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| -rwxr-xr-x | Fundamental_Electrical_and_Electronic_Principles/ch4.ipynb | 709 | ||||
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| -rwxr-xr-x | Fundamental_Electrical_and_Electronic_Principles/ch6.ipynb | 721 | ||||
| -rwxr-xr-x | Fundamental_Electrical_and_Electronic_Principles/ch7.ipynb | 78 | ||||
| -rwxr-xr-x | Fundamental_Electrical_and_Electronic_Principles/ch8.ipynb | 165 | ||||
| -rwxr-xr-x | Fundamental_Electrical_and_Electronic_Principles/ch9.ipynb | 206 | ||||
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diff --git a/Fundamental_Electrical_and_Electronic_Principles/ch1.ipynb b/Fundamental_Electrical_and_Electronic_Principles/ch1.ipynb new file mode 100755 index 00000000..ec327a6f --- /dev/null +++ b/Fundamental_Electrical_and_Electronic_Principles/ch1.ipynb @@ -0,0 +1,818 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:d510303f7e7e2853fb2cf56867915796aa62fb3bbb679ce357903afade2bd69c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1: Fundamentals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.1, Page 3" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I =.000018; # Electric current, A\n", + "V = 15000; # Electric potential, V\n", + "P = 250000000 # Electric Power, W\n", + "\n", + "#Calculations&Results\n", + "# Display standard form \n", + "print \"Standard form:\"\n", + "print \"==============\"\n", + "print \"%f A = %3.1e A\"%(I, I)\n", + "print \"%5.0f V = %3.1e V\"%(V, V)\n", + "print \"%9.0f W = %3.1e W\"%(P, P)\n", + "# Display scientific notation \n", + "print \"\\n\\nScientific form:\"\n", + "print \"================\"\n", + "print \"%f A = %2d micro-ampere\"%(I, I/1e-06)\n", + "print \"%5.0f V = %2d kilo-volt\"%(V, V/1e+03)\n", + "print \"%9.0f W = %3d mega-watt\"%(P, P/1e+06)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Standard form:\n", + "==============\n", + "0.000018 A = 1.8e-05 A\n", + "15000 V = 1.5e+04 V\n", + "250000000 W = 2.5e+08 W\n", + "\n", + "\n", + "Scientific form:\n", + "================\n", + "0.000018 A = 18 micro-ampere\n", + "15000 V = 15 kilo-volt\n", + "250000000 W = 250 mega-watt\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.2, Page 3" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I = 25e-05; # Electric Current,A\n", + "P = 3e-04; # Electric Power, W\n", + "W = 850000.; # Work done, J\n", + "V = 0.0016; # Electric Potential, V\n", + "\n", + "#Calculations&Results\n", + "print \"Scientific (Engineering) notation:\";\n", + "print \"===================================\";\n", + "print \"%2e A = %3d micro-ampere = %3.2f mA\"%(I, I/1e-06, I/1e-03);\n", + "print \"%1.0e W = %.e milli-watt\"%(P, P/1e-03);\n", + "print \"%6d J = %3d kJ = %3.2f MJ\"%(W, W/1e+03, W/1e+06);\n", + "print \"%5.4f V = %3.1f milli-volt\"%(V, V/1e-03)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Scientific (Engineering) notation:\n", + "===================================\n", + "2.500000e-04 A = 250 micro-ampere = 0.25 mA\n", + "3e-04 W = 3e-01 milli-watt\n", + "850000 J = 850 kJ = 0.85 MJ\n", + "0.0016 V = 1.6 milli-volt\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3, Page 5" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "m = 750/1e+03; # Mass of the body, kg\n", + "F = 2; # Force acting on the mass, N\n", + "\n", + "#Calculations\n", + "# Since F = m * a, (Newton's Second Law of motion), solving for a\n", + "a = F/m; # Acceleration produced in the body, metre per second square\n", + "\n", + "# Result\n", + "print \"The acceleration produced in the body = %5.3f metre per second square\"%a\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The acceleration produced in the body = 2.667 metre per second square\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.4, Page 9" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Q = 35e-03; # Electric charge, C\n", + "t = 20e-03; # Time for transference of charge between two points, s\n", + "\n", + "#Calculations\n", + "# Since Q = I * t, solving for I\n", + "I = Q/t; # Electric current flowing between the two points, A\n", + "\n", + "# Result\n", + "print \"The value of electric current flowing = %4.2f A\"%I\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of electric current flowing = 1.75 A\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5, Page 9" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I = 120e-06; # Electric current, A\n", + "t = 15; # Time for transference of charge between two points, s\n", + "\n", + "#Calculations\n", + "# Since I = Q/t, solving for Q\n", + "Q = I*t; # Electric charge transferred, C\n", + "\n", + "# Result\n", + "print \"The value of electric charge transferred = %3.1f mC\"%(Q/1e-03)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of electric charge transferred = 1.8 mC\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6, Page 10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Q = 80; # Electric charge, C\n", + "I = 0.5; # Electric current, A\n", + "\n", + "#Calculations\n", + "# Since Q = I*t, solving for t\n", + "t = Q/I; # Time for transference of charge between two points, s\n", + "\n", + "# Result\n", + "print \"The duration of time for which the current flowed = %3d s\"%t\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The duration of time for which the current flowed = 160 s\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.7, Page 13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I = 5.5e-03; # Electric current, A\n", + "R = 33000; # Resistance, ohms\n", + "\n", + "#Calculations\n", + "# From Ohm's law, V = I*R\n", + "V = I*R; # Potential difference across resistor, V\n", + "\n", + "# Result\n", + "print \"The potential difference developed across resistor = %5.1f V\"%V\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The potential difference developed across resistor = 181.5 V\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.8, Page 14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V = 24.; # Potential difference,V\n", + "R = 15; # Resistance, ohms\n", + "\n", + "#Calculations\n", + "# From Ohm's law, V = I*R, then solving for I\n", + "I = V/R; # Electric current, A\n", + "\n", + "# Result\n", + "print \"The current flowing through the resistor = %3.1f A\"%I\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current flowing through the resistor = 1.6 A\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.9, Page 16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "E = 6; # E.m.f of battery, V\n", + "r = 0.15; # Internal resistance of battery, ohm\n", + "I_1 = .5; # Electric current, A\n", + "I_2 = 2; # Electric current, A\n", + "I_3 = 10; # Electric current, A\n", + "\n", + "#Calculations\n", + "# Using relation V = E - I*R and substituting the values of I_1, I_2 and I_3 one by one in it\n", + "V_1 = E - I_1*r; # Terminal potential difference, V\n", + "V_2 = E - I_2*r; # Terminal potential difference, V\n", + "V_3 = E - I_3*r; # Terminal potential difference, V\n", + "\n", + "# Results\n", + "print \"The terminal potential difference developed across resistor for a current of %3.1f A = %5.3f V\"%(I_1,V_1)\n", + "print \"The terminal potential difference developed across resistor for a current of %1d A = %3.1f V\"%(I_2,V_2)\n", + "print \"The terminal potential difference developed across resistor for a current of %2d A = %3.1f V\"%(I_3,V_3)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The terminal potential difference developed across resistor for a current of 0.5 A = 5.925 V\n", + "The terminal potential difference developed across resistor for a current of 2 A = 5.7 V\n", + "The terminal potential difference developed across resistor for a current of 10 A = 4.5 V\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.10, Page 16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "E = 12; # E.m.f, V\n", + "I = 5; # Electric current, A\n", + "V = 11.5; # Terminal potential difference, V\n", + "\n", + "#Calculations\n", + "# Using relation V = E - I*r, solving for r\n", + "r = ( E - V )/I; # Internal resistance of battery, ohm\n", + "# From Ohm's law, V = I*R, then solving for R\n", + "R = V/I; # Resistance, ohms\n", + "\n", + "# Results\n", + "print \"The internal resistance of battery = %3.1f ohm\"%r\n", + "print \"The resistance of external circuit = %3.1f ohm\"%R\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The internal resistance of battery = 0.1 ohm\n", + "The resistance of external circuit = 2.3 ohm\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.11, Page 17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I = 200e-03; # Electric current, A\n", + "t = 300; # Time for which current flows, s\n", + "R = 750; # Resistance, ohms\n", + "\n", + "#Calculations\n", + "# Using Ohm's law, V = I*R\n", + "V = I*R; # Electric potential difference, V\n", + "W = I**2*R*t; # Energy dissipated, joule\n", + "\n", + "# Result\n", + "print \"The potential difference developed across the resistor = %3d V\\nThe energy dissipated across the resistor = %4.0f J or %1d kJ\"%(V, W, W*1e-03)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The potential difference developed across the resistor = 150 V\n", + "The energy dissipated across the resistor = 9000 J or 9 kJ\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.12, Page 18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "R = 680; # Resistance, ohms\n", + "P = 85e-03; # Electric power, W\n", + "\n", + "#Calculations\n", + "# Using P = V**2/R, solving for V\n", + "V = math.sqrt( P*R ); # Potential difference, V\n", + "# Using P = I**2*R, solving for I\n", + "I = math.sqrt( P/R ); # Electric current, A\n", + "\n", + "# Result\n", + "print \"The potential difference developed across the resistance = %3.1f V\\nThe current flowing through the resistor = %5.2f mA\"%(V, I/1e-03)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The potential difference developed across the resistance = 7.6 V\n", + "The current flowing through the resistor = 11.18 mA\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.13, Page 19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I = 1.4; # Electric current, A\n", + "t = 900; # Time for which current flows, s\n", + "W = 200000; # Energy dissipated, J\n", + "\n", + "#Calculations\n", + "# Using relation W = V*I*t, solving for V\n", + "V = W/( I*t ); # Potential difference, V\n", + "# Using relation P = V*I\n", + "P = V*I; # Electric power, W\n", + "# From Ohm's law, V = I*R, solving for R\n", + "R = V/I; # Resistance, ohm\n", + "\n", + "# Result\n", + "print \"The potential difference developed = %5.1f V\\nThe power dissipated = %5.1f W\\nThe resistance of the circuit = %5.1f ohm\"%(V, P, R)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The potential difference developed = 158.7 V\n", + "The power dissipated = 222.2 W\n", + "The resistance of the circuit = 113.4 ohm\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14, Page 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "P = 12.5; # Power of the machine, kW\n", + "t = 8.5; # Time for which the machine is operated, h\n", + "\n", + "#Calculations\n", + "W = P*t; # Electric energy, kWh\n", + "# Cost per unit = 7.902 p, therefore calculating the cost of 106.25 units\n", + "cost = ( W*7.902 ); # Cost for operating machine, p\n", + "\n", + "# Result\n", + "print \"The cost of operating the machine = %4.2f pounds\"%(cost*1e-02)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The cost of operating the machine = 8.40 pounds\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15, Page 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Total_bill = 78.75; # pounds\n", + "Standing_charge = 15.00; # pounds\n", + "Units_used = 750; # kWh\n", + "\n", + "#Calculations\n", + "Cost_per_unit = ( Total_bill - Standing_charge )/Units_used; # p\n", + "Cost_of_energy_used = 67.50; # pounds\n", + "Total_bill = Cost_of_energy_used + Standing_charge; # pounds\n", + "\n", + "# Result\n", + "print \"The cost per unit = %5.3f pounds or %3.1f p\\nTotal bill = %5.2f pounds\"%(Cost_per_unit,Cost_per_unit/1e-02,Total_bill)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The cost per unit = 0.085 pounds or 8.5 p\n", + "Total bill = 82.50 pounds\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.16, Page 22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "l = 200; # Length of Cu wire, metre\n", + "rho = 2e-08; # Resistivity of Cu, ohm-metre\n", + "A = 8e-07; # Cross sectional area of Cu wire, metre square\n", + "\n", + "#Calculations\n", + "# Using relation R = ( rho*l )/A\n", + "R = ( rho*l )/A; # Resistance, ohm\n", + "\n", + "# Result\n", + "print \"The resistance of the coil = %1d ohm\"%R\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance of the coil = 5 ohm\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.17, Page 22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "l = 250; # Length of Cu wire, metre\n", + "d = 5e-04; # Diameter of Cu wire, metre\n", + "rho = 1.8e-08; # Resistivity of Cu wire, ohm-metre\n", + "\n", + "#Calculations\n", + "A = (math.pi*d**2 )/4; # Cross sectional area of Cu wire, metre square\n", + "# Using relation R = rho*l/A\n", + "R = rho*l/A; # Resistance, ohm\n", + "\n", + "# Result\n", + "print \"The resistance of the coil = %5.2f ohm\"%R\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance of the coil = 22.92 ohm\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.18, Page 23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_1 = 250; # Resistance of field coil, ohm\n", + "Theta_1 = 15; # Initial temperature of motor, degree celcius \n", + "Theta_2 = 45; # Final temperature of motor, degree celcius\n", + "Alpha = 4.28e-03; # Temperature coefficient of resistance, per degree celcius\n", + "\n", + "#Calculations\n", + "# Using relation, R_1/R_2 = ( 1 + Alpha*Theta_1 )/( 1 + Alpha*Theta_2 ), solving for R_2\n", + "R_2 = R_1 * (( 1 + Alpha*Theta_2 )/( 1 + Alpha*Theta_1 )); # Resistance, ohms\n", + "\n", + "# Result\n", + "print \"The resistance of field coil at %2d degree celcius = %5.1f ohm\"%(Theta_2, R_2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance of field coil at 45 degree celcius = 280.2 ohm\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.19, Page 24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_0 = 350; # Resistance, ohms\n", + "Theta_1 = 60; # Temperature, degree celcius \n", + "Alpha = 4.26e-03; # Temperature coefficient, per degree celcius\n", + "\n", + "#Calculations\n", + "# Using relation R_1 = R_0 * ( 1 + Alpha*Theta_1 )\n", + "R_1 = R_0 * ( 1 + Alpha*Theta_1 ); # Resistance, ohms\n", + "\n", + "# Result\n", + "print \"The resistance of the wire at %2d degree celcius = %5.1f ohm\"%(Theta_1, R_1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance of the wire at 60 degree celcius = 439.5 ohm\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.20, Page 24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_1 = 120; # Resistance, ohms\n", + "Theta_1 = 16; # Temperature, degree celcius \n", + "Theta_2 = 32; # Temperature, degree celcius\n", + "Alpha = -4.8e-04; # Temperature coefficient, per degree celcius\n", + "\n", + "#Calculations\n", + "# Using relation, R_1/R_2 = ( 1 + Alpha*Theta_1 )/( 1 + Alpha*Theta_2 ), solving for R_2\n", + "R_2 = R_1 * (( 1 + Alpha*Theta_2 )/( 1 + Alpha*Theta_1 )); # Resistance, ohm\n", + "\n", + "# Result\n", + "print \"The resistance of carbon resistor at %2d degree celcius = %5.1f ohm\"%(Theta_2, R_2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance of carbon resistor at 32 degree celcius = 119.1 ohm\n" + ] + } + ], + "prompt_number": 20 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamental_Electrical_and_Electronic_Principles/ch2.ipynb b/Fundamental_Electrical_and_Electronic_Principles/ch2.ipynb new file mode 100755 index 00000000..3c5ca034 --- /dev/null +++ b/Fundamental_Electrical_and_Electronic_Principles/ch2.ipynb @@ -0,0 +1,788 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e72fe3ba0215215547973defaa93d1e04f3f859c30d61d6e8cd806a66ccb1f3d" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2: D.C. Circuits" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1, Page 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "E = 24.; # E.m.f of battery,V\n", + "R1 = 330; # Resistance, ohms\n", + "R2 = 1500; # Resistance, ohms\n", + "R3 = 470; # Resistance, ohms\n", + "\n", + "#Calculations\n", + "# As resistances R1, R2 & R3 are joined end-to-end hence, they are in series & in