Added(A)/Deleted(D) following books

 A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter11.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter12.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter13.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter14.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter15.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter16.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter18.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter19.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter2.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter20.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter21.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter22.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter23.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter5.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter6.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter8.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/Chapter9.ipynb
A  Electrical_&_Electronic_Systems_by_Neil_Storey/screenshots/pic1.png
A  Electrical_&_Electronic_Systems_by_Neil_Storey/screenshots/pic2.png
A  Electrical_&_Electronic_Systems_by_Neil_Storey/screenshots/pic3.png
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter10_67BmtXx.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter10_LGPoR7F.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter10_PDCS2qh.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter10_lNjLAte.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter10_n5s4jXl.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter10_npy5vv0.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter10_pjpRgex.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter11_IE4byFL.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter11_RAQoou9.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter11_VcicGy5.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter11_hoqMFBX.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter11_iBTGbp8.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter11_k1rTVhh.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter11_nGrFEGY.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter11_pfzLlc6.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter1_1FJHa67.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter1_2wryDDQ.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter1_BuZVTIP.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter1_UCsno1y.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter1_Yd7uN6t.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter1_bGP3Wsd.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter1_pFJzyF5.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter2_242icmu.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter2_Rcy1wii.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter2_TvxYpxT.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter2_VU0Vuul.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter2_Xiz19Ms.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter2_rAG8C10.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter2_xUN650b.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter3_0WKL8dM.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter3_BsRbQCe.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter3_IF7BAbT.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter3_Qh2uphO.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter3_WS64jkH.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter3_aM1BqRM.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter3_tEQK8Lr.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter4_0E18xYL.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter4_4P7HQZZ.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter4_8GR2j3i.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter4_KE5ki12.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter4_LmrwpIC.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter4_ZuWrQKN.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter4_wUtoaip.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter5_0Ke5Vhq.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter5_23aeJMe.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter5_4a14Khd.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter5_eftppGy.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter5_jOQv5Ua.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter5_m75AW4e.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter5_mZ9916M.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter6_2OboaPO.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter6_Cf1Ae70.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter6_IZI7GIy.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter6_TRm6El0.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter6_WfVJQIc.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter6_Z5dEk19.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter6_z9FlTqT.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter8_6BR5OK6.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter8_FV7Jgrb.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter8_Q79Sp1O.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter8_Qhj9HPA.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter8_fBoE3j3.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter8_mFOxHz2.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter8_zNY63b3.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter9_039FJN6.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter9_0AhnrOb.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter9_8K1gGKx.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter9_E4M5MRA.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter9_Ivqhz8T.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter9_ZSvbidk.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter9_u7ekl0N.ipynb
