Added(A)/Deleted(D) following books

 A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/Chapter11_1.ipynb
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/Chapter12_1.ipynb
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/Chapter14_1.ipynb
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/Chapter1_1.ipynb
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/Chapter2_1.ipynb
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/Chapter4_1.ipynb
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/Chapter5_1.ipynb
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/Chapter6_1.ipynb
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/Chapter7_1.ipynb
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/Chapter9_1.ipynb
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/screenshots/11.1new.png
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/screenshots/5.1new.png
A  Modern_Electronic_Instrumentation_And_Measurement_Techniques_by_A._D._Helfrick_And_W._D._Cooper/screenshots/5.4new.png
A  sample_notebooks/ChandraShiva/CHAPTER09.ipynb
A  "sample_notebooks/Jaya PratyushaKothuri/Chapter2.ipynb"
A  sample_notebooks/TarunikaBoyapati/CHAPTER02.ipynb

3 files changed, 975 insertions, 0 deletions

diff --git a/sample_notebooks/ChandraShiva/CHAPTER09.ipynb b/sample_notebooks/ChandraShiva/CHAPTER09.ipynb
new file mode 100644
index 00000000..6549f9c6
--- /dev/null
+++ b/sample_notebooks/ChandraShiva/CHAPTER09.ipynb
@@ -0,0 +1,325 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:2fe5d303c17886b921f7fce80c636f86023d9235c3c8b0a7cac768a73d655baa"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "CHAPTER09 : NETWORK FUNCTION"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E24 - Pg 608"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "                             # EXAMPLE 9-24         PG NO-608-609\n",

+      "import math \n",

+      "L=20.;                          # INDUCTANCE\n",

+      "R=2.*L;                        # RESISTANCE\n",

+      "print '%s %.2f %s' %('i)  Resistance (R) is    =',R,'ohm');\n",

+      "Wo=math.sqrt(101.);\n",

+      "print '%s %.2f %s' %('ii)  Wo (Wo) is          =',Wo,'rad/sec');\n",

+      "Q=(Wo*L)/R;\n",

+      "print '%s %.2f' %('iii) Q  is               = ',Q);\n",

+      "BW=Wo/Q;\n",

+      "print '%s %.2f %s' %('iv) BANDWIDTH (BW) is    =  ',BW,' rad/sec  ');\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "i)  Resistance (R) is    = 40.00 ohm\n",

+        "ii)  Wo (Wo) is          = 10.05 rad/sec\n",

+        "iii) Q  is               =  5.02\n",

+        "iv) BANDWIDTH (BW) is    =   2.00  rad/sec  \n"

+       ]

+      }

+     ],

+     "prompt_number": 1

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E26 - Pg 609"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "                           # EXAMPLE 9-26            PG NO-609-610\n",

+      "C=10.** -6.;\n",

+      "X=5.*10.** 6.;\n",

+      "L=1./(C*X);\n",

+      "print '%s %.2f %s' %('i) INDUCTAR (L) is     =',L,' H  ');\n",

+      "R=10.*L;\n",

+      "print '%s %.2f %s' %('ii) Resistance (R) is  =',R,' ohm  ');\n",

+      "W=2.236*10.** 3.;\n",

+      "Q=(W*L)/R;\n",

+      "print '%s %.2f' %('iii) (Q) is            =  ',Q);\n",

+      "BW=W/Q;\n",

+      "print '%s %.2f %s' %('iv) Band Width (BW) is =',BW,' rad/sec  ');\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "i) INDUCTAR (L) is     = 0.20  H  \n",

+        "ii) Resistance (R) is  = 2.00  ohm  \n",

+        "iii) (Q) is            =   223.60\n",

+        "iv) Band Width (BW) is = 10.00  rad/sec  \n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E32 - Pg 618"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "                           # Example 9-32         PG NO 618-619\n",

+      "import math, cmath\n",

+      "P1=1-1j*50;\n",

+      "P2=1+1j*150;\n",

+      "Z1=0+0j*50;\n",

+      "I=(0.2*Z1)/(P1*P2);\n",

+      "print 'i) Current (I) is    =  ',I,'A'\n",

+      "L=5.; \n",

+      "R=10.;\n",

+      "C=2.*10.** -5.;\n",

+      "Wo=1/math.sqrt(L*C);\n",

+      "print 'ii) Wo (Wo) is    =  ',Wo,' rad/sec  '\n",

+      "Q=(Wo*L)/R;\n",

+      "print 'iii) Q (Q) is    =  ',Q;\n",

+      "BW=Wo/Q;\n",

+      "print 'ii) Band Width (BW) is    =  ',BW,' rad/sec  '\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "i) Current (I) is    =   0j A\n",

+        "ii) Wo (Wo) is    =   100.0  rad/sec  \n",

+        "iii) Q (Q) is    =   50.0\n",

+        "ii) Band Width (BW) is    =   2.0  rad/sec  \n"

+       ]

+      }

+     ],

+     "prompt_number": 3

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E37 - Pg 623"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "                             # EXAMPLE 9-37             PG NO 623-624\n",

+      "C=1./8.5;                                  # Capacitor\n",

+      "L=1./(17.*C);                              # Inductar\n",

+      "print '%s %.2f %s' %('ii) Inductar (L) is    =  ',L,'H');\n",

+      "R=2.*L;                                  # Resistance\n",

+      "print '%s %.2f %s' %('ii) Resistance (R) is  =  ',R,' ohm  ');\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "ii) Inductar (L) is    =   0.50 H\n",

+        "ii) Resistance (R) is  =   1.00  ohm  \n"

+       ]

+      }

+     ],

+     "prompt_number": 4

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E38 - Pg 624"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "                         # EXAMPLE 9-38   PG NO=624-625\n",

+      "C=1./9.;                # CAPACITOR\n",

+      "X=2.;              # R/L=X\n",

+      "Y=6-X;             # G/C\n",

+      "G=4.*C;\n",

+      "print '%s %.2f %s' %('i) G (G)   =  ',G,' ohm')\n",

+      "L=0.9;\n",

+      "R=1.8;\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "i) G (G)   =   0.44  ohm\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E46 - Pg 630"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "                           # EXAMPLE 9-46            PG NO 630-631\n",

+      "import cmath\n",

+      "ZA=5+1j*3;\n",

+      "YA=1./ZA;\n",

+      "print 'i) Admittance (YA) is    =  ',YA,' siemens  ';\n",

+      "V=100.;                       # VOLTAGE\n",

+      "IA=V*YA;\n",

+      "print 'ii) Current (IA) is    =  ',IA,' A  ';\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "i) Admittance (YA) is    =   (0.147058823529-0.0882352941176j)  siemens  \n",

+        "ii) Current (IA) is    =   (14.7058823529-8.82352941176j)  A  \n"

+       ]

+      }

+     ],

+     "prompt_number": 6

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E50 - Pg 632"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "                 # EXAMPLE 9-50            PG NO-632\n",

+      "I1=17.39-1j*4.66;         # CURRENT\n",

+      "I2=9+1j*15.68;           # CURRENT\n",

+      "I3=-1j*10.95;            # CURRENT\n",

+      "I=I1+I2+I3;\n",

+      "print 'i)CURRENT (I)   =  ',I,' A'\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "i)CURRENT (I)   =   (26.39+0.07j)  A\n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E56 - Pg 636"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "                # example 9-56   pg no-636\n",

+      "Z1=8.05+1j*2.156;             # IMPEDANCE\n",

+      "XL=2.155;\n",

+      "W=5000;\n",

+      "L=XL/W;\n",

+      "print 'i)INDUCTANCE (L)   =  ',L,' H'\n",

+      "Z2=4.166-1j*7.216;          # IMPEDANCE\n",

+      "Xc=7.216;\n",

+      "C=1/(W*Xc);\n",

+      "print 'ii)CAPACITOR (C)   =  ',C,' F'\n",

+      "D=11.708;             # DIAMETER\n",

+      "XL1=12.81;\n",

+      "L1=XL1/W;\n",

+      "print 'i) INDUCTANCE (L1)   =  ',L1,' H'\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "i)INDUCTANCE (L)   =   0.000431  H\n",

