From 37d315828bbfc0f5cabee669d2b9dd8cd17b5154 Mon Sep 17 00:00:00 2001
From: hardythe1
Date: Wed, 17 Jun 2015 11:14:34 +0530
Subject: add books

---
 .../chapter1.ipynb                                 | 1596 ++++++++++++++++++++
 1 file changed, 1596 insertions(+)
 create mode 100755 Electronic_Devices_and_Circuits_By_I.JNagrath/chapter1.ipynb

(limited to 'Electronic_Devices_and_Circuits_By_I.JNagrath/chapter1.ipynb')

diff --git a/Electronic_Devices_and_Circuits_By_I.JNagrath/chapter1.ipynb b/Electronic_Devices_and_Circuits_By_I.JNagrath/chapter1.ipynb
new file mode 100755
index 00000000..41def6eb
--- /dev/null
+++ b/Electronic_Devices_and_Circuits_By_I.JNagrath/chapter1.ipynb
@@ -0,0 +1,1596 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a9ed36d73f9d154839d703d9a4c0fe9425b38f6baf4e4351a575d82599eff181"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1: SEMICONDUCTORS,DIODE AND DIODE CIRCUITS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1, Page number 5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "A=6.022*10**23     #avagadro's number(/m^3)\n",
+      "d=2.7*10**6        #density of aluminium conductor(g/m^3)\n",
+      "a=26.98            # atomic weight aluminium conductor(g/g-atom)\n",
+      "D=10**4.           #current density(A/m^2)\n",
+      "e=1.6*10**-19      #electronic charge(C)\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a\n",
+      "n=A*d/a            #number of atoms(n/m^3)\n",
+      "\n",
+      "#Part b\n",
+      "u=D/(n*e)          #drift velocity (m/s)\n",
+      "\n",
+      "#Results\n",
+      "print \"number of atoms per cubic meter is \",round(n/1e+28,3),\"*10^28 /m^3\"\n",
+      "print \"drift velocity is\",round(u/1e-6,2),\"*10^-6 m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of atoms per cubic meter is  6.026 *10^28 /m^3\n",
+        "drift velocity is 1.04 *10^-6 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2, Page number 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "n=10**23        #number of electrons(n/m^3)\n",
+      "e=1.6*10**-19   #electronic charge(C) \n",
+      "u=0.4           #mobility(m^2/Vs)           \n",
+      "a=10**-7        #cross sectional area(m^2) \n",
+      "l=15*10**-2     #conductor length(m)\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a\n",
+      "G=n*e*u         #conductivity(S/m)\n",
+      "\n",
+      "#Part b\n",
+      "R=l/(a*G)       #resistance(ohm)\n",
+      "\n",
+      "#Results\n",
+      "print\"conductivity of the conductor is\",round((G/1E+3),1),\"*10**3 S/m\"\n",
+      "print\"resistance of the conductor is\",round(R,1),\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conductivity of the conductor is 6.4 *10**3 S/m\n",
+        "resistance of the conductor is 234.4 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3, Page number 9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "A=6.022*10**23   #avagadro's number\n",
+      "d=5.32*10**6     #density of Ge at 300k(g/m^3)\n",
+      "a=72.60          #atomic weight of Ge(g/g-atom)\n",
+      "e=1.6*10**-19    #electronic charge(C)\n",
+      "ni=2.4*10**19    #intrinsic concentration(electron-hole pairs/m^3)\n",
+      "un=0.39          #electron mobility(m^2/V.s)\n",
+      "up=0.19          #hole mobility(m^2/V.s)\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a\n",
+      "nA=A*d/a         #number of atoms(nA/m^3)using avagadro's law\n",
+      "x=nA/ni          #Germanium atoms/electron hole pair\n",
+      "\n",
+      "#Part b\n",
+      "g=(un+up)*e*ni   #intrinsic conductivity(S/m)\n",
+      "r=1/g            #intrinsic resistivity(ohm.m)\n",
+      "\n",
+      "#Results\n",
+      "print\"the relative concentration of Ge and electron hole pairs is\",round((x/1E9),2),\"*10^9 atoms/electron-hole pair\"\n",
+      "print\"the intrinsic resistivity of Ge is\",round(r,3),\"ohm.m\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the relative concentration of Ge and electron hole pairs is 1.84 *10^9 atoms/electron-hole pair\n",
+        "the intrinsic resistivity of Ge is 0.449 ohm.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4,Page number 13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "ni=1.5*10**16    #intrinsic concentration(electron-hole pairs/m^3)\n",
+      "n=4.99*10**28    #number of Si atoms(atoms/m^3)\n",
+      "un=0.13          #electron mobility(m^2/V.s)\n",
+      "up=0.05          #hole mobility(m^2/V.s)\n",
+      "e=1.6*10**-19    #electronic charge(c)\n",
+      "\n",
+      "#Calculation\n",
+      "#Part a\n",
+      "g=e*ni*(un+up)  #intrinsic conductivity(S/m)\n",
+      "r=1/g           #interinsic resistivity(ohm.m)\n",
+      "Nd=n/10**8      #doped silicon(atoms/m^3)=nn,majority carriers\n",
+      "pn=ni**2/Nd     #minority carrier density(holes/m^3)\n",
+      "\n",
+      "#Part b\n",
+      "k=e*un*Nd    #conductivity(S/m)\n",
+      "             #using Nd in place of nn as Nd=nn\n",
+      "rho=1/k      #resistivity(ohm.m)\n",
