximum input signal voltage =600 mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3: Page number 200-201"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=9.0;                  #Colector supply voltage in V\n",
-      "VBB=2.0;                  #Base supply voltage in V\n",
-      "R_B=100.0;                #Base resistor's resistance in k\u2126\n",
-      "R_C=2.0;                  #Collector load in k\u2126\n",
-      "beta=50.0;                #base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Case (i):\n",
-      "\n",
-      "#Applying Kirchhoff's law to the input circuit\n",
-      "#We get, IB*RB +VBE =VBB.\n",
-      "#Neglecting the small base-emitter voltage, we get:\n",
-      "I_B=VBB/R_B;                #Base current in mA\n",
-      "I_C=beta*I_B;               #Collector current in mA\n",
-      "\n",
-      "print(\"Collector current = %dmA\"%I_C);\n",
-      "\n",
-      "#Applying Kirchhoff's law to the output ciruit\n",
-      "#We get, IC*RC + VCE= VCC.\n",
-      "#From the above equation, we get:\n",
-      "V_CE=VCC-I_C*R_C;               #Collector emitter voltage in V\n",
-      "\n",
-      "print(\"Collector emitter voltage =%dV.\"%V_CE);\n",
-      "\n",
-      "\n",
-      "#Case (ii):\n",
-      "\n",
-      "R_B=50.0;\n",
-      "I_B=VBB/R_B;\n",
-      "I_C=beta*I_B;\n",
-      "V_CE=VCC  - I_C*R_C;\n",
-      "\n",
-      "print(\"The new operating point for base resistor RB=50 k\u2126 is, VCE=%dV and IC=%dmA.\"%(V_CE,I_C));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current = 1mA\n",
-        "Collector emitter voltage =7V.\n",
-        "The new operating point for base resistor RB=50 k\u2126 is, VCE=5V and IC=2mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4: Page number  201-202"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#variable declaration\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "VCC=6.0;                  #Collector suply voltagein V\n",
-      "VBE=0.7                   #Base emitter voltage in V\n",
-      "R_B=530.0;                #Base resistor's resistance in k\u2126 .\n",
-      "R_C=2.0;                  #Collector resistor's resistance in k\u2126 .\n",
-      "\n",
-      "#Calculation\n",
-      "#D.C load line equation : VCE=VCC-IC*RC;\n",
-      "#Calculating maximum VCE  ,by IC=0;\n",
-      "I_C_Vce_max=0;                      #Collector current for maximum collector-emitter voltage, in mA\n",
-      "VCE_max=VCC;-I_C_Vce_max*R_C;       #Maximum collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculating maximum collector current IC,by VCE=0;\n",
-      "V_CE_IC_max=0;             #Collector-emitter voltage for maximum collector current, in V         \n",
-      "I_C_max=(VCC-V_CE_IC_max)/R_C;        #Maximum collector current in mA\n",
-      "\n",
-      "\n",
-      "#Operating point:\n",
-      "#For input circuit, applying Kirchhoff's law, We get,\n",
-      "#VCC=IB*RB  + VBE.\n",
-      "#From the above equation,\n",
-      "IB=(VCC-VBE)/R_B;                #Base current in mA\n",
-      "IC=beta*IB;                    #Collector current\n",
-      "\n",
-      "#From the output circuit, applying Kirchhoff's law, we get:\n",
-      "VCE=VCC-IC*R_C;                 #Collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Stability factor\n",
-      "SF=beta+1;              \n",
-      "\n",
-      "#Result\n",
-      "print(\"Operating point: VCE= %dV and IC=%d mA\"%(VCE,IC));\n",
-      "print(\"Stability factor= %d.\"%SF);\n",
-      "\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,10])\n",
-      "limit.set_ylim([0,5])\n",
-      "VCE=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC=[((VCC-i)/(R_C)) for i in (VCE[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE,IC);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: VCE= 4V and IC=1 mA\n",
-        "Stability factor= 101.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8eea67df10>"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5: Page number 202"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage in V\n",
-      "beta=100.0;              #base current amplification factor\n",
-      "I_C_zero_signal=1.0;      #zero signal collector current in mA\n",
-      "VBE=0.3;                  #Base-emitter voltage of Ge transistor in V\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#Case(i)\n",
-      "I_B_zero_signal=I_C_zero_signal/beta;               #Zero signal base current in  mA\n",
-      "\n",
-      "#applying the Kirchhoff's law along input circuit:\n",
-      "#We get, VCC=IB*RB +VBE\n",
-      "#From the above equation we get,\n",
-      "R_B=(VCC-VBE)/I_B_zero_signal;              #Required base resistor's resistance in k\u2126\n",
-      "\n",
-      "print(\"Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = %d k\u2126\"%R_B);\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Case(ii)\n",
-      "beta=50;\n",
-      "I_B=(VCC-VBE)/R_B;                  #Base current of another transistor with beta=50, in mA\n",
-      "I_C_zero_signal=beta*I_B;           #Zero signal collector current for beta=50 , in mA\n",
-      "\n",
-      "print(\"The new value of zero signal collector current =%.1fmA\"%I_C_zero_signal);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = 1170 k\u2126\n",
-        "The new value of zero signal collector current =0.5mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6:Page number 202-203"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Varaible declaration\n",
-      "VCC=10.0;                     #Collector supply voltage in V\n",
-      "VBE=0;                        #Base emitter voltage in V(considering itas zero due to it's small value)\n",
-      "R_B=1.0;                      #Base resistor's resistance in M\u2126\n",
-      "R_C=2.0;                      #Collector resistor's resistance in k\u2126                  \n",
-      "R_E=1.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#using Kirchhoff's law in the input circuit, we get:\n",
-      "#VCC=IB*RB +VBE +IE*RE\n",
-      "#Since, IE=(beta +1)*I_B\n",
-      "#From the above equation we get:\n",
-      "I_B=round((VCC-VBE)/((beta + 1)*R_E + R_B*1000),4);           #Base current in mA\n",
-      "I_C=round(beta*I_B,2);                                   #Collector current in mA\n",
-      "I_E=I_B+I_C;                                    #Emitter current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Base current =%.4f mA\"%I_B);\n",
-      "print(\"Collector current =%.2f mA\"%I_C);\n",
-      "print(\"Emitter current =%.3f mA\"%I_E);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current =0.0091 mA\n",
-        "Collector current =0.91 mA\n",
-        "Emitter current =0.919 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.7: Page number 203-204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=8.0;                      #Collector-emitter voltage at operating point in V\n",
-      "IC=2.0;                       #Colector current at operating point in mA\n",
-      "VCC=15.0;                     #Collector supply voltagein V\n",
-      "beta=100.0;                   #base current amplification factor\n",
-      "VBE=0.6;                    #base emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC=VCE+IC*RC.\n",
-      "#So, from above equation we get:\n",
-      "RC=(VCC-VCE)/IC;                 #Collector resistor's resistance in k\u2126 .\n",
-      "IB=IC/beta;                      #Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the input circuit,\n",
-      "#we get, VCC=IB*RB + VBE\n",
-      "#So, from the above equation:\n",
-      "RB=(VCC-VBE)/IB;                #Base resistor's resistance in k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector load =%.1f k\u2126 .\"%RC);\n",
-      "print(\"Base resistor=%d k\u2126 .\"%RB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector load =3.5 k\u2126 .\n",
-        "Base resistor=720 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8: Page number 204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                #Collector supply voltage in V\n",
-      "VBE=0.7;                 #Base-emitter voltage in V\n",
-      "RB=100.0;                #Base resistor's resistance in k\u2126\n",
-      "RC=560.0;                #Collector resistor's resistance in \u2126\n",
-      "beta_25=100.0;           #base current amplification factor at 25 degree celsius\n",
-      "beta_75=150.0;           #base current amplification factor at 25 degree celsius\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit, we get\n",
-      "#VCC=IB*RB+VBE\n",
-      "IB=(VCC-VBE)/RB;                #Base current at 25 degree celsius, in mA\n",
-      "\n",
-      "\n",
-      "#For temperature 25 degree celsius\n",
-      "IC_25=beta_25*IB;                  #Collector current at 25 degree celsius, in mA\n",
-      "\n",
-      "\n",
-      "#Applying Kirchhoff's alw at the output circuit,\n",
-      "#we get: VCC=IC*RC + VCE\n",
-      "#From the above equation,\n",
-      "VCE_25=round(VCC-(IC_25/1000)*RC,2);                #Collector emitter voltage  at 25 degree celsius, in V\n",
-      "\n",
-      "\n",
-      "#For temperature 75 degree celsius\n",
-      "IC_75=round(beta_75*IB,0);                  #Collector current at 75 degree celsius, in mA\n",
-      "\n",
-      "#Applying Kirchhoff's alw at the output circuit,\n",
-      "#we get: VCC=IC*RC + VCE\n",
-      "#From the above equation,\n",
-      "VCE_75=round(VCC-(IC_75/1000)*RC,2);                #Collector emitter voltage at 75 degree celsius, in V\n",
-      "\n",
-      "\n",
-      "change_IC=(IC_75-IC_25)*100.0/IC_25;        #percentage change in collector current\n",
-      "change_VCE=(VCE_75-VCE_25)*100.0/VCE_25;    #Percentage change in collector-emitter voltage \n",
-      "\n",
-      "#Results\n",
-      "print(\"The percentage change in collector current =%d%%\"%change_IC);\n",
-      "print(\"The percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage change in collector current =50%\n",
-        "The percentage change in collector-emitter voltage =-56.3%\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.10: Page number 205"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE_max=20.0;             #Maximum collector-emitter voltage in V\n",
-      "VBE=0.7;                  #Base-emitter voltage in V\n",
-      "IC_max=8.0;               #Maximum collector current in mA\n",
-      "IB=40.0;                  #Base current in microampere\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#During cut off state the collector-emitter voltage is maximum and equal to collector supply voltage\n",
-      "VCC=VCE_max;                #Collector supply voltage in V\n",
-      "\n",
-      "#Maximum collector current IC_max=collector supply voltage(VCC)/collector load(RC)\n",
-      "#Collector load(RC)=VCC*IC_max\n",
-      "RC=VCC/IC_max;                  #Collector load in k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC=IB*RB +VBE.\n",
-      "#From the above equation, we get:\n",
-      "RB=(VCC-VBE)/(IB/1000);    #Base resistor's resistance in k\u2126 .\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector supply voltage = %dV\"%VCC);\n",
-      "print(\"Collector load=%.1f k\u2126 .\"%RC);\n",
-      "print(\"Base resistor's resistance=%.1f k\u2126 .\"%RB);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector supply voltage = 20V\n",
-        "Collector load=2.5 k\u2126 .\n",
-        "Base resistor's resistance=482.5 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.12: Page number 208"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage in V\n",
-      "VEE=-20.0;                     #Emitter supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=4.7;                       #Collector resistor's resistance in k\u2126\n",
-      "RE=10.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "VBE=0.7;                      #Base-emitter voltage in V\n",
-      "beta=85.0;                    #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n",
-      "#we get,IB*RB  +IE*RE +VBE -VEE=0.\n",
-      "#Since IB=IC/beta and IC~IE,\n",
-      "#(IE/beta)*RB + IE*RE + VBE + VEE =0.\n",
-      "IE=(-VEE-VBE)/(RE + RB/beta);                        #Emitter current in mA\n",
-      "IC=IE;                                              #Collector current (approximately equal to emitter current) in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law from VCC till collector terminal,\n",
-      "#we get, VCC - IC*RC =VC\n",
-      "VC=VCC-IC*RC;                               #voltage at collector terminal in V\n",
-      "\n",
-      "#Applying Kirchhoff's law from emitter terminal to VEE\n",
-      "#we get, VE -IE*RE =VEE\n",
-      "VE=VEE + IE*RE;                             #Voltage at emitter treminal in V\n",
-      "\n",
-      "VCE=VC-VE;                                  #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The collector current = %.2f mA\"%IC);\n",
-      "print(\"The emitter current = %.2f mA\"%IE);\n",
-      "print(\"The voltage at collector terminal = %.1f V\"%VC);\n",
-      "print(\"The collector-emitter voltage = %.1f V\"%VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The collector current = 1.73 mA\n",
-        "The emitter current = 1.73 mA\n",
-        "The voltage at collector terminal = 11.9 V\n",
-        "The collector-emitter voltage = 14.6 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.13: Page number 208-209\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage in V\n",
-      "VEE=-20.0;                     #Emitter supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=4.7;                       #Collector resistor's resistance in k\u2126\n",
-      "RE=10.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "beta1=85.0;                   #Base current amplification factor for case 1       \n",
-      "beta2=100.0;                  #Base current amplification factor for case 1\n",
-      "VBE_1=0.7;                    #Base emitter voltage for case 1 in V\n",
-      "VBE_2=0.6;                    #Base emitter voltage for case 2 in V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#For beta=85 and VBE=0.7,\n",
-      "#As calculated in the previous question,\n",
-      "IC_1=1.73;                          #Collector current in mA.\n",
-      "VCE_1=14.6;                           #Collector-emitter voltage in V.\n",
-      "\n",
-      "\n",
-      "#For case (ii)\n",
-      "#beta=100 and VBE=0.6\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n",
-      "#we get,IB*RB  +IE*RE +VBE -VEE=0.\n",
-      "#Since IB=IC/beta and IC~IE,\n",
-      "#(IE/beta)*RB + IE*RE + VBE +VEE =0.\n",
-      "IE_2=round((-VEE-VBE_2)/(RE + RB/beta2),2);                        #Emitter current in mA\n",
-      "IC_2=IE_2;                                                #Collector current (approximately equal to emitter current) in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law from VCC till collector terminal,\n",
-      "#we get, VCC - IC*RC =VC\n",
-      "VC=round(VCC-IC_2*RC,1);                               #voltage at collector terminal in V\n",
-      "\n",
-      "#Applying Kirchhoff's law from emitter terminal to VEE\n",
-      "#we get, VE -IE*RE =VEE\n",
-      "VE=round(VEE + IE_2*RE,1);                             #Voltage at emitter treminal in V\n",
-      "\n",
-      "VCE_2=VC-VE;                                #Collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "change_IC= (IC_2-IC_1)*100/IC_1;                    #%age change in collector current\n",
-      "\n",
-      "change_VCE=(VCE_2-VCE_1)*100/VCE_2;                 #%age change in collector-emitter voltage\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Percentage change in collector current =%.1f%%\"%change_IC);\n",
-      "print(\"Percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage change in collector current =1.7%\n",
-        "Percentage change in collector-emitter voltage =-3.5%\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.14: Page number 210\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                 #Collector supply voltage in V\n",
-      "VBE=0.7                   #Base-emitter voltage in V\n",
-      "RB=100.0;                 #Base resistor's resistance in k\u2126\n",
-      "RC=1.0;                   #Collector resistor's resistance in k\u2126\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC -IC*RC -IB*RB -VBE=0.\n",
-      "#since IC= beta*IB,\n",
-      "#We get,\n",
-      "IB=(VCC-VBE)/(RB + beta*RC);            #Base current in mA\n",
-      "IC=beta*IB;                             #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC=0.\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*RC;                          #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point : VCE=10.35V and IC=9.65mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.15: Page number 210-211\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                         #Collector supply voltage in V\n",
-      "VBE=0.3;                          #Base emitter voltage in V\n",
-      "IC=1.0;                           #Collector current in mA\n",
-      "VCE=8.0;                          #Collector emitter voltage in V\n",
-      "beta=100.0;                         #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Case(i)\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-IC*RC-VCE=0.\n",
-      "#from the above equation we get,\n",
-      "RC=(VCC-VCE)/IC;             #Collector load in kilo ohm\n",
-      "IB=IC/beta;                  #Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit\n",
-      "#we get, VCC-VBE-(beta*IB*RC)-IB*RB=0.\n",
-      "#From the above equation we get,\n",
-      "RB=round((VCC-VBE-beta*IB*RC)/IB,0);         #Base resistor's resistance in k\u2126\n",
-      "\n",
-      "#Results\n",
-      "print(\"The resistance value of base resistor=%d k\u2126 and collector load= %d k\u2126.\"%(RB,RC));\n",
-      "\n",
-      "#Case(ii)\n",
-      "\n",
-      "beta=50;\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC -IC*RC -IB*RB -VBE=0.\n",
-      "#since IC= beta*IB,\n",
-      "#We get,\n",
-      "IB=(VCC-VBE)/(RB + beta*RC);            #Base current in mA\n",
-      "IC=beta*IB;                             #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC=0.\n",
-      "#From the above equation,\n",
-      "VCE=round(VCC-IC*RC,1);                          #Collector emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.1fV and IC=%.1fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resistance value of base resistor=770 k\u2126 and collector load= 4 k\u2126.\n",
-        "The operating point : VCE=9.6V and IC=0.6mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.16 : Page number 211"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=2.0;                      #Collector-emitter voltage at operating point in V\n",
-      "VBE=0.7;                      #Base-emitter voltage in V            \n",
-      "IC=1.0;                       #Collector current at operating point in mA\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "IB=IC/beta;                     #Base current in mA\n",
-      "\n",
-      "#As, VCE=VCB +VBE\n",
-      "#we get,\n",
-      "VCB=VCE-VBE;                    #Collector-base voltage in V\n",
-      "RB=VCB/IB;                      #Base resistor's resistance in k\u2126\n",
-      "\n",
-      "#Results\n",
-      "print(\"Value of base resistor's resistance=%d k\u2126.\"%RB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of base resistor's resistance=130 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.17 : Page number 211-212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage in V\n",
-      "VBE=0.7                   #Base-emitter voltage in V\n",
-      "RB=400.0;                 #Base resistor's resistance in k\u2126\n",
-      "RC=4.0;                   #Collector resistor's resistance in k\u2126\n",
-      "RE=1.0;                   #Emitter resistor's resistance in k\u2126\n",
-      "beta=100.0;               #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along outut circuit,\n",
-      "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n",
-      "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n",
-      "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n",
-      "IC=(VCC-VBE)/(RB/beta + RC + RE);       #Collector current current in mA.\n",
-      "IE=IC;                                                #Emitter current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC -IE*RE=0. (IE~IC)\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*(RC+RE);                          #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.1fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point : VCE=5.7V and IC=1.26mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.18 : Page number 212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=10.0;                      #Collector resistor's resistance in k\u2126\n",
-      "RE=0;                         #Emitter resistor's  resistance in k\u2126\n",
-      "VBE=0.7;                      #Base-emitter voltage in V\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along outut circuit,\n",
-      "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n",
-      "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n",
-      "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n",
-      "IC=(VCC-VBE)/(RC +RB/beta + RE);                #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC =0. (IE~IC)\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*RC;                                  #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The d.c bias values are: VCE=%.2fV and IC=%.3fmA\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The d.c bias values are: VCE=1.55V and IC=0.845mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.19: Page number 214-215\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage in V\n",
-      "R1=10.0;                            #Resistor R1's resistance in k\u2126\n",
-      "R2=5.0;                             #Resistor R2's resistance in k\u2126\n",
-      "RC=1.0;                             #Collector resistor's resistance in k\u2126              \n",
-      "RE=2.0;                             #Emitter resistor's resistance in k\u2126\n",
-      "VBE=0.7;                            #Base-emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along output circuit\n",
-      "#VCE=VCC-IC*(RC+RE);\n",
-      "#IC=0, for VCE_max\n",
-      "VCE_max=VCC;                        #Maximum collector-emitter voltage in V\n",
-      "#VCE=0, for IC_max\n",
-      "IC_max=VCC/(RC+RE);                 #Maximum collector current in mA\n",
-      "\n",
-      "#Operating point\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage across R2 resistor V\n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "IC=IE;                              #Collector current(Approx. equal to emitter current) in mA\n",
-      "VCE=VCC-IC*(RC+RE);                 #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n",
-      "print(\"Collector current at operating point = %.2fmA\"%IC);\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,20])\n",
-      "limit.set_ylim([0,6])\n",
-      "VCE=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC=[((VCC-i)/(RC+RE)) for i in (VCE[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE,IC);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector-emitter voltage at operating point=8.55V\n",
-        "Collector current at operating point = 2.15mA\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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hKB07Mu+WiAStzuwdHBwwZswYrF69GiNHjgQADBgwAGfPnjXP4pzZi65xtPM//wPMny/s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\nAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8eeacf1ad0>"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.20: Page number 215-216\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage in V\n",
-      "R1=10.0;                            #Resistor R1's resistance in k\u2126 .\n",
-      "R2=5.0;                             #Resistor R2's resistance in k\u2126 .\n",
-      "RC=1.0;                             #Collector resistor's resistance in k\u2126 .              \n",
-      "RE=2.0;                             #Emitter resistor's resistance in k\u2126 .\n",
-      "VBE=0.7;                            #Base-emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's Theorem for replacing circuit consisting of VCC,R1,R2\n",
-      "E0=(VCC*R2)/(R1+R2);                #Thevenin's voltage in V\n",
-      "R0=(R1*R2)/(R1+R2);                 #Thevenin's equivalent resistance in k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "#IC=(E0-VBE)/(R0/beta +RE);\n",
-      "IC=(E0-VBE)/RE;                         #(Since R0/beta << RE) collector current in mA\n",
-      "VCE=VCC-IC*(RC+RE);                     #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n",
-      "print(\"Collector current at operating point = %.2fmA\"%IC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector-emitter voltage at operating point=8.55V\n",
-        "Collector current at operating point = 2.15mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.21: Page number 216-217\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                           #Collector  supply voltage in V\n",
-      "RE=1.0;                             #Emitter resistor, k\u2126 .\n",
-      "R1=50.0;                            #Resistor R1, k\u2126 .\n",
-      "R2=10.0;                            #Resistor R2, k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "VBE=0.1;                            #Base-emitter voltage in V\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V \n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "\n",
-      "print(\"(i)Emitter current= %.1fmA\"%IE);\n",
-      "\n",
-      "#(ii)\n",
-      "VBE=0.3;                            #Base-emitter voltage in V\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V\n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "\n",
-      "print(\"(ii)Emitter current= %.1fmA\"%IE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Emitter current= 1.9mA\n",
-        "(ii)Emitter current= 1.7mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.22: Page number 217\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                   #Collector supply voltage, V\n",
-      "R1=10.0;                    #Resistor R1, k\u2126\n",
-      "R2=10.0;                    #Resistor R2, k\u2126 .\n",
-      "RC=1.0;                     #Collector resistor, k\u2126 .\n",
-      "RE=5.0;                     #Emitter resistor, k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V\n",
-      "\n",
-      "#Applying kirchhoff's law from base terminal to emitter resistor\n",
-      "#V2=VBE+IE*RE\n",
-      "#VBE is neglected due to its small value\n",
-      "\n",
-      "IE=V2/RE;                     #Emitter current in mA\n",
-      "IC=IE;                              #Collector current (approx. equal to emitter current), mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along output circuit\n",
-      "VCE=VCC-IC*(RC+RE);                 #Collector-emitter voltage , V\n",
-      "VC=VCC-IC*RC;                       #Voltage at collector terminal,V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Emitter current =%dmA\"%IE);\n",
-      "print(\"Collector-emitter voltage=%dV\"%VCE);\n",
-      "print(\"Collector terminal's voltage=%dV\"%VC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Emitter current =2mA\n",
-        "Collector-emitter voltage=8V\n",
-        "Collector terminal's voltage=18V\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.23: Page number 219-220\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage, V\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "beta=50;                  #Base current amplification factor\n",
-      "R1=150;                   #Resistor R1, k\u2126 .\n",
-      "R2=100;                   #Resistor R2, k\u2126 .\n",
-      "RC=4.7;                   #Collector resistor, k\u2126 .\n",
-      "RE=2.2;                   #Emitter resistor, k\u2126  .\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "IB=round((E0-VBE)/(R0+beta*RE),3);              #Base current in mA\n",
-      "IC=round(beta*IB,1);                             #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-IC*(RC+RE);                     #Collector-emitter voltage, V\n",
-      "\n",
-      "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE);               #Stability factor\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n",
-      "print(\"Stability factor=%.1f\"%S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 3.72V and IC=1.2mA\n",
-        "Stability factor=18.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.24 : Page number 220\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                 #Collector supply voltage, V\n",
-      "VBE=0.7;                  #Base-emitter voltage , V\n",
-      "beta=100.0;                 #Base current amplification factor\n",
-      "R1=6.0;                     #Resistor R1, k\u2126 .\n",
-      "R2=3.0;                     #Resistor R2, k\u2126 .\n",
-      "RC=470.0;                   #Collector resistor, \u2126.\n",
-      "RE=1.0;                     #Emitter resistor, k\u2126 .\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "IB=round((E0-VBE)/(R0+beta*RE),3);              #Base current in mA\n",
-      "IC=round(beta*IB,1);                             #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-IC*(RC/1000+RE);                     #Collector-emitter voltage, V\n",
-      "\n",
-      "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE);               #Stability factor\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n",
-      "print(\"Stability factor=%.2f\"%S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 8.83V and IC=4.2mA\n",
-        "Stability factor=2.94\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.25 : Page number 221-222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "VCC=9;                    #Collector supply voltage, V\n",
-      "VCE=3;                    #Collector-emitter voltage, V\n",
-      "VBE=0.3;                    #Base-emitter voltage in V\n",
-      "RC=2.2;                     #Collector resistor , k\u2126 .\n",
-      "IC=2;                     #Collector current, mA\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "IB=IC/beta;                 #Base current in mA\n",
-      "\n",
-      "#According to given relation, I1=10*IB\n",
-      "I1=IB*10;                   #Current through the resistor R1, mA\n",
-      "\n",
-      "#I1=VCC/(R1+R2), .'s LAW\n",
-      "R1_R2_sum=VCC/I1;               #Sum of the resistor's R1 and R2, k\u2126 (OHM'S LAW).\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "#VCC=IC*RC+VCE+IE*RE\n",
-      "#IC~IE\n",
-      "RE=(VCC-IC*RC-VCE)/IC;      #Emitter resistor, k\u2126 .\n",
-      "RE=round(RE*1000,0);                 #Emittter resistor, \u2126 .\n",
-      "\n",
-      "IE=IC;                      #Emittter current(approximately equal to collector current), mA\n",
-      "VE=IE*(RE/1000);                   #Voltage at emitter terminal (OHM's LAW), V\n",
-      "V2=VBE+VE;                  #Voltage drop across resistor R2, V\n",
-      "\n",
-      "R2=V2/I1;                   #Resistor R2,(OHM's LAW), k\u2126 .\n",
-      "R1=R1_R2_sum-R2;            #Resistor R1, k\u2126 .\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"RE=%d \u2126., R1=%.2f k\u2126 . and R2=%.2f k\u2126 .\"%(RE,R1,R2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RE=800 \u2126., R1=17.75 k\u2126 . and R2=4.75 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.26 : Page number 222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=16.0;                     #Collector supply voltage, V\n",
-      "R2=20.0;                      #Resistor R2, k\u2126\n",
-      "RE=2.0;                       #Emitter resistor, k\u2126\n",
-      "VCE=6.0;                      #Collector-emitter voltage, V\n",
-      "IC=2.0;                       #Collector current , mA\n",
-      "VBE=0.3;                      #Base-emitter voltage,V\n",
-      "alpha=0.985;                  #Current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "beta=alpha/(1-alpha);               #Base current amplificatioon factor\n",
-      "IE=IC;                              #Emitter current, mA\n",
-      "IB=IC/beta;                         #Base current, mA\n",
-      "VE=IE*RE;                           #Emitter voltage,(OHM's LAW) V\n",
-      "V2=VBE+VE;                          #Voltage drop across resistor R2,(Kirchhoff's law) V\n",
-      "V_R1=VCC-V2;                        #Voltage drop across resistor R1, V\n",
-      "I1=V2/R2;                           #Current through resistor R2 an R1,(OHM's LAW) mA\n",
-      "R1=V_R1/I1;                         #Resistor R1,(OHM's LAW) k\u2126\n",
-      "\n",
-      "V_RC=(VCC-VCE-VE);                  #Voltage across collector resistor, V\n",
-      "RC=V_RC/IC;                         #Collector resistor,(OHM's LAW) k\u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"R1=%.1f k\u2126 and RC=%d k\u2126.\"%(R1,RC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R1=54.4 k\u2126 and RC=3 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.27 :Page number 222-223\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                     #Collector supply voltage, V\n",
-      "R1=10.0;                      #Resistor R1, k\u2126 \n",
-      "R2=5.0;                       #Resistor R2, k\u2126 \n",
-      "RC=1.0;                       #Collector resistor, k\u2126 \u007f\n",
-      "RE=2.0;                       #Emitter resistor, k\u2126 \n",
-      "VBE=0.7;                      #Base-emitter voltage, V\n",
-      "beta=100;                     #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 \n",
-      "\n",
-      "#Applying Kirchhoff' law along Thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IB=IE/beta,\n",
-      "IE=(E0-VBE)/(R0/beta + RE);                 #Emitter current , mA\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"The exact value of emitter current in the circuit = %.2fmA.\"%IE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The exact value of emitter current in the circuit = 2.11mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.28: Page number 223-224\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "IE=2.0;                     #Emitter current, mA\n",
-      "IB=50.0;                    #Base current, mA\n",
-      "VCC=10.0;                   #Collector supply voltage, V\n",
-      "VBE=0.2;                    #Base-emitter voltage, V\n",
-      "R2=10.0;                    #Resistor R2, k\u2126\n",
-      "RE=1.0;                     #Emitter resistance, k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law from the base to the emitter resistor,\n",
-      "V2=VBE+IE*RE;                   #Voltage at base terminal, V\n",
-      "I2=V2/R2;                       #Current through the resistor R2, mA\n",
-      "I1=I2+IB/1000;                  #Current through the resistor R2, mA\n",
-      "V1=VCC-V2;                      #Voltage drop across the resistor R2\n",
-      "R1=V1/I1;                       #Resistor R1, k\u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The value of the resistor R1=%.2f k\u2126.\"%R1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of the resistor R1=28.89 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.30 :Page number 225-226\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=8.0;                  #Collector supply voltage, V\n",
-      "RB=360.0;                 #Base resistor, k\u2126\n",
-      "RC=2.0;                   #Collector resistor, k\u2126\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "IC_max=VCC/RC;              #Maximum collector current, mA\n",
-      "VCE_max=VCC;                #Maximum collector voltage, V\n",
-      "\n",
-      "#Operating point\n",
-      "#Applying Kirchhoff's law along the input circuit\n",
-      "IB=(VCC-VBE)/RB;                    #Base current, mA\n",
-      "IC=beta*IB;                         #Collector current, mA\n",
-      "\n",
-      "#Kirchhoff' law along the output circuit\n",
-      "VCE=VCC-IC*RC;                      #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VCE=3.94V, is approximately half of VCC=8V \n",
-        " therefore it is mid-point biased.\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.31: page number 226\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage, V\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "beta=50.0;                     #Base current amplification factor\n",
-      "R1=12.0;                     #Resistor R1, k\u2126 \n",
-      "R2=2.7;                      #Resistor R2, k\u2126 \n",
-      "RC=620.0;                    #Collector resistor, \u2126 \n",
-      "RE=180.0;                    #Emitter resistor, \u2126\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Voltage divder rule across R1 and R2\n",
-      "V2=round((VCC*R2)/(R1+R2),2);            #Voltage drop across resistor R2, V\n",
-      "IE=round(((V2-VBE)/RE)*1000,2);                 #Emitter current, mA\n",
-      "IC=IE;                          #Collector current(Approximately equal to emitter current), mA\n",
-      "print(\"IC~IE=%.2fmA.\"%IC);\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-(IC/1000)*(RC+RE);                 #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IC~IE=6.33mA.\n",