+        "ii)CAPACITOR (C)   =   2.77161862528e-05  F\n",

+        "i) INDUCTANCE (L1)   =   0.002562  H\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file
diff --git a/sample_notebooks/Jaya PratyushaKothuri/Chapter2.ipynb b/sample_notebooks/Jaya PratyushaKothuri/Chapter2.ipynb
new file mode 100644
index 00000000..7b0498ca
--- /dev/null
+++ b/sample_notebooks/Jaya PratyushaKothuri/Chapter2.ipynb
@@ -0,0 +1,231 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2 - Ionization"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1: pg 22"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the breakdown strength of air for 0.1mm air gap is (kV/cm.) =  43.447\n",
+      "\n",
+      "the breakdown strength of air for 20 cm air gap is (kV/cm.) =  25.58\n"
+     ]
+    }
+   ],
+   "source": [
+    "#example 2.1\n",
+    "#calculation of breakdown strength of air\n",
+    "\n",
+    "#given data\n",
+    "d1=0.1#length(in cm) of the gap\n",
+    "d2=20#length(in cm) of the gap\n",
+    "\n",
+    "#calculation\n",
+    "#from equation of breakdown strength\n",
+    "E1=24.22+(6.08/(d1**(1./2)))#for gap d1\n",
+    "E2=24.22+(6.08/(d2**(1./2)))#for gap d2\n",
+    "#results\n",
+    "print 'the breakdown strength of air for 0.1mm air gap is (kV/cm.) = ',round(E1,3)\n",
+    "print '\\nthe breakdown strength of air for 20 cm air gap is (kV/cm.) = ',round(E2,3)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2: pg 23"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Townsend primary ioniztion coefficient is (/cm torr) =  7.675\n"
+     ]
+    }
+   ],
+   "source": [
+    "#example 2.2\n",
+    "#calculation of Townsend primary ionization coefficient\n",
+    "from math import log\n",
+    "#given data\n",
+    "d1=0.4#gap distance(in cm)\n",
+    "d2=0.1#gap distance(in cm)\n",
+    "I1=5.5*10**-8#value of current(in A)\n",
+    "I2=5.5*10**-9#value of current(in A)\n",
+    "\n",
+    "#calculation\n",
+    "#from equation of current at anode I=I0*exp(alpha*d)\n",
+    "alpha=(log(I1/I2))*(1/(d1-d2))\n",
+    "#results\n",
+    "print 'Townsend primary ioniztion coefficient is (/cm torr) = ',round(alpha,3)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3: pg 25"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the value of Townsend secondary ionization coefficient is  9.994e-04\n"
+     ]
+    }
+   ],
+   "source": [
+    "#example 2.3\n",
+    "#calculation of Townsend secondary ionization coefficient\n",
+    "from math import exp\n",
+    "#given data\n",
+    "d=0.9#gap distance(in cm)\n",
+    "alpha=7.676#value of alpha\n",
+    "\n",
+    "#calculation\n",
+    "#from condition of breakdown.....gama*exp(alpha*d)=1\n",
+    "gama=1/(exp(d*alpha))\n",
+    "#results\n",
+    "print '%s %.3e' %('the value of Townsend secondary ionization coefficient is ',gama)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4: pg 26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the value of breakdown voltage of the spark gap is (V) =  5626.0\n",
+      "The answer is a bit different due to rounding off error in textbook\n"
+     ]
+    }
+   ],
+   "source": [
+    "#example 2.4\n",
+    "#calculation of breakdown voltage of a spark gap\n",
+    "from math import log\n",
+    "#given data\n",
+    "A=15#value of A(in per cm)\n",
+    "B=360#value of B(in per cm)\n",
+    "d=0.1#spark gap(in cm)\n",
+    "gama=1.5*10**-4#value of gama\n",
+    "p=760#value of pressure of gas(in torr)\n",
+    "\n",
+    "#calculation\n",
+    "#from equation of breakdown voltage\n",
+    "V=(B*p*d)/(log((A*p*d)/(log(1+(1/gama)))))\n",
+    "\n",
+    "#results\n",
+    "print 'the value of breakdown voltage of the spark gap is (V) = ',round(V)\n",
+    "print 'The answer is a bit different due to rounding off error in textbook'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5: pg 26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the value of minimum spark over voltage is (V) =  481.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#example 2.5\n",
+    "#calculation of minimum spark over voltage\n",
+    "from math import log\n",
+    "#given data\n",
+    "A=15#value of A(in per cm)\n",
+    "B=360#value of B(in per cm)\n",
+    "gama=10**-4#value of gama\n",
+    "e=2.178#value of constant\n",
+    "\n",
+    "#calculation\n",
+    "Vbmin=(B*e/A)*(log(1+(1/gama)))\n",
+    "\n",
+    "#results\n",
+    "print 'the value of minimum spark over voltage is (V) = ',round(Vbmin)\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.11"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/TarunikaBoyapati/CHAPTER02.ipynb b/sample_notebooks/TarunikaBoyapati/CHAPTER02.ipynb
new file mode 100644
index 00000000..809dc7f6
--- /dev/null
+++ b/sample_notebooks/TarunikaBoyapati/CHAPTER02.ipynb
@@ -0,0 +1,419 @@
+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:896619f98735f380fd7764ca92373e979d96248275ce12cd21f86086d05b2351"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "CHAPTER02 : THE CIRCUIT ELEMENTS"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E1a - Pg 21"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#1a\n",