+      "\n",
+      "#Results\n",
+      "print\"the minority carrier density of Si is\",round(pn/1e+11,2),\"*10^11 holes/m^3\"\n",
+      "print\"the resistivity of Si is\",round((rho/1E-2),2),\"*10**-2 ohm.m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the minority carrier density of Si is 4.51 *10^11 holes/m^3\n",
+        "the resistivity of Si is 9.63 *10**-2 ohm.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5,Page number 17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Vo=0.7                  #contact potential(V)\n",
+      "Vf=0.4                  #forward biasing voltage(V)    \n",
+      "\n",
+      "#Calculation\n",
+      "x=math.exp(-20*(Vo-Vf))/math.exp(-20*Vo)   #increase in probability of majority carriers\n",
+      "\n",
+      "#Result\n",
+      "print\"increase in probability of majority carriers is\",round(x),\"times\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "increase in probability of majority carriers is 2981.0 times\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6,Page number 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "I=10    #Ge diode carries current(mA)\n",
+      "V=0.2   #forward bias voltage(V)\n",
+      "\n",
+      "#Calculation\n",
+      "#Part a\n",
+      "Is=I/(math.expm1(40*V)-1)      #reverse current(mA)\n",
+      "\n",
+      "#part b\n",
+      "I1=1*10**-3                     #current(mA)                     \n",
+      "V1=(math.log((I1/Is)+1))/40      #voltage(V)\n",
+      "I2=100*10**-3                  #current(mA) \n",
+      "V2=(math.log((I2/Is)+1))/40     #voltage(V) \n",
+      "\n",
+      "#Part c\n",
+      "Is1=4*Is                 #reverse saturation current doubles for every 10 degree celcius temp rise,so for 20 degree rise it will be 4 timese/   \n",
+      "x=37.44                  #let x=e/kT\n",
+      "I3=Is1*(math.expm1(x*V)) #current when temp doubles(mA)\n",
+      "\n",
+      "#Results\n",
+      "print\"the reverse current is\",round(Is/1e-3,3),\"mA\"  #incorrect units given in the textbook\n",
+      "print\"bias voltages are\",round(V1,3),\"V and\", round(V2,3),\"V resp\"\n",
+      "print\"Is at 20 degree is\",round(Is1/1e-3,2),\"uA and diode current at 0.2 V is\",round(I3,2),\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the reverse current is 3.357 mA\n",
+        "bias voltages are 0.007 V and 0.086 V resp\n",
+        "Is at 20 degree is 13.43 uA and diode current at 0.2 V is 23.97 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7,Page number 21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "V=3.            #Voltage(V)\n",
+      "Req=300.        #total resistance as per circuit(ohm)\n",
+      "Rfa=20           #forward resistance(ohm)    \n",
+      "Vt=0.7          #Thevinine's voltage(V)\n",
+      "Rfb=0            #forward resistance(ohm)\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a\n",
+      "I=V/Req          #current(A)\n",
+      "\n",
+      "#Part b\n",
+      "Id=(V-Vt)/Req   #diode current(mA)\n",
+      "\n",
+      "#Part c\n",
+      "Rf=20               #forward resistance(ohms) \n",
+      "Id1=(V-Vt)/(Req+Rfa) #diode current(mA)\n",
+      "\n",
+      "#Results\n",
+      "print\"current in this case is\",round(I,2),\"A\"\n",
+      "print\"diode current is\",round((Id/1E-3),2),\"mA\"\n",
+      "print\"diode current is\",round((Id1/1E-3),2),\"mA\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "current in this case is 0.01 A\n",
+        "diode current is 7.67 mA\n",
+        "diode current is 7.19 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.9,Page number 22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Vx=1.4              #voltage at point X(V) \n",
+      "Vt=0.7              #diode voltage(V)\n",
+      "Vcc=5               #cathode voltage(V)  \n",
+      "R=1                 #circuit resistance(ohm) \n",
+      "Vs=Vx-Vt            #supply voltage(V)\n",
+      "\n",
+      "#Calculations\n",
+      "I1=(Vcc-Vt-Vs)/R      #current throgh D1(mA) for 0<Vs<0.7\n",
+      "I2=0                  #current through D2 and D3\n",
+      "I1=I2=(Vcc-Vt-Vs)/R   #for Vs>0.7 as D2 and D3 conducts\n",
+      "\n",
+      "#Results\n",
+      "print\"I1 for 0<Vs<0.7 is\",I1,\"mA\"\n",
+      "print\"I2 for 0<Vs<0.7 is\",I2,\"mA\"\n",
+      "print\"I1 and I2 for Vs>0.7 is\",I1,\"mA\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "I1 for 0<Vs<0.7 is 3.6 mA\n",
+        "I2 for 0<Vs<0.7 is 0 mA\n",
+        "I1 and I2 for Vs>0.7 is 3.6 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.11,Page number 23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Vz=100          #zener voltage(V)\n",
+      "Rz=25           #diode resistance(ohm)\n",
+      "Il=0.05         #load current(A)\n",
+      "Iz=0.01         #zener diode current(A)\n",
+      "Rs=250          #supply resistance(ohm)\n",
+      "\n",
+      "#Calculations\n",
+      "Vl=Vz+(Iz*Rz)     #load voltage(V)\n",
+      "Vs=Vl+(Il+Iz)*Rs  #supply voltage(V)\n",