-        "VCE=4.94V, is approximately half of VCC=10V \n",
-        " therefore it is mid-point biased.\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.32 : Page number 227\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=10.0;                    #Collector supply voltage, V\n",
-      "IC=10.0;                     #Collector  current, mA   \n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "R1=1.5;                      #Resistor R1, k\u2126 \n",
-      "R2=680.0;                    #Resistor R2, \u2126  \n",
-      "RC=260.0;                    #Collector resistor, \u2126 \n",
-      "RE=240.0;                    #Emitter resistor, \u2126 \n",
-      "beta_min=100;                #Minimum value of base current amplification factor\n",
-      "beta_max=400;                #Maximum value of base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Voltage divder rule across R1 and R2\n",
-      "V2=round((VCC*R2/1000)/(R1+R2/1000),2);               #Voltage drop across resistor R2, V\n",
-      "IE=round((V2-VBE)/(RE/1000),0);                         #OHM' LAW, Emitter current, mA\n",
-      "IC=IE;                                  #Collector current(approx. equal to emitter current),mA\n",
-      "beta_avg=sqrt(beta_min*beta_max);       #Average value of base current amplification factor\n",
-      "IB=IE/(beta_avg +1);                    #Base current, mA\n",
-      "IB=IB*1000;                             #Base current, \ud835\udf07A\n",
-      "\n",
-      "#Results\n",
-      "print(\"Base current= %.2f \ud835\udf07A\"%IB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current= 49.75 \ud835\udf07A\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.33 : Page number 227-228\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VEE=12.0;                    #Emitter supply voltage, V\n",
-      "RC=1.5;                     #Collector resistor, k\u2126\n",
-      "RB=120.0;                   #Base resistor k\u2126\n",
-      "RE=510.0;                   #Emitter resistor, \u2126                 \n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=60.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law,\n",
-      "#IB*RB - VBE - IE*RE +VEE=0\n",
-      "#Since IE~IC and IC=beta*IB,\n",
-      "IB=(VEE-VBE)/(RB + beta*RE/1000);           #Base current , mA\n",
-      "IC=round(beta*IB,1);                                 #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along output circuit,\n",
-      "VCE=VEE-IC*(RC + RE/1000);                        #Collector-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA.\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 2.96V and IC=4.5mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.34 : Page number 228-229\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "VEE=9.0;                    #Emitter supply voltage, V\n",
-      "RC=1.2;                     #Collector resistor, k\u2126\n",
-      "RB=100.0;                   #Base resistor ,k\u2126\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=45.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law,\n",
-      "#IB*RB + VBE=VEE\n",
-      "#Since IE~IC and IC=beta*IB,\n",
-      "IB=round((VEE-VBE)/RB,3);           #Base current , mA\n",
-      "IC=floor(beta*IB*100)/100;                                 #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along output circuit,\n",
-      "VCE=VEE-IC*RC;                        #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.2fmA.\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 4.52V and IC=3.73mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 68
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.35 : Page number 229\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=16.0;                   #Collector supply voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "IC=1.0;                     #Collector current, mA\n",
-      "VCE=6.0;                    #Collector-emitter voltage, V\n",
-      "beta=150.0;                 #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#For a good design, VE=VCC/10;\n",
-      "VE=VCC/10;                  #Emitter terminal's voltage, V\n",
-      "#OHM's Law\n",
-      "#And, taking IE~IC\n",
-      "RE=VE/IC;                   #Emitter resistor, k\u2126\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law alog output circuit:\n",
-      "#VCC=IC*RC + VCE + VE\n",
-      "RC=(VCC-VCE-VE)/IC;                       #Collector resistor, k\u2126\n",
-      "V2=VE+VBE;                                #Voltage drop across resistor R2,V\n",
-      "#From the relation I1=10*IB\n",
-      "R2=(beta*RE)/10;                          #Resistor R2, kilo ohm\n",
-      "\n",
-      "#From voltage divider rule across R1 and R2,\n",
-      "#V2=(VCC*R2)/(R1+R2)\n",
-      "R1=(VCC-V2)*R2/V2;                          #Resistor R1, k\u2126            \n",
-      "\n",
-      "#Results\n",
-      "print(\"RE=%.1f k\u2126 , RC=%.1f k\u2126, R1=%.0f k\u2126 and R2=%d k\u2126.\"%(RE,RC,R1,R2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RE=1.6 k\u2126 , RC=8.4 k\u2126, R1=143 k\u2126 and R2=24 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 69
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.36 : Page number 230-231\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ICBO=5.0;                          #Collector to base leakage current, microampere\n",
-      "beta=40.0;                         #Base current amplification factor\n",
-      "IC_zero_signal=2.0;                #Zero signal collector current, mA\n",
-      "op_temp=25.0;                      #operating temperature, degree celsius\n",
-      "temp_risen=55.0;                   #Temperature risen, degree celsius\n",
-      "temp_ICBO_doubles=10.0;            #Temperature after which ICBO doubles, degree celsius\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "ICEO=(beta+1)*ICBO;                 #Collector to emitter leakage current, microampere\n",
-      "\n",
-      "#(ii)\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;  #Number of times ICBO doubles\n",
-      "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled;                        #Final value of collector to base leakage current, microampere\n",
-      "ICEO_final=ICBO_final*(beta + 1);                                       #Final value of collector to emitter leakage current, microampere\n",
-      "\n",
-      "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal;                    #Zero signal collector current at 55 degree celius\n",
-      "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal;           #Percentage change in zero signal collector current\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The percentage change in the zero signal collector current=%.0f%%. \"%change)\n",
-      "\n",
-      "#(iii)\n",
-      "#For the silicon transistor\n",
-      "ICBO=0.1;                          #Collector to base leakage current, microampere\n",
-      "\n",
-      "ICEO=(beta+1)*ICBO;                 #Collector to emitter leakage current, microampere\n",
-      "\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;  #Number of times ICBO doubles\n",
-      "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled;                        #Final value of collector to base leakage current, microampere\n",
-      "ICEO_final=ICBO_final*(beta + 1);                                       #Final value of collector to emitter leakage current, microampere\n",
-      "\n",
-      "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal;                    #Zero signal collector current at 55 degree celius\n",
-      "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal;           #Percentage change in zero signal collector current\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) The percentage change in the zero signal collector current=%.1f%%. \"%change)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The percentage change in the zero signal collector current=82%. \n",
-        "(ii) The percentage change in the zero signal collector current=1.6%. \n"
-       ]
-      }
-     ],
-     "prompt_number": 70
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.37 : Page number 231\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ICBO=0.02                          #Collector to base leakage current, \ud835\udf07A\n",
-      "alpha=0.99;                        #Current amplification factor\n",
-      "IE=1.0;                            #Emitter current, mA\n",
-      "op_temp=27.0;                      #operating temperature, degree celsius\n",
-      "temp_risen=57.0;                   #Temperature risen, degree celsius\n",
-      "temp_ICBO_doubles=6.0;             #Temperature after which ICBO doubles, degree celsius\n",
-      "\n",
-      "#Calculations\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;       #Number of times ICBO doubles\n",
-      "ICBO_55=ICBO*2**Number_of_times_ICBO_doubled;                                #collector to base leakage current at 55 degree celsius, \ud835\udf07A\n",
-      "IC=alpha*IE + ICBO_55/1000;                                                     #Collector current, mA\n",
-      "IB=IE-IC;                                                                    #Base current, mA\n",
-      "IB=IB*1000;                                                                  #Base current,\ud835\udf07A\n",
-      "\n",
-      "#Results\n",
-      "print(\"Base current at 57 degree celsius=%.1f \ud835\udf07A \"%IB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current at 57 degree celsius=9.4 \ud835\udf07A \n"
-       ]
-      }
-     ],
-     "prompt_number": 71
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_4.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_4.ipynb
deleted file mode 100755
index e8168ab8..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_4.ipynb
+++ /dev/null
@@ -1,1907 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1e5463fc2e8c67a2f9099cf3f6c078bc1c9dccccd63589a75ad6ce5024fe6432"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 9 : TRANSISTOR BIASING"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1: Page number 195-196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=6.0;                     #Collector supply voltage\n",
-      "R_C=2.5;                      #Collector load in k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "#For faithful amplification Vce (collector-emitter voltage)> 1V for Si transistor.\n",
-      "V_CE_max=1;                                   #Maximum allowed collector-emitter voltage for faithful amplification, in V.\n",
-      "V_Rc_max=V_CC-V_CE_max;                       #maximum voltage drop across collector load in V.\n",
-      "I_C_max=V_Rc_max/R_C;                         #Maximum allowed collector current in mA\n",
-      "\n",
-      "#(ii)\n",
-      "IC_min_zero_signal=I_C_max/2;                            #Minimum zero signal collector current in mA\n",
-      "\n",
-      "#Results\n",
-      "print(\"The maximum allowed collector current during application of signal for faithful amplification = %d mA.\"%I_C_max);\n",
-      "print(\"The minimum zero signal collector current required = %d mA.\"%IC_min_zero_signal);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowed collector current during application of signal for faithful amplification = 2 mA.\n",
-        "The minimum zero signal collector current required = 1 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2: Page number 196\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=13.0;                     #Collector supply voltage in V\n",
-      "V_knee=1.0;                   #Knee voltage in V\n",
-      "R_C=4.0;                      #Collector load  in k\u2126\n",
-      "rate_IC_VBE=5.0;              #Rate of change of collector current IC with base-emitter voltage VBE in mA/V.\n",
-      "beta=100.0;                   #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V_Rc_max=VCC-V_knee;                    #Maximum allowed voltage across collector load in V\n",
-      "I_C_max=V_Rc_max/R_C;                   #Maximum allowed collector current in mA\n",
-      "I_B_max=I_C_max/beta;                   #Maximum base current in mA\n",
-      "I_B_max=I_B_max*1000;                   #Maximum base current in \ud835\udf07A\n",
-      "\n",
-      "V_B_max=I_C_max/rate_IC_VBE;            #Maximum base voltage signal in V\n",
-      "V_B_max=V_B_max*1000;                   #Maximum base voltage signal in mV\n",
-      "\n",
-      "#Results\n",
-      "print(\"Maximum base current =%d \ud835\udf07A.\"%I_B_max);\n",
-      "print(\"Maximum input signal voltage =%d mV.\"%V_B_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum base current =30 \ud835\udf07A.\n",
-        "Maximum input signal voltage =600 mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3: Page number 200-201"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=9.0;                  #Colector supply voltage in V\n",
-      "VBB=2.0;                  #Base supply voltage in V\n",
-      "R_B=100.0;                #Base resistor's resistance in k\u2126\n",
-      "R_C=2.0;                  #Collector load in k\u2126\n",
-      "beta=50.0;                #base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Case (i):\n",
-      "\n",
-      "#Applying Kirchhoff's law to the input circuit\n",
-      "#We get, IB*RB +VBE =VBB.\n",
-      "#Neglecting the small base-emitter voltage, we get:\n",
-      "I_B=VBB/R_B;                #Base current in mA\n",
-      "I_C=beta*I_B;               #Collector current in mA\n",
-      "\n",
-      "print(\"Collector current = %dmA\"%I_C);\n",
-      "\n",
-      "#Applying Kirchhoff's law to the output ciruit\n",
-      "#We get, IC*RC + VCE= VCC.\n",
-      "#From the above equation, we get:\n",
-      "V_CE=VCC-I_C*R_C;               #Collector emitter voltage in V\n",
-      "\n",
-      "print(\"Collector emitter voltage =%dV.\"%V_CE);\n",
-      "\n",
-      "\n",
-      "#Case (ii):\n",
-      "\n",
-      "R_B=50.0;\n",
-      "I_B=VBB/R_B;\n",
-      "I_C=beta*I_B;\n",
-      "V_CE=VCC  - I_C*R_C;\n",
-      "\n",
-      "print(\"The new operating point for base resistor RB=50 k\u2126 is, VCE=%dV and IC=%dmA.\"%(V_CE,I_C));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current = 1mA\n",
-        "Collector emitter voltage =7V.\n",
-        "The new operating point for base resistor RB=50 k\u2126 is, VCE=5V and IC=2mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4: Page number  201-202"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#variable declaration\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "VCC=6.0;                  #Collector suply voltagein V\n",
-      "VBE=0.7                   #Base emitter voltage in V\n",
-      "R_B=530.0;                #Base resistor's resistance in k\u2126 .\n",
-      "R_C=2.0;                  #Collector resistor's resistance in k\u2126 .\n",
-      "\n",
-      "#Calculation\n",
-      "#D.C load line equation : VCE=VCC-IC*RC;\n",
-      "#Calculating maximum VCE  ,by IC=0;\n",
-      "I_C_Vce_max=0;                      #Collector current for maximum collector-emitter voltage, in mA\n",
-      "VCE_max=VCC;-I_C_Vce_max*R_C;       #Maximum collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculating maximum collector current IC,by VCE=0;\n",
-      "V_CE_IC_max=0;             #Collector-emitter voltage for maximum collector current, in V         \n",
-      "I_C_max=(VCC-V_CE_IC_max)/R_C;        #Maximum collector current in mA\n",
-      "\n",
-      "\n",
-      "#Operating point:\n",
-      "#For input circuit, applying Kirchhoff's law, We get,\n",
-      "#VCC=IB*RB  + VBE.\n",
-      "#From the above equation,\n",
-      "IB=(VCC-VBE)/R_B;                #Base current in mA\n",
-      "IC=beta*IB;                    #Collector current\n",
-      "\n",
-      "#From the output circuit, applying Kirchhoff's law, we get:\n",
-      "VCE=VCC-IC*R_C;                 #Collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Stability factor\n",
-      "SF=beta+1;              \n",
-      "\n",
-      "#Result\n",
-      "print(\"Operating point: VCE= %dV and IC=%d mA\"%(VCE,IC));\n",
-      "print(\"Stability factor= %d.\"%SF);\n",
-      "\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,10])\n",
-      "limit.set_ylim([0,5])\n",
-      "VCE=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC=[((VCC-i)/(R_C)) for i in (VCE[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE,IC);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: VCE= 4V and IC=1 mA\n",
-        "Stability factor= 101.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8eea67df10>"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5: Page number 202"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage in V\n",
-      "beta=100.0;              #base current amplification factor\n",
-      "I_C_zero_signal=1.0;      #zero signal collector current in mA\n",
-      "VBE=0.3;                  #Base-emitter voltage of Ge transistor in V\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#Case(i)\n",
-      "I_B_zero_signal=I_C_zero_signal/beta;               #Zero signal base current in  mA\n",
-      "\n",
-      "#applying the Kirchhoff's law along input circuit:\n",
-      "#We get, VCC=IB*RB +VBE\n",
-      "#From the above equation we get,\n",
-      "R_B=(VCC-VBE)/I_B_zero_signal;              #Required base resistor's resistance in k\u2126\n",
-      "\n",
-      "print(\"Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = %d k\u2126\"%R_B);\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Case(ii)\n",
-      "beta=50;\n",
-      "I_B=(VCC-VBE)/R_B;                  #Base current of another transistor with beta=50, in mA\n",
-      "I_C_zero_signal=beta*I_B;           #Zero signal collector current for beta=50 , in mA\n",
-      "\n",
-      "print(\"The new value of zero signal collector current =%.1fmA\"%I_C_zero_signal);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = 1170 k\u2126\n",
-        "The new value of zero signal collector current =0.5mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6:Page number 202-203"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Varaible declaration\n",
-      "VCC=10.0;                     #Collector supply voltage in V\n",
-      "VBE=0;                        #Base emitter voltage in V(considering itas zero due to it's small value)\n",
-      "R_B=1.0;                      #Base resistor's resistance in M\u2126\n",
-      "R_C=2.0;                      #Collector resistor's resistance in k\u2126                  \n",
-      "R_E=1.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#using Kirchhoff's law in the input circuit, we get:\n",
-      "#VCC=IB*RB +VBE +IE*RE\n",
-      "#Since, IE=(beta +1)*I_B\n",
-      "#From the above equation we get:\n",
-      "I_B=round((VCC-VBE)/((beta + 1)*R_E + R_B*1000),4);           #Base current in mA\n",
-      "I_C=round(beta*I_B,2);                                   #Collector current in mA\n",
-      "I_E=I_B+I_C;                                    #Emitter current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Base current =%.4f mA\"%I_B);\n",
-      "print(\"Collector current =%.2f mA\"%I_C);\n",
-      "print(\"Emitter current =%.3f mA\"%I_E);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current =0.0091 mA\n",
-        "Collector current =0.91 mA\n",
-        "Emitter current =0.919 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.7: Page number 203-204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=8.0;                      #Collector-emitter voltage at operating point in V\n",
-      "IC=2.0;                       #Colector current at operating point in mA\n",
-      "VCC=15.0;                     #Collector supply voltagein V\n",
-      "beta=100.0;                   #base current amplification factor\n",
-      "VBE=0.6;                    #base emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC=VCE+IC*RC.\n",
-      "#So, from above equation we get:\n",
-      "RC=(VCC-VCE)/IC;                 #Collector resistor's resistance in k\u2126 .\n",
-      "IB=IC/beta;                      #Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the input circuit,\n",
-      "#we get, VCC=IB*RB + VBE\n",
-      "#So, from the above equation:\n",
-      "RB=(VCC-VBE)/IB;                #Base resistor's resistance in k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector load =%.1f k\u2126 .\"%RC);\n",
-      "print(\"Base resistor=%d k\u2126 .\"%RB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector load =3.5 k\u2126 .\n",
-        "Base resistor=720 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8: Page number 204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                #Collector supply voltage in V\n",
-      "VBE=0.7;                 #Base-emitter voltage in V\n",
-      "RB=100.0;                #Base resistor's resistance in k\u2126\n",
-      "RC=560.0;                #Collector resistor's resistance in \u2126\n",
-      "beta_25=100.0;           #base current amplification factor at 25 degree celsius\n",
-      "beta_75=150.0;           #base current amplification factor at 25 degree celsius\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit, we get\n",
-      "#VCC=IB*RB+VBE\n",
-      "IB=(VCC-VBE)/RB;                #Base current at 25 degree celsius, in mA\n",
-      "\n",
-      "\n",
-      "#For temperature 25 degree celsius\n",
-      "IC_25=beta_25*IB;                  #Collector current at 25 degree celsius, in mA\n",
-      "\n",
-      "\n",
-      "#Applying Kirchhoff's alw at the output circuit,\n",
-      "#we get: VCC=IC*RC + VCE\n",
-      "#From the above equation,\n",
-      "VCE_25=round(VCC-(IC_25/1000)*RC,2);                #Collector emitter voltage  at 25 degree celsius, in V\n",
-      "\n",
-      "\n",
-      "#For temperature 75 degree celsius\n",
-      "IC_75=round(beta_75*IB,0);                  #Collector current at 75 degree celsius, in mA\n",
-      "\n",
-      "#Applying Kirchhoff's alw at the output circuit,\n",
-      "#we get: VCC=IC*RC + VCE\n",
-      "#From the above equation,\n",
-      "VCE_75=round(VCC-(IC_75/1000)*RC,2);                #Collector emitter voltage at 75 degree celsius, in V\n",
-      "\n",
-      "\n",
-      "change_IC=(IC_75-IC_25)*100.0/IC_25;        #percentage change in collector current\n",
-      "change_VCE=(VCE_75-VCE_25)*100.0/VCE_25;    #Percentage change in collector-emitter voltage \n",
-      "\n",
-      "#Results\n",
-      "print(\"The percentage change in collector current =%d%%\"%change_IC);\n",
-      "print(\"The percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage change in collector current =50%\n",
-        "The percentage change in collector-emitter voltage =-56.3%\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.10: Page number 205"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE_max=20.0;             #Maximum collector-emitter voltage in V\n",
-      "VBE=0.7;                  #Base-emitter voltage in V\n",
-      "IC_max=8.0;               #Maximum collector current in mA\n",
-      "IB=40.0;                  #Base current in microampere\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#During cut off state the collector-emitter voltage is maximum and equal to collector supply voltage\n",
-      "VCC=VCE_max;                #Collector supply voltage in V\n",
-      "\n",
-      "#Maximum collector current IC_max=collector supply voltage(VCC)/collector load(RC)\n",
-      "#Collector load(RC)=VCC*IC_max\n",
-      "RC=VCC/IC_max;                  #Collector load in k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC=IB*RB +VBE.\n",
-      "#From the above equation, we get:\n",
-      "RB=(VCC-VBE)/(IB/1000);    #Base resistor's resistance in k\u2126 .\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector supply voltage = %dV\"%VCC);\n",
-      "print(\"Collector load=%.1f k\u2126 .\"%RC);\n",
-      "print(\"Base resistor's resistance=%.1f k\u2126 .\"%RB);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector supply voltage = 20V\n",
-        "Collector load=2.5 k\u2126 .\n",
-        "Base resistor's resistance=482.5 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.12: Page number 208"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage in V\n",
-      "VEE=-20.0;                     #Emitter supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=4.7;                       #Collector resistor's resistance in k\u2126\n",
-      "RE=10.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "VBE=0.7;                      #Base-emitter voltage in V\n",
-      "beta=85.0;                    #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n",
-      "#we get,IB*RB  +IE*RE +VBE -VEE=0.\n",
-      "#Since IB=IC/beta and IC~IE,\n",
-      "#(IE/beta)*RB + IE*RE + VBE + VEE =0.\n",
-      "IE=(-VEE-VBE)/(RE + RB/beta);                        #Emitter current in mA\n",
-      "IC=IE;                                              #Collector current (approximately equal to emitter current) in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law from VCC till collector terminal,\n",
-      "#we get, VCC - IC*RC =VC\n",
-      "VC=VCC-IC*RC;                               #voltage at collector terminal in V\n",
-      "\n",
-      "#Applying Kirchhoff's law from emitter terminal to VEE\n",
-      "#we get, VE -IE*RE =VEE\n",
-      "VE=VEE + IE*RE;                             #Voltage at emitter treminal in V\n",
-      "\n",
-      "VCE=VC-VE;                                  #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The collector current = %.2f mA\"%IC);\n",
-      "print(\"The emitter current = %.2f mA\"%IE);\n",
-      "print(\"The voltage at collector terminal = %.1f V\"%VC);\n",
-      "print(\"The collector-emitter voltage = %.1f V\"%VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The collector current = 1.73 mA\n",
-        "The emitter current = 1.73 mA\n",
-        "The voltage at collector terminal = 11.9 V\n",
-        "The collector-emitter voltage = 14.6 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.13: Page number 208-209\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage in V\n",
-      "VEE=-20.0;                     #Emitter supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=4.7;                       #Collector resistor's resistance in k\u2126\n",
-      "RE=10.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "beta1=85.0;                   #Base current amplification factor for case 1       \n",
-      "beta2=100.0;                  #Base current amplification factor for case 1\n",
-      "VBE_1=0.7;                    #Base emitter voltage for case 1 in V\n",
-      "VBE_2=0.6;                    #Base emitter voltage for case 2 in V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#For beta=85 and VBE=0.7,\n",
-      "#As calculated in the previous question,\n",
-      "IC_1=1.73;                          #Collector current in mA.\n",
-      "VCE_1=14.6;                           #Collector-emitter voltage in V.\n",
-      "\n",
-      "\n",
-      "#For case (ii)\n",
-      "#beta=100 and VBE=0.6\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n",
-      "#we get,IB*RB  +IE*RE +VBE -VEE=0.\n",
-      "#Since IB=IC/beta and IC~IE,\n",
-      "#(IE/beta)*RB + IE*RE + VBE +VEE =0.\n",
-      "IE_2=round((-VEE-VBE_2)/(RE + RB/beta2),2);                        #Emitter current in mA\n",
-      "IC_2=IE_2;                                                #Collector current (approximately equal to emitter current) in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law from VCC till collector terminal,\n",
-      "#we get, VCC - IC*RC =VC\n",
-      "VC=round(VCC-IC_2*RC,1);                               #voltage at collector terminal in V\n",
-      "\n",
-      "#Applying Kirchhoff's law from emitter terminal to VEE\n",
-      "#we get, VE -IE*RE =VEE\n",
-      "VE=round(VEE + IE_2*RE,1);                             #Voltage at emitter treminal in V\n",
-      "\n",
-      "VCE_2=VC-VE;                                #Collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "change_IC= (IC_2-IC_1)*100/IC_1;                    #%age change in collector current\n",
-      "\n",
-      "change_VCE=(VCE_2-VCE_1)*100/VCE_2;                 #%age change in collector-emitter voltage\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Percentage change in collector current =%.1f%%\"%change_IC);\n",
-      "print(\"Percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage change in collector current =1.7%\n",
-        "Percentage change in collector-emitter voltage =-3.5%\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.14: Page number 210\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                 #Collector supply voltage in V\n",
-      "VBE=0.7                   #Base-emitter voltage in V\n",
-      "RB=100.0;                 #Base resistor's resistance in k\u2126\n",
-      "RC=1.0;                   #Collector resistor's resistance in k\u2126\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC -IC*RC -IB*RB -VBE=0.\n",
-      "#since IC= beta*IB,\n",
-      "#We get,\n",
-      "IB=(VCC-VBE)/(RB + beta*RC);            #Base current in mA\n",
-      "IC=beta*IB;                             #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC=0.\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*RC;                          #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point : VCE=10.35V and IC=9.65mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.15: Page number 210-211\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                         #Collector supply voltage in V\n",
-      "VBE=0.3;                          #Base emitter voltage in V\n",
-      "IC=1.0;                           #Collector current in mA\n",
-      "VCE=8.0;                          #Collector emitter voltage in V\n",
-      "beta=100.0;                         #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Case(i)\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-IC*RC-VCE=0.\n",
-      "#from the above equation we get,\n",
-      "RC=(VCC-VCE)/IC;             #Collector load in kilo ohm\n",
-      "IB=IC/beta;                  #Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit\n",
-      "#we get, VCC-VBE-(beta*IB*RC)-IB*RB=0.\n",
-      "#From the above equation we get,\n",
-      "RB=round((VCC-VBE-beta*IB*RC)/IB,0);         #Base resistor's resistance in k\u2126\n",
-      "\n",
-      "#Results\n",
-      "print(\"The resistance value of base resistor=%d k\u2126 and collector load= %d k\u2126.\"%(RB,RC));\n",
-      "\n",
-      "#Case(ii)\n",
-      "\n",
-      "beta=50;\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC -IC*RC -IB*RB -VBE=0.\n",
-      "#since IC= beta*IB,\n",
-      "#We get,\n",
-      "IB=(VCC-VBE)/(RB + beta*RC);            #Base current in mA\n",
-      "IC=beta*IB;                             #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC=0.\n",
-      "#From the above equation,\n",
-      "VCE=round(VCC-IC*RC,1);                          #Collector emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.1fV and IC=%.1fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resistance value of base resistor=770 k\u2126 and collector load= 4 k\u2126.\n",
-        "The operating point : VCE=9.6V and IC=0.6mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.16 : Page number 211"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=2.0;                      #Collector-emitter voltage at operating point in V\n",
-      "VBE=0.7;                      #Base-emitter voltage in V            \n",
-      "IC=1.0;                       #Collector current at operating point in mA\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "IB=IC/beta;                     #Base current in mA\n",
-      "\n",
-      "#As, VCE=VCB +VBE\n",
-      "#we get,\n",
-      "VCB=VCE-VBE;                    #Collector-base voltage in V\n",
-      "RB=VCB/IB;                      #Base resistor's resistance in k\u2126\n",
-      "\n",
-      "#Results\n",
-      "print(\"Value of base resistor's resistance=%d k\u2126.\"%RB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of base resistor's resistance=130 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.17 : Page number 211-212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage in V\n",
-      "VBE=0.7                   #Base-emitter voltage in V\n",
-      "RB=400.0;                 #Base resistor's resistance in k\u2126\n",
-      "RC=4.0;                   #Collector resistor's resistance in k\u2126\n",
-      "RE=1.0;                   #Emitter resistor's resistance in k\u2126\n",
-      "beta=100.0;               #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along outut circuit,\n",
-      "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n",
-      "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n",
-      "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n",
-      "IC=(VCC-VBE)/(RB/beta + RC + RE);       #Collector current current in mA.\n",
-      "IE=IC;                                                #Emitter current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC -IE*RE=0. (IE~IC)\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*(RC+RE);                          #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.1fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point : VCE=5.7V and IC=1.26mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.18 : Page number 212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=10.0;                      #Collector resistor's resistance in k\u2126\n",
-      "RE=0;                         #Emitter resistor's  resistance in k\u2126\n",
-      "VBE=0.7;                      #Base-emitter voltage in V\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along outut circuit,\n",
-      "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n",
-      "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n",
-      "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n",
-      "IC=(VCC-VBE)/(RC +RB/beta + RE);                #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC =0. (IE~IC)\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*RC;                                  #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The d.c bias values are: VCE=%.2fV and IC=%.3fmA\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The d.c bias values are: VCE=1.55V and IC=0.845mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.19: Page number 214-215\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage in V\n",
-      "R1=10.0;                            #Resistor R1's resistance in k\u2126\n",
-      "R2=5.0;                             #Resistor R2's resistance in k\u2126\n",
-      "RC=1.0;                             #Collector resistor's resistance in k\u2126              \n",
-      "RE=2.0;                             #Emitter resistor's resistance in k\u2126\n",
-      "VBE=0.7;                            #Base-emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along output circuit\n",
-      "#VCE=VCC-IC*(RC+RE);\n",
-      "#IC=0, for VCE_max\n",
-      "VCE_max=VCC;                        #Maximum collector-emitter voltage in V\n",
-      "#VCE=0, for IC_max\n",
-      "IC_max=VCC/(RC+RE);                 #Maximum collector current in mA\n",
-      "\n",
-      "#Operating point\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage across R2 resistor V\n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "IC=IE;                              #Collector current(Approx. equal to emitter current) in mA\n",
-      "VCE=VCC-IC*(RC+RE);                 #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n",
-      "print(\"Collector current at operating point = %.2fmA\"%IC);\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,20])\n",
-      "limit.set_ylim([0,6])\n",
-      "VCE=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC=[((VCC-i)/(RC+RE)) for i in (VCE[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE,IC);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector-emitter voltage at operating point=8.55V\n",
-        "Collector current at operating point = 2.15mA\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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hKB07Mu+WiAStzuwdHBwwZswYrF69GiNHjgQADBgwAGfPnjXP4pzZi65xtPM//wPMny/s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\nAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8eeacf1ad0>"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.20: Page number 215-216\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage in V\n",
-      "R1=10.0;                            #Resistor R1's resistance in k\u2126 .\n",
-      "R2=5.0;                             #Resistor R2's resistance in k\u2126 .\n",
-      "RC=1.0;                             #Collector resistor's resistance in k\u2126 .              \n",
-      "RE=2.0;                             #Emitter resistor's resistance in k\u2126 .\n",
-      "VBE=0.7;                            #Base-emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's Theorem for replacing circuit consisting of VCC,R1,R2\n",
-      "E0=(VCC*R2)/(R1+R2);                #Thevenin's voltage in V\n",
-      "R0=(R1*R2)/(R1+R2);                 #Thevenin's equivalent resistance in k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "#IC=(E0-VBE)/(R0/beta +RE);\n",
-      "IC=(E0-VBE)/RE;                         #(Since R0/beta << RE) collector current in mA\n",