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/screenshots/ch1_mHmqCLQ.png
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/screenshots/ch_9_UicjBqW.png
A  MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/screenshots/ch_xk3ylh6.png
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter2_eh02mMg.ipynb
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter4_QnODdtI.ipynb
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter5_IwMblyq.ipynb
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter11_nHcyQSN.ipynb
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter1_ivpTi0v.ipynb
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter3.ipynb
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter6_7Vcvq3x.ipynb
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter7_Iro5ijO.ipynb
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter8_dRxKPQv.ipynb
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter9_QP9exWK.ipynb
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/screenshots/chap1_gG9EuuG.png
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/screenshots/chapter2_RW9JXUw.png
A  basic_electrical_engineering_by_nagsarkar_and_sukhija/screenshots/chapter6_kNIRtlM.png

1 files changed, 1146 insertions, 0 deletions

diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter2_eh02mMg.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter2_eh02mMg.ipynb
new file mode 100644
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--- /dev/null
+++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter2_eh02mMg.ipynb
@@ -0,0 +1,1146 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2:Network Analysis And Network Theorems"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.1:Page number-50"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "i= 2.5 A\n",
+      "voltage across 6 ohm resistor= 6.0 V\n",
+      "voltage across 4 ohm resistor= 4.0 V\n",
+      "voltage when 4 ohm resistor is connected= 40.0 V\n",
+      "voltage when both resistors are in series= 100.0 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "v=10\n",
+    "r=4\n",
+    "#case a\n",
+    "i=v/float(r)\n",
+    "print \"i=\",format(i,'.1f'),\"A\"\n",
+    "#case b\n",
+    "#6ohm resistor is in series with 4 ohm resistor\n",
+    "i=v/(6+4)\n",
+    "v1=i*6\n",
+    "v2=i*4\n",
+    "print \"voltage across 6 ohm resistor=\",format(v1,'.1f'),\"V\"\n",
+    "print \"voltage across 4 ohm resistor=\",format(v2,'.1f'),\"V\"\n",
+    "#case c\n",
+    "i=10 #constant in both cases\n",
+    "v4=i*4\n",
+    "print \"voltage when 4 ohm resistor is connected=\",format(v4,'.1f'),\"V\"\n",
+    "v6=i*6\n",
+    "v=v4+v6\n",
+    "print \"voltage when both resistors are in series=\",format(v,'.1f'),\"V\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.2:Page number-53"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "rs= 0.5 ohm\n",
+      "the load voltage is expressed as 36rl/(0.5+rl)\n"
+     ]
+    },
+    {
+     "data": {
+      "image/png": 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+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x7f9cd722d810>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "import math\n",
+    "i=72\n",
+    "v=36\n",
+    "rs=v/float(i)\n",
+    "print \"rs=\",format(rs,'.1f'),\"ohm\"\n",
+    "print \"the load voltage is expressed as 36rl/(0.5+rl)\"\n",
+    "%matplotlib inline\n",
+    "import matplotlib.pyplot as plt\n",
+    "x=[40,50,60,72]\n",
+    "y=[36,34,32,30]\n",
+    "plt.plot(x,y)\n",
+    "plt.xlabel('il(A)')\n",
+    "plt.ylabel('vl(V)')\n",
+    "plt.show()\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.3:Page number-55"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "ir= 32.0 A\n",
+      "il= 2.23 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "v=24\n",
+    "r=0.75\n",
+    "ir=v/r\n",
+    "print \"ir=\",format(ir,'.1f'),\"A\"\n",
+    "il=v/(10+r) #since 10 is in series with r\n",
+    "print \"il=\",format(il,'.2f'),\"A\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.4:Page number-56"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "power= 120.0 W\n",
+      "power dissipated= 30.0 W\n",
+      "total power supplied by practical source is= 90.0 W\n",
+      "current source= 40.0 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "vs=12\n",
+    "rs=0.3\n",
+    "il=10\n",
+    "#case a\n",
+    "p=vs*il\n",
+    "print \"power=\",format(p,'.1f'),\"W\"\n",
+    "#case b\n",
+    "power=il**2*rs\n",
+    "print \"power dissipated=\",format(power,'.1f'),\"W\"\n",
+    "#case c\n",
+    "totpow=(vs-il*rs)*il\n",
+    "print \"total power supplied by practical source is=\",format(totpow,'.1f'),\"W\"\n",
+    "i=vs/rs\n",
+    "print \"current source=\",format(i,'.1f'),\"A\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.5:Page number-58"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "r2= 15.0 ohm\n",
+      "req= 15.0 ohm\n",
+      "0.0291666666667\n",
+      "req= 15.0 ohm\n",
+      "0.230833333333\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#case a\n",
+    "#v0/vs=r2/(r1+r2)=0.4r2=0.6r1\n",
+    "r1=10\n",
+    "r2=(0.6*r1)/float(0.4)\n",
+    "print \"r2=\",format(r2,'.1f'),\"ohm\"\n",
+    "#case b\n",
+    "#when r2 is parallel to r3\n",
+    "r3=200000\n",
+    "req=(r2*r3)/(r2+r3)\n",
+    "print \"req=\",format(req,'.1f'),\"ohm\"\n",
+    "#v0/vs=0.5825\n",
+    "change=(0.6-0.5825)/float(0.6)\n",
+    "print change\n",
+    "r3=20000\n",
+    "req=(r2*r3)/(r3+r2)\n",
+    "print \"req=\",format(req,'.1f'),\"ohm\"\n",
+    "#v0/vs=0.4615\n",
+    "change=(0.6-0.4615)/0.6\n",
+    "print change"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.6:Page number-60"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "req= 1.09 ohm\n",
+      "vs= 7.66 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "r=2\n",
+    "i=2\n",
+    "i3=3 #obtained by applying current divider rule to figure\n",
+    "i4=1\n",
+    "req=1/float(0.5+0.25+0.166) #1/2,1/4,1/6 values are converted to decimal form\n",
+    "print \"req=\",format(req,'.2f'),\"ohm\"\n",
+    "i2=(4*i4/float(6))\n",
+    "i1=(6*i2)/float(req)\n",
+    "#tracing circuit cabc via 6 ohm resistor and applying ohms law,\n",
+    "vs=i1*i4+i2*6\n",
+    "print \"vs=\",format(vs,'.2f'),\"V\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.7:Page number-61"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the value of series parallel resistances is 10 ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#combining series parallel series\n",
+    "#[(2+2+2)||(6+5+2)||10]+5\n",
+    "#[[6*6/6+6]+7]||10]+5=[10+10/10*10]+5=5+5=10\n",
+    "print \"the value of series parallel resistances is 10 ohm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "rab= 54.55 ohm\n",
+      "rab= 54.286 ohm\n",
+      "rcd= 50.91 ohm\n",
+      "rab= 50.67 ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#case a\n",
+    "#rab=(80+40)||(60+40)\n",
+    "rab=(120*100)/float(120+100)\n",
+    "print \"rab=\",format(rab,'.2f'),\"ohm\"\n",
+    "#rab=(80||60)+(40||40)\n",
+    "rab=(4800/float(140))+(1600/80)\n",
+    "print \"rab=\",format(rab,'.3f'),\"ohm\"\n",
+    "#case b\n",
+    "#(60+80)||(40+40)\n",
+    "rcd=(140*80)/float(140+80)\n",
+    "print \"rcd=\",format(rcd,'.2f'),\"ohm\"\n",
+    "#(60||40)+(80||40)\n",
+    "rab=float(2400/float(100))+(3200/float(120))\n",
+    "print \"rab=\",format(rab,'.2f'),\"ohm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 2.9:Page number-65"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "ceq= 0.83402836 F\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#simplifying the circuit \n",
+    "ceq=1/float(0.333+0.666+0.2) #converted to decimal form\n",
+    "print \"ceq=\",format(ceq,'.8f'),\"F\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.10:Page number-67"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 29,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "i= 1.200000 A\n",
+      "i1= 0.800000 A\n",
+      "i2= 0.400000 A\n",
+      "power consumed by 2 ohm resistor= 2.88 W\n",
+      "power consumed by 12 ohm resistor= 7.68 W\n",
+      "power consumed by 2 ohm resistor= 3.84 W\n",
+      "voltage drop= 2.4 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#case a\n",
+    "I=12/(2+((12*24)/float(36))) #values taken from circuit\n",
+    "I1=I*(24/float(36))\n",
+    "I2=I*(12/float(36))\n",
+    "print \"i=\",format(I,'1f'),\"A\"\n",
+    "print \"i1=\",format(I1,'1f'),\"A\"\n",
+    "print \"i2=\",format(I2,'1f'),\"A\"\n",
+    "#case b\n",
+    "power=(I**2)*2\n",
+    "print \"power consumed by 2 ohm resistor=\",format(power,'.2f'),\"W\"\n",
+    "power=(I1**2)*12\n",
+    "print \"power consumed by 12 ohm resistor=\",format(power,'.2f'),\"W\"\n",
+    "power=(I2**2)*24\n",
+    "print \"power consumed by 2 ohm resistor=\",format(power,'.2f'),\"W\"\n",
+    "#case c\n",
+    "v=I*2\n",
+    "print \"voltage drop=\",format(v,'.1f'),\"V\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.11:Page number-69"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 30,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "rab=3.12ohm\n",
+      "ran=6 ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#case a\n",
+    "#values taken and calculated from figure\n",
+    "r1=6\n",
+    "r2=12\n",
+    "r3=18\n",
+    "rab=3.21 #calculating similar to above using parallel in series resistances\n",
+    "print \"rab=3.12ohm\"\n",
+    "#case b\n",
+    "r4=30\n",
+    "r5=15\n",
+    "r6=30\n",
+    "ran=6 #similar as above\n",
+    "print \"ran=6 ohm\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.12:Page number-73"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 31,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "v1=0.0769 V\n",
+      "v2=-0.3846V\n",
+      "current in 0.5ohm resistance is 0.154A,0.25ohm resistance is 1.846,0.66ohm resistor is -1.154A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#eqns derived from figure\n",
+    "#6v1-4v2=2-->1\n",
+    "#-4v1+7v2=-3-->2\n",
+    "#eqn 1 and 2 are written in matrix form and solved using cramers rule\n",
+    "print \"v1=0.0769 V\"\n",
+    "print \"v2=-0.3846V\"\n",