+      "VL=Vl*1.01        #increase in Vl(V)\n",
+      "IZ=(VL-Vz)/Rz     #increase in zener current\n",
+      "VS=Vl+(Il+IZ)*Rs  #increase in supply voltage(V)\n",
+      "Vss=(VS-Vs)/Vs    #%increase in supply voltage(V)\n",
+      "P=Il*VL           #power consumed(W) \n",
+      "\n",
+      "#Results\n",
+      "print\"load voltage is\",Vl,\"V\"\n",
+      "print\"supply voltage is\",Vs,\"V\"\n",
+      "print\"increase in supply voltage is\",VS,\"V\"\n",
+      "print\"power consumed is\",round(P,2),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " load voltage is 100.25 V\n",
+        "supply voltage is 115.25 V\n",
+        "increase in supply voltage is 125.275 V\n",
+        "power consumed is 5.06 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.12,Page number 25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Vbb=5           #bias voltage(V)\n",
+      "Rl=1            #resistance(ohm)\n",
+      "Id=4.4          #from the figure(mA)\n",
+      "\n",
+      "#Part a\n",
+      "i=Vbb/Rl        #load line intercepts the Id axis at i(mA)\n",
+      "Vl=Id*Rl        #load voltage(V)\n",
+      "\n",
+      "#Part b\n",
+      "Vd=Vbb-Vl       #diode voltage(V)\n",
+      "P=Vd*Id         #power absorbed in diode(mW)\n",
+      "\n",
+      "#Part c        \n",
+      "Ida=1.42         #diode current(mA)for 2V\n",
+      "Idb=7.35         #diode current(mA)for 8V\n",
+      "\n",
+      "#Part d\n",
+      "Idc=8.7         #diode current(mA)for Rl=0.5k ohm \n",
+      "Idd=2.2         #diode current(mA)for Rl=2k ohm\n",
+      "\n",
+      "#Results\n",
+      "print\"diode current is\",Id,\"mA and voltage across the load is\",Vl,\"V\"\n",
+      "print\"power absorbed in diode is\",P,\"mW\"\n",
+      "print\"diode current for Vbb=2V is\",Ida,\"mA\",\"and for Vbb=8V is\",Idb,\"mA\"\n",
+      "print\"diode current for Rl=0.5kohm is\",Idc,\"mA\",\"and for Rl=2kohm is\",Idd,\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "diode current is 4.4 mA and voltage across the load is 4.4 V\n",
+        "power absorbed in diode is 2.64 mW\n",
+        "diode current for Vbb=2V is 1.42 mA and for Vbb=8V is 7.35 mA\n",
+        "diode current for Rl=0.5kohm is 8.7 mA and for Rl=2kohm is 2.2 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.13,Page number 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "T=300             #temperature(k)\n",
+      "Ig=100*10**-3     #current(mA)\n",
+      "Is=1*10**-9       #current(nA)\n",
+      "x=0.0259          #x=kT/e\n",
+      "\n",
+      "#Calculations\n",
+      "Voc=x*math.log(Ig/Is+1)     #as Voc=kT/e*ln((Ig/Is)+1) where ln((Ig/Is)+1)=18.42 after solving \n",
+      "Isc=Ig\n",
+      "\n",
+      "#Result\n",
+      "print\"for a solar cell Voc is\",round(Voc,3),\"V and Isc is\",round(Isc/1E-3),\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "for a solar cell Voc is 0.477 V and Isc is 100.0 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.14,Page number 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Idc=0.1           #dc current(A)\n",
+      "Rf=0.5            #forward resistance(ohms)\n",
+      "Rl=20             #load resistance(ohm)\n",
+      "Rs=1              #secondary resistance of transformer(ohm)\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a\n",
+      "Vdc=Idc*Rl                       #dc voltage(V)\n",
+      "Vm=(math.pi/2)*(Vdc+Idc*(Rs+Rf)) #mean voltage(V)\n",
+      "Vrms=Vm/math.sqrt(2)             #rms value of voltage(V)   \n",
+      "\n",
+      "#Part b\n",
+      "Pdc=Idc**2*Rl                  #dc power supplied to the load\n",
+      "\n",
+      "#Part c\n",
+      "PIV=2*Vm                       #PIV rating for each diode(V)\n",
+      "\n",
+      "#Part d\n",
+      "Im=(math.pi/2)*Idc            #peak value of current(mA)\n",
+      "Irms=Im/math.sqrt(2)          #rms calue of current(A)\n",
+      "Pac=Irms**2*(Rs+Rf+Rl)        #ac power input(W)\n",
+      "\n",
+      "#Part e\n",
+      "eta=(Pdc/Pac)*100             #conversion efficiency\n",
+      "\n",
+      "#Part f\n",
+      "Vr=((Rs+Rf)/Rl)*100           #voltage regulation(V)\n",
+      "\n",
+      "#results\n",
+      "print\"rms value of voltage is\",round(Vrms,2),\"V\"\n",
+      "print\"dc power supplied to load is\",Pdc,\"W\"\n",
+      "print\"PIV rating for each diode\",round(PIV,2),\"V\"\n",
+      "print\"ac input power is\",round(Pac,3),\"W\"\n",
+      "print\"conversion efficiency\",round(eta,1),\"%\"\n",
+      "print\"voltage regulation\",Vr,\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rms value of voltage is 2.39 V\n",
+        "dc power supplied to load is 0.2 W\n",
+        "PIV rating for each diode 6.75 V\n",
+        "ac input power is 0.265 W\n",
+        "conversion efficiency 75.4 %\n",
+        "voltage regulation 7.5 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.15,Page number 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from scipy import integrate\n",