-      "VCE=VCC-IC*(RC+RE);                     #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n",
-      "print(\"Collector current at operating point = %.2fmA\"%IC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector-emitter voltage at operating point=8.55V\n",
-        "Collector current at operating point = 2.15mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.21: Page number 216-217\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                           #Collector  supply voltage in V\n",
-      "RE=1.0;                             #Emitter resistor, k\u2126 .\n",
-      "R1=50.0;                            #Resistor R1, k\u2126 .\n",
-      "R2=10.0;                            #Resistor R2, k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "VBE=0.1;                            #Base-emitter voltage in V\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V \n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "\n",
-      "print(\"(i)Emitter current= %.1fmA\"%IE);\n",
-      "\n",
-      "#(ii)\n",
-      "VBE=0.3;                            #Base-emitter voltage in V\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V\n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "\n",
-      "print(\"(ii)Emitter current= %.1fmA\"%IE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Emitter current= 1.9mA\n",
-        "(ii)Emitter current= 1.7mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.22: Page number 217\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                   #Collector supply voltage, V\n",
-      "R1=10.0;                    #Resistor R1, k\u2126\n",
-      "R2=10.0;                    #Resistor R2, k\u2126 .\n",
-      "RC=1.0;                     #Collector resistor, k\u2126 .\n",
-      "RE=5.0;                     #Emitter resistor, k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V\n",
-      "\n",
-      "#Applying kirchhoff's law from base terminal to emitter resistor\n",
-      "#V2=VBE+IE*RE\n",
-      "#VBE is neglected due to its small value\n",
-      "\n",
-      "IE=V2/RE;                     #Emitter current in mA\n",
-      "IC=IE;                              #Collector current (approx. equal to emitter current), mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along output circuit\n",
-      "VCE=VCC-IC*(RC+RE);                 #Collector-emitter voltage , V\n",
-      "VC=VCC-IC*RC;                       #Voltage at collector terminal,V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Emitter current =%dmA\"%IE);\n",
-      "print(\"Collector-emitter voltage=%dV\"%VCE);\n",
-      "print(\"Collector terminal's voltage=%dV\"%VC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Emitter current =2mA\n",
-        "Collector-emitter voltage=8V\n",
-        "Collector terminal's voltage=18V\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.23: Page number 219-220\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage, V\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "beta=50;                  #Base current amplification factor\n",
-      "R1=150;                   #Resistor R1, k\u2126 .\n",
-      "R2=100;                   #Resistor R2, k\u2126 .\n",
-      "RC=4.7;                   #Collector resistor, k\u2126 .\n",
-      "RE=2.2;                   #Emitter resistor, k\u2126  .\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "IB=round((E0-VBE)/(R0+beta*RE),3);              #Base current in mA\n",
-      "IC=round(beta*IB,1);                             #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-IC*(RC+RE);                     #Collector-emitter voltage, V\n",
-      "\n",
-      "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE);               #Stability factor\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n",
-      "print(\"Stability factor=%.1f\"%S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 3.72V and IC=1.2mA\n",
-        "Stability factor=18.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.24 : Page number 220\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                 #Collector supply voltage, V\n",
-      "VBE=0.7;                  #Base-emitter voltage , V\n",
-      "beta=100.0;                 #Base current amplification factor\n",
-      "R1=6.0;                     #Resistor R1, k\u2126 .\n",
-      "R2=3.0;                     #Resistor R2, k\u2126 .\n",
-      "RC=470.0;                   #Collector resistor, \u2126.\n",
-      "RE=1.0;                     #Emitter resistor, k\u2126 .\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "IB=round((E0-VBE)/(R0+beta*RE),3);              #Base current in mA\n",
-      "IC=round(beta*IB,1);                             #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-IC*(RC/1000+RE);                     #Collector-emitter voltage, V\n",
-      "\n",
-      "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE);               #Stability factor\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n",
-      "print(\"Stability factor=%.2f\"%S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 8.83V and IC=4.2mA\n",
-        "Stability factor=2.94\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.25 : Page number 221-222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "VCC=9;                    #Collector supply voltage, V\n",
-      "VCE=3;                    #Collector-emitter voltage, V\n",
-      "VBE=0.3;                    #Base-emitter voltage in V\n",
-      "RC=2.2;                     #Collector resistor , k\u2126 .\n",
-      "IC=2;                     #Collector current, mA\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "IB=IC/beta;                 #Base current in mA\n",
-      "\n",
-      "#According to given relation, I1=10*IB\n",
-      "I1=IB*10;                   #Current through the resistor R1, mA\n",
-      "\n",
-      "#I1=VCC/(R1+R2), .'s LAW\n",
-      "R1_R2_sum=VCC/I1;               #Sum of the resistor's R1 and R2, k\u2126 (OHM'S LAW).\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "#VCC=IC*RC+VCE+IE*RE\n",
-      "#IC~IE\n",
-      "RE=(VCC-IC*RC-VCE)/IC;      #Emitter resistor, k\u2126 .\n",
-      "RE=round(RE*1000,0);                 #Emittter resistor, \u2126 .\n",
-      "\n",
-      "IE=IC;                      #Emittter current(approximately equal to collector current), mA\n",
-      "VE=IE*(RE/1000);                   #Voltage at emitter terminal (OHM's LAW), V\n",
-      "V2=VBE+VE;                  #Voltage drop across resistor R2, V\n",
-      "\n",
-      "R2=V2/I1;                   #Resistor R2,(OHM's LAW), k\u2126 .\n",
-      "R1=R1_R2_sum-R2;            #Resistor R1, k\u2126 .\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"RE=%d \u2126., R1=%.2f k\u2126 . and R2=%.2f k\u2126 .\"%(RE,R1,R2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RE=800 \u2126., R1=17.75 k\u2126 . and R2=4.75 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.26 : Page number 222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=16.0;                     #Collector supply voltage, V\n",
-      "R2=20.0;                      #Resistor R2, k\u2126\n",
-      "RE=2.0;                       #Emitter resistor, k\u2126\n",
-      "VCE=6.0;                      #Collector-emitter voltage, V\n",
-      "IC=2.0;                       #Collector current , mA\n",
-      "VBE=0.3;                      #Base-emitter voltage,V\n",
-      "alpha=0.985;                  #Current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "beta=alpha/(1-alpha);               #Base current amplificatioon factor\n",
-      "IE=IC;                              #Emitter current, mA\n",
-      "IB=IC/beta;                         #Base current, mA\n",
-      "VE=IE*RE;                           #Emitter voltage,(OHM's LAW) V\n",
-      "V2=VBE+VE;                          #Voltage drop across resistor R2,(Kirchhoff's law) V\n",
-      "V_R1=VCC-V2;                        #Voltage drop across resistor R1, V\n",
-      "I1=V2/R2;                           #Current through resistor R2 an R1,(OHM's LAW) mA\n",
-      "R1=V_R1/I1;                         #Resistor R1,(OHM's LAW) k\u2126\n",
-      "\n",
-      "V_RC=(VCC-VCE-VE);                  #Voltage across collector resistor, V\n",
-      "RC=V_RC/IC;                         #Collector resistor,(OHM's LAW) k\u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"R1=%.1f k\u2126 and RC=%d k\u2126.\"%(R1,RC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R1=54.4 k\u2126 and RC=3 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.27 :Page number 222-223\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                     #Collector supply voltage, V\n",
-      "R1=10.0;                      #Resistor R1, k\u2126 \n",
-      "R2=5.0;                       #Resistor R2, k\u2126 \n",
-      "RC=1.0;                       #Collector resistor, k\u2126 \u007f\n",
-      "RE=2.0;                       #Emitter resistor, k\u2126 \n",
-      "VBE=0.7;                      #Base-emitter voltage, V\n",
-      "beta=100;                     #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 \n",
-      "\n",
-      "#Applying Kirchhoff' law along Thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IB=IE/beta,\n",
-      "IE=(E0-VBE)/(R0/beta + RE);                 #Emitter current , mA\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"The exact value of emitter current in the circuit = %.2fmA.\"%IE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The exact value of emitter current in the circuit = 2.11mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.28: Page number 223-224\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "IE=2.0;                     #Emitter current, mA\n",
-      "IB=50.0;                    #Base current, mA\n",
-      "VCC=10.0;                   #Collector supply voltage, V\n",
-      "VBE=0.2;                    #Base-emitter voltage, V\n",
-      "R2=10.0;                    #Resistor R2, k\u2126\n",
-      "RE=1.0;                     #Emitter resistance, k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law from the base to the emitter resistor,\n",
-      "V2=VBE+IE*RE;                   #Voltage at base terminal, V\n",
-      "I2=V2/R2;                       #Current through the resistor R2, mA\n",
-      "I1=I2+IB/1000;                  #Current through the resistor R2, mA\n",
-      "V1=VCC-V2;                      #Voltage drop across the resistor R2\n",
-      "R1=V1/I1;                       #Resistor R1, k\u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The value of the resistor R1=%.2f k\u2126.\"%R1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of the resistor R1=28.89 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.30 :Page number 225-226\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=8.0;                  #Collector supply voltage, V\n",
-      "RB=360.0;                 #Base resistor, k\u2126\n",
-      "RC=2.0;                   #Collector resistor, k\u2126\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "IC_max=VCC/RC;              #Maximum collector current, mA\n",
-      "VCE_max=VCC;                #Maximum collector voltage, V\n",
-      "\n",
-      "#Operating point\n",
-      "#Applying Kirchhoff's law along the input circuit\n",
-      "IB=(VCC-VBE)/RB;                    #Base current, mA\n",
-      "IC=beta*IB;                         #Collector current, mA\n",
-      "\n",
-      "#Kirchhoff' law along the output circuit\n",
-      "VCE=VCC-IC*RC;                      #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VCE=3.94V, is approximately half of VCC=8V \n",
-        " therefore it is mid-point biased.\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.31: page number 226\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage, V\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "beta=50.0;                     #Base current amplification factor\n",
-      "R1=12.0;                     #Resistor R1, k\u2126 \n",
-      "R2=2.7;                      #Resistor R2, k\u2126 \n",
-      "RC=620.0;                    #Collector resistor, \u2126 \n",
-      "RE=180.0;                    #Emitter resistor, \u2126\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Voltage divder rule across R1 and R2\n",
-      "V2=round((VCC*R2)/(R1+R2),2);            #Voltage drop across resistor R2, V\n",
-      "IE=round(((V2-VBE)/RE)*1000,2);                 #Emitter current, mA\n",
-      "IC=IE;                          #Collector current(Approximately equal to emitter current), mA\n",
-      "print(\"IC~IE=%.2fmA.\"%IC);\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-(IC/1000)*(RC+RE);                 #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IC~IE=6.33mA.\n",
-        "VCE=4.94V, is approximately half of VCC=10V \n",
-        " therefore it is mid-point biased.\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.32 : Page number 227\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=10.0;                    #Collector supply voltage, V\n",
-      "IC=10.0;                     #Collector  current, mA   \n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "R1=1.5;                      #Resistor R1, k\u2126 \n",
-      "R2=680.0;                    #Resistor R2, \u2126  \n",
-      "RC=260.0;                    #Collector resistor, \u2126 \n",
-      "RE=240.0;                    #Emitter resistor, \u2126 \n",
-      "beta_min=100;                #Minimum value of base current amplification factor\n",
-      "beta_max=400;                #Maximum value of base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Voltage divder rule across R1 and R2\n",
-      "V2=round((VCC*R2/1000)/(R1+R2/1000),2);               #Voltage drop across resistor R2, V\n",
-      "IE=round((V2-VBE)/(RE/1000),0);                         #OHM' LAW, Emitter current, mA\n",
-      "IC=IE;                                  #Collector current(approx. equal to emitter current),mA\n",
-      "beta_avg=sqrt(beta_min*beta_max);       #Average value of base current amplification factor\n",
-      "IB=IE/(beta_avg +1);                    #Base current, mA\n",
-      "IB=IB*1000;                             #Base current, \ud835\udf07A\n",
-      "\n",
-      "#Results\n",
-      "print(\"Base current= %.2f \ud835\udf07A\"%IB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current= 49.75 \ud835\udf07A\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.33 : Page number 227-228\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VEE=12.0;                    #Emitter supply voltage, V\n",
-      "RC=1.5;                     #Collector resistor, k\u2126\n",
-      "RB=120.0;                   #Base resistor k\u2126\n",
-      "RE=510.0;                   #Emitter resistor, \u2126                 \n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=60.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law,\n",
-      "#IB*RB - VBE - IE*RE +VEE=0\n",
-      "#Since IE~IC and IC=beta*IB,\n",
-      "IB=(VEE-VBE)/(RB + beta*RE/1000);           #Base current , mA\n",
-      "IC=round(beta*IB,1);                                 #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along output circuit,\n",
-      "VCE=VEE-IC*(RC + RE/1000);                        #Collector-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA.\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 2.96V and IC=4.5mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.34 : Page number 228-229\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "VEE=9.0;                    #Emitter supply voltage, V\n",
-      "RC=1.2;                     #Collector resistor, k\u2126\n",
-      "RB=100.0;                   #Base resistor ,k\u2126\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=45.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law,\n",
-      "#IB*RB + VBE=VEE\n",
-      "#Since IE~IC and IC=beta*IB,\n",
-      "IB=round((VEE-VBE)/RB,3);           #Base current , mA\n",
-      "IC=floor(beta*IB*100)/100;                                 #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along output circuit,\n",
-      "VCE=VEE-IC*RC;                        #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.2fmA.\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 4.52V and IC=3.73mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 68
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.35 : Page number 229\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=16.0;                   #Collector supply voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "IC=1.0;                     #Collector current, mA\n",
-      "VCE=6.0;                    #Collector-emitter voltage, V\n",
-      "beta=150.0;                 #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#For a good design, VE=VCC/10;\n",
-      "VE=VCC/10;                  #Emitter terminal's voltage, V\n",
-      "#OHM's Law\n",
-      "#And, taking IE~IC\n",
-      "RE=VE/IC;                   #Emitter resistor, k\u2126\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law alog output circuit:\n",
-      "#VCC=IC*RC + VCE + VE\n",
-      "RC=(VCC-VCE-VE)/IC;                       #Collector resistor, k\u2126\n",
-      "V2=VE+VBE;                                #Voltage drop across resistor R2,V\n",
-      "#From the relation I1=10*IB\n",
-      "R2=(beta*RE)/10;                          #Resistor R2, kilo ohm\n",
-      "\n",
-      "#From voltage divider rule across R1 and R2,\n",
-      "#V2=(VCC*R2)/(R1+R2)\n",
-      "R1=(VCC-V2)*R2/V2;                          #Resistor R1, k\u2126            \n",
-      "\n",
-      "#Results\n",
-      "print(\"RE=%.1f k\u2126 , RC=%.1f k\u2126, R1=%.0f k\u2126 and R2=%d k\u2126.\"%(RE,RC,R1,R2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RE=1.6 k\u2126 , RC=8.4 k\u2126, R1=143 k\u2126 and R2=24 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 69
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.36 : Page number 230-231\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ICBO=5.0;                          #Collector to base leakage current, microampere\n",
-      "beta=40.0;                         #Base current amplification factor\n",
-      "IC_zero_signal=2.0;                #Zero signal collector current, mA\n",
-      "op_temp=25.0;                      #operating temperature, degree celsius\n",
-      "temp_risen=55.0;                   #Temperature risen, degree celsius\n",
-      "temp_ICBO_doubles=10.0;            #Temperature after which ICBO doubles, degree celsius\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "ICEO=(beta+1)*ICBO;                 #Collector to emitter leakage current, microampere\n",
-      "\n",
-      "#(ii)\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;  #Number of times ICBO doubles\n",