series connection, circuit resistance is the sum of individual resistances present in the circuit\n", + "R = R1 + R2 + R3; # Resistance of circuit, ohms\n", + "I = E/R; # Circuit current, A\n", + "# As the resistances are in series so same current flows through each resistor & potential drop across each resistor is equal to the product of circuit current & its respective resistance( from Ohm's law, V = I*R )\n", + "V1 = I*R1; # Potential difference developed across resistance R1, V\n", + "V2 = I*R2; # Potential difference developed across resistance R2, V\n", + "V3 = I*R3; # Potential difference developed across resistance R3, V\n", + "P = E*I; # Electric power dissipated by the complete circuit, W\n", + "\n", + "# Results\n", + "print \"The circuit resistance = %4d ohms or %3.1f kilo-ohms\"%(R, R*1e-03);\n", + "print \"The circuit current = %5.2f milli-ampere\"%(I/1e-03);\n", + "print \"The potential drop across resisatnce R1 = %4.2f volts\\nThe potential drop across resistance R2 = %5.2f volts\\nThe potential drop across resistance R3 = %4.2f volts\"%(V1, V2, V3);\n", + "print \"The power dissipated by the complete circuit = %4.2f watt or %3d milli-watt\"%(P,P/1e-03)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The circuit resistance = 2300 ohms or 2.3 kilo-ohms\n", + "The circuit current = 10.43 milli-ampere\n", + "The potential drop across resisatnce R1 = 3.44 volts\n", + "The potential drop across resistance R2 = 15.65 volts\n", + "The potential drop across resistance R3 = 4.90 volts\n", + "The power dissipated by the complete circuit = 0.25 watt or 250 milli-watt\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2, Page 34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "E = 12.; # E.m.f of battery, V\n", + "R_BC = 16.; # Resistance across branch BC, ohms\n", + "P_BC = 4; # Electric power dissipated by resistance R_BC, W\n", + "\n", + "#Calculations\n", + "# using relation P = I^2/R, solving for I\n", + "I = math.sqrt( P_BC/R_BC); # Electric current,A\n", + "R = E/I; # Total circuit resistance, ohms\n", + "R_AB = R - R_BC; # Resistance across branch AB, ohms\n", + "\n", + "# Result\n", + "print \"The circuit current = %3.1f A\\nThe value of other resistor = %1d ohms\"%(I, R_AB)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The circuit current = 0.5 A\n", + "The value of other resistor = 8 ohms\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3, Page 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "E = 24.; # E.m.f of battery, V\n", + "R_1 = 330.; # Resistance, ohms\n", + "R_2 = 1500.; #Resistance, ohms\n", + "R_3 = 470.; #Resistance, ohms\n", + "\n", + "#Calculations\n", + "# Since one end of each resistor is connected to positive terminal of battery and the other end to the negative terminal, therefore, the resistors are in parallel & in parallel connection the equivalent resistance of the circuit is equal to the reciprocal of the sum of conductances of individual resistances present in the circuit i.e 1/R = 1/R_1 + 1/R_2 + 1/R_3, solving for R \n", + "R = (R_1*R_2*R_3)/( R_1*R_2 + R_2*R_3 + R_3*R_1); # Equivalent resisance of circuit, ohms\n", + "# Since the resistances are in parallel so potetial difference across each resistor is same & in our case is equal to e.m.f of battery & from Ohm's law, V = I*R, solving for I\n", + "I_1 = E/R_1; # Current through resistor R_1, A\n", + "I_2 = E/R_2; # Current through resistor R_2, A\n", + "I_3 = E/R_3; # Current through resistance R_3, A\n", + "# Current drawn from battery is equal to the sum of branch currents\n", + "I = I_1 + I_2 + I_3; # Current drawn from battery, A\n", + "\n", + "# Results\n", + "print \"The total resistance of the circuit = %6.2f ohms\"%(R);\n", + "print \"The branch current I1 = %5.2f mA\\nThe branch current I2 = %2d mA\\nThe branch current I3 = %5.2f mA\"%(I_1/1e-03, I_2/1e-03, I_3/1e-03);\n", + "print \"The current drawn from the battery = %5.1f mA\"%(I/1e-03)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total resistance of the circuit = 171.68 ohms\n", + "The branch current I1 = 72.73 mA\n", + "The branch current I2 = 16 mA\n", + "The branch current I3 = 51.06 mA\n", + "The current drawn from the battery = 139.8 mA\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.4, Page 38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "E = 12; # E.m.f of battery, V\n", + "R1 = 6; # Resistance, ohms\n", + "R2 = 3; # Resistance, ohms\n", + "\n", + "#Calculations\n", + "# Since the two resistances are in parallel, therefore effective resistance of the circuit is equal to the reciprocal \n", + "#of the sum of conductances ( 1/Ressistance) of individual resistances present in the circuit i.e 1/R = 1/R1 + 1/R2, \n", + "#simplifying for R \n", + "R = ( R1*R2)/(R1 + R2); # Effective resistance of the circuit, ohms\n", + "# Fron Ohm's law, V = I*R, solving for I\n", + "I = E/R; # Circuit current, A\n", + "I1 = E/R1; # Current through resistance R1, A\n", + "I2 = E/R2; # Current thrugh resistance R2, A\n", + "\n", + "# Results\n", + "print \"Effective resistance of the circuit = %1d ohms\"%R\n", + "print \"The current drawn from the battery = %1d A\"%I\n", + "print \"The current through resistor R1 = %1d A\"%I1\n", + "print \"The current through R2 resistor = %1d A\"%I2\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Effective resistance of the circuit = 2 ohms\n", + "The current drawn from the battery = 6 A\n", + "The current through resistor R1 = 2 A\n", + "The current through R2 resistor = 4 A\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5, Page 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1 = 10.; # Resistance, ohm\n", + "R2 = 20.; # Resistance, ohm\n", + "R3 = 30.; # Resistance, ohm\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Since in series combination, the equivalent resistance of the circuit is the sum of the individual resistances \n", + "#present in the circuit i.e R = R1 + R2 + R3\n", + "R_s = R1 + R2 + R3; # Equivalent series resistance of the circuit, ohms\n", + "# Part (b)\n", + "# Since in parallel combination, the equivalent resistance of the circuit is the reciprocal of the sum of the \n", + "#conductances of the individual resistances present in the circuit i.e 1/R = 1/R1 + 1/R2 + 1/R3, solving for R;\n", + "R_p = ( R1*R2*R3 )/( R1*R2 + R2*R3 + R3*R1 ); # Equivalent parallel resistance of the circuit, ohms\n", + "\n", + "# Results\n", + "print \"Equivalent series resistance of the circuit = %2d ohm\"%R_s\n", + "print \"Equivalent parallel resistance of the circuit = %4.2f ohm\"%R_p\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent series resistance of the circuit = 60 ohm\n", + "Equivalent parallel resistance of the circuit = 5.45 ohm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.6, Page 44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "E = 64.; # E.m.f of battery, V\n", + "R1 = 6.; # Resistance, ohm\n", + "R2 = 4.; # Resistance, ohm\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Since R1 & R2 are parallel to one another hence, their equivalent resistance is equal to the sum of reciprocal of \n", + "#their individual resistances\n", + "R_BC = ( R1*R2)/( R1 + R2 ); # Equivalent resistance across branch BC, ohm\n", + "R_AB = 5.6; # Resistance across branch AB, ohm\n", + "# Since R_AB & R_BC are in series, therefore, their equivalent resistance is equal to the sum of their individual \n", + "#resistances\n", + "R_AC = R_AB + R_BC; # Total circuit resistance, ohm\n", + "# From Ohm's law, V = I*R, solving for I\n", + "I = E/R_AC; # Total circuit current, A\n", + "# Part (b)\n", + "V_BC = I*R_BC; # Potential difference across branch BC, V\n", + "I1 = V_BC/R1; # Electric current through resistor R1, A\n", + "# Part (c)\n", + "# Since P = I^2*R\n", + "P_AB = I**2*R_AB; # Power dissipated by 5.6 ohm resistance, W\n", + "\n", + "# Results\n", + "print \"The current drawn from the supply = %1d A \"%I;\n", + "print \"The current through %1d ohm resistor = %3.1f A\"%(R1, I1);\n", + "print \"The power dissipated by %3.1f ohm resistor = %5.1f W\"%(R_AB, P_AB);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current drawn from the supply = 8 A \n", + "The current through 6 ohm resistor = 3.2 A\n", + "The power dissipated by 5.6 ohm resistor = 358.4 W\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7, Page 46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "E = 18.; # E.m.f of battery, V\n", + "R1 = 4.; # Resistance, ohm\n", + "R2 = 6.; # Resistance, ohm\n", + "R3 = 5.; # Resistance, ohm\n", + "R4 = 3.; # Resistance, ohm\n", + "R5 = 6.; # Resistance, ohm\n", + "R6 = 8.; # Resistance, ohm\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Since resistance R1 & R2 are in parallel, therefore, equivalent resistance across branch AB will be equal to the \n", + "#reciprocal of the sum of conductances ( 1/Ressistance) of individual resistances present in the circuit \n", + "#i.e 1/R_AB = 1/R1 + 1/R2, simplifying for R_AB \n", + "R_AB = ( R1*R2 )/( R1 + R2); # Resistance, ohm\n", + "R_BC = R3; # Resistance across branch BC, ohm\n", + "# Since resistance R4, R5 & R6 are in parallel, therefore, equivalent resistance across branch CD will be equal to \n", + "#the reciprocal of the sum of conductances ( 1/Ressistance) of individual resistances present in the circuit \n", + "#i.e 1/R_CD = 1/R4 + 1/R5 + 1/R6, simplifying for R _CD\n", + "R_CD = ( R4*R5*R6 )/( R4*R5 + R5*R6 + R6*R4 ); # Resistance, ohm\n", + "# Since R_AB, R_BC & R_CD forms series combination, therefore circuit resistance will be their series sum\n", + "R = R_AB + R_BC + R_CD; # Circuit resistance, ohm\n", + "I = E/R; # Supply current, A\n", + "# Part (b)\n", + "# AS resistances R1 & R2 are parallel, therefore tere will be same potential difference across them, denoted by V_AB\n", + "V_AB = I*R_AB; # Potential difference, V\n", + "# AS resistances R4, R5 & R6 are parallel, therefore tere will be same potential difference across them, denoted by V_CD\n", + "V_CD = I*R_CD; # Potential difference, V\n", + "V_BC = I*R_BC; # Potential difference, V\n", + "# Part (c)\n", + "I1 = V_AB/R1; # Current through R1 resistor, A\n", + "I2 = V_AB/R2; # Current through R2 resistor, A\n", + "I4 = V_CD/R4; # Current through R4 resistor, A\n", + "I5 = V_CD/R5; # Current through R5 resistor, A\n", + "I6 = V_CD/R6; # Current through R6 resistor, A\n", + "# Part (d)\n", + "P3= I**2*R3; # Power dissipated, W\n", + "\n", + "# Results\n", + "print \"The current drawn from the source = %1d A\"%I\n", + "print \"The p.d. across resistor %1d ohm & %1d ohm = %3.1f V\"%(R1, R2, V_AB);\n", + "print \"The p.d. across resistor %1d ohm, %1d ohm & %1d ohm = %3.1f V\"%(R4, R5, R6, V_CD);\n", + "print \"The p.d. across resistor %1d ohm = %2d V\"%(R3, V_BC);\n", + "print \"The current through resistor %1d ohm = %3.1f A\"%(R1, I1);\n", + "print \"The current through resistor %1d ohm = %3.1f A\"%(R2, I2);\n", + "print \"The current through resistor %1d ohm = %1d A\"%(R3, I);\n", + "print \"The current through resistor %1d ohm = %5.3f A\"%(R4, I4);\n", + "print \"The current through resistor %1d ohm = %5.3f A\"%(R5,I5);\n", + "print \"The current through resistor %1d ohm = %3.1f A\"%(R6, I6);\n", + "print \"The power dissipated by the %1d ohm resistor = %2d W\"%(R3, P3);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current drawn from the source = 2 A\n", + "The p.d. across resistor 4 ohm & 6 ohm = 4.8 V\n", + "The p.d. across resistor 3 ohm, 6 ohm & 8 ohm = 3.2 V\n", + "The p.d. across resistor 5 ohm = 10 V\n", + "The current through resistor 4 ohm = 1.2 A\n", + "The current through resistor 6 ohm = 0.8 A\n", + "The current through resistor 5 ohm = 2 A\n", + "The current through resistor 3 ohm = 1.067 A\n", + "The current through resistor 6 ohm = 0.533 A\n", + "The current through resistor 8 ohm = 0.4 A\n", + "The power dissipated by the 5 ohm resistor = 20 W\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.8, Page 49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration&Calculations\n", + "# Applying Kirchhoff's current law (the sum of the currents arriving at a junction is equal to the sum of the \n", + "#currents leaving that junction) at junction A\n", + "I2 = 40 + 10; # Electric current, A\n", + "# Applying Kirchhoff's current law at junction C\n", + "I1 = 80 - I2; # Electric current, A\n", + "# Applying Kirchhoff's current law at junction D\n", + "I3 = 80 + 30; # Electric current, A\n", + "# Applying Kirchhoff's current law at junction E\n", + "I4 = I3 - 25; # Electric current, A\n", + "# Applying Kirchhoff's current law at junction F\n", + "I5 = 30 - 85; # Electric current, A\n", + "\n", + "# Result\n", + "print \"Current I1 = %2d A\\nCurrent I2 = %2d A\\nCurrent I3 = %3d A\\nCurrent I4 = %2d A\\nCurrent I5 = %2d A,\"%(I1, I2, I3, I4, I5);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current I1 = 30 A\n", + "Current I2 = 50 A\n", + "Current I3 = 110 A\n", + "Current I4 = 85 A\n", + "Current I5 = -55 A,\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.9, Page 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import numpy as np\n", + "from numpy import linalg\n", + "from numpy.linalg import inv\n", + "\n", + "#Variable declaration\n", + "R1 = 3; # Resistance, ohms\n", + "R2 = 2; # Resistance, ohms\n", + "R3 = 10; # Resistance, ohms\n", + "E1 = 10; # E.m.f, V\n", + "E2 = 4; # E.m.f, V\n", + "\n", + "#Calculations\n", + "# Applying Kirchhoff's Current Law(the sum of the currents arriving at a junction is equal to the sum of the \n", + "#currents leaving that junction) \n", + "A = np.array([[3.,-2.],[13.,10.]])\n", + "B = np.array([6.,10.])\n", + "I1,I2 = np.linalg.solve(A,B)\n", + "I3 = ( I1 + I2 ); # Electric current through branch CD, A\n", + "V_CD = R3*I3; # P.d.across R3 resistor, V\n", + "\n", + "# Results\n", + "print \"The current through branch FA = %6.3f A\"%I1\n", + "print \"The current through branch EB = %5.3f A\"%I2\n", + "print \"The current through branch CD = %5.3f A\"%I3\n", + "print \"p.d.across %2d resistor = %4.2f V\"%(R3,V_CD)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through branch FA = 1.429 A\n", + "The current through branch EB = -0.857 A\n", + "The current through branch CD = 0.571 A\n", + "p.d.across 10 resistor = 5.71 V\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.10, Page 53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import numpy as np\n", + "from numpy import linalg\n", + "\n", + "#Variable declaration\n", + "E1 = 6; # E.m.f of battery, V\n", + "E2 = 4.5; # E.m.f of battery, V\n", + "R1 = 1.5; # Resistance, ohm\n", + "R2 = 2; # Resistance, ohm\n", + "R3 = 5; # Resistance, ohm\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Using matrix method for solving set of equations\n", + "A = np.array([[6.5, 5], [5, 7]])\n", + "B = np.array([6, 4.5]);\n", + "#X = inv(A)*B;\n", + "I1,I2 = np.linalg.solve(A,B) # Electric current through branch FA & DC, A\n", + "I3 = ( I1 + I2); # Electric current through branch BE, A\n", + "# Part (b)\n", + "V_BE = I3*R3; # P.d across resistor R3, V\n", + "\n", + "# Results\n", + "print \"Electric current through branch FA = %5.3f A\"%I1\n", + "print \"Electric current through branch DC = %6.4f A\"%I2\n", + "print \"Electric current through branch BE = %5.3f A\"%I3\n", + "print \"p.d across resistor %1d ohms = %5.3f V\"%(R3, V_BE);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Electric current through branch FA = 0.951 A\n", + "Electric current through branch DC = -0.0366 A\n", + "Electric current through branch BE = 0.915 A\n", + "p.d across resistor 5 ohms = 4.573 V\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.11, Page 57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import numpy as np\n", + "from numpy import linalg\n", + "\n", + "#Variable declaration\n", + "R_AB = 6; # Resistance, ohm\n", + "R_BC = 4; # Resistance, ohm\n", + "R_DC = 1; # Resistance, ohm\n", + "R_AD = 3; # Resistance, ohm\n", + "R_BD = 5; # Resistance, ohm\n", + "\n", + "#Calculations\n", + "# Using matrix method for solving the set of equations \n", + "A = np.array([[6, -3, 5], [-4, 1, 10], [0, 4, 1]]);\n", + "B = ([0, 0, 10]);\n", + "#X = inv(A)*B;\n", + "I1,I2,I3 = np.linalg.solve(A,B); # Electric current, A\n", + "I_BC = I1 - I3; # Electric current, A\n", + "I_DC = I2 + I3; # Electric current, A\n", + "I = I1 + I2; # Supply current, A\n", + "\n", + "# Results\n", + "print \"The current through %1d ohm resistor = %5.3f A\"%(R_AB, I1)\n", + "print \"The current