+      "\n",
+      "#Variable declaration\n",
+      "Vt=1                 \n",
+      "Vl=12\n",
+      "Vm=63.63                      #peak voltage(V) as Vm=sqr root of 2*45\n",
+      "Idc=8.                        #charging current(A)\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a\n",
+      "theta1=math.degrees(math.asin((Vt+Vl)/Vm))\n",
+      "theta2=180-theta1\n",
+      "Rl=((2*Vm*math.cos(theta1))-(2*(math.pi-2*theta1)*(Vt+Vl)))/(Idc*math.pi)\n",
+      "\n",
+      "#Part b\n",
+      "wt = lambda wt: (((((math.sqrt(2)*45*math.sin(wt))-(Vt+Vl))/Rl)*wt)**2)\n",
+      "integ,err = integrate.quad(wt, theta1 , theta2)\n",
+      "print integ\n",
+      "Irms = (integ/math.pi)**0.5\n",
+      "Pl=Irms**2*Rl               #power loss in resistance(W)\n",
+      "\n",
+      "#Part c\n",
+      "P=Vl*Idc                   #power supplied to battery(W)\n",
+      "\n",
+      "#results\n",
+      "print\"Resistance to be added is\",round(Rl,2),\"Ohms\"\n",
+      "print\"\",Pl\n",
+      "print\"power supplied to battery is\",P,\"W\"\n",
+      "print\"\",Irms  "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5703935.44277\n",
+        "Resistance to be added is 24.75 Ohms\n",
+        " 44936628.7032\n",
+        "power supplied to battery is 96.0 W\n",
+        " 1347.44908683\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.16,Page number 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Rf=5                    #forward resistance(ohms)\n",
+      "Vo=20                   #output voltage(V)\n",
+      "Rs=10                   #secondary resistance of transformer(ohm)\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a\n",
+      "Idc=0.1                 #dc current(A)                  \n",
+      "Vm=Vo*(math.sqrt(2))    #mean voltage(V)\n",
+      "Vdc=(2*Vm/(math.pi))-Idc*(Rs+2*Rf)  #dc voltage(V)\n",
+      "\n",
+      "#Part b\n",
+      "Idc1=0.2                #full load dc current(A)\n",
+      "Vdc2=((2*(math.sqrt(2))*Vo)/(math.pi))-Idc1*(Rs+2*Rf) #full load dc voltage(V)\n",
+      "Rl=Vdc2/Idc1              #load resistance(ohm)\n",
+      "x=((2*Rf+Rs)/Rl)*100      #% regulation \n",
+      "\n",
+      "#Part c\n",
+      "Idc=0.2                  #dc current(A)\n",
+      "Im=(math.pi)*Idc/2       #peak current(mA)\n",
+      "Ilrms=Im/math.sqrt(2)    #rms current(mA)\n",
+      "Vlrms=Ilrms*Rl           #load rms voltage(V) \n",
+      "\n",
+      "#Part d\n",
+      "Vldc=14                               #load dc voltage(V)\n",
+      "Vlacrms=math.sqrt(Vlrms**2-Vldc**2)   #rms value of ac component(V)\n",
+      "\n",
+      "#Results\n",
+      "print\"dc voltage\",round(Vdc),\"V\"\n",
+      "print\"regulation is\",round(x,2),\"%\"\n",
+      "print\"rms value of output voltage at dc load current is\",round(Vlrms,2),\"V\"\n",
+      "print\"rms value of ac component of voltage\",round(Vlacrms,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "dc voltage 16.0 V\n",
+        "regulation is 28.56 %\n",
+        "rms value of output voltage at dc load current is 15.56 V\n",
+        "rms value of ac component of voltage 6.78 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.17,Page number 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Vh=60.           #higher output voltage(V)\n",
+      "Vl=45.           #lower output voltage(V)  \n",
+      "fz=50.           #frequency(Hz)\n",
+      "Vr=15.           #peak to peak ripple voltage(V)\n",
+      "Rl=600.          #resistance(ohms)\n",
+      "     \n",
+      "#Calculations\n",
+      "Vldc=(Vh+Vl)/2   #avg load dc voltage(V) as voltage drops from 60 to 45\n",
+      "Idc=Vldc/Rl      #dc current(A)\n",
+      "T=1/fz           #discharging time(ms)\n",
+      "C=(Idc*T)/Vr     #linear discharge rate(uF)\n",
+      "C1=C*2           #new capacitance(uF)\n",
+      "'''\n",
+      "         CVr(p-p)       234Vr(p-p)*10^3\n",
+      "Idc = -------------- = ---------------    ----(1)\n",
+      "          T                   20\n",
+      "    \n",
+      "      60+[60-Vr(p-p)     120-Vr(p-p)\n",
+      "Idc = --------------- = ------------*1000   ----(2)\n",
+      "         2Rl               2*600\n",
+      "         \n",
+      "Equating equations 1 & 2, we get,    \n",
+      "'''\n",
+      "\n",
+      "Vr1 = (20*120*1000)/(1200*254)\n",
+      "Idc1=(Vh-(Vr1/2))/Rl               #dc load current(mA)\n",
+      "\n",
+      "#Results\n",
+      "print\"value of capacitance is\",round(C/1E-6),\"uF\" \n",
+      "print\"Vr1 is\",Vr1,\"V\" \n",
+      "print\"dc load current Idc is\",round(Idc1/1E-3),\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of capacitance is 117.0 uF\n",
+        "Vr1 is 7 V\n",
+        "dc load current Idc is 95.0 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.18,Page number 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Vdc=30                          #dc voltage(V)\n",
+      "V1=220                          #source voltage(V)\n",
+      "f=50                            #frequency(Hz)\n",
+      "Rl=1000                         #load resistance(k ohms)\n",
+      "\n",
+      "#Calculations\n",
+      "C=100/f*Rl                      #as Vdc/Vr=100\n",
+      "Vm=Vdc+(Vr/2)                   #peak voltage(V)\n",
+      "V2=Vm/(math.sqrt(2))            #secondary voltage(V)\n",
+      "r=V1/V2                         #transformer turn ratio\n",
+      "\n",
+      "#Results\n",
+      "print\"capacitor filtor is\",C,\"uF\"\n",
+      "print\"transformer turn ratio is\",round(r,2),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "capacitor filtor is 2000 uF\n",
+        "transformer turn ratio is 10.37 \n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.19,Page number 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Idc=60*10**-3                      #dc current(A)\n",
+      "Vm=60                              #peak volage(V)\n",
+      "f=50                               #frequency(Hz)\n",
+      "C=120*10**-6                       #capacitance(F)\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a\n",
+      "Vrms=Idc/(4*(math.sqrt(3))*f*C*Vm) #rms voltage(V)\n",
+      "Vr=2*(math.sqrt(3))*Vrms           #ripple factor(V)\n",
+      "\n",
+      "#Part b\n",
+      "Vdc=Vm-(Vr/2)                      #by simplifying\n",
+      "\n",
+      "#Part c\n",
+      "r=(Vrms/Vdc)*100                   #ripple factor\n",
+      "\n",
+      "#Results\n",
+      "print\"ripple factor is\",round(Vr,3),\"Vdc\"\n",
+      "print\"dc voltage is\",round(Vdc),\"V\"\n",
+      "print\"ripple factor\",round(r,4),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ripple factor is 0.083 Vdc\n",
+        "dc voltage is 60.0 V\n",
+        "ripple factor 0.0401 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.20,Page number 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a\n",
+      "'''     200*1.141      4\n",
+      "v1(t)=-------------(1- - cos628t) \n",
+      "          3.14         3\n",
+      "       200*1.141      800*1.141 \n",
+      "v2(t)=----------- - ------------  cos(628t+<(V2/V1))\n",
+      "         3.14          3*3.14\n",
+      "\n",
+      "V2/V1|w=0 =0.8;V2/V1|w=628 =6.43*10^-4 <V2/V1|w=628 =180\n",
+      "v2(t)=72.02+0.0538 cos628t\n",
+      "'''\n",
+      "#Part b\n",
+      "vrms=0.0538\n",
+      "vdc=math.sqrt(2)*72.02\n",
+      "r=vrms/vdc\n",
+      "\n",
+      "#Results\n",
+      "print\"ripple factor is\",round((r/1E-4),2),\"*10**-4\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ripple factor is 5.28 *10**-4\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.24,Page number 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Vz=2                    #zener voltage(V)\n",
+      "r1=10                   #resistance after reducing circuit by thevinin(ohms)\n",
+      "r2=20                   #resistance after reducing circuit by thevinin(ohms)\n",
+      "V1=7.5                  #voltage after circuit reduction(V)\n",
+      "V2=15                   #voltage after circuit reduction(V)\n",
+      "Rz=100/3                #zener resistance(ohms)\n",
+      "\n",
+      "#Calculations\n",
+      "Vab=V2-(((V2-V1)/(r1+r2))*r2)         #thevinin voltage at ab(V)\n",
+      "Rth=(Vab*r2)/(Vab+r2)                 #thevinin resistance at ab(ohms)\n",
+      "Vd=Vab-Vz                             #diode voltage(V)\n",
+      "Id=Vd/(Rth+Rz)                        #diode current(A)\n",
+      "\n",
+      "#Results\n",
+      "print\"diode current is\",round(Id,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "diode current is 0.2 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.25,Page number 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Vd=0.7                      #diode voltage(V)\n",
+      "Ro=18                       #output resistance(k ohms)\n",
+      "R1=2                        #diode1 resistance(k ohms)\n",
+      "R2=2                        #diode2 resistance(k ohms) \n",
+      "\n",
+      "#Calculations\n",
+      "#Part a\n",
+      "V1=10                       #voltage to D1(V)\n",
+      "V2=0                        #voltage to D2(V)\n",
+      "Io=(V1-Vd)/(R1+Ro)          #output current(mA)   \n",
+      "Vo=Io*Ro                    #output voltage(V)\n",
+      "\n",
+      "#Part b\n",
+      "V1=5                        #voltage to D1(V)\n",
+      "V2=0                        #voltage to D2(V)\n",
+      "Io=(V1-Vd)/(R1+Ro)          #output current(mA)   \n",
+      "Vo1=Io*Ro                   #output voltage(V)\n",
+      "\n",
+      "#Part c\n",
+      "V1=10                      #voltage to D1(V)\n",
+      "V2=5                       #voltage to D2(V)\n",
+      "Vo=8.37                    #as D1 only conducts,so,Vo is same as in part a\n",
+      "Vd1=V2-Vo                  #assume D1 conducts\n",
+      "Vo2=8.37                   #D2 does not conduct as as Vd1 is negative\n",
+      "\n",
+      "#Part d\n",
+      "V1=V2=5                  #voltage to D1 and D2(V) \n",
+      "Id1=(V1-Vd-Vo)/2         #diode1 current(mA)   \n",
+      "Io=Vo/Ro                 #output current(mA)   \n",
+      "Vo3=(Ro*(V1-Vd))/(Ro+1)  #output voltage(V)\n",
+      "\n",
+      "print\"a)output voltage is\",Vo,\"V\"\n",
+      "print\"b)output voltage is\",Vo1,\"V\"\n",
+      "print\"c)output voltage is\",Vo2,\"V\"\n",
+      "print\"d)output voltage is\",round(Vo3,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)output voltage is 8.37 V\n",
+        "b)output voltage is 3.87 V\n",
+        "c)output voltage is 8.37 V\n",
+        "d)output voltage is 4.07 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 71
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.26,Page number 62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Vs=10.                       #supply voltage(V)  \n",
+      "Rs=1                         #supply resistane(ohm)\n",
+      "Vl=10.                       #load voltage(V)\n",
+      "Vi=50.                       #nput voltage(V)\n",
+      "Iz=32                        #zener diode current(mA)\n",
+      "Is=40                        #supply current(mA) \n",
+      "\n",
+      "#Calculations\n",
+      "#Part a (Rl is min when Iz=0)  \n",
+      "Is=(Vi-Vs)/Rs               #source current(mA)       \n",
+      "Rlmin=Vl/(Vi-Vs)            #load resistance minimum(ohm) \n",
+      "\n",
+      "#Part b(Rl is maximum when Iz=32 mA)  \n",
+      "Il=(Is-Iz)*10**-3          #load current(A) \n",
+      "Rlmax=Vl/Il                #maximum load resistance(k ohms)\n",
+      "P=Vl*Iz                    #max diode wattage consumed(mW)\n",
+      "\n",
+      "#Results\n",
+      "print\"Range of Rl is\",round((Rlmin/1E-3),3),\"ohm\",\"to\", round((Rlmax/1E+3),2),\"k ohm\" \n",
+      "print \"Il = \",(Il/1E-3),\"*10**-3 A\"\n",
+      "print\"max power consumed is\",P,\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Range of Rl is 250.0 ohm to 1.25 k ohm\n",
+        "Il =  8.0 *10**-3 A\n",
+        "max power consumed is 320.0 mW\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.27,Page number 63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Vz=20           #zener voltage(V)\n",
+      "Izmax=50        #maximum zener current(mA)\n",
+      "Rz=0            #zener resistance(ohms)\n",
+      "Rl=2.           #load resistance(ohm)\n",
+      "Vl=20.          #as Vz=Vl(V)\n",
+      "Rs=0.25         #source resistance(k ohms)\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a\n",
+      "Il=Vl/Rl           #load current(mA)      \n",
+      "Vsmin=(Rs+Rl)*Il   #as Iz is floating so Iz=0\n",
+      "\n",
+      "#Part b\n",
+      "Is=Izmax+Il        #source current(mA)   \n",
+      "Vsmax=Vz+(Is*Rs)   #maximum source voltage(V)\n",
+      "\n",
+      "#Results\n",
+      "print\"Vsmin\",round(Vsmin,1),\"V\"\n",
+      "print\"Range of input voltage is\",round(Vsmin,1),\"to\", Vsmax,\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Il is 10.0 mA\n",
+        "Vsmin 22.5 V\n",
+        "Range of input voltage is 22.5 to 35.0 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.28,Page number 63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Ilmax=100                #load maximum current(mA)\n",
+      "Ilmin=0                  #load minimum current(mA)\n",
+      "Rz=0.05                  #zener diode resistance(ohms)\n",
+      "Rs=10.                    #source resistance(k ohms)\n",
+      "Vl=16.015                #load voltage(V)\n",
+      "Vl1=16.                  #nominal load voltage(V) \n",
+      "Vs=20                    #source voltage(V)\n",
+      "Vz=16                    #zener diode voltage(V)\n",
+      "\n",
+      "#Calculations\n",
+      "#Case 1 (i)\n",
+      "Iz=(Vl-Vl1)/Rz           #zener current(mA)\n",
+      "Is=Iz+Ilmax              #supply current(A)\n",
+      "\n",
+      "#Case 1 (ii)\n",
+      "Is1=(Vs-Vz)/(Rs+Rz)           #supply current(mA)\n",
+      "Vl2=Vl1+(Is1*Rz)              #voltage(V)  \n",
+      "Vr=((Vl2-Vl)/Vl1)*100        #voltage regulation\n",
+      "\n",
+      "#Case 2 (i)\n",
+      "Vs=18                        #supply voltage(V)\n",
+      "Ilmax=0.1                    #load current max(A)\n",
+      "Vl=16.005                    #load voltage(V)\n",
+      "Iz=(Vl-Vl1)/Rz               #zener current(mA)\n",
+      "Is2=Ilmax+Iz                 #supply current(A)\n",
+      "\n",
+      "#Case 2 (ii)\n",
+      "Ilmin=0\n",
+      "Iz1=(Vs-Vl1)/(Rs+Rz)        #minimum diode current(mA) \n",
+      "Vl=Vl1+(Iz*Rz)              #load voltage at Ilmin(V)          \n",
+      "\n",
+      "#Part a\n",
+      "#Variable declaration\n",
+      "Is=0.4                  #supply current(A)\n",
+      "Vs=20                    #supply voltage(V)   \n",
+      "Vl=16.015               #load voltage(V)\n",
+      "Iz=0.3                   #zener current(mA)\n",
+      "\n",
+      "#Calculations\n",
+      "P=Is**2*Rs               #power dissipated by Rs(W)\n",
+      "\n",
+      "#Part b\n",
+      "Pd=Vl*Iz                  #power dissipated(W)\n",
+      "Po=(Vs**2)/Rs             #output power(W)\n",
+      "\n",
+      "print\"maximum power dissipated by Rs is\",P,\"W\"\n",
+      "print\"maximum power dissipated by diode is\",round(Pd,3),\"W\"\n",
+      "print\"minimum diode current is\",round(Iz1,3),\"A\"\n",
+      "print\"voltage regulation is\",round(Vr,2),\"%\"\n",
+      "print\"output shorted will be\",Po,\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum power dissipated by Rs is 1.6 W\n",
+        "maximum power dissipated by diode is 4.804 W\n",
+        "minimum diode current is 0.199 A\n",
+        "voltage regulation is 0.03 %\n",
+        "output shorted will be 40.0 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.29,Page number 65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Vrms=20                   #secondary voltage(V)\n",
+      "Rs=10                     #Winding resistance(ohm)\n",
+      "Rf=5                      #diode has forward resistance(ohms)\n",
+      "Idc=2*10**-3              #load current(mA)     \n",
+      "\n",
+      "#Calculations\n",
+      "#Part a \n",
+      "Vdc=(Vrms*(math.sqrt(2)))/(math.pi)       #no load Vdc\n",
+      "\n",
+      "#Part b \n",
+      "Vldc=Vdc-(Idc*(Rs+Rf))                    #dc output voltage when load is 20mA\n",
+      "\n",
+      "#Part c\n",
+      "Rl=Vldc/Idc                                #load resistance(ohms)\n",
+      "r=((Rs+Rf)/Rl)*100                         #percentage regulation(%)\n",
+      "\n",
+      "#Part d\n",
+      "Im=Idc*(math.pi)                           #peak current(mA)\n",
+      "Ilrms=Im/2                                 #rms load current(mA) \n",
+      "Vlrms=Ilrms*Rl                             #rms load voltage(V)    \n",
+      "Vlrmsac=math.sqrt((Vlrms**2)-(Vldc**2))    #Ripple voltage rms(V)\n",
+      "f=50*2                                     #rippLe frequency(Hz)\n",
+      "\n",
+      "#Part e\n",
+      "eta=(((2*(math.pi))**2)/(1+((Rs+Rf)/Rl)))*100  #efficiency\n",
+      "\n",
+      "#Part f\n",
+      "PIV=Vm=Vrms*(math.sqrt(2))                     #peak inverse voltage(V)   \n",
+      "\n",
+      "#Results\n",
+      "print\"no load dc voltage is\",round(Vdc),\"V\" \n",
+      "print\"dc output voltage when the load is drawing 20 mA is\",round(Vldc,2),\"V\"\n",
+      "print\"percentage regulation at this load is\",round((r/1E-1),2),\"%\"\n",
+      "print\"ripple voltage rms is\",round(Vlrmsac,2),\"V and ripple frequency is\",f,\"Hz\"\n",
+      "print\"power conversion efficiency is\",round((eta/1E+2),1),\"%\"\n",
+      "print\"PIV is\",round(PIV),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "no load dc voltage is 9.0 V\n",
+        "dc output voltage when the load is drawing 20 mA is 8.97 V\n",
+        "percentage regulation at this load is 3.34 %\n",
+        "ripple voltage rms is 10.87 V and ripple frequency is 100 Hz\n",
+        "power conversion efficiency is 39.3 %\n",
+        "PIV is 28.0 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.30,Page number 66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Vl=24                       #battery voltage(V)\n",
+      "Vm=60*(math.sqrt(2))        #peak voltage(V)\n",
+      "Ip=2.5                      #peak current(A)\n",
+      "c=20                        #charge(Ah)\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a\n",
+      "theta=math.asin(Vl/Vm)     #angle at which conduction begins\n",
+      "Rs=(Vm-Vl)/Ip              #source resistance(ohms) \n",
+      "\n",
+      "#Part b\n",
+      "Idc=(Vm/(math.pi)*Rs)*(math.cos(theta))-(((math.pi)-(2*theta))/2*math.pi)*(Vl/Rs)   #load current(A)\n",
+      "T=c/Idc                                                                             #time to deliver 20Ah(h)\n",
+      "\n",
+      "#Results\n",
+      "print\"resistance connected in series is\",round(Rs,1),\"ohm\"\n",
+      "print\"time required to deliver a charge of 20 Ah is\",round((T/1E-3),1),\"h\" \n",
+      "print\"Idc\",round((Idc/1E+3),2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance connected in series is 24.3 ohm\n",
+        "time required to deliver a charge of 20 Ah is 31.9 h\n",
+        "Idc 0.63 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.32,Page number 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "R=25.                #external resistance(ohms)\n",
+      "Vm=200.              #peak value of voltage(V) as vs=200 sinwt\n",
+      "Rf=50.               #forward resistance(ohms)\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a \n",
+      "Id=Vm/(2*Rf+R)      #diode current(peak)\n",
+      "\n",
+      "#Part b \n",
+      "Idc=(2*Id)/math.pi    #dc current(A)\n",
+      "\n",
+      "#Part c\n",
+      "PIV=Vm/2               #peak value of voltage across D1\n",
+      "PIVac=100/math.pi      #average value of voltage across D1\n",
+      "\n",
+      "#Part d\n",
+      "Im=Id                  #peak value of current(A)\n",
+      "Irms=Im/(math.sqrt(2)) #rms value of current(A)\n",
+      "\n",
+      "#Results\n",
+      "print\"peak value of current is\",Id,\"A\"\n",
+      "print\"dc currect is\",round(Idc,2),\"A\"\n",
+      "print\"across D1 are peak voltage is\",PIV,\"V and average voltage is\",round(PIVac,1),\"V\"\n",
+      "print\"Irms is\",round(Irms,2),\"A\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "peak value of current is 1.6 A\n",
+        "dc currect is 1.02 A\n",
+        "across D1 are peak voltage is 100.0 V and average voltage is 31.8 V\n",
+        "Irms is 1.13 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.33,Page number 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "f=50.              #frequency(Hz)\n",
+      "dv=7.              #difference between maximum and minimum(25-18)voltages across the load(V)\n",
+      "Ic=100.            #load current(mA)\n",
+      "\n",
+      "#Calculations\n",
+      "dt=1/(2*f)        #time of discharge(seconds)\n",
+      "C=Ic/(dv/dt)      #capacitance(uF) \n",
+      "\n",
+      "#Results\n",
+      "print\"value of capacitor is\",round((C/1E-3),2),\"uF\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of capacitor is 142.86 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 56
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.34,Page number 68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "Vr=10.                          #peak to peak ripple voltage(V)\n",
+      "Vm=50.                          #peak output voltage(V)  \n",
+      "C=300.                          #Capacitance(uF)\n",
+      "Rl=470.                         #load resistance(ohms)\n",
+      "f=50.                           #frequency(Hz)\n",
+      "\n",
+      "#Calculations\n",
+      "#Part a \n",
+      "Vdc=Vm-(Vr/2)                   #dc voltage(V)      \n",
+      "C=Vdc/(f*Vr*Rl)                 #capacitance(mF)\n",
+      "\n",
+      "#Part b\n",
+      "C1=300*10**-6                    #capacitance is increased(uF)\n",
+      "Vr=2*Vm/((2*f*C1*Rl)+1)\n",
+      "Vdc=Vm-Vr/2                     #load voltage ripple(V)\n",
+      "Idc=Vdc/Rl                      #average load current(mA)\n",
+      "\n",
+      "#Results\n",
+      "print\"value of capacitor is\",round((C/1E-6),1),\"mF\" \n",
+      "print\"load voltage ripple is\",round(Vdc,2),\"V and average load current is\",round((Idc/1E-4),1),\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of capacitor is 191.5 mF\n",
+        "load voltage ripple is 46.69 V and average load current is 993.4 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.35,Page number 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "vo=7.5                  #instantaneous voltage(V)\n",
+      "R1=15                   #resistance(k ohms)\n",
+      "Von=0.5                 #voltage of diode when on(V)\n",
+      "\n",
+      "#Calculations\n",
+      "Rth=(R1*vo)/(R1+vo)                 #equivalent resistance(V)\n",
+      "T=2*(math.pi)/10**4                 #time period(ms)\n",
+      "t1=(math.asin(Von/2.5))/10**4       #timimgs when D1 conducts(ms)\n",
+      "t2=(T/2)-t1\n",
+      "\n",
+      "#Results\n",
+      "print\"time period is\",round((T/1E-3),3),\"ms\"\n",
+      "print\"t1 is\",t1,\"ms\"\n",
+      "print \"t2 is\",round((t2/1E-3),3),\"ms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time period is 0.628 ms\n",
+        "t1 is 2.0135792079e-05 ms\n",
+        "t2 is 0.294 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.36,Page number 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declarations\n",
+      "v=12                         #output voltage(V)\n",
+      "vm=20.                       #peak voltage(V)\n",
+      "v1=8                         #output voltage(V) for negative half cycle\n",
+      "vm1=20.                      #peak voltage(V) for negative half cycle \n",
+      "\n",
+      "#Calculations\n",
+      "t1=(math.asin(v/vm))/10**4         #for positive half cycle when D1 conducts\n",
+      "t2=(0.1*math.pi)-t1/1e-3           \n",
+      "t3=(math.asin(v1/vm1))/10**4       #for negative half cycle when D2 conducts\n",
+      "t4=(0.1*(math.pi))+t3/1e-3           \n",
+      "t5=(0.2*(math.pi))-t3/1e-3\n",
+      "\n",
+      "#Results\n",
+      "print\"t1 is\",round(t1/1e-3,3),\"ms\"\n",
+      "print\"t2 is\",round(t2,2),\"ms\"\n",
+      "print\"t3 is\",round(t3/1e-3,3),\"ms\"\n",
+      "print\"t4 is\",round(t4,3),\"ms\"\n",
+      "print\"t5 is\",round(t5,3),\"ms\"\n",
+      "print\"vo is\",\"-5.33+6.66*sin(10**4*.15)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "t1 is 0.064 ms\n",
+        "t2 is 0.25 ms\n",
+        "t3 is 0.041 ms\n",
+        "t4 is 0.355 ms\n",
+        "t5 is 0.587 ms\n",
+        "vo is -5.33+6.66*sin(10**4*.15)\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
-- 
cgit