-      "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled;                        #Final value of collector to base leakage current, microampere\n",
-      "ICEO_final=ICBO_final*(beta + 1);                                       #Final value of collector to emitter leakage current, microampere\n",
-      "\n",
-      "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal;                    #Zero signal collector current at 55 degree celius\n",
-      "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal;           #Percentage change in zero signal collector current\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The percentage change in the zero signal collector current=%.0f%%. \"%change)\n",
-      "\n",
-      "#(iii)\n",
-      "#For the silicon transistor\n",
-      "ICBO=0.1;                          #Collector to base leakage current, microampere\n",
-      "\n",
-      "ICEO=(beta+1)*ICBO;                 #Collector to emitter leakage current, microampere\n",
-      "\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;  #Number of times ICBO doubles\n",
-      "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled;                        #Final value of collector to base leakage current, microampere\n",
-      "ICEO_final=ICBO_final*(beta + 1);                                       #Final value of collector to emitter leakage current, microampere\n",
-      "\n",
-      "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal;                    #Zero signal collector current at 55 degree celius\n",
-      "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal;           #Percentage change in zero signal collector current\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) The percentage change in the zero signal collector current=%.1f%%. \"%change)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The percentage change in the zero signal collector current=82%. \n",
-        "(ii) The percentage change in the zero signal collector current=1.6%. \n"
-       ]
-      }
-     ],
-     "prompt_number": 70
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.37 : Page number 231\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ICBO=0.02                          #Collector to base leakage current, \ud835\udf07A\n",
-      "alpha=0.99;                        #Current amplification factor\n",
-      "IE=1.0;                            #Emitter current, mA\n",
-      "op_temp=27.0;                      #operating temperature, degree celsius\n",
-      "temp_risen=57.0;                   #Temperature risen, degree celsius\n",
-      "temp_ICBO_doubles=6.0;             #Temperature after which ICBO doubles, degree celsius\n",
-      "\n",
-      "#Calculations\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;       #Number of times ICBO doubles\n",
-      "ICBO_55=ICBO*2**Number_of_times_ICBO_doubled;                                #collector to base leakage current at 55 degree celsius, \ud835\udf07A\n",
-      "IC=alpha*IE + ICBO_55/1000;                                                     #Collector current, mA\n",
-      "IB=IE-IC;                                                                    #Base current, mA\n",
-      "IB=IB*1000;                                                                  #Base current,\ud835\udf07A\n",
-      "\n",
-      "#Results\n",
-      "print(\"Base current at 57 degree celsius=%.1f \ud835\udf07A \"%IB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current at 57 degree celsius=9.4 \ud835\udf07A \n"
-       ]
-      }
-     ],
-     "prompt_number": 71
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_5.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_5.ipynb
deleted file mode 100755
index e8168ab8..00000000
--- a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_5.ipynb
+++ /dev/null
@@ -1,1907 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1e5463fc2e8c67a2f9099cf3f6c078bc1c9dccccd63589a75ad6ce5024fe6432"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "CHAPTER 9 : TRANSISTOR BIASING"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.1: Page number 195-196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "V_CC=6.0;                     #Collector supply voltage\n",
-      "R_C=2.5;                      #Collector load in k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "#For faithful amplification Vce (collector-emitter voltage)> 1V for Si transistor.\n",
-      "V_CE_max=1;                                   #Maximum allowed collector-emitter voltage for faithful amplification, in V.\n",
-      "V_Rc_max=V_CC-V_CE_max;                       #maximum voltage drop across collector load in V.\n",
-      "I_C_max=V_Rc_max/R_C;                         #Maximum allowed collector current in mA\n",
-      "\n",
-      "#(ii)\n",
-      "IC_min_zero_signal=I_C_max/2;                            #Minimum zero signal collector current in mA\n",
-      "\n",
-      "#Results\n",
-      "print(\"The maximum allowed collector current during application of signal for faithful amplification = %d mA.\"%I_C_max);\n",
-      "print(\"The minimum zero signal collector current required = %d mA.\"%IC_min_zero_signal);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum allowed collector current during application of signal for faithful amplification = 2 mA.\n",
-        "The minimum zero signal collector current required = 1 mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.2: Page number 196\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=13.0;                     #Collector supply voltage in V\n",
-      "V_knee=1.0;                   #Knee voltage in V\n",
-      "R_C=4.0;                      #Collector load  in k\u2126\n",
-      "rate_IC_VBE=5.0;              #Rate of change of collector current IC with base-emitter voltage VBE in mA/V.\n",
-      "beta=100.0;                   #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V_Rc_max=VCC-V_knee;                    #Maximum allowed voltage across collector load in V\n",
-      "I_C_max=V_Rc_max/R_C;                   #Maximum allowed collector current in mA\n",
-      "I_B_max=I_C_max/beta;                   #Maximum base current in mA\n",
-      "I_B_max=I_B_max*1000;                   #Maximum base current in \ud835\udf07A\n",
-      "\n",
-      "V_B_max=I_C_max/rate_IC_VBE;            #Maximum base voltage signal in V\n",
-      "V_B_max=V_B_max*1000;                   #Maximum base voltage signal in mV\n",
-      "\n",
-      "#Results\n",
-      "print(\"Maximum base current =%d \ud835\udf07A.\"%I_B_max);\n",
-      "print(\"Maximum input signal voltage =%d mV.\"%V_B_max);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum base current =30 \ud835\udf07A.\n",
-        "Maximum input signal voltage =600 mV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.3: Page number 200-201"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=9.0;                  #Colector supply voltage in V\n",
-      "VBB=2.0;                  #Base supply voltage in V\n",
-      "R_B=100.0;                #Base resistor's resistance in k\u2126\n",
-      "R_C=2.0;                  #Collector load in k\u2126\n",
-      "beta=50.0;                #base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Case (i):\n",
-      "\n",
-      "#Applying Kirchhoff's law to the input circuit\n",
-      "#We get, IB*RB +VBE =VBB.\n",
-      "#Neglecting the small base-emitter voltage, we get:\n",
-      "I_B=VBB/R_B;                #Base current in mA\n",
-      "I_C=beta*I_B;               #Collector current in mA\n",
-      "\n",
-      "print(\"Collector current = %dmA\"%I_C);\n",
-      "\n",
-      "#Applying Kirchhoff's law to the output ciruit\n",
-      "#We get, IC*RC + VCE= VCC.\n",
-      "#From the above equation, we get:\n",
-      "V_CE=VCC-I_C*R_C;               #Collector emitter voltage in V\n",
-      "\n",
-      "print(\"Collector emitter voltage =%dV.\"%V_CE);\n",
-      "\n",
-      "\n",
-      "#Case (ii):\n",
-      "\n",
-      "R_B=50.0;\n",
-      "I_B=VBB/R_B;\n",
-      "I_C=beta*I_B;\n",
-      "V_CE=VCC  - I_C*R_C;\n",
-      "\n",
-      "print(\"The new operating point for base resistor RB=50 k\u2126 is, VCE=%dV and IC=%dmA.\"%(V_CE,I_C));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector current = 1mA\n",
-        "Collector emitter voltage =7V.\n",
-        "The new operating point for base resistor RB=50 k\u2126 is, VCE=5V and IC=2mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.4: Page number  201-202"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#variable declaration\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "VCC=6.0;                  #Collector suply voltagein V\n",
-      "VBE=0.7                   #Base emitter voltage in V\n",
-      "R_B=530.0;                #Base resistor's resistance in k\u2126 .\n",
-      "R_C=2.0;                  #Collector resistor's resistance in k\u2126 .\n",
-      "\n",
-      "#Calculation\n",
-      "#D.C load line equation : VCE=VCC-IC*RC;\n",
-      "#Calculating maximum VCE  ,by IC=0;\n",
-      "I_C_Vce_max=0;                      #Collector current for maximum collector-emitter voltage, in mA\n",
-      "VCE_max=VCC;-I_C_Vce_max*R_C;       #Maximum collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Calculating maximum collector current IC,by VCE=0;\n",
-      "V_CE_IC_max=0;             #Collector-emitter voltage for maximum collector current, in V         \n",
-      "I_C_max=(VCC-V_CE_IC_max)/R_C;        #Maximum collector current in mA\n",
-      "\n",
-      "\n",
-      "#Operating point:\n",
-      "#For input circuit, applying Kirchhoff's law, We get,\n",
-      "#VCC=IB*RB  + VBE.\n",
-      "#From the above equation,\n",
-      "IB=(VCC-VBE)/R_B;                #Base current in mA\n",
-      "IC=beta*IB;                    #Collector current\n",
-      "\n",
-      "#From the output circuit, applying Kirchhoff's law, we get:\n",
-      "VCE=VCC-IC*R_C;                 #Collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Stability factor\n",
-      "SF=beta+1;              \n",
-      "\n",
-      "#Result\n",
-      "print(\"Operating point: VCE= %dV and IC=%d mA\"%(VCE,IC));\n",
-      "print(\"Stability factor= %d.\"%SF);\n",
-      "\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,10])\n",
-      "limit.set_ylim([0,5])\n",
-      "VCE=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC=[((VCC-i)/(R_C)) for i in (VCE[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE,IC);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point: VCE= 4V and IC=1 mA\n",
-        "Stability factor= 101.\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8eea67df10>"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.5: Page number 202"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage in V\n",
-      "beta=100.0;              #base current amplification factor\n",
-      "I_C_zero_signal=1.0;      #zero signal collector current in mA\n",
-      "VBE=0.3;                  #Base-emitter voltage of Ge transistor in V\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#Case(i)\n",
-      "I_B_zero_signal=I_C_zero_signal/beta;               #Zero signal base current in  mA\n",
-      "\n",
-      "#applying the Kirchhoff's law along input circuit:\n",
-      "#We get, VCC=IB*RB +VBE\n",
-      "#From the above equation we get,\n",
-      "R_B=(VCC-VBE)/I_B_zero_signal;              #Required base resistor's resistance in k\u2126\n",
-      "\n",
-      "print(\"Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = %d k\u2126\"%R_B);\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Case(ii)\n",
-      "beta=50;\n",
-      "I_B=(VCC-VBE)/R_B;                  #Base current of another transistor with beta=50, in mA\n",
-      "I_C_zero_signal=beta*I_B;           #Zero signal collector current for beta=50 , in mA\n",
-      "\n",
-      "print(\"The new value of zero signal collector current =%.1fmA\"%I_C_zero_signal);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = 1170 k\u2126\n",
-        "The new value of zero signal collector current =0.5mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.6:Page number 202-203"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Varaible declaration\n",
-      "VCC=10.0;                     #Collector supply voltage in V\n",
-      "VBE=0;                        #Base emitter voltage in V(considering itas zero due to it's small value)\n",
-      "R_B=1.0;                      #Base resistor's resistance in M\u2126\n",
-      "R_C=2.0;                      #Collector resistor's resistance in k\u2126                  \n",
-      "R_E=1.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#using Kirchhoff's law in the input circuit, we get:\n",
-      "#VCC=IB*RB +VBE +IE*RE\n",
-      "#Since, IE=(beta +1)*I_B\n",
-      "#From the above equation we get:\n",
-      "I_B=round((VCC-VBE)/((beta + 1)*R_E + R_B*1000),4);           #Base current in mA\n",
-      "I_C=round(beta*I_B,2);                                   #Collector current in mA\n",
-      "I_E=I_B+I_C;                                    #Emitter current in mA\n",
-      "\n",
-      "#Result\n",
-      "print(\"Base current =%.4f mA\"%I_B);\n",
-      "print(\"Collector current =%.2f mA\"%I_C);\n",
-      "print(\"Emitter current =%.3f mA\"%I_E);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current =0.0091 mA\n",
-        "Collector current =0.91 mA\n",
-        "Emitter current =0.919 mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.7: Page number 203-204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=8.0;                      #Collector-emitter voltage at operating point in V\n",
-      "IC=2.0;                       #Colector current at operating point in mA\n",
-      "VCC=15.0;                     #Collector supply voltagein V\n",
-      "beta=100.0;                   #base current amplification factor\n",
-      "VBE=0.6;                    #base emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC=VCE+IC*RC.\n",
-      "#So, from above equation we get:\n",
-      "RC=(VCC-VCE)/IC;                 #Collector resistor's resistance in k\u2126 .\n",
-      "IB=IC/beta;                      #Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the input circuit,\n",
-      "#we get, VCC=IB*RB + VBE\n",
-      "#So, from the above equation:\n",
-      "RB=(VCC-VBE)/IB;                #Base resistor's resistance in k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector load =%.1f k\u2126 .\"%RC);\n",
-      "print(\"Base resistor=%d k\u2126 .\"%RB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector load =3.5 k\u2126 .\n",
-        "Base resistor=720 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.8: Page number 204"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                #Collector supply voltage in V\n",
-      "VBE=0.7;                 #Base-emitter voltage in V\n",
-      "RB=100.0;                #Base resistor's resistance in k\u2126\n",
-      "RC=560.0;                #Collector resistor's resistance in \u2126\n",
-      "beta_25=100.0;           #base current amplification factor at 25 degree celsius\n",
-      "beta_75=150.0;           #base current amplification factor at 25 degree celsius\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit, we get\n",
-      "#VCC=IB*RB+VBE\n",
-      "IB=(VCC-VBE)/RB;                #Base current at 25 degree celsius, in mA\n",
-      "\n",
-      "\n",
-      "#For temperature 25 degree celsius\n",
-      "IC_25=beta_25*IB;                  #Collector current at 25 degree celsius, in mA\n",
-      "\n",
-      "\n",
-      "#Applying Kirchhoff's alw at the output circuit,\n",
-      "#we get: VCC=IC*RC + VCE\n",
-      "#From the above equation,\n",
-      "VCE_25=round(VCC-(IC_25/1000)*RC,2);                #Collector emitter voltage  at 25 degree celsius, in V\n",
-      "\n",
-      "\n",
-      "#For temperature 75 degree celsius\n",
-      "IC_75=round(beta_75*IB,0);                  #Collector current at 75 degree celsius, in mA\n",
-      "\n",
-      "#Applying Kirchhoff's alw at the output circuit,\n",
-      "#we get: VCC=IC*RC + VCE\n",
-      "#From the above equation,\n",
-      "VCE_75=round(VCC-(IC_75/1000)*RC,2);                #Collector emitter voltage at 75 degree celsius, in V\n",
-      "\n",
-      "\n",
-      "change_IC=(IC_75-IC_25)*100.0/IC_25;        #percentage change in collector current\n",
-      "change_VCE=(VCE_75-VCE_25)*100.0/VCE_25;    #Percentage change in collector-emitter voltage \n",
-      "\n",
-      "#Results\n",
-      "print(\"The percentage change in collector current =%d%%\"%change_IC);\n",
-      "print(\"The percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage change in collector current =50%\n",
-        "The percentage change in collector-emitter voltage =-56.3%\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.10: Page number 205"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE_max=20.0;             #Maximum collector-emitter voltage in V\n",
-      "VBE=0.7;                  #Base-emitter voltage in V\n",
-      "IC_max=8.0;               #Maximum collector current in mA\n",
-      "IB=40.0;                  #Base current in microampere\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#During cut off state the collector-emitter voltage is maximum and equal to collector supply voltage\n",
-      "VCC=VCE_max;                #Collector supply voltage in V\n",
-      "\n",
-      "#Maximum collector current IC_max=collector supply voltage(VCC)/collector load(RC)\n",
-      "#Collector load(RC)=VCC*IC_max\n",
-      "RC=VCC/IC_max;                  #Collector load in k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC=IB*RB +VBE.\n",
-      "#From the above equation, we get:\n",
-      "RB=(VCC-VBE)/(IB/1000);    #Base resistor's resistance in k\u2126 .\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector supply voltage = %dV\"%VCC);\n",
-      "print(\"Collector load=%.1f k\u2126 .\"%RC);\n",
-      "print(\"Base resistor's resistance=%.1f k\u2126 .\"%RB);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector supply voltage = 20V\n",
-        "Collector load=2.5 k\u2126 .\n",
-        "Base resistor's resistance=482.5 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.12: Page number 208"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage in V\n",
-      "VEE=-20.0;                     #Emitter supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=4.7;                       #Collector resistor's resistance in k\u2126\n",
-      "RE=10.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "VBE=0.7;                      #Base-emitter voltage in V\n",
-      "beta=85.0;                    #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n",
-      "#we get,IB*RB  +IE*RE +VBE -VEE=0.\n",
-      "#Since IB=IC/beta and IC~IE,\n",
-      "#(IE/beta)*RB + IE*RE + VBE + VEE =0.\n",
-      "IE=(-VEE-VBE)/(RE + RB/beta);                        #Emitter current in mA\n",
-      "IC=IE;                                              #Collector current (approximately equal to emitter current) in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law from VCC till collector terminal,\n",
-      "#we get, VCC - IC*RC =VC\n",
-      "VC=VCC-IC*RC;                               #voltage at collector terminal in V\n",
-      "\n",
-      "#Applying Kirchhoff's law from emitter terminal to VEE\n",
-      "#we get, VE -IE*RE =VEE\n",
-      "VE=VEE + IE*RE;                             #Voltage at emitter treminal in V\n",
-      "\n",
-      "VCE=VC-VE;                                  #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The collector current = %.2f mA\"%IC);\n",
-      "print(\"The emitter current = %.2f mA\"%IE);\n",
-      "print(\"The voltage at collector terminal = %.1f V\"%VC);\n",
-      "print(\"The collector-emitter voltage = %.1f V\"%VCE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The collector current = 1.73 mA\n",
-        "The emitter current = 1.73 mA\n",
-        "The voltage at collector terminal = 11.9 V\n",
-        "The collector-emitter voltage = 14.6 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.13: Page number 208-209\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                     #Collector supply voltage in V\n",
-      "VEE=-20.0;                     #Emitter supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=4.7;                       #Collector resistor's resistance in k\u2126\n",
-      "RE=10.0;                      #Emitter resistor's resistance in k\u2126\n",
-      "beta1=85.0;                   #Base current amplification factor for case 1       \n",
-      "beta2=100.0;                  #Base current amplification factor for case 1\n",
-      "VBE_1=0.7;                    #Base emitter voltage for case 1 in V\n",
-      "VBE_2=0.6;                    #Base emitter voltage for case 2 in V\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#For beta=85 and VBE=0.7,\n",
-      "#As calculated in the previous question,\n",
-      "IC_1=1.73;                          #Collector current in mA.\n",
-      "VCE_1=14.6;                           #Collector-emitter voltage in V.\n",
-      "\n",
-      "\n",
-      "#For case (ii)\n",
-      "#beta=100 and VBE=0.6\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n",
-      "#we get,IB*RB  +IE*RE +VBE -VEE=0.\n",
-      "#Since IB=IC/beta and IC~IE,\n",
-      "#(IE/beta)*RB + IE*RE + VBE +VEE =0.\n",
-      "IE_2=round((-VEE-VBE_2)/(RE + RB/beta2),2);                        #Emitter current in mA\n",
-      "IC_2=IE_2;                                                #Collector current (approximately equal to emitter current) in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law from VCC till collector terminal,\n",
-      "#we get, VCC - IC*RC =VC\n",
-      "VC=round(VCC-IC_2*RC,1);                               #voltage at collector terminal in V\n",
-      "\n",
-      "#Applying Kirchhoff's law from emitter terminal to VEE\n",
-      "#we get, VE -IE*RE =VEE\n",
-      "VE=round(VEE + IE_2*RE,1);                             #Voltage at emitter treminal in V\n",
-      "\n",
-      "VCE_2=VC-VE;                                #Collector-emitter voltage in V\n",
-      "\n",
-      "\n",
-      "change_IC= (IC_2-IC_1)*100/IC_1;                    #%age change in collector current\n",
-      "\n",
-      "change_VCE=(VCE_2-VCE_1)*100/VCE_2;                 #%age change in collector-emitter voltage\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Percentage change in collector current =%.1f%%\"%change_IC);\n",
-      "print(\"Percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage change in collector current =1.7%\n",
-        "Percentage change in collector-emitter voltage =-3.5%\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.14: Page number 210\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                 #Collector supply voltage in V\n",
-      "VBE=0.7                   #Base-emitter voltage in V\n",
-      "RB=100.0;                 #Base resistor's resistance in k\u2126\n",
-      "RC=1.0;                   #Collector resistor's resistance in k\u2126\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC -IC*RC -IB*RB -VBE=0.\n",
-      "#since IC= beta*IB,\n",
-      "#We get,\n",
-      "IB=(VCC-VBE)/(RB + beta*RC);            #Base current in mA\n",
-      "IC=beta*IB;                             #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC=0.\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*RC;                          #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point : VCE=10.35V and IC=9.65mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.15: Page number 210-211\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                         #Collector supply voltage in V\n",
-      "VBE=0.3;                          #Base emitter voltage in V\n",
-      "IC=1.0;                           #Collector current in mA\n",
-      "VCE=8.0;                          #Collector emitter voltage in V\n",
-      "beta=100.0;                         #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#Case(i)\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-IC*RC-VCE=0.\n",
-      "#from the above equation we get,\n",
-      "RC=(VCC-VCE)/IC;             #Collector load in kilo ohm\n",
-      "IB=IC/beta;                  #Base current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit\n",
-      "#we get, VCC-VBE-(beta*IB*RC)-IB*RB=0.\n",
-      "#From the above equation we get,\n",
-      "RB=round((VCC-VBE-beta*IB*RC)/IB,0);         #Base resistor's resistance in k\u2126\n",
-      "\n",
-      "#Results\n",
-      "print(\"The resistance value of base resistor=%d k\u2126 and collector load= %d k\u2126.\"%(RB,RC));\n",
-      "\n",
-      "#Case(ii)\n",
-      "\n",
-      "beta=50;\n",
-      "\n",
-      "#Applying Kirchhoff's law along input circuit,\n",
-      "#we get, VCC -IC*RC -IB*RB -VBE=0.\n",
-      "#since IC= beta*IB,\n",
-      "#We get,\n",
-      "IB=(VCC-VBE)/(RB + beta*RC);            #Base current in mA\n",
-      "IC=beta*IB;                             #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC=0.\n",
-      "#From the above equation,\n",
-      "VCE=round(VCC-IC*RC,1);                          #Collector emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.1fV and IC=%.1fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The resistance value of base resistor=770 k\u2126 and collector load= 4 k\u2126.\n",
-        "The operating point : VCE=9.6V and IC=0.6mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.16 : Page number 211"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCE=2.0;                      #Collector-emitter voltage at operating point in V\n",
-      "VBE=0.7;                      #Base-emitter voltage in V            \n",
-      "IC=1.0;                       #Collector current at operating point in mA\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "IB=IC/beta;                     #Base current in mA\n",
-      "\n",
-      "#As, VCE=VCB +VBE\n",
-      "#we get,\n",
-      "VCB=VCE-VBE;                    #Collector-base voltage in V\n",
-      "RB=VCB/IB;                      #Base resistor's resistance in k\u2126\n",
-      "\n",
-      "#Results\n",
-      "print(\"Value of base resistor's resistance=%d k\u2126.\"%RB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Value of base resistor's resistance=130 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.17 : Page number 211-212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage in V\n",
-      "VBE=0.7                   #Base-emitter voltage in V\n",
-      "RB=400.0;                 #Base resistor's resistance in k\u2126\n",
-      "RC=4.0;                   #Collector resistor's resistance in k\u2126\n",
-      "RE=1.0;                   #Emitter resistor's resistance in k\u2126\n",
-      "beta=100.0;               #Base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along outut circuit,\n",
-      "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n",
-      "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n",
-      "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n",
-      "IC=(VCC-VBE)/(RB/beta + RC + RE);       #Collector current current in mA.\n",
-      "IE=IC;                                                #Emitter current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC -IE*RE=0. (IE~IC)\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*(RC+RE);                          #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The operating point : VCE=%.1fV and IC=%.2fmA.\"%(VCE,IC));\n",
-      "\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The operating point : VCE=5.7V and IC=1.26mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.18 : Page number 212"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage in V\n",
-      "RB=100.0;                     #Base resistor's resistance in k\u2126\n",
-      "RC=10.0;                      #Collector resistor's resistance in k\u2126\n",
-      "RE=0;                         #Emitter resistor's  resistance in k\u2126\n",
-      "VBE=0.7;                      #Base-emitter voltage in V\n",
-      "beta=100.0;                   #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along outut circuit,\n",
-      "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n",
-      "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n",
-      "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n",
-      "IC=(VCC-VBE)/(RC +RB/beta + RE);                #Collector current in mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit,\n",
-      "#we get, VCC-VCE - IC*RC =0. (IE~IC)\n",
-      "#From the above equation,\n",
-      "VCE=VCC-IC*RC;                                  #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"The d.c bias values are: VCE=%.2fV and IC=%.3fmA\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The d.c bias values are: VCE=1.55V and IC=0.845mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.19: Page number 214-215\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import matplotlib.pyplot as plt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage in V\n",
-      "R1=10.0;                            #Resistor R1's resistance in k\u2126\n",
-      "R2=5.0;                             #Resistor R2's resistance in k\u2126\n",
-      "RC=1.0;                             #Collector resistor's resistance in k\u2126              \n",
-      "RE=2.0;                             #Emitter resistor's resistance in k\u2126\n",
-      "VBE=0.7;                            #Base-emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law along output circuit\n",
-      "#VCE=VCC-IC*(RC+RE);\n",
-      "#IC=0, for VCE_max\n",
-      "VCE_max=VCC;                        #Maximum collector-emitter voltage in V\n",
-      "#VCE=0, for IC_max\n",
-      "IC_max=VCC/(RC+RE);                 #Maximum collector current in mA\n",
-      "\n",
-      "#Operating point\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage across R2 resistor V\n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "IC=IE;                              #Collector current(Approx. equal to emitter current) in mA\n",
-      "VCE=VCC-IC*(RC+RE);                 #Collector-emitter voltage in V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n",
-      "print(\"Collector current at operating point = %.2fmA\"%IC);\n",
-      "\n",
-      "#plot\n",
-      "limit = plt.gca()\n",
-      "limit.set_xlim([0,20])\n",
-      "limit.set_ylim([0,6])\n",
-      "VCE=[i for i in range(0,(int)(VCC+1))];          #Plot variable for V_CE\n",
-      "IC=[((VCC-i)/(RC+RE)) for i in (VCE[:])];      #Plot variable for I_C\n",
-      "\n",
-      "p=plot(VCE,IC);\n",
-      "xlabel(\"VCE(V)\");\n",
-      "ylabel(\"IC(mA)\");\n",
-      "title(\"d.c load line\");\n",
-      "show(p);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector-emitter voltage at operating point=8.55V\n",
-        "Collector current at operating point = 2.15mA\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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hKB07Mu+WiAStzuwdHBwwZswYrF69GiNHjgQADBgwAGfPnjXP4pzZi65xtPM//wPMny/s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\nAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x7f8eeacf1ad0>"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.20: Page number 215-216\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                           #Collector supply voltage in V\n",
-      "R1=10.0;                            #Resistor R1's resistance in k\u2126 .\n",
-      "R2=5.0;                             #Resistor R2's resistance in k\u2126 .\n",
-      "RC=1.0;                             #Collector resistor's resistance in k\u2126 .              \n",
-      "RE=2.0;                             #Emitter resistor's resistance in k\u2126 .\n",
-      "VBE=0.7;                            #Base-emitter voltage in V\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's Theorem for replacing circuit consisting of VCC,R1,R2\n",
-      "E0=(VCC*R2)/(R1+R2);                #Thevenin's voltage in V\n",
-      "R0=(R1*R2)/(R1+R2);                 #Thevenin's equivalent resistance in k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "#IC=(E0-VBE)/(R0/beta +RE);\n",
-      "IC=(E0-VBE)/RE;                         #(Since R0/beta << RE) collector current in mA\n",
-      "VCE=VCC-IC*(RC+RE);                     #Collector emitter voltage in V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n",
-      "print(\"Collector current at operating point = %.2fmA\"%IC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Collector-emitter voltage at operating point=8.55V\n",
-        "Collector current at operating point = 2.15mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.21: Page number 216-217\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=12.0;                           #Collector  supply voltage in V\n",
-      "RE=1.0;                             #Emitter resistor, k\u2126 .\n",
-      "R1=50.0;                            #Resistor R1, k\u2126 .\n",
-      "R2=10.0;                            #Resistor R2, k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "VBE=0.1;                            #Base-emitter voltage in V\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V \n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "\n",
-      "print(\"(i)Emitter current= %.1fmA\"%IE);\n",
-      "\n",
-      "#(ii)\n",
-      "VBE=0.3;                            #Base-emitter voltage in V\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V\n",
-      "IE=(V2-VBE)/RE;                     #Emitter current in mA\n",
-      "\n",
-      "print(\"(ii)Emitter current= %.1fmA\"%IE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Emitter current= 1.9mA\n",
-        "(ii)Emitter current= 1.7mA\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.22: Page number 217\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=20.0;                   #Collector supply voltage, V\n",
-      "R1=10.0;                    #Resistor R1, k\u2126\n",
-      "R2=10.0;                    #Resistor R2, k\u2126 .\n",
-      "RC=1.0;                     #Collector resistor, k\u2126 .\n",
-      "RE=5.0;                     #Emitter resistor, k\u2126 .\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "V2=(VCC*R2)/(R1+R2);                #Voltage drop across resistor R2, V\n",
-      "\n",
-      "#Applying kirchhoff's law from base terminal to emitter resistor\n",
-      "#V2=VBE+IE*RE\n",
-      "#VBE is neglected due to its small value\n",
-      "\n",
-      "IE=V2/RE;                     #Emitter current in mA\n",
-      "IC=IE;                              #Collector current (approx. equal to emitter current), mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along output circuit\n",
-      "VCE=VCC-IC*(RC+RE);                 #Collector-emitter voltage , V\n",
-      "VC=VCC-IC*RC;                       #Voltage at collector terminal,V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Emitter current =%dmA\"%IE);\n",
-      "print(\"Collector-emitter voltage=%dV\"%VCE);\n",
-      "print(\"Collector terminal's voltage=%dV\"%VC);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Emitter current =2mA\n",
-        "Collector-emitter voltage=8V\n",
-        "Collector terminal's voltage=18V\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.23: Page number 219-220\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=12.0;                 #Collector supply voltage, V\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "beta=50;                  #Base current amplification factor\n",
-      "R1=150;                   #Resistor R1, k\u2126 .\n",
-      "R2=100;                   #Resistor R2, k\u2126 .\n",
-      "RC=4.7;                   #Collector resistor, k\u2126 .\n",
-      "RE=2.2;                   #Emitter resistor, k\u2126  .\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "IB=round((E0-VBE)/(R0+beta*RE),3);              #Base current in mA\n",
-      "IC=round(beta*IB,1);                             #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-IC*(RC+RE);                     #Collector-emitter voltage, V\n",
-      "\n",
-      "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE);               #Stability factor\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n",
-      "print(\"Stability factor=%.1f\"%S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 3.72V and IC=1.2mA\n",
-        "Stability factor=18.4\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.24 : Page number 220\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "VCC=15.0;                 #Collector supply voltage, V\n",
-      "VBE=0.7;                  #Base-emitter voltage , V\n",
-      "beta=100.0;                 #Base current amplification factor\n",
-      "R1=6.0;                     #Resistor R1, k\u2126 .\n",
-      "R2=3.0;                     #Resistor R2, k\u2126 .\n",
-      "RC=470.0;                   #Collector resistor, \u2126.\n",
-      "RE=1.0;                     #Emitter resistor, k\u2126 .\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 .\n",
-      "\n",
-      "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IC=beta*IB\n",
-      "IB=round((E0-VBE)/(R0+beta*RE),3);              #Base current in mA\n",
-      "IC=round(beta*IB,1);                             #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-IC*(RC/1000+RE);                     #Collector-emitter voltage, V\n",
-      "\n",
-      "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE);               #Stability factor\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n",
-      "print(\"Stability factor=%.2f\"%S);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 8.83V and IC=4.2mA\n",
-        "Stability factor=2.94\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.25 : Page number 221-222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Varaible declaration\n",
-      "VCC=9;                    #Collector supply voltage, V\n",
-      "VCE=3;                    #Collector-emitter voltage, V\n",
-      "VBE=0.3;                    #Base-emitter voltage in V\n",
-      "RC=2.2;                     #Collector resistor , k\u2126 .\n",
-      "IC=2;                     #Collector current, mA\n",
-      "beta=50.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "IB=IC/beta;                 #Base current in mA\n",
-      "\n",
-      "#According to given relation, I1=10*IB\n",
-      "I1=IB*10;                   #Current through the resistor R1, mA\n",
-      "\n",
-      "#I1=VCC/(R1+R2), .'s LAW\n",
-      "R1_R2_sum=VCC/I1;               #Sum of the resistor's R1 and R2, k\u2126 (OHM'S LAW).\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "#VCC=IC*RC+VCE+IE*RE\n",
-      "#IC~IE\n",
-      "RE=(VCC-IC*RC-VCE)/IC;      #Emitter resistor, k\u2126 .\n",
-      "RE=round(RE*1000,0);                 #Emittter resistor, \u2126 .\n",
-      "\n",
-      "IE=IC;                      #Emittter current(approximately equal to collector current), mA\n",
-      "VE=IE*(RE/1000);                   #Voltage at emitter terminal (OHM's LAW), V\n",
-      "V2=VBE+VE;                  #Voltage drop across resistor R2, V\n",
-      "\n",
-      "R2=V2/I1;                   #Resistor R2,(OHM's LAW), k\u2126 .\n",
-      "R1=R1_R2_sum-R2;            #Resistor R1, k\u2126 .\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"RE=%d \u2126., R1=%.2f k\u2126 . and R2=%.2f k\u2126 .\"%(RE,R1,R2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RE=800 \u2126., R1=17.75 k\u2126 . and R2=4.75 k\u2126 .\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.26 : Page number 222\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=16.0;                     #Collector supply voltage, V\n",
-      "R2=20.0;                      #Resistor R2, k\u2126\n",
-      "RE=2.0;                       #Emitter resistor, k\u2126\n",
-      "VCE=6.0;                      #Collector-emitter voltage, V\n",
-      "IC=2.0;                       #Collector current , mA\n",
-      "VBE=0.3;                      #Base-emitter voltage,V\n",
-      "alpha=0.985;                  #Current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "beta=alpha/(1-alpha);               #Base current amplificatioon factor\n",
-      "IE=IC;                              #Emitter current, mA\n",
-      "IB=IC/beta;                         #Base current, mA\n",
-      "VE=IE*RE;                           #Emitter voltage,(OHM's LAW) V\n",
-      "V2=VBE+VE;                          #Voltage drop across resistor R2,(Kirchhoff's law) V\n",
-      "V_R1=VCC-V2;                        #Voltage drop across resistor R1, V\n",
-      "I1=V2/R2;                           #Current through resistor R2 an R1,(OHM's LAW) mA\n",
-      "R1=V_R1/I1;                         #Resistor R1,(OHM's LAW) k\u2126\n",
-      "\n",
-      "V_RC=(VCC-VCE-VE);                  #Voltage across collector resistor, V\n",
-      "RC=V_RC/IC;                         #Collector resistor,(OHM's LAW) k\u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"R1=%.1f k\u2126 and RC=%d k\u2126.\"%(R1,RC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "R1=54.4 k\u2126 and RC=3 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.27 :Page number 222-223\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=15.0;                     #Collector supply voltage, V\n",
-      "R1=10.0;                      #Resistor R1, k\u2126 \n",
-      "R2=5.0;                       #Resistor R2, k\u2126 \n",
-      "RC=1.0;                       #Collector resistor, k\u2126 \u007f\n",
-      "RE=2.0;                       #Emitter resistor, k\u2126 \n",
-      "VBE=0.7;                      #Base-emitter voltage, V\n",
-      "beta=100;                     #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n",
-      "E0=(VCC*R2)/(R1+R2);        #Thevenin's voltage, V\n",
-      "R0=(R1*R2)/(R1+R2);         #Thevenin's resistance, k\u2126 \n",
-      "\n",
-      "#Applying Kirchhoff' law along Thevenin's equivalent circuit,\n",
-      "#E0=IB*R0+VBE+IE*RE;\n",
-      "#Since IE~IC and IB=IE/beta,\n",
-      "IE=(E0-VBE)/(R0/beta + RE);                 #Emitter current , mA\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "print(\"The exact value of emitter current in the circuit = %.2fmA.\"%IE);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The exact value of emitter current in the circuit = 2.11mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2.28: Page number 223-224\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "IE=2.0;                     #Emitter current, mA\n",
-      "IB=50.0;                    #Base current, mA\n",
-      "VCC=10.0;                   #Collector supply voltage, V\n",
-      "VBE=0.2;                    #Base-emitter voltage, V\n",
-      "R2=10.0;                    #Resistor R2, k\u2126\n",
-      "RE=1.0;                     #Emitter resistance, k\u2126\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's law from the base to the emitter resistor,\n",
-      "V2=VBE+IE*RE;                   #Voltage at base terminal, V\n",
-      "I2=V2/R2;                       #Current through the resistor R2, mA\n",
-      "I1=I2+IB/1000;                  #Current through the resistor R2, mA\n",
-      "V1=VCC-V2;                      #Voltage drop across the resistor R2\n",
-      "R1=V1/I1;                       #Resistor R1, k\u2126\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"The value of the resistor R1=%.2f k\u2126.\"%R1);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of the resistor R1=28.89 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.30 :Page number 225-226\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=8.0;                  #Collector supply voltage, V\n",
-      "RB=360.0;                 #Base resistor, k\u2126\n",
-      "RC=2.0;                   #Collector resistor, k\u2126\n",
-      "VBE=0.7;                  #Base-emitter voltage, V\n",
-      "beta=100.0;               #base current amplification factor\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "IC_max=VCC/RC;              #Maximum collector current, mA\n",
-      "VCE_max=VCC;                #Maximum collector voltage, V\n",
-      "\n",
-      "#Operating point\n",
-      "#Applying Kirchhoff's law along the input circuit\n",
-      "IB=(VCC-VBE)/RB;                    #Base current, mA\n",
-      "IC=beta*IB;                         #Collector current, mA\n",
-      "\n",
-      "#Kirchhoff' law along the output circuit\n",
-      "VCE=VCC-IC*RC;                      #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "VCE=3.94V, is approximately half of VCC=8V \n",
-        " therefore it is mid-point biased.\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.31: page number 226\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=10.0;                     #Collector supply voltage, V\n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "beta=50.0;                     #Base current amplification factor\n",
-      "R1=12.0;                     #Resistor R1, k\u2126 \n",
-      "R2=2.7;                      #Resistor R2, k\u2126 \n",
-      "RC=620.0;                    #Collector resistor, \u2126 \n",
-      "RE=180.0;                    #Emitter resistor, \u2126\n",
-      "\n",
-      "\n",
-      "#Calculations\n",
-      "#Voltage divder rule across R1 and R2\n",
-      "V2=round((VCC*R2)/(R1+R2),2);            #Voltage drop across resistor R2, V\n",
-      "IE=round(((V2-VBE)/RE)*1000,2);                 #Emitter current, mA\n",
-      "IC=IE;                          #Collector current(Approximately equal to emitter current), mA\n",
-      "print(\"IC~IE=%.2fmA.\"%IC);\n",
-      "\n",
-      "#Applying Kirchhoff's law along the output circuit\n",
-      "VCE=VCC-(IC/1000)*(RC+RE);                 #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "IC~IE=6.33mA.\n",
-        "VCE=4.94V, is approximately half of VCC=10V \n",
-        " therefore it is mid-point biased.\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.32 : Page number 227\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import sqrt\n",
-      "\n",
-      "#Variable declaration\n",
-      "VCC=10.0;                    #Collector supply voltage, V\n",
-      "IC=10.0;                     #Collector  current, mA   \n",
-      "VBE=0.7;                     #Base-emitter voltage, V\n",
-      "R1=1.5;                      #Resistor R1, k\u2126 \n",
-      "R2=680.0;                    #Resistor R2, \u2126  \n",
-      "RC=260.0;                    #Collector resistor, \u2126 \n",
-      "RE=240.0;                    #Emitter resistor, \u2126 \n",
-      "beta_min=100;                #Minimum value of base current amplification factor\n",
-      "beta_max=400;                #Maximum value of base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Voltage divder rule across R1 and R2\n",
-      "V2=round((VCC*R2/1000)/(R1+R2/1000),2);               #Voltage drop across resistor R2, V\n",
-      "IE=round((V2-VBE)/(RE/1000),0);                         #OHM' LAW, Emitter current, mA\n",
-      "IC=IE;                                  #Collector current(approx. equal to emitter current),mA\n",
-      "beta_avg=sqrt(beta_min*beta_max);       #Average value of base current amplification factor\n",
-      "IB=IE/(beta_avg +1);                    #Base current, mA\n",
-      "IB=IB*1000;                             #Base current, \ud835\udf07A\n",
-      "\n",
-      "#Results\n",
-      "print(\"Base current= %.2f \ud835\udf07A\"%IB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current= 49.75 \ud835\udf07A\n"
-       ]
-      }
-     ],
-     "prompt_number": 51
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.33 : Page number 227-228\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VEE=12.0;                    #Emitter supply voltage, V\n",
-      "RC=1.5;                     #Collector resistor, k\u2126\n",
-      "RB=120.0;                   #Base resistor k\u2126\n",
-      "RE=510.0;                   #Emitter resistor, \u2126                 \n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=60.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law,\n",
-      "#IB*RB - VBE - IE*RE +VEE=0\n",
-      "#Since IE~IC and IC=beta*IB,\n",
-      "IB=(VEE-VBE)/(RB + beta*RE/1000);           #Base current , mA\n",
-      "IC=round(beta*IB,1);                                 #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along output circuit,\n",
-      "VCE=VEE-IC*(RC + RE/1000);                        #Collector-emitter voltage, V\n",
-      "\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.1fmA.\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 2.96V and IC=4.5mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.34 : Page number 228-229\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from math import floor\n",
-      "\n",
-      "#Variable declaration\n",
-      "VEE=9.0;                    #Emitter supply voltage, V\n",
-      "RC=1.2;                     #Collector resistor, k\u2126\n",
-      "RB=100.0;                   #Base resistor ,k\u2126\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "beta=45.0;                  #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#Applying Kirchhoff's voltage law,\n",
-      "#IB*RB + VBE=VEE\n",
-      "#Since IE~IC and IC=beta*IB,\n",
-      "IB=round((VEE-VBE)/RB,3);           #Base current , mA\n",
-      "IC=floor(beta*IB*100)/100;                                 #Collector current, mA\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law along output circuit,\n",
-      "VCE=VEE-IC*RC;                        #Collector-emitter voltage, V\n",
-      "\n",
-      "#Results\n",
-      "print(\"Operating point : VCE= %.2fV and IC=%.2fmA.\"%(VCE,IC));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Operating point : VCE= 4.52V and IC=3.73mA.\n"
-       ]
-      }
-     ],
-     "prompt_number": 68
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.35 : Page number 229\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "VCC=16.0;                   #Collector supply voltage, V\n",
-      "VBE=0.7;                    #Base-emitter voltage, V\n",
-      "IC=1.0;                     #Collector current, mA\n",
-      "VCE=6.0;                    #Collector-emitter voltage, V\n",
-      "beta=150.0;                 #Base current amplification factor\n",
-      "\n",
-      "#Calculations\n",
-      "#For a good design, VE=VCC/10;\n",
-      "VE=VCC/10;                  #Emitter terminal's voltage, V\n",
-      "#OHM's Law\n",
-      "#And, taking IE~IC\n",
-      "RE=VE/IC;                   #Emitter resistor, k\u2126\n",
-      "\n",
-      "#Applying Kirchhoff's voltage law alog output circuit:\n",
-      "#VCC=IC*RC + VCE + VE\n",
-      "RC=(VCC-VCE-VE)/IC;                       #Collector resistor, k\u2126\n",
-      "V2=VE+VBE;                                #Voltage drop across resistor R2,V\n",
-      "#From the relation I1=10*IB\n",
-      "R2=(beta*RE)/10;                          #Resistor R2, kilo ohm\n",
-      "\n",
-      "#From voltage divider rule across R1 and R2,\n",
-      "#V2=(VCC*R2)/(R1+R2)\n",
-      "R1=(VCC-V2)*R2/V2;                          #Resistor R1, k\u2126            \n",
-      "\n",
-      "#Results\n",
-      "print(\"RE=%.1f k\u2126 , RC=%.1f k\u2126, R1=%.0f k\u2126 and R2=%d k\u2126.\"%(RE,RC,R1,R2));\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "RE=1.6 k\u2126 , RC=8.4 k\u2126, R1=143 k\u2126 and R2=24 k\u2126.\n"
-       ]
-      }
-     ],
-     "prompt_number": 69
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.36 : Page number 230-231\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ICBO=5.0;                          #Collector to base leakage current, microampere\n",
-      "beta=40.0;                         #Base current amplification factor\n",
-      "IC_zero_signal=2.0;                #Zero signal collector current, mA\n",
-      "op_temp=25.0;                      #operating temperature, degree celsius\n",
-      "temp_risen=55.0;                   #Temperature risen, degree celsius\n",
-      "temp_ICBO_doubles=10.0;            #Temperature after which ICBO doubles, degree celsius\n",
-      "\n",
-      "#Calculations\n",
-      "\n",
-      "#(i)\n",
-      "ICEO=(beta+1)*ICBO;                 #Collector to emitter leakage current, microampere\n",
-      "\n",
-      "#(ii)\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;  #Number of times ICBO doubles\n",
-      "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled;                        #Final value of collector to base leakage current, microampere\n",
-      "ICEO_final=ICBO_final*(beta + 1);                                       #Final value of collector to emitter leakage current, microampere\n",
-      "\n",
-      "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal;                    #Zero signal collector current at 55 degree celius\n",
-      "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal;           #Percentage change in zero signal collector current\n",
-      "\n",
-      "#Result\n",
-      "print(\"(i) The percentage change in the zero signal collector current=%.0f%%. \"%change)\n",
-      "\n",
-      "#(iii)\n",
-      "#For the silicon transistor\n",
-      "ICBO=0.1;                          #Collector to base leakage current, microampere\n",
-      "\n",
-      "ICEO=(beta+1)*ICBO;                 #Collector to emitter leakage current, microampere\n",
-      "\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;  #Number of times ICBO doubles\n",
-      "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled;                        #Final value of collector to base leakage current, microampere\n",
-      "ICEO_final=ICBO_final*(beta + 1);                                       #Final value of collector to emitter leakage current, microampere\n",
-      "\n",
-      "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal;                    #Zero signal collector current at 55 degree celius\n",
-      "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal;           #Percentage change in zero signal collector current\n",
-      "\n",
-      "\n",
-      "#Result\n",
-      "print(\"(ii) The percentage change in the zero signal collector current=%.1f%%. \"%change)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The percentage change in the zero signal collector current=82%. \n",
-        "(ii) The percentage change in the zero signal collector current=1.6%. \n"
-       ]
-      }
-     ],
-     "prompt_number": 70
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 9.37 : Page number 231\n"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration\n",
-      "ICBO=0.02                          #Collector to base leakage current, \ud835\udf07A\n",
-      "alpha=0.99;                        #Current amplification factor\n",
-      "IE=1.0;                            #Emitter current, mA\n",
-      "op_temp=27.0;                      #operating temperature, degree celsius\n",
-      "temp_risen=57.0;                   #Temperature risen, degree celsius\n",
-      "temp_ICBO_doubles=6.0;             #Temperature after which ICBO doubles, degree celsius\n",
-      "\n",
-      "#Calculations\n",
-      "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles;       #Number of times ICBO doubles\n",
-      "ICBO_55=ICBO*2**Number_of_times_ICBO_doubled;                                #collector to base leakage current at 55 degree celsius, \ud835\udf07A\n",
-      "IC=alpha*IE + ICBO_55/1000;                                                     #Collector current, mA\n",
-      "IB=IE-IC;                                                                    #Base current, mA\n",
-      "IB=IB*1000;                                                                  #Base current,\ud835\udf07A\n",
-      "\n",
-      "#Results\n",
-      "print(\"Base current at 57 degree celsius=%.1f \ud835\udf07A \"%IB);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Base current at 57 degree celsius=9.4 \ud835\udf07A \n"
-       ]
-      }
-     ],
-     "prompt_number": 71
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file