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diff --git a/BASIC_ELECTRICAL_ENGINEERING_by_D_C_KULSHRESHTHA/Chapter7.ipynb b/BASIC_ELECTRICAL_ENGINEERING_by_D_C_KULSHRESHTHA/Chapter7.ipynb new file mode 100755 index 00000000..acf4ea53 --- /dev/null +++ b/BASIC_ELECTRICAL_ENGINEERING_by_D_C_KULSHRESHTHA/Chapter7.ipynb @@ -0,0 +1,975 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7: SELF AND MUTUAL INDUCTANCES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1,Page number: 184\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the emf induced in a coil.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "L=4 #Self inductance of the coil(in Henry) \n",
+ "di=4-10 #Change in current(in Amperes)\n",
+ "dt=0.1 #Time interval(in seconds)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "e=-L*(di/dt)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The emf induced in the coil is %.2f V.\" %(e)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The emf induced in the coil is 240.00 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2,Page number: 184"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the inductance of a coil.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "N=150 #Number of turns in the coil \n",
+ "flux=0.01 #Flux linked with the coil(in Webers)\n",
+ "I=10 #Current in the coil(in Amperes) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "L=(N*flux)/I\n",
+ "di=-10-(10)\n",
+ "dt=0.01\n",
+ "e=-L*(di/dt)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The inductance of the coil is %.2f H.\" %(L)\n",
+ "print \"The induced emf is %.2f V.\" %(e)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The inductance of the coil is 0.15 H.\n",
+ "The induced emf is 300.00 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3,Page number: 185"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the inductance of the coil and the emf induced.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "N=100 #Number of turns in the coil\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "dflux=-0.4-0.4\n",
+ "di=-10-10\n",
+ "L=N*(dflux/di)\n",
+ "dt=0.01\n",
+ "e=-(L*(di/dt))/1000\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The inductance of the coil is %.2f mH.\" %(L)\n",
+ "print \"The induced emf is %.2f V.\" %(e)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The inductance of the coil is 4.00 mH.\n",
+ "The induced emf is 8.00 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4,Page number: 185"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the energy stored in an air-cored solenoid.\"\"\"\n",
+ "\n",
+ "from math import pi,pow\n",
+ "\n",
+ "\"\"\" All quantities expresssed in SI System.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "l=0.30 #Length of the solenoid(in metres) \n",
+ "d=0.015 #Internal diameter of the solenoid(in metres) \n",
+ "r=0.0075 #Internal radius of the solenoid(in metres)\n",
+ "N=900 #Number of turns in the coil \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "A=pi*pow(r,2)\n",
+ "L=(pow(N,2)*4*pi*A)/(0.30*10000000)\n",
+ "I=5\n",
+ "W=0.5*L*pow(I,2)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The inductance of the air-cored solenoid is %.2f mH.\" %(L*1000)\n",
+ "print \"The amount of energy stored in the air-cored solenoid is %.2f mJ.\" %(W*1000)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The inductance of the air-cored solenoid is 0.60 mH.\n",
+ "The amount of energy stored in the air-cored solenoid is 7.49 mJ.\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.5,Page number: 185"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the relative permeability of iron and the inductance of a coil.\"\"\"\n",
+ "\n",
+ "from math import pow,pi\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "r=0.01 #Radius of circular ring(in metres)\n",
+ "A=pi*pow(r,2) #Area of circular ring(in square metres) \n",
+ "N=3000 #Number of turns in the coil \n",
+ "I=0.5 #Current in the coil(in Amperes)\n",
+ "l=20.0/100 #Length of the iron rod(in metres) \n",
+ "B=1.2 #Magnitude of magnetic field(in Tesla)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "H=(N*I)/l\n",
+ "per=B/H\n",
+ "rel_per=(per*10000000)/(4*pi)\n",
+ "L=(N*B*A)/I\n",
+ "dflux=(0.1-1)*A*B\n",
+ "dt=0.01\n",
+ "e=-N*(dflux/dt)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The permeability of iron is %e Tm/A.\" %(per)\n",
+ "print \"(b)The relative permeability of iron is %d.\" %(rel_per)\n",
+ "print \"(c)The inductance of the coil is %.2f H.\" %(L)\n",
+ "print \"(d)The voltage in the coil is %.2f V.\" %(e)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The permeability of iron is 1.600000e-04 Tm/A.\n",
+ "(b)The relative permeability of iron is 127.\n",
+ "(c)The inductance of the coil is 2.26 H.\n",