+    "print \"current in 0.5ohm resistance is 0.154A,0.25ohm resistance is 1.846,0.66ohm resistor is -1.154A\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.13:Page number-74"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 32,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "v1=3.6V\n",
+      "v2=2.2V\n",
+      "the current in 0.6 ohm resistor is 10.8A,0.2 ohm resistor is 7A,0.16ohm resistor is 13.2 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#from the figure the eqns are written in matrix form and using cramers rule the value of v1 and v2 can be found\n",
+    "print \"v1=3.6V\"\n",
+    "print \"v2=2.2V\"\n",
+    "print \"the current in 0.6 ohm resistor is 10.8A,0.2 ohm resistor is 7A,0.16ohm resistor is 13.2 A\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.14:Page number-76 "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 33,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the voltages of nodes 1 and 3 are 50.29 and 57.71 respectively\n",
+      "current through 16 ohm resistor is 1.64A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#kcl is applied to the circuit and the eqns obtained are solved using cramer's rule\n",
+    "print \"the voltages of nodes 1 and 3 are 50.29 and 57.71 respectively\"\n",
+    "#i3=v/r\n",
+    "print \"current through 16 ohm resistor is 1.64A\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.15:Page number-78"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 34,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "voltage across 3 ohm resistor is= 5.832 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#the eqns obtained are converted to matrix form for solving using cramer's rule values are found\n",
+    "i1=5.224\n",
+    "i2=0.7463\n",
+    "i3=3.28\n",
+    "v=(i1-i3)*3\n",
+    "print \"voltage across 3 ohm resistor is=\",format(v,'.3f'),\"V\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.16:page number-79"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 35,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "currents obtained are i1=2.013 and i2=1.273\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#kvl eqns are obtained from figure which are solved to obtain currents\n",
+    "print \"currents obtained are i1=2.013 and i2=1.273\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.17:Page number-80"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 37,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "voltage at node D= 5.68 v\n",
+      "current in 4 ohm resistor is= 1.47 A\n",
+      "power supplied by 18V source is= 27.72 W\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#the currents are obtained by solving the eqns\n",
+    "i1=5.87\n",
+    "i2=-0.13\n",
+    "i3=-1.54\n",
+    "v=18-1.54*8\n",
+    "print \"voltage at node D=\",format(v,'.2f'),\"v\"\n",
+    "i=5.86/float(4)\n",
+    "print \"current in 4 ohm resistor is=\",format(i,'.2f'),\"A\"\n",
+    "power=18*1.54\n",
+    "print \"power supplied by 18V source is=\",format(power,'.2f'),\"W\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.18:Page number-82"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 38,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "va=8.33V and vb=4.17V\n",
+      "current through 8 ohm resistor is= 1.04 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#node eqns are obtained form the figure\n",
+    "print \"va=8.33V and vb=4.17V\"\n",
+    "i=8.33/float(8)\n",
+    "print \"current through 8 ohm resistor is=\",format(i,'.2f'),\"A\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.19:Page number-83 "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 39,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "i1=-1.363A and i2=-3.4A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#eqns obtained are calculated just like above problems and are aolved for i1 and i2\n",
+    "print \"i1=-1.363A and i2=-3.4A\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.20:Page number-84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 40,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "current supplied by dependent source is= -6.0 A\n",
+      "power supplied by voltage source is= 41.34 W\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#eqns are obtained from the figure and are solved for currents\n",
+    "i1=6.89\n",
+    "i2=3.89\n",
+    "i3=-2.12\n",
+    "i=2*(i2-i1)\n",
+    "print \"current supplied by dependent source is=\",format(i,'.1f'),\"A\"\n",
+    "power=6*i1\n",
+    "print \"power supplied by voltage source is=\",format(power,'.2f'),\"W\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.21:Page number-86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "i8= 0.667 A\n",
+      "i8'= 1.333 A\n",
+      "total current= 2.0 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#the following problem is based on usage of superposition theorem\n",
+    "i8=12/float(6+4+8) #current for 8 ohm resistor.the resistances are in series with each other.Hence 6+4+8\n",
+    "#next when voltage source is short circuited (8+4) total of resistance is obtained.The 4A is distributed in parallel branches as per current divider rule\n",
+    "i=(4*6)/float(6+12)\n",
+    "print \"i8=\",format(i8,'.3f'),\"A\"\n",
+    "print \"i8'=\",format(i,'.3f'),\"A\"\n",
+    "tot=i8+i\n",
+    "print \"total current=\",format(tot,'.1f'),\"A\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exampe 2.22:Page number-88"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.972972972973 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#kvl is applied to circuit\n",
+    "i=1\n",
+    "vth=12-(1*4) #12 is voltage 1 is current and 4 is resistance\n",
+    "rth=(4*5)/float(4+5)\n",
+    "i6=vth/float(rth+6) #since current passes through 6 ohm resistor\n",
+    "print i6,\"A\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.23:Page number-89"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "current through 2 ohm resistor is= 2.45 A\n",
+      "Note that the same problem is again solved using superposition theorem and hence ignored \n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#thevenin's theorem and superposition theorem used here\n",
+    "#applying mesh eqns to the 2 circuits and after getting the eqns they are solved using cramer's rule to obtain i1 and i2\n",
+    "i1=-0.6\n",
+    "i2=-1.2\n",
+    "#the value of currents indicates that they have assumed to be flowing in directions opposite to the assumed direction\n",
+    "vth=12-1.2*3 #voltage eqn\n",
+    "rth=1.425 #(1+2||12)||3=(1+(2*12)/(2+12))||3=19/7||3=19/7*3/19/7+3=1.425\n",
+    "i2=vth/(rth+2)\n",
+    "print \"current through 2 ohm resistor is=\",format(i2,'.2f'),\"A\"\n",
+    "print \"Note that the same problem is again solved using superposition theorem and hence ignored \" "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.24:Page number-91"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "current through 5 ohm resistor is= 1.327 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#using thevenin's theorem\n",
+    "#applying kcl at node a va is obtained\n",
+    "va=12\n",
+    "rth=1.33 #2||4\n",
+    "i5=vth/(rth+5)\n",
+    "print \"current through 5 ohm resistor is=\",format(i5,'.3f'),\"A\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.25:Page number-92"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "rth= 4.997 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#applying kvl to circuit\n",
+    "i=0.414\n",
+    "vth=12-4*0.414 #using vth formula\n",
+    "#when terminals a and b are short circuited applying kcl to node a gives isc=5*i\n",
+    "isc=2.07\n",
+    "rth=vth/isc\n",
+    "print \"rth=\",format(rth,'.3f'),\"A\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.26:Page number-93"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "iab= 1.5 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#norton's theorem\n",
+    "v=10\n",
+    "#applying kvl to closed circuit \n",
+    "isc=12/float(2+2) \n",
+    "rn=4 #resistance obtained by short circuiting v and opening i\n",
+    "iab=(4*3)/float(4+4) #current through 4 ohm connected across AB\n",
+    "print \"iab=\",format(iab,'.1f'),\"A\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.27:Page number-103"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "natural frequency= -0.91668 secinverse\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#natural frequency needs to be determined\n",
+    "#req=[(6+6)||4]+[1||2]=3.6666\n",
+    "req=3.6667\n",
+    "l=4 #inductance\n",
+    "s=-req/float(l)\n",
+    "print \"natural frequency=\",format(s,'.5f'),\"secinverse\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.28"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "natural frequency= -0.15873 secinverse\n",
+      "time constant= 6.3 sec\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#req=[10+2+(5||15)]=15.75\n",
+    "#case a\n",
+    "c=0.4\n",
+    "req=15.75\n",
+    "s=-1/float(c*req)\n",
+    "print \"natural frequency=\",format(s,'.5f'),\"secinverse\"\n",
+    "#case b\n",
+    "tc=req*0.4 #time constant\n",
+    "print \"time constant=\",format(tc,'.1f'),\"sec\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.30:Page number-109"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "voltage= 1560.0 v\n",
+      "r=20 ohm\n",
+      "tc= 0.1667 sec\n",
+      "balance energy= 2.25 J\n",
+      "t=0.25 sec\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "v=120\n",
+    "r=40\n",
+    "i=v/float(r)\n",
+    "#applying kvl to the closed loop\n",
+    "v=3*520\n",
+    "print \"voltage=\",format(v,'.1f'),\"v\"\n",
+    "#when v=120,R can be found by I*(r+20)=120-->r=20\n",
+    "r=20\n",
+    "print \"r=20 ohm\"\n",
+    "#when r=20 total r=20+20+20=60\n",
+    "r=60\n",
+    "l=10\n",
+    "tc=l/float(r) #time constant\n",
+    "print \"tc=\",format(tc,'.4f'),\"sec\"\n",
+    "#i=I0*e^-(t/tc)=3*e^(-6t)\n",
+    "energy=(10*9)/float(2)\n",
+    "benergy=0.05*energy\n",
+    "print \"balance energy=\",format(benergy,'.2f'),\"J\"\n",
+    "#(L*i^2)/2=2.25-->hence i=0.6708\n",
+    "#3*e^-6t=0.6708-->e^-6t=0.2236-->applying log on both sides we get t=0.25\n",
+    "print \"t=0.25 sec\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.34:Page number-116"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "R=2.72Mohm\n",
+      "t=9.16 sec\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "v=120\n",
+    "V=200\n",
+    "#v=V(1-e^-5/2R)\n",
+    "#120=200*(1-e^-5/2R)\n",
+    "#applying log on both sides and solving we get R=2.72 Mohm\n",
+    "print \"R=2.72Mohm\"\n",
+    "R=5 \n",
+    "tc=10\n",
+    "#applying in the above eqn and solving lograthmically we get t=9.16\n",
+    "print \"t=9.16 sec\""
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "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": "2.7.5"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}