through %1d ohm resistor = %4.2f A\"%(R_BC, I_BC);\n", + "print \"The current through %1d ohm resistor = %5.3f A\"%(R_DC, I_DC);\n", + "print \"The current through %1d ohm resistor = %5.3f A\"%(R_AD, I2);\n", + "print \"The current through %1d ohm resistor = %5.3f A\"%(R_BD, I3);\n", + "print \"The supply current = %5.3f A\"%I\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through 6 ohm resistor = 1.074 A\n", + "The current through 4 ohm resistor = 0.89 A\n", + "The current through 1 ohm resistor = 2.638 A\n", + "The current through 3 ohm resistor = 2.454 A\n", + "The current through 5 ohm resistor = 0.184 A\n", + "The supply current = 3.528 A\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.12, Page 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_AB = 6; # Resistance across branch AB, ohm \n", + "R_AD = 3; # Resistance across branch AD, ohm\n", + "R_BC = 4; # Resistance across branch BC, ohm\n", + "R_DC = 2; # Resistance across branch DC, ohm\n", + "\n", + "#Calculations\n", + "# Since R_AB/R_AD = R_BC/R_DC, so the wheatstone bridge is balanced hence no current flows through branch BD\n", + "I3 = 0;\n", + "\n", + "# Result\n", + "print \"The current through branch BD i.e I3 = %1d A\"%I3\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through branch BD i.e I3 = 0 A\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.13, Page 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import numpy as np\n", + "from numpy import linalg\n", + "\n", + "#Variable declaration\n", + "R1 = 20; # Resistance, ohm\n", + "R2 = 10; # Resistance, ohm\n", + "R3 = 8; # Resistance, ohm\n", + "R4 = 5; # Resistance, ohm\n", + "R5 = 2; # Resistance, ohm\n", + "\n", + "#Calculations\n", + "A = np.array([[20, -10, 8], [-5, 2, 15], [0, 12, 2]]);\n", + "B = np.array([0, 0, 10]);\n", + "I1,I2,I3 = np.linalg.solve(A,B) # Electric current through BD, A\n", + "V_BD = I3*R3; # P.d across branch BD, V\n", + "# For balance conditions i.e I3 = 0, R1/R2 = R4/R5, solving for R4\n", + "R_4 = ( R1*R5 )/R2; # Resistance, ohm\n", + "\n", + "# Results\n", + "print \"The p.d between terminals B and D = %5.3f V\"%V_BD\n", + "print \"The value to which %1d ohm resistor must be adjusted in order to reduce the current through %1d ohm resistor to zero = %1d ohm\"%(R4, R3, R_4);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The p.d between terminals B and D = 0.195 V\n", + "The value to which 5 ohm resistor must be adjusted in order to reduce the current through 8 ohm resistor to zero = 4 ohm\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.14, Page 64" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "# For part (a)\n", + "Rm = 1000.; # Resistance, ohm\n", + "Rd = 1; # Resistance, ohm\n", + "Rv = 3502.; # Resistance, ohm\n", + "# Using Wheatstone bridge balanced condition i.e Rx/Rv = Rm/Rd , solving for Rx\n", + "Rx = ( Rm/Rd) * Rv; # Resistance,ohm\n", + "print \"The value of the resistance being measured = %5.3f mega-ohm\"%(Rx*1e-06);\n", + "\n", + "# Part (b)\n", + "Rm = 1.; # Resistance, ohm\n", + "Rd = 1000; # Resistance, ohm\n", + "Rv = 296; # Resistance, ohm\n", + "# Using Wheatstone bridge balanced condition i.e Rx/Rv = Rm/Rd , solving for Rx\n", + "Rx = ( Rm/Rd )*Rv; # Resistance,ohm\n", + "print \"The value of the resistance being measured = %5.3f ohm\"%Rx\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of the resistance being measured = 3.502 mega-ohm\n", + "The value of the resistance being measured = 0.296 ohm\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.15, Page 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "l1 = 600e-03; # Scale reading, metre\n", + "l2 = 745e-03; # Scale reading, metre\n", + "l_s = 509.3e-03; # Total scale length, metre\n", + "E_s = 1.0186; # Source voltage, V\n", + "\n", + "#Calculations\n", + "E1 = ( l1/l_s )*E_s; # Voltage drop across length l1, V\n", + "E2 = ( l2/l_s)*E_s; # Voltage drop across length l2, V\n", + "\n", + "# Results\n", + "print \"The emf of the first cell = %3.1f V \"%E1\n", + "print \"The emf of the second cell = %3.2f V \"%E2\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emf of the first cell = 1.2 V \n", + "The emf of the second cell = 1.49 V \n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamental_Electrical_and_Electronic_Principles/ch3.ipynb b/Fundamental_Electrical_and_Electronic_Principles/ch3.ipynb new file mode 100755 index 00000000..492adb89 --- /dev/null +++ b/Fundamental_Electrical_and_Electronic_Principles/ch3.ipynb @@ -0,0 +1,877 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:6afa9a01ae55a53807ec01423714e6c6f113931ad8642d6e60f895f0e33b2225" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3: Electric fields and Capacitors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1, Page 79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Q = 50e-03; # Electric charge, C\n", + "A = 600e-06; # Area of plate, m^2\n", + "\n", + "#Calculations\n", + "# Solving for electric field density, D\n", + "D = Q/A; # Electric field density, C/m^2\n", + "\n", + "#Result\n", + "print \"The density of the electric field existing between the plates = %4.1f C/m-square\"%D\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The density of the electric field existing between the plates = 83.3 C/m-square\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2, Page 80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A = 400e-06; # Cross-sectional area of plate, m^2\n", + "I = 50e-06; # Source current, A\n", + "t = 3; # Flow time of current, s\n", + "\n", + "#Calculations\n", + "# Since electric current is the rate of flow of charge i.e I = Q/t, solving for Q \n", + "Q = I*t; # Amount of charge on plates, C\n", + "#Solving for density of the electric field between the plates\n", + "D = Q/A; # Electric field density, C/m^2\n", + "\n", + "#Results\n", + "print \"The charge on the plates = %3d micro-coloumb\"%(Q/1e-06);\n", + "print \"The density of the electric field between the plates = %5.3f C/m-square\"%D\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The charge on the plates = 150 micro-coloumb\n", + "The density of the electric field between the plates = 0.375 C/m-square\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3, Page 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "d = 3e-03; # Thickness of dielectric, m\n", + "Q = 35e-03; # Electric charge on plates, C\n", + "V = 150; # Supply voltage, V\n", + "A = 144e-06; # Cross-sectional area of plates, m^2\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Since electric field strength(E) = potential gradient therefore we have\n", + "E = V/d; # Electric field strength, V/m\n", + "# Part (b)\n", + "# Solving for electric field density, D\n", + "D = Q/A; # Electric field density, C/m^2\n", + "\n", + "#Results\n", + "print \"The electric field strength = %2d kV/m\"%(E*1e-03);\n", + "print \"The flux density = %5.1f C/m^2\"%D\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electric field strength = 50 kV/m\n", + "The flux density = 243.1 C/m^2\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4, Page 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "d = 4e-03; # Thickness of air, m\n", + "Q = 2e-04; # Electric charge on plates, C\n", + "V = 125; # Supply voltage, V\n", + "D = 15; # Electric field density, coulomb-per-metre-square\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Since electric field strength(E) = potential gradient, therefore we have \n", + "E = V/d; # Electric field strength, V/m\n", + "# Part (b)\n", + "# Since D = Q/A, solving for A\n", + "A = Q/D; # Cross-sectional area of plates, m^2\n", + "# Part (c)\n", + "# Since Q = C*V, solving for C\n", + "C = Q/V; # Capacitance of the plates, F\n", + "\n", + "#Results\n", + "print \"The electric field strength between the plates = %5.2f kV/m\"%(E*1e-03);\n", + "print \"The csa of the field between the plates = %4.1f mm^2\"%(A/1e-06);\n", + "print \"The capacitance of the plates = %3.1f micro-coulomb\"%(C/1e-06);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electric field strength between the plates = 31.25 kV/m\n", + "The csa of the field between the plates = 13.3 mm^2\n", + "The capacitance of the plates = 1.6 micro-coulomb\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5, Page 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A = 6e-04; # Cross-sectional area of plates, m^2\n", + "d = 5e-04; # Thickness of mica sheet, m\n", + "Epsilon_r = 5.8; # Relative permittivity, unitless\n", + "Epsilon_0 = 8.854e-12; # Permittivity of Free Space\n", + "V = 200; # Potential difference, V\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for d & putting Epsilon = Epsilon_0*Epsilon_r \n", + "C = ( Epsilon_r*Epsilon_0*A )/d; # Capacitance, F\n", + "# Part (b)\n", + "# Since electric field strength(E) = potential gradient, therefore we have \n", + "E = V/d; # Electric field strength, V/m\n", + "\n", + "#Results\n", + "print \"The capacitance of the capacitor = %5.2f pF\"%(C/1e-12);\n", + "print \"Electric field strength = %3d kV/m\"%(E*1e-03);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The capacitance of the capacitor = 61.62 pF\n", + "Electric field strength = 400 kV/m\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6, Page 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "C = 0.224e-09; #Capacitance, F\n", + "A = 5625e-06; # Cross-sectional area of plates, m^2\n", + "Epsilon_r = 2.5; # Relative permittivity\n", + "Epsilon_0 = 8.854e-12; # Permittivity of Free Space\n", + "\n", + "#Calculations\n", + "# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for d & putting Epsilon = Epsilon_0*Epsilon_r\n", + "d = ( Epsilon_r*Epsilon_0*A )/C; # Thickness of waxed paper dielectric, m\n", + "\n", + "#Result\n", + "print \"The thickness of paper required = %3.2f mm\"%(d/1e-03);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The thickness of paper required = 0.56 mm\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7, Page 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "C = 4.7e-08; #Capacitance, F\n", + "A = 4e-04; # Cross-sectional area of plates, m^2\n", + "d = 1e-04; # Thickness of dielectric, m\n", + "Epsilon_0 = 8.854e-12; # Permittivity of Free Space\n", + "\n", + "#Calculations\n", + "# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for Epsilon_r & putting Epsilon = Epsilon_0*Epsilon_r\n", + "Epsilon_r = (C*d)/(Epsilon_0*A); # Relative permittivity\n", + "\n", + "#Result\n", + "print \"Relative permittivity = %4d\"%Epsilon_r\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Relative permittivity = 1327\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8, Page 87" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V = 180; # Potential difference, V\n", + "d = 3e-03; # Thickness of dielectric, m\n", + "A = 4.2e-04; # Cross-sectional area of plates, m^2\n", + "Epsilon_r = 3.5; # Relative permittivity\n", + "Epsilon_0 = 8.854e-12; # Permittivity of Free Space\n", + "\n", + "#Calculations\n", + "# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for C & putting Epsilon = Epsilon_0*Epsilon_r \n", + "C = ( Epsilon_r*Epsilon_0*A )/d; # Capacitance, F\n", + "# Since C = Q/V, solving for Q\n", + "Q = C*V; # Electric charge, C\n", + "# Using D = Q/A,\n", + "D = Q/A; # Electric field density, C/m^2\n", + "\n", + "#Results\n", + "print \"The flux thus produced = %3.2f nC.\"%(Q/1e-09)\n", + "print \"The flux density thus produced. = %3.2f micro-coulomb-per-metre-square\"%(D/1e-06); \n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The flux thus produced = 0.78 nC.\n", + "The flux density thus produced. = 1.86 micro-coulomb-per-metre-square\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9, Page 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "C_1 = 4.7e-06; #Capacitance, F\n", + "C_2 = 3.9e-06; #Capacitance, F\n", + "C_3 = 2.2e-06; #Capacitance, F\n", + "\n", + "#Calculations\n", + "# The resulting capacitance of parallerly connected capacitors is the sum of the individual capacitance present \n", + "#in the circuit\n", + "C = C_1 + C_2 + C_3; # Resulting capacitance of the circuit, F\n", + "\n", + "#Result\n", + "print \"The resulting capacitance of the combination = %4.1f micro-farad\"%(C/1e-06);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resulting capacitance of the combination = 10.8 micro-farad\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.10, Page 90" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "C_1 = 6e-06; #Capacitance, F\n", + "C_2 = 4e-06; #Capacitance, F\n", + "V = 150; # Supply voltage, V\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# The reciprocal of the resulting capacitance of capacitors connected in series is the sum of the reciprocal \n", + "#of the individual capacitances present in the circuit i.e 1/C = 1/C1 + 1/C2, solving for C\n", + "C = ( C_1*C_2 )/(C_1 + C_2); # Resulting capacitance, F\n", + "# Part (b)\n", + "Q = V*C; # Electric charge on the capacitors, C\n", + "# Part (c)\n", + "V_1 = Q/C_1; # P.d across capacitor C_1, V\n", + "V_2 = Q/C_2; # P.d across capacitor C_2, V\n", + "\n", + "#Results\n", + "print \"The total capacitance of the combination = %3.1f micro-farad\"%(C/1e-06);\n", + "print \"The charge on each capacitor = %3d micro-coulomb\"%(Q/1e-06);\n", + "print \"The p.d. developed across %1d micro-farad capacitor = %2d V\"%(C_1/1e-06, V_1);\n", + "print \"The p.d. developed across %1d micro-farad capacitor = %2d V\"%(C_2/1e-06, V_2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total capacitance of the combination = 2.4 micro-farad\n", + "The charge on each capacitor = 360 micro-coulomb\n", + "The p.d. developed across 6 micro-farad capacitor = 60 V\n", + "The p.d. developed across 4 micro-farad capacitor = 90 V\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11, Page 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "C_1 = 3e-06; #Capacitance, F\n", + "C_3 = 12e-06; #Capacitance, F\n", + "C_2 = 6e-06; #Capacitance, F\n", + "V = 400; # Supply voltage, V\n", + "\n", + "#Calculations\n", + "# The reciprocal of the resulting capacitance of capacitors connected in series is the sum of the reciprocal of the \n", + "#individual capacitances present in the circuit i.e 1/C = 1/C1 + 1/C2 + 1/C_3, solving for C\n", + "C = (C_1 * C_2 * C_3)/( C_1*C_2 + C_2*C_3 + C_3*C_1); # Resulting capacitance, F\n", + "Q = V*C; # Electric charge on the capacitors, C\n", + "# Part (c)\n", + "V_1 = Q/C_1; # P.d across capacitor C_1, V\n", + "V_2 = Q/C_2; # P.d across capacitor C_2, V\n", + "V_3 =Q/C_3; # P.d across capacitor C_2, V\n", + "\n", + "#Results\n", + "print \"P.d across capacitor %1d micro-farad = %5.1f V\"%(C_1/1e-06, V_1);\n", + "print \"P.d across capacitor %1d micro-farad = %5.1f V\"%(C_2/1e-06, V_2);\n", + "print \"P.d across capacitor %2d micro-farad = %4.1f V\"%(C_3/1e-06, V_3);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P.d across capacitor 3 micro-farad = 228.6 V\n", + "P.d across capacitor 6 micro-farad = 114.3 V\n", + "P.d across capacitor 12 micro-farad = 57.1 V\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12, Page 92" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V = 200; # Supply voltage, voltage\n", + "C_AB = 2.; # Capacitance across branch AB, micro-farad\n", + "C_BC = 3.; # Capacitance across branch BC, micro-farad\n", + "C_CD = 6.; # Capacitance across branch CD, micro-farad\n", + "C_EF = 8.; # Capacitance across branch EF, micro-farad\n", + "C_BD = 4.; # Capacitance across branch EF, micro-farad\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Since 3-micro-farad & 6-micro-farad capacitors are in series & the reciprocal of the resulting capacitance of \n", + "#capacitors connected in series is the sum of the reciprocal of the individual capacitances present in the circuit, \n", + "#therefore i.e 1/C = 1/C1 + 1/C2 \n", + "C_BCD = ( C_BC*C_CD )/(C_BC+C_CD); # Resulting capacitance across branch BCD, micro-farad\n", + "#Since C_BCD & 4-micro-farad capacitors are in parallel & the resulting capacitance of parallerly connected capacitors \n", + "#is the sum of the individual capacitance present in the circuit\n", + "C_BD = C_BCD + C_BD; # Resulting capacitance across branch BD, micro-farad\n", + "# Since 2-micro-farad & C_BD capacitors are in series & the reciprocal of the resulting capacitance of capacitors \n", + "#connected in series is the sum of the reciprocal of the individual capacitances present in the circuit, therefore, \n", + "#we have\n", + "C_AD = (C_BD*C_AB)/(C_BD+C_AB); # Resulting capacitance across branch AD, micro-farad\n", + "#Since C_AD & C_EF capacitors are in parallel & the resulting capacitance of parallerly connected capacitors is the \n", + "#sum of the individual capacitance present in the circuit\n", + "C = C_AD + C_EF; # Resulting capacitance of the circuit, micro-farad\n", + "Q = V*C; # Electric charge drawn from the supply, C\n", + "\n", + "# Part (b)\n", + "Q_EF = V*C_EF; # The charge on the 8 micro-farad capacitor, micro-coulomb\n", + "\n", + "# Part (c)\n", + "Q_AD = Q - Q_EF; # The charge on the 4 micro-farad capacitor, C\n", + "Q_BD = Q_AD; # Charge in series combination of capacitors, micro-farad\n", + "# Since Q = C*V, solving for V\n", + "V_BD = Q_BD/C_BD; # The p.d. across the 4F capacitor,V\n", + "\n", + "# Part(d)\n", + "Q_BCD = V_BD*C_BCD; # Electric charge across branch BCD, C\n", + "Q_BC = Q_BCD; # Electric charge, C\n", + "V_BC = Q_BC/C_BC; # The p.d. across the 3 micro-farad capacitor\n", + "\n", + "#Results\n", + "print \"The charge drawn from the supply = %3.1f mC\"%(Q/1e+03);\n", + "print \"The charge on the %1d micro-farad capacitor = %3.1f mC\"%(C_EF, Q_EF/1e+03);\n", + "print \"The p.d. across the %1d micro-farad capacitor= %2d V\"%(C_BD, V_BD);\n", + "print \"The p.d. across the %1d micro-farad capacitor = %5.2f V\"%(Q_BC, V_BC);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The charge drawn from the supply = 1.9 mC\n", + "The charge on the 8 micro-farad capacitor = 1.6 mC\n", + "The p.d. across the 6 micro-farad capacitor= 50 V\n", + "The p.d. across the 100 micro-farad capacitor = 33.33 V\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13, Page 96" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "N = 20; # Number of plates in a capacitor\n", + "A = 6400e-06; # Cross - sectional area of plate, m^2\n", + "d = 1.5e-03; # Distance between plates, m\n", + "epsilon_r = 6.4; # Relative permittivity for mica\n", + "epsilon_o = 8.854e-12; # Relative permittivity for free space\n", + "\n", + "#Calculations\n", + "# Calculating the capacitance of the capacitor\n", + "C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d; # Capacitance, F\n", + "\n", + "#Result\n", + "print \" The capacitance of the capacitor = %3.1f nF\"%(C/1e-09);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The capacitance of the capacitor = 4.6 nF\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.14, Page 96" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "N = 9; # Number of plates in a capacitor\n", + "A = 1200e-06; # Cross - sectional area of plate, m^2\n", + "C = 3e-10; # Capacitance, F\n", + "epsilon_r = 5; # Relative permittivity for mica\n", + "epsilon_o = 8.854e-12; # Relative permittivity for free space\n", + "\n", + "#Calculations\n", + "# Using the formula of capacitance, C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d and solving for d, we have\n", + "d = ((epsilon_o)*(epsilon_r)*A*(N-1))/C; # Distance between plates, m\n", + "\n", + "#Result\n", + "print \"The thickness of mica between parallel plates of a capacitor = %4.2f mm\"%(d/1e-03);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The thickness of mica between parallel plates of a capacitor = 1.42 mm\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15, Page 97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "N = 11; # Number of plates in a capacitor\n", + "r = 25e-03; # Radius of circular plate, m\n", + "A = (math.pi*r**2); # Cross - sectional area of plate, m^2\n", + "d = 5e-04; # Distance between plates, m\n", + "epsilon_r = 1; # Relative permittivity for air\n", + "epsilon_o = 8.854e-12; # Relative permittivity for free space\n", + "\n", + "#Calculations\n", + "# Calculating the capacitance of the capacitor\n", + "C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d; # Capacitance, F\n", + "\n", + "#Result\n", + "print \" The capacitance of the capacitor = %3.2f pF\"%(C/1e-10);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The capacitance of the capacitor = 3.48 pF\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16, Page 99" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "C_1 = 3e-06; # Capacitance, F\n", + "C_2 = 6e-06; # Capacitance, F\n", + "V_1 = 250; # Voltage across capacitor C_1, V\n", + "\n", + "#Calculations\n", + "# Since each capacitor will take charge according to its capacitance, so we have\n", + "Q = C_1*V_1; # Charge on first capacitor C_1, C\n", + "W_1 = 0.5*C_1*(V_1**2); # Energy stored, J\n", + "# When the two capacitors are connected in parallel the 3 micro-farad will share its charge with 6 micro-farad capacitor. Thus the total charge in the system will remain unchanged, but the total capacitance will now be different\n", + "C = C_1 + C_2; # Total capacitance, F\n", + "# Since Q = C*V, solving for V\n", + "V = Q/C; # Voltage across capacitor C_2, V\n", + "W = 0.5*C*(V**2); # Total energy stored by the combination, J\n", + "\n", + "#Results\n", + "print \"The charge and energy stored by %1d micro-F capacitor are %3.2f mC and %5.2f mJ respectively \"%(C_1/1e-06, Q/1e-03 , W_1/1e-03);\n", + "print \"The p.d. between the plates = %5.2f V\"%V\n", + "print \"The energy stored by the combination of %1d micro-F and %1d micro-F capacitors = %5.2f mJ\"%(C_1/1e-06, C_2/1e-06, W/1e-03);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The charge and energy stored by 3 micro-F capacitor are 0.75 mC and 93.75 mJ respectively \n", + "The p.d. between the plates = 83.33 V\n", + "The energy stored by the combination of 3 micro-F and 6 micro-F capacitors = 31.25 mJ\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.17, Page 100" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V = 200; # Supply voltage, V\n", + "C_1 = 10e-06; # Capacitance, farad\n", + "C_2 = 6.8e-06; # Capacitance, farad\n", + "C_3 = 4.7e-06; # Capacitance, farad\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Since each capacitor will take charge according to its capacitance, so we have\n", + "Q_1 = V*C_1; # Charge stored on capacitor C_1, C\n", + "W_1 = 0.5*C_1*(V**2); # Energy stored on capacitor C_1, J\n", + "# Part (b)\n", + "# Since C_2 and C_3 are in series and hence, their equivalent capacitance is given by their series combination\n", + "C_4 = (C_2 * C_3)/(C_2 + C_3); # Equivalent capacitance of C_2 and C_3, F\n", + "# Since C_1 and C_4 are in parallel and hence, their equivalent capacitance is given by their parallel combination\n", + "C = C_1 + C_4; # Total capacitance of circuit, F\n", + "# Since Q = C*V, solving for V\n", + "V_1 = Q_1/C; # New p.d across C_1, V\n", + "W = 0.5*C*(V_1**2); # Total energy remaining in the circuit, J\n", + "energy_used = W_1 - W; # Energy, J\n", + "\n", + "#Results\n", + "print \"The charge and energy stored by %2d micro-F capacitor are %1d mC and %2.1f J respectively \"%(C_1/1e-06, Q_1/1e-03, W_1);\n", + "print \"The new p.d across %2d micro-F capacitor = %5.1f V\"%(C_1/1e-06,V_1);\n", + "print \"The amount of energy used in charging %3.1f micro-F and %3.2f micro-F capacitors from %2d micro-F capacitor = %4.3f J\"%(C_2/1e-06, C_3/1e-06, C_1/1e-06, energy_used/1e-03);\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The charge and energy stored by 10 micro-F capacitor are 2 mC and 0.2 J respectively \n", + "The new p.d across 10 micro-F capacitor = 156.5 V\n", + "The amount of energy used in charging 6.8 micro-F and 4.70 micro-F capacitors from 10 micro-F capacitor = 43.495 J\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.18, Page 101" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V = 400; # Supply voltage, V\n", + "E = 0.5e06; # Dielectric strength, V/m\n", + "\n", + "#Calculations\n", + "# Since E = V/d, solving for d\n", + "d = V/E; # Thickness of dielectric, m\n", + "\n", + "#Result\n", + "print \"The minimum thickness of dielectric required = %3.1fmm\"%(d/1e-03);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum thickness of dielectric required = 0.8mm\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.19, Page 101" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "C = 270e-12; # Capacitance, F\n", + "A = 60e-04; # Cross-sectional area of plate, m^2\n", + "E = 350e03; # Dielectric strength, V/m\n", + "epsilon_r = 2.1; # Relative permittivity\n", + "epsilon_o = 8.854e-12; # Permittivity of free space\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Since formula for capacitance, C = ((epsilon_o)*(eplison_r)*A)/d, solving for d\n", + "d = ((epsilon_o)*(epsilon_r)*A)/C; # Thickness of dielectric, m\n", + "# Part (b)\n", + "# Since E = V/d, solving for V\n", + "V = E*d; # Maximum possible working voltage, V\n", + "\n", + "#Results\n", + "print \"The thickness of Teflon sheet required = %5.4f mm\"%(d/1e-03);\n", + "print \"The maximum possible working voltage for the capacitor = %5.1f V\"%V;\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The thickness of Teflon sheet required = 0.4132 mm\n", + "The maximum possible working voltage for the capacitor = 144.6 V\n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamental_Electrical_and_Electronic_Principles/ch4.ipynb b/Fundamental_Electrical_and_Electronic_Principles/ch4.ipynb new file mode 100755 index 00000000..ffd2b0b9 --- /dev/null +++ b/Fundamental_Electrical_and_Electronic_Principles/ch4.ipynb @@ -0,0 +1,709 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:354118d2fca7350165cb8c5d000bd36293e8da69f0cbb274034a02640d69b2e6" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4: Magnetic Fields and Circuits" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1, Page 116" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A = 6e-04; # Cross-sectional area of pole face, metre-square\n", + "phi = 30e-06; # Flux, Wb\n", + "\n", + "#Calculations\n", + "B = phi/A; # Flux density, T\n", + "\n", + "#Result\n", + "print \"The flux density at the pole face = %2d mT\"%(B/1e-03)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The flux density at the pole face = 50 mT\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2, Page 116" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A = 45e-06; # Cross sectional area of pole face, metre-square\n", + "B = 0.6; # Flux density, T\n", + "\n", + "#Calculations\n", + "# Using formula B = phi/A, solving for phi\n", + "phi = B*A; # Flux, Wb\n", + "\n", + "#Result\n", + "print \"The flux produced by pole face = %2d micro-wWb\"%(phi/1e-06);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The flux produced by pole face = 27 micro-wWb\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3, Page 117" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "N = 1500; # Number of turns in a coil\n", + "A = 5e-04; # Cross- sectional area of of coil, metre-square\n", + "phi = 0.2e-03; # Flux, Wb\n", + "I = 0.75; # Coil-current, A\n", + "\n", + "#Calculations\n", + "# Since m.m.f is the product of the current and the number of turns, therefore, we have\n", + "F = N*I; # Magnetomotive force, At\n", + "B = phi/A; # Flux density, T\n", + "\n", + "#Result\n", + "print \"The m.m.f and flux density produced are %4d At and %3.1f T respectively\"%(F, B);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The m.m.f and flux density produced are 1125 At and 0.4 T respectively\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4, Page 117" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "N = 600; # Number of turns in a coil\n", + "F = 1500.; # Magnetomotive force, At\n", + "\n", + "#Calculations\n", + "# Since magnetomotive force,F = N*I, solving for I\n", + "I = F/N; # Excitation-current, A\n", + "\n", + "#Result\n", + "print \"The excitation current required = %3.1f A\"%I;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The excitation current required = 2.5 A\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5, Page 118" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "I = 0.4; # Current, A\n", + "N = 550; # Number of turns in a coil\n", + "d = 8e-02; # Diameter, m\n", + "l = (math.pi*d); # Average length of the magnetic circuit, m\n", + "\n", + "#Calculations\n", + "# Since magnetic field strength is defined as the mmf per metre length of the magnetic circuit, therefore, we have\n", + "H = (N*I)/l; # Magnetic field strength, At/m\n", + "\n", + "#Result\n", + "print \"The magnetic field strength inside the toroid = %6.2f At/m\"%H\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The magnetic field strength inside the toroid = 875.35 At/m\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6, Page 120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A = 15e-04; # Cross-sectional area of core, metre-square\n", + "mew_r1 = 65; # Relative permeability of core\n", + "phi_1 = 2e-04; # Flux, Wb\n", + "mew_r2 = 800.; # Changed relative permeability of core\n", + "\n", + "#Calculations\n", + "B_1 = phi_1/A; # Flux density, T\n", + "mew_r = mew_r2/mew_r1; # Relative permeability of core\n", + "# Since cross-sectional area of core A remains constant, therefore, we have mew_r = B_1/B_2 , solving for B_2\n", + "B_2 = mew_r*B_1; # New flux density, T\n", + "# Since B_2 = phi_2/A, solving for phi_2\n", + "phi_2 = B_2*A; # New flux, Wb\n", + "\n", + "#Result\n", + "print \"The new flux and flux density are %5.3f mWb and %5.3f T respectively\"%(phi_2/1e-03, B_2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The new flux and flux density are 2.462 mWb and 1.641 T respectively\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7, Page 120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "r = 0.04; # Mean radius of torod, m\n", + "A = 3e-04; # Csa of toroid, m^2\n", + "mew_o = 4*(math.pi)*1e-07; # Permeability of free space\n", + "mew_r = 150; # Relative permeability of toroid\n", + "N = 900; # Number of turns on coil\n", + "I = 1.5; # Coil current, A\n", + "l = 2*(math.pi)*r; # Effective length of toroid, m\n", + "\n", + "#Calculations&Results\n", + "# Part (a)\n", + "# Since m.m.f is the product of the current and the number of turns, therefore, we have\n", + "F = N*I; # Magnetomotive force, At\n", + "print \"The m.m.f of toroid = %4d At\"%F\n", + "\n", + "# Part (b)\n", + "# Since magnetic field strength is defined as the mmf per metre length of the magnetic circuit, therefore, we have\n", + "H = F/l; # Magnetic field strength, At/m\n", + "print \"The magnetic field strength = %6.1f At/m\"%H;\n", + "\n", + "# Part (c)\n", + "B = (mew_r*mew_o*H); # Flux density, T\n", + "phi = B*A; # Flux, Wb\n", + "print \"The flux and flux density are %6.2f micro-weber and %6.4f T respectively\"%(phi/1e-06,B)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The m.m.f of toroid = 1350 At\n", + "The magnetic field strength = 5371.5 At/m\n", + "The flux and flux density are 303.75 micro-weber and 1.0125 T respectively\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8, Page 120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "r = 3e-02; # Radius of toroid, m\n", + "A = 4.5e-04; # Cross-sectional area of toroid, metre-square\n", + "N = 500; # Number of turns \n", + "phi = 250e-06; # Flux, Wb\n", + "mew_o = 4*(math.pi)*(1e-07); # Permeability of free space\n", + "mew_r = 300; # Relative permeability\n", + "\n", + "#Calculations\n", + "l = 2*(math.pi)*r; # Effective length, m\n", + "B = phi/A; # Flux density, T\n", + "# Since B = (mew_r)*(mew_o)*H, solving for H\n", + "H = B /((mew_r)*(mew_o)); # Magnetic field strength, At/m\n", + "# Since H = F/l, solving for F\n", + "F = H*l; # Magnetomotive force, At\n", + "# Since mmf,F = N*I, solving for I\n", + "I = F/N; # Electric current, A\n", + "\n", + "#Result\n", + "print \"The value of current needs to be passed through the coil = %4.2f A\"%I\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of current needs to be passed through the coil = 0.56 A\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9, Page 121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "# Part (a)\n", + "I = 0.2; # Electric current, A\n", + "l = 5e-02; # Effective length, m\n", + "A = 7e-04; # Cross-sectional area, metre-square\n", + "d = 0.5e-03; # Diameter, m\n", + "mew_r = 1; #Relative permeability for wood\n", + "\n", + "#Calculations\n", + "mew_o = 4*(math.pi)*1e-07; # Permeability for free space\n", + "N = l/d; # Number of turns \n", + "# Since mmf is the product of the current and the number of turns, therefore, we have\n", + "F = N*I; # Magnetomotive force, At\n", + "# Part (b)\n", + "# Since magnetic field strength is defined as the mmf per metre length of the magnetic circuit, therefore, we have\n", + "H = F/l; # Magnetic field strength, At/m\n", + "B = ( mew_r * mew_o * H ); # Flux density, T\n", + "# Part (c)\n", + "phi = B * A; # Flux, Wb\n", + "\n", + "#Result\n", + "print \"The mmf produced = %2d At\"%F\n", + "print \"The flux density produced = %3d micro-tesla\"%(B/1e-06);\n", + "print \"The flux produced = %5.3f micro-weber\"%(phi/1e-06);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mmf produced = 20 At\n", + "The flux density produced = 502 micro-tesla\n", + "The flux produced = 0.352 micro-weber\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.10, Page 125" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "N = 1000; # Number of turns on coil\n", + "r = 0.1; # Mean radius of toroid, m\n", + "phi = 0.1775e-03; # Flux density(value from graph), Wb\n", + "A = math.pi*1e-04; # Csa of toroid, m^2\n", + "H = 88; # Magnetic field strength(value from graph), At/m\n", + "B = phi/A; # Flux density, T\n", + "\n", + "#Calculations&Results\n", + "# Part (a)\n", + "l = 2*math.pi*r; # Effective length of toroid, m\n", + "# Since H = (N*I)/l, solving for I\n", + "I = (H*l)/N ; # Electric current in coil, A\n", + "print \"Coil current = %4.1f mA\"%(I/1e-03);\n", + "\n", + "# Part (b)\n", + "mew_o = 4*(math.pi)*1e-07; # Permeability for free space\n", + "# Since B = mew_o * mew_r * H, solving for mew_r\n", + "mew_r = B/(mew_o*H); #Relative permeability of toroid\n", + "print \"The relative permeability of toroid = %4d\"%mew_r;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Coil current = 55.3 mA\n", + "The relative permeability of toroid = 5109\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11, Page 125" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "mew_o = 4*(math.pi)*1e-07; # Permeability for free space\n", + "l = 0.15; # Mean length, m\n", + "N = 2500; # Number of turns\n", + "I = 0.3; # Electric current, A\n", + "\n", + "#Calculations\n", + "# Since magnetic field strength is defined as the mmf per metre length of the magnetic circuit, therefore, we have\n", + "H = (N*I)/l; # Magnetic field strength, At/m\n", + "B = 0.75; # Flux density( value taken from graph ), T\n", + "# Since B = ( mew_r * mew_o * H ), solving for mew_r\n", + "mew_r = B/(mew_o * H); # Relative permeability\n", + "\n", + "#Results\n", + "print \"The flux density of given toroid = %3.2f T \"%B\n", + "print \"The relative permeability of given toroid = %5.1f\"%mew_r\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The flux density of given toroid = 0.75 T \n", + "The relative permeability of given toroid = 119.4\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.12, Page 126" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "mew_o = 4*(math.pi)*1e-07; # Permeability for free space\n", + "l = 0.1875; # Mean length, m\n", + "A = 8e-05; # Cross- sectional area of of coil, metre-square\n", + "N = 750; # Number of turns\n", + "phi = 112e-06; # Flux, Wb\n", + "l_gap = 0.5e-03; # Average length of the magnetic circuit,m\n", + "B = phi/A; # Flux density, Wb\n", + "H = 2000; # Magnetic field strength( value taken from graph ), At/m\n", + "\n", + "#Calculations\n", + "F_Fe = H*l; # The m.m.f in the iron part of the circuit, At\n", + "# Since F = I*N, solving for I\n", + "I = F_Fe/N; # Coil current under normal conditions, A\n", + "# Since B = mew_o * H_gap, solving for H_gap\n", + "H_gap = B/mew_o; # Magnetic field strength, At/m\n", + "# Since H_gap = F_gap/l_gap, solving for F_gap\n", + "F_gap = H_gap * l_gap; # The mmf in the air part of the circuit, At\n", + "F = F_Fe + F_gap; # Total circuit mmf, At\n", + "I_new = F/N; # Current required to maintain the flux at its original value, A\n", + "\n", + "#Results\n", + "print \"The coil current required to produce a flux of %3d micro-weber in the toroid = %3.1f A \"%(phi/1e-06, I);\n", + "print \"Current required to maintain the flux at its original value = %5.3f A\"%(I_new);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The coil current required to produce a flux of 112 micro-weber in the toroid = 0.5 A \n", + "Current required to maintain the flux at its original value = 1.243 A\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.13, Page 127" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "l_A = 0.25; # Mean length of circuit A, m\n", + "l_B = 0.15; # Mean length of circuit A, m\n", + "A_A = 11.5e-04; # Cross-sectional area of circuit A, metre-square\n", + "A_B = 12e-04; # Cross-sectional area of circuit B, metre-square\n", + "phi = 1.5e-03; # Flux, Wb\n", + "N = 1000; # Number of turns\n", + "\n", + "#Calculations\n", + "B_A = phi/A_A; # Flux density linked with circuit A, T\n", + "B_B = phi/A_B; # Flux density linked with circuit B, T\n", + "H_A = 1470; # Magnetic field strength of circuit A( value taken from graph ), At/m\n", + "H_B = 845; # Magnetic field strength of circuit B( value taken from graph ), At/m\n", + "# Since H = F/l, solving for F \n", + "F_A = H_A * l_A; # Magnetic field strength of circuit A, At/m\n", + "F_B = H_B * l_B; # Magnetic field strength of circuit B, At/m\n", + "F = F_A + F_B; # Total circuit m.m.f, At/m\n", + "I = F/N; # Coil current, A\n", + "\n", + "#Result\n", + "print \"Coil current in the magnetic circuit = %5.3f A\"%I\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Coil current in the magnetic circuit = 0.494 A\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.14, Page 130" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "A = 8e-04; # Cross-sectional area, metre-square\n", + "d = 24e-02; # Mean diameter of iron ring, m \n", + "phi = 1.2e-03; # Flux, Wb\n", + "mew_r = 1200; # Relative permeability\n", + "mew_o = 4*(math.pi)*1e-07; # Permeability for free space\n", + "mew_air = 1; # Permeability for air\n", + "l_gap = 3e-03; # Mean length, m\n", + "\n", + "#Calculations\n", + "l_Fe = (math.pi) * d; # Mean length of iron circuit, m\n", + "S_Fe = l_Fe/(mew_r * mew_o *A); # Reluctance of iron circuit, At/Wb\n", + "S_gap = l_gap/(mew_air * mew_o *A); # Reluctance of gap, At/Wb\n", + "S = S_Fe + S_gap; # Total circuit reluctance, At/Wb\n", + "# Since phi = F/S, solving for F\n", + "F = phi*S; # Magnetomotive force, At\n", + "\n", + "#Result\n", + "print \"The required mmf = %5.1f At\"%F\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required mmf = 4331.0 At\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.15, Page 130" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "N = 500; # Number of turns on first section's coil\n", + "phi = 2e-03; # Flux produced by first section, Wb\n", + "l_1 = 85e-02; # Length of first section, m\n", + "l_2 = 65e-02; # Length of second section, m\n", + "l_3 = 0.1e-02; # Length of third section, m\n", + "A_1 = 10e-04; # Csa of first section, m^2\n", + "A_2 = 15e-04; # Csa of second section, m^2\n", + "A_3 = 12.5e-04; # Csa of second section, m^2\n", + "mew_o = 4*(math.pi)*1e-07; # Permeability for free space\n", + "mew_r1 = 600; # Relative permeability of first section\n", + "mew_r2 = 950; # Relative permeability of second section\n", + "mew_r3 = 1; # Relative permeability of third section\n", + "\n", + "#Calculations&Results\n", + "# Part (a)\n", + "S_1 = l_1/(mew_r1 * mew_o * A_1); # Reluctance of first section, At/Wb\n", + "S_2 = l_2/(mew_r2 * mew_o * A_2); # Reluctance of first section, At/Wb\n", + "S_3 = l_3/(mew_r3 * mew_o * A_3); # Reluctance of first section, At/Wb\n", + "S = S_1 + S_2 + S_3; # Total reluctance of the circuit, At/Wb\n", + "print \"Total reluctance of the circuit = %4.2fe+06 At/Wb\"%(S*1e-06);\n", + "\n", + "# Part (b)\n", + "# Since phi = F/S, solving for F\n", + "F = phi*S; # Magnetomotive force, At\n", + "# Since F = N*I, solving for I\n", + "I = F/N; # Electric current in first section, A\n", + "print \"Electric current in first section = %4.2f A\"%I\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total reluctance of the circuit = 2.13e+06 At/Wb\n", + "Electric current in first section = 8.51 A\n" + ] + } + ], + "prompt_number": 16 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamental_Electrical_and_Electronic_Principles/ch5.ipynb b/Fundamental_Electrical_and_Electronic_Principles/ch5.ipynb new file mode 100755 index 00000000..f256e262 --- /dev/null +++ b/Fundamental_Electrical_and_Electronic_Principles/ch5.ipynb @@ -0,0 +1,1322 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4381350f1f5861e39c88a88906555fe70746a56edfa9636dec1dfc702e44302a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5: Electromagnetism" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1, Page 145" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "N = 100; # Number of turns\n", + "delta_phi = 10e-03; # Flux linked with coil, Wb\n", + "delta_t = 2e-03; # Time during which flux changes, s\n", + "\n", + "#Calculations\n", + "e =((-N)*delta_phi)/delta_t; # Average induced emf, V\n", + "\n", + "#Result\n", + "print \"The average emf induced in the coil = %3d V\"%e\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The average emf induced in the coil = -500 V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2, Page 146" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "N = 250; # Number of turns\n", + "delta_phi1 = 20e-03; # Flux linked with coil, Wb\n", + "delta_phi2 = -16e-03; # Flux linked with coil, Wb\n", + "delta_t1 = 0.05; # Time, s\n", + "delta_t2 = 0.01; # Time, s\n", + "\n", + "#Calculations\n", + "e_1 =((-N)*delta_phi1)/delta_t1; # Average induced emf, V\n", + "e_2 =((-N)*delta_phi2)/delta_t2; # Average induced emf, V\n", + "\n", + "#Results\n", + "print \"Change in flux in first case = %4.2f weber\"%delta_phi1\n", + "print \"Emf induced in first case = %3d volts\"%e_1\n", + "print \"Change in flux in second case = %4.2f weber\"%delta_phi2\n", + "print \"Emf induced in second case = %3d volts\"%e_2\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in flux in first case = 0.02 weber\n", + "Emf induced in first case = -100 volts\n", + "Change in flux in second case = -0.02 weber\n", + "Emf induced in second case = 400 volts\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3, Page 147" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "e = 100; # Induced emf, V\n", + "\n", + "#Calculations\n", + "# For simplification let (delta_phi)/(delta_t) = k\n", + "k = 0.1; # Rate of chage of flux linked with coil, Wb/s\n", + "# Since e =((-N)*delta_phi)/delta_t, soling for N\n", + "N = (e)/k; # Number of turns\n", + "\n", + "#Result\n", + "print \"The number of turns on the coil = %4d\"%N" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of turns on the coil = 1000\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4, Page 149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "v = 5; # Velocity, m^2\n", + "theta =(math.pi/3); # Angle, degrees\n", + "phi = 1.6e-03; # Flux, Wb\n", + "l = 0.1; # Length of pole face, m\n", + "d = 0.4; # Breadth of pole face, m\n", + "\n", + "#Calculations\n", + "A = l*d; # Cross-sectional area of pole face, m^2\n", + "B = phi/ A; # Flux density, T\n", + "e =( B*l*v)*math.sin(theta); # Induced emf, V\n", + "\n", + "#Result\n", + "print \"The emf induced = %5.4f V\"%e\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emf induced = 0.0173 V\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5, Page 149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "l = 0.15; # Effective length of conductor, m\n", + "v = 8; # Velocity, m^2\n", + "theta = 55; # Angle, degrees\n", + "e = 2.5; # Induced emf, V\n", + "\n", + "#Calculations\n", + "# Since e = B*l*v*sin(theta), solving for B\n", + "B = e/(l*v*math.sin(theta*math.pi/180)); # Flux density, T\n", + "\n", + "#Result\n", + "print \"The density of the field = %5.3f Tesla\"%B\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The density of the field = 2.543 Tesla\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6, Page 149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "l = 2.2; # Effective length of conductor, m\n", + "B =38e-06; # Flux density, T\n", + "theta = (math.pi/2); # Angle, degrees\n", + "v = 800/36; # Velocity, m^2\n", + "\n", + "#Calculations\n", + "e = B*l*v*math.sin(theta); # Induced emf, V\n", + "\n", + "#Result\n", + "print \"The emf induced in the axle = %4.2f mV\"%(e/1e-03)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emf induced in the axle = 1.84 mV\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7, Page 152" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "l = 0.22; # Effective length of conductor, m\n", + "B = 0.35; # Flux density, T\n", + "I = 3; # Current, A\n", + "theta = (math.pi/2); # Angle, degrees\n", + "\n", + "#Calculations\n", + "# Since the force exerted on the conductor placed in magnetic field is directly proportional to the flux density , \n", + "#the value of current flowing through the conductor, and the length of conductor lying inside the field, therefore\n", + "F = B*I*l*math.sin(theta); # Force, N\n", + "\n", + "#Result\n", + "print \"The force exerted on the conductor = %5.3f N\"%F\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The force exerted on the conductor = 0.231 N\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.8, Page 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "phi = 2.5e-03; # Flux, Wb\n", + "l = 0.05; # Effective length of pole, m\n", + "d = 0.03; # Effective width of pole, m\n", + "F = 1.25; # Force exerted on conductor, N\n", + "A = l*d; # Cross-sectional area of pole face, m^2\n", + "B = phi/A; # Flux density, T\n", + "theta = (math.pi/2); # Angle, degrees\n", + "\n", + "#Calculations\n", + "# Since F = B*I*l*sin(theta), solving for I\n", + "I = F/(B*l*math.sin(theta)); # Current in conductor, A\n", + "theta_2 = (math.pi/4); # New angle, degrees\n", + "F_2 = B*I*l*math.sin(theta_2); # Force exerted on conductor, N\n", + "\n", + "#Results\n", + "print \"The value of the current = %2d A\"%I\n", + "print \"The force exerted on conductor when placed at 45 degrees to the field = %5.3f newton\"%F_2\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of the current = 14 A\n", + "The force exerted on conductor when placed at 45 degrees to the field = 0.884 newton\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9, Page 154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "l = 0.015; # Length of coil, m\n", + "d = 0.006; # Width of coil, m\n", + "B = 1.2; # Flux density, T\n", + "I = 1e-02; # Current, a\n", + "r = d/2; # Radius of rotation, m\n", + "\n", + "#Calculations\n", + "# Since torque is given by the product of force and distance, therefore, we have\n", + "T = 2*B*I*l*r; # Torque, Nm\n", + "\n", + "#Result\n", + "print \"The torque exerted on the coil = %4.2f micro-Nm\"%(T/1e-06)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The torque exerted on the coil = 1.08 micro-Nm\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.10, Page 155" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "N = 80; # Number of turns\n", + "l = 0.02; # Length of coil, m\n", + "r = 0.012; # Radius of coil, m\n", + "I = 45e-06; # Current in coil, A\n", + "T = 1.4e-06; # Torque exerted on coil, Nm\n", + "A = l*r; # Cross-sectional area of coil, m^2\n", + "\n", + "#Calculations\n", + "# Since T = 2*B*I*l*r, solving for B\n", + "B = T/(2*A*N*I); # Flux density, T\n", + "\n", + "#Result\n", + "print \"The flux density produced by the pole pieces = %4.2f T\"%B\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The flux density produced by the pole pieces = 0.81 T\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.11, Page 158" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "d = 0.035; # Distance between two parallel conductors, m\n", + "I_1 = 50; # Electric current in first coil, A\n", + "I_2 = 40; # Electric current in second coil, A\n", + "\n", + "#Calculations\n", + "F = ((2e-07)*I_1*I_2)/d; # Force exerted by conductors, N\n", + "\n", + "#Result\n", + "print \"The force exerted between the conductors = %4.1f mN\"%(F/1e-03);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The force exerted between the conductors = 11.4 mN\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.12, Page 158" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "d = 2; # Distance between two parallel conductors, m\n", + "I_1 = 1000; # Electric current in first coil, A\n", + "I_2 = 300; # Electric current in second coil, A\n", + "\n", + "#Calculations\n", + "mew_o = 4*(math.pi)*1e-07; # Permeability for free space\n", + "B = (mew_o*I_1)/d; # Flux density due to first coil, T\n", + "F = ((2e-07)*I_1*I_2)/d; # Force exerted by conductors, N\n", + "\n", + "#Result\n", + "print \"The flux density at a distance of %1d m from the centre of a conductor carrying a current of %4d A = %5.3f mT\"%(d, I_1, B/1e-03);\n", + "print \"Force exerted by conductors = %2d mN\"%(F/1e-03);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The flux density at a distance of 2 m from the centre of a conductor carrying a current of 1000 A = 0.628 mT\n", + "Force exerted by conductors = 30 mN\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.13, Page 163" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_c = 40; # Resistance of coil, ohm\n", + "I_fsd = 5e-04; # Full-scale deflection current, A\n", + "I = 3; # Current reading, A\n", + "\n", + "#Calculations\n", + "V_c = I_fsd*R_c; # Potential difference, V\n", + "# Since I = I_s + I_fsd, solving for I_s\n", + "I_s = I-I_fsd; # Shunt current, A\n", + "# From Ohm's law, V_c = I_s*R_s, solving for R_s\n", + "R_s = V_c/I_s; # Shunt resistance, ohm\n", + "\n", + "#Result\n", + "print \"The value of required shunt resistance = %4.2f milli-ohm\"%(R_s/1e-03); \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of required shunt resistance = 6.67 milli-ohm\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.14, Page 163" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_c = 40; # Resistance of coil, ohm\n", + "I_fsd = 5e-04; # Full-scale deflection current, A\n", + "I_fsd = 5e-04; # Full-scale deflection current, A\n", + "V = 10; # Voltage reading range, V\n", + "V_c = 0.02; # Potential difference across coil resistance, V\n", + "\n", + "#Calculations\n", + "# From Ohm's law, V = I_fsd*R, solving for R\n", + "R = V/I_fsd; # Total resistance, ohm\n", + "# Since R = R_m + R_c, solving R_m\n", + "R_m = R - R_c; # Multiplier resistance, ohm\n", + "\n", + "#Result\n", + "print \"The required value of multiplier resistance = %5.2f kilo-ohms\"%(R_m*1e-03);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required value of multiplier resistance = 19.96 kilo-ohms\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.15, Page 164" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_c = 1500; # Coil resistance, ohm\n", + "I_fsd = 75e-06; # Full-scale deflection current, A\n", + "I = 5; # Current range, A\n", + "V = 10; # Voltage range, V\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Using Ohm's law,\n", + "V_c = I_fsd*R_c; # Potential difference across coil resistance, V\n", + "# Since I = I_s + I_fsd, solving for I_s\n", + "I_s = I-I_fsd; # Shunt current, A\n", + "# From Ohm's law, V_c = I_s*R_s, solving for R_s\n", + "R_s = V_c/I_s; # Shunt resistance, ohm\n", + "\n", + "# Part (b)\n", + "# Since = V = V_m + V_c, solving for V_m\n", + "V_m = V - V_c; # Potential difference across multiplier resistance, V\n", + "# From Ohm's law, V_m = I_fsd*R_m, solving for R_m\n", + "R_m = V_m/I_fsd # Multiplier resistance, ohm\n", + "\n", + "#Results\n", + "print \"The required value of shunt resistance = %4.1f mega-ohm\"%(R_s/1e-03);\n", + "print \"The required value of multiplier resistance = %4.1f mega-ohm\"%(R_m*1e-03);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required value of shunt resistance = 22.5 mega-ohm\n", + "The required value of multiplier resistance = 131.8 mega-ohm\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.16, Page 166" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_1 = 30.; # Resistance, ohm\n", + "R_2 = 70.; # Resistance, ohm\n", + "R_in = 200.; # Internal resistance of meter, ohm\n", + "V = 12.; # Supply voltage, V\n", + "\n", + "#Calculations\n", + "# Using voltage divider rule, we have\n", + "V_2t = (R_2 /(R_1 + R_2))*V # True value of p.d across resistance R_2, V\n", + "# Since the resistances R_2 and R-in are parallel, so their equivalent resistance is given their parallel combination\n", + "R_BC = (R_2 * R_in)/(R_2 + R_in); # Resistance, ohms\n", + "# Using the potential divider technique, \n", + "V_2i = (R_BC / ( R_BC + R_1 ))*V # Indicated value of p.d across by voltmetre, volts\n", + "err = (( V_2i-V_2t ) / V_2t)*100 # Percentage error in the reading\n", + "\n", + "#Results\n", + "print \"The p.d. indicated by the meter = %3.1f V\"%V_2i\n", + "print \"The percentage error in the reading = %4.2f percent\"%err\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The p.d. indicated by the meter = 7.6 V\n", + "The percentage error in the reading = -9.50 percent\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.17, Page 168" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_in = 200.; # Internal resistance of meter, kilo-ohms\n", + "V = 10.; # Supply voltage, volts\n", + "R_1 = 10.; # Resistance, kilo-ohms\n", + "R_2 = 47.; # Resistance, kilo-ohms\n", + "\n", + "#Calculations\n", + "V_1 = R_1/(R_1+R_2)*V # P.d across resistance R_1, V\n", + "V_2 = R_2/(R_1+R_2)*V # P.d across resistance R_2, V\n", + "\n", + "# Part (a)\n", + "R_AB = (R_1 * R_in)/(R_1 + R_in); # Resistance, kilo-ohms\n", + "V_AB = (R_AB / ( R_AB + R_2 ))*V # True value of p.d across by voltmetre, V\n", + "R_BC = (R_2 * R_in)/(R_2 + R_in); # Resistance, kilo-ohms\n", + "V_BC = (R_BC / ( R_BC + R_1 ))*V # Indicated value of p.d across by voltmetre, V\n", + "\n", + "# Part (b)\n", + "# Error for V_1 measurement\n", + "error_AB = (V_AB - V_1)/V_1*100 # Percentage error in the reading\n", + "#Error for V_2 measurement\n", + "error_BC = (V_BC-V_2)/V_2*100 # Percentage error in the reading \n", + "\n", + "#Results\n", + "print \"The p.d. indicated by the meter across first resistor = %4.2f V\"%V_AB\n", + "print \"The p.d. indicated by the meter across second resistor = %4.2f V\"%V_BC\n", + "print \"Percentage error for V_1 measurement = %4.2f percent\"%error_AB\n", + "print \"Percentage error for V_2 measurement = %4.2f percent\"%error_BC\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The p.d. indicated by the meter across first resistor = 1.68 V\n", + "The p.d. indicated by the meter across second resistor = 7.92 V\n", + "Percentage error for V_1 measurement = -3.96 percent\n", + "Percentage error for V_2 measurement = -3.96 percent\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.18, Page 176" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "L = 0.25; # Self-inductance, H\n", + "delta_I = 250e-03; # Change in current, A\n", + "delta_t = 25e-03; # Time, s\n", + "\n", + "#Calculations\n", + "e = ((L)*delta_I)/(delta_t); # Induced emf, V\n", + "\n", + "#Result\n", + "print \"The value of emf induced = %3.1f V\"%e\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of emf induced = 2.5 V\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.19, Page 176" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "e = 30.; # Induced emf, V\n", + "\n", + "#Calculations\n", + "# For simplicity, let rate of change of current i.e delta_I/delta_t = k\n", + "k = 200; # Rate of change of current, ampere-second\n", + "# Since e = ((-L)*delta_I)/(delta_t), solving for L\n", + "L = e/k; # Self-inductance, H\n", + "\n", + "#Result\n", + "print \"The inductance of the circuit = %4.2f H\"%L\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The inductance of the circuit = 0.15 H\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.20, Page 176" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "L = 50e-03; # Self-inductance, H\n", + "e = 8; # Induced emf, V\n", + "\n", + "#Calculations\n", + "# Since e = ((-L)*delta_I)/(delta_t), solving for delta_I/delta_t,and for simplicity letting the rate of change of \n", + "#current i.e delta_I/delta_t = k\n", + "k = e/L; # Rate of change of current, As\n", + "\n", + "#Result\n", + "print \"The rate of change of current = %3d A/s\"%k\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rate of change of current = 160 A/s\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.21, Page 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "N = 150; # Number of turns in a coil\n", + "I = 10; # Electric current flowing through coil, A\n", + "phi = 0.10; # Flux, Wb\n", + "delta_t = 0.1; # Time, s\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "L = (N * phi)/I # Self-inductance, H\n", + "delta_I = 20; # Change in current, A\n", + "# Part (b)\n", + "e = abs((-L*delta_I)/(delta_t)); # Induced emf, V\n", + "\n", + "#Results\n", + "print \"The inductance of the coi = %3.1f H\"%L\n", + "print \"The emf induced in the coil = %2d V\"%e\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The inductance of the coi = 1.5 H\n", + "The emf induced in the coil = 300 V\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.22, Page 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I_1 = 8; # Electric current, A\n", + "I_2 = 2; # Electric current, A\n", + "N = 3000; # Number of turns in a coil\n", + "phi_1 = 4e-03; # Flux, Wb\n", + "delta_t = 0.1; # Reversal time of current, s\n", + "\n", + "#Calculations\n", + "L = (N * phi_1)/I_1; # Self-inductance, H\n", + "delta_I = I_1 - I_2; # Change in current, A\n", + "e = ((L)*delta_I)/(delta_t); # Induced emf, V\n", + "\n", + "#Result\n", + "print \"The emf induced in the coil = %2d volts\"%e\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emf induced in the coil = 90 volts\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.23, Page 179" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "N_1 = 600; # Number of turns in a coil in first case\n", + "N_2 = 900; # Number of turns in a coil in second case \n", + "N_3 = 900; # Number of turns in a coil in third case\n", + "l = 45e-03; # Effective length of coil, m\n", + "A = 4e-04; # Cross-sectional area of coil, m^2\n", + "mew_o = 4*(math.pi)*1e-07; # Permeability for free space\n", + "mew_r1 = 1; # Relative permeability in first case\n", + "mew_r2 = 1; # Relative permeability in second case\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "mew_r3 = 75; # Relative permeability in third case\n", + "L_1 = (mew_o*mew_r1*(N_1**2)*A)/l; # Self-inductance of coil in first case, H\n", + "\n", + "# Part (b)\n", + "# Since self-inductance of a coil is directly proportional to the number of turns in a coil, therefore, \n", + "#we have L_2/L_1 = (N_2^2)/(N_1^2), solving for L_2\n", + "L_2 = (L_1*(N_2**2))/(N_1**2); # Self-inductance of coil in second case, H\n", + "\n", + "# Part (c)\n", + "# Since mew_r3 = 75*mew_r2, keeping all other quantities same we have\n", + "L_3 = mew_r3*L_2; # Self-inductance of coil in third case, H\n", + "\n", + "#Results\n", + "print \"Self-inductance of coil in first case = %4.2f mH\"%(L_1/1e-03);\n", + "print \"Self-inductance of coil in second case = %5.3f mH\"%(L_2/1e-03);\n", + "print \"Self-inductance of coil in third case = %5.3f H\"%L_3\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Self-inductance of coil in first case = 4.02 mH\n", + "Self-inductance of coil in second case = 9.048 mH\n", + "Self-inductance of coil in third case = 0.679 H\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.24, Page 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "N_A = 2000; # Number of turns in a coil A\n", + "N_B = 1500; # Number of turns in a coil B\n", + "I_A = 0.5; # Electric current in coil A, A\n", + "phi_A = 60e-06; # Flux linked with coil A, Wb\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "L_A = (N_A*phi_A)/I_A; # Self-inductance of coil A\n", + "phi_B = 0.83*(60e-06); # Flux linked with coil B, Wb\n", + "\n", + "# Part (b)\n", + "M = (N_B*phi_B)/I_A; # Mutual inductance of the two coils, H\n", + "\n", + "#Results\n", + "print \"Self-inductance of coil A = %4.2f H\"%L_A\n", + "print \"Mutual inductance of the two coils = %5.3f H\"%M\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Self-inductance of coil A = 0.24 H\n", + "Mutual inductance of the two coils = 0.149 H\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.25, Page 183" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "N = 400; # Number of turns in a coil\n", + "l = 0.25; # Effective length of coil, m\n", + "A = 4.5e-04; # Cross-sectional area, m^2\n", + "mew_r = 180; # Relative permeability \n", + "\n", + "#Calculations\n", + "mew_o = 4*(math.pi)*1e-07; # Pemeability for free space\n", + "L = (mew_o*mew_r*(N**2)*A)/l # Self-inductance of coil, H\n", + "\n", + "#Result\n", + "print \"The self inductance of the coil = %2d milli-henry\"%(L/1e-03);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The self inductance of the coil = 65 milli-henry\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.26, Page 183" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "L_1 = 65e-03; # Self-inductance of first coil, H\n", + "delta_I = 1.5; # Change in current, A\n", + "delta_t = 3e-03; # Time, s\n", + "k = 0.95; # 95 percent of flux produced\n", + "N_1 = 400; # Number of turns in a coil A\n", + "N_2 = 650; # Number of turns in a coil B \n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Since self-inductance of a coil is directly proportional to the number of turns in a coil, therefore, \n", + "#we have L_2/L_1 = (N_2^2)/(N_1^2), solving for L_2\n", + "L_2 = (L_1*(N_2**2))/(N_1**2) # Self-inductance of second coil , H\n", + "\n", + "# Part (b)\n", + "M = k*math.sqrt(L_1*L_2); # Mutual inductance of two coils, H\n", + "\n", + "# Part (c)\n", + "e_1 = ((L_1)*delta_I)/(delta_t); # Induced emf in first coil, V \n", + "\n", + "# Part (d)\n", + "e_2 = (M*delta_I)/delta_t; # Induced emf in second coil, V \n", + "\n", + "#Results\n", + "print \"The self-inductance of coil 2 = %3d mH\"%(L_2/1e-03)\n", + "print \"The value of mutual inductance = %3d mH\"%(M/1e-03)\n", + "print \"The self-induced emf in coil 1 = %4.1f V\"%e_1\n", + "print \"The mutually induced emf in coil 2 = %2d V\"%e_2\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The self-inductance of coil 2 = 171 mH\n", + "The value of mutual inductance = 100 mH\n", + "The self-induced emf in coil 1 = 32.5 V\n", + "The mutually induced emf in coil 2 = 50 V\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.27, Page 185" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "L = 50e-03; # Self-inductance of coil, H\n", + "I = 0.75; # Electric current in coil, A\n", + "\n", + "#Calculations\n", + "W = (L*(I**2))/2 # Energy stored, J\n", + "\n", + "#Result\n", + "print \"Energy stored in the inductor = %4.1f mJ\"%(W/1e-03)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy stored in the inductor = 14.1 mJ\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.28, Page 185" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "L_1 = 25e-03; # Self-inductance of first coil, H\n", + "L_2 = 40e-03; # Self-inductance of second coil, H\n", + "I = 0.25; # Electric current in coils, A\n", + "k =0.8; # Coupling coefficient\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "W_1 = (L_1*(I**2))/2; # Energy stored in first coil, J\n", + "W_2 = (L_2*(I**2))/2; # Energy stored in second coil, J\n", + "M = k*math.sqrt(L_1*L_2); # Mutual inductance of coils\n", + "\n", + "# Part (b)\n", + "W_M = M*(I)*(I); # Energy stored due to mutual inductance of coils, J\n", + "W_sa = W_1 + W_2 + W_M; # Energy stored by two inductors when connected in series aiding, J\n", + "W_so = W_1 + W_2 - W_M; # Energy stored by two inductors when connected in series opposition, J\n", + "\n", + "#Results\n", + "print \"Energy stored in first coil = %4.2f mJ\"%(W_1/1e-03)\n", + "print \"Energy stored in second coil = %4.2f mJ\"%(W_2/1e-03)\n", + "print \"Energy stored by two inductors when connected in series aiding = %3.1f mJ\"%(W_sa/1e-03)\n", + "print \"Energy stored by two inductors when connected in series opposition = %4.2f mJ\"%(W_so/1e-03)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy stored in first coil = 0.78 mJ\n", + "Energy stored in second coil = 1.25 mJ\n", + "Energy stored by two inductors when connected in series aiding = 3.6 mJ\n", + "Energy stored by two inductors when connected in series opposition = 0.45 mJ\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.29, Page 189" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_2 = 60; # Output voltage, V\n", + "V_1 = 240; # Input voltage, V\n", + "N_2 = 500; # Secondary turns\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# For simplicity let V_1/V_2 = N_1/N_2 = k\n", + "k = V_1/V_2 # Turns ratio\n", + "\n", + "# Part (b)\n", + "# Since V_1/V_2 = N_1/N_2, solving for N_1\n", + "N_1 = k*N_2; # Primary turns\n", + "\n", + "#Results\n", + "print \"The required turns ratio = %1d:1\"%k\n", + "print \"The number of primary turns = %4d\"%N_1\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required turns ratio = 4:1\n", + "The number of primary turns = 2000\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.30, Page 190" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_L = 15; # Load resistor, ohms\n", + "V_2 = 240.; # Terminal p.d at secondary, V\n", + "V_1 = 600; # Supply voltage, V\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Since V_1/V_2 = N_1/N_2 = k\n", + "k = V_1/V_2; # Turns ratio\n", + "\n", + "# Part (b)\n", + "I_2 = V_2/R_L; # Current drawn by the load, A\n", + "P_2 = V_2*I_2; # Power drawn by the load, W\n", + "\n", + "# Part (c)\n", + "I_1 = P_2/V_1 # Current drawn from the supply, A\n", + "\n", + "#Results\n", + "print \"The transformer turns ratio = %3.1f:1\"%k\n", + "print \"The current drawn by the load = %2d A\"%I_2\n", + "print \"The power drawn by the load = %4.2f W\"%(P_2*1e-03);\n", + "print \"The current drawn from the supply = %3.1f A\"%I_1\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The transformer turns ratio = 2.5:1\n", + "The current drawn by the load = 16 A\n", + "The power drawn by the load = 3.84 W\n", + "The current drawn from the supply = 6.4 A\n" + ] + } + ], + "prompt_number": 35 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamental_Electrical_and_Electronic_Principles/ch6.ipynb b/Fundamental_Electrical_and_Electronic_Principles/ch6.ipynb new file mode 100755 index 00000000..d96755d0 --- /dev/null +++ b/Fundamental_Electrical_and_Electronic_Principles/ch6.ipynb @@ -0,0 +1,721 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:b84f65369d0a7456fd5ada25a94ec2301f3722e12d3bfd972d934901296bd409" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6: Alternating Quantities" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1, Page 202" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Comparing alternating voltage v = 35*sin(314.2*t) with the standard Eq.\n", + "# Part (a)\n", + "V_m = 35; # Maximum value of alternating voltage, volt\n", + "\n", + "# Part (b)\n", + "#We know that v= vm8sin(2*pi*f*t)\n", + "#Comparing the alternating voltage equation with the above one, we get,\n", + "f = 314.2/(2*math.pi) # Frequency of waveform, Hz\n", + "\n", + "# Part (c)\n", + "T = 1/f; # Time period of waveform, sec\n", + "\n", + "# Part (d)\n", + "t = 3.5; # Time with reference to zero crossing, sec\n", + "v = 35*math.sin(2*math.pi*50*3.5*1e-03); # Volatge value after the waveform passes through zero, going positive\n", + "\n", + "#Results\n", + "print \"The maximum value of alternating voltage = %2d volt\"%V_m\n", + "print \"The frequency of alternating voltage = %2d Hz\"%f\n", + "print \"The time period of alternating voltage = %3.1f ms\"%(T/1e-03);\n", + "print \"The volatge value after the waveform passes through zero = %5.2f volt\"%v\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum value of alternating voltage = 35 volt\n", + "The frequency of alternating voltage = 50 Hz\n", + "The time period of alternating voltage = 20.0 ms\n", + "The volatge value after the waveform passes through zero = 31.19 volt\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2, Page 202" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Part (a)\n", + "#I = Im*sin(2*pi*f*t)\n", + "# Given i = 75*sin(200*%pi*t) mA which on comparing with the general expression gives\n", + "#Comparing the above two equations, we get,\n", + "f = (200*math.pi)/(2*math.pi) # Frequency of alternating current, Hz\n", + "\n", + "# Part(b)\n", + "i = 35.; # Alternating current after passing through zero, mA\n", + "t = math.asin(i/75)/(200*math.pi*1e-03); # Time taken for current to reach 35 mA, ms\n", + "\n", + "#Results\n", + "print \"The frequency of alternating current = %2d Hz\"%f\n", + "print \"The time taken for current to reach 35 mA = %5.3f mA\"%t\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of alternating current = 100 Hz\n", + "The time taken for current to reach 35 mA = 0.773 mA\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3, Page 204" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_av = 3.5; # Average value of sinusoidal alternating voltage, V\n", + "T = 6.67e-03; # Time period of alternating current, s\n", + "\n", + "#Calculations\n", + "V_m = V_av/0.637; # Peak value of alternating current, V\n", + "f = 1/T; # Frequency of alternating volatge, Hz\n", + "\n", + "#Result\n", + "print \"The standard expression for %3.1f voltage = %3.1f sin(%3d*pi*t) volt\"%(V_av, V_m,round(2*f))\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The standard expression for 3.5 voltage = 5.5 sin(300*pi*t) volt\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4, Page 204" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "V_av = 3.5; # Average value of sinusoidal alternating voltage, V\n", + "T = 6.67e-03; # Time period of alternating voltage, s\n", + "\n", + "#Calculations&Results\n", + "V_m = V_av/0.637; # Peak value of alternating voltage, V\n", + "f = 1/T; # Frequency of alternating volatge, Hz\n", + "\n", + "# Part (a)\n", + "t = 0.5e-03; # Time taken by the waveform after passing through zero, s\n", + "v = V_m*math.sin(2*math.pi*f*t); # Instantaneous value of alternating voltage, s\n", + "print \"The instantaneous value of alternating voltage after %3.1f ms = %3.1f volt\"%(t/1e-03, v)\n", + "\n", + "# Part (b)\n", + "t = 4.5e-03; # Time taken by the waveform after passing through zero, s\n", + "v = V_m*math.sin(2*math.pi*f*t); # Instantaneous value of alternating voltage, s\n", + "print \"The instantaneous value of alternating voltage after %3.1f ms = %3.1f volt\"%(t/1e-03, v);\n", + "\n", + "# Part (c)\n", + "v = 3; # Alternating voltage after passing through zero, mA\n", + "t = math.asin(v/V_m)/(2*math.pi*f); # Time taken for current to reach 3 V, s\n", + "print \"The time taken for voltage to reach %1d volt = %5.3f ms\"%(v, t/1e-03);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The instantaneous value of alternating voltage after 0.5 ms = 2.5 volt\n", + "The instantaneous value of alternating voltage after 4.5 ms = -4.9 volt\n", + "The time taken for voltage to reach 3 volt = 0.613 ms\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5, Page 206" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "V = 240; # Rms vlaue of alternating voltage, volt\n", + "\n", + "#Calculations\n", + "V_m = math.sqrt(2)*V; # Peak value of alternating voltage, volt\n", + "\n", + "#Result\n", + "print \"The amplitude of household %3d volt supply = %5.1f volt\"%(V, V_m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The amplitude of household 240 volt supply = 339.4 volt\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6, Page 207" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "pf = 2.5; # Peak factor of non-sinusoidal alternating voltage\n", + "V = 240; # Rms vlaue of alternating voltage, volt\n", + "\n", + "#Calculations\n", + "V_m = pf*V; # Peak value of alternating voltage, volt\n", + "\n", + "#Result\n", + "print \"The absolute minimum working voltage = %3d volt\"%V_m\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The absolute minimum working voltage = 600 volt\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7, Page 207" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "l = 0.25; # Length of the rectangular coil, m\n", + "d = 0.2; # Width of rectangular coil, m\n", + "N = 80; # Number of turns of the rectangular coil\n", + "B = 0.075; # Magnetic flux density, tesla\n", + "n = 3000/60; # Frequency of revolution of the coil, rev/s\n", + "v = n*math.pi*d; # Linear speed with which the coil sides move, m/s\n", + "t = 2e-03; # Time after the emf crosses zero, s\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# As e = 2*N*B*l*v*sin(2*pi*f*t) volt, and for maximum value of sin(2*pi*f*t) = 1\n", + "E_m = 2*N*B*l*v*(1); # Amplitude of emf, volt\n", + "E = 0.707*E_m; # rms value of emf, volt\n", + "E_av = 0.637*E_m; # Average value of emf, volt\n", + "# For a two pole field system, \n", + "f = n; # Frequency of generated waveform, Hz\n", + "\n", + "# Part (b)\n", + "T = 1./f; # Time period of generated waveform, Hz\n", + "\n", + "# Part (c)\n", + "e = E_m*math.sin(2*math.pi*f*t); # Instantaneous value at time 2 ms after zero, volt\n", + "\n", + "#Results\n", + "print \"The amplitude, rms and average value of emf = %5.2f V, %5.2f V and %5.2f V resp.\"%(E_m, E, E_av);\n", + "print \"The frequency and time period of generated waveform = %2d Hz and %2d ms resp.\"%(f, T/1e-03);\n", + "print \"The instantaneous value of emf at time 2 ms after crossing zero = %4.1f V\"%e\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The amplitude, rms and average value of emf = 94.25 V, 66.63 V and 60.04 V resp.\n", + "The frequency and time period of generated waveform = 50 Hz and 20 ms resp.\n", + "The instantaneous value of emf at time 2 ms after crossing zero = 55.4 V\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8, Page 212" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "R_c = 50; # Resistance of the coil of meter, ohm\n", + "K = 10e+03; # Figure of merit of the moving coil meter, ohm per volt\n", + "V = 10; # d.c. range of coil meter, volt\n", + "\n", + "#Calculations&Results\n", + "# Part (a)\n", + "I_fsd = 1/K; # Full scale deflection for moving coil meter, ampere\n", + "R = V/I_fsd; # Total meter resistance, ohm\n", + "# As R = R_m + R_c, solvign for R_m\n", + "R_m = R - R_c; # Multiplier resistance required by the meter, ohm\n", + "print \"The multiplier resistance required for 10 V d.c. range = %5.2f k-ohm\"%(R_m/1e+03)\n", + "\n", + "# Part(b)\n", + "I_av = I_fsd; # Average value of ac current, A\n", + "I_rms = math.pi/(2*math.sqrt(2))*I_av; # rms value of ac current, A\n", + "V = 10 ; # a.c. range of coil meter, volt\n", + "R = V/I_rms; # Total meter resistance, ohm\n", + "# As R = R_m + R_c, solvign for R_m\n", + "R_m = R - R_c; # Multiplier resistance required by the meter, ohm\n", + "print \"The multiplier resistance required for 10 V a.c. range = %5.2f k-ohm\"%(R_m/1e+03);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The multiplier resistance required for 10 V d.c. range = 99.95 k-ohm\n", + "The multiplier resistance required for 10 V a.c. range = 89.98 k-ohm\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9, Page 213" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Case_I: Square_wave\n", + "ff = 1.11; # Form factor of calibrated meter\n", + "ff_square = 1; # Form factor for square wave\n", + "V_apparent = 5; # Meter reading for sqaure wave, volt\n", + "V_true = V_apparent*1*(ff_square/ff); # True rms value of square wave voltage, volt\n", + "print \"The true rms value of square wave voltage = %5.3f V\"%V_true\n", + "\n", + "# Case_II: Triangular_wave\n", + "ff_triangle = 1.15; # Form factor for triangular wave\n", + "V_apparent = 5; # Meter reading for triangular wave, volt\n", + "V_true = V_apparent*(ff_triangle/ff); # True rms value of triangular wave voltage, volt\n", + "print \"The true rms value of triangular wave voltage = %4.2f V\"%V_true\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The true rms value of square wave voltage = 4.505 V\n", + "The true rms value of triangular wave voltage = 5.18 V\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10, Page 215" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "# The general expression for alternating current is I = Io*sin(2*pi*f*t + phi)\n", + "#Comparing the given equations with the above, we get,\n", + "f = (80*math.pi)/(2*math.pi) # Frequency of alternating current, Hz\n", + "\n", + "#Calculations\n", + "# I2 is the reference waveform with zero phase angle, so that\n", + "phi2 = 0; # Phase angle for reference waveform I2, degrees\n", + "Im2 = 3; # Current amplitude of reference waveform I2, A\n", + "Im1 = 5; # Current amplitude of reference waveform I1, A\n", + "Im3 = 6; # Current amplitude of reference waveform I3, A\n", + "phi1 = math.pi/6*(180/math.pi); # Phase angle for reference waveform I1, degrees\n", + "phi3 = math.pi/4*(180/math.pi); # Phase angle for reference waveform I3, degrees\n", + "\n", + "#Results\n", + "print \"The frequency of all three waveforms = %2d Hz\"%f\n", + "print \"I1 leads I2 by = %2.0f degrees\"%(phi1-phi2);\n", + "print \"I3 lags I2 by = %2d degrees\"%(phi3-phi2);\n", + "print \"Current amplitude of reference waveform I1 = %1d A\"%Im1\n", + "print \"Current amplitude of reference waveform I2 = %1d A\"%Im2\n", + "print \"Current amplitude of reference waveform I3 = %1d A\"%Im3\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of all three waveforms = 40 Hz\n", + "I1 leads I2 by = 30 degrees\n", + "I3 lags I2 by = 45 degrees\n", + "Current amplitude of reference waveform I1 = 5 A\n", + "Current amplitude of reference waveform I2 = 3 A\n", + "Current amplitude of reference waveform I3 = 6 A\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12, Page 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "Im1 = 7; # Current amplitude of reference waveform I1, A\n", + "Im2 = 6; # Current amplitude of reference waveform I2, A\n", + "Im3 = 5; # Current amplitude of reference waveform I3, A\n", + "Im4 = 4; # Current amplitude of reference waveform I4, A\n", + "\n", + "#Calculations\n", + "phi1 = 70*math.pi/180; # Phase angle for reference waveform I1, rad\n", + "phi2 = 0*math.pi/180; # Phase angle for reference waveform I2, rad\n", + "phi3 = -50*math.pi/180; # Phase angle for reference waveform I3, rad\n", + "phi4 = -90*math.pi/180; # Phase angle for reference waveform I4, rad\n", + "\n", + "#Results\n", + "print \"i1 = %dsin(wt + %4.2f) amp\"%(Im1, phi1)\n", + "print \"i2 = %dsin wt amp\"%Im2;\n", + "print \"i3 = %dsin(wt + %4.2f) amp\"%(Im3, phi3);\n", + "print \"i4 = %dsin(wt + %4.2f) amp\"%(Im4, phi4);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i1 = 7sin(wt + 1.22) amp\n", + "i2 = 6sin wt amp\n", + "i3 = 5sin(wt + -0.87) amp\n", + "i4 = 4sin(wt + -1.57) amp\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13, Page 221" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "omega = 314.; # Angular frequency of voltage, rad per sec\n", + "Vm1 = 25.; # Peak value of first phasor, V\n", + "Vm2 = 15.; # Peak value of second phasor, V\n", + "\n", + "#Calculations\n", + "H_C = Vm1*math.cos(math.pi/3)+Vm2*math.cos(-math.pi/6); # Horizontal component of phasor sum, V\n", + "V_C = Vm1*math.sin(math.pi/3)+Vm2*math.sin(-math.pi/6); # Vertical component of phasor sum, V\n", + "Vm = math.sqrt(H_C**2+V_C**2); # Peak value of phasor sum, V\n", + "phi = math.atan(V_C/H_C); # Phase angle, degrees\n", + "print \"v = %5.2fsin(%3dt + %5.3f) volt\"%(Vm, omega, phi);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "v = 29.15sin(314t + 0.507) volt\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.14, Page 222" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "Im1 = 6; # Peak value of first phasor, A\n", + "Im2 = 8; # Peak value of second phasor, A\n", + "Im3 = 4; # Peak value of third phasor, A\n", + "\n", + "#Calculations\n", + "H_C = Im1*math.cos(0)+Im2*math.cos(-math.pi/2)+Im3*math.cos(math.pi/6); # Horizontal component of phasor sum, A\n", + "V_C = Im1*math.sin(0)+Im2*math.sin(-math.pi/2)+Im3*math.sin(math.pi/6); # Vertical component of phasor sum, A\n", + "Im = math.sqrt(H_C**2+V_C**2); # Peak value of phasor sum, V\n", + "phi = math.atan(V_C/H_C); # Phase angle, rad\n", + "\n", + "\n", + "print \"i = %4.1fsin(wt%5.3f) amp\"%(Im, phi);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i = 11.2sin(wt-0.565) amp\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.15, Page 222" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Part (a)\n", + "omega = 628; # Angular frequency of voltage, rad per sec\n", + "f = omega/(2*math.pi); # Frequency of the waveforms, Hz\n", + "Vm1 = 10.; # Peak value of first phasor, V\n", + "Vm2 = 8.; # Peak value of second phasor, V\n", + "Vm3 = 12.; # Peak value of third phasor, V\n", + "phi1 = -math.pi/6*180/math.pi; # Phase angle for first voltage, degrees\n", + "phi2 = math.pi/3*180/math.pi; # Phase angle for second voltage, degrees\n", + "phi3 = math.pi/4*180/math.pi; # Phase angle for third voltage, degrees\n", + "print \"The frequency of all three waveforms = %3d Hz\"%f\n", + "print \"The phase angle and frequency of first voltage : %2d degrees, %2d V\"%(phi1, Vm1);\n", + "print \"The phase angle and frequency of second voltage : %2d degrees, %2d V\"%(phi2, Vm2);\n", + "print \"The phase angle and frequency of third voltage : %2d degrees, %2d V\"%(phi3, Vm3);\n", + "\n", + "# Part (b)\n", + "H_C = Vm1*math.cos(phi1)+Vm2*math.cos(phi2)+Vm3*math.cos(phi3); # Horizontal component of phasor sum, V\n", + "V_C = Vm1*math.sin(phi1)+Vm2*math.sin(phi2)+Vm3*math.sin(phi3); # Horizontal component of phasor sum, V\n", + "Vm = math.sqrt(H_C**2+V_C**2); # Peak value of phasor sum, V\n", + "phi = math.atan(V_C/H_C); # Phase angle, rad\n", + "print \"v = %5.2fsin(%3dt + %5.3f) volt\"%(Vm, omega, phi);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of all three waveforms = 99 Hz\n", + "The phase angle and frequency of first voltage : -29 degrees, 10 V\n", + "The phase angle and frequency of second voltage : 59 degrees, 8 V\n", + "The phase angle and frequency of third voltage : 45 degrees, 12 V\n", + "v = 17.65sin(628t + 1.558) volt\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.16, Page 228" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "tb1 = 0.1e-03; # Timebase of channel 1, s/cm\n", + "tb2 = 10e-06; # Timebase of channel 2, s/cm\n", + "Y_amp1 = 5.; # Y-amp setting for channel 1, V/cm\n", + "Y_amp2 = 0.5; # Y-amp setting for channel 2, V/cm\n", + "\n", + "#Calculations&Results\n", + "# Channel 1\n", + "V_pp = 3*Y_amp1; # Peak-to-peak value of waveform in channel 1, V\n", + "Vm = V_pp/2; # Amplitude of waveform in channel 1, V\n", + "V = Vm/math.sqrt(2); # rms value of sine wave in channel 1, V\n", + "T = 4*tb1; # Time period of sine wave, second\n", + "f = 1./(T*1000); # Frequency of sine wave, kHz\n", + "print \"The amplitude of sine waveform in channel 1 = %3.1f V\"%Vm\n", + "print \"The rms value of sine wave in channel 1 = %3.1f V\"%V\n", + "print \"The frequency of sine wave in channel 1 = %3.1f kHz\"%f\n", + "\n", + "# Channel 2\n", + "V_pp = 2*Y_amp2; # Peak-to-peak value of waveform in channel 2, V\n", + "Vm = V_pp/2; # Amplitude of waveform in channel 2, V\n", + "V = Vm; # rms value of square wave in channel 2, V\n", + "T = 2./3*tb2; # Time period of square wave, second\n", + "f = 1./(T*1000); # Frequency of square wave, kHz\n", + "print \"The amplitude of square waveform in channel 2 = %3.1f V\"%Vm\n", + "print \"The rms value of square wave in channel 2 = %3.1f V\"%V\n", + "print \"The frequency of square wave in channel 2 = %3d kHz\"%f\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The amplitude of sine waveform in channel 1 = 7.5 V\n", + "The rms value of sine wave in channel 1 = 5.3 V\n", + "The frequency of sine wave in channel 1 = 2.5 kHz\n", + "The amplitude of square waveform in channel 2 = 0.5 V\n", + "The rms value of square wave in channel 2 = 0.5 V\n", + "The frequency of square wave in channel 2 = 150 kHz\n" + ] + } + ], + "prompt_number": 27 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamental_Electrical_and_Electronic_Principles/ch7.ipynb b/Fundamental_Electrical_and_Electronic_Principles/ch7.ipynb new file mode 100755 index 00000000..91ced7ef --- /dev/null +++ b/Fundamental_Electrical_and_Electronic_Principles/ch7.ipynb @@ -0,0 +1,78 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:63415a109c6989d49d84f5199fa0cd1f338b44376604f4b24df855330c3cb38a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7: D.C. Machines" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1, Page 243" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Rf = 200; # The resistance of field winding of a shunt generator, ohm\n", + "Po = 80e+03; # Power delivered by the machine, watt\n", + "V = 450; # The terminal voltage, volt\n", + "E = 475; # The generated emf, volt\n", + "\n", + "\n", + "#Calculations\n", + "# Part (a)\n", + "# Po = V*I_L, solving for I_L\n", + "I_L = Po/V; # Load current, A\n", + "I_f = V/Rf; # Current through field resistor, A\n", + "I_a = I_L + I_f; # Current through armature resistance, A\n", + "# As I_a*Ra = E - V, solving for Ra\n", + "Ra = 25/I_a; # Armature resistance, ohm\n", + "\n", + "# Part (b)\n", + "Po = 50e+03; # Output power delivered by the machine, watt\n", + "V = 460; # The terminal voltage, volt\n", + "I_L = Po/V; # Load current, A\n", + "I_f = V/Rf; # Current through field resistor, A\n", + "I_a = I_L + I_f; # Current through armature resistance, A\n", + "# As I_a*Ra = E - V, solving for E\n", + "E = V + I_a*Ra; # The generated voltage, volt\n", + "\n", + "#Results\n", + "print \"The armature resistance = %5.3f ohm\"%Ra\n", + "print \"The value of generated emf = %5.1f V\"%E\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The armature resistance = 0.139 ohm\n", + "The value of generated emf = 475.4 V\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamental_Electrical_and_Electronic_Principles/ch8.ipynb b/Fundamental_Electrical_and_Electronic_Principles/ch8.ipynb new file mode 100755 index 00000000..a3af808b --- /dev/null +++ b/Fundamental_Electrical_and_Electronic_Principles/ch8.ipynb @@ -0,0 +1,165 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:8e01926cf42f8b82c73cff46263ef90c672622041c85ff2d1b1ef7a3962ce2cc" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8: D.C. Transients" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1, 253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "C = 8e-06; # Value of capacitance of capacitor, farad\n", + "R = 0.5e+06; # Value of series resistor, ohm\n", + "E = 200; # Value of d.c. voltage supply, volt\n", + "\n", + "#Calculations&Results\n", + "# Part (a)\n", + "tau = C*R; # Time constant of the R-C circuit while charging, s\n", + "print \"The circuit time constant while charging = %1d s\"%tau\n", + "\n", + "# Part (b)\n", + "I_0 = E/R; # Initial charging current through capacitor, A\n", + "print \"The initial charging current through capacitor = %3d micro-ampere\"%(I_0/1e-06);\n", + "\n", + "# Part (c)\n", + "t = 4; # Time after the supply is connected, s\n", + "v_C = 0.632*E; # p.d. across the capacitor 4s after the supply is connected, V\n", + "v_R = E - v_C; # p.d. across the resistor 4s after the supply is connected, V\n", + "print \"The p.d. across resistor and capacitor %d s after the supply is connected = %5.1f V and %4.1f V respectively\"%(t, v_C, v_R);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The circuit time constant while charging = 4 s\n", + "The initial charging current through capacitor = 400 micro-ampere\n", + "The p.d. across resistor and capacitor 4 s after the supply is connected = 126.4 V and 73.6 V respectively\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.2, Page 255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "C = 0.5e-06; # Value of capacitance of capacitor, farad\n", + "R1 = 220e+03; # Value of series resistor, ohm\n", + "R2 = 110e+03; # Value of parallel resistor, ohm\n", + "E = 150; # Value of d.c. voltage supply, volt\n", + "\n", + "#Calculations&Results\n", + "# Part (a)\n", + "tau = C*R1; # Time constant of the R1-C circuit while charging, s\n", + "print \"The circuit time constant while charging = %4.2f s\"%tau\n", + "I_0 = E/R1; # Initial charging current through capacitor, A\n", + "print \"The initial charging current through capacitor = %3d micro-ampere\"%(I_0/1e-06)\n", + "\n", + "# Part (b)\n", + "tau = C*(R1+R2); # Time constant of the R1-C-R2 circuit while discharging, s\n", + "print \"The circuit time constant while discharging = %4.2f s\"%tau\n", + "I_0 = E/(R1 + R2); # Initial discharging current through capacitor, ampere\n", + "i = 0.368*I_0; # Discharge current after one time constant, ampere\n", + "V_R2 = i*R2; # Potential difference across R2 after one time constant, volt\n", + "print \"The p.d. across R2 after one time constant while discharging = %4.1f volt\"%V_R2\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The circuit time constant while charging = 0.11 s\n", + "The initial charging current through capacitor = 681 micro-ampere\n", + "The circuit time constant while discharging = 0.16 s\n", + "The p.d. across R2 after one time constant while discharging = 18.4 volt\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3, Page 258" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "E = 110.; # Value of d.c. voltage supply, volt\n", + "L = 1.5; # Inductor value, henry\n", + "R = 220; # Value of series resistor, ohm\n", + "\n", + "#Calculations&Results\n", + "# Part (a)\n", + "di_dt = E/L; # The initial rate of change of current through inductor, H\n", + "print \"The initial rate of change of current through inductor = %5.2f A/s\"%di_dt\n", + "\n", + "# Part (b)\n", + "I = E/R; # The final steady current, A\n", + "print \"The final steady current through inductor = %3.1f A\"%I\n", + "\n", + "# Part (c)\n", + "tau = L/R; # The time taken for the current to reach its fi nal steady value, s\n", + "print \"The time taken for the current to reach its final steady value = %4.1f ms\"%(5*tau/1e-03);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The initial rate of change of current through inductor = 73.33 A/s\n", + "The final steady current through inductor = 0.5 A\n", + "The time taken for the current to reach its final steady value = 34.1 ms\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamental_Electrical_and_Electronic_Principles/ch9.ipynb b/Fundamental_Electrical_and_Electronic_Principles/ch9.ipynb new file mode 100755 index 00000000..0bbdaca8 --- /dev/null +++ b/Fundamental_Electrical_and_Electronic_Principles/ch9.ipynb @@ -0,0 +1,206 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:49f87810817ddb9615b6e59a9caa61ca07b12c7b29a4202a634fda7cd35deb7f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9: Semiconductor Theory and Devices" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1, Page 277" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Part (a)\n", + "V_Z = 9.1; # Zener voltage of zener diode, volt\n", + "P_Z = 0.5; # Power rating of zener diode at V_Z, W\n", + "r_Z = 1.5; # Slope resistance of zener diode, ohm\n", + "V = 12; # Nominal value of input voltage, volt\n", + "R_L = 2.5e+03; # Load resistance across zener diode, ohm\n", + "\n", + "\n", + "I_Z = P_Z/V_Z*1e+03; # Zener current, mA\n", + "I_S = I_Z; # Current through series resistor, mA\n", + "V_S = V - V_Z; # Voltage drop across series resistor, volt\n", + "R_S = V_S/I_S*1e+03; # Value of series resistance, ohm\n", + "P_max = (I_S*1e-03)**2*R_S; # Maximum power rating of series resistance, W\n", + "print \"The value of series resistance = %5.2f ohm\"%R_S\n", + "print \"The value of maximum power rating of series resistance = %4.2f W\"%P_max\n", + "print \"(a) The suitable value of R_S should be 54 ohm, 0.25 W\"\n", + "\n", + "# Part (b)\n", + "V_o = V_Z; # Output voltage across zener, volt\n", + "I_L = V_o/R_L*1e+03; # Load current, mA\n", + "I_Z = I_S - I_L; # Zener current, mA\n", + "print \"(b) The value of diode current with load resistance across zener = %5.2f mA\"%I_Z \n", + "\n", + "# Part (c)\n", + "V = 12 - (0.1*12); # Final value of input voltage after falling below 12 V, volt\n", + "R_S = 56; # Standard value of series resistance, ohm\n", + "I_S = (V - V_Z)/R_S*1e+03; # Current through series resistance, mA\n", + "I_Z = I_S - I_L; # Resulting diode current, mA\n", + "delta_I_Z = 51.36 - I_Z; # Change in zener current, mA\n", + "delta_V_Z = delta_I_Z*1e-03*r_Z; # Change in zener voltage, V\n", + "change = delta_V_Z/V_Z*100; # %age change in zener voltage\n", + "print \"(c) The percentage change in the p.d. across the load = %4.2f percent\"%change\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of series resistance = 52.78 ohm\n", + "The value of maximum power rating of series resistance = 0.16 W\n", + "(a) The suitable value of R_S should be 54 ohm, 0.25 W\n", + "(b) The value of diode current with load resistance across zener = 51.31 mA\n", + "(c) The percentage change in the p.d. across the load = 0.41 percent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2, Page 279" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import numpy as np\n", + "\n", + "#Variable declaration\n", + "Diode = np.array([3, 1]); # Declare a diode cell\n", + "Diode_1 = [1, 15, 30, 0.5, 0.007]; # Data for Ist diode\n", + "Diode_2 = [2, 15, 15, 1.3, 0.20]; # Data for 2nd diode\n", + "Diode_3 = [1, 15, 2.5, 5.0, 0.67]; # Data for 3rd diode\n", + "Resistor = np.array([5, 1]) # Declare a resistor cell\n", + "Resistor_1 = [0.25, 0.026]; # Data for Ist resistor\n", + "Resistor_2 = [0.5, 0.038]; # Data for 2nd resistor\n", + "Resistor_3 = [1.0, 0.055]; # Data for 3rd resistor\n", + "Resistor_4 = [2.5, 0.260]; # Data for 4th resistor\n", + "Resistor_5 = [7.5, 0.280]; # Data for 5th resistor\n", + "V = 24; # Input voltage, volt\n", + "\n", + "#Calculations&Results\n", + "V_Z = Diode_1[1]; # Zener voltage for Ist diode, volt\n", + "V_S = V - V_Z; # Voltage drop across series resistor for all the three diodes, volt\n", + "\n", + "#Part (a)\n", + "# Diode 1\n", + "P_Z = Diode_1[3]; # Power rating of Ist diode, W\n", + "I_Z = P_Z/V_Z*1e+03; # Zener current, mA\n", + "R_S = V_S/I_Z*1e+03; # Value of series resistance, ohm\n", + "P_S = V_S**2/R_S; # Power dissipation across series resistor, watt\n", + "print \"Diode 1:\";\n", + "print \"========\";\n", + "print \"The value of series resistance = %3d ohm\"%R_S;\n", + "print \"The value of power rating of series resistance = %3.1f W\"%P_S\n", + "R_S = 270; # Chosen value of series resistor, ohm\n", + "P_S = 0.3; # Chosen value of power rating, ohm\n", + "print \"The suitable value of R_S should be %3d ohm, %3.1f W\"%(R_S, P_S);\n", + "print \"Total unit cost = %5.3f pounds\\n\"%(Diode_1[4]+Resistor_2[1]);\n", + "\n", + "# Diode 2\n", + "print \"Diode 2:\";\n", + "print \"========\";\n", + "P_Z = Diode_2[3]; # Power rating of 2nd diode, W\n", + "I_Z = P_Z/V_Z*1e+03; # Zener current, mA\n", + "R_S = V_S/I_Z*1e+03; # Value of series resistance, ohm\n", + "P_S = V_S**2/R_S; # Power dissipation across series resistor, watt\n", + "print \"The value of series resistance = %5.2f ohm\"%R_S\n", + "print \"The value of power rating of series resistance = %4.2f W\"%P_S\n", + "R_S = 120; # Chosen value of series resistor, ohm\n", + "P_S = 1.0; # Chosen value of power rating, ohm\n", + "print \"The suitable value of R_S should be %3d ohm, %3.1f W\"%(R_S, P_S);\n", + "print \"Total unit cost = %4.2f pounds\"%(Diode_2[4]+Resistor_3[1]);\n", + "\n", + "# Diode 3\n", + "print \"\\nDiode 3:\";\n", + "print \"========\";\n", + "P_Z = Diode_3[3]; # Power rating of 3rd diode, W\n", + "I_Z = P_Z/V_Z*1e+03; # Zener current, mA\n", + "R_S = V_S/I_Z*1e+03; # Value of series resistance, ohm\n", + "P_S = V_S**2/R_S; # Power dissipation across series resistor, watt\n", + "print \"The value of series resistance = %3d ohm\"%R_S;\n", + "print \"The value of power rating of series resistance = %3.1f W\"%P_S\n", + "R_S = 27; # Chosen value of series resistor, ohm\n", + "P_S = 7.5; # Chosen value of power rating, ohm\n", + "print \"The suitable value of R_S should be %3d ohm, %3.1f W\"%(R_S, P_S);\n", + "print \"Total unit cost = %4.2f pounds\"%(Diode_3[4]+Resistor_5[1]);\n", + "\n", + "# Part (b)\n", + "delta_V_Z = (5*15)/100; # Allowable change in V_Z, volt\n", + "delta_I_Z = 30e-03; # Allowable change in zener current, A\n", + "delta_VZ = np.zeros(3);\n", + "delta_VZ_1 = 30e-03*30; # Change in zener voltage dor diode 1, V\n", + "delta_VZ_2 = 30e-03*15; # Change in zener voltage dor diode 2, V\n", + "delta_VZ_3 = 30e-03*2.5; # Change in zener voltage dor diode 3, V\n", + "print \"\\nThe maximum value of zener voltage change = %4.2f V\"%(max(delta_VZ_2, delta_VZ_3));\n", + "print \"To meet the specification at lowest cost, circuit 2 would be adopted\";\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diode 1:\n", + "========\n", + "The value of series resistance = 269 ohm\n", + "The value of power rating of series resistance = 0.3 W\n", + "The suitable value of R_S should be 270 ohm, 0.3 W\n", + "Total unit cost = 0.045 pounds\n", + "\n", + "Diode 2:\n", + "========\n", + "The value of series resistance = 103.85 ohm\n", + "The value of power rating of series resistance = 0.78 W\n", + "The suitable value of R_S should be 120 ohm, 1.0 W\n", + "Total unit cost = 0.26 pounds\n", + "\n", + "Diode 3:\n", + "========\n", + "The value of series resistance = 27 ohm\n", + "The value of power rating of series resistance = 3.0 W\n", + "The suitable value of R_S should be 27 ohm, 7.5 W\n", + "Total unit cost = 0.95 pounds\n", + "\n", + "The maximum value of zener voltage change = 0.45 V\n", + "To meet the specification at lowest cost, circuit 2 would be adopted\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
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