+ "(d)The voltage in the coil is 101.79 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.6,Page number: 186"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the potential difference across the terminals of a coil.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "R=3 #Resistance of the coil(in Ohms)\n",
+ "i=1 #Current in the coil(in Amperes)\n",
+ "di=10000 #Change in current(in Amperes) \n",
+ "dt=1 #Time interval(in seconds)\n",
+ "L=0.1/1000 #Self inductance of the coil(in Henry)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "V=(i*R)+(L*(di/dt))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The potential difference that exists across the terminals of the coil is %.2f V.\" %(V) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential difference that exists across the terminals of the coil is 4.00 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7,Page number: 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the mutual inductance and emf induced in a search coil.\"\"\"\n",
+ "\n",
+ "from math import pi,pow\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "N1=2000 #Number of turns in the solenoid \n",
+ "N2=500 #Number of turns in the search coil\n",
+ "l=0.70 #Length of the solenoid(in metres) \n",
+ "k=1 #Coefficient of coupling\n",
+ "A=30.0/10000 #Mean area of the search coil(in square metres)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "per=(4*pi)/10000000.0\n",
+ "M=(k*N1*N2*per*A)/l\n",
+ "di1=260.0\n",
+ "dt=1\n",
+ "e=M*(di1/dt)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The mutual inductance is %.4f mH.\" %(M*1000)\n",
+ "print \"(b)The emf induced in the search coil is %.2f V.\" %(e)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The mutual inductance is 5.3856 mH.\n",
+ "(b)The emf induced in the search coil is 1.40 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.8,Page number: 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the mutual inductance and the coefficient of coupling between two coils.\"\"\"\n",
+ "\n",
+ "from math import pow,sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "N1=600.0 #Number of turns in the first coil\n",
+ "N2=1700.0 #Number of turns in the second coil \n",
+ "flux2=0.8/1000 #Magnetic flux produced in the second coil(in Webers) \n",
+ "I2=6 #Current in the second coil(in Amperes)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "L2=(N2*flux2)/I2\n",
+ "L1=L2*pow((N1/N2),2)\n",
+ "flux21=0.5/1000\n",
+ "k=flux21/flux2\n",
+ "M=k*sqrt(L1*L2)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"L1=%.4f H.\" %(L1)\n",
+ "print \"L2=%.4f H.\" %(L2)\n",
+ "print \"The coefficient of coupling(k)=%.4f.\" %(k) \n",
+ "print \"M=%.4f H.\" %(M)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "L1=0.0282 H.\n",
+ "L2=0.2267 H.\n",
+ "The coefficient of coupling(k)=0.6250.\n",
+ "M=0.0500 H.\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.9,Page number: 189 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the mutual inductance and the coefficient of coupling between two coils.\"\"\"\n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "N1=1200.0 #Number of turns in the first coil \n",
+ "flux1=0.25/1000 #Magnetic flux produced in the first coil(in Webers) \n",
+ "I1=5 #Current in the first coil(in Amperes)\n",
+ "N2=800.0 #Number of turns in the second coil\n",
+ "flux2=0.15/1000 #Magnetic flux produced in the second coil(in Webers)\n",
+ "I2=5 #Current in the second coil(in Amperes)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "L1=(N1*flux1)/I1\n",
+ "L2=(N2*flux2)/I2\n",
+ "k=0.6\n",
+ "flux12=k*flux1\n",
+ "M=(N2*flux12)/I1\n",
+ "k_new=M/sqrt(L1*L2)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The mutual inductance(M) is %.4f H.\" %(M)\n",
+ "print \"The coefficient of coupling is %.4f.\" %(k_new)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mutual inductance(M) is 0.0240 H.\n",
+ "The coefficient of coupling is 0.6325.\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.10,Page number: 192 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the mutual inductance and the coefficient of coupling between two coils.\"\"\"\n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "Lsa=1.4/1000 #Net inductance in series-aiding connections(in Henry) \n",
+ "Lso=0.6/1000 #Net inductance in series-opposing connections(in Henry) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "M=(Lsa-Lso)/4\n",
+ "\"\"\"Lsa=L1+L2+2M \n",
+ " L1+L2=1 mH; As the two coils are similar L1=L2=0.5mH \"\"\"\n",
+ "L1=0.5/1000\n",
+ "L2=0.5/1000\n",
+ "k=M/sqrt(L1*L2)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The mutual inductance is %.2f mH.\" %(M*1000)\n",
+ "print \"The coefficient of coupling(k) is %.2f.\" %(k)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mutual inductance is 0.20 mH.\n",
+ "The coefficient of coupling(k) is 0.40.\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.11,Page number: 193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the mutual inductance and the self-inductances of two coils. \"\"\"\n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "\"\"\" Equation 1 is L1+L2+(2*M)=1.8;\n",