+      "V = 1.; #  voltage supply \n",

+      "R = 10.; #  resistance in ohms \n",

+      "I = V/R # current flowing through R\n",

+      "print '%s' %(\"a)\")\n",

+      "print '%s %.f' %(\"voltage across the resistor (in volts)=\",V)\n",

+      "print '%s %.2f' %(\"current flowing through the resistor (in amps) =\",I)\n",

+      "\n",

+      "#1b\n",

+      "V = 1.; #  voltage supply \n",

+      "R1 = 10.; #  first resistance in ohms \n",

+      "R2 = 5.; # resistance of the second resistor \n",

+      "Vr1 = V * (R1/(R1 + R2)); # voltage across R1\n",

+      "Vr2 = V - Vr1; # voltage across R2\n",

+      "Ir = Vr1/R1; # current flowing through R\n",

+      "print '%s' %(\"b)\")\n",

+      "print '%s %.2f' %(\"voltage across the first resistor (in volts)=\",Vr1)\n",

+      "print '%s %.2f' %(\"voltage across the second resistor (in volts)=\",Vr2)\n",

+      "print '%s %.2f' %(\"current flowing through the resistor (in amps) =\",Ir)\n",

+      "\n",

+      "#1c\n",

+      "# c - a\n",

+      "R1 = 10.; #  first resistance in ohms\n",

+      "R2 = 10.;\n",

+      "I = 1.; #  current source \n",

+      "V = I*R1; #  voltage across R\n",

+      "print '%s' %(\"c - a)\")\n",

+      "print '%s %.f' %(\"voltage across the resistor (in volts)=\",V)\n",

+      "print '%s %.f' %(\"current flowing through the resistor (in amps) =\",I)\n",

+      "# c - b\n",

+      "Vr1 = I*R1; #  voltage across R1\n",

+      "Vr2 = I*R2; # voltage across R2\n",

+      "Vr=Vr1+Vr2;\n",

+      "print '%s' %(\"c - b)\")\n",

+      "print '%s %.f' %(\"voltage across the resistor (in volts)=\",Vr)\n",

+      "print '%s %.f' %(\"current flowing through the resistor (in amps) =\",I)\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "a)\n",

+        "voltage across the resistor (in volts)= 1\n",

+        "current flowing through the resistor (in amps) = 0.10\n",

+        "b)\n",

+        "voltage across the first resistor (in volts)= 0.67\n",

+        "voltage across the second resistor (in volts)= 0.33\n",

+        "current flowing through the resistor (in amps) = 0.07\n",

+        "c - a)\n",

+        "voltage across the resistor (in volts)= 10\n",

+        "current flowing through the resistor (in amps) = 1\n",

+        "c - b)\n",

+        "voltage across the resistor (in volts)= 20\n",

+        "current flowing through the resistor (in amps) = 1\n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E2 - Pg 25"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "R = 100.; #  resistance in ohms\n",

+      "I = 0.3; #  current in amps \n",

+      "P = I**2 * R; #  power \n",

+      "# power specification of the resistors available in the stock \n",

+      "Pa = 5.;\n",

+      "Pb = 7.5;\n",

+      "Pc = 10.;\n",

+      "\n",

+      "if Pa > P :\n",

+      "    print '%s' %(\"we should select resistor a\")\n",

+      "if Pb > P :\n",

+      "    print '%s' %(\"we should select resistor b\")\n",

+      "if Pc > P :\n",

+      "    print '%s' %(\"we should select resistor c\")"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "we should select resistor c\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E3 - Pg 26"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "L = 1.; # length of the copper wire in meters\n",

+      "A = 1. * 10.**-4.; #  cross sectional area of the wire in meter square \n",

+      "rho = 1.724 * 10.**-8.; #  resistivity of copper in ohm meter\n",

+      "R = rho*L / A; #  resistance of the wire in ohm \n",

+      "\n",

+      "print '%s %.2e' %(\"resistance of the wire (in ohms)=\",R) "

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "resistance of the wire (in ohms)= 1.72e-04\n"

+       ]

+      }

+     ],

+     "prompt_number": 9

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E4 - Pg 27"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# 1 inches = 0.0254meters\n",

+      "# 1 foot = 0.3048 meters\n",

+      "import math \n",

+      "d = 0.1*0.0254; #  diameter of the wire in meters\n",

+      "L = 10.*0.3048; # length of the wire in meters \n",

+      "rho = 1.724*10.**-8.; #  resistivity of the wire in ohm-meter\n",

+      "A = math.pi*(d/2.)**2.; #  cross sectional area of the wire \n",

+      "R = rho*L/A; # resistance of the wire in ohm \n",

+      "print '%s %.2f' %(\"resistance of the wire (in ohm)=\",R)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "resistance of the wire (in ohm)= 0.01\n"

+       ]

+      }

+     ],

+     "prompt_number": 10

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E5 - Pg 29"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "%matplotlib inline\n",

+      "import math\n",

+      "import numpy as np\n",

+      "from matplotlib import pyplot\n",

+      "L = 0.1; #  inductance of the coil in henry \n",

+      "t1= np.linspace(0,0.1, num=101)\n",

+      "t2= np.linspace(0.101,0.3, num=201)\n",

+      "t3= np.linspace(0.301,0.6,num=301)\n",

+      "t4= np.linspace(0.601,0.7,num=101)\n",

+      "t5= np.linspace(0.701,0.9,num=201)\n",

+      "# current variation as a function of time \n",

+      "i1 = 100.*t1;\n",

+      "i2 = (-50.*t2) + 15.;\n",

+      "i3 = np.zeros(301)\n",

+      "for i in range(0,301):\n",

+      "\ti3[i] = -100.*math.sin(math.pi*(t3[i]-0.3)/0.3);\n",

+      "\n",

+      "i4 = (100.*t4) - 60.;\n",

+      "i5 = (-50.*t5) + 45.;\n",

+      "\n",

+      "t = ([t1,t2,t3,t4,t5]);\n",

+      "i = ([i1,i2,i3,i4,i5]);\n",

+      "pyplot.plot(t1, i1);\n",

+      "pyplot.plot(t2, i2);\n",

+      "pyplot.plot(t3, i3);\n",

+      "pyplot.plot(t4, i4);\n",

+      "pyplot.plot(t5, i5);\n",

+      "\n",

+      "dt = 0.001;\n",

+      "di1 = np.diff(i1);\n",

+      "di2 = np.diff(i2);\n",

+      "di3 = np.diff(i3);\n",

+      "di4 = np.diff(i4);\n",

+      "di5 = np.diff(i5);\n",

+      "V1 =np.array((L/dt)*di1); # voltage drop appearing across the inductor terminals\n",

+      "V2 =np.array((L/dt)*di2); # voltage drop appearing across the inductor terminals\n",

+      "V3 =np.array((L/dt)*di3); # voltage drop appearing across the inductor terminals\n",

+      "V4 = np.array((L/dt)*di4); # voltage drop appearing across the inductor terminals\n",

+      "V5 = np.array((L/dt)*di5); # voltage drop appearing across the inductor terminals\n",

+      "print(V2)\n",

+      "Tv = np.linspace(0,0.899,num=900);\n",

+      "V = []\n",

+      "V.extend(V1)\n",

+      "V.extend(V2)\n",

+      "V.extend(V3)\n",

+      "V.extend(V4)\n",

+      "V.extend(V5)\n",

+      "print(len(V))\n",

+      "pyplot.plot(Tv, V)\n",

+      "pyplot.show();"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "[-4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",

+        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975]\n",

+        "900\n"

+       ]

+      },

+      {

+       "metadata": {},

+       "output_type": "display_data",

+       "png": 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+       "text": [

+        "<matplotlib.figure.Figure at 0x5cdc0b0>"

+       ]

+      }

+     ],

+     "prompt_number": 11

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E7 - Pg 31"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "# a\n",

+      "Ri = 1.; \n",

+      "Rf = 39.;\n",

+      "A = 10.**5.; # open loop gain of the op-amp\n",

+      "G = A/(1. + (A*Ri/(Ri+Rf))); # actual voltage gain of the circuit \n",

+      "print '%s' %(\"a\")\n",

+      "print '%s %.2f' %(\"actual voltage of the circuit =\",G)\n",

+      "\n",

+      "# b\n",

+      "G1 = 1 + (Rf/Ri); #  voltage gain of the circuit with infinite open loop gain\n",

+      "print '%s' %(\"b\")\n",

+      "print '%s %.f' %(\"for ideal case the voltage gain =\",G1)\n",

+      "\n",

+      "# c\n",

+      "er = ((G1 - G)/G)*100.; # percent error \n",

+      "print '%s' %(\"c\")\n",

+      "print '%s %.2f' %(\"percent error of the ideal value compared to the actual value=\",er)"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "a\n",

+        "actual voltage of the circuit = 39.98\n",

+        "b\n",

+        "for ideal case the voltage gain = 40\n",

+        "c\n",

+        "percent error of the ideal value compared to the actual value= 0.04\n"

+       ]

+      }

+     ],

+     "prompt_number": 12

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E8 - Pg 33"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "G = 4.; #  voltage gain of the circuit \n",

+      "r = G -1.; #  ratio of the resistances in the non-inverting op-amp circuit\n",

+      "print '%s %.2f' %(\"Rf/Ri =\",r)\n",

+      "# Result:\n",

+      "# A suitable choice for R1 is 10K, Hence Rf = 30K\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Rf/Ri = 3.00\n"

+       ]

+      }

+     ],

+     "prompt_number": 15

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E9 - Pg 34"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "G = 4.;\n",

+      "r = G; #  ratio of the resistances in the inverting op-amp circuit\n",

+      "print '%s %.f' %(\"Rf/Ri\",r)\n",

+      "# Result;\n",

+      "# A suitable choice for Rf=30K and R1=7.5K\n",

+      "# therefore input resistance R1 = 7.5K\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Rf/Ri 4\n"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file