diff options
Diffstat (limited to 'Fundamental_Electrical_and_Electronic_Principles/ch1.ipynb')
| -rwxr-xr-x | Fundamental_Electrical_and_Electronic_Principles/ch1.ipynb | 818 |
1 files changed, 818 insertions, 0 deletions
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 |