+ " \n",
+ " Equation 2 is L1+L2-(2*M)=0.8. \"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "k=0.6 #Coefficient of coupling\n",
+ "eq1=1.8 #Net inductance when fluxes are in same direction(in Henry)\n",
+ "eq2=0.8 #Net inductance when fluxes are in opposite direction(in Henry)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "M=(eq1-eq2)/4\n",
+ "sum=(eq1+eq2)/2\n",
+ "product=(M*M)/(k*k)\n",
+ "diff=sqrt((sum*sum)-(4*product))\n",
+ "L1=(sum+diff)/2\n",
+ "L2=(sum-diff)/2\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The mutual inductance of the two coils is %.3f H.\" %(M)\n",
+ "print \"The self inducatnce of the first coil is %.3f H and the self inductance of the second coil is %.3f H.\" %(L1,L2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mutual inductance of the two coils is 0.250 H.\n",
+ "The self inducatnce of the first coil is 1.149 H and the self inductance of the second coil is 0.151 H.\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.12,Page number:195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the equivalent inductance of a combination of inductances connected in parallel.\"\"\"\n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "k=0.433 #Coefficient of coupling \n",
+ "L1=8 #Self-inductance of the first coil \n",
+ "L2=6 #Self-inductance of the second coil\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "M=k*sqrt(L1*L2)\n",
+ "Lpa=((L1*L2)-(M*M))/(L1+L2-(2*M))\n",
+ "Lpo=((L1*L2)-(M*M))/(L1+L2+(2*M))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The equivalent inductance such that the mutual induction assists the self induction is %.3f H.\" %(Lpa)\n",
+ "print \"(b)The equivalent inductance such that the mutual induction opposes the self induction is %.3f H.\" %(Lpo) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The equivalent inductance such that the mutual induction assists the self induction is 4.875 H.\n",
+ "(b)The equivalent inductance such that the mutual induction opposes the self induction is 1.950 H.\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.13,Page number: 196 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the number of turns in an air-cored coil.\"\"\"\n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "l=2.5e-02 #Length of the coil(in metres)\n",
+ "A=2e-04 #Average cross-sectional area of the coil(in square-metres)\n",
+ "L=400e-06 #Self-inductance of the coil(in Henry)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "abs_per=(4*pi)/(1e07)\n",
+ "N=sqrt((L*l)/(abs_per*A))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The number of turns in the air-cored coil is %d.\" %(round(N,0)) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of turns in the air-cored coil is 199.\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.14,Page number: 196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the mutual inductance between two coils and their self inductances.\"\"\"\n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "k=0.75 #Coefficient of coupling between two coils\n",
+ "I1=3.0 #Current in the first coil(in Amperes)\n",
+ "N1=250.0 #Number of turns in the first coil\n",
+ "flux1=4e-03 #Flux produced in the first coil(in Webers)\n",
+ "V2=70.0 #Voltage induced in the second coil due to first coil(in Volts)\n",
+ "di1=3.0 #Change in current in the first coil(in Amperes)\n",
+ "dt=3e-03 #Time interval(in seconds)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "L1=N1*(flux1/I1)\n",
+ "M=(V2*dt)/di1\n",
+ "L2=(M*M)/(k*k*L1)\n",
+ "N2=N1*sqrt(L2/L1)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"L1=%.4f H.\" %(L1)\n",
+ "print \"L2=%.4f H.\" %(L2)\n",
+ "print \"M=%.4f H.\" %(M)\n",
+ "print \"N2=%d.\" %(round(N2,0))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "L1=0.3333 H.\n",
+ "L2=0.0261 H.\n",
+ "M=0.0700 H.\n",
+ "N2=70.\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.15,Page number: 197"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the mean value of self inductance of a coil.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "N=1000.0 #Number of turns in the coil\n",
+ "A=20e-04 #Cross-sectional area of the coil(in square-metre)\n",
+ "I1=4.0 #First current(in Amperes)\n",
+ "B1=1.0 #Flux density associated with the first current(in Weber per sqyare-metre) \n",
+ "I2=9.0 #Second current(in Amperes)\n",
+ "B2=1.4 #Flux density associated with the first current(in Weber per sqyare-metre)\n",
+ "dt=0.05 #Time interval(in seconds)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "L1=(N*B1*A)/I1\n",
+ "L2=(N*B2*A)/I2\n",
+ "L=(L1+L2)/2.0\n",
+ "di=I2-I1\n",
+ "e=L*(di/dt)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The mean value of inductance between the given current limits is %.4f H.\" %(L)\n",
+ "print \"The emf induced in the coil is %.2f V.\" %(e)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mean value of inductance between the given current limits is 0.4056 H.\n",
+ "The emf induced in the coil is 40.56 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.16,Page number: 197"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the mutual inductance between two coils and their respective self-inductances.\"\"\"\n",
+ "\n",
+ "from math import pi,sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "N1=100.0 #Number of turns in the first coil\n",
+ "N2=150.0 #Number of turns in the second coil\n",
+ "A=125e-04 #Area of cross-section(in square-metres)\n",
+ "l=200e-02 #Mean length(in metres)\n",
+ "rel_per=2000.0 #Relative permeability of iron\n",
+ "k=1 #Coefficient of coupling\n",
+ "\n",
+ "\"\"\" NOTE: As the two coils are wound side by side,there is tight coupling. Therefore, k=1. \"\"\"\n",
+ "\n",
+ "#Calculations:\n",
+ "abs_per=(4*pi)/(1e07)\n",
+ "L1=(N1*N1*rel_per*abs_per*A)/l\n",
+ "L2=(N2*N2*rel_per*abs_per*A)/l\n",
+ "M=k*sqrt(L1*L2)\n",
+ "di1=5.0\n",
+ "dt=0.02\n",
+ "e2=M*(di1/dt)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The self inductances of the tow coils are: L1=%.3f mH and L2=%.3f mH.\" %((L1*1000.0),(L2*1000.0))\n",
+ "print \"(b)The mutual inductance between the two coils is %.3f mH.\" %(M*1000.0)\n",
+ "print \"(c)The emf induced in the second coil is %.2f V.\" %(e2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The self inductances of the tow coils are: L1=157.080 mH and L2=353.429 mH.\n",
+ "(b)The mutual inductance between the two coils is 235.619 mH.\n",
+ "(c)The emf induced in the second coil is 58.90 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.17,Page number: 198"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the coeffcient of coupling and the self-inductance of two coils.\"\"\" \n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "Lsa=4.0 #Equivalent inductance of series aiding(in Henry)\n",
+ "Lso=0.8 #Equivalent inductance of series opposing(in Henry)\n",
+ "\n",
+ "\"\"\" NOTE: Lsa=L+L+(2*M);\n",
+ " Lso=L+L-(2*M); \"\"\"\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "L=(Lsa+Lso)/4.0\n",
+ "M=(Lsa-Lso)/4.0\n",
+ "k=M/sqrt(L*L)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The self inductance of each coil is %.2f H.\" %(L)\n",
+ "print \"The coefficient of coupling is %.3f.\" %(round(k,3))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The self inductance of each coil is 1.20 H.\n",
+ "The coefficient of coupling is 0.667.\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.18,Page number: 198"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the equivalent inductance of different combinations of two coils.\"\"\"\n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "L1=200e-03 #Self-inductance of the first coil(in Henry) \n",
+ "L2=800e-03 #Self-inductance of the second coil(in Henry)\n",
+ "k=0.5 #Coefficient of coupling between two coils\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "M=k*sqrt(L1*L2)\n",
+ "Lsa=L1+L2+(2*M)\n",
+ "Lso=L1+L2-(2*M)\n",
+ "Lpa=((L1*L2)-(M*M))/Lso\n",
+ "Lpo=((L1*L2)-(M*M))/Lsa\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The equivalent inductance of series aiding is %.3f mH.\" %(Lsa*1000.0)\n",
+ "print \"(b)The equivalent inductance of series opposing is %.3f mH.\" %(Lso*1000.0)\n",
+ "print \"(c)The equivalent inductance of parallel aiding is %.3f mH.\" %(Lpa*1000.0)\n",
+ "print \"(d)The equivalent inductance of parallel opposing is %.3f mH.\" %(Lpo*1000.0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The equivalent inductance of series aiding is 1400.000 mH.\n",
+ "(b)The equivalent inductance of series opposing is 600.000 mH.\n",
+ "(c)The equivalent inductance of parallel aiding is 200.000 mH.\n",
+ "(d)The equivalent inductance of parallel opposing is 85.714 mH.\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.19,Page number: 198"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the exciting current for a horse-shoe magnet.\"\"\"\n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "l=45e-02 #Length of the iron path(in metres)\n",
+ "A=6e-04 #Cross-sectional area of the wrought iron bar(in square-metres)\n",
+ "N=500.0 #Number of turns in exciting coil\n",
+ "load=60.0 #Load to be lifted(in kilograms)\n",
+ "rel_per=800.0 #Relative permeability of iron\n",
+ "g=9.8 #Accelaration due to gravity(in metre per square-seconds) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "abs_per=(4*pi)/(1e07)\n",
+ "F=(load/2.0)*g\n",
+ "B=sqrt((2*abs_per*F)/A)\n",
+ "H=B/(abs_per*rel_per)\n",
+ "At=H*l\n",
+ "I=At/(N*2)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The exciting current needed for the magnet is %.5f A.\" %(I)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The exciting current needed for the magnet is 0.49674 A.\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |