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Diffstat (limited to 'basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter2.ipynb')
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diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter2.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter2.ipynb new file mode 100644 index 00000000..4f463827 --- /dev/null +++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter2.ipynb @@ -0,0 +1,1385 @@ +{ + "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", + "\n", + "v=10\n", + "r=4\n", + "\n", + "#case a\n", + "\n", + "i=v/float(r)\n", + "\n", + "print \"i=\",format(i,'.1f'),\"A\"\n", + "\n", + "#case b\n", + "\n", + "#6ohm resistor is in series with 4 ohm resistor\n", + "\n", + "i=v/(6+4)\n", + "\n", + "v1=i*6\n", + "v2=i*4\n", + "\n", + "print \"voltage across 6 ohm resistor=\",format(v1,'.1f'),\"V\"\n", + "\n", + "print \"voltage across 4 ohm resistor=\",format(v2,'.1f'),\"V\"\n", + "\n", + "#case c\n", + "\n", + "i=10 #constant in both cases\n", + "\n", + "v4=i*4\n", + "\n", + "print \"voltage when 4 ohm resistor is connected=\",format(v4,'.1f'),\"V\"\n", + "\n", + "v6=i*6\n", + "\n", + "v=v4+v6\n", + "\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", + "\n", + "i=72\n", + "v=36\n", + "\n", + "rs=v/float(i)\n", + "\n", + "print \"rs=\",format(rs,'.1f'),\"ohm\"\n", + "\n", + "print \"the load voltage is expressed as 36rl/(0.5+rl)\"\n", + "\n", + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "x=[40,50,60,72]\n", + "y=[36,34,32,30]\n", + "\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", + "\n", + "v=24\n", + "r=0.75\n", + "\n", + "ir=v/r\n", + "\n", + "print \"ir=\",format(ir,'.1f'),\"A\"\n", + "\n", + "il=v/(10+r) #since 10 is in series with r\n", + "\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", + "\n", + "vs=12\n", + "rs=0.3\n", + "il=10\n", + "\n", + "#case a\n", + "\n", + "p=vs*il\n", + "\n", + "print \"power=\",format(p,'.1f'),\"W\"\n", + "\n", + "#case b\n", + "\n", + "power=il**2*rs\n", + "\n", + "print \"power dissipated=\",format(power,'.1f'),\"W\"\n", + "\n", + "#case c\n", + "\n", + "totpow=(vs-il*rs)*il\n", + "\n", + "print \"total power supplied by practical source is=\",format(totpow,'.1f'),\"W\"\n", + "\n", + "i=vs/rs\n", + "\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", + "\n", + "#case a\n", + "\n", + "#v0/vs=r2/(r1+r2)=0.4r2=0.6r1\n", + "\n", + "r1=10\n", + "\n", + "r2=(0.6*r1)/float(0.4)\n", + "\n", + "print \"r2=\",format(r2,'.1f'),\"ohm\"\n", + "\n", + "#case b\n", + "\n", + "#when r2 is parallel to r3\n", + "r3=200000\n", + "req=(r2*r3)/(r2+r3)\n", + "\n", + "print \"req=\",format(req,'.1f'),\"ohm\"\n", + "\n", + "#v0/vs=0.5825\n", + "\n", + "change=(0.6-0.5825)/float(0.6)\n", + "\n", + "print change\n", + "\n", + "r3=20000\n", + "\n", + "req=(r2*r3)/(r3+r2)\n", + "\n", + "print \"req=\",format(req,'.1f'),\"ohm\"\n", + "\n", + "#v0/vs=0.4615\n", + "\n", + "change=(0.6-0.4615)/0.6\n", + "\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", + "\n", + "r=2\n", + "i=2\n", + "\n", + "i3=3 #obtained by applying current divider rule to figure\n", + "\n", + "i4=1\n", + "\n", + "req=1/float(0.5+0.25+0.166) #1/2,1/4,1/6 values are converted to decimal form\n", + "\n", + "print \"req=\",format(req,'.2f'),\"ohm\"\n", + "\n", + "i2=(4*i4/float(6))\n", + "\n", + "i1=(6*i2)/float(req)\n", + "\n", + "#tracing circuit cabc via 6 ohm resistor and applying ohms law,\n", + "\n", + "vs=i1*i4+i2*6\n", + "\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", + "\n", + "#combining series parallel series\n", + "\n", + "#[(2+2+2)||(6+5+2)||10]+5\n", + "\n", + "#[[6*6/6+6]+7]||10]+5=[10+10/10*10]+5=5+5=10\n", + "\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", + "\n", + "#case a\n", + "\n", + "#rab=(80+40)||(60+40)\n", + "\n", + "rab=(120*100)/float(120+100)\n", + "\n", + "print \"rab=\",format(rab,'.2f'),\"ohm\"\n", + "\n", + "#rab=(80||60)+(40||40)\n", + "\n", + "rab=(4800/float(140))+(1600/80)\n", + "print \"rab=\",format(rab,'.3f'),\"ohm\"\n", + "\n", + "#case b\n", + "\n", + "#(60+80)||(40+40)\n", + "\n", + "rcd=(140*80)/float(140+80)\n", + "\n", + "print \"rcd=\",format(rcd,'.2f'),\"ohm\"\n", + "\n", + "#(60||40)+(80||40)\n", + "\n", + "rab=float(2400/float(100))+(3200/float(120))\n", + "\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", + "\n", + "#simplifying the circuit \n", + "\n", + "ceq=1/float(0.333+0.666+0.2) #converted to decimal form\n", + "\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", + "\n", + "#case a\n", + "\n", + "I=12/(2+((12*24)/float(36))) #values taken from circuit\n", + "\n", + "I1=I*(24/float(36))\n", + "\n", + "I2=I*(12/float(36))\n", + "\n", + "print \"i=\",format(I,'1f'),\"A\"\n", + "\n", + "print \"i1=\",format(I1,'1f'),\"A\"\n", + "\n", + "print \"i2=\",format(I2,'1f'),\"A\"\n", + "\n", + "#case b\n", + "\n", + "power=(I**2)*2\n", + "\n", + "print \"power consumed by 2 ohm resistor=\",format(power,'.2f'),\"W\"\n", + "\n", + "\n", + "power=(I1**2)*12\n", + "\n", + "print \"power consumed by 12 ohm resistor=\",format(power,'.2f'),\"W\"\n", + "\n", + "power=(I2**2)*24\n", + "\n", + "print \"power consumed by 2 ohm resistor=\",format(power,'.2f'),\"W\"\n", + "\n", + "#case c\n", + "\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", + "\n", + "#case a\n", + "\n", + "#values taken and calculated from figure\n", + "\n", + "r1=6\n", + "r2=12\n", + "r3=18\n", + "\n", + "rab=3.21 #calculating similar to above using parallel in series resistances\n", + "\n", + "print \"rab=3.12ohm\"\n", + "\n", + "#case b\n", + "\n", + "r4=30\n", + "r5=15\n", + "r6=30\n", + "\n", + "ran=6 #similar as above\n", + "\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", + "\n", + "#eqns derived from figure\n", + "\n", + "#6v1-4v2=2-->1\n", + "#-4v1+7v2=-3-->2\n", + "\n", + "#eqn 1 and 2 are written in matrix form and solved using cramers rule\n", + "\n", + "print \"v1=0.0769 V\"\n", + "\n", + "print \"v2=-0.3846V\"\n", + "\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", + "\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", + "\n", + "print \"v1=3.6V\"\n", + "\n", + "print \"v2=2.2V\"\n", + "\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", + "\n", + "#kcl is applied to the circuit and the eqns obtained are solved using cramer's rule\n", + "\n", + "print \"the voltages of nodes 1 and 3 are 50.29 and 57.71 respectively\"\n", + "\n", + "#i3=v/r\n", + "\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", + "\n", + "#the eqns obtained are converted to matrix form for solving using cramer's rule values are found\n", + "\n", + "i1=5.224\n", + "i2=0.7463\n", + "i3=3.28\n", + "\n", + "v=(i1-i3)*3\n", + "\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", + "\n", + "#kvl eqns are obtained from figure which are solved to obtain currents\n", + "\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", + "\n", + "#the currents are obtained by solving the eqns\n", + "\n", + "i1=5.87\n", + "i2=-0.13\n", + "i3=-1.54\n", + "\n", + "v=18-1.54*8\n", + "\n", + "print \"voltage at node D=\",format(v,'.2f'),\"v\"\n", + "\n", + "i=5.86/float(4)\n", + "\n", + "print \"current in 4 ohm resistor is=\",format(i,'.2f'),\"A\"\n", + "\n", + "power=18*1.54\n", + "\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", + "\n", + "#node eqns are obtained form the figure\n", + "\n", + "print \"va=8.33V and vb=4.17V\"\n", + "\n", + "i=8.33/float(8)\n", + "\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", + "\n", + "#eqns obtained are calculated just like above problems and are aolved for i1 and i2\n", + "\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", + "\n", + "#eqns are obtained from the figure and are solved for currents\n", + "\n", + "i1=6.89\n", + "i2=3.89\n", + "i3=-2.12\n", + "\n", + "i=2*(i2-i1)\n", + "\n", + "print \"current supplied by dependent source is=\",format(i,'.1f'),\"A\"\n", + "\n", + "power=6*i1\n", + "\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", + "\n", + "#the following problem is based on usage of superposition theorem\n", + "\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", + "\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", + "\n", + "i=(4*6)/float(6+12)\n", + "\n", + "print \"i8=\",format(i8,'.3f'),\"A\"\n", + "\n", + "print \"i8'=\",format(i,'.3f'),\"A\"\n", + "\n", + "tot=i8+i\n", + "\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", + "\n", + "#kvl is applied to circuit\n", + "\n", + "i=1\n", + "\n", + "vth=12-(1*4) #12 is voltage 1 is current and 4 is resistance\n", + "\n", + "rth=(4*5)/float(4+5)\n", + "\n", + "i6=vth/float(rth+6) #since current passes through 6 ohm resistor\n", + "\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", + "\n", + "#thevenin's theorem and superposition theorem used here\n", + "\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", + "\n", + "i1=-0.6\n", + "i2=-1.2\n", + "\n", + "#the value of currents indicates that they have assumed to be flowing in directions opposite to the assumed direction\n", + "\n", + "vth=12-1.2*3 #voltage eqn\n", + "\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", + "\n", + "i2=vth/(rth+2)\n", + "\n", + "print \"current through 2 ohm resistor is=\",format(i2,'.2f'),\"A\"\n", + "\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", + "\n", + "#using thevenin's theorem\n", + "\n", + "#applying kcl at node a va is obtained\n", + "\n", + "va=12\n", + "\n", + "rth=1.33 #2||4\n", + "\n", + "i5=vth/(rth+5)\n", + "\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", + "\n", + "#applying kvl to circuit\n", + "\n", + "i=0.414\n", + "\n", + "vth=12-4*0.414 #using vth formula\n", + "\n", + "#when terminals a and b are short circuited applying kcl to node a gives isc=5*i\n", + "\n", + "isc=2.07\n", + "\n", + "rth=vth/isc\n", + "\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", + "\n", + "#norton's theorem\n", + "\n", + "v=10\n", + "\n", + "#applying kvl to closed circuit \n", + "\n", + "isc=12/float(2+2) \n", + "\n", + "rn=4 #resistance obtained by short circuiting v and opening i\n", + "\n", + "iab=(4*3)/float(4+4) #current through 4 ohm connected across AB\n", + "\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", + "\n", + "#natural frequency needs to be determined\n", + "\n", + "#req=[(6+6)||4]+[1||2]=3.6666\n", + "\n", + "req=3.6667\n", + "\n", + "l=4 #inductance\n", + "\n", + "s=-req/float(l)\n", + "\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", + "\n", + "#req=[10+2+(5||15)]=15.75\n", + "\n", + "#case a\n", + "\n", + "c=0.4\n", + "req=15.75\n", + "s=-1/float(c*req)\n", + "\n", + "print \"natural frequency=\",format(s,'.5f'),\"secinverse\"\n", + "\n", + "#case b\n", + "\n", + "tc=req*0.4 #time constant\n", + "\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", + "\n", + "v=120\n", + "r=40\n", + "\n", + "i=v/float(r)\n", + "\n", + "#applying kvl to the closed loop\n", + "\n", + "v=3*520\n", + "\n", + "print \"voltage=\",format(v,'.1f'),\"v\"\n", + "\n", + "#when v=120,R can be found by I*(r+20)=120-->r=20\n", + "\n", + "r=20\n", + "\n", + "print \"r=20 ohm\"\n", + "\n", + "#when r=20 total r=20+20+20=60\n", + "\n", + "r=60\n", + "\n", + "l=10\n", + "\n", + "tc=l/float(r) #time constant\n", + "\n", + "print \"tc=\",format(tc,'.4f'),\"sec\"\n", + "\n", + "#i=I0*e^-(t/tc)=3*e^(-6t)\n", + "\n", + "energy=(10*9)/float(2)\n", + "\n", + "benergy=0.05*energy\n", + "\n", + "print \"balance energy=\",format(benergy,'.2f'),\"J\"\n", + "\n", + "#(L*i^2)/2=2.25-->hence i=0.6708\n", + "\n", + "#3*e^-6t=0.6708-->e^-6t=0.2236-->applying log on both sides we get t=0.25\n", + "\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", + "\n", + "v=120\n", + "\n", + "V=200\n", + "\n", + "#v=V(1-e^-5/2R)\n", + "\n", + "#120=200*(1-e^-5/2R)\n", + "\n", + "#applying log on both sides and solving we get R=2.72 Mohm\n", + "\n", + "print \"R=2.72Mohm\"\n", + "\n", + "R=5 \n", + "tc=10\n", + "\n", + "#applying in the above eqn and solving lograthmically we get t=9.16\n", + "\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": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |