diff options
184 files changed, 0 insertions, 95985 deletions
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint.ipynb deleted file mode 100755 index b33db331..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint.ipynb +++ /dev/null @@ -1,329 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 1: Units Associated with Basic Electrical Quantities</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 4</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the quantity of electricity transferred.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 5; # in Ampere\n",
- "t = 120; # in sec\n",
- "\n",
- "#calculation:\n",
- "Q = I*t\n",
- "\n",
- "#results\n",
- "print \"Charge, Q = \", Q,\"coulomb(C)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charge, Q = 600 coulomb(C)"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 5</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the force needed.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "M = 5; # in Kg\n",
- "a = 2; # in m/s2\n",
- "\n",
- "#calculation:\n",
- "F = M*a\n",
- "\n",
- "#results\n",
- "print \"Force:\", F,\"Newton(N)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force: 10 Newton(N)"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 5</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the force acting vertically downwards\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "M = 0.2; # in Kg\n",
- "g = 9.81; # in m/s2\n",
- "\n",
- "#calculation:\n",
- "F = M*g\n",
- "\n",
- "#results\n",
- "print \"Force:\", F,\"Newton(N)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force: 1.962 Newton(N)"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 6</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate Work Done and Average Power?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "F = 200; # in Newton\n",
- "d = 20; # in m\n",
- "t = 25; # in sec\n",
- "\n",
- "#calculation:\n",
- "W = F*d\n",
- "P = W/t\n",
- "\n",
- "#results\n",
- "print \"Power:\", P,\"watt(W)\"\n",
- "print \"Work Done:\", W,\"Nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power: 160.0 watt(W)\n",
- "Work Done: 4000 Nm"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 6</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What is (a) the work done and (b) the power developed?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "M = 1000; # in Kg\n",
- "h = 10; # in m\n",
- "t = 20; # in sec\n",
- "g = 9.81 # in m/s2\n",
- "\n",
- "#calculation:\n",
- "W = M*g*h\n",
- "P = W/t\n",
- "\n",
- "#results\n",
- "print \"Work Done:\", W,\"Joule(J)\"\n",
- "print \"Power:\", P,\"watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Work Done: 98100.0 Joule(J)\n",
- "Power: 4905.0 watt(W)"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 7</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the conductance of a conductor of resistances\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 10; # in ohm\n",
- "R2 = 5000; # in ohm\n",
- "R3 = 0.1; # in ohm\n",
- "#calculation:\n",
- "G1 = 1/R1\n",
- "G2 = 1/R2\n",
- "G3 = 1/R3\n",
- "\n",
- "#results\n",
- "print \"conductance(G1):\", G1,\"seimen(S)\"\n",
- "print \"conductance(G2):\", G2,\"seimen(S)\"\n",
- "print \"conductance(G3):\", G3,\"seimen(S)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "conductance(G1): 0.1 seimen(S)\n",
- "conductance(G2): 0.0002 seimen(S)\n",
- "conductance(G3): 10.0 seimen(S)"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 7</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#How much energy is provided in this time?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 5; # in Volts\n",
- "I = 3; # in Ampere\n",
- "t = 600; # in sec\n",
- "#calculation:\n",
- "P = V*I\n",
- "E = P*t\n",
- "\n",
- "#results\n",
- "print \"Energy(E):\", E,\"Joule(J)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy(E): 9000 Joule(J)"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 8</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the power rating of the heater and the current taken from the supply.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "E = 18E5; # in Joule\n",
- "V = 250; # in Volts\n",
- "t = 1800; # in sec\n",
- "\n",
- "#calculation:\n",
- "P = E/t\n",
- "I = P/V\n",
- "\n",
- "#results\n",
- "print \"Power(P):\", P,\"Watt(W)\"\n",
- "print \"Current(I):\", I,\"Ampere(A)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power(P): 1000.0 Watt(W)\n",
- "Current(I): 4.0 Ampere(A)"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_1.ipynb deleted file mode 100755 index b33db331..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_1.ipynb +++ /dev/null @@ -1,329 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 1: Units Associated with Basic Electrical Quantities</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 4</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the quantity of electricity transferred.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 5; # in Ampere\n",
- "t = 120; # in sec\n",
- "\n",
- "#calculation:\n",
- "Q = I*t\n",
- "\n",
- "#results\n",
- "print \"Charge, Q = \", Q,\"coulomb(C)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charge, Q = 600 coulomb(C)"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 5</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the force needed.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "M = 5; # in Kg\n",
- "a = 2; # in m/s2\n",
- "\n",
- "#calculation:\n",
- "F = M*a\n",
- "\n",
- "#results\n",
- "print \"Force:\", F,\"Newton(N)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force: 10 Newton(N)"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 5</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the force acting vertically downwards\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "M = 0.2; # in Kg\n",
- "g = 9.81; # in m/s2\n",
- "\n",
- "#calculation:\n",
- "F = M*g\n",
- "\n",
- "#results\n",
- "print \"Force:\", F,\"Newton(N)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force: 1.962 Newton(N)"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 6</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate Work Done and Average Power?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "F = 200; # in Newton\n",
- "d = 20; # in m\n",
- "t = 25; # in sec\n",
- "\n",
- "#calculation:\n",
- "W = F*d\n",
- "P = W/t\n",
- "\n",
- "#results\n",
- "print \"Power:\", P,\"watt(W)\"\n",
- "print \"Work Done:\", W,\"Nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power: 160.0 watt(W)\n",
- "Work Done: 4000 Nm"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 6</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What is (a) the work done and (b) the power developed?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "M = 1000; # in Kg\n",
- "h = 10; # in m\n",
- "t = 20; # in sec\n",
- "g = 9.81 # in m/s2\n",
- "\n",
- "#calculation:\n",
- "W = M*g*h\n",
- "P = W/t\n",
- "\n",
- "#results\n",
- "print \"Work Done:\", W,\"Joule(J)\"\n",
- "print \"Power:\", P,\"watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Work Done: 98100.0 Joule(J)\n",
- "Power: 4905.0 watt(W)"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 7</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the conductance of a conductor of resistances\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 10; # in ohm\n",
- "R2 = 5000; # in ohm\n",
- "R3 = 0.1; # in ohm\n",
- "#calculation:\n",
- "G1 = 1/R1\n",
- "G2 = 1/R2\n",
- "G3 = 1/R3\n",
- "\n",
- "#results\n",
- "print \"conductance(G1):\", G1,\"seimen(S)\"\n",
- "print \"conductance(G2):\", G2,\"seimen(S)\"\n",
- "print \"conductance(G3):\", G3,\"seimen(S)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "conductance(G1): 0.1 seimen(S)\n",
- "conductance(G2): 0.0002 seimen(S)\n",
- "conductance(G3): 10.0 seimen(S)"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 7</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#How much energy is provided in this time?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 5; # in Volts\n",
- "I = 3; # in Ampere\n",
- "t = 600; # in sec\n",
- "#calculation:\n",
- "P = V*I\n",
- "E = P*t\n",
- "\n",
- "#results\n",
- "print \"Energy(E):\", E,\"Joule(J)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy(E): 9000 Joule(J)"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 8</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the power rating of the heater and the current taken from the supply.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "E = 18E5; # in Joule\n",
- "V = 250; # in Volts\n",
- "t = 1800; # in sec\n",
- "\n",
- "#calculation:\n",
- "P = E/t\n",
- "I = P/V\n",
- "\n",
- "#results\n",
- "print \"Power(P):\", P,\"Watt(W)\"\n",
- "print \"Current(I):\", I,\"Ampere(A)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power(P): 1000.0 Watt(W)\n",
- "Current(I): 4.0 Ampere(A)"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_2.ipynb deleted file mode 100755 index b33db331..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_2.ipynb +++ /dev/null @@ -1,329 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 1: Units Associated with Basic Electrical Quantities</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 4</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the quantity of electricity transferred.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 5; # in Ampere\n",
- "t = 120; # in sec\n",
- "\n",
- "#calculation:\n",
- "Q = I*t\n",
- "\n",
- "#results\n",
- "print \"Charge, Q = \", Q,\"coulomb(C)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charge, Q = 600 coulomb(C)"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 5</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the force needed.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "M = 5; # in Kg\n",
- "a = 2; # in m/s2\n",
- "\n",
- "#calculation:\n",
- "F = M*a\n",
- "\n",
- "#results\n",
- "print \"Force:\", F,\"Newton(N)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force: 10 Newton(N)"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 5</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the force acting vertically downwards\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "M = 0.2; # in Kg\n",
- "g = 9.81; # in m/s2\n",
- "\n",
- "#calculation:\n",
- "F = M*g\n",
- "\n",
- "#results\n",
- "print \"Force:\", F,\"Newton(N)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force: 1.962 Newton(N)"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 6</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate Work Done and Average Power?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "F = 200; # in Newton\n",
- "d = 20; # in m\n",
- "t = 25; # in sec\n",
- "\n",
- "#calculation:\n",
- "W = F*d\n",
- "P = W/t\n",
- "\n",
- "#results\n",
- "print \"Power:\", P,\"watt(W)\"\n",
- "print \"Work Done:\", W,\"Nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power: 160.0 watt(W)\n",
- "Work Done: 4000 Nm"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 6</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What is (a) the work done and (b) the power developed?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "M = 1000; # in Kg\n",
- "h = 10; # in m\n",
- "t = 20; # in sec\n",
- "g = 9.81 # in m/s2\n",
- "\n",
- "#calculation:\n",
- "W = M*g*h\n",
- "P = W/t\n",
- "\n",
- "#results\n",
- "print \"Work Done:\", W,\"Joule(J)\"\n",
- "print \"Power:\", P,\"watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Work Done: 98100.0 Joule(J)\n",
- "Power: 4905.0 watt(W)"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 7</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the conductance of a conductor of resistances\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 10; # in ohm\n",
- "R2 = 5000; # in ohm\n",
- "R3 = 0.1; # in ohm\n",
- "#calculation:\n",
- "G1 = 1/R1\n",
- "G2 = 1/R2\n",
- "G3 = 1/R3\n",
- "\n",
- "#results\n",
- "print \"conductance(G1):\", G1,\"seimen(S)\"\n",
- "print \"conductance(G2):\", G2,\"seimen(S)\"\n",
- "print \"conductance(G3):\", G3,\"seimen(S)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "conductance(G1): 0.1 seimen(S)\n",
- "conductance(G2): 0.0002 seimen(S)\n",
- "conductance(G3): 10.0 seimen(S)"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 7</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#How much energy is provided in this time?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 5; # in Volts\n",
- "I = 3; # in Ampere\n",
- "t = 600; # in sec\n",
- "#calculation:\n",
- "P = V*I\n",
- "E = P*t\n",
- "\n",
- "#results\n",
- "print \"Energy(E):\", E,\"Joule(J)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy(E): 9000 Joule(J)"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 8</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the power rating of the heater and the current taken from the supply.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "E = 18E5; # in Joule\n",
- "V = 250; # in Volts\n",
- "t = 1800; # in sec\n",
- "\n",
- "#calculation:\n",
- "P = E/t\n",
- "I = P/V\n",
- "\n",
- "#results\n",
- "print \"Power(P):\", P,\"Watt(W)\"\n",
- "print \"Current(I):\", I,\"Ampere(A)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power(P): 1000.0 Watt(W)\n",
- "Current(I): 4.0 Ampere(A)"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint.ipynb deleted file mode 100755 index 3ab84705..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint.ipynb +++ /dev/null @@ -1,806 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 2: An Introduction to Electric Circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 12</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What current must flow if 0.24 coulombs is to be transferred in 15 ms?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "Q = 0.24; # in Coulomb\n",
- "t = 0.015; # in sec\n",
- "\n",
- "#calculation:\n",
- "I = Q/t\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 16.0 Ampere(A)\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 12</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the quantity of electricity transferred.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 10; # in Ampere\n",
- "t = 240; # in sec\n",
- "\n",
- "#calculation:\n",
- "Q = I*t\n",
- "\n",
- "#resuts\n",
- "print \"Charge(Q):\", Q,\"Coulomb(C)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charge(Q): 2400 Coulomb(C)\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 14</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the resistance\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 0.8; # in Ampere\n",
- "V = 20; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R = V/I\n",
- "\n",
- "#results\n",
- "print \"Resistance(R):\", R,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R): 25.0 Ohms"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 15/h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the p.d.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 0.010; # in Ampere\n",
- "R = 2000; # in ohms\n",
- "\n",
- "#calculation:\n",
- "V = I*R\n",
- "\n",
- "#results\n",
- "print \"p.d.(V):\", V,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "p.d.(V): 20.0 Volts(V)"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 15/h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What is the resistance of the coil?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 0.050; # in Ampere\n",
- "V = 12; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R = V/I\n",
- "\n",
- "#results\n",
- "print \"Resistance(R):\", R,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R): 240.0 Ohms"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 15/h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#what will be the new value of the current flowing?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I1 = 0.005; # in Ampere\n",
- "V1 = 100; # in Volts\n",
- "V2 = 25; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R = V1/I1\n",
- "I2 = V2/R\n",
- "\n",
- "#results\n",
- "print \"Resistance(R):\", R,\"Ohms\"\n",
- "print \"Current(I):\", I2,\"Ampere(A)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R): 20000.0 Ohms\n",
- "Current(I): 0.00125 Ampere(A)"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 15/h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What is the resistance of a coil\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I1 = 0.050; # in Ampere\n",
- "I2 = 200E-6; # in Ampere\n",
- "V = 120; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R1 = V/I1\n",
- "R2 = V/I2\n",
- "\n",
- "#results\n",
- "print \"Resistance(R1):\", R1,\"Ohms\"\n",
- "print \"Resistance(R2):\", R2,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R1): 2400.0 Ohms\n",
- "Resistance(R2): 600000.0 Ohms"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 16</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current flowing in the bulb, and (b) the resistance of the bulb.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "P = 100; # in Watt\n",
- "V = 250; # in Volts\n",
- "\n",
- "#calculation:\n",
- "I = P/V\n",
- "R = V/I\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\"\n",
- "print \"Resistance(R):\", R,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 0.4 Ampere(A)\n",
- "Resistance(R): 625.0 Ohms"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 17</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the power dissipated\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 0.004; # in ampere\n",
- "R = 5000; # in ohms\n",
- "\n",
- "#calculation:\n",
- "P = I*I*R\n",
- "\n",
- "#results\n",
- "print \"Power(P):\", P,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power(P): 0.08 Watt(W)"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 17</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What current will flow when it is connected to a 240 V supply?\n",
- "#Find also the power rating of the kettle.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 240; # in Volts\n",
- "R = 30; # in ohms\n",
- "\n",
- "#calculation:\n",
- "I = V/R\n",
- "P = V*I\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\"\n",
- "print \"Power(P):\", P,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 8.0 Ampere(A)\n",
- "Power(P): 1920.0 Watt(W)"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 17</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the p.d. across the winding, and (b) the power dissipated by the coil.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 5; # in ampere\n",
- "R = 100; # in ohms\n",
- "\n",
- "#calculation:\n",
- "V = I*R\n",
- "P = I*R*I\n",
- "\n",
- "#results\n",
- "print \"p.d(V):\", V,\"Volts(V)\"\n",
- "print \"Power(P):\", P,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "p.d(V): 500 Volts(V)\n",
- "Power(P): 2500 Watt(W)"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 17</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the resistance of each resistor.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I1 = 0.020; # in ampere\n",
- "V1 = 20; # in Volts\n",
- "I2 = 0.005; # in ampere\n",
- "V2 = 16; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R1 = V1/I1\n",
- "R2 = V2/I2\n",
- "\n",
- "#results\n",
- "print \"Resistance(R1):\", R1,\"Ohms\"\n",
- "print \"Resistance(R2):\", R2,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R1): 1000.0 Ohms\n",
- "Resistance(R2): 3200.0 Ohms"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 18</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the current taken by the lamp and its power rating\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 240; # in Volts\n",
- "R = 960; # in ohms\n",
- "\n",
- "#calculation:\n",
- "I = V/R\n",
- "P = I*V\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\"\n",
- "print \"Power(P):\", P,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 0.25 Ampere(A)\n",
- "Power(P): 60.0 Watt(W)"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 18</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the load, the power consumed and the energy dissipated in 2 minutes\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 12; # in Volts\n",
- "R = 40; # in ohms\n",
- "t = 120; # in sec\n",
- "\n",
- "#calculation:\n",
- "I = V/R\n",
- "P = I*V\n",
- "E = P*t\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\"\n",
- "print \"Power(P):\", P,\"Watt(W)\"\n",
- "print \"Energy(E):\", E,\"Joule(J)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 0.3 Ampere(A)\n",
- "Power(P): 3.6 Watt(W)\n",
- "Energy(E): 432.0 Joule(J)"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 18</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#How much energy is provided in this time?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 15; # in Volts\n",
- "I = 2; # in ampere\n",
- "t = 360; # in sec\n",
- "\n",
- "#calculation:\n",
- "E = V*I*t\n",
- "\n",
- "#results\n",
- "print \"Energy(E):\", E,\"Joule(J)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy(E): 10800 Joule(J)"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 18</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Estimate the cost per week of electricity\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 240; # in Volts\n",
- "I = 13; # in ampere\n",
- "t = 30; # in hours\n",
- "p = 7; # in paise per kWh\n",
- "\n",
- "#calculation:\n",
- "P = V*I\n",
- "E = P*t/1000 # in kWh\n",
- "C = E*p\n",
- "\n",
- "#results\n",
- "print \"Cost per week:\", C,\" Paise(p)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cost per week: 655.2 Paise(p)"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 19</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the power rating of the heater and the current taken from the supply.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 250; # in Volts\n",
- "E = 3.6E6; # energy in J\n",
- "t = 2400; # in sec\n",
- "\n",
- "#calculation:\n",
- "P = E/t\n",
- "I = P/V\n",
- "\n",
- "#results\n",
- "print \"Power(P):\", P,\"Watt(W)\"\n",
- "print \"Current(I):\", I,\"Ampere(A)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power(P): 1500.0 Watt(W)\n",
- "Current(I): 6.0 Ampere(A)"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 19</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the power dissipated\n",
- "#determine the energy used and the cost\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 20; # in ohms\n",
- "I = 10; # in ampere\n",
- "t = 6; # in hours\n",
- "p = 7; # in paise per kWh\n",
- "\n",
- "#calculation:\n",
- "P = I*I*R\n",
- "E = P*t/1000 # in kWh\n",
- "C = E*p\n",
- "\n",
- "#results\n",
- "print \"Power(P):\", P,\"Watt(W)\"\n",
- "print \"Cost per week:\", C,\"Paise(p)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power(P): 2000 Watt(W)\n",
- "Cost per week: 84.0 Paise(p)"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 19</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the weekly cost of electricity to the business.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "P1 = 3; # in kW\n",
- "P2 = 150; # in Watt\n",
- "n1 = 2; # no. of P1 Equips\n",
- "n2 = 6; # no. of P2 Equips\n",
- "t1 = 20; # in hours each per week\n",
- "t2 = 30; # in hours each per week\n",
- "p = 7; # in paise per kWh\n",
- "\n",
- "#calculation:\n",
- "E1 = P1*t1*n1 # in kWh by two P1 eqips\n",
- "E2 = P2*t2*n2/1000 # in kWh by six P2 eqips\n",
- "Et = E1 + E2\n",
- "C = Et * 7\n",
- "\n",
- "#results\n",
- "print \"Cost per week:\", C,\"Paise(p)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cost per week: 1029.0 Paise(p)"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 20</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#state which is most appropriate for the following appliances which are both connected to a 240 V supply \n",
- "#(a) Electric toaster having a power rating of 1 kW (b) Electric fire having a power rating of 3 kW\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I1 = 5; # in Amp\n",
- "I2 = 10; # in Amp\n",
- "I3 = 13; # in Amp\n",
- "P1 = 1000; # in Watts\n",
- "P2 = 3000; # in Watts\n",
- "V = 240; #in Volts\n",
- "\n",
- "#calculation:\n",
- "It = P1/V\n",
- "If = P2/V\n",
- "\n",
- "#results\n",
- "print \"For the toaster,\", I1,\"A fuse is most appropriate\"\n",
- "print \"For the Fire,\", I3,\"A fuse is most appropriate\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For the toaster, 5 A fuse is most appropriate\n",
- "For the Fire, 13 A fuse is most appropriate"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_1.ipynb deleted file mode 100755 index 3ab84705..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_1.ipynb +++ /dev/null @@ -1,806 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 2: An Introduction to Electric Circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 12</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What current must flow if 0.24 coulombs is to be transferred in 15 ms?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "Q = 0.24; # in Coulomb\n",
- "t = 0.015; # in sec\n",
- "\n",
- "#calculation:\n",
- "I = Q/t\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 16.0 Ampere(A)\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 12</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the quantity of electricity transferred.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 10; # in Ampere\n",
- "t = 240; # in sec\n",
- "\n",
- "#calculation:\n",
- "Q = I*t\n",
- "\n",
- "#resuts\n",
- "print \"Charge(Q):\", Q,\"Coulomb(C)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charge(Q): 2400 Coulomb(C)\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 14</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the resistance\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 0.8; # in Ampere\n",
- "V = 20; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R = V/I\n",
- "\n",
- "#results\n",
- "print \"Resistance(R):\", R,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R): 25.0 Ohms"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 15/h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the p.d.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 0.010; # in Ampere\n",
- "R = 2000; # in ohms\n",
- "\n",
- "#calculation:\n",
- "V = I*R\n",
- "\n",
- "#results\n",
- "print \"p.d.(V):\", V,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "p.d.(V): 20.0 Volts(V)"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 15/h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What is the resistance of the coil?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 0.050; # in Ampere\n",
- "V = 12; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R = V/I\n",
- "\n",
- "#results\n",
- "print \"Resistance(R):\", R,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R): 240.0 Ohms"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 15/h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#what will be the new value of the current flowing?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I1 = 0.005; # in Ampere\n",
- "V1 = 100; # in Volts\n",
- "V2 = 25; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R = V1/I1\n",
- "I2 = V2/R\n",
- "\n",
- "#results\n",
- "print \"Resistance(R):\", R,\"Ohms\"\n",
- "print \"Current(I):\", I2,\"Ampere(A)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R): 20000.0 Ohms\n",
- "Current(I): 0.00125 Ampere(A)"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 15/h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What is the resistance of a coil\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I1 = 0.050; # in Ampere\n",
- "I2 = 200E-6; # in Ampere\n",
- "V = 120; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R1 = V/I1\n",
- "R2 = V/I2\n",
- "\n",
- "#results\n",
- "print \"Resistance(R1):\", R1,\"Ohms\"\n",
- "print \"Resistance(R2):\", R2,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R1): 2400.0 Ohms\n",
- "Resistance(R2): 600000.0 Ohms"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 16</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current flowing in the bulb, and (b) the resistance of the bulb.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "P = 100; # in Watt\n",
- "V = 250; # in Volts\n",
- "\n",
- "#calculation:\n",
- "I = P/V\n",
- "R = V/I\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\"\n",
- "print \"Resistance(R):\", R,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 0.4 Ampere(A)\n",
- "Resistance(R): 625.0 Ohms"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 17</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the power dissipated\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 0.004; # in ampere\n",
- "R = 5000; # in ohms\n",
- "\n",
- "#calculation:\n",
- "P = I*I*R\n",
- "\n",
- "#results\n",
- "print \"Power(P):\", P,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power(P): 0.08 Watt(W)"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 17</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What current will flow when it is connected to a 240 V supply?\n",
- "#Find also the power rating of the kettle.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 240; # in Volts\n",
- "R = 30; # in ohms\n",
- "\n",
- "#calculation:\n",
- "I = V/R\n",
- "P = V*I\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\"\n",
- "print \"Power(P):\", P,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 8.0 Ampere(A)\n",
- "Power(P): 1920.0 Watt(W)"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 17</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the p.d. across the winding, and (b) the power dissipated by the coil.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 5; # in ampere\n",
- "R = 100; # in ohms\n",
- "\n",
- "#calculation:\n",
- "V = I*R\n",
- "P = I*R*I\n",
- "\n",
- "#results\n",
- "print \"p.d(V):\", V,\"Volts(V)\"\n",
- "print \"Power(P):\", P,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "p.d(V): 500 Volts(V)\n",
- "Power(P): 2500 Watt(W)"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 17</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the resistance of each resistor.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I1 = 0.020; # in ampere\n",
- "V1 = 20; # in Volts\n",
- "I2 = 0.005; # in ampere\n",
- "V2 = 16; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R1 = V1/I1\n",
- "R2 = V2/I2\n",
- "\n",
- "#results\n",
- "print \"Resistance(R1):\", R1,\"Ohms\"\n",
- "print \"Resistance(R2):\", R2,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R1): 1000.0 Ohms\n",
- "Resistance(R2): 3200.0 Ohms"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 18</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the current taken by the lamp and its power rating\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 240; # in Volts\n",
- "R = 960; # in ohms\n",
- "\n",
- "#calculation:\n",
- "I = V/R\n",
- "P = I*V\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\"\n",
- "print \"Power(P):\", P,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 0.25 Ampere(A)\n",
- "Power(P): 60.0 Watt(W)"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 18</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the load, the power consumed and the energy dissipated in 2 minutes\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 12; # in Volts\n",
- "R = 40; # in ohms\n",
- "t = 120; # in sec\n",
- "\n",
- "#calculation:\n",
- "I = V/R\n",
- "P = I*V\n",
- "E = P*t\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\"\n",
- "print \"Power(P):\", P,\"Watt(W)\"\n",
- "print \"Energy(E):\", E,\"Joule(J)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 0.3 Ampere(A)\n",
- "Power(P): 3.6 Watt(W)\n",
- "Energy(E): 432.0 Joule(J)"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 18</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#How much energy is provided in this time?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 15; # in Volts\n",
- "I = 2; # in ampere\n",
- "t = 360; # in sec\n",
- "\n",
- "#calculation:\n",
- "E = V*I*t\n",
- "\n",
- "#results\n",
- "print \"Energy(E):\", E,\"Joule(J)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy(E): 10800 Joule(J)"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 18</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Estimate the cost per week of electricity\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 240; # in Volts\n",
- "I = 13; # in ampere\n",
- "t = 30; # in hours\n",
- "p = 7; # in paise per kWh\n",
- "\n",
- "#calculation:\n",
- "P = V*I\n",
- "E = P*t/1000 # in kWh\n",
- "C = E*p\n",
- "\n",
- "#results\n",
- "print \"Cost per week:\", C,\" Paise(p)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cost per week: 655.2 Paise(p)"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 19</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the power rating of the heater and the current taken from the supply.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 250; # in Volts\n",
- "E = 3.6E6; # energy in J\n",
- "t = 2400; # in sec\n",
- "\n",
- "#calculation:\n",
- "P = E/t\n",
- "I = P/V\n",
- "\n",
- "#results\n",
- "print \"Power(P):\", P,\"Watt(W)\"\n",
- "print \"Current(I):\", I,\"Ampere(A)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power(P): 1500.0 Watt(W)\n",
- "Current(I): 6.0 Ampere(A)"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 19</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the power dissipated\n",
- "#determine the energy used and the cost\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 20; # in ohms\n",
- "I = 10; # in ampere\n",
- "t = 6; # in hours\n",
- "p = 7; # in paise per kWh\n",
- "\n",
- "#calculation:\n",
- "P = I*I*R\n",
- "E = P*t/1000 # in kWh\n",
- "C = E*p\n",
- "\n",
- "#results\n",
- "print \"Power(P):\", P,\"Watt(W)\"\n",
- "print \"Cost per week:\", C,\"Paise(p)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power(P): 2000 Watt(W)\n",
- "Cost per week: 84.0 Paise(p)"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 19</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the weekly cost of electricity to the business.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "P1 = 3; # in kW\n",
- "P2 = 150; # in Watt\n",
- "n1 = 2; # no. of P1 Equips\n",
- "n2 = 6; # no. of P2 Equips\n",
- "t1 = 20; # in hours each per week\n",
- "t2 = 30; # in hours each per week\n",
- "p = 7; # in paise per kWh\n",
- "\n",
- "#calculation:\n",
- "E1 = P1*t1*n1 # in kWh by two P1 eqips\n",
- "E2 = P2*t2*n2/1000 # in kWh by six P2 eqips\n",
- "Et = E1 + E2\n",
- "C = Et * 7\n",
- "\n",
- "#results\n",
- "print \"Cost per week:\", C,\"Paise(p)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cost per week: 1029.0 Paise(p)"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 20</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#state which is most appropriate for the following appliances which are both connected to a 240 V supply \n",
- "#(a) Electric toaster having a power rating of 1 kW (b) Electric fire having a power rating of 3 kW\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I1 = 5; # in Amp\n",
- "I2 = 10; # in Amp\n",
- "I3 = 13; # in Amp\n",
- "P1 = 1000; # in Watts\n",
- "P2 = 3000; # in Watts\n",
- "V = 240; #in Volts\n",
- "\n",
- "#calculation:\n",
- "It = P1/V\n",
- "If = P2/V\n",
- "\n",
- "#results\n",
- "print \"For the toaster,\", I1,\"A fuse is most appropriate\"\n",
- "print \"For the Fire,\", I3,\"A fuse is most appropriate\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For the toaster, 5 A fuse is most appropriate\n",
- "For the Fire, 13 A fuse is most appropriate"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_2.ipynb deleted file mode 100755 index 3ab84705..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_2.ipynb +++ /dev/null @@ -1,806 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 2: An Introduction to Electric Circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 12</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What current must flow if 0.24 coulombs is to be transferred in 15 ms?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "Q = 0.24; # in Coulomb\n",
- "t = 0.015; # in sec\n",
- "\n",
- "#calculation:\n",
- "I = Q/t\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 16.0 Ampere(A)\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 12</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the quantity of electricity transferred.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 10; # in Ampere\n",
- "t = 240; # in sec\n",
- "\n",
- "#calculation:\n",
- "Q = I*t\n",
- "\n",
- "#resuts\n",
- "print \"Charge(Q):\", Q,\"Coulomb(C)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charge(Q): 2400 Coulomb(C)\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 14</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the resistance\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 0.8; # in Ampere\n",
- "V = 20; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R = V/I\n",
- "\n",
- "#results\n",
- "print \"Resistance(R):\", R,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R): 25.0 Ohms"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 15/h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the p.d.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 0.010; # in Ampere\n",
- "R = 2000; # in ohms\n",
- "\n",
- "#calculation:\n",
- "V = I*R\n",
- "\n",
- "#results\n",
- "print \"p.d.(V):\", V,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "p.d.(V): 20.0 Volts(V)"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 15/h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What is the resistance of the coil?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 0.050; # in Ampere\n",
- "V = 12; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R = V/I\n",
- "\n",
- "#results\n",
- "print \"Resistance(R):\", R,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R): 240.0 Ohms"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 15/h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#what will be the new value of the current flowing?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I1 = 0.005; # in Ampere\n",
- "V1 = 100; # in Volts\n",
- "V2 = 25; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R = V1/I1\n",
- "I2 = V2/R\n",
- "\n",
- "#results\n",
- "print \"Resistance(R):\", R,\"Ohms\"\n",
- "print \"Current(I):\", I2,\"Ampere(A)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R): 20000.0 Ohms\n",
- "Current(I): 0.00125 Ampere(A)"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 15/h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What is the resistance of a coil\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I1 = 0.050; # in Ampere\n",
- "I2 = 200E-6; # in Ampere\n",
- "V = 120; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R1 = V/I1\n",
- "R2 = V/I2\n",
- "\n",
- "#results\n",
- "print \"Resistance(R1):\", R1,\"Ohms\"\n",
- "print \"Resistance(R2):\", R2,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R1): 2400.0 Ohms\n",
- "Resistance(R2): 600000.0 Ohms"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 16</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current flowing in the bulb, and (b) the resistance of the bulb.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "P = 100; # in Watt\n",
- "V = 250; # in Volts\n",
- "\n",
- "#calculation:\n",
- "I = P/V\n",
- "R = V/I\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\"\n",
- "print \"Resistance(R):\", R,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 0.4 Ampere(A)\n",
- "Resistance(R): 625.0 Ohms"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 17</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the power dissipated\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 0.004; # in ampere\n",
- "R = 5000; # in ohms\n",
- "\n",
- "#calculation:\n",
- "P = I*I*R\n",
- "\n",
- "#results\n",
- "print \"Power(P):\", P,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power(P): 0.08 Watt(W)"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 17</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What current will flow when it is connected to a 240 V supply?\n",
- "#Find also the power rating of the kettle.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 240; # in Volts\n",
- "R = 30; # in ohms\n",
- "\n",
- "#calculation:\n",
- "I = V/R\n",
- "P = V*I\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\"\n",
- "print \"Power(P):\", P,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 8.0 Ampere(A)\n",
- "Power(P): 1920.0 Watt(W)"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 17</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the p.d. across the winding, and (b) the power dissipated by the coil.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I = 5; # in ampere\n",
- "R = 100; # in ohms\n",
- "\n",
- "#calculation:\n",
- "V = I*R\n",
- "P = I*R*I\n",
- "\n",
- "#results\n",
- "print \"p.d(V):\", V,\"Volts(V)\"\n",
- "print \"Power(P):\", P,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "p.d(V): 500 Volts(V)\n",
- "Power(P): 2500 Watt(W)"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 17</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the resistance of each resistor.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I1 = 0.020; # in ampere\n",
- "V1 = 20; # in Volts\n",
- "I2 = 0.005; # in ampere\n",
- "V2 = 16; # in Volts\n",
- "\n",
- "#calculation:\n",
- "R1 = V1/I1\n",
- "R2 = V2/I2\n",
- "\n",
- "#results\n",
- "print \"Resistance(R1):\", R1,\"Ohms\"\n",
- "print \"Resistance(R2):\", R2,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance(R1): 1000.0 Ohms\n",
- "Resistance(R2): 3200.0 Ohms"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 18</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the current taken by the lamp and its power rating\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 240; # in Volts\n",
- "R = 960; # in ohms\n",
- "\n",
- "#calculation:\n",
- "I = V/R\n",
- "P = I*V\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\"\n",
- "print \"Power(P):\", P,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 0.25 Ampere(A)\n",
- "Power(P): 60.0 Watt(W)"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 18</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the load, the power consumed and the energy dissipated in 2 minutes\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 12; # in Volts\n",
- "R = 40; # in ohms\n",
- "t = 120; # in sec\n",
- "\n",
- "#calculation:\n",
- "I = V/R\n",
- "P = I*V\n",
- "E = P*t\n",
- "\n",
- "#results\n",
- "print \"Current(I):\", I,\"Ampere(A)\"\n",
- "print \"Power(P):\", P,\"Watt(W)\"\n",
- "print \"Energy(E):\", E,\"Joule(J)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current(I): 0.3 Ampere(A)\n",
- "Power(P): 3.6 Watt(W)\n",
- "Energy(E): 432.0 Joule(J)"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 18</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#How much energy is provided in this time?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 15; # in Volts\n",
- "I = 2; # in ampere\n",
- "t = 360; # in sec\n",
- "\n",
- "#calculation:\n",
- "E = V*I*t\n",
- "\n",
- "#results\n",
- "print \"Energy(E):\", E,\"Joule(J)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy(E): 10800 Joule(J)"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 18</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Estimate the cost per week of electricity\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 240; # in Volts\n",
- "I = 13; # in ampere\n",
- "t = 30; # in hours\n",
- "p = 7; # in paise per kWh\n",
- "\n",
- "#calculation:\n",
- "P = V*I\n",
- "E = P*t/1000 # in kWh\n",
- "C = E*p\n",
- "\n",
- "#results\n",
- "print \"Cost per week:\", C,\" Paise(p)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cost per week: 655.2 Paise(p)"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 19</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the power rating of the heater and the current taken from the supply.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V = 250; # in Volts\n",
- "E = 3.6E6; # energy in J\n",
- "t = 2400; # in sec\n",
- "\n",
- "#calculation:\n",
- "P = E/t\n",
- "I = P/V\n",
- "\n",
- "#results\n",
- "print \"Power(P):\", P,\"Watt(W)\"\n",
- "print \"Current(I):\", I,\"Ampere(A)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power(P): 1500.0 Watt(W)\n",
- "Current(I): 6.0 Ampere(A)"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 19</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the power dissipated\n",
- "#determine the energy used and the cost\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 20; # in ohms\n",
- "I = 10; # in ampere\n",
- "t = 6; # in hours\n",
- "p = 7; # in paise per kWh\n",
- "\n",
- "#calculation:\n",
- "P = I*I*R\n",
- "E = P*t/1000 # in kWh\n",
- "C = E*p\n",
- "\n",
- "#results\n",
- "print \"Power(P):\", P,\"Watt(W)\"\n",
- "print \"Cost per week:\", C,\"Paise(p)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power(P): 2000 Watt(W)\n",
- "Cost per week: 84.0 Paise(p)"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 19</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the weekly cost of electricity to the business.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "P1 = 3; # in kW\n",
- "P2 = 150; # in Watt\n",
- "n1 = 2; # no. of P1 Equips\n",
- "n2 = 6; # no. of P2 Equips\n",
- "t1 = 20; # in hours each per week\n",
- "t2 = 30; # in hours each per week\n",
- "p = 7; # in paise per kWh\n",
- "\n",
- "#calculation:\n",
- "E1 = P1*t1*n1 # in kWh by two P1 eqips\n",
- "E2 = P2*t2*n2/1000 # in kWh by six P2 eqips\n",
- "Et = E1 + E2\n",
- "C = Et * 7\n",
- "\n",
- "#results\n",
- "print \"Cost per week:\", C,\"Paise(p)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cost per week: 1029.0 Paise(p)"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 20</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#state which is most appropriate for the following appliances which are both connected to a 240 V supply \n",
- "#(a) Electric toaster having a power rating of 1 kW (b) Electric fire having a power rating of 3 kW\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "I1 = 5; # in Amp\n",
- "I2 = 10; # in Amp\n",
- "I3 = 13; # in Amp\n",
- "P1 = 1000; # in Watts\n",
- "P2 = 3000; # in Watts\n",
- "V = 240; #in Volts\n",
- "\n",
- "#calculation:\n",
- "It = P1/V\n",
- "If = P2/V\n",
- "\n",
- "#results\n",
- "print \"For the toaster,\", I1,\"A fuse is most appropriate\"\n",
- "print \"For the Fire,\", I3,\"A fuse is most appropriate\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For the toaster, 5 A fuse is most appropriate\n",
- "For the Fire, 13 A fuse is most appropriate"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_07_1.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_07_1.ipynb deleted file mode 100755 index e3a59d72..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_07_1.ipynb +++ /dev/null @@ -1,23 +0,0 @@ -{
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- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_07_2.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_07_2.ipynb deleted file mode 100755 index e3a59d72..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_07_2.ipynb +++ /dev/null @@ -1,23 +0,0 @@ -{
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\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_07_3.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_07_3.ipynb deleted file mode 100755 index e3a59d72..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_07_3.ipynb +++ /dev/null @@ -1,23 +0,0 @@ -{
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\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_11_1.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_11_1.ipynb deleted file mode 100755 index 3454218e..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_11_1.ipynb +++ /dev/null @@ -1,23 +0,0 @@ -{
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- "Semiconductor diodes"
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- }
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- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_11_2.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_11_2.ipynb deleted file mode 100755 index 3454218e..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_11_2.ipynb +++ /dev/null @@ -1,23 +0,0 @@ -{
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\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_11_3.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_11_3.ipynb deleted file mode 100755 index 3454218e..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_11_3.ipynb +++ /dev/null @@ -1,23 +0,0 @@ -{
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- "metadata": {}
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- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_37_1.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_37_1.ipynb deleted file mode 100755 index de8c7108..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_37_1.ipynb +++ /dev/null @@ -1,23 +0,0 @@ -{
- "metadata": {
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- "source": [
- "A numerical method of harmonic analysis"
- ]
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_37_2.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_37_2.ipynb deleted file mode 100755 index de8c7108..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_37_2.ipynb +++ /dev/null @@ -1,23 +0,0 @@ -{
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-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_37_3.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_37_3.ipynb deleted file mode 100755 index de8c7108..00000000 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_37_3.ipynb +++ /dev/null @@ -1,23 +0,0 @@ -{
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- "source": [
- "A numerical method of harmonic analysis"
- ]
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/README.txt b/Electrical_Circuit_Theory_And_Technology/README.txt index e94e265f..e94e265f 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/README.txt +++ b/Electrical_Circuit_Theory_And_Technology/README.txt diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_01.ipynb index 52fbd3f7..52fbd3f7 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_01.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_02.ipynb index 8bc22a1d..8bc22a1d 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_02.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint.ipynb deleted file mode 100755 index f79d1396..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint.ipynb +++ /dev/null @@ -1,525 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 3: Resistance variation</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 24</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resistance of an 8 m length of the same wire,\n",
- "#and (b) the length of the same wire when the resistance is 420\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 600; # in ohms\n",
- "L = 5; # in meter\n",
- "L1 = 8; # in meter\n",
- "R2 = 420; # in ohms\n",
- "\n",
- "#calculation:\n",
- "R1 = R*L1/L\n",
- "L2 = R2*L/R\n",
- "\n",
- "#results\n",
- "print \"a)Resistance\", R1,\"Ohms\"\n",
- "print \"b)Length:\", L2,\"meters(m)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a)Resistance 960.0 Ohms\n",
- "b)Length: 3.5 meters(m)"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 24</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the resistance of a wire of the same length and material if the cross-sectional area is 5 mm2,\n",
- "#(b) the cross-sectional area of a wire of the same length and material of resistance 750 \n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 300; # in ohms\n",
- "A = 2; # in mm2\n",
- "A1 = 5; # in mm2\n",
- "R2 = 750; # in ohms\n",
- "\n",
- "#calculation:\n",
- "R1 = R*A/A1\n",
- "A2 = R*A/R2\n",
- "\n",
- "#results\n",
- "print \"(a)Resistance\", R1,\"Ohms\"\n",
- "print \"(b)C.S.A:\", A2,\"mm^2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Resistance 120.0 Ohms\n",
- "(b)C.S.A: 0.8 mm^2"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 25</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the resistance of the wire.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 0.16; # in ohms\n",
- "A = 3; # in mm2\n",
- "L = 8; # in m\n",
- "A1 = 1; # in mm2\n",
- "\n",
- "#calculation:\n",
- "L1 = L*3\n",
- "R1 = R*A*L1/(A1*L)\n",
- "\n",
- "#results\n",
- "print \"Resistance\", R1,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 1.44 Ohms"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 25</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the resistance.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "A = 100E-6; # in m2\n",
- "L = 2000; # in m\n",
- "p = 0.03E-6; # in ohm m\n",
- "\n",
- "#calculation:\n",
- "R = p*L/A\n",
- "\n",
- "#results\n",
- "print \"Resistance\", R,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 0.6 Ohms"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 25</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the cross-sectional area\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 0.25; # in ohms\n",
- "L = 40; # in m\n",
- "p = 0.02E-6; # in ohm m\n",
- "\n",
- "#calculation:\n",
- "A = p*L*1E6/R\n",
- "\n",
- "#results\n",
- "print \"C.S.A \", A,\"mm^2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "C.S.A 3.2 mm^2"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 25</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistivity of the wire.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 150; # in ohms\n",
- "L = 1500; # in m\n",
- "A = 0.17E-6; # in m2\n",
- "\n",
- "#calculation:\n",
- "p = R*A*1E6/L\n",
- "\n",
- "#results\n",
- "print \"resistivity\", p,\"uOhm.m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistivity 0.017 uOhm.m"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 26</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "d = 0.012; # in m\n",
- "L = 1200; # in m\n",
- "p = 1.7E-8; # in ohm m\n",
- "\n",
- "#calculation:\n",
- "A = math.pi*d*d/4\n",
- "R = p*L/A\n",
- "\n",
- "#results\n",
- "print \"resistance\", round(R,3),\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance 0.18 Ohm"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 27</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its resistance\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R0 = 100; # in ohms\n",
- "T0 = 0; # in \u00b0C\n",
- "T1 = 70; # in \u00b0C\n",
- "a0 = 0.0043; # in per\u00b0C\n",
- "\n",
- "#calculation:\n",
- "R70 = R0*(1 + (a0*T1))\n",
- "\n",
- "#results\n",
- "print \"Resistance\", R70,\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 130.1 Ohm"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 27</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its resistance\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 27; # in ohms\n",
- "T0 = 0; # in \u00b0C\n",
- "T1 = 35; # in \u00b0C\n",
- "a0 = 0.0038; # in per\u00b0C\n",
- "\n",
- "#calculation:\n",
- "R0 = R1/(1 + (a0*T1))\n",
- "\n",
- "#results\n",
- "print \"resistance\", round(R0,2),\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance 23.83 Ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 27</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its resistance\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R0 = 1000; # in ohms\n",
- "T0 = 0; # in \u00b0C\n",
- "T1 = 80; # in \u00b0C\n",
- "a0 = -0.0005; # in per\u00b0C\n",
- "\n",
- "#calculation:\n",
- "R80 = R0*(1 + (a0*T1))\n",
- "\n",
- "#results\n",
- "print \"resistance\", R80,\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance 960.0 Ohm"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 28</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the resistance of the coil\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R20 = 10; # in ohms\n",
- "T0 = 20; # in \u00b0C\n",
- "T1 = 100; # in \u00b0C\n",
- "a20 = 0.004; # in per\u00b0C\n",
- "\n",
- "\n",
- "#calculation:\n",
- "R100 = R20*(1 + (a20)*(T1 - T0))\n",
- "\n",
- "#results\n",
- "print \"resistance\", R100,\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance 13.2 Ohm"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 28</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the temperature to which the coil has risen\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R18 = 200; # in ohms\n",
- "R1 = 240; # in ohms\n",
- "T0 = 18; # in \u00b0C\n",
- "a18 = 0.0039; # in per\u00b0C\n",
- "\n",
- "\n",
- "#calculation:\n",
- "T1 = (((R1/R18)-1)/a18) + T0\n",
- "\n",
- "#results\n",
- "print \"Temperature\", round(T1,2),\"degC\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Temperature 69.28 degC"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 29</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance of the wire\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R20 = 200; # in ohms\n",
- "T0 = 20; # in \u00b0C\n",
- "T1 = 90; # in \u00b0C\n",
- "a0 = 0.004; # in per\u00b0C\n",
- "\n",
- "\n",
- "#calculation:\n",
- "R90 = R20*(1 + (a0*T1))/(1 + (a0*T0))\n",
- "\n",
- "#results\n",
- "print \"Resistance\", round(R90,0),\"ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 252.0 ohms"
- ]
- }
- ],
- "prompt_number": 7
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_1.ipynb deleted file mode 100755 index f79d1396..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_1.ipynb +++ /dev/null @@ -1,525 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 3: Resistance variation</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 24</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resistance of an 8 m length of the same wire,\n",
- "#and (b) the length of the same wire when the resistance is 420\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 600; # in ohms\n",
- "L = 5; # in meter\n",
- "L1 = 8; # in meter\n",
- "R2 = 420; # in ohms\n",
- "\n",
- "#calculation:\n",
- "R1 = R*L1/L\n",
- "L2 = R2*L/R\n",
- "\n",
- "#results\n",
- "print \"a)Resistance\", R1,\"Ohms\"\n",
- "print \"b)Length:\", L2,\"meters(m)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a)Resistance 960.0 Ohms\n",
- "b)Length: 3.5 meters(m)"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 24</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the resistance of a wire of the same length and material if the cross-sectional area is 5 mm2,\n",
- "#(b) the cross-sectional area of a wire of the same length and material of resistance 750 \n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 300; # in ohms\n",
- "A = 2; # in mm2\n",
- "A1 = 5; # in mm2\n",
- "R2 = 750; # in ohms\n",
- "\n",
- "#calculation:\n",
- "R1 = R*A/A1\n",
- "A2 = R*A/R2\n",
- "\n",
- "#results\n",
- "print \"(a)Resistance\", R1,\"Ohms\"\n",
- "print \"(b)C.S.A:\", A2,\"mm^2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Resistance 120.0 Ohms\n",
- "(b)C.S.A: 0.8 mm^2"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 25</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the resistance of the wire.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 0.16; # in ohms\n",
- "A = 3; # in mm2\n",
- "L = 8; # in m\n",
- "A1 = 1; # in mm2\n",
- "\n",
- "#calculation:\n",
- "L1 = L*3\n",
- "R1 = R*A*L1/(A1*L)\n",
- "\n",
- "#results\n",
- "print \"Resistance\", R1,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 1.44 Ohms"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 25</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the resistance.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "A = 100E-6; # in m2\n",
- "L = 2000; # in m\n",
- "p = 0.03E-6; # in ohm m\n",
- "\n",
- "#calculation:\n",
- "R = p*L/A\n",
- "\n",
- "#results\n",
- "print \"Resistance\", R,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 0.6 Ohms"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 25</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the cross-sectional area\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 0.25; # in ohms\n",
- "L = 40; # in m\n",
- "p = 0.02E-6; # in ohm m\n",
- "\n",
- "#calculation:\n",
- "A = p*L*1E6/R\n",
- "\n",
- "#results\n",
- "print \"C.S.A \", A,\"mm^2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "C.S.A 3.2 mm^2"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 25</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistivity of the wire.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 150; # in ohms\n",
- "L = 1500; # in m\n",
- "A = 0.17E-6; # in m2\n",
- "\n",
- "#calculation:\n",
- "p = R*A*1E6/L\n",
- "\n",
- "#results\n",
- "print \"resistivity\", p,\"uOhm.m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistivity 0.017 uOhm.m"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 26</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "d = 0.012; # in m\n",
- "L = 1200; # in m\n",
- "p = 1.7E-8; # in ohm m\n",
- "\n",
- "#calculation:\n",
- "A = math.pi*d*d/4\n",
- "R = p*L/A\n",
- "\n",
- "#results\n",
- "print \"resistance\", round(R,3),\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance 0.18 Ohm"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 27</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its resistance\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R0 = 100; # in ohms\n",
- "T0 = 0; # in \u00b0C\n",
- "T1 = 70; # in \u00b0C\n",
- "a0 = 0.0043; # in per\u00b0C\n",
- "\n",
- "#calculation:\n",
- "R70 = R0*(1 + (a0*T1))\n",
- "\n",
- "#results\n",
- "print \"Resistance\", R70,\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 130.1 Ohm"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 27</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its resistance\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 27; # in ohms\n",
- "T0 = 0; # in \u00b0C\n",
- "T1 = 35; # in \u00b0C\n",
- "a0 = 0.0038; # in per\u00b0C\n",
- "\n",
- "#calculation:\n",
- "R0 = R1/(1 + (a0*T1))\n",
- "\n",
- "#results\n",
- "print \"resistance\", round(R0,2),\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance 23.83 Ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 27</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its resistance\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R0 = 1000; # in ohms\n",
- "T0 = 0; # in \u00b0C\n",
- "T1 = 80; # in \u00b0C\n",
- "a0 = -0.0005; # in per\u00b0C\n",
- "\n",
- "#calculation:\n",
- "R80 = R0*(1 + (a0*T1))\n",
- "\n",
- "#results\n",
- "print \"resistance\", R80,\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance 960.0 Ohm"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 28</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the resistance of the coil\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R20 = 10; # in ohms\n",
- "T0 = 20; # in \u00b0C\n",
- "T1 = 100; # in \u00b0C\n",
- "a20 = 0.004; # in per\u00b0C\n",
- "\n",
- "\n",
- "#calculation:\n",
- "R100 = R20*(1 + (a20)*(T1 - T0))\n",
- "\n",
- "#results\n",
- "print \"resistance\", R100,\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance 13.2 Ohm"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 28</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the temperature to which the coil has risen\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R18 = 200; # in ohms\n",
- "R1 = 240; # in ohms\n",
- "T0 = 18; # in \u00b0C\n",
- "a18 = 0.0039; # in per\u00b0C\n",
- "\n",
- "\n",
- "#calculation:\n",
- "T1 = (((R1/R18)-1)/a18) + T0\n",
- "\n",
- "#results\n",
- "print \"Temperature\", round(T1,2),\"degC\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Temperature 69.28 degC"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 29</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance of the wire\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R20 = 200; # in ohms\n",
- "T0 = 20; # in \u00b0C\n",
- "T1 = 90; # in \u00b0C\n",
- "a0 = 0.004; # in per\u00b0C\n",
- "\n",
- "\n",
- "#calculation:\n",
- "R90 = R20*(1 + (a0*T1))/(1 + (a0*T0))\n",
- "\n",
- "#results\n",
- "print \"Resistance\", round(R90,0),\"ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 252.0 ohms"
- ]
- }
- ],
- "prompt_number": 7
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_2.ipynb deleted file mode 100755 index f79d1396..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_2.ipynb +++ /dev/null @@ -1,525 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 3: Resistance variation</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 24</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resistance of an 8 m length of the same wire,\n",
- "#and (b) the length of the same wire when the resistance is 420\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 600; # in ohms\n",
- "L = 5; # in meter\n",
- "L1 = 8; # in meter\n",
- "R2 = 420; # in ohms\n",
- "\n",
- "#calculation:\n",
- "R1 = R*L1/L\n",
- "L2 = R2*L/R\n",
- "\n",
- "#results\n",
- "print \"a)Resistance\", R1,\"Ohms\"\n",
- "print \"b)Length:\", L2,\"meters(m)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a)Resistance 960.0 Ohms\n",
- "b)Length: 3.5 meters(m)"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 24</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the resistance of a wire of the same length and material if the cross-sectional area is 5 mm2,\n",
- "#(b) the cross-sectional area of a wire of the same length and material of resistance 750 \n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 300; # in ohms\n",
- "A = 2; # in mm2\n",
- "A1 = 5; # in mm2\n",
- "R2 = 750; # in ohms\n",
- "\n",
- "#calculation:\n",
- "R1 = R*A/A1\n",
- "A2 = R*A/R2\n",
- "\n",
- "#results\n",
- "print \"(a)Resistance\", R1,\"Ohms\"\n",
- "print \"(b)C.S.A:\", A2,\"mm^2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Resistance 120.0 Ohms\n",
- "(b)C.S.A: 0.8 mm^2"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 25</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the resistance of the wire.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 0.16; # in ohms\n",
- "A = 3; # in mm2\n",
- "L = 8; # in m\n",
- "A1 = 1; # in mm2\n",
- "\n",
- "#calculation:\n",
- "L1 = L*3\n",
- "R1 = R*A*L1/(A1*L)\n",
- "\n",
- "#results\n",
- "print \"Resistance\", R1,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 1.44 Ohms"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 25</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the resistance.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "A = 100E-6; # in m2\n",
- "L = 2000; # in m\n",
- "p = 0.03E-6; # in ohm m\n",
- "\n",
- "#calculation:\n",
- "R = p*L/A\n",
- "\n",
- "#results\n",
- "print \"Resistance\", R,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 0.6 Ohms"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 25</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the cross-sectional area\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 0.25; # in ohms\n",
- "L = 40; # in m\n",
- "p = 0.02E-6; # in ohm m\n",
- "\n",
- "#calculation:\n",
- "A = p*L*1E6/R\n",
- "\n",
- "#results\n",
- "print \"C.S.A \", A,\"mm^2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "C.S.A 3.2 mm^2"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 25</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistivity of the wire.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 150; # in ohms\n",
- "L = 1500; # in m\n",
- "A = 0.17E-6; # in m2\n",
- "\n",
- "#calculation:\n",
- "p = R*A*1E6/L\n",
- "\n",
- "#results\n",
- "print \"resistivity\", p,\"uOhm.m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistivity 0.017 uOhm.m"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 26</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "d = 0.012; # in m\n",
- "L = 1200; # in m\n",
- "p = 1.7E-8; # in ohm m\n",
- "\n",
- "#calculation:\n",
- "A = math.pi*d*d/4\n",
- "R = p*L/A\n",
- "\n",
- "#results\n",
- "print \"resistance\", round(R,3),\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance 0.18 Ohm"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 27</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its resistance\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R0 = 100; # in ohms\n",
- "T0 = 0; # in \u00b0C\n",
- "T1 = 70; # in \u00b0C\n",
- "a0 = 0.0043; # in per\u00b0C\n",
- "\n",
- "#calculation:\n",
- "R70 = R0*(1 + (a0*T1))\n",
- "\n",
- "#results\n",
- "print \"Resistance\", R70,\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 130.1 Ohm"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 27</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its resistance\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 27; # in ohms\n",
- "T0 = 0; # in \u00b0C\n",
- "T1 = 35; # in \u00b0C\n",
- "a0 = 0.0038; # in per\u00b0C\n",
- "\n",
- "#calculation:\n",
- "R0 = R1/(1 + (a0*T1))\n",
- "\n",
- "#results\n",
- "print \"resistance\", round(R0,2),\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance 23.83 Ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 27</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its resistance\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R0 = 1000; # in ohms\n",
- "T0 = 0; # in \u00b0C\n",
- "T1 = 80; # in \u00b0C\n",
- "a0 = -0.0005; # in per\u00b0C\n",
- "\n",
- "#calculation:\n",
- "R80 = R0*(1 + (a0*T1))\n",
- "\n",
- "#results\n",
- "print \"resistance\", R80,\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance 960.0 Ohm"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 28</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the resistance of the coil\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R20 = 10; # in ohms\n",
- "T0 = 20; # in \u00b0C\n",
- "T1 = 100; # in \u00b0C\n",
- "a20 = 0.004; # in per\u00b0C\n",
- "\n",
- "\n",
- "#calculation:\n",
- "R100 = R20*(1 + (a20)*(T1 - T0))\n",
- "\n",
- "#results\n",
- "print \"resistance\", R100,\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance 13.2 Ohm"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 28</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the temperature to which the coil has risen\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R18 = 200; # in ohms\n",
- "R1 = 240; # in ohms\n",
- "T0 = 18; # in \u00b0C\n",
- "a18 = 0.0039; # in per\u00b0C\n",
- "\n",
- "\n",
- "#calculation:\n",
- "T1 = (((R1/R18)-1)/a18) + T0\n",
- "\n",
- "#results\n",
- "print \"Temperature\", round(T1,2),\"degC\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Temperature 69.28 degC"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 29</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance of the wire\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R20 = 200; # in ohms\n",
- "T0 = 20; # in \u00b0C\n",
- "T1 = 90; # in \u00b0C\n",
- "a0 = 0.004; # in per\u00b0C\n",
- "\n",
- "\n",
- "#calculation:\n",
- "R90 = R20*(1 + (a0*T1))/(1 + (a0*T0))\n",
- "\n",
- "#results\n",
- "print \"Resistance\", round(R90,0),\"ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 252.0 ohms"
- ]
- }
- ],
- "prompt_number": 7
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_03.ipynb index d9b6cd03..d9b6cd03 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_03.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint.ipynb deleted file mode 100755 index 0c38cd30..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint.ipynb +++ /dev/null @@ -1,185 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 4: Chemical effects of\n",
- "electricity</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 35</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. and the internal resistance of the batteries formed by (a)Series arrangement and (b)Parallel Arrangement.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 0.2; # in ohms\n",
- "n = 8; # no. of cells\n",
- "e = 2.2; # in volts\n",
- "\n",
- "#calculation:\n",
- "es = n*e\n",
- "ep = e\n",
- "Rs = n*R\n",
- "Rp = R/n\n",
- "\n",
- "#results\n",
- "print \"(a)Resistance\", Rs,\"ohms; e.m.f\", es,\"Volts(V)\"\n",
- "print \"(b)Resistance\", Rp,\"ohms; e.m.f\", ep,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Resistance 1.6 ohms; e.m.f 17.6 Volts(V)\n",
- "(b)Resistance 0.025 ohms; e.m.f 2.2 Volts(V)"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 35</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate its terminal p.d. if it delivers (a) 5 A, (b) 50 A\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "r = 0.02; # in ohms\n",
- "e = 2; # in volts\n",
- "I1 = 5; # in Amperes\n",
- "I2 = 50; # in Amperes\n",
- "\n",
- "#calculation:\n",
- "pd1 = e - (I1*r)\n",
- "pd2 = e - (I2*r)\n",
- "\n",
- "#results\n",
- "print \"(a)p.d\", pd1,\"Volts(V)\"\n",
- "print \"(b)p.d\", pd2,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)p.d 1.9 Volts(V)\n",
- "(b)p.d 1.0 Volts(V)"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 36</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the internal resistance of the battery.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "e1 = 25; # in volts\n",
- "e2 = 24; # in volts\n",
- "I2 = 10; # in Amperes\n",
- "\n",
- "#calculation:\n",
- "r = (e1 - e2)/I2\n",
- "\n",
- "#results\n",
- "print \"Resistance\", r,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 0.1 Ohms"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 36</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current flowing in the circuit and (b) the p.d. at the battery terminals.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "r = 0.2; # in ohms\n",
- "n = 10; # no. of cells\n",
- "e = 1.5; # in volts\n",
- "R = 58; # in ohms\n",
- "\n",
- "#calculation:\n",
- "es = n*e\n",
- "rs = n*r\n",
- "I = es/(rs + R)\n",
- "pd = es - (I*rs)\n",
- "\n",
- "#results\n",
- "print \"(a)Current\", I,\"Amperes(A)\"\n",
- "print \"(b)p.d\", pd,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Current 0.25 Amperes(A)\n",
- "(b)p.d 14.5 Volts(V)"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_1.ipynb deleted file mode 100755 index 0c38cd30..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_1.ipynb +++ /dev/null @@ -1,185 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 4: Chemical effects of\n",
- "electricity</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 35</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. and the internal resistance of the batteries formed by (a)Series arrangement and (b)Parallel Arrangement.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 0.2; # in ohms\n",
- "n = 8; # no. of cells\n",
- "e = 2.2; # in volts\n",
- "\n",
- "#calculation:\n",
- "es = n*e\n",
- "ep = e\n",
- "Rs = n*R\n",
- "Rp = R/n\n",
- "\n",
- "#results\n",
- "print \"(a)Resistance\", Rs,\"ohms; e.m.f\", es,\"Volts(V)\"\n",
- "print \"(b)Resistance\", Rp,\"ohms; e.m.f\", ep,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Resistance 1.6 ohms; e.m.f 17.6 Volts(V)\n",
- "(b)Resistance 0.025 ohms; e.m.f 2.2 Volts(V)"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 35</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate its terminal p.d. if it delivers (a) 5 A, (b) 50 A\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "r = 0.02; # in ohms\n",
- "e = 2; # in volts\n",
- "I1 = 5; # in Amperes\n",
- "I2 = 50; # in Amperes\n",
- "\n",
- "#calculation:\n",
- "pd1 = e - (I1*r)\n",
- "pd2 = e - (I2*r)\n",
- "\n",
- "#results\n",
- "print \"(a)p.d\", pd1,\"Volts(V)\"\n",
- "print \"(b)p.d\", pd2,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)p.d 1.9 Volts(V)\n",
- "(b)p.d 1.0 Volts(V)"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 36</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the internal resistance of the battery.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "e1 = 25; # in volts\n",
- "e2 = 24; # in volts\n",
- "I2 = 10; # in Amperes\n",
- "\n",
- "#calculation:\n",
- "r = (e1 - e2)/I2\n",
- "\n",
- "#results\n",
- "print \"Resistance\", r,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 0.1 Ohms"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 36</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current flowing in the circuit and (b) the p.d. at the battery terminals.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "r = 0.2; # in ohms\n",
- "n = 10; # no. of cells\n",
- "e = 1.5; # in volts\n",
- "R = 58; # in ohms\n",
- "\n",
- "#calculation:\n",
- "es = n*e\n",
- "rs = n*r\n",
- "I = es/(rs + R)\n",
- "pd = es - (I*rs)\n",
- "\n",
- "#results\n",
- "print \"(a)Current\", I,\"Amperes(A)\"\n",
- "print \"(b)p.d\", pd,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Current 0.25 Amperes(A)\n",
- "(b)p.d 14.5 Volts(V)"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_2.ipynb deleted file mode 100755 index 0c38cd30..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_2.ipynb +++ /dev/null @@ -1,185 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 4: Chemical effects of\n",
- "electricity</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 35</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. and the internal resistance of the batteries formed by (a)Series arrangement and (b)Parallel Arrangement.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R = 0.2; # in ohms\n",
- "n = 8; # no. of cells\n",
- "e = 2.2; # in volts\n",
- "\n",
- "#calculation:\n",
- "es = n*e\n",
- "ep = e\n",
- "Rs = n*R\n",
- "Rp = R/n\n",
- "\n",
- "#results\n",
- "print \"(a)Resistance\", Rs,\"ohms; e.m.f\", es,\"Volts(V)\"\n",
- "print \"(b)Resistance\", Rp,\"ohms; e.m.f\", ep,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Resistance 1.6 ohms; e.m.f 17.6 Volts(V)\n",
- "(b)Resistance 0.025 ohms; e.m.f 2.2 Volts(V)"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 35</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate its terminal p.d. if it delivers (a) 5 A, (b) 50 A\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "r = 0.02; # in ohms\n",
- "e = 2; # in volts\n",
- "I1 = 5; # in Amperes\n",
- "I2 = 50; # in Amperes\n",
- "\n",
- "#calculation:\n",
- "pd1 = e - (I1*r)\n",
- "pd2 = e - (I2*r)\n",
- "\n",
- "#results\n",
- "print \"(a)p.d\", pd1,\"Volts(V)\"\n",
- "print \"(b)p.d\", pd2,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)p.d 1.9 Volts(V)\n",
- "(b)p.d 1.0 Volts(V)"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 36</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the internal resistance of the battery.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "e1 = 25; # in volts\n",
- "e2 = 24; # in volts\n",
- "I2 = 10; # in Amperes\n",
- "\n",
- "#calculation:\n",
- "r = (e1 - e2)/I2\n",
- "\n",
- "#results\n",
- "print \"Resistance\", r,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance 0.1 Ohms"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 36</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current flowing in the circuit and (b) the p.d. at the battery terminals.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "r = 0.2; # in ohms\n",
- "n = 10; # no. of cells\n",
- "e = 1.5; # in volts\n",
- "R = 58; # in ohms\n",
- "\n",
- "#calculation:\n",
- "es = n*e\n",
- "rs = n*r\n",
- "I = es/(rs + R)\n",
- "pd = es - (I*rs)\n",
- "\n",
- "#results\n",
- "print \"(a)Current\", I,\"Amperes(A)\"\n",
- "print \"(b)p.d\", pd,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Current 0.25 Amperes(A)\n",
- "(b)p.d 14.5 Volts(V)"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_04.ipynb index 860dc4f0..860dc4f0 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_04.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint.ipynb deleted file mode 100755 index b881dd41..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint.ipynb +++ /dev/null @@ -1,763 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 5: Series and parallel\n",
- "networks</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 43</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the battery voltage V, (b) the total resistance of the circuit, and \n",
- "#(c) the values of resistance of resistors R1, R2 and R3,\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V1 = 5; # in volts\n",
- "V2 = 2; # in volts\n",
- "V3 = 6; # in volts\n",
- "I = 4; # in Amperes\n",
- "\n",
- "#calculation:\n",
- "Vt = V1 + V2 + V3\n",
- "Rt = Vt/I\n",
- "R1 = V1/I\n",
- "R2 = V2/I\n",
- "R3 = V3/I\n",
- "\n",
- "#results\n",
- "print \"(a) Total Voltage\", Vt,\"Volts(V)\"\n",
- "print \"(b)Total Resistance\", Rt,\"Ohms\"\n",
- "print \"(c)Resistance(R1)\", R1,\"Ohms; Resistance(R2)\", R2,\"Ohms and\"\n",
- "print \"Resistance(R3)\", R3,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Total Voltage 13 Volts(V)\n",
- "(b)Total Resistance 3.25 Ohms\n",
- "(c)Resistance(R1) 1.25 Ohms; Resistance(R2) 0.5 Ohms and\n",
- "Resistance(R3) 1.5 Ohms"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 43</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the p.d. across resistor R3.\n",
- "#Find value of resistor R2\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V1 = 10; # in volts\n",
- "V2 = 4; # in volts\n",
- "Vt = 25; # in volts\n",
- "Rt = 100; # in ohms\n",
- "\n",
- "#calculation:\n",
- "V3 = Vt - V1 - V2\n",
- "I = Vt/Rt\n",
- "R2 = V2/I\n",
- "\n",
- "#results\n",
- "print \"(a)Voltage(V3)\", V3,\"Volts(V)\"\n",
- "print \"(b)current\", I,\"Amperes(A)\"\n",
- "print \"(c)Resistance(R2)\", R2,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Voltage(V3) 11 Volts(V)\n",
- "(b)current 0.25 Amperes(A)\n",
- "(c)Resistance(R2) 16.0 Ohms"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 44</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing through, and the p.d. across the 9 ohm resistor. \n",
- "#Find also the power dissipated in the 11 ohm resistor.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "Vt = 12; # in volts\n",
- "R1 = 4; # in ohms\n",
- "R2 = 9; # in ohms\n",
- "R3 = 11; # in ohms\n",
- "\n",
- "#calculation:\n",
- "Rt = R1 + R2 + R3\n",
- "I = Vt/Rt\n",
- "V9 = I*R2\n",
- "P11 = I*I*R3\n",
- "#results\n",
- "print \"a)current\", I,\"Amperes(A)\\n\"\n",
- "print \"b)Voltage(V2)\", V9,\"Volts(V)\\n\"\n",
- "print \"c)Power\", P11,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a)current 0.5 Amperes(A)\n",
- "\n",
- "b)Voltage(V2) 4.5 Volts(V)\n",
- "\n",
- "c)Power 2.75 Watt(W)"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 44</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of voltage V\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "Vt = 50; # in volts\n",
- "R1 = 4; # in ohms\n",
- "R2 = 6; # in ohms\n",
- "\n",
- "#calculation:\n",
- "Rt = R1 + R2\n",
- "I = Vt/Rt\n",
- "V2 = I*R2\n",
- "\n",
- "#results\n",
- "print \"Voltage(V)\", V2,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage(V) 30.0 Volts(V)"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 45</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the value of the other resistor, and \n",
- "#(b) the p.d. across the 2 \u0006 resistor. \n",
- "#If the circuit is connected for 50 hours, how much energy is used?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "Vt = 24; # in volts\n",
- "R1 = 2; # in ohms\n",
- "I = 3; # in Amperes\n",
- "t = 50; # in hrs\n",
- "\n",
- "#calculation:\n",
- "V1 = I*R1\n",
- "R2 = (Vt-(I*R1))/I\n",
- "E = Vt*I*t\n",
- "\n",
- "#results\n",
- "print \"a)Voltage(V1)\", V1,\"Volts(V)\\n\"\n",
- "print \"b)Resistance(R2)\", R2,\"Ohms\\n\"\n",
- "print \"c)Energy(E)\", E/1000,\"kWh\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a)Voltage(V1) 6 Volts(V)\n",
- "\n",
- "b)Resistance(R2) 6.0 Ohms\n",
- "\n",
- "c)Energy(E) 3.6 kWh"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 46</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the reading on the ammeter, and \n",
- "# (b) the value of resistor R2\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 5; # in ohms\n",
- "R3 = 20; # in ohms\n",
- "I1 = 8; # in Amperes\n",
- "It = 11; # in Amperes\n",
- "\n",
- "#calculation:\n",
- "Vt = I1*R1\n",
- "I3 = Vt/R3\n",
- "R2 = Vt/(It - I1 - I3)\n",
- "\n",
- "#results\n",
- "print \"a)Ammeter Reading\", I3,\"Amperes(A)\\n\"\n",
- "print \"b)Resistance(R2)\", R2,\"Ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a)Ammeter Reading 2.0 Amperes(A)\n",
- "\n",
- "b)Resistance(R2) 40.0 Ohms"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 46</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the total circuit resistance and (b) the current flowing in the 3 ohm resistor.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 3; # in ohms\n",
- "R2 = 6; # in ohms\n",
- "Vt = 12; # in volts\n",
- "\n",
- "#calculation:\n",
- "Rt = R1*R2/(R1 + R2)\n",
- "I1 = (Vt/R1)\n",
- "\n",
- "#Result\n",
- "print \"(a)Total Resistance\", Rt,\"Ohms\\n\"\n",
- "print \"(b)Current(I1)\", I1,\"Amperes(A)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Total Resistance 2.0 Ohms\n",
- "\n",
- "(b)Current(I1) 4.0 Amperes(A)"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 47</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the value of the supply voltage V and (b) the value of current I.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 10;# in ohms\n",
- "R2 = 20;# in ohms\n",
- "R3 = 60;# in ohms\n",
- "I2 = 3;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "Vt = I2*R2\n",
- "I1 = Vt/R1\n",
- "I3 = Vt/R3\n",
- "I = I1 +I2 + I3\n",
- "\n",
- "print \"\\nResult\\n\"\n",
- "print \"\\n(a)Voltage(V) \",Vt,\" Volts(V)\\n\"\n",
- "print \"\\n(b)Total Current(I) \",I,\" Amperes(A)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Result\n",
- "\n",
- "\n",
- "(a)Voltage(V) 60 Volts(V)\n",
- "\n",
- "\n",
- "(b)Total Current(I) 10.0 Amperes(A)"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 47</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#state how they must be connected to give an overall resistance of \n",
- "#(a) 1/4 ohm (b) 1 ohm (c) 4/3 ohm (d)2.5 ohm, all four resistors being connected in each case\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 1;# in ohms\n",
- "\n",
- "#calculation\n",
- "R1 = 1/(1/R + 1/R + 1/R + 1/R)\n",
- "R2 = 2*R*2*R/(4*R)\n",
- "R3 = 1/(1/R + 1/R + 1/R) + 1\n",
- "R4 = R*R/(2*R) + 2*R\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n(a)All four in parallel for \",R1,\" ohm\\n\"\n",
- "print \"\\n(b)Two in series, in parallel with another two in series for\",R2,\" ohm\\n\"\n",
- "print \"\\n(c)Three in parallel, in series with one for \",round(R3,2),\" ohm\\n\"\n",
- "print \"\\n(d)Two in parallel, in series with two in series for \",R4,\" ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- "(a)All four in parallel for 0.25 ohm\n",
- "\n",
- "\n",
- "(b)Two in series, in parallel with another two in series for 1.0 ohm\n",
- "\n",
- "\n",
- "(c)Three in parallel, in series with one for 1.33 ohm\n",
- "\n",
- "\n",
- "(d)Two in parallel, in series with two in series for 2.5 ohm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 48</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the equivalent resistance for the circuit\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 1;# in ohms\n",
- "R2 = 2.2;# in ohms\n",
- "R3 = 3;# in ohms\n",
- "R4 = 6;# in ohms\n",
- "R5 = 18;# in ohms\n",
- "R6 = 4;# in ohms\n",
- "\n",
- "\n",
- "#calculation:\n",
- "R0 = 1/((1/3) + (1/6) + (1/18))\n",
- "Rt = R1 + R2 + R0 + R6\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Equivalent Resistance \",Rt,\" Ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Equivalent Resistance 9.0 Ohms"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 48</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the supply current, (b) the current flowing through each resistor and (c) the p.d. across each resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 2.5;# in ohms\n",
- "R2 = 6;# in ohms\n",
- "R3 = 2;# in ohms\n",
- "R4 = 4;# in ohms\n",
- "Vt = 200;# in volts\n",
- "\n",
- "#calculation:\n",
- "R0 = 1/((1/R2) + (1/R3))\n",
- "Rt = R1 + R0 + R4\n",
- "It = Vt/Rt\n",
- "I1 = It\n",
- "I4 = It\n",
- "I2 = R3*It/(R3+R2)\n",
- "I3 = It - I2\n",
- "V1 = I1*R1\n",
- "V2 = I2*R2\n",
- "V3 = I3*R3\n",
- "V4 = I4*R4\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Total Current Supply \",It,\" Amperes(A)\\n\"\n",
- "print \"\\n (b)Current through resistors (R1, R2, R3, R4)\\n \",I1,\", \", I2,\", \", I3,\", \", I4,\" Amperes(A) respectively\\n\"\n",
- "print \"\\n (c)voltage across resistors (R1, R2, R3, R4)\\n \",V1,\", \", V2,\", \", V3,\", \", V4,\" Volts(V) respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Total Current Supply 25.0 Amperes(A)\n",
- "\n",
- "\n",
- " (b)Current through resistors (R1, R2, R3, R4)\n",
- " 25.0 , 6.25 , 18.75 , 25.0 Amperes(A) respectively\n",
- "\n",
- "\n",
- " (c)voltage across resistors (R1, R2, R3, R4)\n",
- " 62.5 , 37.5 , 37.5 , 100.0 Volts(V) respectively"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 49</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate (a) the value of resistor Rx such that the total power dissipated in the circuit is 2.5 kW, and \n",
- "#(b) the current flowing in each of the four resistors\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 15;# in ohms\n",
- "R2 = 10;# in ohms\n",
- "R3 = 38;# in ohms\n",
- "Vt = 250;# in volts\n",
- "P = 2500;# in Watt\n",
- "\n",
- "#calculation:\n",
- "It = P/Vt\n",
- "I2 = R1*It/(R1+R2)\n",
- "I1 = It - I2\n",
- "Re1 = 1/((1/R1) + (1/R2))\n",
- "Rt = Vt/It\n",
- "Re2 = Rt - Re1\n",
- "Rx = 1/((1/Re2) - (1/R3))\n",
- "I4 = R3*It/(R3+Rx)\n",
- "I3 = It - I4\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Resistance (Rx) \",Rx,\" Ohms\\n\"\n",
- "print \"\\n (b)Current through resistors (R1, R2, R3, R4): \\n \",I1,\", \", I2,\", \", I3,\", \"\n",
- "print \", I4,\" Amperes(A) respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance (Rx) 38.0 Ohms\n",
- "\n",
- "\n",
- " (b)Current through resistors (R1, R2, R3, R4): \n",
- " 4.0 , 6.0 , 5.0 , 5.0 Amperes(A) respectively"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 51</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the current Ix\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 8;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R3 = 1.4;# in ohms\n",
- "R4 = 9;# in ohms\n",
- "R5 = 2;# in ohms\n",
- "Vt = 17;# in volts\n",
- "\n",
- "#calculation:\n",
- "R01 = R1*R2/(R1 + R2)\n",
- "R02 = R01 + R3\n",
- "R03 = R4*R02/(R4 +R02)\n",
- "Rt = R5 + R03\n",
- "It = Vt/Rt\n",
- "I1 = R4*It/(R4 + R02)\n",
- "Ix = R2*I1/(R1 + R2)\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Current(Ix) \",Ix,\" Amperes(A)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Current(Ix) 0.6 Amperes(A)"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 52</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the resistance of one lamp.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rt = 150;# in ohms\n",
- "n = 3;# no. of identical lamp\n",
- "\n",
- "#calculation:\n",
- "R = Rt*3# (1/Rt)=(1/R)+(1/R)+(1/R)\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Resistance \",R,\" Ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Resistance 450 Ohms"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 52</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#State (a) the voltage across each lamp,\n",
- "# and (b) the effect of lamp C failing.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "#series connection\n",
- "n = 3;# no. of identical lamp\n",
- "Vt = 150;# in volts\n",
- "\n",
- "#calculation:\n",
- "V = Vt/3# Since each lamp is identical, then V volts across each.\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n a)Voltage across each resistor = \",V,\" Volts(V)\\n\"\n",
- "print \"\\n b)If lamp C fails, i.e., open circuits, no current will flow and lamps A and B will not operate.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " a)Voltage across each resistor = 50.0 Volts(V)\n",
- "\n",
- "\n",
- " b)If lamp C fails, i.e., open circuits, no current will flow and lamps A and B will not operate."
- ]
- }
- ],
- "prompt_number": 21
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_1.ipynb deleted file mode 100755 index b881dd41..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_1.ipynb +++ /dev/null @@ -1,763 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 5: Series and parallel\n",
- "networks</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 43</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the battery voltage V, (b) the total resistance of the circuit, and \n",
- "#(c) the values of resistance of resistors R1, R2 and R3,\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V1 = 5; # in volts\n",
- "V2 = 2; # in volts\n",
- "V3 = 6; # in volts\n",
- "I = 4; # in Amperes\n",
- "\n",
- "#calculation:\n",
- "Vt = V1 + V2 + V3\n",
- "Rt = Vt/I\n",
- "R1 = V1/I\n",
- "R2 = V2/I\n",
- "R3 = V3/I\n",
- "\n",
- "#results\n",
- "print \"(a) Total Voltage\", Vt,\"Volts(V)\"\n",
- "print \"(b)Total Resistance\", Rt,\"Ohms\"\n",
- "print \"(c)Resistance(R1)\", R1,\"Ohms; Resistance(R2)\", R2,\"Ohms and\"\n",
- "print \"Resistance(R3)\", R3,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Total Voltage 13 Volts(V)\n",
- "(b)Total Resistance 3.25 Ohms\n",
- "(c)Resistance(R1) 1.25 Ohms; Resistance(R2) 0.5 Ohms and\n",
- "Resistance(R3) 1.5 Ohms"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 43</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the p.d. across resistor R3.\n",
- "#Find value of resistor R2\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V1 = 10; # in volts\n",
- "V2 = 4; # in volts\n",
- "Vt = 25; # in volts\n",
- "Rt = 100; # in ohms\n",
- "\n",
- "#calculation:\n",
- "V3 = Vt - V1 - V2\n",
- "I = Vt/Rt\n",
- "R2 = V2/I\n",
- "\n",
- "#results\n",
- "print \"(a)Voltage(V3)\", V3,\"Volts(V)\"\n",
- "print \"(b)current\", I,\"Amperes(A)\"\n",
- "print \"(c)Resistance(R2)\", R2,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Voltage(V3) 11 Volts(V)\n",
- "(b)current 0.25 Amperes(A)\n",
- "(c)Resistance(R2) 16.0 Ohms"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 44</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing through, and the p.d. across the 9 ohm resistor. \n",
- "#Find also the power dissipated in the 11 ohm resistor.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "Vt = 12; # in volts\n",
- "R1 = 4; # in ohms\n",
- "R2 = 9; # in ohms\n",
- "R3 = 11; # in ohms\n",
- "\n",
- "#calculation:\n",
- "Rt = R1 + R2 + R3\n",
- "I = Vt/Rt\n",
- "V9 = I*R2\n",
- "P11 = I*I*R3\n",
- "#results\n",
- "print \"a)current\", I,\"Amperes(A)\\n\"\n",
- "print \"b)Voltage(V2)\", V9,\"Volts(V)\\n\"\n",
- "print \"c)Power\", P11,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a)current 0.5 Amperes(A)\n",
- "\n",
- "b)Voltage(V2) 4.5 Volts(V)\n",
- "\n",
- "c)Power 2.75 Watt(W)"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 44</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of voltage V\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "Vt = 50; # in volts\n",
- "R1 = 4; # in ohms\n",
- "R2 = 6; # in ohms\n",
- "\n",
- "#calculation:\n",
- "Rt = R1 + R2\n",
- "I = Vt/Rt\n",
- "V2 = I*R2\n",
- "\n",
- "#results\n",
- "print \"Voltage(V)\", V2,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage(V) 30.0 Volts(V)"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 45</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the value of the other resistor, and \n",
- "#(b) the p.d. across the 2 \u0006 resistor. \n",
- "#If the circuit is connected for 50 hours, how much energy is used?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "Vt = 24; # in volts\n",
- "R1 = 2; # in ohms\n",
- "I = 3; # in Amperes\n",
- "t = 50; # in hrs\n",
- "\n",
- "#calculation:\n",
- "V1 = I*R1\n",
- "R2 = (Vt-(I*R1))/I\n",
- "E = Vt*I*t\n",
- "\n",
- "#results\n",
- "print \"a)Voltage(V1)\", V1,\"Volts(V)\\n\"\n",
- "print \"b)Resistance(R2)\", R2,\"Ohms\\n\"\n",
- "print \"c)Energy(E)\", E/1000,\"kWh\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a)Voltage(V1) 6 Volts(V)\n",
- "\n",
- "b)Resistance(R2) 6.0 Ohms\n",
- "\n",
- "c)Energy(E) 3.6 kWh"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 46</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the reading on the ammeter, and \n",
- "# (b) the value of resistor R2\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 5; # in ohms\n",
- "R3 = 20; # in ohms\n",
- "I1 = 8; # in Amperes\n",
- "It = 11; # in Amperes\n",
- "\n",
- "#calculation:\n",
- "Vt = I1*R1\n",
- "I3 = Vt/R3\n",
- "R2 = Vt/(It - I1 - I3)\n",
- "\n",
- "#results\n",
- "print \"a)Ammeter Reading\", I3,\"Amperes(A)\\n\"\n",
- "print \"b)Resistance(R2)\", R2,\"Ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a)Ammeter Reading 2.0 Amperes(A)\n",
- "\n",
- "b)Resistance(R2) 40.0 Ohms"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 46</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the total circuit resistance and (b) the current flowing in the 3 ohm resistor.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 3; # in ohms\n",
- "R2 = 6; # in ohms\n",
- "Vt = 12; # in volts\n",
- "\n",
- "#calculation:\n",
- "Rt = R1*R2/(R1 + R2)\n",
- "I1 = (Vt/R1)\n",
- "\n",
- "#Result\n",
- "print \"(a)Total Resistance\", Rt,\"Ohms\\n\"\n",
- "print \"(b)Current(I1)\", I1,\"Amperes(A)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Total Resistance 2.0 Ohms\n",
- "\n",
- "(b)Current(I1) 4.0 Amperes(A)"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 47</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the value of the supply voltage V and (b) the value of current I.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 10;# in ohms\n",
- "R2 = 20;# in ohms\n",
- "R3 = 60;# in ohms\n",
- "I2 = 3;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "Vt = I2*R2\n",
- "I1 = Vt/R1\n",
- "I3 = Vt/R3\n",
- "I = I1 +I2 + I3\n",
- "\n",
- "print \"\\nResult\\n\"\n",
- "print \"\\n(a)Voltage(V) \",Vt,\" Volts(V)\\n\"\n",
- "print \"\\n(b)Total Current(I) \",I,\" Amperes(A)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Result\n",
- "\n",
- "\n",
- "(a)Voltage(V) 60 Volts(V)\n",
- "\n",
- "\n",
- "(b)Total Current(I) 10.0 Amperes(A)"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 47</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#state how they must be connected to give an overall resistance of \n",
- "#(a) 1/4 ohm (b) 1 ohm (c) 4/3 ohm (d)2.5 ohm, all four resistors being connected in each case\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 1;# in ohms\n",
- "\n",
- "#calculation\n",
- "R1 = 1/(1/R + 1/R + 1/R + 1/R)\n",
- "R2 = 2*R*2*R/(4*R)\n",
- "R3 = 1/(1/R + 1/R + 1/R) + 1\n",
- "R4 = R*R/(2*R) + 2*R\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n(a)All four in parallel for \",R1,\" ohm\\n\"\n",
- "print \"\\n(b)Two in series, in parallel with another two in series for\",R2,\" ohm\\n\"\n",
- "print \"\\n(c)Three in parallel, in series with one for \",round(R3,2),\" ohm\\n\"\n",
- "print \"\\n(d)Two in parallel, in series with two in series for \",R4,\" ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- "(a)All four in parallel for 0.25 ohm\n",
- "\n",
- "\n",
- "(b)Two in series, in parallel with another two in series for 1.0 ohm\n",
- "\n",
- "\n",
- "(c)Three in parallel, in series with one for 1.33 ohm\n",
- "\n",
- "\n",
- "(d)Two in parallel, in series with two in series for 2.5 ohm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 48</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the equivalent resistance for the circuit\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 1;# in ohms\n",
- "R2 = 2.2;# in ohms\n",
- "R3 = 3;# in ohms\n",
- "R4 = 6;# in ohms\n",
- "R5 = 18;# in ohms\n",
- "R6 = 4;# in ohms\n",
- "\n",
- "\n",
- "#calculation:\n",
- "R0 = 1/((1/3) + (1/6) + (1/18))\n",
- "Rt = R1 + R2 + R0 + R6\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Equivalent Resistance \",Rt,\" Ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Equivalent Resistance 9.0 Ohms"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 48</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the supply current, (b) the current flowing through each resistor and (c) the p.d. across each resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 2.5;# in ohms\n",
- "R2 = 6;# in ohms\n",
- "R3 = 2;# in ohms\n",
- "R4 = 4;# in ohms\n",
- "Vt = 200;# in volts\n",
- "\n",
- "#calculation:\n",
- "R0 = 1/((1/R2) + (1/R3))\n",
- "Rt = R1 + R0 + R4\n",
- "It = Vt/Rt\n",
- "I1 = It\n",
- "I4 = It\n",
- "I2 = R3*It/(R3+R2)\n",
- "I3 = It - I2\n",
- "V1 = I1*R1\n",
- "V2 = I2*R2\n",
- "V3 = I3*R3\n",
- "V4 = I4*R4\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Total Current Supply \",It,\" Amperes(A)\\n\"\n",
- "print \"\\n (b)Current through resistors (R1, R2, R3, R4)\\n \",I1,\", \", I2,\", \", I3,\", \", I4,\" Amperes(A) respectively\\n\"\n",
- "print \"\\n (c)voltage across resistors (R1, R2, R3, R4)\\n \",V1,\", \", V2,\", \", V3,\", \", V4,\" Volts(V) respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Total Current Supply 25.0 Amperes(A)\n",
- "\n",
- "\n",
- " (b)Current through resistors (R1, R2, R3, R4)\n",
- " 25.0 , 6.25 , 18.75 , 25.0 Amperes(A) respectively\n",
- "\n",
- "\n",
- " (c)voltage across resistors (R1, R2, R3, R4)\n",
- " 62.5 , 37.5 , 37.5 , 100.0 Volts(V) respectively"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 49</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate (a) the value of resistor Rx such that the total power dissipated in the circuit is 2.5 kW, and \n",
- "#(b) the current flowing in each of the four resistors\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 15;# in ohms\n",
- "R2 = 10;# in ohms\n",
- "R3 = 38;# in ohms\n",
- "Vt = 250;# in volts\n",
- "P = 2500;# in Watt\n",
- "\n",
- "#calculation:\n",
- "It = P/Vt\n",
- "I2 = R1*It/(R1+R2)\n",
- "I1 = It - I2\n",
- "Re1 = 1/((1/R1) + (1/R2))\n",
- "Rt = Vt/It\n",
- "Re2 = Rt - Re1\n",
- "Rx = 1/((1/Re2) - (1/R3))\n",
- "I4 = R3*It/(R3+Rx)\n",
- "I3 = It - I4\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Resistance (Rx) \",Rx,\" Ohms\\n\"\n",
- "print \"\\n (b)Current through resistors (R1, R2, R3, R4): \\n \",I1,\", \", I2,\", \", I3,\", \"\n",
- "print \", I4,\" Amperes(A) respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance (Rx) 38.0 Ohms\n",
- "\n",
- "\n",
- " (b)Current through resistors (R1, R2, R3, R4): \n",
- " 4.0 , 6.0 , 5.0 , 5.0 Amperes(A) respectively"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 51</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the current Ix\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 8;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R3 = 1.4;# in ohms\n",
- "R4 = 9;# in ohms\n",
- "R5 = 2;# in ohms\n",
- "Vt = 17;# in volts\n",
- "\n",
- "#calculation:\n",
- "R01 = R1*R2/(R1 + R2)\n",
- "R02 = R01 + R3\n",
- "R03 = R4*R02/(R4 +R02)\n",
- "Rt = R5 + R03\n",
- "It = Vt/Rt\n",
- "I1 = R4*It/(R4 + R02)\n",
- "Ix = R2*I1/(R1 + R2)\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Current(Ix) \",Ix,\" Amperes(A)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Current(Ix) 0.6 Amperes(A)"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 52</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the resistance of one lamp.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rt = 150;# in ohms\n",
- "n = 3;# no. of identical lamp\n",
- "\n",
- "#calculation:\n",
- "R = Rt*3# (1/Rt)=(1/R)+(1/R)+(1/R)\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Resistance \",R,\" Ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Resistance 450 Ohms"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 52</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#State (a) the voltage across each lamp,\n",
- "# and (b) the effect of lamp C failing.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "#series connection\n",
- "n = 3;# no. of identical lamp\n",
- "Vt = 150;# in volts\n",
- "\n",
- "#calculation:\n",
- "V = Vt/3# Since each lamp is identical, then V volts across each.\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n a)Voltage across each resistor = \",V,\" Volts(V)\\n\"\n",
- "print \"\\n b)If lamp C fails, i.e., open circuits, no current will flow and lamps A and B will not operate.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " a)Voltage across each resistor = 50.0 Volts(V)\n",
- "\n",
- "\n",
- " b)If lamp C fails, i.e., open circuits, no current will flow and lamps A and B will not operate."
- ]
- }
- ],
- "prompt_number": 21
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_2.ipynb deleted file mode 100755 index b881dd41..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_2.ipynb +++ /dev/null @@ -1,763 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 5: Series and parallel\n",
- "networks</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 43</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the battery voltage V, (b) the total resistance of the circuit, and \n",
- "#(c) the values of resistance of resistors R1, R2 and R3,\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V1 = 5; # in volts\n",
- "V2 = 2; # in volts\n",
- "V3 = 6; # in volts\n",
- "I = 4; # in Amperes\n",
- "\n",
- "#calculation:\n",
- "Vt = V1 + V2 + V3\n",
- "Rt = Vt/I\n",
- "R1 = V1/I\n",
- "R2 = V2/I\n",
- "R3 = V3/I\n",
- "\n",
- "#results\n",
- "print \"(a) Total Voltage\", Vt,\"Volts(V)\"\n",
- "print \"(b)Total Resistance\", Rt,\"Ohms\"\n",
- "print \"(c)Resistance(R1)\", R1,\"Ohms; Resistance(R2)\", R2,\"Ohms and\"\n",
- "print \"Resistance(R3)\", R3,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Total Voltage 13 Volts(V)\n",
- "(b)Total Resistance 3.25 Ohms\n",
- "(c)Resistance(R1) 1.25 Ohms; Resistance(R2) 0.5 Ohms and\n",
- "Resistance(R3) 1.5 Ohms"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 43</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the p.d. across resistor R3.\n",
- "#Find value of resistor R2\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "V1 = 10; # in volts\n",
- "V2 = 4; # in volts\n",
- "Vt = 25; # in volts\n",
- "Rt = 100; # in ohms\n",
- "\n",
- "#calculation:\n",
- "V3 = Vt - V1 - V2\n",
- "I = Vt/Rt\n",
- "R2 = V2/I\n",
- "\n",
- "#results\n",
- "print \"(a)Voltage(V3)\", V3,\"Volts(V)\"\n",
- "print \"(b)current\", I,\"Amperes(A)\"\n",
- "print \"(c)Resistance(R2)\", R2,\"Ohms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Voltage(V3) 11 Volts(V)\n",
- "(b)current 0.25 Amperes(A)\n",
- "(c)Resistance(R2) 16.0 Ohms"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 44</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing through, and the p.d. across the 9 ohm resistor. \n",
- "#Find also the power dissipated in the 11 ohm resistor.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "Vt = 12; # in volts\n",
- "R1 = 4; # in ohms\n",
- "R2 = 9; # in ohms\n",
- "R3 = 11; # in ohms\n",
- "\n",
- "#calculation:\n",
- "Rt = R1 + R2 + R3\n",
- "I = Vt/Rt\n",
- "V9 = I*R2\n",
- "P11 = I*I*R3\n",
- "#results\n",
- "print \"a)current\", I,\"Amperes(A)\\n\"\n",
- "print \"b)Voltage(V2)\", V9,\"Volts(V)\\n\"\n",
- "print \"c)Power\", P11,\"Watt(W)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a)current 0.5 Amperes(A)\n",
- "\n",
- "b)Voltage(V2) 4.5 Volts(V)\n",
- "\n",
- "c)Power 2.75 Watt(W)"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 44</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of voltage V\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "Vt = 50; # in volts\n",
- "R1 = 4; # in ohms\n",
- "R2 = 6; # in ohms\n",
- "\n",
- "#calculation:\n",
- "Rt = R1 + R2\n",
- "I = Vt/Rt\n",
- "V2 = I*R2\n",
- "\n",
- "#results\n",
- "print \"Voltage(V)\", V2,\"Volts(V)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage(V) 30.0 Volts(V)"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 45</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the value of the other resistor, and \n",
- "#(b) the p.d. across the 2 \u0006 resistor. \n",
- "#If the circuit is connected for 50 hours, how much energy is used?\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "Vt = 24; # in volts\n",
- "R1 = 2; # in ohms\n",
- "I = 3; # in Amperes\n",
- "t = 50; # in hrs\n",
- "\n",
- "#calculation:\n",
- "V1 = I*R1\n",
- "R2 = (Vt-(I*R1))/I\n",
- "E = Vt*I*t\n",
- "\n",
- "#results\n",
- "print \"a)Voltage(V1)\", V1,\"Volts(V)\\n\"\n",
- "print \"b)Resistance(R2)\", R2,\"Ohms\\n\"\n",
- "print \"c)Energy(E)\", E/1000,\"kWh\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a)Voltage(V1) 6 Volts(V)\n",
- "\n",
- "b)Resistance(R2) 6.0 Ohms\n",
- "\n",
- "c)Energy(E) 3.6 kWh"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 46</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the reading on the ammeter, and \n",
- "# (b) the value of resistor R2\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 5; # in ohms\n",
- "R3 = 20; # in ohms\n",
- "I1 = 8; # in Amperes\n",
- "It = 11; # in Amperes\n",
- "\n",
- "#calculation:\n",
- "Vt = I1*R1\n",
- "I3 = Vt/R3\n",
- "R2 = Vt/(It - I1 - I3)\n",
- "\n",
- "#results\n",
- "print \"a)Ammeter Reading\", I3,\"Amperes(A)\\n\"\n",
- "print \"b)Resistance(R2)\", R2,\"Ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "a)Ammeter Reading 2.0 Amperes(A)\n",
- "\n",
- "b)Resistance(R2) 40.0 Ohms"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 46</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the total circuit resistance and (b) the current flowing in the 3 ohm resistor.\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 3; # in ohms\n",
- "R2 = 6; # in ohms\n",
- "Vt = 12; # in volts\n",
- "\n",
- "#calculation:\n",
- "Rt = R1*R2/(R1 + R2)\n",
- "I1 = (Vt/R1)\n",
- "\n",
- "#Result\n",
- "print \"(a)Total Resistance\", Rt,\"Ohms\\n\"\n",
- "print \"(b)Current(I1)\", I1,\"Amperes(A)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Total Resistance 2.0 Ohms\n",
- "\n",
- "(b)Current(I1) 4.0 Amperes(A)"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 47</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the value of the supply voltage V and (b) the value of current I.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 10;# in ohms\n",
- "R2 = 20;# in ohms\n",
- "R3 = 60;# in ohms\n",
- "I2 = 3;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "Vt = I2*R2\n",
- "I1 = Vt/R1\n",
- "I3 = Vt/R3\n",
- "I = I1 +I2 + I3\n",
- "\n",
- "print \"\\nResult\\n\"\n",
- "print \"\\n(a)Voltage(V) \",Vt,\" Volts(V)\\n\"\n",
- "print \"\\n(b)Total Current(I) \",I,\" Amperes(A)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Result\n",
- "\n",
- "\n",
- "(a)Voltage(V) 60 Volts(V)\n",
- "\n",
- "\n",
- "(b)Total Current(I) 10.0 Amperes(A)"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 47</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#state how they must be connected to give an overall resistance of \n",
- "#(a) 1/4 ohm (b) 1 ohm (c) 4/3 ohm (d)2.5 ohm, all four resistors being connected in each case\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 1;# in ohms\n",
- "\n",
- "#calculation\n",
- "R1 = 1/(1/R + 1/R + 1/R + 1/R)\n",
- "R2 = 2*R*2*R/(4*R)\n",
- "R3 = 1/(1/R + 1/R + 1/R) + 1\n",
- "R4 = R*R/(2*R) + 2*R\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n(a)All four in parallel for \",R1,\" ohm\\n\"\n",
- "print \"\\n(b)Two in series, in parallel with another two in series for\",R2,\" ohm\\n\"\n",
- "print \"\\n(c)Three in parallel, in series with one for \",round(R3,2),\" ohm\\n\"\n",
- "print \"\\n(d)Two in parallel, in series with two in series for \",R4,\" ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- "(a)All four in parallel for 0.25 ohm\n",
- "\n",
- "\n",
- "(b)Two in series, in parallel with another two in series for 1.0 ohm\n",
- "\n",
- "\n",
- "(c)Three in parallel, in series with one for 1.33 ohm\n",
- "\n",
- "\n",
- "(d)Two in parallel, in series with two in series for 2.5 ohm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 48</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the equivalent resistance for the circuit\n",
- "from __future__ import division\n",
- "#initializing the variables:\n",
- "R1 = 1;# in ohms\n",
- "R2 = 2.2;# in ohms\n",
- "R3 = 3;# in ohms\n",
- "R4 = 6;# in ohms\n",
- "R5 = 18;# in ohms\n",
- "R6 = 4;# in ohms\n",
- "\n",
- "\n",
- "#calculation:\n",
- "R0 = 1/((1/3) + (1/6) + (1/18))\n",
- "Rt = R1 + R2 + R0 + R6\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Equivalent Resistance \",Rt,\" Ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Equivalent Resistance 9.0 Ohms"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 48</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the supply current, (b) the current flowing through each resistor and (c) the p.d. across each resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 2.5;# in ohms\n",
- "R2 = 6;# in ohms\n",
- "R3 = 2;# in ohms\n",
- "R4 = 4;# in ohms\n",
- "Vt = 200;# in volts\n",
- "\n",
- "#calculation:\n",
- "R0 = 1/((1/R2) + (1/R3))\n",
- "Rt = R1 + R0 + R4\n",
- "It = Vt/Rt\n",
- "I1 = It\n",
- "I4 = It\n",
- "I2 = R3*It/(R3+R2)\n",
- "I3 = It - I2\n",
- "V1 = I1*R1\n",
- "V2 = I2*R2\n",
- "V3 = I3*R3\n",
- "V4 = I4*R4\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Total Current Supply \",It,\" Amperes(A)\\n\"\n",
- "print \"\\n (b)Current through resistors (R1, R2, R3, R4)\\n \",I1,\", \", I2,\", \", I3,\", \", I4,\" Amperes(A) respectively\\n\"\n",
- "print \"\\n (c)voltage across resistors (R1, R2, R3, R4)\\n \",V1,\", \", V2,\", \", V3,\", \", V4,\" Volts(V) respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Total Current Supply 25.0 Amperes(A)\n",
- "\n",
- "\n",
- " (b)Current through resistors (R1, R2, R3, R4)\n",
- " 25.0 , 6.25 , 18.75 , 25.0 Amperes(A) respectively\n",
- "\n",
- "\n",
- " (c)voltage across resistors (R1, R2, R3, R4)\n",
- " 62.5 , 37.5 , 37.5 , 100.0 Volts(V) respectively"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 49</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate (a) the value of resistor Rx such that the total power dissipated in the circuit is 2.5 kW, and \n",
- "#(b) the current flowing in each of the four resistors\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 15;# in ohms\n",
- "R2 = 10;# in ohms\n",
- "R3 = 38;# in ohms\n",
- "Vt = 250;# in volts\n",
- "P = 2500;# in Watt\n",
- "\n",
- "#calculation:\n",
- "It = P/Vt\n",
- "I2 = R1*It/(R1+R2)\n",
- "I1 = It - I2\n",
- "Re1 = 1/((1/R1) + (1/R2))\n",
- "Rt = Vt/It\n",
- "Re2 = Rt - Re1\n",
- "Rx = 1/((1/Re2) - (1/R3))\n",
- "I4 = R3*It/(R3+Rx)\n",
- "I3 = It - I4\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Resistance (Rx) \",Rx,\" Ohms\\n\"\n",
- "print \"\\n (b)Current through resistors (R1, R2, R3, R4): \\n \",I1,\", \", I2,\", \", I3,\", \"\n",
- "print \", I4,\" Amperes(A) respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance (Rx) 38.0 Ohms\n",
- "\n",
- "\n",
- " (b)Current through resistors (R1, R2, R3, R4): \n",
- " 4.0 , 6.0 , 5.0 , 5.0 Amperes(A) respectively"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 51</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the current Ix\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 8;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R3 = 1.4;# in ohms\n",
- "R4 = 9;# in ohms\n",
- "R5 = 2;# in ohms\n",
- "Vt = 17;# in volts\n",
- "\n",
- "#calculation:\n",
- "R01 = R1*R2/(R1 + R2)\n",
- "R02 = R01 + R3\n",
- "R03 = R4*R02/(R4 +R02)\n",
- "Rt = R5 + R03\n",
- "It = Vt/Rt\n",
- "I1 = R4*It/(R4 + R02)\n",
- "Ix = R2*I1/(R1 + R2)\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Current(Ix) \",Ix,\" Amperes(A)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Current(Ix) 0.6 Amperes(A)"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 52</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the resistance of one lamp.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rt = 150;# in ohms\n",
- "n = 3;# no. of identical lamp\n",
- "\n",
- "#calculation:\n",
- "R = Rt*3# (1/Rt)=(1/R)+(1/R)+(1/R)\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Resistance \",R,\" Ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Resistance 450 Ohms"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 52</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#State (a) the voltage across each lamp,\n",
- "# and (b) the effect of lamp C failing.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "#series connection\n",
- "n = 3;# no. of identical lamp\n",
- "Vt = 150;# in volts\n",
- "\n",
- "#calculation:\n",
- "V = Vt/3# Since each lamp is identical, then V volts across each.\n",
- "\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n a)Voltage across each resistor = \",V,\" Volts(V)\\n\"\n",
- "print \"\\n b)If lamp C fails, i.e., open circuits, no current will flow and lamps A and B will not operate.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " a)Voltage across each resistor = 50.0 Volts(V)\n",
- "\n",
- "\n",
- " b)If lamp C fails, i.e., open circuits, no current will flow and lamps A and B will not operate."
- ]
- }
- ],
- "prompt_number": 21
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_05.ipynb index 0b10490d..0b10490d 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_05.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint.ipynb deleted file mode 100755 index 772a8bc5..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint.ipynb +++ /dev/null @@ -1,873 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 6: Capacitors and capacitance</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 58</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the p.d. across a 4 \u03bcF capacitor when charged with 5 mC.\n",
- "# (b) Find the charge on a 50 pF capacitor when the voltage applied to it is 2 kV.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 4E-6;# in Farad\n",
- "C2 = 50E-12;# in Farad\n",
- "Q1 = 5E-3;# in Coulomb\n",
- "V2 = 2000;# in volts\n",
- "\n",
- "#calculation:\n",
- "V1 = Q1/C1\n",
- "Q2 = C2*V2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)P.d \",V1,\" Volts(V)\\n\"\n",
- "print \"\\n (b)Charge(Q) \",(Q2/1E-6),\" micro-Coulomb\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)P.d 1250.0 Volts(V)\n",
- "\n",
- "\n",
- " (b)Charge(Q) 0.1 micro-Coulomb"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 58</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the pd between the plates.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 4;# in amperes\n",
- "C = 20E-6;# in Farad\n",
- "t = 3E-3;# in sec\n",
- "\n",
- "#calculation:\n",
- "Q = I*t\n",
- "V = Q/C\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)P.d \",V,\" Volts(V)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)P.d 600.0 Volts(V)"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 59</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate how long the capacitor can provide an average discharge current of 2 mA.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 2E-3;# in amperes\n",
- "C = 5E-6;# in Farad\n",
- "V = 800;# in volts\n",
- "\n",
- "#calculation:\n",
- "Q = C*V\n",
- "t = Q/I\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n capacitor can provide an average discharge current for \",t,\" Sec\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " capacitor can provide an average discharge current for 2.0 Sec"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 60</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the electric flux density.\n",
- "#determine the electric field strength.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Q = 0.2E-6;# in Coulomb\n",
- "A = 800E-4;# in m2\n",
- "d = 0.005;# in m\n",
- "V = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- "D = Q/A\n",
- "E = V/d\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Electric flux density D \",(D/1E-6),\" uC/m2\\n\"\n",
- "print \"\\n (b)Electric field strength E \",(E/1000),\" kV/m\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Electric flux density D 2.5 uC/m2\n",
- "\n",
- "\n",
- " (b)Electric field strength E 50.0 kV/m"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 60</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the voltage gradient between the plates.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "D = 2E-6;# in micro-C/m2\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 5;\n",
- "\n",
- "#calculation:\n",
- "E = D/(e0*er)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Electric field strength E \",round((E/1000),2),\" kV/m\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Electric field strength E 45.2 kV/m"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 60</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What is the electric field strength?\n",
- "#Find also the flux density when the dielectric between the plates is\n",
- "#(a) air, and (b) polythene of relative permittivity 2.3\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "d = 0.8E-3;# in m\n",
- "e0 = 8.85E-12;# in F/m\n",
- "era = 1;# for air\n",
- "erp = 2.3;# for polythene\n",
- "V =200;# in Volts\n",
- "\n",
- "#calculation:\n",
- "E = V/d\n",
- "#for air\n",
- "Da = E*e0*era\n",
- "#for polythene\n",
- "Dp = E*e0*erp\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Electric flux density for air \",round((Da/1E-6),2),\" micro-C/m2\\n\"\n",
- "print \"\\n (b)Electric flux density for polythene \",round((Dp/1E-6),2),\" micro-C/m2\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Electric flux density for air 2.21 micro-C/m2\n",
- "\n",
- "\n",
- " (b)Electric flux density for polythene 5.09 micro-C/m2"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 62</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)Calculate the capacitance of the capacitor in picofarads. \n",
- "#(b)what will be the pd between the plates?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "A = 4E-4;# in m2\n",
- "d = 0.1E-3;# in m\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 100;\n",
- "Q = 1.2E-6;# in coulomb\n",
- "\n",
- "#calculation:\n",
- "C = e0*er*A/d\n",
- "V = Q/C\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Capacitance \",(C/1E-12),\" pF\\n\"\n",
- "print \"\\n (b)P.d.= \",round(V,2),\" Volts(V)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Capacitance 3540.0 pF\n",
- "\n",
- "\n",
- " (b)P.d.= 338.98 Volts(V)"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 62</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the effective thickness of the paper if its relative permittivity is 2.5\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "A = 800E-4;# in m2\n",
- "C = 4425E-12;# in Farads\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 2.5;\n",
- "\n",
- "#calculation:\n",
- "d = e0*er*A/C\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Thickness \",(d/1E-3),\" mm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Thickness 0.4 mm"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 62</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the capacitance of the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "n = 19;# no. of plates\n",
- "L = 75E-3;# in m\n",
- "B = 75E-3;# in m\n",
- "d = 0.2E-3;# in m\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 5;\n",
- "#calculation:\n",
- "A = L*B\n",
- "C = e0*er*A*(n-1)/d\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Capacitance \",round((C/1E-9),2),\" nF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Capacitance 22.4 nF"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 65</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the equivalent capacitance of two capacitors of 6 \u03bcF and 4 \u03bcF connected (a) in parallel and (b) in series\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 6E-6;# in Farads\n",
- "C2 = 4E-6;# in Farads\n",
- "\n",
- "#calculation:\n",
- "# in Parallel\n",
- "Cp = C1 + C2\n",
- "# in Series\n",
- "Cs = 1/((1/C1) + (1/C2))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Capacitance in parallel \",(Cp/1E-6),\" uF\\n\"\n",
- "print \"\\n (b)Capacitance in Series \",(Cs/1E-6),\" uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Capacitance in parallel 10.0 uF\n",
- "\n",
- "\n",
- " (b)Capacitance in Series 2.4 uF"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 65</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What capacitance must be connected in series\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 30E-6;# in Farads\n",
- "Cs = 12E-6;# in Farads\n",
- "\n",
- "#calculation:\n",
- "# in Series\n",
- "C2 = 1/((1/Cs) - (1/C1))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Capacitance in series \",(C2/1E-6),\" uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Capacitance in series 20.0 uF"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 65</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the equivalent circuit capacitance, \n",
- "#(b) the total charge and\n",
- "#(c) the charge on each capacitor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 1E-6;# in Farads\n",
- "C2 = 3E-6;# in Farads\n",
- "C3 = 5E-6;# in Farads\n",
- "C4 = 6E-6;# in Farads\n",
- "Vt = 100;# in Volts\n",
- "\n",
- "#calculation:\n",
- "# in Parallel\n",
- "Cp = C1 + C2 + C3 + C4\n",
- "Qt = Vt*Cp\n",
- "Q1 = C1*Vt\n",
- "Q2 = C2*Vt\n",
- "Q3 = C3*Vt\n",
- "Q4 = C4*Vt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Equivalent Capacitance in Parallel \",(Cp/1E-6),\" uF\\n\"\n",
- "print \"\\n (b)Total charge \",(Qt/1E-3),\" mC\\n\"\n",
- "print \"\\n (c)Charge on each capacitors (C1, C2, C3, C4)\\n \",(Q1/1E-3),\", \",(Q2/1E-3),\", \"\n",
- "print \",(Q3/1E-3),\", \",(Q4/1E-3),\" mC respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "ename": "SyntaxError",
- "evalue": "invalid syntax (<ipython-input-1-0c99632e8192>, line 27)",
- "output_type": "pyerr",
- "traceback": [
- "\u001b[1;36m File \u001b[1;32m\"<ipython-input-1-0c99632e8192>\"\u001b[1;36m, line \u001b[1;32m27\u001b[0m\n\u001b[1;33m print \",(Q3/1E-3),\", \",(Q4/1E-3),\" mC respectively\\n\"\u001b[0m\n\u001b[1;37m ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 66</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the equivalent circuit capacitance, \n",
- "#(b) the charge on each capacitor and\n",
- "#(c) the pd across each capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 3E-6;# in Farads\n",
- "C2 = 6E-6;# in Farads\n",
- "C3 = 12E-6;# in Farads\n",
- "Vt = 350;# in Volts\n",
- "#calculation:\n",
- "# in series\n",
- "Cs = 1/((1/C1) + (1/C2) + (1/C3))\n",
- "Qt = Vt*Cs\n",
- "V1 = Qt/C1\n",
- "V2 = Qt/C2\n",
- "V3 = Qt/C3\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Equivalent Capacitance in Series \",(Cs/1E-6),\" uF\\n\"\n",
- "print \"\\n (b)Charge on each capacitors (C1, C2, C3) \",(Qt/1E-3),\" mC \\n\"\n",
- "print \"\\n (b)P.d Across each capacitors (C1, C2, C3)\\n \",V1,\" V, \", V2,\" V, \", V3,\" V respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Equivalent Capacitance in Series 1.71428571429 uF\n",
- "\n",
- "\n",
- " (b)Charge on each capacitors (C1, C2, C3) 0.6 mC \n",
- "\n",
- "\n",
- " (b)P.d Across each capacitors (C1, C2, C3)\n",
- " 200.0 V, 100.0 V, 50.0 V respectively"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 67</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the thickness of the mica needed, and \n",
- "#(b) the area of a plate assuming a two-plate construction.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 0.2E-6;# in Farads\n",
- "V = 1250;# in Volts\n",
- "E = 50E6# in V/m\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 6;\n",
- "\n",
- "#calculation:\n",
- "d = V/E\n",
- "A = C*d/e0/er\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Thickness \",(d/1E-3),\" mm\\n\"\n",
- "print \"\\n (b)Area of plate is \",round((A/1E-4),2),\" cm^2 \\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Thickness 0.025 mm\n",
- "\n",
- "\n",
- " (b)Area of plate is 941.62 cm^2 "
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 68</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the energy stored in a 3 \u03bcF capacitor when charged to 400 V.\n",
- "#(b) Find also the average power developed if this energy is dissipated in a time of 10 \u03bcs\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 3E-6;# in Farads\n",
- "V = 400;# in Volts\n",
- "t = 10E-6;# in secs\n",
- "\n",
- "#calculation:\n",
- "W = C*V*V/2\n",
- "P = W/t\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Energy stored \",W,\" J\\n\"\n",
- "print \"\\n (b)Power developed \",(P/1E3),\" kW \\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Energy stored 0.24 J\n",
- "\n",
- "\n",
- " (b)Power developed 24.0 kW "
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 68</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the pd to which the capacitor must be charged.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 12E-6;# in Farads\n",
- "W = 4;# in Joules\n",
- "\n",
- "#calculation:\n",
- "V = (2*W/C)**0.5\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n P.d \",round(V,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " P.d 816.5 V"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 68</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the voltage and (b) the capacitance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "W = 1.2;# in Joules\n",
- "Q = 10E-3;# in Coulomb\n",
- "\n",
- "#calculation:\n",
- "V = 2*W/Q\n",
- "C = Q/V\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)P.d \",V,\" V\\n\"\n",
- "print \"\\n (b)Capacitance \",round((C/1E-6),2),\" uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)P.d 240.0 V\n",
- "\n",
- "\n",
- " (b)Capacitance 41.67 uF"
- ]
- }
- ],
- "prompt_number": 19
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_1.ipynb deleted file mode 100755 index 772a8bc5..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_1.ipynb +++ /dev/null @@ -1,873 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 6: Capacitors and capacitance</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 58</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the p.d. across a 4 \u03bcF capacitor when charged with 5 mC.\n",
- "# (b) Find the charge on a 50 pF capacitor when the voltage applied to it is 2 kV.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 4E-6;# in Farad\n",
- "C2 = 50E-12;# in Farad\n",
- "Q1 = 5E-3;# in Coulomb\n",
- "V2 = 2000;# in volts\n",
- "\n",
- "#calculation:\n",
- "V1 = Q1/C1\n",
- "Q2 = C2*V2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)P.d \",V1,\" Volts(V)\\n\"\n",
- "print \"\\n (b)Charge(Q) \",(Q2/1E-6),\" micro-Coulomb\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)P.d 1250.0 Volts(V)\n",
- "\n",
- "\n",
- " (b)Charge(Q) 0.1 micro-Coulomb"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 58</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the pd between the plates.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 4;# in amperes\n",
- "C = 20E-6;# in Farad\n",
- "t = 3E-3;# in sec\n",
- "\n",
- "#calculation:\n",
- "Q = I*t\n",
- "V = Q/C\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)P.d \",V,\" Volts(V)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)P.d 600.0 Volts(V)"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 59</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate how long the capacitor can provide an average discharge current of 2 mA.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 2E-3;# in amperes\n",
- "C = 5E-6;# in Farad\n",
- "V = 800;# in volts\n",
- "\n",
- "#calculation:\n",
- "Q = C*V\n",
- "t = Q/I\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n capacitor can provide an average discharge current for \",t,\" Sec\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " capacitor can provide an average discharge current for 2.0 Sec"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 60</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the electric flux density.\n",
- "#determine the electric field strength.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Q = 0.2E-6;# in Coulomb\n",
- "A = 800E-4;# in m2\n",
- "d = 0.005;# in m\n",
- "V = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- "D = Q/A\n",
- "E = V/d\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Electric flux density D \",(D/1E-6),\" uC/m2\\n\"\n",
- "print \"\\n (b)Electric field strength E \",(E/1000),\" kV/m\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Electric flux density D 2.5 uC/m2\n",
- "\n",
- "\n",
- " (b)Electric field strength E 50.0 kV/m"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 60</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the voltage gradient between the plates.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "D = 2E-6;# in micro-C/m2\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 5;\n",
- "\n",
- "#calculation:\n",
- "E = D/(e0*er)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Electric field strength E \",round((E/1000),2),\" kV/m\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Electric field strength E 45.2 kV/m"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 60</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What is the electric field strength?\n",
- "#Find also the flux density when the dielectric between the plates is\n",
- "#(a) air, and (b) polythene of relative permittivity 2.3\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "d = 0.8E-3;# in m\n",
- "e0 = 8.85E-12;# in F/m\n",
- "era = 1;# for air\n",
- "erp = 2.3;# for polythene\n",
- "V =200;# in Volts\n",
- "\n",
- "#calculation:\n",
- "E = V/d\n",
- "#for air\n",
- "Da = E*e0*era\n",
- "#for polythene\n",
- "Dp = E*e0*erp\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Electric flux density for air \",round((Da/1E-6),2),\" micro-C/m2\\n\"\n",
- "print \"\\n (b)Electric flux density for polythene \",round((Dp/1E-6),2),\" micro-C/m2\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Electric flux density for air 2.21 micro-C/m2\n",
- "\n",
- "\n",
- " (b)Electric flux density for polythene 5.09 micro-C/m2"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 62</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)Calculate the capacitance of the capacitor in picofarads. \n",
- "#(b)what will be the pd between the plates?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "A = 4E-4;# in m2\n",
- "d = 0.1E-3;# in m\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 100;\n",
- "Q = 1.2E-6;# in coulomb\n",
- "\n",
- "#calculation:\n",
- "C = e0*er*A/d\n",
- "V = Q/C\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Capacitance \",(C/1E-12),\" pF\\n\"\n",
- "print \"\\n (b)P.d.= \",round(V,2),\" Volts(V)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Capacitance 3540.0 pF\n",
- "\n",
- "\n",
- " (b)P.d.= 338.98 Volts(V)"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 62</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the effective thickness of the paper if its relative permittivity is 2.5\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "A = 800E-4;# in m2\n",
- "C = 4425E-12;# in Farads\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 2.5;\n",
- "\n",
- "#calculation:\n",
- "d = e0*er*A/C\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Thickness \",(d/1E-3),\" mm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Thickness 0.4 mm"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 62</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the capacitance of the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "n = 19;# no. of plates\n",
- "L = 75E-3;# in m\n",
- "B = 75E-3;# in m\n",
- "d = 0.2E-3;# in m\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 5;\n",
- "#calculation:\n",
- "A = L*B\n",
- "C = e0*er*A*(n-1)/d\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Capacitance \",round((C/1E-9),2),\" nF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Capacitance 22.4 nF"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 65</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the equivalent capacitance of two capacitors of 6 \u03bcF and 4 \u03bcF connected (a) in parallel and (b) in series\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 6E-6;# in Farads\n",
- "C2 = 4E-6;# in Farads\n",
- "\n",
- "#calculation:\n",
- "# in Parallel\n",
- "Cp = C1 + C2\n",
- "# in Series\n",
- "Cs = 1/((1/C1) + (1/C2))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Capacitance in parallel \",(Cp/1E-6),\" uF\\n\"\n",
- "print \"\\n (b)Capacitance in Series \",(Cs/1E-6),\" uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Capacitance in parallel 10.0 uF\n",
- "\n",
- "\n",
- " (b)Capacitance in Series 2.4 uF"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 65</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What capacitance must be connected in series\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 30E-6;# in Farads\n",
- "Cs = 12E-6;# in Farads\n",
- "\n",
- "#calculation:\n",
- "# in Series\n",
- "C2 = 1/((1/Cs) - (1/C1))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Capacitance in series \",(C2/1E-6),\" uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Capacitance in series 20.0 uF"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 65</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the equivalent circuit capacitance, \n",
- "#(b) the total charge and\n",
- "#(c) the charge on each capacitor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 1E-6;# in Farads\n",
- "C2 = 3E-6;# in Farads\n",
- "C3 = 5E-6;# in Farads\n",
- "C4 = 6E-6;# in Farads\n",
- "Vt = 100;# in Volts\n",
- "\n",
- "#calculation:\n",
- "# in Parallel\n",
- "Cp = C1 + C2 + C3 + C4\n",
- "Qt = Vt*Cp\n",
- "Q1 = C1*Vt\n",
- "Q2 = C2*Vt\n",
- "Q3 = C3*Vt\n",
- "Q4 = C4*Vt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Equivalent Capacitance in Parallel \",(Cp/1E-6),\" uF\\n\"\n",
- "print \"\\n (b)Total charge \",(Qt/1E-3),\" mC\\n\"\n",
- "print \"\\n (c)Charge on each capacitors (C1, C2, C3, C4)\\n \",(Q1/1E-3),\", \",(Q2/1E-3),\", \"\n",
- "print \",(Q3/1E-3),\", \",(Q4/1E-3),\" mC respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "ename": "SyntaxError",
- "evalue": "invalid syntax (<ipython-input-1-0c99632e8192>, line 27)",
- "output_type": "pyerr",
- "traceback": [
- "\u001b[1;36m File \u001b[1;32m\"<ipython-input-1-0c99632e8192>\"\u001b[1;36m, line \u001b[1;32m27\u001b[0m\n\u001b[1;33m print \",(Q3/1E-3),\", \",(Q4/1E-3),\" mC respectively\\n\"\u001b[0m\n\u001b[1;37m ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 66</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the equivalent circuit capacitance, \n",
- "#(b) the charge on each capacitor and\n",
- "#(c) the pd across each capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 3E-6;# in Farads\n",
- "C2 = 6E-6;# in Farads\n",
- "C3 = 12E-6;# in Farads\n",
- "Vt = 350;# in Volts\n",
- "#calculation:\n",
- "# in series\n",
- "Cs = 1/((1/C1) + (1/C2) + (1/C3))\n",
- "Qt = Vt*Cs\n",
- "V1 = Qt/C1\n",
- "V2 = Qt/C2\n",
- "V3 = Qt/C3\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Equivalent Capacitance in Series \",(Cs/1E-6),\" uF\\n\"\n",
- "print \"\\n (b)Charge on each capacitors (C1, C2, C3) \",(Qt/1E-3),\" mC \\n\"\n",
- "print \"\\n (b)P.d Across each capacitors (C1, C2, C3)\\n \",V1,\" V, \", V2,\" V, \", V3,\" V respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Equivalent Capacitance in Series 1.71428571429 uF\n",
- "\n",
- "\n",
- " (b)Charge on each capacitors (C1, C2, C3) 0.6 mC \n",
- "\n",
- "\n",
- " (b)P.d Across each capacitors (C1, C2, C3)\n",
- " 200.0 V, 100.0 V, 50.0 V respectively"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 67</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the thickness of the mica needed, and \n",
- "#(b) the area of a plate assuming a two-plate construction.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 0.2E-6;# in Farads\n",
- "V = 1250;# in Volts\n",
- "E = 50E6# in V/m\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 6;\n",
- "\n",
- "#calculation:\n",
- "d = V/E\n",
- "A = C*d/e0/er\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Thickness \",(d/1E-3),\" mm\\n\"\n",
- "print \"\\n (b)Area of plate is \",round((A/1E-4),2),\" cm^2 \\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Thickness 0.025 mm\n",
- "\n",
- "\n",
- " (b)Area of plate is 941.62 cm^2 "
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 68</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the energy stored in a 3 \u03bcF capacitor when charged to 400 V.\n",
- "#(b) Find also the average power developed if this energy is dissipated in a time of 10 \u03bcs\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 3E-6;# in Farads\n",
- "V = 400;# in Volts\n",
- "t = 10E-6;# in secs\n",
- "\n",
- "#calculation:\n",
- "W = C*V*V/2\n",
- "P = W/t\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Energy stored \",W,\" J\\n\"\n",
- "print \"\\n (b)Power developed \",(P/1E3),\" kW \\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Energy stored 0.24 J\n",
- "\n",
- "\n",
- " (b)Power developed 24.0 kW "
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 68</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the pd to which the capacitor must be charged.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 12E-6;# in Farads\n",
- "W = 4;# in Joules\n",
- "\n",
- "#calculation:\n",
- "V = (2*W/C)**0.5\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n P.d \",round(V,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " P.d 816.5 V"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 68</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the voltage and (b) the capacitance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "W = 1.2;# in Joules\n",
- "Q = 10E-3;# in Coulomb\n",
- "\n",
- "#calculation:\n",
- "V = 2*W/Q\n",
- "C = Q/V\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)P.d \",V,\" V\\n\"\n",
- "print \"\\n (b)Capacitance \",round((C/1E-6),2),\" uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)P.d 240.0 V\n",
- "\n",
- "\n",
- " (b)Capacitance 41.67 uF"
- ]
- }
- ],
- "prompt_number": 19
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_2.ipynb deleted file mode 100755 index 772a8bc5..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_2.ipynb +++ /dev/null @@ -1,873 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 6: Capacitors and capacitance</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 58</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the p.d. across a 4 \u03bcF capacitor when charged with 5 mC.\n",
- "# (b) Find the charge on a 50 pF capacitor when the voltage applied to it is 2 kV.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 4E-6;# in Farad\n",
- "C2 = 50E-12;# in Farad\n",
- "Q1 = 5E-3;# in Coulomb\n",
- "V2 = 2000;# in volts\n",
- "\n",
- "#calculation:\n",
- "V1 = Q1/C1\n",
- "Q2 = C2*V2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)P.d \",V1,\" Volts(V)\\n\"\n",
- "print \"\\n (b)Charge(Q) \",(Q2/1E-6),\" micro-Coulomb\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)P.d 1250.0 Volts(V)\n",
- "\n",
- "\n",
- " (b)Charge(Q) 0.1 micro-Coulomb"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 58</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the pd between the plates.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 4;# in amperes\n",
- "C = 20E-6;# in Farad\n",
- "t = 3E-3;# in sec\n",
- "\n",
- "#calculation:\n",
- "Q = I*t\n",
- "V = Q/C\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)P.d \",V,\" Volts(V)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)P.d 600.0 Volts(V)"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 59</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate how long the capacitor can provide an average discharge current of 2 mA.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 2E-3;# in amperes\n",
- "C = 5E-6;# in Farad\n",
- "V = 800;# in volts\n",
- "\n",
- "#calculation:\n",
- "Q = C*V\n",
- "t = Q/I\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n capacitor can provide an average discharge current for \",t,\" Sec\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " capacitor can provide an average discharge current for 2.0 Sec"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 60</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the electric flux density.\n",
- "#determine the electric field strength.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Q = 0.2E-6;# in Coulomb\n",
- "A = 800E-4;# in m2\n",
- "d = 0.005;# in m\n",
- "V = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- "D = Q/A\n",
- "E = V/d\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Electric flux density D \",(D/1E-6),\" uC/m2\\n\"\n",
- "print \"\\n (b)Electric field strength E \",(E/1000),\" kV/m\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Electric flux density D 2.5 uC/m2\n",
- "\n",
- "\n",
- " (b)Electric field strength E 50.0 kV/m"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 60</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the voltage gradient between the plates.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "D = 2E-6;# in micro-C/m2\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 5;\n",
- "\n",
- "#calculation:\n",
- "E = D/(e0*er)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Electric field strength E \",round((E/1000),2),\" kV/m\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Electric field strength E 45.2 kV/m"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 60</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What is the electric field strength?\n",
- "#Find also the flux density when the dielectric between the plates is\n",
- "#(a) air, and (b) polythene of relative permittivity 2.3\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "d = 0.8E-3;# in m\n",
- "e0 = 8.85E-12;# in F/m\n",
- "era = 1;# for air\n",
- "erp = 2.3;# for polythene\n",
- "V =200;# in Volts\n",
- "\n",
- "#calculation:\n",
- "E = V/d\n",
- "#for air\n",
- "Da = E*e0*era\n",
- "#for polythene\n",
- "Dp = E*e0*erp\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Electric flux density for air \",round((Da/1E-6),2),\" micro-C/m2\\n\"\n",
- "print \"\\n (b)Electric flux density for polythene \",round((Dp/1E-6),2),\" micro-C/m2\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Electric flux density for air 2.21 micro-C/m2\n",
- "\n",
- "\n",
- " (b)Electric flux density for polythene 5.09 micro-C/m2"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 62</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)Calculate the capacitance of the capacitor in picofarads. \n",
- "#(b)what will be the pd between the plates?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "A = 4E-4;# in m2\n",
- "d = 0.1E-3;# in m\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 100;\n",
- "Q = 1.2E-6;# in coulomb\n",
- "\n",
- "#calculation:\n",
- "C = e0*er*A/d\n",
- "V = Q/C\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Capacitance \",(C/1E-12),\" pF\\n\"\n",
- "print \"\\n (b)P.d.= \",round(V,2),\" Volts(V)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Capacitance 3540.0 pF\n",
- "\n",
- "\n",
- " (b)P.d.= 338.98 Volts(V)"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 62</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the effective thickness of the paper if its relative permittivity is 2.5\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "A = 800E-4;# in m2\n",
- "C = 4425E-12;# in Farads\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 2.5;\n",
- "\n",
- "#calculation:\n",
- "d = e0*er*A/C\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Thickness \",(d/1E-3),\" mm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Thickness 0.4 mm"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 62</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the capacitance of the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "n = 19;# no. of plates\n",
- "L = 75E-3;# in m\n",
- "B = 75E-3;# in m\n",
- "d = 0.2E-3;# in m\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 5;\n",
- "#calculation:\n",
- "A = L*B\n",
- "C = e0*er*A*(n-1)/d\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Capacitance \",round((C/1E-9),2),\" nF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Capacitance 22.4 nF"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 65</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the equivalent capacitance of two capacitors of 6 \u03bcF and 4 \u03bcF connected (a) in parallel and (b) in series\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 6E-6;# in Farads\n",
- "C2 = 4E-6;# in Farads\n",
- "\n",
- "#calculation:\n",
- "# in Parallel\n",
- "Cp = C1 + C2\n",
- "# in Series\n",
- "Cs = 1/((1/C1) + (1/C2))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Capacitance in parallel \",(Cp/1E-6),\" uF\\n\"\n",
- "print \"\\n (b)Capacitance in Series \",(Cs/1E-6),\" uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Capacitance in parallel 10.0 uF\n",
- "\n",
- "\n",
- " (b)Capacitance in Series 2.4 uF"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 65</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What capacitance must be connected in series\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 30E-6;# in Farads\n",
- "Cs = 12E-6;# in Farads\n",
- "\n",
- "#calculation:\n",
- "# in Series\n",
- "C2 = 1/((1/Cs) - (1/C1))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Capacitance in series \",(C2/1E-6),\" uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Capacitance in series 20.0 uF"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 65</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the equivalent circuit capacitance, \n",
- "#(b) the total charge and\n",
- "#(c) the charge on each capacitor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 1E-6;# in Farads\n",
- "C2 = 3E-6;# in Farads\n",
- "C3 = 5E-6;# in Farads\n",
- "C4 = 6E-6;# in Farads\n",
- "Vt = 100;# in Volts\n",
- "\n",
- "#calculation:\n",
- "# in Parallel\n",
- "Cp = C1 + C2 + C3 + C4\n",
- "Qt = Vt*Cp\n",
- "Q1 = C1*Vt\n",
- "Q2 = C2*Vt\n",
- "Q3 = C3*Vt\n",
- "Q4 = C4*Vt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Equivalent Capacitance in Parallel \",(Cp/1E-6),\" uF\\n\"\n",
- "print \"\\n (b)Total charge \",(Qt/1E-3),\" mC\\n\"\n",
- "print \"\\n (c)Charge on each capacitors (C1, C2, C3, C4)\\n \",(Q1/1E-3),\", \",(Q2/1E-3),\", \"\n",
- "print \",(Q3/1E-3),\", \",(Q4/1E-3),\" mC respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "ename": "SyntaxError",
- "evalue": "invalid syntax (<ipython-input-1-0c99632e8192>, line 27)",
- "output_type": "pyerr",
- "traceback": [
- "\u001b[1;36m File \u001b[1;32m\"<ipython-input-1-0c99632e8192>\"\u001b[1;36m, line \u001b[1;32m27\u001b[0m\n\u001b[1;33m print \",(Q3/1E-3),\", \",(Q4/1E-3),\" mC respectively\\n\"\u001b[0m\n\u001b[1;37m ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 66</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the equivalent circuit capacitance, \n",
- "#(b) the charge on each capacitor and\n",
- "#(c) the pd across each capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C1 = 3E-6;# in Farads\n",
- "C2 = 6E-6;# in Farads\n",
- "C3 = 12E-6;# in Farads\n",
- "Vt = 350;# in Volts\n",
- "#calculation:\n",
- "# in series\n",
- "Cs = 1/((1/C1) + (1/C2) + (1/C3))\n",
- "Qt = Vt*Cs\n",
- "V1 = Qt/C1\n",
- "V2 = Qt/C2\n",
- "V3 = Qt/C3\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Equivalent Capacitance in Series \",(Cs/1E-6),\" uF\\n\"\n",
- "print \"\\n (b)Charge on each capacitors (C1, C2, C3) \",(Qt/1E-3),\" mC \\n\"\n",
- "print \"\\n (b)P.d Across each capacitors (C1, C2, C3)\\n \",V1,\" V, \", V2,\" V, \", V3,\" V respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Equivalent Capacitance in Series 1.71428571429 uF\n",
- "\n",
- "\n",
- " (b)Charge on each capacitors (C1, C2, C3) 0.6 mC \n",
- "\n",
- "\n",
- " (b)P.d Across each capacitors (C1, C2, C3)\n",
- " 200.0 V, 100.0 V, 50.0 V respectively"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 67</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the thickness of the mica needed, and \n",
- "#(b) the area of a plate assuming a two-plate construction.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 0.2E-6;# in Farads\n",
- "V = 1250;# in Volts\n",
- "E = 50E6# in V/m\n",
- "e0 = 8.85E-12;# in F/m\n",
- "er = 6;\n",
- "\n",
- "#calculation:\n",
- "d = V/E\n",
- "A = C*d/e0/er\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Thickness \",(d/1E-3),\" mm\\n\"\n",
- "print \"\\n (b)Area of plate is \",round((A/1E-4),2),\" cm^2 \\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Thickness 0.025 mm\n",
- "\n",
- "\n",
- " (b)Area of plate is 941.62 cm^2 "
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 68</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the energy stored in a 3 \u03bcF capacitor when charged to 400 V.\n",
- "#(b) Find also the average power developed if this energy is dissipated in a time of 10 \u03bcs\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 3E-6;# in Farads\n",
- "V = 400;# in Volts\n",
- "t = 10E-6;# in secs\n",
- "\n",
- "#calculation:\n",
- "W = C*V*V/2\n",
- "P = W/t\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Energy stored \",W,\" J\\n\"\n",
- "print \"\\n (b)Power developed \",(P/1E3),\" kW \\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Energy stored 0.24 J\n",
- "\n",
- "\n",
- " (b)Power developed 24.0 kW "
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 68</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the pd to which the capacitor must be charged.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 12E-6;# in Farads\n",
- "W = 4;# in Joules\n",
- "\n",
- "#calculation:\n",
- "V = (2*W/C)**0.5\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n P.d \",round(V,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " P.d 816.5 V"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 68</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the voltage and (b) the capacitance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "W = 1.2;# in Joules\n",
- "Q = 10E-3;# in Coulomb\n",
- "\n",
- "#calculation:\n",
- "V = 2*W/Q\n",
- "C = Q/V\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)P.d \",V,\" V\\n\"\n",
- "print \"\\n (b)Capacitance \",round((C/1E-6),2),\" uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)P.d 240.0 V\n",
- "\n",
- "\n",
- " (b)Capacitance 41.67 uF"
- ]
- }
- ],
- "prompt_number": 19
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_06.ipynb index 72813a43..72813a43 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_06.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_07.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_07.ipynb index e3a59d72..e3a59d72 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_07.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_07.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint.ipynb deleted file mode 100755 index 1d749dd1..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint.ipynb +++ /dev/null @@ -1,273 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 8: Electromagnetism</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 93</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the force acting on the conductor.\n",
- "#Determine also the value of the force\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "B = 0.9;# in tesla\n",
- "I = 20;# in Amperes\n",
- "l = 0.30;# in m\n",
- "alpha = 30;# in degree\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "F1 = B*I*l\n",
- "F2 = B*I*l*math.sin(alpha*math.pi/180)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Force when the conductor is at right angles to the field = \",F1,\" N\\n\"\n",
- "print \"\\n (b)Force when the conductor is at 30\u00b0 angle to the field = \",F2,\" N\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Force when the conductor is at right angles to the field = 5.4 N\n",
- "\n",
- "\n",
- " (b)Force when the conductor is at 30\u00c2\u00b0 angle to the field = 2.7 N"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 94</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current required?\n",
- "#what is the direction of the force?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "F = 1.92;# in newton\n",
- "B = 1.2;# in tesla\n",
- "l = 0.40;# in m\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "I = F/(B*l)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Current I = \",I,\" Amperes(A)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Current I = 4.0 Amperes(A)"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 95</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the magnitude of the force exerted on the conductor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "r = 0.06;# in m\n",
- "I = 10;# in Amperes\n",
- "l = 0.35;# in m\n",
- "Phi = 0.5E-3;# in Wb\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "A = math.pi*r*r\n",
- "B = Phi/A\n",
- "F = B*I*l\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Force F = \",round(F,2),\" N\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Force F = 0.15 N"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 95</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the force on each coil side (a) for a single-turn coil,\n",
- "#(b) for a coil wound with 300 turns.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 1;# for a single-turn coil\n",
- "N2 = 300;# no. of turns\n",
- "b = 0.024;# in m\n",
- "B = 0.8;# in Tesla\n",
- "I = 0.05;# in Amperes\n",
- "l = 0.030;# in m\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "#For a single-turn coil,\n",
- "F1 = N1*B*I*l\n",
- "#for a coil wound with 300 turns.\n",
- "F2 = N2*B*I*l\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)For a single-turn coil, force on each coil side = \",F1,\" N\\n\"\n",
- "print \"\\n (b)For a 300-turn coil, force on each coil side = \",F2,\" N\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)For a single-turn coil, force on each coil side = 0.0012 N\n",
- "\n",
- "\n",
- " (b)For a 300-turn coil, force on each coil side = 0.36 N"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 98</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the force exerted on the electron in the field.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Q = 1.6E-19;# in Coulomb\n",
- "v = 3E7;# in m/s\n",
- "B = 18.5E-6;# in Tesla\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "F = Q*v*B\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Force exerted on the electron in the field. = \",(F/1E-17),\"E-17 N\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Force exerted on the electron in the field. = 8.88 E-17 N"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_1.ipynb deleted file mode 100755 index 1d749dd1..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_1.ipynb +++ /dev/null @@ -1,273 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 8: Electromagnetism</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 93</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the force acting on the conductor.\n",
- "#Determine also the value of the force\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "B = 0.9;# in tesla\n",
- "I = 20;# in Amperes\n",
- "l = 0.30;# in m\n",
- "alpha = 30;# in degree\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "F1 = B*I*l\n",
- "F2 = B*I*l*math.sin(alpha*math.pi/180)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Force when the conductor is at right angles to the field = \",F1,\" N\\n\"\n",
- "print \"\\n (b)Force when the conductor is at 30\u00b0 angle to the field = \",F2,\" N\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Force when the conductor is at right angles to the field = 5.4 N\n",
- "\n",
- "\n",
- " (b)Force when the conductor is at 30\u00c2\u00b0 angle to the field = 2.7 N"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 94</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current required?\n",
- "#what is the direction of the force?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "F = 1.92;# in newton\n",
- "B = 1.2;# in tesla\n",
- "l = 0.40;# in m\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "I = F/(B*l)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Current I = \",I,\" Amperes(A)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Current I = 4.0 Amperes(A)"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 95</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the magnitude of the force exerted on the conductor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "r = 0.06;# in m\n",
- "I = 10;# in Amperes\n",
- "l = 0.35;# in m\n",
- "Phi = 0.5E-3;# in Wb\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "A = math.pi*r*r\n",
- "B = Phi/A\n",
- "F = B*I*l\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Force F = \",round(F,2),\" N\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Force F = 0.15 N"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 95</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the force on each coil side (a) for a single-turn coil,\n",
- "#(b) for a coil wound with 300 turns.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 1;# for a single-turn coil\n",
- "N2 = 300;# no. of turns\n",
- "b = 0.024;# in m\n",
- "B = 0.8;# in Tesla\n",
- "I = 0.05;# in Amperes\n",
- "l = 0.030;# in m\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "#For a single-turn coil,\n",
- "F1 = N1*B*I*l\n",
- "#for a coil wound with 300 turns.\n",
- "F2 = N2*B*I*l\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)For a single-turn coil, force on each coil side = \",F1,\" N\\n\"\n",
- "print \"\\n (b)For a 300-turn coil, force on each coil side = \",F2,\" N\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)For a single-turn coil, force on each coil side = 0.0012 N\n",
- "\n",
- "\n",
- " (b)For a 300-turn coil, force on each coil side = 0.36 N"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 98</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the force exerted on the electron in the field.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Q = 1.6E-19;# in Coulomb\n",
- "v = 3E7;# in m/s\n",
- "B = 18.5E-6;# in Tesla\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "F = Q*v*B\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Force exerted on the electron in the field. = \",(F/1E-17),\"E-17 N\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Force exerted on the electron in the field. = 8.88 E-17 N"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_2.ipynb deleted file mode 100755 index 1d749dd1..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_2.ipynb +++ /dev/null @@ -1,273 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 8: Electromagnetism</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 93</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the force acting on the conductor.\n",
- "#Determine also the value of the force\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "B = 0.9;# in tesla\n",
- "I = 20;# in Amperes\n",
- "l = 0.30;# in m\n",
- "alpha = 30;# in degree\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "F1 = B*I*l\n",
- "F2 = B*I*l*math.sin(alpha*math.pi/180)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Force when the conductor is at right angles to the field = \",F1,\" N\\n\"\n",
- "print \"\\n (b)Force when the conductor is at 30\u00b0 angle to the field = \",F2,\" N\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Force when the conductor is at right angles to the field = 5.4 N\n",
- "\n",
- "\n",
- " (b)Force when the conductor is at 30\u00c2\u00b0 angle to the field = 2.7 N"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 94</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current required?\n",
- "#what is the direction of the force?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "F = 1.92;# in newton\n",
- "B = 1.2;# in tesla\n",
- "l = 0.40;# in m\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "I = F/(B*l)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Current I = \",I,\" Amperes(A)\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Current I = 4.0 Amperes(A)"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 95</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the magnitude of the force exerted on the conductor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "r = 0.06;# in m\n",
- "I = 10;# in Amperes\n",
- "l = 0.35;# in m\n",
- "Phi = 0.5E-3;# in Wb\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "A = math.pi*r*r\n",
- "B = Phi/A\n",
- "F = B*I*l\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)Force F = \",round(F,2),\" N\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)Force F = 0.15 N"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 95</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the force on each coil side (a) for a single-turn coil,\n",
- "#(b) for a coil wound with 300 turns.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 1;# for a single-turn coil\n",
- "N2 = 300;# no. of turns\n",
- "b = 0.024;# in m\n",
- "B = 0.8;# in Tesla\n",
- "I = 0.05;# in Amperes\n",
- "l = 0.030;# in m\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "#For a single-turn coil,\n",
- "F1 = N1*B*I*l\n",
- "#for a coil wound with 300 turns.\n",
- "F2 = N2*B*I*l\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)For a single-turn coil, force on each coil side = \",F1,\" N\\n\"\n",
- "print \"\\n (b)For a 300-turn coil, force on each coil side = \",F2,\" N\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)For a single-turn coil, force on each coil side = 0.0012 N\n",
- "\n",
- "\n",
- " (b)For a 300-turn coil, force on each coil side = 0.36 N"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 98</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the force exerted on the electron in the field.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Q = 1.6E-19;# in Coulomb\n",
- "v = 3E7;# in m/s\n",
- "B = 18.5E-6;# in Tesla\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "F = Q*v*B\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Force exerted on the electron in the field. = \",(F/1E-17),\"E-17 N\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Force exerted on the electron in the field. = 8.88 E-17 N"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_08.ipynb index 876584cc..876584cc 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_08.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint.ipynb deleted file mode 100755 index 02f3e6f4..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint.ipynb +++ /dev/null @@ -1,759 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 9: Electromagnetic induction</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 102</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the conductor when \n",
- "#(a) its ends are open-circuited, \n",
- "#(b) its ends are connected to a load of 20 ohm resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "l = 0.3;# in m\n",
- "v = 4;# in m/s\n",
- "B = 1.25;# in Tesla\n",
- "R = 20;# in ohms\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "E = B*l*v\n",
- "I2 = E/R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)If the ends of the conductor are open circuited \"\n",
- "print \"no current will flow even though \",E,\" V has been induced.\\n\"\n",
- "print \"\\n (b)From Ohms law, I = \",I2,\" Ampere\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)If the ends of the conductor are open circuited no current will flow even though 1.5 V has been induced.\n",
- "\n",
- "\n",
- " (b)From Ohms law, I = 0.075 Ampere"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 103</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#At what velocity must a conductor 75 mm long cut a magnetic field of flux density 0.6 T\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "l = 0.075;# in m\n",
- "E = 9;# in Volts\n",
- "B = 0.6;# in Tesla\n",
- "R = 20;# in ohms\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "v = E/(B*l)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n velocity v = \",v,\" m/s\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " velocity v = 200.0 m/s"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 103</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the magnitude of the induced e.m.f. in each case.\n",
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#initializing the variables:\n",
- "l = 0.02;# in m\n",
- "b = 0.02;# in m\n",
- "v = 15;# in m/s\n",
- "R = 20;# in ohms\n",
- "Phi = 5E-6;# in Wb\n",
- "u0 = 4*math.pi*1E-7;\n",
- "a1 = 90;# in degrees\n",
- "a2 = 60;# in degrees\n",
- "a3 = 30;# in degrees\n",
- "\n",
- "#calculation:\n",
- "A = l*b\n",
- "B = Phi/A\n",
- "E90 = B*l*v*math.sin(a1*math.pi/180)\n",
- "E60 = B*l*v*math.sin(a2*math.pi/180)\n",
- "E30 = B*l*v*math.sin(a3*math.pi/180)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. at angles 90\u00b0, 60\u00b0, 30\u00b0 are \",(E90/1E-3),\" V, \",round((E60/1E-3),2),\" V, \"\n",
- "print \"(E30/1E-3),\" V respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. at angles 90\u00c2\u00b0, 60\u00c2\u00b0, 30\u00c2\u00b0 are 3.75 V, 3.25 V, 1.875 V respectively"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 103</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the e.m.f. induced between its wing tips\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "s = 36;# in m\n",
- "v = 400;# in km/h\n",
- "u0 = 4*math.pi*1E-7;\n",
- "B = 40E-6;# in Tesla\n",
- "\n",
- "#calculation:\n",
- "v0 = v*5/18\n",
- "E = B*s*v0\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. = 0.16 V"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 105</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. induced in a coil\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N = 200;# no. of turns\n",
- "dt = 0.050;# change of time in sec\n",
- "u0 = 4*math.pi*1E-7;\n",
- "dPhi = 0.025;# change of flux in Wb\n",
- "\n",
- "#calculation:\n",
- "E = -1*N*dPhi/dt\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. = -100.0 V"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 105</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the average e.m.f. induced.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N = 150;# no. of turns\n",
- "dt = 0.040;# change of time in sec\n",
- "u0 = 4*math.pi*1E-7;\n",
- "dPhi = 800E-6;# change of flux in Wb\n",
- "\n",
- "#calculation:\n",
- "#Since the flux reverses, the flux changes from C400 \u03bcWb to \u0003400 \u03bcWb, a total change of flux of 800 \u03bcWb\n",
- "E = -1*N*dPhi/dt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. = -3.0 V"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 105</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the e.m.f. induced in a coil\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 12;# in Henry\n",
- "u0 = 4*math.pi*1E-7;\n",
- "dIdt = 4;# change of current with change in time in A/s\n",
- "\n",
- "#calculation:\n",
- "E = -1*L*dIdt\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. = -48 V"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 106</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find inductance of the coil.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E = 1500;# in Volts\n",
- "dt = 0.008;# Change of time in sec\n",
- "dI = 4;# change of current in A/s\n",
- "\n",
- "#calculation:\n",
- "L = abs(E)*dt/dI\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Inductance L= \",L,\" H\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Inductance L= 3.0 H"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 107</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#How much energy is stored in the magnetic field of the inductor?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 8;# in Henry\n",
- "I = 3;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "W = L*I*I/2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Energy stored, W = \",W,\" J\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Energy stored, W = 36.0 J"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 107</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the coil inductance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 4;# in Amperes\n",
- "N = 800;#turns\n",
- "Phi = 0.005;# in Wb\n",
- "\n",
- "#calculation:\n",
- "L = N*Phi/I\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Inductance L = \",L,\" H\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Inductance L = 1.0 H"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 107</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the inductance of the coil, \n",
- "#(b) the energy stored in the magnetic field, and \n",
- "#(c) the average e.m.f. induced if the current falls to zero in 150 ms.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I1 = 3;# in Amperes\n",
- "I2 = 0;# in Amperes\n",
- "dt = 0.150;# in secs\n",
- "N = 1500;#turns\n",
- "Phi = 0.025;# in Wb\n",
- "\n",
- "#calculation:\n",
- "L = N*Phi/I1\n",
- "W = L*I1*I1/2\n",
- "dI = I1 - I2\n",
- "E = -1*L*dI/dt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductance L = \",L,\" H\\n\"\n",
- "print \"\\n (b)energy stored W = \",W,\" J\\n\"\n",
- "print \"\\n (c)e.m.f. induced = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductance L = 12.5 H\n",
- "\n",
- "\n",
- " (b)energy stored W = 56.25 J\n",
- "\n",
- "\n",
- " (c)e.m.f. induced = -250.0 V"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 108</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the flux linking the coil and the e.m.f. induced in the coil when the current collapses to zero in 20 ms\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I1 = 2;# in Amperes\n",
- "I2 = 0;# in Amperes\n",
- "dt = 0.020;# in secs\n",
- "N = 750;#turns\n",
- "L = 3;# in Henry\n",
- "\n",
- "#calculation:\n",
- "Phi = L*I1/N\n",
- "dI = I1 - I2\n",
- "E = -1*L*dI/dt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Flux = \",Phi,\" Wb\\n\"\n",
- "print \"\\n (b)e.m.f. induced = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Flux = 0.008 Wb\n",
- "\n",
- "\n",
- " (b)e.m.f. induced = -300.0 V"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 108</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the mutual inductance between two coils\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "dI1dt = 200;# change of current with change in time in A/s\n",
- "N = 2;# no. of coils\n",
- "E2 = 1.5;# in Volts\n",
- "\n",
- "#calculation:\n",
- "M = abs(E2)/dI1dt\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance, M = \", M,\" H\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance, M = 0.0075 H"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 109</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the steady rate of change of current in one coil to induce an e.m.f. of 0.72 V in the other.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "M = 0.018;# in Henry\n",
- "N = 2;# no. of coils\n",
- "E2 = 0.72;# in Volts\n",
- "\n",
- "#calculation:\n",
- "dI1dt = abs(E2)/M\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n rate of change of current dI1/dt = \", dI1dt,\" A/s\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " rate of change of current dI1/dt = 40.0 A/s"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 109</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate (a) the average induced e.m.f. in the second coil, \n",
- "#(b) the change of flux linked with the second coil if it is wound with 500 turns.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "M = 0.2;# in Henry\n",
- "I1 = 10;# in Amperes\n",
- "I2 = 4;# in Amperes\n",
- "dt = 0.010;# in secs\n",
- "N = 500;# turns\n",
- "\n",
- "#calculation:\n",
- "dI1dt = (I1 -I2)/dt \n",
- "E2 = -1*dI1dt*M\n",
- "dPhi = abs(E2)*dt/N\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Induced e.m.f. E2 = \", E2,\" V\\n\"\n",
- "print \"\\n (b)change of flux = \", dPhi,\" Wb\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Induced e.m.f. E2 = -120.0 V\n",
- "\n",
- "\n",
- " (b)change of flux = 0.0024 Wb"
- ]
- }
- ],
- "prompt_number": 15
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_1.ipynb deleted file mode 100755 index 02f3e6f4..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_1.ipynb +++ /dev/null @@ -1,759 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 9: Electromagnetic induction</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 102</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the conductor when \n",
- "#(a) its ends are open-circuited, \n",
- "#(b) its ends are connected to a load of 20 ohm resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "l = 0.3;# in m\n",
- "v = 4;# in m/s\n",
- "B = 1.25;# in Tesla\n",
- "R = 20;# in ohms\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "E = B*l*v\n",
- "I2 = E/R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)If the ends of the conductor are open circuited \"\n",
- "print \"no current will flow even though \",E,\" V has been induced.\\n\"\n",
- "print \"\\n (b)From Ohms law, I = \",I2,\" Ampere\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)If the ends of the conductor are open circuited no current will flow even though 1.5 V has been induced.\n",
- "\n",
- "\n",
- " (b)From Ohms law, I = 0.075 Ampere"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 103</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#At what velocity must a conductor 75 mm long cut a magnetic field of flux density 0.6 T\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "l = 0.075;# in m\n",
- "E = 9;# in Volts\n",
- "B = 0.6;# in Tesla\n",
- "R = 20;# in ohms\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "v = E/(B*l)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n velocity v = \",v,\" m/s\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " velocity v = 200.0 m/s"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 103</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the magnitude of the induced e.m.f. in each case.\n",
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#initializing the variables:\n",
- "l = 0.02;# in m\n",
- "b = 0.02;# in m\n",
- "v = 15;# in m/s\n",
- "R = 20;# in ohms\n",
- "Phi = 5E-6;# in Wb\n",
- "u0 = 4*math.pi*1E-7;\n",
- "a1 = 90;# in degrees\n",
- "a2 = 60;# in degrees\n",
- "a3 = 30;# in degrees\n",
- "\n",
- "#calculation:\n",
- "A = l*b\n",
- "B = Phi/A\n",
- "E90 = B*l*v*math.sin(a1*math.pi/180)\n",
- "E60 = B*l*v*math.sin(a2*math.pi/180)\n",
- "E30 = B*l*v*math.sin(a3*math.pi/180)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. at angles 90\u00b0, 60\u00b0, 30\u00b0 are \",(E90/1E-3),\" V, \",round((E60/1E-3),2),\" V, \"\n",
- "print \"(E30/1E-3),\" V respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. at angles 90\u00c2\u00b0, 60\u00c2\u00b0, 30\u00c2\u00b0 are 3.75 V, 3.25 V, 1.875 V respectively"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 103</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the e.m.f. induced between its wing tips\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "s = 36;# in m\n",
- "v = 400;# in km/h\n",
- "u0 = 4*math.pi*1E-7;\n",
- "B = 40E-6;# in Tesla\n",
- "\n",
- "#calculation:\n",
- "v0 = v*5/18\n",
- "E = B*s*v0\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. = 0.16 V"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 105</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. induced in a coil\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N = 200;# no. of turns\n",
- "dt = 0.050;# change of time in sec\n",
- "u0 = 4*math.pi*1E-7;\n",
- "dPhi = 0.025;# change of flux in Wb\n",
- "\n",
- "#calculation:\n",
- "E = -1*N*dPhi/dt\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. = -100.0 V"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 105</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the average e.m.f. induced.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N = 150;# no. of turns\n",
- "dt = 0.040;# change of time in sec\n",
- "u0 = 4*math.pi*1E-7;\n",
- "dPhi = 800E-6;# change of flux in Wb\n",
- "\n",
- "#calculation:\n",
- "#Since the flux reverses, the flux changes from C400 \u03bcWb to \u0003400 \u03bcWb, a total change of flux of 800 \u03bcWb\n",
- "E = -1*N*dPhi/dt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. = -3.0 V"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 105</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the e.m.f. induced in a coil\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 12;# in Henry\n",
- "u0 = 4*math.pi*1E-7;\n",
- "dIdt = 4;# change of current with change in time in A/s\n",
- "\n",
- "#calculation:\n",
- "E = -1*L*dIdt\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. = -48 V"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 106</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find inductance of the coil.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E = 1500;# in Volts\n",
- "dt = 0.008;# Change of time in sec\n",
- "dI = 4;# change of current in A/s\n",
- "\n",
- "#calculation:\n",
- "L = abs(E)*dt/dI\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Inductance L= \",L,\" H\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Inductance L= 3.0 H"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 107</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#How much energy is stored in the magnetic field of the inductor?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 8;# in Henry\n",
- "I = 3;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "W = L*I*I/2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Energy stored, W = \",W,\" J\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Energy stored, W = 36.0 J"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 107</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the coil inductance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 4;# in Amperes\n",
- "N = 800;#turns\n",
- "Phi = 0.005;# in Wb\n",
- "\n",
- "#calculation:\n",
- "L = N*Phi/I\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Inductance L = \",L,\" H\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Inductance L = 1.0 H"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 107</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the inductance of the coil, \n",
- "#(b) the energy stored in the magnetic field, and \n",
- "#(c) the average e.m.f. induced if the current falls to zero in 150 ms.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I1 = 3;# in Amperes\n",
- "I2 = 0;# in Amperes\n",
- "dt = 0.150;# in secs\n",
- "N = 1500;#turns\n",
- "Phi = 0.025;# in Wb\n",
- "\n",
- "#calculation:\n",
- "L = N*Phi/I1\n",
- "W = L*I1*I1/2\n",
- "dI = I1 - I2\n",
- "E = -1*L*dI/dt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductance L = \",L,\" H\\n\"\n",
- "print \"\\n (b)energy stored W = \",W,\" J\\n\"\n",
- "print \"\\n (c)e.m.f. induced = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductance L = 12.5 H\n",
- "\n",
- "\n",
- " (b)energy stored W = 56.25 J\n",
- "\n",
- "\n",
- " (c)e.m.f. induced = -250.0 V"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 108</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the flux linking the coil and the e.m.f. induced in the coil when the current collapses to zero in 20 ms\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I1 = 2;# in Amperes\n",
- "I2 = 0;# in Amperes\n",
- "dt = 0.020;# in secs\n",
- "N = 750;#turns\n",
- "L = 3;# in Henry\n",
- "\n",
- "#calculation:\n",
- "Phi = L*I1/N\n",
- "dI = I1 - I2\n",
- "E = -1*L*dI/dt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Flux = \",Phi,\" Wb\\n\"\n",
- "print \"\\n (b)e.m.f. induced = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Flux = 0.008 Wb\n",
- "\n",
- "\n",
- " (b)e.m.f. induced = -300.0 V"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 108</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the mutual inductance between two coils\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "dI1dt = 200;# change of current with change in time in A/s\n",
- "N = 2;# no. of coils\n",
- "E2 = 1.5;# in Volts\n",
- "\n",
- "#calculation:\n",
- "M = abs(E2)/dI1dt\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance, M = \", M,\" H\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance, M = 0.0075 H"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 109</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the steady rate of change of current in one coil to induce an e.m.f. of 0.72 V in the other.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "M = 0.018;# in Henry\n",
- "N = 2;# no. of coils\n",
- "E2 = 0.72;# in Volts\n",
- "\n",
- "#calculation:\n",
- "dI1dt = abs(E2)/M\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n rate of change of current dI1/dt = \", dI1dt,\" A/s\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " rate of change of current dI1/dt = 40.0 A/s"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 109</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate (a) the average induced e.m.f. in the second coil, \n",
- "#(b) the change of flux linked with the second coil if it is wound with 500 turns.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "M = 0.2;# in Henry\n",
- "I1 = 10;# in Amperes\n",
- "I2 = 4;# in Amperes\n",
- "dt = 0.010;# in secs\n",
- "N = 500;# turns\n",
- "\n",
- "#calculation:\n",
- "dI1dt = (I1 -I2)/dt \n",
- "E2 = -1*dI1dt*M\n",
- "dPhi = abs(E2)*dt/N\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Induced e.m.f. E2 = \", E2,\" V\\n\"\n",
- "print \"\\n (b)change of flux = \", dPhi,\" Wb\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Induced e.m.f. E2 = -120.0 V\n",
- "\n",
- "\n",
- " (b)change of flux = 0.0024 Wb"
- ]
- }
- ],
- "prompt_number": 15
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_2.ipynb deleted file mode 100755 index 02f3e6f4..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_2.ipynb +++ /dev/null @@ -1,759 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 9: Electromagnetic induction</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 102</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the conductor when \n",
- "#(a) its ends are open-circuited, \n",
- "#(b) its ends are connected to a load of 20 ohm resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "l = 0.3;# in m\n",
- "v = 4;# in m/s\n",
- "B = 1.25;# in Tesla\n",
- "R = 20;# in ohms\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "E = B*l*v\n",
- "I2 = E/R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n (a)If the ends of the conductor are open circuited \"\n",
- "print \"no current will flow even though \",E,\" V has been induced.\\n\"\n",
- "print \"\\n (b)From Ohms law, I = \",I2,\" Ampere\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " (a)If the ends of the conductor are open circuited no current will flow even though 1.5 V has been induced.\n",
- "\n",
- "\n",
- " (b)From Ohms law, I = 0.075 Ampere"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 103</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#At what velocity must a conductor 75 mm long cut a magnetic field of flux density 0.6 T\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "l = 0.075;# in m\n",
- "E = 9;# in Volts\n",
- "B = 0.6;# in Tesla\n",
- "R = 20;# in ohms\n",
- "u0 = 4*math.pi*1E-7;\n",
- "\n",
- "#calculation:\n",
- "v = E/(B*l)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n velocity v = \",v,\" m/s\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " velocity v = 200.0 m/s"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 103</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the magnitude of the induced e.m.f. in each case.\n",
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#initializing the variables:\n",
- "l = 0.02;# in m\n",
- "b = 0.02;# in m\n",
- "v = 15;# in m/s\n",
- "R = 20;# in ohms\n",
- "Phi = 5E-6;# in Wb\n",
- "u0 = 4*math.pi*1E-7;\n",
- "a1 = 90;# in degrees\n",
- "a2 = 60;# in degrees\n",
- "a3 = 30;# in degrees\n",
- "\n",
- "#calculation:\n",
- "A = l*b\n",
- "B = Phi/A\n",
- "E90 = B*l*v*math.sin(a1*math.pi/180)\n",
- "E60 = B*l*v*math.sin(a2*math.pi/180)\n",
- "E30 = B*l*v*math.sin(a3*math.pi/180)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. at angles 90\u00b0, 60\u00b0, 30\u00b0 are \",(E90/1E-3),\" V, \",round((E60/1E-3),2),\" V, \"\n",
- "print \"(E30/1E-3),\" V respectively\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. at angles 90\u00c2\u00b0, 60\u00c2\u00b0, 30\u00c2\u00b0 are 3.75 V, 3.25 V, 1.875 V respectively"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 103</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the e.m.f. induced between its wing tips\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "s = 36;# in m\n",
- "v = 400;# in km/h\n",
- "u0 = 4*math.pi*1E-7;\n",
- "B = 40E-6;# in Tesla\n",
- "\n",
- "#calculation:\n",
- "v0 = v*5/18\n",
- "E = B*s*v0\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. = 0.16 V"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 105</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. induced in a coil\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N = 200;# no. of turns\n",
- "dt = 0.050;# change of time in sec\n",
- "u0 = 4*math.pi*1E-7;\n",
- "dPhi = 0.025;# change of flux in Wb\n",
- "\n",
- "#calculation:\n",
- "E = -1*N*dPhi/dt\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. = -100.0 V"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 105</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the average e.m.f. induced.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N = 150;# no. of turns\n",
- "dt = 0.040;# change of time in sec\n",
- "u0 = 4*math.pi*1E-7;\n",
- "dPhi = 800E-6;# change of flux in Wb\n",
- "\n",
- "#calculation:\n",
- "#Since the flux reverses, the flux changes from C400 \u03bcWb to \u0003400 \u03bcWb, a total change of flux of 800 \u03bcWb\n",
- "E = -1*N*dPhi/dt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. = -3.0 V"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 105</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the e.m.f. induced in a coil\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 12;# in Henry\n",
- "u0 = 4*math.pi*1E-7;\n",
- "dIdt = 4;# change of current with change in time in A/s\n",
- "\n",
- "#calculation:\n",
- "E = -1*L*dIdt\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\nResult\\n\\n\"\n",
- "print \"\\n Induced e.m.f. = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Result\n",
- "\n",
- "\n",
- "\n",
- " Induced e.m.f. = -48 V"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 106</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find inductance of the coil.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E = 1500;# in Volts\n",
- "dt = 0.008;# Change of time in sec\n",
- "dI = 4;# change of current in A/s\n",
- "\n",
- "#calculation:\n",
- "L = abs(E)*dt/dI\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Inductance L= \",L,\" H\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Inductance L= 3.0 H"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 107</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#How much energy is stored in the magnetic field of the inductor?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 8;# in Henry\n",
- "I = 3;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "W = L*I*I/2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Energy stored, W = \",W,\" J\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Energy stored, W = 36.0 J"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 107</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the coil inductance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 4;# in Amperes\n",
- "N = 800;#turns\n",
- "Phi = 0.005;# in Wb\n",
- "\n",
- "#calculation:\n",
- "L = N*Phi/I\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Inductance L = \",L,\" H\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Inductance L = 1.0 H"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 107</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the inductance of the coil, \n",
- "#(b) the energy stored in the magnetic field, and \n",
- "#(c) the average e.m.f. induced if the current falls to zero in 150 ms.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I1 = 3;# in Amperes\n",
- "I2 = 0;# in Amperes\n",
- "dt = 0.150;# in secs\n",
- "N = 1500;#turns\n",
- "Phi = 0.025;# in Wb\n",
- "\n",
- "#calculation:\n",
- "L = N*Phi/I1\n",
- "W = L*I1*I1/2\n",
- "dI = I1 - I2\n",
- "E = -1*L*dI/dt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductance L = \",L,\" H\\n\"\n",
- "print \"\\n (b)energy stored W = \",W,\" J\\n\"\n",
- "print \"\\n (c)e.m.f. induced = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductance L = 12.5 H\n",
- "\n",
- "\n",
- " (b)energy stored W = 56.25 J\n",
- "\n",
- "\n",
- " (c)e.m.f. induced = -250.0 V"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 108</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the flux linking the coil and the e.m.f. induced in the coil when the current collapses to zero in 20 ms\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I1 = 2;# in Amperes\n",
- "I2 = 0;# in Amperes\n",
- "dt = 0.020;# in secs\n",
- "N = 750;#turns\n",
- "L = 3;# in Henry\n",
- "\n",
- "#calculation:\n",
- "Phi = L*I1/N\n",
- "dI = I1 - I2\n",
- "E = -1*L*dI/dt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Flux = \",Phi,\" Wb\\n\"\n",
- "print \"\\n (b)e.m.f. induced = \",E,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Flux = 0.008 Wb\n",
- "\n",
- "\n",
- " (b)e.m.f. induced = -300.0 V"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 108</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the mutual inductance between two coils\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "dI1dt = 200;# change of current with change in time in A/s\n",
- "N = 2;# no. of coils\n",
- "E2 = 1.5;# in Volts\n",
- "\n",
- "#calculation:\n",
- "M = abs(E2)/dI1dt\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance, M = \", M,\" H\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance, M = 0.0075 H"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 109</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the steady rate of change of current in one coil to induce an e.m.f. of 0.72 V in the other.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "M = 0.018;# in Henry\n",
- "N = 2;# no. of coils\n",
- "E2 = 0.72;# in Volts\n",
- "\n",
- "#calculation:\n",
- "dI1dt = abs(E2)/M\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n rate of change of current dI1/dt = \", dI1dt,\" A/s\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " rate of change of current dI1/dt = 40.0 A/s"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 109</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate (a) the average induced e.m.f. in the second coil, \n",
- "#(b) the change of flux linked with the second coil if it is wound with 500 turns.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "M = 0.2;# in Henry\n",
- "I1 = 10;# in Amperes\n",
- "I2 = 4;# in Amperes\n",
- "dt = 0.010;# in secs\n",
- "N = 500;# turns\n",
- "\n",
- "#calculation:\n",
- "dI1dt = (I1 -I2)/dt \n",
- "E2 = -1*dI1dt*M\n",
- "dPhi = abs(E2)*dt/N\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Induced e.m.f. E2 = \", E2,\" V\\n\"\n",
- "print \"\\n (b)change of flux = \", dPhi,\" Wb\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Induced e.m.f. E2 = -120.0 V\n",
- "\n",
- "\n",
- " (b)change of flux = 0.0024 Wb"
- ]
- }
- ],
- "prompt_number": 15
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_09.ipynb index 0248550c..0248550c 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_09.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint.ipynb deleted file mode 100755 index c63a225f..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint.ipynb +++ /dev/null @@ -1,1066 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 10: Electrical measuring instruments and measurements</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 116</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the value of the shunt to be connected in parallel with the meter\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia = 0.040;# in Amperes\n",
- "I = 50;# in Amperes\n",
- "ra = 25;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Is = I - Ia\n",
- "V = Ia*ra\n",
- "Rs = V/Is\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n value of the shunt to be connected in parallel = \", round((Rs/1E-3),2),\" mohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " value of the shunt to be connected in parallel = 20.02 mohms"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 116</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the value of the multiplier to be connected in series with the instrument\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 100;# in volts\n",
- "I = 0.008;# in Amperes\n",
- "ra = 10;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Rm = (V/I) - ra\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n value of the multiplier to be connected in series = \", (Rm/1E3),\" kohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " value of the multiplier to be connected in series = 12.49 kohms"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 119</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the power dissipated by the voltmeter and by resistor R\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fsd = 200;# in volts\n",
- "R1 = 250;# in ohms\n",
- "R2 = 2E6;# in ohms\n",
- "sensitivity = 10000;# in ohms/V\n",
- "V = 100; # in volts\n",
- "\n",
- "#calculation:\n",
- "Rv = sensitivity*fsd\n",
- "Iv = V/Rv\n",
- "Pv = V*Iv\n",
- "I1 = V/R1\n",
- "P1 = V*I1\n",
- "I2 = V/R2\n",
- "P2 = V*I2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the power dissipated by the voltmeter = \", (Pv/1E-3),\" mW\\n\"\n",
- "print \"\\n (b)the power dissipated by resistor 250 ohm = \", P1,\" W\\n\"\n",
- "print \"\\n (c)the power dissipated by resistor 2 Mohm = \", (P2/1E-3),\" mW\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the power dissipated by the voltmeter = 5.0 mW\n",
- "\n",
- "\n",
- " (b)the power dissipated by resistor 250 ohm = 40.0 W\n",
- "\n",
- "\n",
- " (c)the power dissipated by resistor 2 Mohm = 5.0 mW"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 119</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the ammeter reading expected (neglecting its resistance),\n",
- "#(b) the actual current in the circuit, \n",
- "#(c) the power dissipated in the ammeter, and\n",
- "#(d) the power dissipated in the load\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 10;# in volts\n",
- "fsd = 0.1;# in Amperes\n",
- "ra = 50;# in ohms\n",
- "R = 500;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Ie = V/R\n",
- "Ia = V/(R + ra)\n",
- "Pa = Ia*Ia*ra\n",
- "PR = Ia*Ia*R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)expected ammeter reading = \", (Ie/1E-3),\" mA\\n\"\n",
- "print \"\\n (b)Actual ammeter reading = \",round((Ia/1E-3),2),\" mA\\n\"\n",
- "print \"\\n (c)Power dissipated in the ammeter = \",round((Pa/1E-3),2),\" mW\\n\"\n",
- "print \"\\n (d)Power dissipated in the load resistor = \", round((PR/1E-3),2),\" mW\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)expected ammeter reading = 20.0 mA\n",
- "\n",
- "\n",
- " (b)Actual ammeter reading = 18.18 mA\n",
- "\n",
- "\n",
- " (c)Power dissipated in the ammeter = 16.53 mW\n",
- "\n",
- "\n",
- " (d)Power dissipated in the load resistor = 165.29 mW"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 120</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of voltage V1 with the voltmeter not connected, and \n",
- "#(b) the voltage indicated by the voltmeter when connected between A and B.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fsd = 100;# in volts\n",
- "R1 = 40E3;# in ohms\n",
- "R2 = 60E3;# in ohms\n",
- "sensitivity = 1600;# in ohms/V\n",
- "\n",
- "#calculation:\n",
- "V1 = (R1/(R1 + R2))*fsd\n",
- "Rv = fsd*sensitivity\n",
- "Rep = R1*Rv/(R1 + Rv)\n",
- "V1n = (Rep/(Rep + R2))*fsd\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the value of voltage V1 with the voltmeter6 not connected = \", V1,\" V\\n\"\n",
- "print \"\\n (b)the voltage indicated by the voltmeter when connected between A and B = \",round(V1n,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the value of voltage V1 with the voltmeter6 not connected = 40.0 V\n",
- "\n",
- "\n",
- " (b)the voltage indicated by the voltmeter when connected between A and B = 34.78 V"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 120</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the power dissipated in the load.\n",
- "#(b) A wattmeter, whose current coil has a resistance of 0.01 ohm\n",
- "# Determine the wattmeter reading.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 20;# in amperes\n",
- "R = 2;# in ohms\n",
- "Rw = 0.01;# in ohms\n",
- "\n",
- "#calculation:\n",
- "PR = I*I*R\n",
- "Rt = R + Rw\n",
- "Pw = I*I*Rt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the power dissipated in the load = \", PR,\" W\\n\"\n",
- "print \"\\n (b)the wattmeter reading. = \",Pw,\" W\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the power dissipated in the load = 800 W\n",
- "\n",
- "\n",
- " (b)the wattmeter reading. = 804.0 W"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 122</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the periodic time, (b) the frequency and (c) the peak-to-peak voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tc = 100E-6;# in s/cm\n",
- "Vc = 20;# in V/cm\n",
- "w = 5.2;# in cm ( width of one complete cycle )\n",
- "h = 3.6; # in cm ( peak-to-peak height of the display )\n",
- "\n",
- "#calculation:\n",
- "T = w*tc\n",
- "f = 1/T\n",
- "ptpv = h*Vc\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the periodic time, T = \", (T/1E-3),\" msec\\n\"\n",
- "print \"\\n (b)Frequency, f = \",round(f,2),\" Hz\\n\"\n",
- "print \"\\n (c)the peak-to-peak voltage = \",ptpv,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the periodic time, T = 0.52 msec\n",
- "\n",
- "\n",
- " (b)Frequency, f = 1923.08 Hz\n",
- "\n",
- "\n",
- " (c)the peak-to-peak voltage = 72.0 V"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 123</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the periodic time, (b) the frequency, (c) the magnitude of the pulse voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tc = 50E-3;# in s/cm\n",
- "Vc = 0.2;# in V/cm\n",
- "w = 3.5;# in cm ( width of one complete cycle )\n",
- "h = 3.4; # in cm ( peak-to-peak height of the display )\n",
- "\n",
- "#calculation:\n",
- "T = w*tc\n",
- "f = 1/T\n",
- "ptpv = h*Vc\n",
- "\n",
- "#Results \n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the periodic time, T = \", (T/1E-3),\" msec\\n\"\n",
- "print \"\\n (b)Frequency, f = \",round(f,2),\" Hz\\n\"\n",
- "print \"\\n (c)the peak-to-peak voltage = \",ptpv,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the periodic time, T = 175.0 msec\n",
- "\n",
- "\n",
- " (b)Frequency, f = 5.71 Hz\n",
- "\n",
- "\n",
- " (c)the peak-to-peak voltage = 0.68 V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 123</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find, for the waveform, (a) the frequency, \n",
- "#(b) the peak-to-peak voltage, (c) the amplitude, (d) the r.m.s. value.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tc = 500E-6;# in s/cm\n",
- "Vc = 5;# in V/cm\n",
- "w = 4;# in cm ( width of one complete cycle )\n",
- "h = 5; # in cm ( peak-to-peak height of the display )\n",
- "\n",
- "#calculation:\n",
- "T = w*tc\n",
- "f = 1/T\n",
- "ptpv = h*Vc\n",
- "Amp = ptpv/2\n",
- "Vrms = Amp/(2**0.5)\n",
- "\n",
- "#Results \n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Frequency, f = \",f,\" Hz\\n\"\n",
- "print \"\\n (b)the peak-to-peak voltage = \",ptpv,\" V\\n\"\n",
- "print \"\\n (c)Amplitude = \",Amp,\" V\\n\"\n",
- "print \"\\n (d)r.m.s voltage = \",round(Vrms,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Frequency, f = 500.0 Hz\n",
- "\n",
- "\n",
- " (b)the peak-to-peak voltage = 25 V\n",
- "\n",
- "\n",
- " (c)Amplitude = 12.5 V\n",
- "\n",
- "\n",
- " (d)r.m.s voltage = 8.84 V"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 123</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) their frequency, (b) their r.m.s. values, (c) their phase difference.\n",
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#initializing the variables:\n",
- "tc = 100E-6;# in s/cm\n",
- "Vc = 2;# in V/cm\n",
- "w = 5;# in cm ( width of one complete cycle for both waveform )\n",
- "h1 = 2; # in cm ( peak-to-peak height of the display )\n",
- "h2 = 2.5; # in cm ( peak-to-peak height of the display\n",
- "\n",
- "#calculation:\n",
- "T = w*tc\n",
- "f = 1/T\n",
- "ptpv1 = h1*Vc\n",
- "Vrms1 = ptpv1/(2**0.5)\n",
- "ptpv2 = h2*Vc\n",
- "Vrms2 = ptpv2/(2**0.5)\n",
- "phi = 0.5*360/w\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Frequency, f = \",f,\" Hz\\n\"\n",
- "print \"\\n (b1)r.m.s voltage of 1st waveform = \",round(Vrms1,2),\" V\\n\"\n",
- "print \"\\n (b2)r.m.s voltage of 2nd waveform = \",round(Vrms2,2),\" V\\n\"\n",
- "print \"\\n (c)Phase difference = \",phi,\"deg\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Frequency, f = 2000.0 Hz\n",
- "\n",
- "\n",
- " (b1)r.m.s voltage of 1st waveform = 2.83 V\n",
- "\n",
- "\n",
- " (b2)r.m.s voltage of 2nd waveform = 3.54 V\n",
- "\n",
- "\n",
- " (c)Phase difference = 36.0 deg"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 127</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the decibel power ratio in each case.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "rP1 = 3;# ratio of two powers\n",
- "rP2 = 20;# ratio of two powers\n",
- "rP3 = 400;# ratio of two powers\n",
- "rP4 = 1/20;# ratio of two powers\n",
- "\n",
- "#calculation:\n",
- "X1 = 10*(1/2.303)*math.log(3)\n",
- "X2 = 10*(1/2.303)*math.log(20)\n",
- "X3 = 10*(1/2.303)*math.log(400)\n",
- "X4 = 10*(1/2.303)*math.log(1/20)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)decibel power ratio for power ratio 3 = \",round(X1,2),\" dB\\n\"\n",
- "print \"\\n (b)decibel power ratio for power ratio 20 = \",round(X2,2),\" dB\\n\"\n",
- "print \"\\n (c)decibel power ratio for power ratio 400 = \",round(X3,2),\" dB\\n\"\n",
- "print \"\\n (d)decibel power ratio for power ratio 1/20 = \",round(X4,2),\" dB\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)decibel power ratio for power ratio 3 = 4.77 dB\n",
- "\n",
- "\n",
- " (b)decibel power ratio for power ratio 20 = 13.01 dB\n",
- "\n",
- "\n",
- " (c)decibel power ratio for power ratio 400 = 26.02 dB\n",
- "\n",
- "\n",
- " (d)decibel power ratio for power ratio 1/20 = -13.01 dB"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 127</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the decibel current ratio assuming the input and load resistances of the system are equal.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I2 = 0.020;# in ampere\n",
- "I1 = 0.005;# in ampere\n",
- "\n",
- "#calculation:\n",
- "X = 20*math.log10(math.e)*math.log(I2/I1)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n decibel current ratio = \",round(X,2),\" dB\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " decibel current ratio = 12.04 dB\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 128</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the power loss in decibels.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "rP = 0.06;# power ratios rP = P2/P1\n",
- "\n",
- "#calculation:\n",
- "X = 10*math.log10(math.e)*math.log(rP)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n decibel Power ratios = \",round(X,2),\" dB\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " decibel Power ratios = -12.22 dB\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 128</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find its output power\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "X = 14;# decibal power ratio in dB\n",
- "P1 = 0.008;# in Watt\n",
- "\n",
- "#calculation:\n",
- "rP = 10**(X/10)\n",
- "P2 = rP*P1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output power P2 = \",round(P2,2),\" W\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output power P2 = 0.2 W"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 128</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of the input voltage\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "X = 27;# Voltage gain in decibels\n",
- "V2 = 4;# output voltage in Volts\n",
- "\n",
- "#calculation:\n",
- "V1 = V2/(10**(27/20))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n input Voltage V1 = \",round(V1,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " input Voltage V1 = 0.18 V"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 129</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the value of the unknown resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R2 = 100;# in ohms\n",
- "R3 = 400;# in ohms\n",
- "R4 = 10;# in ohms\n",
- "\n",
- "#calculation:\n",
- "R1 = R2*R3/R4\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n unknown resistance, R1 = \",R1,\" Ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " unknown resistance, R1 = 4000.0 Ohms"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 130</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. of a dry cell if balance is obtained with a length of 650 mm\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 1.0186;# in Volts\n",
- "l1 = 0.400;# in m\n",
- "l2 = 0.650;# in m\n",
- "\n",
- "#calculation:\n",
- "E2 = (l2/l1)*E1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the e.m.f. of a dry cell = \",round(E2,2),\" Volts\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the e.m.f. of a dry cell = 1.66 Volts"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 132</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the nominal value of the voltage across the resistor and its accuracy.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 0.0025;# in Amperes\n",
- "R = 5000;# in ohms\n",
- "e1 = 0.4;# in %\n",
- "e2 = 0.5;# in %\n",
- "\n",
- "#calculation:\n",
- "V = I*R\n",
- "em = e1 + e2\n",
- "Ve = em*V/100\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage V = \",V,\"V(+-)\",Ve,\"V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage V = 12.5 V(+-) 0.1125 V"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 132</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance of the resistor, and its accuracy of measurement\n",
- "#if both instruments have a limit of error of 2% of f.s.d.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 6.25;# in Amperes\n",
- "Im = 10;# max in Amperes\n",
- "V = 36.5;# in volts\n",
- "Vm = 50;# max in volts\n",
- "e = 2;# in %\n",
- "\n",
- "#calculation:\n",
- "R = V/I\n",
- "Ve = e*Vm/100 \n",
- "Ve1 = Ve*100/V# in %\n",
- "Ie = e*Im/100\n",
- "Ie1 = Ie*100/I# in %\n",
- "em = Ve1 + Ie1\n",
- "Re = em*R/100\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resistance R = \",R,\" ohms(+-)\",Re,\" ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resistance R = 5.84 ohms(+-) 0.34688 ohms"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 133</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the unknown resistance and its accuracy of measurement.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 1000;# in ohms\n",
- "R2 = 100;# in ohms\n",
- "R3 = 432.5;# in ohms\n",
- "e1 = 1;# in %\n",
- "e2 = 0.5;# in %\n",
- "e3 = 0.2;# in %\n",
- "\n",
- "#calculation:\n",
- "Rx = R2*R3/R1\n",
- "em = e1 + e2 + e3\n",
- "Re = em*Rx/100\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resistance R = \",Rx,\" ohms(+-)\",Re,\" ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resistance R = 43.25 ohms(+-) 0.73525 ohms"
- ]
- }
- ],
- "prompt_number": 21
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_1.ipynb deleted file mode 100755 index c63a225f..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_1.ipynb +++ /dev/null @@ -1,1066 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 10: Electrical measuring instruments and measurements</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 116</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the value of the shunt to be connected in parallel with the meter\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia = 0.040;# in Amperes\n",
- "I = 50;# in Amperes\n",
- "ra = 25;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Is = I - Ia\n",
- "V = Ia*ra\n",
- "Rs = V/Is\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n value of the shunt to be connected in parallel = \", round((Rs/1E-3),2),\" mohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " value of the shunt to be connected in parallel = 20.02 mohms"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 116</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the value of the multiplier to be connected in series with the instrument\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 100;# in volts\n",
- "I = 0.008;# in Amperes\n",
- "ra = 10;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Rm = (V/I) - ra\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n value of the multiplier to be connected in series = \", (Rm/1E3),\" kohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " value of the multiplier to be connected in series = 12.49 kohms"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 119</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the power dissipated by the voltmeter and by resistor R\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fsd = 200;# in volts\n",
- "R1 = 250;# in ohms\n",
- "R2 = 2E6;# in ohms\n",
- "sensitivity = 10000;# in ohms/V\n",
- "V = 100; # in volts\n",
- "\n",
- "#calculation:\n",
- "Rv = sensitivity*fsd\n",
- "Iv = V/Rv\n",
- "Pv = V*Iv\n",
- "I1 = V/R1\n",
- "P1 = V*I1\n",
- "I2 = V/R2\n",
- "P2 = V*I2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the power dissipated by the voltmeter = \", (Pv/1E-3),\" mW\\n\"\n",
- "print \"\\n (b)the power dissipated by resistor 250 ohm = \", P1,\" W\\n\"\n",
- "print \"\\n (c)the power dissipated by resistor 2 Mohm = \", (P2/1E-3),\" mW\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the power dissipated by the voltmeter = 5.0 mW\n",
- "\n",
- "\n",
- " (b)the power dissipated by resistor 250 ohm = 40.0 W\n",
- "\n",
- "\n",
- " (c)the power dissipated by resistor 2 Mohm = 5.0 mW"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 119</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the ammeter reading expected (neglecting its resistance),\n",
- "#(b) the actual current in the circuit, \n",
- "#(c) the power dissipated in the ammeter, and\n",
- "#(d) the power dissipated in the load\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 10;# in volts\n",
- "fsd = 0.1;# in Amperes\n",
- "ra = 50;# in ohms\n",
- "R = 500;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Ie = V/R\n",
- "Ia = V/(R + ra)\n",
- "Pa = Ia*Ia*ra\n",
- "PR = Ia*Ia*R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)expected ammeter reading = \", (Ie/1E-3),\" mA\\n\"\n",
- "print \"\\n (b)Actual ammeter reading = \",round((Ia/1E-3),2),\" mA\\n\"\n",
- "print \"\\n (c)Power dissipated in the ammeter = \",round((Pa/1E-3),2),\" mW\\n\"\n",
- "print \"\\n (d)Power dissipated in the load resistor = \", round((PR/1E-3),2),\" mW\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)expected ammeter reading = 20.0 mA\n",
- "\n",
- "\n",
- " (b)Actual ammeter reading = 18.18 mA\n",
- "\n",
- "\n",
- " (c)Power dissipated in the ammeter = 16.53 mW\n",
- "\n",
- "\n",
- " (d)Power dissipated in the load resistor = 165.29 mW"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 120</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of voltage V1 with the voltmeter not connected, and \n",
- "#(b) the voltage indicated by the voltmeter when connected between A and B.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fsd = 100;# in volts\n",
- "R1 = 40E3;# in ohms\n",
- "R2 = 60E3;# in ohms\n",
- "sensitivity = 1600;# in ohms/V\n",
- "\n",
- "#calculation:\n",
- "V1 = (R1/(R1 + R2))*fsd\n",
- "Rv = fsd*sensitivity\n",
- "Rep = R1*Rv/(R1 + Rv)\n",
- "V1n = (Rep/(Rep + R2))*fsd\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the value of voltage V1 with the voltmeter6 not connected = \", V1,\" V\\n\"\n",
- "print \"\\n (b)the voltage indicated by the voltmeter when connected between A and B = \",round(V1n,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the value of voltage V1 with the voltmeter6 not connected = 40.0 V\n",
- "\n",
- "\n",
- " (b)the voltage indicated by the voltmeter when connected between A and B = 34.78 V"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 120</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the power dissipated in the load.\n",
- "#(b) A wattmeter, whose current coil has a resistance of 0.01 ohm\n",
- "# Determine the wattmeter reading.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 20;# in amperes\n",
- "R = 2;# in ohms\n",
- "Rw = 0.01;# in ohms\n",
- "\n",
- "#calculation:\n",
- "PR = I*I*R\n",
- "Rt = R + Rw\n",
- "Pw = I*I*Rt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the power dissipated in the load = \", PR,\" W\\n\"\n",
- "print \"\\n (b)the wattmeter reading. = \",Pw,\" W\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the power dissipated in the load = 800 W\n",
- "\n",
- "\n",
- " (b)the wattmeter reading. = 804.0 W"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 122</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the periodic time, (b) the frequency and (c) the peak-to-peak voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tc = 100E-6;# in s/cm\n",
- "Vc = 20;# in V/cm\n",
- "w = 5.2;# in cm ( width of one complete cycle )\n",
- "h = 3.6; # in cm ( peak-to-peak height of the display )\n",
- "\n",
- "#calculation:\n",
- "T = w*tc\n",
- "f = 1/T\n",
- "ptpv = h*Vc\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the periodic time, T = \", (T/1E-3),\" msec\\n\"\n",
- "print \"\\n (b)Frequency, f = \",round(f,2),\" Hz\\n\"\n",
- "print \"\\n (c)the peak-to-peak voltage = \",ptpv,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the periodic time, T = 0.52 msec\n",
- "\n",
- "\n",
- " (b)Frequency, f = 1923.08 Hz\n",
- "\n",
- "\n",
- " (c)the peak-to-peak voltage = 72.0 V"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 123</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the periodic time, (b) the frequency, (c) the magnitude of the pulse voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tc = 50E-3;# in s/cm\n",
- "Vc = 0.2;# in V/cm\n",
- "w = 3.5;# in cm ( width of one complete cycle )\n",
- "h = 3.4; # in cm ( peak-to-peak height of the display )\n",
- "\n",
- "#calculation:\n",
- "T = w*tc\n",
- "f = 1/T\n",
- "ptpv = h*Vc\n",
- "\n",
- "#Results \n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the periodic time, T = \", (T/1E-3),\" msec\\n\"\n",
- "print \"\\n (b)Frequency, f = \",round(f,2),\" Hz\\n\"\n",
- "print \"\\n (c)the peak-to-peak voltage = \",ptpv,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the periodic time, T = 175.0 msec\n",
- "\n",
- "\n",
- " (b)Frequency, f = 5.71 Hz\n",
- "\n",
- "\n",
- " (c)the peak-to-peak voltage = 0.68 V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 123</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find, for the waveform, (a) the frequency, \n",
- "#(b) the peak-to-peak voltage, (c) the amplitude, (d) the r.m.s. value.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tc = 500E-6;# in s/cm\n",
- "Vc = 5;# in V/cm\n",
- "w = 4;# in cm ( width of one complete cycle )\n",
- "h = 5; # in cm ( peak-to-peak height of the display )\n",
- "\n",
- "#calculation:\n",
- "T = w*tc\n",
- "f = 1/T\n",
- "ptpv = h*Vc\n",
- "Amp = ptpv/2\n",
- "Vrms = Amp/(2**0.5)\n",
- "\n",
- "#Results \n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Frequency, f = \",f,\" Hz\\n\"\n",
- "print \"\\n (b)the peak-to-peak voltage = \",ptpv,\" V\\n\"\n",
- "print \"\\n (c)Amplitude = \",Amp,\" V\\n\"\n",
- "print \"\\n (d)r.m.s voltage = \",round(Vrms,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Frequency, f = 500.0 Hz\n",
- "\n",
- "\n",
- " (b)the peak-to-peak voltage = 25 V\n",
- "\n",
- "\n",
- " (c)Amplitude = 12.5 V\n",
- "\n",
- "\n",
- " (d)r.m.s voltage = 8.84 V"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 123</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) their frequency, (b) their r.m.s. values, (c) their phase difference.\n",
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#initializing the variables:\n",
- "tc = 100E-6;# in s/cm\n",
- "Vc = 2;# in V/cm\n",
- "w = 5;# in cm ( width of one complete cycle for both waveform )\n",
- "h1 = 2; # in cm ( peak-to-peak height of the display )\n",
- "h2 = 2.5; # in cm ( peak-to-peak height of the display\n",
- "\n",
- "#calculation:\n",
- "T = w*tc\n",
- "f = 1/T\n",
- "ptpv1 = h1*Vc\n",
- "Vrms1 = ptpv1/(2**0.5)\n",
- "ptpv2 = h2*Vc\n",
- "Vrms2 = ptpv2/(2**0.5)\n",
- "phi = 0.5*360/w\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Frequency, f = \",f,\" Hz\\n\"\n",
- "print \"\\n (b1)r.m.s voltage of 1st waveform = \",round(Vrms1,2),\" V\\n\"\n",
- "print \"\\n (b2)r.m.s voltage of 2nd waveform = \",round(Vrms2,2),\" V\\n\"\n",
- "print \"\\n (c)Phase difference = \",phi,\"deg\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Frequency, f = 2000.0 Hz\n",
- "\n",
- "\n",
- " (b1)r.m.s voltage of 1st waveform = 2.83 V\n",
- "\n",
- "\n",
- " (b2)r.m.s voltage of 2nd waveform = 3.54 V\n",
- "\n",
- "\n",
- " (c)Phase difference = 36.0 deg"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 127</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the decibel power ratio in each case.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "rP1 = 3;# ratio of two powers\n",
- "rP2 = 20;# ratio of two powers\n",
- "rP3 = 400;# ratio of two powers\n",
- "rP4 = 1/20;# ratio of two powers\n",
- "\n",
- "#calculation:\n",
- "X1 = 10*(1/2.303)*math.log(3)\n",
- "X2 = 10*(1/2.303)*math.log(20)\n",
- "X3 = 10*(1/2.303)*math.log(400)\n",
- "X4 = 10*(1/2.303)*math.log(1/20)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)decibel power ratio for power ratio 3 = \",round(X1,2),\" dB\\n\"\n",
- "print \"\\n (b)decibel power ratio for power ratio 20 = \",round(X2,2),\" dB\\n\"\n",
- "print \"\\n (c)decibel power ratio for power ratio 400 = \",round(X3,2),\" dB\\n\"\n",
- "print \"\\n (d)decibel power ratio for power ratio 1/20 = \",round(X4,2),\" dB\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)decibel power ratio for power ratio 3 = 4.77 dB\n",
- "\n",
- "\n",
- " (b)decibel power ratio for power ratio 20 = 13.01 dB\n",
- "\n",
- "\n",
- " (c)decibel power ratio for power ratio 400 = 26.02 dB\n",
- "\n",
- "\n",
- " (d)decibel power ratio for power ratio 1/20 = -13.01 dB"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 127</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the decibel current ratio assuming the input and load resistances of the system are equal.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I2 = 0.020;# in ampere\n",
- "I1 = 0.005;# in ampere\n",
- "\n",
- "#calculation:\n",
- "X = 20*math.log10(math.e)*math.log(I2/I1)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n decibel current ratio = \",round(X,2),\" dB\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " decibel current ratio = 12.04 dB\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 128</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the power loss in decibels.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "rP = 0.06;# power ratios rP = P2/P1\n",
- "\n",
- "#calculation:\n",
- "X = 10*math.log10(math.e)*math.log(rP)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n decibel Power ratios = \",round(X,2),\" dB\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " decibel Power ratios = -12.22 dB\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 128</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find its output power\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "X = 14;# decibal power ratio in dB\n",
- "P1 = 0.008;# in Watt\n",
- "\n",
- "#calculation:\n",
- "rP = 10**(X/10)\n",
- "P2 = rP*P1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output power P2 = \",round(P2,2),\" W\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output power P2 = 0.2 W"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 128</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of the input voltage\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "X = 27;# Voltage gain in decibels\n",
- "V2 = 4;# output voltage in Volts\n",
- "\n",
- "#calculation:\n",
- "V1 = V2/(10**(27/20))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n input Voltage V1 = \",round(V1,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " input Voltage V1 = 0.18 V"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 129</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the value of the unknown resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R2 = 100;# in ohms\n",
- "R3 = 400;# in ohms\n",
- "R4 = 10;# in ohms\n",
- "\n",
- "#calculation:\n",
- "R1 = R2*R3/R4\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n unknown resistance, R1 = \",R1,\" Ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " unknown resistance, R1 = 4000.0 Ohms"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 130</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. of a dry cell if balance is obtained with a length of 650 mm\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 1.0186;# in Volts\n",
- "l1 = 0.400;# in m\n",
- "l2 = 0.650;# in m\n",
- "\n",
- "#calculation:\n",
- "E2 = (l2/l1)*E1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the e.m.f. of a dry cell = \",round(E2,2),\" Volts\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the e.m.f. of a dry cell = 1.66 Volts"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 132</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the nominal value of the voltage across the resistor and its accuracy.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 0.0025;# in Amperes\n",
- "R = 5000;# in ohms\n",
- "e1 = 0.4;# in %\n",
- "e2 = 0.5;# in %\n",
- "\n",
- "#calculation:\n",
- "V = I*R\n",
- "em = e1 + e2\n",
- "Ve = em*V/100\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage V = \",V,\"V(+-)\",Ve,\"V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage V = 12.5 V(+-) 0.1125 V"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 132</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance of the resistor, and its accuracy of measurement\n",
- "#if both instruments have a limit of error of 2% of f.s.d.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 6.25;# in Amperes\n",
- "Im = 10;# max in Amperes\n",
- "V = 36.5;# in volts\n",
- "Vm = 50;# max in volts\n",
- "e = 2;# in %\n",
- "\n",
- "#calculation:\n",
- "R = V/I\n",
- "Ve = e*Vm/100 \n",
- "Ve1 = Ve*100/V# in %\n",
- "Ie = e*Im/100\n",
- "Ie1 = Ie*100/I# in %\n",
- "em = Ve1 + Ie1\n",
- "Re = em*R/100\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resistance R = \",R,\" ohms(+-)\",Re,\" ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resistance R = 5.84 ohms(+-) 0.34688 ohms"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 133</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the unknown resistance and its accuracy of measurement.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 1000;# in ohms\n",
- "R2 = 100;# in ohms\n",
- "R3 = 432.5;# in ohms\n",
- "e1 = 1;# in %\n",
- "e2 = 0.5;# in %\n",
- "e3 = 0.2;# in %\n",
- "\n",
- "#calculation:\n",
- "Rx = R2*R3/R1\n",
- "em = e1 + e2 + e3\n",
- "Re = em*Rx/100\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resistance R = \",Rx,\" ohms(+-)\",Re,\" ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resistance R = 43.25 ohms(+-) 0.73525 ohms"
- ]
- }
- ],
- "prompt_number": 21
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_2.ipynb deleted file mode 100755 index c63a225f..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_2.ipynb +++ /dev/null @@ -1,1066 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 10: Electrical measuring instruments and measurements</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 116</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the value of the shunt to be connected in parallel with the meter\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia = 0.040;# in Amperes\n",
- "I = 50;# in Amperes\n",
- "ra = 25;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Is = I - Ia\n",
- "V = Ia*ra\n",
- "Rs = V/Is\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n value of the shunt to be connected in parallel = \", round((Rs/1E-3),2),\" mohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " value of the shunt to be connected in parallel = 20.02 mohms"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 116</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the value of the multiplier to be connected in series with the instrument\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 100;# in volts\n",
- "I = 0.008;# in Amperes\n",
- "ra = 10;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Rm = (V/I) - ra\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n value of the multiplier to be connected in series = \", (Rm/1E3),\" kohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " value of the multiplier to be connected in series = 12.49 kohms"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 119</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the power dissipated by the voltmeter and by resistor R\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fsd = 200;# in volts\n",
- "R1 = 250;# in ohms\n",
- "R2 = 2E6;# in ohms\n",
- "sensitivity = 10000;# in ohms/V\n",
- "V = 100; # in volts\n",
- "\n",
- "#calculation:\n",
- "Rv = sensitivity*fsd\n",
- "Iv = V/Rv\n",
- "Pv = V*Iv\n",
- "I1 = V/R1\n",
- "P1 = V*I1\n",
- "I2 = V/R2\n",
- "P2 = V*I2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the power dissipated by the voltmeter = \", (Pv/1E-3),\" mW\\n\"\n",
- "print \"\\n (b)the power dissipated by resistor 250 ohm = \", P1,\" W\\n\"\n",
- "print \"\\n (c)the power dissipated by resistor 2 Mohm = \", (P2/1E-3),\" mW\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the power dissipated by the voltmeter = 5.0 mW\n",
- "\n",
- "\n",
- " (b)the power dissipated by resistor 250 ohm = 40.0 W\n",
- "\n",
- "\n",
- " (c)the power dissipated by resistor 2 Mohm = 5.0 mW"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 119</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the ammeter reading expected (neglecting its resistance),\n",
- "#(b) the actual current in the circuit, \n",
- "#(c) the power dissipated in the ammeter, and\n",
- "#(d) the power dissipated in the load\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 10;# in volts\n",
- "fsd = 0.1;# in Amperes\n",
- "ra = 50;# in ohms\n",
- "R = 500;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Ie = V/R\n",
- "Ia = V/(R + ra)\n",
- "Pa = Ia*Ia*ra\n",
- "PR = Ia*Ia*R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)expected ammeter reading = \", (Ie/1E-3),\" mA\\n\"\n",
- "print \"\\n (b)Actual ammeter reading = \",round((Ia/1E-3),2),\" mA\\n\"\n",
- "print \"\\n (c)Power dissipated in the ammeter = \",round((Pa/1E-3),2),\" mW\\n\"\n",
- "print \"\\n (d)Power dissipated in the load resistor = \", round((PR/1E-3),2),\" mW\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)expected ammeter reading = 20.0 mA\n",
- "\n",
- "\n",
- " (b)Actual ammeter reading = 18.18 mA\n",
- "\n",
- "\n",
- " (c)Power dissipated in the ammeter = 16.53 mW\n",
- "\n",
- "\n",
- " (d)Power dissipated in the load resistor = 165.29 mW"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 120</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of voltage V1 with the voltmeter not connected, and \n",
- "#(b) the voltage indicated by the voltmeter when connected between A and B.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fsd = 100;# in volts\n",
- "R1 = 40E3;# in ohms\n",
- "R2 = 60E3;# in ohms\n",
- "sensitivity = 1600;# in ohms/V\n",
- "\n",
- "#calculation:\n",
- "V1 = (R1/(R1 + R2))*fsd\n",
- "Rv = fsd*sensitivity\n",
- "Rep = R1*Rv/(R1 + Rv)\n",
- "V1n = (Rep/(Rep + R2))*fsd\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the value of voltage V1 with the voltmeter6 not connected = \", V1,\" V\\n\"\n",
- "print \"\\n (b)the voltage indicated by the voltmeter when connected between A and B = \",round(V1n,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the value of voltage V1 with the voltmeter6 not connected = 40.0 V\n",
- "\n",
- "\n",
- " (b)the voltage indicated by the voltmeter when connected between A and B = 34.78 V"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 120</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the power dissipated in the load.\n",
- "#(b) A wattmeter, whose current coil has a resistance of 0.01 ohm\n",
- "# Determine the wattmeter reading.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 20;# in amperes\n",
- "R = 2;# in ohms\n",
- "Rw = 0.01;# in ohms\n",
- "\n",
- "#calculation:\n",
- "PR = I*I*R\n",
- "Rt = R + Rw\n",
- "Pw = I*I*Rt\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the power dissipated in the load = \", PR,\" W\\n\"\n",
- "print \"\\n (b)the wattmeter reading. = \",Pw,\" W\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the power dissipated in the load = 800 W\n",
- "\n",
- "\n",
- " (b)the wattmeter reading. = 804.0 W"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 122</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the periodic time, (b) the frequency and (c) the peak-to-peak voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tc = 100E-6;# in s/cm\n",
- "Vc = 20;# in V/cm\n",
- "w = 5.2;# in cm ( width of one complete cycle )\n",
- "h = 3.6; # in cm ( peak-to-peak height of the display )\n",
- "\n",
- "#calculation:\n",
- "T = w*tc\n",
- "f = 1/T\n",
- "ptpv = h*Vc\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the periodic time, T = \", (T/1E-3),\" msec\\n\"\n",
- "print \"\\n (b)Frequency, f = \",round(f,2),\" Hz\\n\"\n",
- "print \"\\n (c)the peak-to-peak voltage = \",ptpv,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the periodic time, T = 0.52 msec\n",
- "\n",
- "\n",
- " (b)Frequency, f = 1923.08 Hz\n",
- "\n",
- "\n",
- " (c)the peak-to-peak voltage = 72.0 V"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 123</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the periodic time, (b) the frequency, (c) the magnitude of the pulse voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tc = 50E-3;# in s/cm\n",
- "Vc = 0.2;# in V/cm\n",
- "w = 3.5;# in cm ( width of one complete cycle )\n",
- "h = 3.4; # in cm ( peak-to-peak height of the display )\n",
- "\n",
- "#calculation:\n",
- "T = w*tc\n",
- "f = 1/T\n",
- "ptpv = h*Vc\n",
- "\n",
- "#Results \n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the periodic time, T = \", (T/1E-3),\" msec\\n\"\n",
- "print \"\\n (b)Frequency, f = \",round(f,2),\" Hz\\n\"\n",
- "print \"\\n (c)the peak-to-peak voltage = \",ptpv,\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the periodic time, T = 175.0 msec\n",
- "\n",
- "\n",
- " (b)Frequency, f = 5.71 Hz\n",
- "\n",
- "\n",
- " (c)the peak-to-peak voltage = 0.68 V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 123</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find, for the waveform, (a) the frequency, \n",
- "#(b) the peak-to-peak voltage, (c) the amplitude, (d) the r.m.s. value.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tc = 500E-6;# in s/cm\n",
- "Vc = 5;# in V/cm\n",
- "w = 4;# in cm ( width of one complete cycle )\n",
- "h = 5; # in cm ( peak-to-peak height of the display )\n",
- "\n",
- "#calculation:\n",
- "T = w*tc\n",
- "f = 1/T\n",
- "ptpv = h*Vc\n",
- "Amp = ptpv/2\n",
- "Vrms = Amp/(2**0.5)\n",
- "\n",
- "#Results \n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Frequency, f = \",f,\" Hz\\n\"\n",
- "print \"\\n (b)the peak-to-peak voltage = \",ptpv,\" V\\n\"\n",
- "print \"\\n (c)Amplitude = \",Amp,\" V\\n\"\n",
- "print \"\\n (d)r.m.s voltage = \",round(Vrms,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Frequency, f = 500.0 Hz\n",
- "\n",
- "\n",
- " (b)the peak-to-peak voltage = 25 V\n",
- "\n",
- "\n",
- " (c)Amplitude = 12.5 V\n",
- "\n",
- "\n",
- " (d)r.m.s voltage = 8.84 V"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 123</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) their frequency, (b) their r.m.s. values, (c) their phase difference.\n",
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#initializing the variables:\n",
- "tc = 100E-6;# in s/cm\n",
- "Vc = 2;# in V/cm\n",
- "w = 5;# in cm ( width of one complete cycle for both waveform )\n",
- "h1 = 2; # in cm ( peak-to-peak height of the display )\n",
- "h2 = 2.5; # in cm ( peak-to-peak height of the display\n",
- "\n",
- "#calculation:\n",
- "T = w*tc\n",
- "f = 1/T\n",
- "ptpv1 = h1*Vc\n",
- "Vrms1 = ptpv1/(2**0.5)\n",
- "ptpv2 = h2*Vc\n",
- "Vrms2 = ptpv2/(2**0.5)\n",
- "phi = 0.5*360/w\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Frequency, f = \",f,\" Hz\\n\"\n",
- "print \"\\n (b1)r.m.s voltage of 1st waveform = \",round(Vrms1,2),\" V\\n\"\n",
- "print \"\\n (b2)r.m.s voltage of 2nd waveform = \",round(Vrms2,2),\" V\\n\"\n",
- "print \"\\n (c)Phase difference = \",phi,\"deg\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Frequency, f = 2000.0 Hz\n",
- "\n",
- "\n",
- " (b1)r.m.s voltage of 1st waveform = 2.83 V\n",
- "\n",
- "\n",
- " (b2)r.m.s voltage of 2nd waveform = 3.54 V\n",
- "\n",
- "\n",
- " (c)Phase difference = 36.0 deg"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 127</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the decibel power ratio in each case.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "rP1 = 3;# ratio of two powers\n",
- "rP2 = 20;# ratio of two powers\n",
- "rP3 = 400;# ratio of two powers\n",
- "rP4 = 1/20;# ratio of two powers\n",
- "\n",
- "#calculation:\n",
- "X1 = 10*(1/2.303)*math.log(3)\n",
- "X2 = 10*(1/2.303)*math.log(20)\n",
- "X3 = 10*(1/2.303)*math.log(400)\n",
- "X4 = 10*(1/2.303)*math.log(1/20)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)decibel power ratio for power ratio 3 = \",round(X1,2),\" dB\\n\"\n",
- "print \"\\n (b)decibel power ratio for power ratio 20 = \",round(X2,2),\" dB\\n\"\n",
- "print \"\\n (c)decibel power ratio for power ratio 400 = \",round(X3,2),\" dB\\n\"\n",
- "print \"\\n (d)decibel power ratio for power ratio 1/20 = \",round(X4,2),\" dB\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)decibel power ratio for power ratio 3 = 4.77 dB\n",
- "\n",
- "\n",
- " (b)decibel power ratio for power ratio 20 = 13.01 dB\n",
- "\n",
- "\n",
- " (c)decibel power ratio for power ratio 400 = 26.02 dB\n",
- "\n",
- "\n",
- " (d)decibel power ratio for power ratio 1/20 = -13.01 dB"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 127</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the decibel current ratio assuming the input and load resistances of the system are equal.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I2 = 0.020;# in ampere\n",
- "I1 = 0.005;# in ampere\n",
- "\n",
- "#calculation:\n",
- "X = 20*math.log10(math.e)*math.log(I2/I1)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n decibel current ratio = \",round(X,2),\" dB\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " decibel current ratio = 12.04 dB\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 128</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the power loss in decibels.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "rP = 0.06;# power ratios rP = P2/P1\n",
- "\n",
- "#calculation:\n",
- "X = 10*math.log10(math.e)*math.log(rP)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n decibel Power ratios = \",round(X,2),\" dB\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " decibel Power ratios = -12.22 dB\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 128</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find its output power\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "X = 14;# decibal power ratio in dB\n",
- "P1 = 0.008;# in Watt\n",
- "\n",
- "#calculation:\n",
- "rP = 10**(X/10)\n",
- "P2 = rP*P1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output power P2 = \",round(P2,2),\" W\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output power P2 = 0.2 W"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 128</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of the input voltage\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "X = 27;# Voltage gain in decibels\n",
- "V2 = 4;# output voltage in Volts\n",
- "\n",
- "#calculation:\n",
- "V1 = V2/(10**(27/20))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n input Voltage V1 = \",round(V1,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " input Voltage V1 = 0.18 V"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 129</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the value of the unknown resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R2 = 100;# in ohms\n",
- "R3 = 400;# in ohms\n",
- "R4 = 10;# in ohms\n",
- "\n",
- "#calculation:\n",
- "R1 = R2*R3/R4\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n unknown resistance, R1 = \",R1,\" Ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " unknown resistance, R1 = 4000.0 Ohms"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 130</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. of a dry cell if balance is obtained with a length of 650 mm\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 1.0186;# in Volts\n",
- "l1 = 0.400;# in m\n",
- "l2 = 0.650;# in m\n",
- "\n",
- "#calculation:\n",
- "E2 = (l2/l1)*E1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the e.m.f. of a dry cell = \",round(E2,2),\" Volts\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the e.m.f. of a dry cell = 1.66 Volts"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 132</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the nominal value of the voltage across the resistor and its accuracy.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 0.0025;# in Amperes\n",
- "R = 5000;# in ohms\n",
- "e1 = 0.4;# in %\n",
- "e2 = 0.5;# in %\n",
- "\n",
- "#calculation:\n",
- "V = I*R\n",
- "em = e1 + e2\n",
- "Ve = em*V/100\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage V = \",V,\"V(+-)\",Ve,\"V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage V = 12.5 V(+-) 0.1125 V"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 132</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance of the resistor, and its accuracy of measurement\n",
- "#if both instruments have a limit of error of 2% of f.s.d.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 6.25;# in Amperes\n",
- "Im = 10;# max in Amperes\n",
- "V = 36.5;# in volts\n",
- "Vm = 50;# max in volts\n",
- "e = 2;# in %\n",
- "\n",
- "#calculation:\n",
- "R = V/I\n",
- "Ve = e*Vm/100 \n",
- "Ve1 = Ve*100/V# in %\n",
- "Ie = e*Im/100\n",
- "Ie1 = Ie*100/I# in %\n",
- "em = Ve1 + Ie1\n",
- "Re = em*R/100\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resistance R = \",R,\" ohms(+-)\",Re,\" ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resistance R = 5.84 ohms(+-) 0.34688 ohms"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 133</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the unknown resistance and its accuracy of measurement.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 1000;# in ohms\n",
- "R2 = 100;# in ohms\n",
- "R3 = 432.5;# in ohms\n",
- "e1 = 1;# in %\n",
- "e2 = 0.5;# in %\n",
- "e3 = 0.2;# in %\n",
- "\n",
- "#calculation:\n",
- "Rx = R2*R3/R1\n",
- "em = e1 + e2 + e3\n",
- "Re = em*Rx/100\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resistance R = \",Rx,\" ohms(+-)\",Re,\" ohms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resistance R = 43.25 ohms(+-) 0.73525 ohms"
- ]
- }
- ],
- "prompt_number": 21
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_10.ipynb index 443b9513..443b9513 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_10.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_11.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_11.ipynb index 3454218e..3454218e 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_11.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_11.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint.ipynb deleted file mode 100755 index 8adc225b..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint.ipynb +++ /dev/null @@ -1,85 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 12: Transistors</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 157</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Estimate the voltage gain Av, the current gain Ai and \n",
- "#the power gain Ap when an input current of 20 \u03bcA peak varies sinusoidally about a mean bias of 50 \u03bcA.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia = 0.040;# in Amperes\n",
- "I = 50;# in Amperes\n",
- "Rl = 1.2;# in Kohms\n",
- "Vc = 7; # in Volts\n",
- "R = 1; # in Kohms\n",
- "Ib = 50*1E-6; # in Amps\n",
- "Iic = 20*1E-6; #in Amps\n",
- "\n",
- "#calculation:\n",
- "Vi = (70 - 30)*1E-6 * 1000\n",
- "Vo = 3.6\n",
- "Av = Vo/Vi\n",
- "\n",
- "Io = 3*1E-3\n",
- "Ii = (70 - 30)*1E-6\n",
- "Ai = Io/Ii\n",
- "\n",
- "Ap = Av*Ai\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage gain, Av = \", round(Av,2)\n",
- "print \"\\n Current gain, Av = \", round(Ai,2)\n",
- "print \"\\n Power gain, Av = \", round(Ap,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage gain, Av = 90.0\n",
- "\n",
- " Current gain, Av = 75.0\n",
- "\n",
- " Power gain, Av = 6750.0"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_1.ipynb deleted file mode 100755 index 8adc225b..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_1.ipynb +++ /dev/null @@ -1,85 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 12: Transistors</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 157</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Estimate the voltage gain Av, the current gain Ai and \n",
- "#the power gain Ap when an input current of 20 \u03bcA peak varies sinusoidally about a mean bias of 50 \u03bcA.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia = 0.040;# in Amperes\n",
- "I = 50;# in Amperes\n",
- "Rl = 1.2;# in Kohms\n",
- "Vc = 7; # in Volts\n",
- "R = 1; # in Kohms\n",
- "Ib = 50*1E-6; # in Amps\n",
- "Iic = 20*1E-6; #in Amps\n",
- "\n",
- "#calculation:\n",
- "Vi = (70 - 30)*1E-6 * 1000\n",
- "Vo = 3.6\n",
- "Av = Vo/Vi\n",
- "\n",
- "Io = 3*1E-3\n",
- "Ii = (70 - 30)*1E-6\n",
- "Ai = Io/Ii\n",
- "\n",
- "Ap = Av*Ai\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage gain, Av = \", round(Av,2)\n",
- "print \"\\n Current gain, Av = \", round(Ai,2)\n",
- "print \"\\n Power gain, Av = \", round(Ap,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage gain, Av = 90.0\n",
- "\n",
- " Current gain, Av = 75.0\n",
- "\n",
- " Power gain, Av = 6750.0"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_2.ipynb deleted file mode 100755 index 8adc225b..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_2.ipynb +++ /dev/null @@ -1,85 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 12: Transistors</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 157</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Estimate the voltage gain Av, the current gain Ai and \n",
- "#the power gain Ap when an input current of 20 \u03bcA peak varies sinusoidally about a mean bias of 50 \u03bcA.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia = 0.040;# in Amperes\n",
- "I = 50;# in Amperes\n",
- "Rl = 1.2;# in Kohms\n",
- "Vc = 7; # in Volts\n",
- "R = 1; # in Kohms\n",
- "Ib = 50*1E-6; # in Amps\n",
- "Iic = 20*1E-6; #in Amps\n",
- "\n",
- "#calculation:\n",
- "Vi = (70 - 30)*1E-6 * 1000\n",
- "Vo = 3.6\n",
- "Av = Vo/Vi\n",
- "\n",
- "Io = 3*1E-3\n",
- "Ii = (70 - 30)*1E-6\n",
- "Ai = Io/Ii\n",
- "\n",
- "Ap = Av*Ai\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage gain, Av = \", round(Av,2)\n",
- "print \"\\n Current gain, Av = \", round(Ai,2)\n",
- "print \"\\n Power gain, Av = \", round(Ap,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage gain, Av = 90.0\n",
- "\n",
- " Current gain, Av = 75.0\n",
- "\n",
- " Power gain, Av = 6750.0"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_12.ipynb index 1cd2cf8d..1cd2cf8d 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_12.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint.ipynb deleted file mode 100755 index a468bf64..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint.ipynb +++ /dev/null @@ -1,1203 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 13: D.c. circuit theory</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 168</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#a) Find the unknown currents (b) Determine the value of e.m.f. E\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Iab = 50;# in ampere\n",
- "Ibc = 20;# in ampere\n",
- "Iec = 15;# in ampere\n",
- "Idf = 120;# in ampere\n",
- "Ifg = 40;# in ampere\n",
- "Iab = 50;# in ampere\n",
- "I = 2;# in ampere\n",
- "V1 = 4;# in volts\n",
- "V2 = 3;# in volts\n",
- "V3 = 6;# in volts\n",
- "R1 = 1;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R3 = 2.5;# in ohms\n",
- "R4 = 1.5;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = Iab - Ibc\n",
- "I2 = Ibc + Iec\n",
- "I3 = I1 - Idf\n",
- "I4 = Iec - I3\n",
- "I5 = Idf - Ifg\n",
- "# Applying Kirchhoff\u2019s voltage law and moving clockwise around the loop of Figure 13.3(b) starting at point A:\n",
- "E = I*R2 + I*R3 + I*R4 + I*R1 - V2 - V3 + V1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) unknown currents I1, I2, I3, I4, I5 are \",I1,\"A, \", I2,\"A, \", I3,\"A, \", I4,\"A, \", I5,\"A respetively\\n\"\n",
- "print \"\\n (b) value of e.m.f. E = \",E,\" Volts\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) unknown currents I1, I2, I3, I4, I5 are 30 A, 35 A, -90 A, 105 A, 80 A respetively\n",
- "\n",
- "\n",
- " (b) value of e.m.f. E = 9.0 Volts\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 168</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 2;# in volts\n",
- "r1 = 2;# in ohms\n",
- "r2 = 1;# in ohms\n",
- "R = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "#E1 = I1*(r1 + R) + I2*R,\n",
- "#E2 = I1*R + (R + r2)*I2,\n",
- "I2 = (E1*R - E2*(r1 + R))/(R**2 - (R+r1)*(R + r2))\n",
- "I1 = (E1 - I2*R)/(r1 + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n currents flowing are, I1 = \",round(I1,3),\"A, and I2 = \",round(-1*I2,3),\"A, \"\n",
- "print \"and current flowing in middle branch is I1 - I2 = \", round(I1 + I2, 3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " currents flowing are, I1 = 0.857 A, and I2 = 0.286 A, and current flowing in middle branch is I1 - I2 = 0.571 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 169</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 12;# in volts\n",
- "R1 = 0.5;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R3 = 5;# in ohms\n",
- "\n",
- "#calculation:\n",
- "#E1 + E2 = I1*R1 + I2*R2\n",
- "#E2 = - I1*R3 + I2*(R2 + R3) \n",
- "I2 = ((E1 + E2)*R3 + E2*R1)/(R2*R3 + (R2+R3)*R1)\n",
- "I1 = (E1 + E2 - I2*R2)/R1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n currents flowing are, I1 = \",round(I1,2),\"A, and I2 = \",round(I2,2),\"A\"\n",
- "print \" and in R3 branch is I1 - I2\", round(I1 - I2,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " currents flowing are, I1 = 6.52 A, and I2 = 6.37 A and in R3 branch is I1 - I2 0.15 A"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 170</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the currents in each of the resistors\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 54;# in volts\n",
- "I = 8;# in Amps\n",
- "R1 = 2;# in ohms\n",
- "R2 = 11;# in ohms\n",
- "R3 = 14;# in ohms\n",
- "R4 = 3;# in ohms\n",
- "R5 = 32;# in ohms\n",
- "\n",
- "#calculation:\n",
- "#V = (R1 + R2)*I1 - R2*I2\n",
- "#0 = (R1 + R3)*I1 - R5*I2 - R3*I\n",
- "I1 = V*R5/((R1 + R2)*R5 - (R1 + R3)*R2 + R3*I)\n",
- "I2 = -1*(V - I1*(R2 + R1))/R2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 2 ohm resistor = \",round(I1,0),\"A,\"\n",
- "print \"\\n the current flowing in the 14 ohm resistor = \",round(I - I1,0),\"A,\"\n",
- "print \"\\n the current flowing in the 32 ohm resistor = \",round(I2,0),\"A,\"\n",
- "print \"\\n the current flowing in the 11 ohm resistor = \",round(I1 - I2,0),\"A,\"\n",
- "print \"\\n the current flowing in the 3 ohm resistor = \",round(I - I1 + I2,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 2 ohm resistor = 5.0 A,\n",
- "\n",
- " the current flowing in the 14 ohm resistor = 3.0 A,\n",
- "\n",
- " the current flowing in the 32 ohm resistor = 1.0 A,\n",
- "\n",
- " the current flowing in the 11 ohm resistor = 4.0 A,\n",
- "\n",
- " the current flowing in the 3 ohm resistor = 4.0 A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 171</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current in each branch of the network by using the superposition theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 2;# in Volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 1;# in ohms\n",
- "R = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
- "I2 = (R2/(R + R2))*I1\n",
- "I3 = (R/(R + R2))*I1\n",
- "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
- "I5 = (R1/(R + R1))*I4\n",
- "I6 = (R/(R + R1))*I4\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resultant current flowing through source 1 = \",round(I1 - I6,3),\"A,\"\n",
- "print \"\\n Resultant current flowing through source 2 = \",round(I4 - I3,3),\"A,\"\n",
- "print \"\\n Resultant current flowing through resistor R, = \",round(I2 + I5,3),\"A,\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resultant current flowing through source 1 = 0.857 A,\n",
- "\n",
- " Resultant current flowing through source 2 = -0.286 A,\n",
- "\n",
- " Resultant current flowing through resistor R, = 0.571 A,"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 173</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find, using the superposition theorem, (a) the current flowing in and the pd across the 18 ohm resistor, \n",
- "#(b) the current in the 8 V battery and (c) the current in the 3 V battery.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 8;# in volts\n",
- "E2 = 3;# in Volts\n",
- "R1 = 3;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R = 18;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
- "I2 = (R/(R + R2))*I1\n",
- "I3 = (R2/(R + R2))*I1\n",
- "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
- "I5 = (R/(R + R1))*I4\n",
- "I6 = (R1/(R + R1))*I4\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resultant current in the 18 ohm resistor= \",round(I3 - I6,3),\"A \"\n",
- "print \"and P.d. across the 18 ohm resistor\",round((I3-I6)*R,3),\"V\"\n",
- "print \"\\n (b)the current in the 8 V battery= \",round(I1 + I5,3),\"A\"\n",
- "print \"\\n (c)current in the 3 V battery = \",round(I2 + I4,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resultant current in the 18 ohm resistor= 0.073 A and P.d. across the 18 ohm resistor 1.313 V\n",
- "\n",
- " (b)the current in the 8 V battery= 2.229 A\n",
- "\n",
- " (c)current in the 3 V battery = 2.156 A"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 176</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use Th\u00b4evenin\u2019s theorem to find the current flowing in the 10 ohm resistor for the circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 10;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 8;# in ohms\n",
- "R3 = 5;# in ohms\n",
- "R = 10;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = V/(R1 + R2)\n",
- "E = I1*R2\n",
- "r = R3 + R1*R2/(R1 + R2)\n",
- "I = E/(R + r)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 10 ohm resistor = 0.482 A"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 177</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 0.8 ohm resistor using Th\u00b4evenin\u2019s theorem.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 12;# in volts\n",
- "R1 = 1;# in ohms\n",
- "R2 = 4;# in ohms\n",
- "R3 = 5;# in ohms\n",
- "R = 0.8;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = V/(R1 + R2 + R3)\n",
- "E = I1*R2\n",
- "r = R2*(R1 + R3)/(R1 + R2 + R3)\n",
- "I = E/(R + r)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 0.8 ohm resistor = \",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 0.8 ohm resistor = 1.5 A"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 178</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 4 ohm resistor using Th\u00b4evenin\u2019s theorem.Find also the power dissipated in the 4 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 2;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 1;# in ohms\n",
- "R = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = (E1 - E2)/(R1 + R2)\n",
- "E = E1 - I1*R1\n",
- "r = R2*R1/(R1 + R2)\n",
- "I = E/(R + r)\n",
- "P = R*I**2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A and power dissipated in the 4 ohm resistor = \",round(P,3),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 4 ohm resistor = 0.571 A and power dissipated in the 4 ohm resistor = 1.306 W"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 178</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 3 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 24;# in volts\n",
- "R = 3;# in ohms\n",
- "R1 = 20;# in ohms\n",
- "R2 = 5;# in ohms\n",
- "R3 = 10;# in ohms\n",
- "R4 = 5/3;# in ohms\n",
- "\n",
- "#calculation:\n",
- "E = R3*V/(R3 + R2)\n",
- "r = R4 + R3*R2/(R3 + R2)\n",
- "I = E/(R + r)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 3 ohm resistor = 2.0 A"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 179</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 32 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E = 54;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 14;# in ohms\n",
- "R3 = 3;# in ohms\n",
- "R4 = 11;# in ohms\n",
- "R5 = 32;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Vac = R1*E/(R1 + R4)\n",
- "Vbc = R2*E/(R2 + R3)\n",
- "V = Vbc - Vac\n",
- "r = R4*R1/(R1 + R4) + R3*R2/(R3 + R2)\n",
- "I = V/(R5 + r)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 32 ohm resistor = \",round(I,0),\"A flowing from A to B\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 32 ohm resistor = 1.0 A flowing from A to B"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 181</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 10 ohm resistor using Norton's theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E = 10;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 8;# in ohms\n",
- "R3 = 5;# in ohms\n",
- "R = 10;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = E/R1\n",
- "r = R1*R2/(R1 + R2)\n",
- "I = r*Isc/(r + R3 + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 10 ohm resistor = 0.482 A"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 182</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 4 ohm resistor using Norton's theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 2;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 1;# in ohms\n",
- "R = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = E1/R1 + E2/R2\n",
- "r = R1*R2/(R1 + R2)\n",
- "I = r*Isc/(r + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 4 ohm resistor = 0.571 A"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 182</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 3 ohm resistor using Norton's theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 24;# in volts\n",
- "R = 3;# in ohms\n",
- "R1 = 20;# in ohms\n",
- "R2 = 5;# in ohms\n",
- "R3 = 10;# in ohms\n",
- "R4 = 5/3;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = V/R2\n",
- "r = R3*R2/(R3 + R2)\n",
- "I = r*Isc/(r + R4 + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 3 ohm resistor = 2.0 A"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 183</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 2 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 15;# in Amps\n",
- "R = 2;# in ohms\n",
- "R1 = 6;# in ohms\n",
- "R2 = 4;# in ohms\n",
- "R3 = 8;# in ohms\n",
- "R4 = 7;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = R1*I/(R1 + R2)\n",
- "r = (R1 + R2)*(R3 + R4)/(R3 + R1 + R4 + R2)\n",
- "I = r*Isc/(r + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 2 ohm resistor = \",round(I,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 2 ohm resistor = 6.75 A"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 185</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Convert the circuit to an equivalent Norton network.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 10;# in Volts\n",
- "R = 2;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = V/R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n equivalent Norton network contains Current Source of amp = \",round(Isc,0),\"A and a resistor of \",R,\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " equivalent Norton network contains Current Source of amp = 5.0 A and a resistor of 2 ohm"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 185</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Convert the circuit to an equivalent Thevenin network.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Isc = 4;# in Amps\n",
- "R = 3;# in ohms\n",
- "\n",
- "#calculation:\n",
- "E = Isc*R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,0),\"V and a resistor of \",R,\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " equivalent Thevenin network contains Voltage Source of 12.0 V and a resistor of 3 ohm"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 185</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)Convert the circuit to an equivalent Thevenin networkby initially converting to a Norton equivalent circuit.\n",
- "#(b)Determine the current flowing in the 1.8 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 12;# in Volts\n",
- "E2 = 24;# in Volts\n",
- "R1 = 3;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R = 1.8;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc1 = E1/R1\n",
- "Isc2 = E2/R2\n",
- "I1 = Isc1 + Isc2\n",
- "r = R1*R2/(R1 + R2)\n",
- "E = I1*r\n",
- "I = E/(r + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (A)equivalent Norton network contains Current Source of \",round(I1,1),\"A and a resistor of \",r,\" ohm\"\n",
- "print \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",r,\" ohm\"\n",
- "print \"\\n (B)the current flowing in the 1.8 ohm resistor is \",round(I,1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (A)equivalent Norton network contains Current Source of 16.0 A and a resistor of 1.2 ohm\n",
- "\n",
- " equivalent Thevenin network contains Voltage Source of 19.2 V and a resistor of 1.2 ohm\n",
- "\n",
- " (B)the current flowing in the 1.8 ohm resistor is 6.4 A"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 186</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)Determine by successive conversions between Th\u00b4evenin and Norton equivalent networks \n",
- "#a Thevenin equivalent circuit for terminals AB.\n",
- "#(b)determine the current flowing in the 200 ohm resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 10;# in Volts\n",
- "E2 = 6;# in Volts\n",
- "I1 = 0.001;#in Amp\n",
- "R1 = 2000;# in ohms\n",
- "R2 = 3000;# in ohms\n",
- "R3 = 600;# in ohms\n",
- "R = 200;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc1 = E1/R1\n",
- "Isc2 = E2/R2\n",
- "I2 = Isc1 + Isc2\n",
- "r1 = R1*R2/(R1 + R2)\n",
- "E = I2*r1 - I1*R3\n",
- "r = r1 + R3\n",
- "I = E/(r + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (A)equivalent Norton network contains Current Source of \",round(I2*1000,0),\"mA and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
- "print \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
- "print \"\\n (B)the current flowing in the 200 ohm resistor is \",round(I*1000,1),\"mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (A)equivalent Norton network contains Current Source of 7.0 mA and a resistor of 1.2 Kohm\n",
- "\n",
- " equivalent Thevenin network contains Voltage Source of 7.8 V and a resistor of 1.2 Kohm\n",
- "\n",
- " (B)the current flowing in the 200 ohm resistor is 3.9 mA"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 188</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the power dissipated by the load in each case.\n",
- "#Plot a graph of RL (horizontally) against power (vertically) and determine the maximum power dissipated.\n",
- "from __future__ import division\n",
- "import math\n",
- "#from pylab import *\n",
- "%pylab inline\n",
- "#initializing the variables:\n",
- "E = 6;# in Volts\n",
- "R = 2.5;# in ohms\n",
- "\n",
- "#calculation:\n",
- "RL = []\n",
- "P = []\n",
- "k = []\n",
- "for h in range(6):\n",
- " RL.append(h - 0.5)\n",
- " k = h - 0.5\n",
- " P.append(k*((E/(R + k))**2))\n",
- "fig = plt.figure()\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(RL,P,'-')\n",
- "#plot(RL,P,'-')\n",
- "xlabel('RL(ohm)')\n",
- "ylabel('Power(W)')\n",
- "show()\n",
- "Pmax= R*(E/(2*R))**2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n A graph of RL against P is shown in Figure\" \n",
- "print \" The maximum value of power is\", Pmax,\"W which occurs when RL =\",R,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
- "For more information, type 'help(pylab)'."
- ]
- },
- {
- "output_type": "display_data",
- "png": 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p6emiS1PMwIEDAQBBQUGYOnUqPvzwQ8ybN6/N57rM3z5q6c/dcsstAOQVP6Wl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\nZLqKDYkAAAAASUVORK5CYII=\n"
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " A graph of RL against P is shown in Figure\n",
- " The maximum value of power is 3.6 W which occurs when RL = 2.5 ohm"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 188</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#State the value of load resistance that gives maximum power dissipation and determine the value of this power.\n",
- "from __future__ import division\n",
- "import math\n",
- "from pylab import *\n",
- "#initializing the variables:\n",
- "E = 30;# in Volts\n",
- "R = 1.5;# in ohms\n",
- "\n",
- "#calculation:\n",
- "RL = R\n",
- "I = E/(R + RL)\n",
- "P = I**2*RL\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \" The maximum value of power is\", P,\"W which occurs when RL =\",RL,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- " The maximum value of power is 150.0 W which occurs when RL = 1.5 ohm"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 189</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the value of the load resistor RL that gives maximum power dissipation and determine the value of this power.\n",
- "from __future__ import division\n",
- "import math\n",
- "from pylab import *\n",
- "#initializing the variables:\n",
- "V = 15;# in Volts\n",
- "R1 = 3;# in ohms\n",
- "R2 = 12;# in ohms\n",
- "\n",
- "#calculation:\n",
- "E = (R2/(R2+ R1))*V\n",
- "r = R1*R2/(R1 + R2)\n",
- "RL = r\n",
- "I = E/(r + RL)\n",
- "P = I**2*RL\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \" The maximum value of power is\", P,\"W which occurs when Total Load RL =\",RL,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- " The maximum value of power is 15.0 W which occurs when Total Load RL = 2.4 ohm"
- ]
- }
- ],
- "prompt_number": 22
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_1.ipynb deleted file mode 100755 index a468bf64..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_1.ipynb +++ /dev/null @@ -1,1203 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 13: D.c. circuit theory</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 168</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#a) Find the unknown currents (b) Determine the value of e.m.f. E\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Iab = 50;# in ampere\n",
- "Ibc = 20;# in ampere\n",
- "Iec = 15;# in ampere\n",
- "Idf = 120;# in ampere\n",
- "Ifg = 40;# in ampere\n",
- "Iab = 50;# in ampere\n",
- "I = 2;# in ampere\n",
- "V1 = 4;# in volts\n",
- "V2 = 3;# in volts\n",
- "V3 = 6;# in volts\n",
- "R1 = 1;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R3 = 2.5;# in ohms\n",
- "R4 = 1.5;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = Iab - Ibc\n",
- "I2 = Ibc + Iec\n",
- "I3 = I1 - Idf\n",
- "I4 = Iec - I3\n",
- "I5 = Idf - Ifg\n",
- "# Applying Kirchhoff\u2019s voltage law and moving clockwise around the loop of Figure 13.3(b) starting at point A:\n",
- "E = I*R2 + I*R3 + I*R4 + I*R1 - V2 - V3 + V1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) unknown currents I1, I2, I3, I4, I5 are \",I1,\"A, \", I2,\"A, \", I3,\"A, \", I4,\"A, \", I5,\"A respetively\\n\"\n",
- "print \"\\n (b) value of e.m.f. E = \",E,\" Volts\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) unknown currents I1, I2, I3, I4, I5 are 30 A, 35 A, -90 A, 105 A, 80 A respetively\n",
- "\n",
- "\n",
- " (b) value of e.m.f. E = 9.0 Volts\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 168</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 2;# in volts\n",
- "r1 = 2;# in ohms\n",
- "r2 = 1;# in ohms\n",
- "R = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "#E1 = I1*(r1 + R) + I2*R,\n",
- "#E2 = I1*R + (R + r2)*I2,\n",
- "I2 = (E1*R - E2*(r1 + R))/(R**2 - (R+r1)*(R + r2))\n",
- "I1 = (E1 - I2*R)/(r1 + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n currents flowing are, I1 = \",round(I1,3),\"A, and I2 = \",round(-1*I2,3),\"A, \"\n",
- "print \"and current flowing in middle branch is I1 - I2 = \", round(I1 + I2, 3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " currents flowing are, I1 = 0.857 A, and I2 = 0.286 A, and current flowing in middle branch is I1 - I2 = 0.571 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 169</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 12;# in volts\n",
- "R1 = 0.5;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R3 = 5;# in ohms\n",
- "\n",
- "#calculation:\n",
- "#E1 + E2 = I1*R1 + I2*R2\n",
- "#E2 = - I1*R3 + I2*(R2 + R3) \n",
- "I2 = ((E1 + E2)*R3 + E2*R1)/(R2*R3 + (R2+R3)*R1)\n",
- "I1 = (E1 + E2 - I2*R2)/R1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n currents flowing are, I1 = \",round(I1,2),\"A, and I2 = \",round(I2,2),\"A\"\n",
- "print \" and in R3 branch is I1 - I2\", round(I1 - I2,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " currents flowing are, I1 = 6.52 A, and I2 = 6.37 A and in R3 branch is I1 - I2 0.15 A"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 170</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the currents in each of the resistors\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 54;# in volts\n",
- "I = 8;# in Amps\n",
- "R1 = 2;# in ohms\n",
- "R2 = 11;# in ohms\n",
- "R3 = 14;# in ohms\n",
- "R4 = 3;# in ohms\n",
- "R5 = 32;# in ohms\n",
- "\n",
- "#calculation:\n",
- "#V = (R1 + R2)*I1 - R2*I2\n",
- "#0 = (R1 + R3)*I1 - R5*I2 - R3*I\n",
- "I1 = V*R5/((R1 + R2)*R5 - (R1 + R3)*R2 + R3*I)\n",
- "I2 = -1*(V - I1*(R2 + R1))/R2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 2 ohm resistor = \",round(I1,0),\"A,\"\n",
- "print \"\\n the current flowing in the 14 ohm resistor = \",round(I - I1,0),\"A,\"\n",
- "print \"\\n the current flowing in the 32 ohm resistor = \",round(I2,0),\"A,\"\n",
- "print \"\\n the current flowing in the 11 ohm resistor = \",round(I1 - I2,0),\"A,\"\n",
- "print \"\\n the current flowing in the 3 ohm resistor = \",round(I - I1 + I2,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 2 ohm resistor = 5.0 A,\n",
- "\n",
- " the current flowing in the 14 ohm resistor = 3.0 A,\n",
- "\n",
- " the current flowing in the 32 ohm resistor = 1.0 A,\n",
- "\n",
- " the current flowing in the 11 ohm resistor = 4.0 A,\n",
- "\n",
- " the current flowing in the 3 ohm resistor = 4.0 A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 171</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current in each branch of the network by using the superposition theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 2;# in Volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 1;# in ohms\n",
- "R = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
- "I2 = (R2/(R + R2))*I1\n",
- "I3 = (R/(R + R2))*I1\n",
- "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
- "I5 = (R1/(R + R1))*I4\n",
- "I6 = (R/(R + R1))*I4\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resultant current flowing through source 1 = \",round(I1 - I6,3),\"A,\"\n",
- "print \"\\n Resultant current flowing through source 2 = \",round(I4 - I3,3),\"A,\"\n",
- "print \"\\n Resultant current flowing through resistor R, = \",round(I2 + I5,3),\"A,\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resultant current flowing through source 1 = 0.857 A,\n",
- "\n",
- " Resultant current flowing through source 2 = -0.286 A,\n",
- "\n",
- " Resultant current flowing through resistor R, = 0.571 A,"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 173</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find, using the superposition theorem, (a) the current flowing in and the pd across the 18 ohm resistor, \n",
- "#(b) the current in the 8 V battery and (c) the current in the 3 V battery.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 8;# in volts\n",
- "E2 = 3;# in Volts\n",
- "R1 = 3;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R = 18;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
- "I2 = (R/(R + R2))*I1\n",
- "I3 = (R2/(R + R2))*I1\n",
- "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
- "I5 = (R/(R + R1))*I4\n",
- "I6 = (R1/(R + R1))*I4\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resultant current in the 18 ohm resistor= \",round(I3 - I6,3),\"A \"\n",
- "print \"and P.d. across the 18 ohm resistor\",round((I3-I6)*R,3),\"V\"\n",
- "print \"\\n (b)the current in the 8 V battery= \",round(I1 + I5,3),\"A\"\n",
- "print \"\\n (c)current in the 3 V battery = \",round(I2 + I4,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resultant current in the 18 ohm resistor= 0.073 A and P.d. across the 18 ohm resistor 1.313 V\n",
- "\n",
- " (b)the current in the 8 V battery= 2.229 A\n",
- "\n",
- " (c)current in the 3 V battery = 2.156 A"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 176</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use Th\u00b4evenin\u2019s theorem to find the current flowing in the 10 ohm resistor for the circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 10;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 8;# in ohms\n",
- "R3 = 5;# in ohms\n",
- "R = 10;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = V/(R1 + R2)\n",
- "E = I1*R2\n",
- "r = R3 + R1*R2/(R1 + R2)\n",
- "I = E/(R + r)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 10 ohm resistor = 0.482 A"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 177</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 0.8 ohm resistor using Th\u00b4evenin\u2019s theorem.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 12;# in volts\n",
- "R1 = 1;# in ohms\n",
- "R2 = 4;# in ohms\n",
- "R3 = 5;# in ohms\n",
- "R = 0.8;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = V/(R1 + R2 + R3)\n",
- "E = I1*R2\n",
- "r = R2*(R1 + R3)/(R1 + R2 + R3)\n",
- "I = E/(R + r)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 0.8 ohm resistor = \",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 0.8 ohm resistor = 1.5 A"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 178</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 4 ohm resistor using Th\u00b4evenin\u2019s theorem.Find also the power dissipated in the 4 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 2;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 1;# in ohms\n",
- "R = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = (E1 - E2)/(R1 + R2)\n",
- "E = E1 - I1*R1\n",
- "r = R2*R1/(R1 + R2)\n",
- "I = E/(R + r)\n",
- "P = R*I**2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A and power dissipated in the 4 ohm resistor = \",round(P,3),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 4 ohm resistor = 0.571 A and power dissipated in the 4 ohm resistor = 1.306 W"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 178</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 3 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 24;# in volts\n",
- "R = 3;# in ohms\n",
- "R1 = 20;# in ohms\n",
- "R2 = 5;# in ohms\n",
- "R3 = 10;# in ohms\n",
- "R4 = 5/3;# in ohms\n",
- "\n",
- "#calculation:\n",
- "E = R3*V/(R3 + R2)\n",
- "r = R4 + R3*R2/(R3 + R2)\n",
- "I = E/(R + r)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 3 ohm resistor = 2.0 A"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 179</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 32 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E = 54;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 14;# in ohms\n",
- "R3 = 3;# in ohms\n",
- "R4 = 11;# in ohms\n",
- "R5 = 32;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Vac = R1*E/(R1 + R4)\n",
- "Vbc = R2*E/(R2 + R3)\n",
- "V = Vbc - Vac\n",
- "r = R4*R1/(R1 + R4) + R3*R2/(R3 + R2)\n",
- "I = V/(R5 + r)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 32 ohm resistor = \",round(I,0),\"A flowing from A to B\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 32 ohm resistor = 1.0 A flowing from A to B"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 181</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 10 ohm resistor using Norton's theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E = 10;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 8;# in ohms\n",
- "R3 = 5;# in ohms\n",
- "R = 10;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = E/R1\n",
- "r = R1*R2/(R1 + R2)\n",
- "I = r*Isc/(r + R3 + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 10 ohm resistor = 0.482 A"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 182</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 4 ohm resistor using Norton's theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 2;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 1;# in ohms\n",
- "R = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = E1/R1 + E2/R2\n",
- "r = R1*R2/(R1 + R2)\n",
- "I = r*Isc/(r + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 4 ohm resistor = 0.571 A"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 182</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 3 ohm resistor using Norton's theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 24;# in volts\n",
- "R = 3;# in ohms\n",
- "R1 = 20;# in ohms\n",
- "R2 = 5;# in ohms\n",
- "R3 = 10;# in ohms\n",
- "R4 = 5/3;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = V/R2\n",
- "r = R3*R2/(R3 + R2)\n",
- "I = r*Isc/(r + R4 + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 3 ohm resistor = 2.0 A"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 183</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 2 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 15;# in Amps\n",
- "R = 2;# in ohms\n",
- "R1 = 6;# in ohms\n",
- "R2 = 4;# in ohms\n",
- "R3 = 8;# in ohms\n",
- "R4 = 7;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = R1*I/(R1 + R2)\n",
- "r = (R1 + R2)*(R3 + R4)/(R3 + R1 + R4 + R2)\n",
- "I = r*Isc/(r + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 2 ohm resistor = \",round(I,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 2 ohm resistor = 6.75 A"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 185</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Convert the circuit to an equivalent Norton network.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 10;# in Volts\n",
- "R = 2;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = V/R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n equivalent Norton network contains Current Source of amp = \",round(Isc,0),\"A and a resistor of \",R,\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " equivalent Norton network contains Current Source of amp = 5.0 A and a resistor of 2 ohm"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 185</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Convert the circuit to an equivalent Thevenin network.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Isc = 4;# in Amps\n",
- "R = 3;# in ohms\n",
- "\n",
- "#calculation:\n",
- "E = Isc*R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,0),\"V and a resistor of \",R,\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " equivalent Thevenin network contains Voltage Source of 12.0 V and a resistor of 3 ohm"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 185</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)Convert the circuit to an equivalent Thevenin networkby initially converting to a Norton equivalent circuit.\n",
- "#(b)Determine the current flowing in the 1.8 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 12;# in Volts\n",
- "E2 = 24;# in Volts\n",
- "R1 = 3;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R = 1.8;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc1 = E1/R1\n",
- "Isc2 = E2/R2\n",
- "I1 = Isc1 + Isc2\n",
- "r = R1*R2/(R1 + R2)\n",
- "E = I1*r\n",
- "I = E/(r + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (A)equivalent Norton network contains Current Source of \",round(I1,1),\"A and a resistor of \",r,\" ohm\"\n",
- "print \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",r,\" ohm\"\n",
- "print \"\\n (B)the current flowing in the 1.8 ohm resistor is \",round(I,1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (A)equivalent Norton network contains Current Source of 16.0 A and a resistor of 1.2 ohm\n",
- "\n",
- " equivalent Thevenin network contains Voltage Source of 19.2 V and a resistor of 1.2 ohm\n",
- "\n",
- " (B)the current flowing in the 1.8 ohm resistor is 6.4 A"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 186</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)Determine by successive conversions between Th\u00b4evenin and Norton equivalent networks \n",
- "#a Thevenin equivalent circuit for terminals AB.\n",
- "#(b)determine the current flowing in the 200 ohm resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 10;# in Volts\n",
- "E2 = 6;# in Volts\n",
- "I1 = 0.001;#in Amp\n",
- "R1 = 2000;# in ohms\n",
- "R2 = 3000;# in ohms\n",
- "R3 = 600;# in ohms\n",
- "R = 200;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc1 = E1/R1\n",
- "Isc2 = E2/R2\n",
- "I2 = Isc1 + Isc2\n",
- "r1 = R1*R2/(R1 + R2)\n",
- "E = I2*r1 - I1*R3\n",
- "r = r1 + R3\n",
- "I = E/(r + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (A)equivalent Norton network contains Current Source of \",round(I2*1000,0),\"mA and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
- "print \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
- "print \"\\n (B)the current flowing in the 200 ohm resistor is \",round(I*1000,1),\"mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (A)equivalent Norton network contains Current Source of 7.0 mA and a resistor of 1.2 Kohm\n",
- "\n",
- " equivalent Thevenin network contains Voltage Source of 7.8 V and a resistor of 1.2 Kohm\n",
- "\n",
- " (B)the current flowing in the 200 ohm resistor is 3.9 mA"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 188</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the power dissipated by the load in each case.\n",
- "#Plot a graph of RL (horizontally) against power (vertically) and determine the maximum power dissipated.\n",
- "from __future__ import division\n",
- "import math\n",
- "#from pylab import *\n",
- "%pylab inline\n",
- "#initializing the variables:\n",
- "E = 6;# in Volts\n",
- "R = 2.5;# in ohms\n",
- "\n",
- "#calculation:\n",
- "RL = []\n",
- "P = []\n",
- "k = []\n",
- "for h in range(6):\n",
- " RL.append(h - 0.5)\n",
- " k = h - 0.5\n",
- " P.append(k*((E/(R + k))**2))\n",
- "fig = plt.figure()\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(RL,P,'-')\n",
- "#plot(RL,P,'-')\n",
- "xlabel('RL(ohm)')\n",
- "ylabel('Power(W)')\n",
- "show()\n",
- "Pmax= R*(E/(2*R))**2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n A graph of RL against P is shown in Figure\" \n",
- "print \" The maximum value of power is\", Pmax,\"W which occurs when RL =\",R,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
- "For more information, type 'help(pylab)'."
- ]
- },
- {
- "output_type": "display_data",
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p6emiS1PMwIEDAQBBQUGYOnUqPvzwQ8ybN6/N57rM3z5q6c/dcsstAOQVP6Wl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\nZLqKDYkAAAAASUVORK5CYII=\n"
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " A graph of RL against P is shown in Figure\n",
- " The maximum value of power is 3.6 W which occurs when RL = 2.5 ohm"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 188</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#State the value of load resistance that gives maximum power dissipation and determine the value of this power.\n",
- "from __future__ import division\n",
- "import math\n",
- "from pylab import *\n",
- "#initializing the variables:\n",
- "E = 30;# in Volts\n",
- "R = 1.5;# in ohms\n",
- "\n",
- "#calculation:\n",
- "RL = R\n",
- "I = E/(R + RL)\n",
- "P = I**2*RL\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \" The maximum value of power is\", P,\"W which occurs when RL =\",RL,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- " The maximum value of power is 150.0 W which occurs when RL = 1.5 ohm"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 189</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the value of the load resistor RL that gives maximum power dissipation and determine the value of this power.\n",
- "from __future__ import division\n",
- "import math\n",
- "from pylab import *\n",
- "#initializing the variables:\n",
- "V = 15;# in Volts\n",
- "R1 = 3;# in ohms\n",
- "R2 = 12;# in ohms\n",
- "\n",
- "#calculation:\n",
- "E = (R2/(R2+ R1))*V\n",
- "r = R1*R2/(R1 + R2)\n",
- "RL = r\n",
- "I = E/(r + RL)\n",
- "P = I**2*RL\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \" The maximum value of power is\", P,\"W which occurs when Total Load RL =\",RL,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- " The maximum value of power is 15.0 W which occurs when Total Load RL = 2.4 ohm"
- ]
- }
- ],
- "prompt_number": 22
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_2.ipynb deleted file mode 100755 index a468bf64..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_2.ipynb +++ /dev/null @@ -1,1203 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 13: D.c. circuit theory</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 168</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#a) Find the unknown currents (b) Determine the value of e.m.f. E\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Iab = 50;# in ampere\n",
- "Ibc = 20;# in ampere\n",
- "Iec = 15;# in ampere\n",
- "Idf = 120;# in ampere\n",
- "Ifg = 40;# in ampere\n",
- "Iab = 50;# in ampere\n",
- "I = 2;# in ampere\n",
- "V1 = 4;# in volts\n",
- "V2 = 3;# in volts\n",
- "V3 = 6;# in volts\n",
- "R1 = 1;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R3 = 2.5;# in ohms\n",
- "R4 = 1.5;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = Iab - Ibc\n",
- "I2 = Ibc + Iec\n",
- "I3 = I1 - Idf\n",
- "I4 = Iec - I3\n",
- "I5 = Idf - Ifg\n",
- "# Applying Kirchhoff\u2019s voltage law and moving clockwise around the loop of Figure 13.3(b) starting at point A:\n",
- "E = I*R2 + I*R3 + I*R4 + I*R1 - V2 - V3 + V1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) unknown currents I1, I2, I3, I4, I5 are \",I1,\"A, \", I2,\"A, \", I3,\"A, \", I4,\"A, \", I5,\"A respetively\\n\"\n",
- "print \"\\n (b) value of e.m.f. E = \",E,\" Volts\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) unknown currents I1, I2, I3, I4, I5 are 30 A, 35 A, -90 A, 105 A, 80 A respetively\n",
- "\n",
- "\n",
- " (b) value of e.m.f. E = 9.0 Volts\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 168</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 2;# in volts\n",
- "r1 = 2;# in ohms\n",
- "r2 = 1;# in ohms\n",
- "R = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "#E1 = I1*(r1 + R) + I2*R,\n",
- "#E2 = I1*R + (R + r2)*I2,\n",
- "I2 = (E1*R - E2*(r1 + R))/(R**2 - (R+r1)*(R + r2))\n",
- "I1 = (E1 - I2*R)/(r1 + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n currents flowing are, I1 = \",round(I1,3),\"A, and I2 = \",round(-1*I2,3),\"A, \"\n",
- "print \"and current flowing in middle branch is I1 - I2 = \", round(I1 + I2, 3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " currents flowing are, I1 = 0.857 A, and I2 = 0.286 A, and current flowing in middle branch is I1 - I2 = 0.571 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 169</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 12;# in volts\n",
- "R1 = 0.5;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R3 = 5;# in ohms\n",
- "\n",
- "#calculation:\n",
- "#E1 + E2 = I1*R1 + I2*R2\n",
- "#E2 = - I1*R3 + I2*(R2 + R3) \n",
- "I2 = ((E1 + E2)*R3 + E2*R1)/(R2*R3 + (R2+R3)*R1)\n",
- "I1 = (E1 + E2 - I2*R2)/R1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n currents flowing are, I1 = \",round(I1,2),\"A, and I2 = \",round(I2,2),\"A\"\n",
- "print \" and in R3 branch is I1 - I2\", round(I1 - I2,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " currents flowing are, I1 = 6.52 A, and I2 = 6.37 A and in R3 branch is I1 - I2 0.15 A"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 170</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the currents in each of the resistors\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 54;# in volts\n",
- "I = 8;# in Amps\n",
- "R1 = 2;# in ohms\n",
- "R2 = 11;# in ohms\n",
- "R3 = 14;# in ohms\n",
- "R4 = 3;# in ohms\n",
- "R5 = 32;# in ohms\n",
- "\n",
- "#calculation:\n",
- "#V = (R1 + R2)*I1 - R2*I2\n",
- "#0 = (R1 + R3)*I1 - R5*I2 - R3*I\n",
- "I1 = V*R5/((R1 + R2)*R5 - (R1 + R3)*R2 + R3*I)\n",
- "I2 = -1*(V - I1*(R2 + R1))/R2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 2 ohm resistor = \",round(I1,0),\"A,\"\n",
- "print \"\\n the current flowing in the 14 ohm resistor = \",round(I - I1,0),\"A,\"\n",
- "print \"\\n the current flowing in the 32 ohm resistor = \",round(I2,0),\"A,\"\n",
- "print \"\\n the current flowing in the 11 ohm resistor = \",round(I1 - I2,0),\"A,\"\n",
- "print \"\\n the current flowing in the 3 ohm resistor = \",round(I - I1 + I2,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 2 ohm resistor = 5.0 A,\n",
- "\n",
- " the current flowing in the 14 ohm resistor = 3.0 A,\n",
- "\n",
- " the current flowing in the 32 ohm resistor = 1.0 A,\n",
- "\n",
- " the current flowing in the 11 ohm resistor = 4.0 A,\n",
- "\n",
- " the current flowing in the 3 ohm resistor = 4.0 A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 171</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current in each branch of the network by using the superposition theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 2;# in Volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 1;# in ohms\n",
- "R = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
- "I2 = (R2/(R + R2))*I1\n",
- "I3 = (R/(R + R2))*I1\n",
- "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
- "I5 = (R1/(R + R1))*I4\n",
- "I6 = (R/(R + R1))*I4\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resultant current flowing through source 1 = \",round(I1 - I6,3),\"A,\"\n",
- "print \"\\n Resultant current flowing through source 2 = \",round(I4 - I3,3),\"A,\"\n",
- "print \"\\n Resultant current flowing through resistor R, = \",round(I2 + I5,3),\"A,\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resultant current flowing through source 1 = 0.857 A,\n",
- "\n",
- " Resultant current flowing through source 2 = -0.286 A,\n",
- "\n",
- " Resultant current flowing through resistor R, = 0.571 A,"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 173</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find, using the superposition theorem, (a) the current flowing in and the pd across the 18 ohm resistor, \n",
- "#(b) the current in the 8 V battery and (c) the current in the 3 V battery.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 8;# in volts\n",
- "E2 = 3;# in Volts\n",
- "R1 = 3;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R = 18;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
- "I2 = (R/(R + R2))*I1\n",
- "I3 = (R2/(R + R2))*I1\n",
- "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
- "I5 = (R/(R + R1))*I4\n",
- "I6 = (R1/(R + R1))*I4\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resultant current in the 18 ohm resistor= \",round(I3 - I6,3),\"A \"\n",
- "print \"and P.d. across the 18 ohm resistor\",round((I3-I6)*R,3),\"V\"\n",
- "print \"\\n (b)the current in the 8 V battery= \",round(I1 + I5,3),\"A\"\n",
- "print \"\\n (c)current in the 3 V battery = \",round(I2 + I4,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resultant current in the 18 ohm resistor= 0.073 A and P.d. across the 18 ohm resistor 1.313 V\n",
- "\n",
- " (b)the current in the 8 V battery= 2.229 A\n",
- "\n",
- " (c)current in the 3 V battery = 2.156 A"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 176</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use Th\u00b4evenin\u2019s theorem to find the current flowing in the 10 ohm resistor for the circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 10;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 8;# in ohms\n",
- "R3 = 5;# in ohms\n",
- "R = 10;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = V/(R1 + R2)\n",
- "E = I1*R2\n",
- "r = R3 + R1*R2/(R1 + R2)\n",
- "I = E/(R + r)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 10 ohm resistor = 0.482 A"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 177</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 0.8 ohm resistor using Th\u00b4evenin\u2019s theorem.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 12;# in volts\n",
- "R1 = 1;# in ohms\n",
- "R2 = 4;# in ohms\n",
- "R3 = 5;# in ohms\n",
- "R = 0.8;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = V/(R1 + R2 + R3)\n",
- "E = I1*R2\n",
- "r = R2*(R1 + R3)/(R1 + R2 + R3)\n",
- "I = E/(R + r)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 0.8 ohm resistor = \",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 0.8 ohm resistor = 1.5 A"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 178</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 4 ohm resistor using Th\u00b4evenin\u2019s theorem.Find also the power dissipated in the 4 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 2;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 1;# in ohms\n",
- "R = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "I1 = (E1 - E2)/(R1 + R2)\n",
- "E = E1 - I1*R1\n",
- "r = R2*R1/(R1 + R2)\n",
- "I = E/(R + r)\n",
- "P = R*I**2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A and power dissipated in the 4 ohm resistor = \",round(P,3),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 4 ohm resistor = 0.571 A and power dissipated in the 4 ohm resistor = 1.306 W"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 178</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 3 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 24;# in volts\n",
- "R = 3;# in ohms\n",
- "R1 = 20;# in ohms\n",
- "R2 = 5;# in ohms\n",
- "R3 = 10;# in ohms\n",
- "R4 = 5/3;# in ohms\n",
- "\n",
- "#calculation:\n",
- "E = R3*V/(R3 + R2)\n",
- "r = R4 + R3*R2/(R3 + R2)\n",
- "I = E/(R + r)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 3 ohm resistor = 2.0 A"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 179</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 32 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E = 54;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 14;# in ohms\n",
- "R3 = 3;# in ohms\n",
- "R4 = 11;# in ohms\n",
- "R5 = 32;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Vac = R1*E/(R1 + R4)\n",
- "Vbc = R2*E/(R2 + R3)\n",
- "V = Vbc - Vac\n",
- "r = R4*R1/(R1 + R4) + R3*R2/(R3 + R2)\n",
- "I = V/(R5 + r)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 32 ohm resistor = \",round(I,0),\"A flowing from A to B\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 32 ohm resistor = 1.0 A flowing from A to B"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 181</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 10 ohm resistor using Norton's theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E = 10;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 8;# in ohms\n",
- "R3 = 5;# in ohms\n",
- "R = 10;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = E/R1\n",
- "r = R1*R2/(R1 + R2)\n",
- "I = r*Isc/(r + R3 + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 10 ohm resistor = 0.482 A"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 182</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 4 ohm resistor using Norton's theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 4;# in volts\n",
- "E2 = 2;# in volts\n",
- "R1 = 2;# in ohms\n",
- "R2 = 1;# in ohms\n",
- "R = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = E1/R1 + E2/R2\n",
- "r = R1*R2/(R1 + R2)\n",
- "I = r*Isc/(r + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 4 ohm resistor = 0.571 A"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 182</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 3 ohm resistor using Norton's theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 24;# in volts\n",
- "R = 3;# in ohms\n",
- "R1 = 20;# in ohms\n",
- "R2 = 5;# in ohms\n",
- "R3 = 10;# in ohms\n",
- "R4 = 5/3;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = V/R2\n",
- "r = R3*R2/(R3 + R2)\n",
- "I = r*Isc/(r + R4 + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 3 ohm resistor = 2.0 A"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 183</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in the 2 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 15;# in Amps\n",
- "R = 2;# in ohms\n",
- "R1 = 6;# in ohms\n",
- "R2 = 4;# in ohms\n",
- "R3 = 8;# in ohms\n",
- "R4 = 7;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = R1*I/(R1 + R2)\n",
- "r = (R1 + R2)*(R3 + R4)/(R3 + R1 + R4 + R2)\n",
- "I = r*Isc/(r + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 2 ohm resistor = \",round(I,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 2 ohm resistor = 6.75 A"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 185</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Convert the circuit to an equivalent Norton network.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 10;# in Volts\n",
- "R = 2;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc = V/R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n equivalent Norton network contains Current Source of amp = \",round(Isc,0),\"A and a resistor of \",R,\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " equivalent Norton network contains Current Source of amp = 5.0 A and a resistor of 2 ohm"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 185</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Convert the circuit to an equivalent Thevenin network.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Isc = 4;# in Amps\n",
- "R = 3;# in ohms\n",
- "\n",
- "#calculation:\n",
- "E = Isc*R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,0),\"V and a resistor of \",R,\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " equivalent Thevenin network contains Voltage Source of 12.0 V and a resistor of 3 ohm"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 185</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)Convert the circuit to an equivalent Thevenin networkby initially converting to a Norton equivalent circuit.\n",
- "#(b)Determine the current flowing in the 1.8 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 12;# in Volts\n",
- "E2 = 24;# in Volts\n",
- "R1 = 3;# in ohms\n",
- "R2 = 2;# in ohms\n",
- "R = 1.8;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc1 = E1/R1\n",
- "Isc2 = E2/R2\n",
- "I1 = Isc1 + Isc2\n",
- "r = R1*R2/(R1 + R2)\n",
- "E = I1*r\n",
- "I = E/(r + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (A)equivalent Norton network contains Current Source of \",round(I1,1),\"A and a resistor of \",r,\" ohm\"\n",
- "print \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",r,\" ohm\"\n",
- "print \"\\n (B)the current flowing in the 1.8 ohm resistor is \",round(I,1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (A)equivalent Norton network contains Current Source of 16.0 A and a resistor of 1.2 ohm\n",
- "\n",
- " equivalent Thevenin network contains Voltage Source of 19.2 V and a resistor of 1.2 ohm\n",
- "\n",
- " (B)the current flowing in the 1.8 ohm resistor is 6.4 A"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 186</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)Determine by successive conversions between Th\u00b4evenin and Norton equivalent networks \n",
- "#a Thevenin equivalent circuit for terminals AB.\n",
- "#(b)determine the current flowing in the 200 ohm resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 10;# in Volts\n",
- "E2 = 6;# in Volts\n",
- "I1 = 0.001;#in Amp\n",
- "R1 = 2000;# in ohms\n",
- "R2 = 3000;# in ohms\n",
- "R3 = 600;# in ohms\n",
- "R = 200;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Isc1 = E1/R1\n",
- "Isc2 = E2/R2\n",
- "I2 = Isc1 + Isc2\n",
- "r1 = R1*R2/(R1 + R2)\n",
- "E = I2*r1 - I1*R3\n",
- "r = r1 + R3\n",
- "I = E/(r + R)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (A)equivalent Norton network contains Current Source of \",round(I2*1000,0),\"mA and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
- "print \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
- "print \"\\n (B)the current flowing in the 200 ohm resistor is \",round(I*1000,1),\"mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (A)equivalent Norton network contains Current Source of 7.0 mA and a resistor of 1.2 Kohm\n",
- "\n",
- " equivalent Thevenin network contains Voltage Source of 7.8 V and a resistor of 1.2 Kohm\n",
- "\n",
- " (B)the current flowing in the 200 ohm resistor is 3.9 mA"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 188</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the power dissipated by the load in each case.\n",
- "#Plot a graph of RL (horizontally) against power (vertically) and determine the maximum power dissipated.\n",
- "from __future__ import division\n",
- "import math\n",
- "#from pylab import *\n",
- "%pylab inline\n",
- "#initializing the variables:\n",
- "E = 6;# in Volts\n",
- "R = 2.5;# in ohms\n",
- "\n",
- "#calculation:\n",
- "RL = []\n",
- "P = []\n",
- "k = []\n",
- "for h in range(6):\n",
- " RL.append(h - 0.5)\n",
- " k = h - 0.5\n",
- " P.append(k*((E/(R + k))**2))\n",
- "fig = plt.figure()\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(RL,P,'-')\n",
- "#plot(RL,P,'-')\n",
- "xlabel('RL(ohm)')\n",
- "ylabel('Power(W)')\n",
- "show()\n",
- "Pmax= R*(E/(2*R))**2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n A graph of RL against P is shown in Figure\" \n",
- "print \" The maximum value of power is\", Pmax,\"W which occurs when RL =\",R,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
- "For more information, type 'help(pylab)'."
- ]
- },
- {
- "output_type": "display_data",
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p6emiS1PMwIEDAQBBQUGYOnUqPvzwQ8ybN6/N57rM3z5q6c/dcsstAOQVP6Wl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\nZLqKDYkAAAAASUVORK5CYII=\n"
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " A graph of RL against P is shown in Figure\n",
- " The maximum value of power is 3.6 W which occurs when RL = 2.5 ohm"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 188</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#State the value of load resistance that gives maximum power dissipation and determine the value of this power.\n",
- "from __future__ import division\n",
- "import math\n",
- "from pylab import *\n",
- "#initializing the variables:\n",
- "E = 30;# in Volts\n",
- "R = 1.5;# in ohms\n",
- "\n",
- "#calculation:\n",
- "RL = R\n",
- "I = E/(R + RL)\n",
- "P = I**2*RL\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \" The maximum value of power is\", P,\"W which occurs when RL =\",RL,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- " The maximum value of power is 150.0 W which occurs when RL = 1.5 ohm"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 189</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the value of the load resistor RL that gives maximum power dissipation and determine the value of this power.\n",
- "from __future__ import division\n",
- "import math\n",
- "from pylab import *\n",
- "#initializing the variables:\n",
- "V = 15;# in Volts\n",
- "R1 = 3;# in ohms\n",
- "R2 = 12;# in ohms\n",
- "\n",
- "#calculation:\n",
- "E = (R2/(R2+ R1))*V\n",
- "r = R1*R2/(R1 + R2)\n",
- "RL = r\n",
- "I = E/(r + RL)\n",
- "P = I**2*RL\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \" The maximum value of power is\", P,\"W which occurs when Total Load RL =\",RL,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- " The maximum value of power is 15.0 W which occurs when Total Load RL = 2.4 ohm"
- ]
- }
- ],
- "prompt_number": 22
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_13.ipynb index 6f69e0f8..6f69e0f8 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_13.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint.ipynb deleted file mode 100755 index 0afe4500..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint.ipynb +++ /dev/null @@ -1,898 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 14: Alternating voltages and currents</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 195</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the periodic time for 50 Hz and 20kHz frequencies.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f1 = 50;# in Hz\n",
- "f2 = 20000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "T1 = 1/f1\n",
- "T2 = 1/f2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) Periodic time T = \",T1,\" secs\\n\"\n",
- "print \"\\n (b) Periodic time T = \",(T2/1E-6),\" usecs\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) Periodic time T = 0.02 secs\n",
- "\n",
- "\n",
- " (b) Periodic time T = 50.0 usecs\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 195</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the frequencies for 4ms and 4us periodic times\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T1 = 0.004;# in secs\n",
- "T2 = 4E-6;# in secs\n",
- "\n",
- "#calculation:\n",
- "f1 = 1/T1\n",
- "f2 = 1/T2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) Frequency f = \",f1,\" Hz\\n\"\n",
- "print \"\\n (b) Frequency f = \",(f2/1E6),\" MHz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) Frequency f = 250.0 Hz\n",
- "\n",
- "\n",
- " (b) Frequency f = 0.25 MHz\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 195</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the frequency of an alternating current which completes 5 cycles in 8 ms\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T = (8E-3)/5;# in secs\n",
- "\n",
- "#calculation:\n",
- "f = 1/T\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Frequency f = \",f,\" Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Frequency f = 625.0 Hz"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 196</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine for each: (i) frequency \n",
- "#(ii) average value over half a cycle \n",
- "#(iii) rms value\n",
- "#(iv) form factor and \n",
- "#(v) peak factor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ta = 0.02;# Time for 1 complete cycle in secs\n",
- "Vamax = 200;# in volts\n",
- "Va1 = 25;# in volts\n",
- "Va2 = 75;# in volts\n",
- "Va3 = 125;# in volts\n",
- "Va4 = 175;# in volts\n",
- "Tb = 0.016;# Time for 1 complete cycle in secs\n",
- "Ibmax = 10;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "#for Triangular waveform (Figure 14.5(a))\n",
- "fa = 1/Ta\n",
- "Aaw = Ta*Vamax/4\n",
- "Vaavg = Aaw*2/Ta\n",
- "Varms = (((Va1**2) + (Va2**2) + (Va3**2) + (Va4**2))/4)**0.5\n",
- "#Note that the greater the number of intervals chosen, the greater the accuracy of the result\n",
- "Ffa = Varms/Vaavg\n",
- "Pfa = Vamax/Varms\n",
- "\n",
- "#for Rectangular waveform (Figure 14.5(b))\n",
- "fb = 1/Tb\n",
- "Abw = Tb*Ibmax/2\n",
- "Ibavg = Abw*2/Tb\n",
- "Ibrms = 10\n",
- "Ffb = Ibrms/Ibavg\n",
- "Pfb = Ibmax/Ibrms\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a1)Frequency f = \",fa,\" Hz\\n\"\n",
- "print \"\\n (a2)average value over half a cycle = \",Vaavg,\" V\\n\"\n",
- "print \"\\n (a3)rms value = \",round(Varms,2),\" V\\n\"\n",
- "print \"\\n (a4)Form factor = \",round(Ffa,2),\"\\n\"\n",
- "print \"\\n (a5)Peak factor = \",round(Pfa,2),\"\\n\"\n",
- "print \"\\n (b1)Frequency f = \",fb,\" Hz\\n\"\n",
- "print \"\\n (b2)average value over half a cycle = \",Ibavg,\" A\\n\"\n",
- "print \"\\n (b3)rms value = \",Ibrms,\" A\\n\"\n",
- "print \"\\n (b4)Form factor = \",Ffb,\"\\n\"\n",
- "print \"\\n (b5)Peak factor = \",Pfb,\"\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a1)Frequency f = 50.0 Hz\n",
- "\n",
- "\n",
- " (a2)average value over half a cycle = 100.0 V\n",
- "\n",
- "\n",
- " (a3)rms value = 114.56 V\n",
- "\n",
- "\n",
- " (a4)Form factor = 1.15 \n",
- "\n",
- "\n",
- " (a5)Peak factor = 1.75 \n",
- "\n",
- "\n",
- " (b1)Frequency f = 62.5 Hz\n",
- "\n",
- "\n",
- " (b2)average value over half a cycle = 10.0 A\n",
- "\n",
- "\n",
- " (b3)rms value = 10 A\n",
- "\n",
- "\n",
- " (b4)Form factor = 1.0 \n",
- "\n",
- "\n",
- " (b5)Peak factor = 1.0 "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 198</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the frequency of the supply, \n",
- "#(b) the instantaneous values of current after 1.25 ms and 3.8 ms, (c) the peak or maximum value,\n",
- "#(d) the mean or average value, and (e) the rms value of the waveform.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy \n",
- "from numpy import mean, sqrt, arange\n",
- "#initializing the variables:\n",
- "Thalf = 5; #in ms\n",
- "Ta = 0.02;# Time for 1 complete cycle in secs\n",
- "\n",
- "#calculation:\n",
- "Tfull = 2*Thalf/1000 # in sec\n",
- "f = 1/Tfull\n",
- "A=[3, 10, 19, 30, 49, 63, 73, 72, 30, 2]\n",
- "Iinst125 = 19\n",
- "Iinst38 = 70\n",
- "sq = 0\n",
- "Ipeak = 76\n",
- "#B=arange(A)\n",
- "Imean = (0.5*1E-3)*numpy.mean(A)*10/(5*1E-3)\n",
- "for h in range(10):\n",
- " sq = sq + A[h]**2\n",
- "\n",
- "Irms = sqrt(sq/10)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Frequency f = \",f,\" Hz\\n\"\n",
- "print \"\\n (b)Instantaneous value of current after 1.25 ms =\",Iinst125,\"A \"\n",
- "print \"and Instantaneous value of current after 3.8 ms\", Iinst38,\"A\\n\"\n",
- "print \"\\n (c)Peak or maximum value = \",Ipeak,\" A\\n\"\n",
- "print \"\\n (d)Mean or average value = \",round(Imean,2),\"A\\n\"\n",
- "print \"\\n (e)rms value = \",round(Irms,1),\"A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Frequency f = 100.0 Hz\n",
- "\n",
- "\n",
- " (b)Instantaneous value of current after 1.25 ms = 19 A and Instantaneous value of current after 3.8 ms 70 A\n",
- "\n",
- "\n",
- " (c)Peak or maximum value = 76 A\n",
- "\n",
- "\n",
- " (d)Mean or average value = 35.1 A\n",
- "\n",
- "\n",
- " (e)rms value = 43.8 A"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 200</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#sinusoidal current has maximum value of 20 A.Calculate its rms value. \n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Imax = 20;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Irms = Imax/(2**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Rms value = \",round(Irms,2),\" A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Rms value = 14.14 A"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 200</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#main supply is of 240V.Determine its peak and mean values.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vrms = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Vmax = Vrms*(2**0.5)\n",
- "Vmean = 0.637*Vmax\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n peak value = \",round(Vmax,2),\" V\\n\"\n",
- "print \"\\n mean value = \",round(Vmean,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " peak value = 339.41 V\n",
- "\n",
- "\n",
- " mean value = 216.2 V"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 200</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its maximum value and its rms value\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vmean = 150;# in Volts\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Vmax = Vmean/0.637\n",
- "Vrms = 0.707*Vmax\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n peak value = \",round(Vmax,2),\" V\\n\"\n",
- "print \"\\n rms value = \",round(Vrms,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " peak value = 235.48 V\n",
- "\n",
- "\n",
- " rms value = 166.48 V"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 201</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the rms voltage, (b) the frequency and (c) the instantaneous value of voltage when t = 4 ms\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vmax = 282.8;# in Volts\n",
- "w = 314;# in rad/sec\n",
- "t = 0.004;# in sec\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Vrms = 0.707*Vmax\n",
- "f = w/(2*math.pi)\n",
- "v = Vmax*math.sin(w*t)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)rms value = \",round(Vrms,2),\" V\\n\"\n",
- "print \"\\n (b)frequency f = \",round(f,2),\" Hz\\n\"\n",
- "print \"\\n (c)instantaneous value of voltage at 4 ms = \",round(v,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)rms value = 199.94 V\n",
- "\n",
- "\n",
- " (b)frequency f = 49.97 Hz\n",
- "\n",
- "\n",
- " (c)instantaneous value of voltage at 4 ms = 268.9 V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 202</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the amplitude, (b) the peak-to-peak value, \n",
- "#(c) the rms value, (d) the periodic time, \n",
- "#(e) the frequency, and (f) the phase angle (in degrees and minutes) relative to 75 sin 200(pi*t)\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vmax = 75;# in Volts\n",
- "w = 200*math.pi;# in rad/sec\n",
- "t = 0.004;# in sec\n",
- "phi = 0.25;# in radians\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Vptp = 2*Vmax\n",
- "Vrms = 0.707*Vmax\n",
- "f = w/(2*math.pi)\n",
- "T = 1/f\n",
- "v = Vmax*math.sin(w*t)\n",
- "phid = phi*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) Amplitude, or peak value = \",Vmax,\" V\\n\"\n",
- "print \"\\n (b) Peak-to-peak value = \",Vptp,\" V\\n\"\n",
- "print \"\\n (c)rms value = \",Vrms,\" V\\n\"\n",
- "print \"\\n (d)periodic time, T = \",T,\" sec\\n\"\n",
- "print \"\\n (e)frequency f = \",f,\" Hz\\n\"\n",
- "print \"\\n (f)phase angle = \",round(phid,2),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) Amplitude, or peak value = 75 V\n",
- "\n",
- "\n",
- " (b) Peak-to-peak value = 150 V\n",
- "\n",
- "\n",
- " (c)rms value = 53.025 V\n",
- "\n",
- "\n",
- " (d)periodic time, T = 0.01 sec\n",
- "\n",
- "\n",
- " (e)frequency f = 100.0 Hz\n",
- "\n",
- "\n",
- " (f)phase angle = 14.32 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 202</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Express the instantaneous voltage in the form v = Vm*sin(wt +- phi)\t\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vmax = 40;# in Volts\n",
- "T = 0.01;# in sec\n",
- "v = -20;# when t = 0sec, in volts\n",
- "t = 0;# in secs\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "w = 2*math.pi/T\n",
- "phir = math.asin(v/Vmax)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n instantaneous voltage v = \", Vmax,\" sin(\",round(w,2),\"t\",round(phir,2),\") V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " instantaneous voltage v = 40 sin( 628.32 t -0.52 ) V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 203</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find:\n",
- "#(a) the peak value, the periodic time, the frequency and phase angle relative to 120 sin 100*pi*t\n",
- "#(b) the value of the current when t = 0\n",
- "#(c) the value of the current when t = 8 ms\n",
- "#(d) the time when the current first reaches 60 A, and\n",
- "#(e) the time when the current is first a maximum\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Imax = 120;# in Amperes\n",
- "w = 100*math.pi;# in rad/sec\n",
- "phi = 0.36;# in rad\n",
- "t1 = 0;# in secs\n",
- "t2 = 0.008;# in secs\n",
- "i = 60;# in amperes\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "f = w/(2*math.pi)\n",
- "T = 1/f\n",
- "phid = phi*180/math.pi\n",
- "i0 = Imax*math.sin((w*t1)+phi)\n",
- "i8 = Imax*math.sin((w*t2)+phi) \n",
- "ti = (math.asin(i/Imax) - phi)/w\n",
- "tm1 = (math.asin(Imax/Imax) - phi)/w\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Peak value = \", Imax,\" A, Periodic time T = \", T,\" sec, \"\n",
- "print \" Frequency, f = \", f,\" Hz Phase angle = \",round(phid,2),\"deg leading\\n\"\n",
- "print \"\\n (b) When t = 0, i = \",round(i0,2),\" A\\n\"\n",
- "print \"\\n (c)When t = 8 ms = \", round(i8,2),\" A\\n\"\n",
- "print \"\\n (d)When i is 60 A, then time t = \",round((ti/1E-3),2),\" ms\\n\"\n",
- "print \"\\n (e)When the current is a maximum, time, t = \", round((tm1/1E-3),2),\" ms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Peak value = 120 A, Periodic time T = 0.02 sec, \n",
- " Frequency, f = 50.0 Hz Phase angle = 20.63 deg leading\n",
- "\n",
- "\n",
- " (b) When t = 0, i = 42.27 A\n",
- "\n",
- "\n",
- " (c)When t = 8 ms = 31.81 A\n",
- "\n",
- "\n",
- " (d)When i is 60 A, then time t = 0.52 ms\n",
- "\n",
- "\n",
- " (e)When the current is a maximum, time, t = 3.85 ms\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 204</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#obtain a sinusoidal expression for i1 + i2\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "i1max = 20;# in Amperes\n",
- "i2max = 10;# in Amperes\n",
- "phi1 = 0;# in rad\n",
- "phi2 = math.pi/3;# in rad\n",
- "\n",
- "#calculation:\n",
- "#Ig = i1 + i2\n",
- "Igmax = 26.5\n",
- "phiIg = 19*math.pi/180\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Current Ig = i1 + i2 =\", Igmax,\"sin(wt + \",round(phiIg,3),\") Amps\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Current Ig = i1 + i2 = 26.5 sin(wt + 0.332 ) Amps"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 205</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find, by calculation, a sinusoidal expression to represent v1 + v2\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1max = 50;# in volts\n",
- "V2max = 100;# in volts\n",
- "phi2 = -1*math.pi/6;# in rad\n",
- "\n",
- "#calculation:\n",
- "#vR2 = v1**2 + v2**2 - 2*v1*v2 cos 150\n",
- "phidiff = math.pi + phi2\n",
- "Vrmax = (V1max**2 + V2max**2 - 2*V1max*V2max*math.cos(phidiff))**0.5\n",
- "#Using the sine rule\n",
- "phi = math.asin(V2max*math.sin(phidiff)/Vrmax)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n VR = \",round(Vrmax,2),\"sin(wt - \",round(phi,2),\") V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " VR = 145.47 sin(wt - 0.35 ) V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 206</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find a sinusoidal expression for \u0005i1 + i2\t of Problem 13, by calculation.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I1max = 20;# in volts\n",
- "I2max = 10;# in volts\n",
- "phi2 = 1*math.pi/3;# in rad\n",
- "\n",
- "#calculation:\n",
- "#iR2 = i1**2 + i2**2 - 2*i1*i2cos150\n",
- "phidiff = math.pi - phi2\n",
- "Irmax = (I1max**2 + I2max**2 - 2*I1max*I2max*math.cos(phidiff))**0.5\n",
- "#Using the sine rule\n",
- "phi = math.asin(I2max*math.sin(phidiff)/Irmax)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n IR = \", round(Irmax,2),\"sin(wt + \",round(phi,2),\") V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " IR = 26.46 sin(wt + 0.33 ) V"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_1.ipynb deleted file mode 100755 index 0afe4500..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_1.ipynb +++ /dev/null @@ -1,898 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 14: Alternating voltages and currents</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 195</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the periodic time for 50 Hz and 20kHz frequencies.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f1 = 50;# in Hz\n",
- "f2 = 20000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "T1 = 1/f1\n",
- "T2 = 1/f2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) Periodic time T = \",T1,\" secs\\n\"\n",
- "print \"\\n (b) Periodic time T = \",(T2/1E-6),\" usecs\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) Periodic time T = 0.02 secs\n",
- "\n",
- "\n",
- " (b) Periodic time T = 50.0 usecs\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 195</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the frequencies for 4ms and 4us periodic times\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T1 = 0.004;# in secs\n",
- "T2 = 4E-6;# in secs\n",
- "\n",
- "#calculation:\n",
- "f1 = 1/T1\n",
- "f2 = 1/T2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) Frequency f = \",f1,\" Hz\\n\"\n",
- "print \"\\n (b) Frequency f = \",(f2/1E6),\" MHz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) Frequency f = 250.0 Hz\n",
- "\n",
- "\n",
- " (b) Frequency f = 0.25 MHz\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 195</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the frequency of an alternating current which completes 5 cycles in 8 ms\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T = (8E-3)/5;# in secs\n",
- "\n",
- "#calculation:\n",
- "f = 1/T\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Frequency f = \",f,\" Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Frequency f = 625.0 Hz"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 196</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine for each: (i) frequency \n",
- "#(ii) average value over half a cycle \n",
- "#(iii) rms value\n",
- "#(iv) form factor and \n",
- "#(v) peak factor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ta = 0.02;# Time for 1 complete cycle in secs\n",
- "Vamax = 200;# in volts\n",
- "Va1 = 25;# in volts\n",
- "Va2 = 75;# in volts\n",
- "Va3 = 125;# in volts\n",
- "Va4 = 175;# in volts\n",
- "Tb = 0.016;# Time for 1 complete cycle in secs\n",
- "Ibmax = 10;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "#for Triangular waveform (Figure 14.5(a))\n",
- "fa = 1/Ta\n",
- "Aaw = Ta*Vamax/4\n",
- "Vaavg = Aaw*2/Ta\n",
- "Varms = (((Va1**2) + (Va2**2) + (Va3**2) + (Va4**2))/4)**0.5\n",
- "#Note that the greater the number of intervals chosen, the greater the accuracy of the result\n",
- "Ffa = Varms/Vaavg\n",
- "Pfa = Vamax/Varms\n",
- "\n",
- "#for Rectangular waveform (Figure 14.5(b))\n",
- "fb = 1/Tb\n",
- "Abw = Tb*Ibmax/2\n",
- "Ibavg = Abw*2/Tb\n",
- "Ibrms = 10\n",
- "Ffb = Ibrms/Ibavg\n",
- "Pfb = Ibmax/Ibrms\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a1)Frequency f = \",fa,\" Hz\\n\"\n",
- "print \"\\n (a2)average value over half a cycle = \",Vaavg,\" V\\n\"\n",
- "print \"\\n (a3)rms value = \",round(Varms,2),\" V\\n\"\n",
- "print \"\\n (a4)Form factor = \",round(Ffa,2),\"\\n\"\n",
- "print \"\\n (a5)Peak factor = \",round(Pfa,2),\"\\n\"\n",
- "print \"\\n (b1)Frequency f = \",fb,\" Hz\\n\"\n",
- "print \"\\n (b2)average value over half a cycle = \",Ibavg,\" A\\n\"\n",
- "print \"\\n (b3)rms value = \",Ibrms,\" A\\n\"\n",
- "print \"\\n (b4)Form factor = \",Ffb,\"\\n\"\n",
- "print \"\\n (b5)Peak factor = \",Pfb,\"\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a1)Frequency f = 50.0 Hz\n",
- "\n",
- "\n",
- " (a2)average value over half a cycle = 100.0 V\n",
- "\n",
- "\n",
- " (a3)rms value = 114.56 V\n",
- "\n",
- "\n",
- " (a4)Form factor = 1.15 \n",
- "\n",
- "\n",
- " (a5)Peak factor = 1.75 \n",
- "\n",
- "\n",
- " (b1)Frequency f = 62.5 Hz\n",
- "\n",
- "\n",
- " (b2)average value over half a cycle = 10.0 A\n",
- "\n",
- "\n",
- " (b3)rms value = 10 A\n",
- "\n",
- "\n",
- " (b4)Form factor = 1.0 \n",
- "\n",
- "\n",
- " (b5)Peak factor = 1.0 "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 198</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the frequency of the supply, \n",
- "#(b) the instantaneous values of current after 1.25 ms and 3.8 ms, (c) the peak or maximum value,\n",
- "#(d) the mean or average value, and (e) the rms value of the waveform.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy \n",
- "from numpy import mean, sqrt, arange\n",
- "#initializing the variables:\n",
- "Thalf = 5; #in ms\n",
- "Ta = 0.02;# Time for 1 complete cycle in secs\n",
- "\n",
- "#calculation:\n",
- "Tfull = 2*Thalf/1000 # in sec\n",
- "f = 1/Tfull\n",
- "A=[3, 10, 19, 30, 49, 63, 73, 72, 30, 2]\n",
- "Iinst125 = 19\n",
- "Iinst38 = 70\n",
- "sq = 0\n",
- "Ipeak = 76\n",
- "#B=arange(A)\n",
- "Imean = (0.5*1E-3)*numpy.mean(A)*10/(5*1E-3)\n",
- "for h in range(10):\n",
- " sq = sq + A[h]**2\n",
- "\n",
- "Irms = sqrt(sq/10)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Frequency f = \",f,\" Hz\\n\"\n",
- "print \"\\n (b)Instantaneous value of current after 1.25 ms =\",Iinst125,\"A \"\n",
- "print \"and Instantaneous value of current after 3.8 ms\", Iinst38,\"A\\n\"\n",
- "print \"\\n (c)Peak or maximum value = \",Ipeak,\" A\\n\"\n",
- "print \"\\n (d)Mean or average value = \",round(Imean,2),\"A\\n\"\n",
- "print \"\\n (e)rms value = \",round(Irms,1),\"A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Frequency f = 100.0 Hz\n",
- "\n",
- "\n",
- " (b)Instantaneous value of current after 1.25 ms = 19 A and Instantaneous value of current after 3.8 ms 70 A\n",
- "\n",
- "\n",
- " (c)Peak or maximum value = 76 A\n",
- "\n",
- "\n",
- " (d)Mean or average value = 35.1 A\n",
- "\n",
- "\n",
- " (e)rms value = 43.8 A"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 200</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#sinusoidal current has maximum value of 20 A.Calculate its rms value. \n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Imax = 20;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Irms = Imax/(2**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Rms value = \",round(Irms,2),\" A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Rms value = 14.14 A"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 200</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#main supply is of 240V.Determine its peak and mean values.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vrms = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Vmax = Vrms*(2**0.5)\n",
- "Vmean = 0.637*Vmax\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n peak value = \",round(Vmax,2),\" V\\n\"\n",
- "print \"\\n mean value = \",round(Vmean,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " peak value = 339.41 V\n",
- "\n",
- "\n",
- " mean value = 216.2 V"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 200</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its maximum value and its rms value\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vmean = 150;# in Volts\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Vmax = Vmean/0.637\n",
- "Vrms = 0.707*Vmax\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n peak value = \",round(Vmax,2),\" V\\n\"\n",
- "print \"\\n rms value = \",round(Vrms,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " peak value = 235.48 V\n",
- "\n",
- "\n",
- " rms value = 166.48 V"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 201</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the rms voltage, (b) the frequency and (c) the instantaneous value of voltage when t = 4 ms\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vmax = 282.8;# in Volts\n",
- "w = 314;# in rad/sec\n",
- "t = 0.004;# in sec\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Vrms = 0.707*Vmax\n",
- "f = w/(2*math.pi)\n",
- "v = Vmax*math.sin(w*t)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)rms value = \",round(Vrms,2),\" V\\n\"\n",
- "print \"\\n (b)frequency f = \",round(f,2),\" Hz\\n\"\n",
- "print \"\\n (c)instantaneous value of voltage at 4 ms = \",round(v,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)rms value = 199.94 V\n",
- "\n",
- "\n",
- " (b)frequency f = 49.97 Hz\n",
- "\n",
- "\n",
- " (c)instantaneous value of voltage at 4 ms = 268.9 V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 202</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the amplitude, (b) the peak-to-peak value, \n",
- "#(c) the rms value, (d) the periodic time, \n",
- "#(e) the frequency, and (f) the phase angle (in degrees and minutes) relative to 75 sin 200(pi*t)\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vmax = 75;# in Volts\n",
- "w = 200*math.pi;# in rad/sec\n",
- "t = 0.004;# in sec\n",
- "phi = 0.25;# in radians\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Vptp = 2*Vmax\n",
- "Vrms = 0.707*Vmax\n",
- "f = w/(2*math.pi)\n",
- "T = 1/f\n",
- "v = Vmax*math.sin(w*t)\n",
- "phid = phi*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) Amplitude, or peak value = \",Vmax,\" V\\n\"\n",
- "print \"\\n (b) Peak-to-peak value = \",Vptp,\" V\\n\"\n",
- "print \"\\n (c)rms value = \",Vrms,\" V\\n\"\n",
- "print \"\\n (d)periodic time, T = \",T,\" sec\\n\"\n",
- "print \"\\n (e)frequency f = \",f,\" Hz\\n\"\n",
- "print \"\\n (f)phase angle = \",round(phid,2),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) Amplitude, or peak value = 75 V\n",
- "\n",
- "\n",
- " (b) Peak-to-peak value = 150 V\n",
- "\n",
- "\n",
- " (c)rms value = 53.025 V\n",
- "\n",
- "\n",
- " (d)periodic time, T = 0.01 sec\n",
- "\n",
- "\n",
- " (e)frequency f = 100.0 Hz\n",
- "\n",
- "\n",
- " (f)phase angle = 14.32 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 202</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Express the instantaneous voltage in the form v = Vm*sin(wt +- phi)\t\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vmax = 40;# in Volts\n",
- "T = 0.01;# in sec\n",
- "v = -20;# when t = 0sec, in volts\n",
- "t = 0;# in secs\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "w = 2*math.pi/T\n",
- "phir = math.asin(v/Vmax)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n instantaneous voltage v = \", Vmax,\" sin(\",round(w,2),\"t\",round(phir,2),\") V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " instantaneous voltage v = 40 sin( 628.32 t -0.52 ) V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 203</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find:\n",
- "#(a) the peak value, the periodic time, the frequency and phase angle relative to 120 sin 100*pi*t\n",
- "#(b) the value of the current when t = 0\n",
- "#(c) the value of the current when t = 8 ms\n",
- "#(d) the time when the current first reaches 60 A, and\n",
- "#(e) the time when the current is first a maximum\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Imax = 120;# in Amperes\n",
- "w = 100*math.pi;# in rad/sec\n",
- "phi = 0.36;# in rad\n",
- "t1 = 0;# in secs\n",
- "t2 = 0.008;# in secs\n",
- "i = 60;# in amperes\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "f = w/(2*math.pi)\n",
- "T = 1/f\n",
- "phid = phi*180/math.pi\n",
- "i0 = Imax*math.sin((w*t1)+phi)\n",
- "i8 = Imax*math.sin((w*t2)+phi) \n",
- "ti = (math.asin(i/Imax) - phi)/w\n",
- "tm1 = (math.asin(Imax/Imax) - phi)/w\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Peak value = \", Imax,\" A, Periodic time T = \", T,\" sec, \"\n",
- "print \" Frequency, f = \", f,\" Hz Phase angle = \",round(phid,2),\"deg leading\\n\"\n",
- "print \"\\n (b) When t = 0, i = \",round(i0,2),\" A\\n\"\n",
- "print \"\\n (c)When t = 8 ms = \", round(i8,2),\" A\\n\"\n",
- "print \"\\n (d)When i is 60 A, then time t = \",round((ti/1E-3),2),\" ms\\n\"\n",
- "print \"\\n (e)When the current is a maximum, time, t = \", round((tm1/1E-3),2),\" ms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Peak value = 120 A, Periodic time T = 0.02 sec, \n",
- " Frequency, f = 50.0 Hz Phase angle = 20.63 deg leading\n",
- "\n",
- "\n",
- " (b) When t = 0, i = 42.27 A\n",
- "\n",
- "\n",
- " (c)When t = 8 ms = 31.81 A\n",
- "\n",
- "\n",
- " (d)When i is 60 A, then time t = 0.52 ms\n",
- "\n",
- "\n",
- " (e)When the current is a maximum, time, t = 3.85 ms\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 204</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#obtain a sinusoidal expression for i1 + i2\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "i1max = 20;# in Amperes\n",
- "i2max = 10;# in Amperes\n",
- "phi1 = 0;# in rad\n",
- "phi2 = math.pi/3;# in rad\n",
- "\n",
- "#calculation:\n",
- "#Ig = i1 + i2\n",
- "Igmax = 26.5\n",
- "phiIg = 19*math.pi/180\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Current Ig = i1 + i2 =\", Igmax,\"sin(wt + \",round(phiIg,3),\") Amps\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Current Ig = i1 + i2 = 26.5 sin(wt + 0.332 ) Amps"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 205</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find, by calculation, a sinusoidal expression to represent v1 + v2\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1max = 50;# in volts\n",
- "V2max = 100;# in volts\n",
- "phi2 = -1*math.pi/6;# in rad\n",
- "\n",
- "#calculation:\n",
- "#vR2 = v1**2 + v2**2 - 2*v1*v2 cos 150\n",
- "phidiff = math.pi + phi2\n",
- "Vrmax = (V1max**2 + V2max**2 - 2*V1max*V2max*math.cos(phidiff))**0.5\n",
- "#Using the sine rule\n",
- "phi = math.asin(V2max*math.sin(phidiff)/Vrmax)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n VR = \",round(Vrmax,2),\"sin(wt - \",round(phi,2),\") V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " VR = 145.47 sin(wt - 0.35 ) V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 206</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find a sinusoidal expression for \u0005i1 + i2\t of Problem 13, by calculation.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I1max = 20;# in volts\n",
- "I2max = 10;# in volts\n",
- "phi2 = 1*math.pi/3;# in rad\n",
- "\n",
- "#calculation:\n",
- "#iR2 = i1**2 + i2**2 - 2*i1*i2cos150\n",
- "phidiff = math.pi - phi2\n",
- "Irmax = (I1max**2 + I2max**2 - 2*I1max*I2max*math.cos(phidiff))**0.5\n",
- "#Using the sine rule\n",
- "phi = math.asin(I2max*math.sin(phidiff)/Irmax)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n IR = \", round(Irmax,2),\"sin(wt + \",round(phi,2),\") V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " IR = 26.46 sin(wt + 0.33 ) V"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_2.ipynb deleted file mode 100755 index 0afe4500..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_2.ipynb +++ /dev/null @@ -1,898 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 14: Alternating voltages and currents</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 195</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the periodic time for 50 Hz and 20kHz frequencies.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f1 = 50;# in Hz\n",
- "f2 = 20000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "T1 = 1/f1\n",
- "T2 = 1/f2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) Periodic time T = \",T1,\" secs\\n\"\n",
- "print \"\\n (b) Periodic time T = \",(T2/1E-6),\" usecs\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) Periodic time T = 0.02 secs\n",
- "\n",
- "\n",
- " (b) Periodic time T = 50.0 usecs\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 195</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the frequencies for 4ms and 4us periodic times\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T1 = 0.004;# in secs\n",
- "T2 = 4E-6;# in secs\n",
- "\n",
- "#calculation:\n",
- "f1 = 1/T1\n",
- "f2 = 1/T2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) Frequency f = \",f1,\" Hz\\n\"\n",
- "print \"\\n (b) Frequency f = \",(f2/1E6),\" MHz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) Frequency f = 250.0 Hz\n",
- "\n",
- "\n",
- " (b) Frequency f = 0.25 MHz\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 195</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the frequency of an alternating current which completes 5 cycles in 8 ms\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T = (8E-3)/5;# in secs\n",
- "\n",
- "#calculation:\n",
- "f = 1/T\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Frequency f = \",f,\" Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Frequency f = 625.0 Hz"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 196</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine for each: (i) frequency \n",
- "#(ii) average value over half a cycle \n",
- "#(iii) rms value\n",
- "#(iv) form factor and \n",
- "#(v) peak factor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ta = 0.02;# Time for 1 complete cycle in secs\n",
- "Vamax = 200;# in volts\n",
- "Va1 = 25;# in volts\n",
- "Va2 = 75;# in volts\n",
- "Va3 = 125;# in volts\n",
- "Va4 = 175;# in volts\n",
- "Tb = 0.016;# Time for 1 complete cycle in secs\n",
- "Ibmax = 10;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "#for Triangular waveform (Figure 14.5(a))\n",
- "fa = 1/Ta\n",
- "Aaw = Ta*Vamax/4\n",
- "Vaavg = Aaw*2/Ta\n",
- "Varms = (((Va1**2) + (Va2**2) + (Va3**2) + (Va4**2))/4)**0.5\n",
- "#Note that the greater the number of intervals chosen, the greater the accuracy of the result\n",
- "Ffa = Varms/Vaavg\n",
- "Pfa = Vamax/Varms\n",
- "\n",
- "#for Rectangular waveform (Figure 14.5(b))\n",
- "fb = 1/Tb\n",
- "Abw = Tb*Ibmax/2\n",
- "Ibavg = Abw*2/Tb\n",
- "Ibrms = 10\n",
- "Ffb = Ibrms/Ibavg\n",
- "Pfb = Ibmax/Ibrms\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a1)Frequency f = \",fa,\" Hz\\n\"\n",
- "print \"\\n (a2)average value over half a cycle = \",Vaavg,\" V\\n\"\n",
- "print \"\\n (a3)rms value = \",round(Varms,2),\" V\\n\"\n",
- "print \"\\n (a4)Form factor = \",round(Ffa,2),\"\\n\"\n",
- "print \"\\n (a5)Peak factor = \",round(Pfa,2),\"\\n\"\n",
- "print \"\\n (b1)Frequency f = \",fb,\" Hz\\n\"\n",
- "print \"\\n (b2)average value over half a cycle = \",Ibavg,\" A\\n\"\n",
- "print \"\\n (b3)rms value = \",Ibrms,\" A\\n\"\n",
- "print \"\\n (b4)Form factor = \",Ffb,\"\\n\"\n",
- "print \"\\n (b5)Peak factor = \",Pfb,\"\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a1)Frequency f = 50.0 Hz\n",
- "\n",
- "\n",
- " (a2)average value over half a cycle = 100.0 V\n",
- "\n",
- "\n",
- " (a3)rms value = 114.56 V\n",
- "\n",
- "\n",
- " (a4)Form factor = 1.15 \n",
- "\n",
- "\n",
- " (a5)Peak factor = 1.75 \n",
- "\n",
- "\n",
- " (b1)Frequency f = 62.5 Hz\n",
- "\n",
- "\n",
- " (b2)average value over half a cycle = 10.0 A\n",
- "\n",
- "\n",
- " (b3)rms value = 10 A\n",
- "\n",
- "\n",
- " (b4)Form factor = 1.0 \n",
- "\n",
- "\n",
- " (b5)Peak factor = 1.0 "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 198</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the frequency of the supply, \n",
- "#(b) the instantaneous values of current after 1.25 ms and 3.8 ms, (c) the peak or maximum value,\n",
- "#(d) the mean or average value, and (e) the rms value of the waveform.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy \n",
- "from numpy import mean, sqrt, arange\n",
- "#initializing the variables:\n",
- "Thalf = 5; #in ms\n",
- "Ta = 0.02;# Time for 1 complete cycle in secs\n",
- "\n",
- "#calculation:\n",
- "Tfull = 2*Thalf/1000 # in sec\n",
- "f = 1/Tfull\n",
- "A=[3, 10, 19, 30, 49, 63, 73, 72, 30, 2]\n",
- "Iinst125 = 19\n",
- "Iinst38 = 70\n",
- "sq = 0\n",
- "Ipeak = 76\n",
- "#B=arange(A)\n",
- "Imean = (0.5*1E-3)*numpy.mean(A)*10/(5*1E-3)\n",
- "for h in range(10):\n",
- " sq = sq + A[h]**2\n",
- "\n",
- "Irms = sqrt(sq/10)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Frequency f = \",f,\" Hz\\n\"\n",
- "print \"\\n (b)Instantaneous value of current after 1.25 ms =\",Iinst125,\"A \"\n",
- "print \"and Instantaneous value of current after 3.8 ms\", Iinst38,\"A\\n\"\n",
- "print \"\\n (c)Peak or maximum value = \",Ipeak,\" A\\n\"\n",
- "print \"\\n (d)Mean or average value = \",round(Imean,2),\"A\\n\"\n",
- "print \"\\n (e)rms value = \",round(Irms,1),\"A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Frequency f = 100.0 Hz\n",
- "\n",
- "\n",
- " (b)Instantaneous value of current after 1.25 ms = 19 A and Instantaneous value of current after 3.8 ms 70 A\n",
- "\n",
- "\n",
- " (c)Peak or maximum value = 76 A\n",
- "\n",
- "\n",
- " (d)Mean or average value = 35.1 A\n",
- "\n",
- "\n",
- " (e)rms value = 43.8 A"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 200</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#sinusoidal current has maximum value of 20 A.Calculate its rms value. \n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Imax = 20;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Irms = Imax/(2**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Rms value = \",round(Irms,2),\" A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Rms value = 14.14 A"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 200</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#main supply is of 240V.Determine its peak and mean values.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vrms = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Vmax = Vrms*(2**0.5)\n",
- "Vmean = 0.637*Vmax\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n peak value = \",round(Vmax,2),\" V\\n\"\n",
- "print \"\\n mean value = \",round(Vmean,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " peak value = 339.41 V\n",
- "\n",
- "\n",
- " mean value = 216.2 V"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 200</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its maximum value and its rms value\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vmean = 150;# in Volts\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Vmax = Vmean/0.637\n",
- "Vrms = 0.707*Vmax\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n peak value = \",round(Vmax,2),\" V\\n\"\n",
- "print \"\\n rms value = \",round(Vrms,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " peak value = 235.48 V\n",
- "\n",
- "\n",
- " rms value = 166.48 V"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 201</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the rms voltage, (b) the frequency and (c) the instantaneous value of voltage when t = 4 ms\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vmax = 282.8;# in Volts\n",
- "w = 314;# in rad/sec\n",
- "t = 0.004;# in sec\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Vrms = 0.707*Vmax\n",
- "f = w/(2*math.pi)\n",
- "v = Vmax*math.sin(w*t)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)rms value = \",round(Vrms,2),\" V\\n\"\n",
- "print \"\\n (b)frequency f = \",round(f,2),\" Hz\\n\"\n",
- "print \"\\n (c)instantaneous value of voltage at 4 ms = \",round(v,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)rms value = 199.94 V\n",
- "\n",
- "\n",
- " (b)frequency f = 49.97 Hz\n",
- "\n",
- "\n",
- " (c)instantaneous value of voltage at 4 ms = 268.9 V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 202</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the amplitude, (b) the peak-to-peak value, \n",
- "#(c) the rms value, (d) the periodic time, \n",
- "#(e) the frequency, and (f) the phase angle (in degrees and minutes) relative to 75 sin 200(pi*t)\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vmax = 75;# in Volts\n",
- "w = 200*math.pi;# in rad/sec\n",
- "t = 0.004;# in sec\n",
- "phi = 0.25;# in radians\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "Vptp = 2*Vmax\n",
- "Vrms = 0.707*Vmax\n",
- "f = w/(2*math.pi)\n",
- "T = 1/f\n",
- "v = Vmax*math.sin(w*t)\n",
- "phid = phi*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) Amplitude, or peak value = \",Vmax,\" V\\n\"\n",
- "print \"\\n (b) Peak-to-peak value = \",Vptp,\" V\\n\"\n",
- "print \"\\n (c)rms value = \",Vrms,\" V\\n\"\n",
- "print \"\\n (d)periodic time, T = \",T,\" sec\\n\"\n",
- "print \"\\n (e)frequency f = \",f,\" Hz\\n\"\n",
- "print \"\\n (f)phase angle = \",round(phid,2),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) Amplitude, or peak value = 75 V\n",
- "\n",
- "\n",
- " (b) Peak-to-peak value = 150 V\n",
- "\n",
- "\n",
- " (c)rms value = 53.025 V\n",
- "\n",
- "\n",
- " (d)periodic time, T = 0.01 sec\n",
- "\n",
- "\n",
- " (e)frequency f = 100.0 Hz\n",
- "\n",
- "\n",
- " (f)phase angle = 14.32 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 202</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Express the instantaneous voltage in the form v = Vm*sin(wt +- phi)\t\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vmax = 40;# in Volts\n",
- "T = 0.01;# in sec\n",
- "v = -20;# when t = 0sec, in volts\n",
- "t = 0;# in secs\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "w = 2*math.pi/T\n",
- "phir = math.asin(v/Vmax)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n instantaneous voltage v = \", Vmax,\" sin(\",round(w,2),\"t\",round(phir,2),\") V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " instantaneous voltage v = 40 sin( 628.32 t -0.52 ) V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 203</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find:\n",
- "#(a) the peak value, the periodic time, the frequency and phase angle relative to 120 sin 100*pi*t\n",
- "#(b) the value of the current when t = 0\n",
- "#(c) the value of the current when t = 8 ms\n",
- "#(d) the time when the current first reaches 60 A, and\n",
- "#(e) the time when the current is first a maximum\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Imax = 120;# in Amperes\n",
- "w = 100*math.pi;# in rad/sec\n",
- "phi = 0.36;# in rad\n",
- "t1 = 0;# in secs\n",
- "t2 = 0.008;# in secs\n",
- "i = 60;# in amperes\n",
- "\n",
- "#calculation:\n",
- "#for a sine wave\n",
- "f = w/(2*math.pi)\n",
- "T = 1/f\n",
- "phid = phi*180/math.pi\n",
- "i0 = Imax*math.sin((w*t1)+phi)\n",
- "i8 = Imax*math.sin((w*t2)+phi) \n",
- "ti = (math.asin(i/Imax) - phi)/w\n",
- "tm1 = (math.asin(Imax/Imax) - phi)/w\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Peak value = \", Imax,\" A, Periodic time T = \", T,\" sec, \"\n",
- "print \" Frequency, f = \", f,\" Hz Phase angle = \",round(phid,2),\"deg leading\\n\"\n",
- "print \"\\n (b) When t = 0, i = \",round(i0,2),\" A\\n\"\n",
- "print \"\\n (c)When t = 8 ms = \", round(i8,2),\" A\\n\"\n",
- "print \"\\n (d)When i is 60 A, then time t = \",round((ti/1E-3),2),\" ms\\n\"\n",
- "print \"\\n (e)When the current is a maximum, time, t = \", round((tm1/1E-3),2),\" ms\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Peak value = 120 A, Periodic time T = 0.02 sec, \n",
- " Frequency, f = 50.0 Hz Phase angle = 20.63 deg leading\n",
- "\n",
- "\n",
- " (b) When t = 0, i = 42.27 A\n",
- "\n",
- "\n",
- " (c)When t = 8 ms = 31.81 A\n",
- "\n",
- "\n",
- " (d)When i is 60 A, then time t = 0.52 ms\n",
- "\n",
- "\n",
- " (e)When the current is a maximum, time, t = 3.85 ms\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 204</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#obtain a sinusoidal expression for i1 + i2\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "i1max = 20;# in Amperes\n",
- "i2max = 10;# in Amperes\n",
- "phi1 = 0;# in rad\n",
- "phi2 = math.pi/3;# in rad\n",
- "\n",
- "#calculation:\n",
- "#Ig = i1 + i2\n",
- "Igmax = 26.5\n",
- "phiIg = 19*math.pi/180\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Current Ig = i1 + i2 =\", Igmax,\"sin(wt + \",round(phiIg,3),\") Amps\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Current Ig = i1 + i2 = 26.5 sin(wt + 0.332 ) Amps"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 205</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find, by calculation, a sinusoidal expression to represent v1 + v2\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1max = 50;# in volts\n",
- "V2max = 100;# in volts\n",
- "phi2 = -1*math.pi/6;# in rad\n",
- "\n",
- "#calculation:\n",
- "#vR2 = v1**2 + v2**2 - 2*v1*v2 cos 150\n",
- "phidiff = math.pi + phi2\n",
- "Vrmax = (V1max**2 + V2max**2 - 2*V1max*V2max*math.cos(phidiff))**0.5\n",
- "#Using the sine rule\n",
- "phi = math.asin(V2max*math.sin(phidiff)/Vrmax)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n VR = \",round(Vrmax,2),\"sin(wt - \",round(phi,2),\") V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " VR = 145.47 sin(wt - 0.35 ) V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 206</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find a sinusoidal expression for \u0005i1 + i2\t of Problem 13, by calculation.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I1max = 20;# in volts\n",
- "I2max = 10;# in volts\n",
- "phi2 = 1*math.pi/3;# in rad\n",
- "\n",
- "#calculation:\n",
- "#iR2 = i1**2 + i2**2 - 2*i1*i2cos150\n",
- "phidiff = math.pi - phi2\n",
- "Irmax = (I1max**2 + I2max**2 - 2*I1max*I2max*math.cos(phidiff))**0.5\n",
- "#Using the sine rule\n",
- "phi = math.asin(I2max*math.sin(phidiff)/Irmax)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n IR = \", round(Irmax,2),\"sin(wt + \",round(phi,2),\") V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " IR = 26.46 sin(wt + 0.33 ) V"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_14.ipynb index d91a1cbb..d91a1cbb 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_14.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint.ipynb deleted file mode 100755 index e7b68d8a..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint.ipynb +++ /dev/null @@ -1,1677 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 15: Single-phase series a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 214</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Calculate the reactance of a coil \n",
- "#(b) Determine the inductance of the coil.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.32;# in Henry\n",
- "f1 = 50;# in Hz\n",
- "f2 = 5000;# in Hz\n",
- "Z = 124;# in ohms\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f1*L\n",
- "L = Z/(2*math.pi*f2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductive reactance, XL = \",round(XL,2),\" ohms \\n\"\n",
- "print \"\\n (b)Inductance L = \",round((L/1E-3),2),\" mH \\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductive reactance, XL = 100.53 ohms \n",
- "\n",
- "\n",
- " (b)Inductance L = 3.95 mH "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 214</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate its inductive reactance and the resulting current if connected to\n",
- "#(a) a 240 V, 50 Hz supply, and (b) a 100 V, 1 kHz supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.040;# in Henry\n",
- "V1 = 240;# in volts\n",
- "V2 = 100;# in volts\n",
- "f1 = 50;# in Hz\n",
- "f2 = 1000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL1 = 2*math.pi*f1*L\n",
- "I1 = V1/XL1\n",
- "XL2 = 2*math.pi*f2*L\n",
- "I2 = V2/XL2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductive reactance, XL = \",round( XL1,2),\" ohms and current I = \",round( I1,2),\" A\\n\"\n",
- "print \"\\n (b)Inductive reactance, XL = \",round( XL2,2),\" ohms and current I = \",round( I2,2),\" A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductive reactance, XL = 12.57 ohms and current I = 19.1 A\n",
- "\n",
- "\n",
- " (b)Inductive reactance, XL = 251.33 ohms and current I = 0.4 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 215</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitive reactance of a capacitor of 10 \u03bcF when connected to a circuit of frequency (a) 50 Hz (b) 20 kHz\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 10E-6;# in Farads\n",
- "f1 = 50;# in Hz\n",
- "f2 = 20000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "Xc1 = 1/(2*math.pi*f1*C)\n",
- "Xc2 = 1/(2*math.pi*f2*C)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Capacitive reactance, Xc = \",round( Xc1,2),\" ohms \"\n",
- "print \"\\n (b)Capacitive reactance, Xc = \",round( Xc2,2),\" ohms \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Capacitive reactance, Xc = 318.31 ohms \n",
- "\n",
- " (b)Capacitive reactance, Xc = 0.8 ohms "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 215</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of its capacitance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 40;# in ohms\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "C = 1/(2*math.pi*f*Z)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Capacitance,C = \",round((C/1E-6),2),\" uF \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Capacitance,C = 79.58 uF "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 215</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the current taken by a 23 \u03bcF capacitor when connected to a 240 V, 50 Hz supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 23E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "V = 240;# in volts\n",
- "\n",
- "#calculation:\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "I = V/Xc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current I = \",round(I,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current I = 1.73 A "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 216</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the supply voltage and the phase angle between current and voltage\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vr = 12;# in volts\n",
- "Vl = 5;# in volts\n",
- "\n",
- "#calculation:\n",
- "V = (Vr**2 + Vl**2)**0.5\n",
- "phi = math.atan(Vl/Vr)\n",
- "phid = phi*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n supply voltage V = \",V,\" V, phase angle between current and voltage is \", round(phid,2),\"deg lagging\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " supply voltage V = 13.0 V, phase angle between current and voltage is 22.62 deg lagging"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 216</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the reactance, (b) the impedance, and (c) the current taken from a 240 V, 50 Hz supply. \n",
- "#Determine also the phase angle between the supply voltage and current\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 240;# in volts\n",
- "R = 4;# in ohms\n",
- "L = 0.00955;# in Henry\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan(XL/R)*180/math.pi\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductive reactance, XL = \",round(XL,2),\" ohms\"\n",
- "print \"\\n (b)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (c)Current, I = \",round(I,2),\" A\"\n",
- "print \"\\n (d)phase angle between the supply voltage and current is \",round(phid,2),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductive reactance, XL = 3.0 ohms\n",
- "\n",
- " (b)Impedance, Z = 5.0 ohms\n",
- "\n",
- " (c)Current, I = 48.0 A\n",
- "\n",
- " (d)phase angle between the supply voltage and current is 36.87 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 217</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the resistance, impedance, inductive reactance and inductance of the coil.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vdc = 12;# in volts\n",
- "Vac = 240;# in volts\n",
- "Iac = 20;# in Amperes\n",
- "Idc = 2;# in Amperes\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "R = Vdc/Idc\n",
- "Z = Vac/Iac\n",
- "XL = (Z**2 - R**2)**0.5\n",
- "L = XL/(2*math.pi*f)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance, R = \",R,\" ohms\"\n",
- "print \"\\n (b)Impedance, Z = \",Z,\" ohms\"\n",
- "print \"\\n (c)Inductive reactance, XL = \",round(XL,2),\" ohms\"\n",
- "print \"\\n (d)Inductance, L = \",round(L,2),\" H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance, R = 6.0 ohms\n",
- "\n",
- " (b)Impedance, Z = 12.0 ohms\n",
- "\n",
- " (c)Inductive reactance, XL = 10.39 ohms\n",
- "\n",
- " (d)Inductance, L = 0.03 H"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 217</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the inductive reactance of the coil, \n",
- "#(b) the impedance of the circuit, \n",
- "#(c) the current in the circuit, \n",
- "#(d) the p.d. across each component, and \n",
- "#(e) the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 200;# in ohms\n",
- "L = 0.3183;# in henry\n",
- "V = 240;# in volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "I = V/Z\n",
- "VL = I*XL\n",
- "VR = I*R\n",
- "phid = math.atan(XL/R)*180/math.pi\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductive reactance, XL = \",round(XL,2),\" ohms\"\n",
- "print \"\\n (b)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (c)current, I = \",round(I,2),\" A\"\n",
- "print \"\\n (d)p.d. across Inductor, VL = \",round(VL,2),\" V and p.d. across resistance, VR = \",round(VR,2),\" V\"\n",
- "print \"\\n (e)circuit phase angle is \",round(phid,2),\" deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductive reactance, XL = 100.0 ohms\n",
- "\n",
- " (b)Impedance, Z = 223.61 ohms\n",
- "\n",
- " (c)current, I = 1.07 A\n",
- "\n",
- " (d)p.d. across Inductor, VL = 107.33 V and p.d. across resistance, VR = 214.66 V\n",
- "\n",
- " (e)circuit phase angle is 26.56 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 218</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate (a) the circuit impedance, \n",
- "#(b) the current flowing, \n",
- "#(c) the p.d. across the resistance, \n",
- "#(d) the p.d. across the inductance and \n",
- "#(e) the phase angle between voltage and current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 100;# in ohms\n",
- "L = 0.2;# in henry\n",
- "Vmax = 200;# in volts\n",
- "w = 500;# in rad/sec\n",
- "\n",
- "#calculation:\n",
- "Vrms = 0.707*Vmax\n",
- "f = w/(2*math.pi)\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "I = Vrms/Z\n",
- "VL = I*XL\n",
- "VR = I*R\n",
- "phid = math.atan(XL/R)*180/math.pi\n",
- "\n",
- "\\\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A\"\n",
- "print \"\\n (c)p.d. across resistance, VR = \",round(VR,2),\" V\"\n",
- "print \"\\n (d)p.d. across Inductor, VL = \",round(VL,2),\" V\"\n",
- "print \"\\n (e)circuit phase angle is \",phid,\"deg\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Impedance, Z = 141.42 ohms\n",
- "\n",
- " (b)current, I = 1.0 A\n",
- "\n",
- " (c)p.d. across resistance, VR = 99.98 V\n",
- "\n",
- " (d)p.d. across Inductor, VL = 99.98 V\n",
- "\n",
- " (e)circuit phase angle is 45.0 deg"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 218</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of the supply voltage and the voltage across the 1.273 mH inductance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "L = 1.2273E-3;# in henry\n",
- "f = 5000;# in Hz\n",
- "VR = 6;# in volts\n",
- "\n",
- "#calculation:\n",
- "I =VR/R\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "V = I*Z\n",
- "VL = I*XL\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)supply voltage = \",round(V,2),\" Volts\"\n",
- "print \"\\n (b)p.d. across Inductor, VL = \",round(VL,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)supply voltage = 9.77 Volts\n",
- "\n",
- " (b)p.d. across Inductor, VL = 7.71 V"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 219</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the impedance of the circuit, \n",
- "#(b) the current in the circuit, \n",
- "#(c) the circuit phase angle, \n",
- "#(d) the p.d. across the 60 ohm resistor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 60;# in ohms\n",
- "Rc = 20;# in ohms\n",
- "L = 159.2E-3;# in henry\n",
- "f = 50;# in Hz\n",
- "V = 240;# in volts\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Rt = R + Rc\n",
- "Z = (Rt**2 + XL**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan(XL/Rt)*180/math.pi\n",
- "VR = I*R\n",
- "Zc = (Rc**2 + XL**2)**0.5\n",
- "Vc = I*Zc\n",
- "VL = I*XL\n",
- "VRc = I*Rc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (b)current, I = \",round(I,3),\" A\"\n",
- "print \"\\n (c)circuit phase angle is \",round(phid,0),\"deg lagging\"\n",
- "print \"\\n (d)p.d. across resistance, VR = \",round( VR,1),\" V\"\n",
- "print \"\\n (e)p.d. across coil, Vc = \",round(Vc,1),\" V\"\n",
- "print \"\\n (f1)p.d. across Inductor, VL = \",round(VL,2),\" V\"\n",
- "print \"\\n (f2)p.d. across coil resistance, VRc = \",round(VRc,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Impedance, Z = 94.35 ohms\n",
- "\n",
- " (b)current, I = 2.544 A\n",
- "\n",
- " (c)circuit phase angle is 32.0 deg lagging\n",
- "\n",
- " (d)p.d. across resistance, VR = 152.6 V\n",
- "\n",
- " (e)p.d. across coil, Vc = 137.0 V\n",
- "\n",
- " (f1)p.d. across Inductor, VL = 127.23 V\n",
- "\n",
- " (f2)p.d. across coil resistance, VRc = 50.88 V"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 220</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the impedance, and (b) the current taken from a 240 V, 50 Hz supply. \n",
- "#Find also the phase angle between the supply voltage and the current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 25;# in ohms\n",
- "C = 45E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "V = 240;# in volts\n",
- "\n",
- "#calculation:\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Z = (R**2 + Xc**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan(Xc/R)*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A\"\n",
- "print \"\\n (c)phase angle between the supply voltage and current is \",round(phid,2),\"deg leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Impedance, Z = 75.02 ohms\n",
- "\n",
- " (b)current, I = 3.2 A\n",
- "\n",
- " (c)phase angle between the supply voltage and current is 70.54 deg leading"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 221</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate: (a) the value of capacitance, C, \n",
- "#(b) the supply voltage, \n",
- "#(c) the phase angle between the supply voltage and current, \n",
- "#(d) the p.d. across the resistor, and\n",
- "#(e) the p.d. across the capacitor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 40;# in ohms\n",
- "f = 60;# in Hz\n",
- "I = 3;#in amperes\n",
- "Z = 50;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Xc = (Z**2 - R**2)**0.5\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- "V = I*Z\n",
- "phid = math.atan(Xc/R)*180/math.pi\n",
- "VR = I*R\n",
- "Vc = I*Xc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)capacitance, C = \",round((C/1E-6),2),\" uF\"\n",
- "print \"\\n (b)Voltage, V = \",V,\" Volts\"\n",
- "print \"\\n (c)phase angle between the supply voltage and current is \",round(phid,2),\"deg leading\"\n",
- "print \"\\n (d)p.d. across resistance, VR = \", VR,\" V\"\n",
- "print \"\\n (e)p.d. across Capacitor, Vc = \",Vc,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)capacitance, C = 88.42 uF\n",
- "\n",
- " (b)Voltage, V = 150 Volts\n",
- "\n",
- " (c)phase angle between the supply voltage and current is 36.87 deg leading\n",
- "\n",
- " (d)p.d. across resistance, VR = 120 V\n",
- "\n",
- " (e)p.d. across Capacitor, Vc = 90.0 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 222</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current flowing, \n",
- "#(b) the phase difference between the supply voltage and current, \n",
- "#(c) the voltage across the coil and \n",
- "#(d) the voltage across the capacitor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 5;# in ohms\n",
- "C = 100E-6;# in Farads\n",
- "L = 0.12;# in Henry\n",
- "f = 50;# in Hz\n",
- "V = 300;# in volts\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "X = XL - Xc\n",
- " #Since XL is greater than Xc, the circuit is inductive.\n",
- "Z = (R**2 + (XL-Xc)**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan((XL-Xc)/R)*180/math.pi\n",
- "Zcl = (R**2 + XL**2)**0.5\n",
- "Vcl = I*Zcl\n",
- "phidc = math.atan(XL/R)*180/math.pi\n",
- "Vc = I*Xc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current,I = \",round(I,2),\" A\"\n",
- "print \"\\n (b)phase angle between the supply voltage and current is \",round(phid,2),\"deg\"\n",
- "print \"\\n (c)Voltage across the coil, Vcoil = \",round(Vcl,0),\" Volts\"\n",
- "print \"\\n (d)p.d. across Capacitor, Vc = \",round(Vc,0),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current,I = 38.91 A\n",
- "\n",
- " (b)phase angle between the supply voltage and current is 49.57 deg\n",
- "\n",
- " (c)Voltage across the coil, Vcoil = 1480.0 Volts\n",
- "\n",
- " (d)p.d. across Capacitor, Vc = 1239.0 V"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 224</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the circuit current, \n",
- "#(b) the circuit phase angle and \n",
- "#(c) the voltage drop across each impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 8;# in ohms\n",
- "C = 0.25E-6;# in Farads\n",
- "L = 130E-6;# in Henry\n",
- "Rc = 5;# in ohms\n",
- "R2 = 10;# in ohms\n",
- "f = 20000;# in Hz\n",
- "V = 40;# in volts\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "X = Xc - XL\n",
- "R = R1 + R2 + Rc\n",
- " #Since Xc is greater than XL, the circuit is capacitive.\n",
- "Z = (R**2 + (Xc-XL)**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan((Xc-XL)/R)*180/math.pi\n",
- "V1 = I*R1\n",
- "V2 = I*((Rc**2 + XL**2)**0.5)\n",
- "V3 = I*((R2**2 + Xc**2)**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current,I = \",round(I,2),\" A\"\n",
- "print \"\\n (b)circuit phase angle is \",round(phid,2),\"deg leading\"\n",
- "print \"\\n (c1)Voltage across the Resistance of 8 ohms = \",round(V1,2),\" Volts\"\n",
- "print \"\\n (c2)Voltage across the coil, Vcoil = \",round(V2,2),\" Volts\"\n",
- "print \"\\n (c3)p.d. across Capacitor resistance circuit = \",round(V3,2),\" Volts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current,I = 1.44 A\n",
- "\n",
- " (b)circuit phase angle is 33.97 deg leading\n",
- "\n",
- " (c1)Voltage across the Resistance of 8 ohms = 11.54 Volts\n",
- "\n",
- " (c2)Voltage across the coil, Vcoil = 24.64 Volts\n",
- "\n",
- " (c3)p.d. across Capacitor resistance circuit = 48.12 Volts"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 224</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the p.d.\u2019s V1 and V2 for the circuit\n",
- "#determine the supply voltage V and the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 4;# in ohms\n",
- "C = 1.273E-6;# in Farads\n",
- "L = 0.286E-3;# in Henry\n",
- "R2 = 8;# in ohms\n",
- "f = 5000;# in Hz\n",
- "I = 5;# in amperes\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "phid1 = math.atan(XL/R1)*180/math.pi\n",
- "V1 = I*((R1**2 + XL**2)**0.5)\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "V2 = I*((R2**2 + Xc**2)**0.5)\n",
- "phid2 = math.atan(Xc/R2)*180/math.pi\n",
- "Z = ((R1+R2)**2 + (Xc-XL)**2)**0.5\n",
- "V = I*Z\n",
- "phid = math.atan((Xc-XL)/(R1+R2))*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Voltage supply, V = \",round(V,2),\" V\"\n",
- "print \"\\n (b)circuit phase angle is \",round(phid,2),\"deg leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Voltage supply, V = 100.08 V\n",
- "\n",
- " (b)circuit phase angle is 53.16 deg leading"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 226</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#At what frequency does resonance occur?\n",
- "#Find the current flowing at the resonant frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "C = 60E-6;# in Farads\n",
- "L = 125E-3;# in Henry\n",
- "V = 120;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = 1/(2*math.pi*(L*C)**0.5)\n",
- " #At resonance, XL = Xc and impedance Z = R\n",
- "I = V/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonance frequency,Fr = \",round(fr,2),\" Hz\"\n",
- "print \"\\n (b)Current, I = \",round(I,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonance frequency,Fr = 58.12 Hz\n",
- "\n",
- " (b)Current, I = 12.0"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 226</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the circuit resistance, and (b) the circuit capacitance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.05E-3;# in Henry\n",
- "fr = 200000;# in Hz\n",
- "V = 0.002;# in Volts\n",
- "I = 0.1E-3;# in amperes\n",
- "#calculation:\n",
- "# L-C-R\n",
- "#At resonance, XL = Xc and impedance Z = R\n",
- "R = V/I\n",
- "C = 1/(L*(2*math.pi*fr)**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance, R = \",round(R,2),\" ohms\"\n",
- "print \"\\n (b)Capacitance, C = \",round((C/1E-9),2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance, R = 20.0 ohms\n",
- "\n",
- " (b)Capacitance, C = 12.67 nF"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 227</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, and \n",
- "#(b) the current at resonance. \n",
- "#How many times greater than the supply voltage is the voltage across the reactances at resonance?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 80E-3;# in Henry\n",
- "C = 0.25E-6;# in Farads\n",
- "R = 12.5;# in ohms\n",
- "V = 100;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- " #At resonance, XL = Xc and impedance Z = R\n",
- "I = V/R\n",
- "VL = I*(2*math.pi*fr*L)\n",
- "Vc = I/(2*math.pi*fr*C)\n",
- "Vm = VL/V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the resonant frequency = \",round(fr,2),\" Hz\"\n",
- "print \"\\n (b)Current, I = \",round(I,2),\"\"\n",
- "print \"\\n (b)Voltage magnification at resonance = \",round(Vm,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the resonant frequency = 1125.4 Hz\n",
- "\n",
- " (b)Current, I = 8.0 \n",
- "\n",
- " (b)Voltage magnification at resonance = 45.25 V"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 228</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the Qfactor of the circuit at resonance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 60E-3;# in Henry\n",
- "C = 30E-6;# in Farads\n",
- "R = 2;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Q = ((L/C)**0.5)/R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n At resonance, Q-factor = \",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " At resonance, Q-factor = 22.36"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 228</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, \n",
- "#(b) the current at resonance,\n",
- "#(c) the voltages across the coil and the capacitor at resonance, and\n",
- "#(d) the Q-factor of the circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 100E-3;# in Henry\n",
- "C = 2E-6;# in Farads\n",
- "R = 10;# in ohms\n",
- "V = 50;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- " #At resonance, XL = Xc and impedance Z = R\n",
- "I = V/R\n",
- "VL = I*(2*math.pi*fr*L)\n",
- "Vc = I/(2*math.pi*fr*C)\n",
- "Q = VL/V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the resonant frequency = \",round(fr,2),\" Hz\"\n",
- "print \"\\n (b)Current, I = \",round(I,2),\"\"\n",
- "print \"\\n (c)Voltage across coil at resonance is \",round(VL,2),\"V \"\n",
- "print \"and Voltage across capacitance at resonance is \",round( Vc,2),\"V\"\n",
- "print \"\\n (d)At resonance, Q-factor = \",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the resonant frequency = 355.88 Hz\n",
- "\n",
- " (b)Current, I = 5.0 \n",
- "\n",
- " (c)Voltage across coil at resonance is 1118.03 V and Voltage across capacitance at resonance is 1118.03 V\n",
- "\n",
- " (d)At resonance, Q-factor = 22.36"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 230</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the bandwidth of the filter.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 20E-3;# in Henry\n",
- "R = 10;# in ohms\n",
- "fr = 5000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "Qr = (2*math.pi*fr)*L/R\n",
- "bw = fr/Qr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Bandwidth, (f2-f1) = \",round(bw,2),\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Bandwidth, (f2-f1) = 79.58 Hz"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 231</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the power dissipated in the resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 5000;# in ohms\n",
- "Imax = 0.250;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "Irms = 0.707*Imax\n",
- "P = Irms*Irms*R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Power, P = \",round(P,2),\" Watts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Power, P = 156.2 Watts"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 25, page no. 231</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the power dissipated.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 60;# in ohms\n",
- "L = 75E-3;# in Henry\n",
- "V = 110;# in Volts\n",
- "f = 60;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "I = V/Z\n",
- "P = I*I*R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Power, P = \",round(P,2),\" Watts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Power, P = 165.02 Watts"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 26, page no. 232</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the value of the inductance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "VI = 300;# in VA\n",
- "V = 150;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "I = VI/V\n",
- "XL = V/I\n",
- "L = XL/(2*math.pi*f)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Inductance = \",round(L,2),\" H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Inductance = 0.24 H"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 27, page no. 232</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the rated power output and the corresponding reactive power.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "VI = 200000;# in VA\n",
- "pf = 0.8;# power factor\n",
- "\n",
- "#calculation:\n",
- "P = VI*pf\n",
- "Q = VI*math.sin(math.acos(pf))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n rated power output is \",round(P/1000,2),\"KW and the corresponding reactive power is \",round(Q/1000,2),\"kvar\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " rated power output is 160.0 KW and the corresponding reactive power is 120.0 kvar"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 28, page no. 233</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the resistance, \n",
- "#(b) the impedance, \n",
- "#(c) the reactance, \n",
- "#(d) the power factor, and \n",
- "#(e) the phase angle between voltage and current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 120;# in Volts\n",
- "f = 50;# in Hz\n",
- "P = 400;# in Watt\n",
- "I = 8;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "R = P/(I*I)\n",
- "Z = V/I\n",
- "XL = (Z**2 - R**2)**0.5\n",
- "pf = P/(V*I)\n",
- "phi = math.acos(pf)*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)resistance = \",round(R,2),\" ohm \"\n",
- "print \"\\n (b)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (c)reactance = \",round(XL,2),\" ohm \"\n",
- "print \"\\n (d)Power factor = \",round(pf,2),\"\"\n",
- "print \"\\n (e)phase angle = \",round(phi,2),\"deg lagging\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)resistance = 6.25 ohm \n",
- "\n",
- " (b)Impedance Z = 15.0 Ohm \n",
- "\n",
- " (c)reactance = 13.64 ohm \n",
- "\n",
- " (d)Power factor = 0.42 \n",
- "\n",
- " (e)phase angle = 65.38 deg lagging"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 29, page no. 233</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the current flowing, (b) the phase angle,\n",
- "#(c) the resistance, (d) the impedance, and (e) the capacitance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 100;# in Volts\n",
- "f = 60;# in Hz\n",
- "P = 100;# in Watt\n",
- "pf = 0.5;# power factor\n",
- "\n",
- "#calculation:\n",
- "I = P/(pf*V)\n",
- "phi = math.acos(pf)*180/math.pi\n",
- "R = P/(I*I)\n",
- "Z = V/I\n",
- "Xc = (Z**2 - R**2)**0.5\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current I = \",round(I,2),\" A \"\n",
- "print \"\\n (b)phase angle = \",round(phi,2),\"deg leading\"\n",
- "print \"\\n (c)resistance = \",round(R,2),\" ohm \"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)capacitance = \",round((C/1E-6),2),\"uF \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current I = 2.0 A \n",
- "\n",
- " (b)phase angle = 60.0 deg leading\n",
- "\n",
- " (c)resistance = 25.0 ohm \n",
- "\n",
- " (d)Impedance Z = 50.0 Ohm \n",
- "\n",
- " (e)capacitance = 61.26 uF "
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_1.ipynb deleted file mode 100755 index e7b68d8a..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_1.ipynb +++ /dev/null @@ -1,1677 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 15: Single-phase series a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 214</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Calculate the reactance of a coil \n",
- "#(b) Determine the inductance of the coil.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.32;# in Henry\n",
- "f1 = 50;# in Hz\n",
- "f2 = 5000;# in Hz\n",
- "Z = 124;# in ohms\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f1*L\n",
- "L = Z/(2*math.pi*f2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductive reactance, XL = \",round(XL,2),\" ohms \\n\"\n",
- "print \"\\n (b)Inductance L = \",round((L/1E-3),2),\" mH \\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductive reactance, XL = 100.53 ohms \n",
- "\n",
- "\n",
- " (b)Inductance L = 3.95 mH "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 214</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate its inductive reactance and the resulting current if connected to\n",
- "#(a) a 240 V, 50 Hz supply, and (b) a 100 V, 1 kHz supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.040;# in Henry\n",
- "V1 = 240;# in volts\n",
- "V2 = 100;# in volts\n",
- "f1 = 50;# in Hz\n",
- "f2 = 1000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL1 = 2*math.pi*f1*L\n",
- "I1 = V1/XL1\n",
- "XL2 = 2*math.pi*f2*L\n",
- "I2 = V2/XL2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductive reactance, XL = \",round( XL1,2),\" ohms and current I = \",round( I1,2),\" A\\n\"\n",
- "print \"\\n (b)Inductive reactance, XL = \",round( XL2,2),\" ohms and current I = \",round( I2,2),\" A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductive reactance, XL = 12.57 ohms and current I = 19.1 A\n",
- "\n",
- "\n",
- " (b)Inductive reactance, XL = 251.33 ohms and current I = 0.4 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 215</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitive reactance of a capacitor of 10 \u03bcF when connected to a circuit of frequency (a) 50 Hz (b) 20 kHz\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 10E-6;# in Farads\n",
- "f1 = 50;# in Hz\n",
- "f2 = 20000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "Xc1 = 1/(2*math.pi*f1*C)\n",
- "Xc2 = 1/(2*math.pi*f2*C)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Capacitive reactance, Xc = \",round( Xc1,2),\" ohms \"\n",
- "print \"\\n (b)Capacitive reactance, Xc = \",round( Xc2,2),\" ohms \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Capacitive reactance, Xc = 318.31 ohms \n",
- "\n",
- " (b)Capacitive reactance, Xc = 0.8 ohms "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 215</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of its capacitance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 40;# in ohms\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "C = 1/(2*math.pi*f*Z)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Capacitance,C = \",round((C/1E-6),2),\" uF \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Capacitance,C = 79.58 uF "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 215</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the current taken by a 23 \u03bcF capacitor when connected to a 240 V, 50 Hz supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 23E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "V = 240;# in volts\n",
- "\n",
- "#calculation:\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "I = V/Xc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current I = \",round(I,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current I = 1.73 A "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 216</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the supply voltage and the phase angle between current and voltage\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vr = 12;# in volts\n",
- "Vl = 5;# in volts\n",
- "\n",
- "#calculation:\n",
- "V = (Vr**2 + Vl**2)**0.5\n",
- "phi = math.atan(Vl/Vr)\n",
- "phid = phi*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n supply voltage V = \",V,\" V, phase angle between current and voltage is \", round(phid,2),\"deg lagging\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " supply voltage V = 13.0 V, phase angle between current and voltage is 22.62 deg lagging"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 216</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the reactance, (b) the impedance, and (c) the current taken from a 240 V, 50 Hz supply. \n",
- "#Determine also the phase angle between the supply voltage and current\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 240;# in volts\n",
- "R = 4;# in ohms\n",
- "L = 0.00955;# in Henry\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan(XL/R)*180/math.pi\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductive reactance, XL = \",round(XL,2),\" ohms\"\n",
- "print \"\\n (b)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (c)Current, I = \",round(I,2),\" A\"\n",
- "print \"\\n (d)phase angle between the supply voltage and current is \",round(phid,2),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductive reactance, XL = 3.0 ohms\n",
- "\n",
- " (b)Impedance, Z = 5.0 ohms\n",
- "\n",
- " (c)Current, I = 48.0 A\n",
- "\n",
- " (d)phase angle between the supply voltage and current is 36.87 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 217</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the resistance, impedance, inductive reactance and inductance of the coil.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vdc = 12;# in volts\n",
- "Vac = 240;# in volts\n",
- "Iac = 20;# in Amperes\n",
- "Idc = 2;# in Amperes\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "R = Vdc/Idc\n",
- "Z = Vac/Iac\n",
- "XL = (Z**2 - R**2)**0.5\n",
- "L = XL/(2*math.pi*f)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance, R = \",R,\" ohms\"\n",
- "print \"\\n (b)Impedance, Z = \",Z,\" ohms\"\n",
- "print \"\\n (c)Inductive reactance, XL = \",round(XL,2),\" ohms\"\n",
- "print \"\\n (d)Inductance, L = \",round(L,2),\" H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance, R = 6.0 ohms\n",
- "\n",
- " (b)Impedance, Z = 12.0 ohms\n",
- "\n",
- " (c)Inductive reactance, XL = 10.39 ohms\n",
- "\n",
- " (d)Inductance, L = 0.03 H"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 217</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the inductive reactance of the coil, \n",
- "#(b) the impedance of the circuit, \n",
- "#(c) the current in the circuit, \n",
- "#(d) the p.d. across each component, and \n",
- "#(e) the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 200;# in ohms\n",
- "L = 0.3183;# in henry\n",
- "V = 240;# in volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "I = V/Z\n",
- "VL = I*XL\n",
- "VR = I*R\n",
- "phid = math.atan(XL/R)*180/math.pi\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductive reactance, XL = \",round(XL,2),\" ohms\"\n",
- "print \"\\n (b)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (c)current, I = \",round(I,2),\" A\"\n",
- "print \"\\n (d)p.d. across Inductor, VL = \",round(VL,2),\" V and p.d. across resistance, VR = \",round(VR,2),\" V\"\n",
- "print \"\\n (e)circuit phase angle is \",round(phid,2),\" deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductive reactance, XL = 100.0 ohms\n",
- "\n",
- " (b)Impedance, Z = 223.61 ohms\n",
- "\n",
- " (c)current, I = 1.07 A\n",
- "\n",
- " (d)p.d. across Inductor, VL = 107.33 V and p.d. across resistance, VR = 214.66 V\n",
- "\n",
- " (e)circuit phase angle is 26.56 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 218</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate (a) the circuit impedance, \n",
- "#(b) the current flowing, \n",
- "#(c) the p.d. across the resistance, \n",
- "#(d) the p.d. across the inductance and \n",
- "#(e) the phase angle between voltage and current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 100;# in ohms\n",
- "L = 0.2;# in henry\n",
- "Vmax = 200;# in volts\n",
- "w = 500;# in rad/sec\n",
- "\n",
- "#calculation:\n",
- "Vrms = 0.707*Vmax\n",
- "f = w/(2*math.pi)\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "I = Vrms/Z\n",
- "VL = I*XL\n",
- "VR = I*R\n",
- "phid = math.atan(XL/R)*180/math.pi\n",
- "\n",
- "\\\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A\"\n",
- "print \"\\n (c)p.d. across resistance, VR = \",round(VR,2),\" V\"\n",
- "print \"\\n (d)p.d. across Inductor, VL = \",round(VL,2),\" V\"\n",
- "print \"\\n (e)circuit phase angle is \",phid,\"deg\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Impedance, Z = 141.42 ohms\n",
- "\n",
- " (b)current, I = 1.0 A\n",
- "\n",
- " (c)p.d. across resistance, VR = 99.98 V\n",
- "\n",
- " (d)p.d. across Inductor, VL = 99.98 V\n",
- "\n",
- " (e)circuit phase angle is 45.0 deg"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 218</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of the supply voltage and the voltage across the 1.273 mH inductance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "L = 1.2273E-3;# in henry\n",
- "f = 5000;# in Hz\n",
- "VR = 6;# in volts\n",
- "\n",
- "#calculation:\n",
- "I =VR/R\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "V = I*Z\n",
- "VL = I*XL\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)supply voltage = \",round(V,2),\" Volts\"\n",
- "print \"\\n (b)p.d. across Inductor, VL = \",round(VL,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)supply voltage = 9.77 Volts\n",
- "\n",
- " (b)p.d. across Inductor, VL = 7.71 V"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 219</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the impedance of the circuit, \n",
- "#(b) the current in the circuit, \n",
- "#(c) the circuit phase angle, \n",
- "#(d) the p.d. across the 60 ohm resistor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 60;# in ohms\n",
- "Rc = 20;# in ohms\n",
- "L = 159.2E-3;# in henry\n",
- "f = 50;# in Hz\n",
- "V = 240;# in volts\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Rt = R + Rc\n",
- "Z = (Rt**2 + XL**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan(XL/Rt)*180/math.pi\n",
- "VR = I*R\n",
- "Zc = (Rc**2 + XL**2)**0.5\n",
- "Vc = I*Zc\n",
- "VL = I*XL\n",
- "VRc = I*Rc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (b)current, I = \",round(I,3),\" A\"\n",
- "print \"\\n (c)circuit phase angle is \",round(phid,0),\"deg lagging\"\n",
- "print \"\\n (d)p.d. across resistance, VR = \",round( VR,1),\" V\"\n",
- "print \"\\n (e)p.d. across coil, Vc = \",round(Vc,1),\" V\"\n",
- "print \"\\n (f1)p.d. across Inductor, VL = \",round(VL,2),\" V\"\n",
- "print \"\\n (f2)p.d. across coil resistance, VRc = \",round(VRc,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Impedance, Z = 94.35 ohms\n",
- "\n",
- " (b)current, I = 2.544 A\n",
- "\n",
- " (c)circuit phase angle is 32.0 deg lagging\n",
- "\n",
- " (d)p.d. across resistance, VR = 152.6 V\n",
- "\n",
- " (e)p.d. across coil, Vc = 137.0 V\n",
- "\n",
- " (f1)p.d. across Inductor, VL = 127.23 V\n",
- "\n",
- " (f2)p.d. across coil resistance, VRc = 50.88 V"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 220</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the impedance, and (b) the current taken from a 240 V, 50 Hz supply. \n",
- "#Find also the phase angle between the supply voltage and the current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 25;# in ohms\n",
- "C = 45E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "V = 240;# in volts\n",
- "\n",
- "#calculation:\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Z = (R**2 + Xc**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan(Xc/R)*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A\"\n",
- "print \"\\n (c)phase angle between the supply voltage and current is \",round(phid,2),\"deg leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Impedance, Z = 75.02 ohms\n",
- "\n",
- " (b)current, I = 3.2 A\n",
- "\n",
- " (c)phase angle between the supply voltage and current is 70.54 deg leading"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 221</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate: (a) the value of capacitance, C, \n",
- "#(b) the supply voltage, \n",
- "#(c) the phase angle between the supply voltage and current, \n",
- "#(d) the p.d. across the resistor, and\n",
- "#(e) the p.d. across the capacitor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 40;# in ohms\n",
- "f = 60;# in Hz\n",
- "I = 3;#in amperes\n",
- "Z = 50;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Xc = (Z**2 - R**2)**0.5\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- "V = I*Z\n",
- "phid = math.atan(Xc/R)*180/math.pi\n",
- "VR = I*R\n",
- "Vc = I*Xc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)capacitance, C = \",round((C/1E-6),2),\" uF\"\n",
- "print \"\\n (b)Voltage, V = \",V,\" Volts\"\n",
- "print \"\\n (c)phase angle between the supply voltage and current is \",round(phid,2),\"deg leading\"\n",
- "print \"\\n (d)p.d. across resistance, VR = \", VR,\" V\"\n",
- "print \"\\n (e)p.d. across Capacitor, Vc = \",Vc,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)capacitance, C = 88.42 uF\n",
- "\n",
- " (b)Voltage, V = 150 Volts\n",
- "\n",
- " (c)phase angle between the supply voltage and current is 36.87 deg leading\n",
- "\n",
- " (d)p.d. across resistance, VR = 120 V\n",
- "\n",
- " (e)p.d. across Capacitor, Vc = 90.0 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 222</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current flowing, \n",
- "#(b) the phase difference between the supply voltage and current, \n",
- "#(c) the voltage across the coil and \n",
- "#(d) the voltage across the capacitor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 5;# in ohms\n",
- "C = 100E-6;# in Farads\n",
- "L = 0.12;# in Henry\n",
- "f = 50;# in Hz\n",
- "V = 300;# in volts\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "X = XL - Xc\n",
- " #Since XL is greater than Xc, the circuit is inductive.\n",
- "Z = (R**2 + (XL-Xc)**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan((XL-Xc)/R)*180/math.pi\n",
- "Zcl = (R**2 + XL**2)**0.5\n",
- "Vcl = I*Zcl\n",
- "phidc = math.atan(XL/R)*180/math.pi\n",
- "Vc = I*Xc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current,I = \",round(I,2),\" A\"\n",
- "print \"\\n (b)phase angle between the supply voltage and current is \",round(phid,2),\"deg\"\n",
- "print \"\\n (c)Voltage across the coil, Vcoil = \",round(Vcl,0),\" Volts\"\n",
- "print \"\\n (d)p.d. across Capacitor, Vc = \",round(Vc,0),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current,I = 38.91 A\n",
- "\n",
- " (b)phase angle between the supply voltage and current is 49.57 deg\n",
- "\n",
- " (c)Voltage across the coil, Vcoil = 1480.0 Volts\n",
- "\n",
- " (d)p.d. across Capacitor, Vc = 1239.0 V"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 224</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the circuit current, \n",
- "#(b) the circuit phase angle and \n",
- "#(c) the voltage drop across each impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 8;# in ohms\n",
- "C = 0.25E-6;# in Farads\n",
- "L = 130E-6;# in Henry\n",
- "Rc = 5;# in ohms\n",
- "R2 = 10;# in ohms\n",
- "f = 20000;# in Hz\n",
- "V = 40;# in volts\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "X = Xc - XL\n",
- "R = R1 + R2 + Rc\n",
- " #Since Xc is greater than XL, the circuit is capacitive.\n",
- "Z = (R**2 + (Xc-XL)**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan((Xc-XL)/R)*180/math.pi\n",
- "V1 = I*R1\n",
- "V2 = I*((Rc**2 + XL**2)**0.5)\n",
- "V3 = I*((R2**2 + Xc**2)**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current,I = \",round(I,2),\" A\"\n",
- "print \"\\n (b)circuit phase angle is \",round(phid,2),\"deg leading\"\n",
- "print \"\\n (c1)Voltage across the Resistance of 8 ohms = \",round(V1,2),\" Volts\"\n",
- "print \"\\n (c2)Voltage across the coil, Vcoil = \",round(V2,2),\" Volts\"\n",
- "print \"\\n (c3)p.d. across Capacitor resistance circuit = \",round(V3,2),\" Volts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current,I = 1.44 A\n",
- "\n",
- " (b)circuit phase angle is 33.97 deg leading\n",
- "\n",
- " (c1)Voltage across the Resistance of 8 ohms = 11.54 Volts\n",
- "\n",
- " (c2)Voltage across the coil, Vcoil = 24.64 Volts\n",
- "\n",
- " (c3)p.d. across Capacitor resistance circuit = 48.12 Volts"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 224</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the p.d.\u2019s V1 and V2 for the circuit\n",
- "#determine the supply voltage V and the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 4;# in ohms\n",
- "C = 1.273E-6;# in Farads\n",
- "L = 0.286E-3;# in Henry\n",
- "R2 = 8;# in ohms\n",
- "f = 5000;# in Hz\n",
- "I = 5;# in amperes\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "phid1 = math.atan(XL/R1)*180/math.pi\n",
- "V1 = I*((R1**2 + XL**2)**0.5)\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "V2 = I*((R2**2 + Xc**2)**0.5)\n",
- "phid2 = math.atan(Xc/R2)*180/math.pi\n",
- "Z = ((R1+R2)**2 + (Xc-XL)**2)**0.5\n",
- "V = I*Z\n",
- "phid = math.atan((Xc-XL)/(R1+R2))*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Voltage supply, V = \",round(V,2),\" V\"\n",
- "print \"\\n (b)circuit phase angle is \",round(phid,2),\"deg leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Voltage supply, V = 100.08 V\n",
- "\n",
- " (b)circuit phase angle is 53.16 deg leading"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 226</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#At what frequency does resonance occur?\n",
- "#Find the current flowing at the resonant frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "C = 60E-6;# in Farads\n",
- "L = 125E-3;# in Henry\n",
- "V = 120;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = 1/(2*math.pi*(L*C)**0.5)\n",
- " #At resonance, XL = Xc and impedance Z = R\n",
- "I = V/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonance frequency,Fr = \",round(fr,2),\" Hz\"\n",
- "print \"\\n (b)Current, I = \",round(I,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonance frequency,Fr = 58.12 Hz\n",
- "\n",
- " (b)Current, I = 12.0"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 226</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the circuit resistance, and (b) the circuit capacitance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.05E-3;# in Henry\n",
- "fr = 200000;# in Hz\n",
- "V = 0.002;# in Volts\n",
- "I = 0.1E-3;# in amperes\n",
- "#calculation:\n",
- "# L-C-R\n",
- "#At resonance, XL = Xc and impedance Z = R\n",
- "R = V/I\n",
- "C = 1/(L*(2*math.pi*fr)**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance, R = \",round(R,2),\" ohms\"\n",
- "print \"\\n (b)Capacitance, C = \",round((C/1E-9),2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance, R = 20.0 ohms\n",
- "\n",
- " (b)Capacitance, C = 12.67 nF"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 227</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, and \n",
- "#(b) the current at resonance. \n",
- "#How many times greater than the supply voltage is the voltage across the reactances at resonance?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 80E-3;# in Henry\n",
- "C = 0.25E-6;# in Farads\n",
- "R = 12.5;# in ohms\n",
- "V = 100;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- " #At resonance, XL = Xc and impedance Z = R\n",
- "I = V/R\n",
- "VL = I*(2*math.pi*fr*L)\n",
- "Vc = I/(2*math.pi*fr*C)\n",
- "Vm = VL/V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the resonant frequency = \",round(fr,2),\" Hz\"\n",
- "print \"\\n (b)Current, I = \",round(I,2),\"\"\n",
- "print \"\\n (b)Voltage magnification at resonance = \",round(Vm,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the resonant frequency = 1125.4 Hz\n",
- "\n",
- " (b)Current, I = 8.0 \n",
- "\n",
- " (b)Voltage magnification at resonance = 45.25 V"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 228</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the Qfactor of the circuit at resonance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 60E-3;# in Henry\n",
- "C = 30E-6;# in Farads\n",
- "R = 2;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Q = ((L/C)**0.5)/R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n At resonance, Q-factor = \",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " At resonance, Q-factor = 22.36"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 228</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, \n",
- "#(b) the current at resonance,\n",
- "#(c) the voltages across the coil and the capacitor at resonance, and\n",
- "#(d) the Q-factor of the circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 100E-3;# in Henry\n",
- "C = 2E-6;# in Farads\n",
- "R = 10;# in ohms\n",
- "V = 50;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- " #At resonance, XL = Xc and impedance Z = R\n",
- "I = V/R\n",
- "VL = I*(2*math.pi*fr*L)\n",
- "Vc = I/(2*math.pi*fr*C)\n",
- "Q = VL/V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the resonant frequency = \",round(fr,2),\" Hz\"\n",
- "print \"\\n (b)Current, I = \",round(I,2),\"\"\n",
- "print \"\\n (c)Voltage across coil at resonance is \",round(VL,2),\"V \"\n",
- "print \"and Voltage across capacitance at resonance is \",round( Vc,2),\"V\"\n",
- "print \"\\n (d)At resonance, Q-factor = \",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the resonant frequency = 355.88 Hz\n",
- "\n",
- " (b)Current, I = 5.0 \n",
- "\n",
- " (c)Voltage across coil at resonance is 1118.03 V and Voltage across capacitance at resonance is 1118.03 V\n",
- "\n",
- " (d)At resonance, Q-factor = 22.36"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 230</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the bandwidth of the filter.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 20E-3;# in Henry\n",
- "R = 10;# in ohms\n",
- "fr = 5000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "Qr = (2*math.pi*fr)*L/R\n",
- "bw = fr/Qr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Bandwidth, (f2-f1) = \",round(bw,2),\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Bandwidth, (f2-f1) = 79.58 Hz"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 231</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the power dissipated in the resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 5000;# in ohms\n",
- "Imax = 0.250;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "Irms = 0.707*Imax\n",
- "P = Irms*Irms*R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Power, P = \",round(P,2),\" Watts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Power, P = 156.2 Watts"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 25, page no. 231</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the power dissipated.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 60;# in ohms\n",
- "L = 75E-3;# in Henry\n",
- "V = 110;# in Volts\n",
- "f = 60;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "I = V/Z\n",
- "P = I*I*R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Power, P = \",round(P,2),\" Watts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Power, P = 165.02 Watts"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 26, page no. 232</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the value of the inductance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "VI = 300;# in VA\n",
- "V = 150;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "I = VI/V\n",
- "XL = V/I\n",
- "L = XL/(2*math.pi*f)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Inductance = \",round(L,2),\" H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Inductance = 0.24 H"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 27, page no. 232</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the rated power output and the corresponding reactive power.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "VI = 200000;# in VA\n",
- "pf = 0.8;# power factor\n",
- "\n",
- "#calculation:\n",
- "P = VI*pf\n",
- "Q = VI*math.sin(math.acos(pf))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n rated power output is \",round(P/1000,2),\"KW and the corresponding reactive power is \",round(Q/1000,2),\"kvar\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " rated power output is 160.0 KW and the corresponding reactive power is 120.0 kvar"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 28, page no. 233</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the resistance, \n",
- "#(b) the impedance, \n",
- "#(c) the reactance, \n",
- "#(d) the power factor, and \n",
- "#(e) the phase angle between voltage and current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 120;# in Volts\n",
- "f = 50;# in Hz\n",
- "P = 400;# in Watt\n",
- "I = 8;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "R = P/(I*I)\n",
- "Z = V/I\n",
- "XL = (Z**2 - R**2)**0.5\n",
- "pf = P/(V*I)\n",
- "phi = math.acos(pf)*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)resistance = \",round(R,2),\" ohm \"\n",
- "print \"\\n (b)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (c)reactance = \",round(XL,2),\" ohm \"\n",
- "print \"\\n (d)Power factor = \",round(pf,2),\"\"\n",
- "print \"\\n (e)phase angle = \",round(phi,2),\"deg lagging\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)resistance = 6.25 ohm \n",
- "\n",
- " (b)Impedance Z = 15.0 Ohm \n",
- "\n",
- " (c)reactance = 13.64 ohm \n",
- "\n",
- " (d)Power factor = 0.42 \n",
- "\n",
- " (e)phase angle = 65.38 deg lagging"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 29, page no. 233</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the current flowing, (b) the phase angle,\n",
- "#(c) the resistance, (d) the impedance, and (e) the capacitance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 100;# in Volts\n",
- "f = 60;# in Hz\n",
- "P = 100;# in Watt\n",
- "pf = 0.5;# power factor\n",
- "\n",
- "#calculation:\n",
- "I = P/(pf*V)\n",
- "phi = math.acos(pf)*180/math.pi\n",
- "R = P/(I*I)\n",
- "Z = V/I\n",
- "Xc = (Z**2 - R**2)**0.5\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current I = \",round(I,2),\" A \"\n",
- "print \"\\n (b)phase angle = \",round(phi,2),\"deg leading\"\n",
- "print \"\\n (c)resistance = \",round(R,2),\" ohm \"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)capacitance = \",round((C/1E-6),2),\"uF \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current I = 2.0 A \n",
- "\n",
- " (b)phase angle = 60.0 deg leading\n",
- "\n",
- " (c)resistance = 25.0 ohm \n",
- "\n",
- " (d)Impedance Z = 50.0 Ohm \n",
- "\n",
- " (e)capacitance = 61.26 uF "
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_2.ipynb deleted file mode 100755 index e7b68d8a..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_2.ipynb +++ /dev/null @@ -1,1677 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 15: Single-phase series a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 214</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Calculate the reactance of a coil \n",
- "#(b) Determine the inductance of the coil.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.32;# in Henry\n",
- "f1 = 50;# in Hz\n",
- "f2 = 5000;# in Hz\n",
- "Z = 124;# in ohms\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f1*L\n",
- "L = Z/(2*math.pi*f2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductive reactance, XL = \",round(XL,2),\" ohms \\n\"\n",
- "print \"\\n (b)Inductance L = \",round((L/1E-3),2),\" mH \\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductive reactance, XL = 100.53 ohms \n",
- "\n",
- "\n",
- " (b)Inductance L = 3.95 mH "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 214</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate its inductive reactance and the resulting current if connected to\n",
- "#(a) a 240 V, 50 Hz supply, and (b) a 100 V, 1 kHz supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.040;# in Henry\n",
- "V1 = 240;# in volts\n",
- "V2 = 100;# in volts\n",
- "f1 = 50;# in Hz\n",
- "f2 = 1000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL1 = 2*math.pi*f1*L\n",
- "I1 = V1/XL1\n",
- "XL2 = 2*math.pi*f2*L\n",
- "I2 = V2/XL2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductive reactance, XL = \",round( XL1,2),\" ohms and current I = \",round( I1,2),\" A\\n\"\n",
- "print \"\\n (b)Inductive reactance, XL = \",round( XL2,2),\" ohms and current I = \",round( I2,2),\" A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductive reactance, XL = 12.57 ohms and current I = 19.1 A\n",
- "\n",
- "\n",
- " (b)Inductive reactance, XL = 251.33 ohms and current I = 0.4 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 215</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitive reactance of a capacitor of 10 \u03bcF when connected to a circuit of frequency (a) 50 Hz (b) 20 kHz\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 10E-6;# in Farads\n",
- "f1 = 50;# in Hz\n",
- "f2 = 20000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "Xc1 = 1/(2*math.pi*f1*C)\n",
- "Xc2 = 1/(2*math.pi*f2*C)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Capacitive reactance, Xc = \",round( Xc1,2),\" ohms \"\n",
- "print \"\\n (b)Capacitive reactance, Xc = \",round( Xc2,2),\" ohms \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Capacitive reactance, Xc = 318.31 ohms \n",
- "\n",
- " (b)Capacitive reactance, Xc = 0.8 ohms "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 215</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of its capacitance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 40;# in ohms\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "C = 1/(2*math.pi*f*Z)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Capacitance,C = \",round((C/1E-6),2),\" uF \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Capacitance,C = 79.58 uF "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 215</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the current taken by a 23 \u03bcF capacitor when connected to a 240 V, 50 Hz supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 23E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "V = 240;# in volts\n",
- "\n",
- "#calculation:\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "I = V/Xc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current I = \",round(I,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current I = 1.73 A "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 216</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the supply voltage and the phase angle between current and voltage\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vr = 12;# in volts\n",
- "Vl = 5;# in volts\n",
- "\n",
- "#calculation:\n",
- "V = (Vr**2 + Vl**2)**0.5\n",
- "phi = math.atan(Vl/Vr)\n",
- "phid = phi*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n supply voltage V = \",V,\" V, phase angle between current and voltage is \", round(phid,2),\"deg lagging\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " supply voltage V = 13.0 V, phase angle between current and voltage is 22.62 deg lagging"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 216</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the reactance, (b) the impedance, and (c) the current taken from a 240 V, 50 Hz supply. \n",
- "#Determine also the phase angle between the supply voltage and current\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 240;# in volts\n",
- "R = 4;# in ohms\n",
- "L = 0.00955;# in Henry\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan(XL/R)*180/math.pi\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductive reactance, XL = \",round(XL,2),\" ohms\"\n",
- "print \"\\n (b)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (c)Current, I = \",round(I,2),\" A\"\n",
- "print \"\\n (d)phase angle between the supply voltage and current is \",round(phid,2),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductive reactance, XL = 3.0 ohms\n",
- "\n",
- " (b)Impedance, Z = 5.0 ohms\n",
- "\n",
- " (c)Current, I = 48.0 A\n",
- "\n",
- " (d)phase angle between the supply voltage and current is 36.87 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 217</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the resistance, impedance, inductive reactance and inductance of the coil.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vdc = 12;# in volts\n",
- "Vac = 240;# in volts\n",
- "Iac = 20;# in Amperes\n",
- "Idc = 2;# in Amperes\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "R = Vdc/Idc\n",
- "Z = Vac/Iac\n",
- "XL = (Z**2 - R**2)**0.5\n",
- "L = XL/(2*math.pi*f)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance, R = \",R,\" ohms\"\n",
- "print \"\\n (b)Impedance, Z = \",Z,\" ohms\"\n",
- "print \"\\n (c)Inductive reactance, XL = \",round(XL,2),\" ohms\"\n",
- "print \"\\n (d)Inductance, L = \",round(L,2),\" H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance, R = 6.0 ohms\n",
- "\n",
- " (b)Impedance, Z = 12.0 ohms\n",
- "\n",
- " (c)Inductive reactance, XL = 10.39 ohms\n",
- "\n",
- " (d)Inductance, L = 0.03 H"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 217</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the inductive reactance of the coil, \n",
- "#(b) the impedance of the circuit, \n",
- "#(c) the current in the circuit, \n",
- "#(d) the p.d. across each component, and \n",
- "#(e) the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 200;# in ohms\n",
- "L = 0.3183;# in henry\n",
- "V = 240;# in volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "I = V/Z\n",
- "VL = I*XL\n",
- "VR = I*R\n",
- "phid = math.atan(XL/R)*180/math.pi\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Inductive reactance, XL = \",round(XL,2),\" ohms\"\n",
- "print \"\\n (b)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (c)current, I = \",round(I,2),\" A\"\n",
- "print \"\\n (d)p.d. across Inductor, VL = \",round(VL,2),\" V and p.d. across resistance, VR = \",round(VR,2),\" V\"\n",
- "print \"\\n (e)circuit phase angle is \",round(phid,2),\" deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Inductive reactance, XL = 100.0 ohms\n",
- "\n",
- " (b)Impedance, Z = 223.61 ohms\n",
- "\n",
- " (c)current, I = 1.07 A\n",
- "\n",
- " (d)p.d. across Inductor, VL = 107.33 V and p.d. across resistance, VR = 214.66 V\n",
- "\n",
- " (e)circuit phase angle is 26.56 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 218</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate (a) the circuit impedance, \n",
- "#(b) the current flowing, \n",
- "#(c) the p.d. across the resistance, \n",
- "#(d) the p.d. across the inductance and \n",
- "#(e) the phase angle between voltage and current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 100;# in ohms\n",
- "L = 0.2;# in henry\n",
- "Vmax = 200;# in volts\n",
- "w = 500;# in rad/sec\n",
- "\n",
- "#calculation:\n",
- "Vrms = 0.707*Vmax\n",
- "f = w/(2*math.pi)\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "I = Vrms/Z\n",
- "VL = I*XL\n",
- "VR = I*R\n",
- "phid = math.atan(XL/R)*180/math.pi\n",
- "\n",
- "\\\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A\"\n",
- "print \"\\n (c)p.d. across resistance, VR = \",round(VR,2),\" V\"\n",
- "print \"\\n (d)p.d. across Inductor, VL = \",round(VL,2),\" V\"\n",
- "print \"\\n (e)circuit phase angle is \",phid,\"deg\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Impedance, Z = 141.42 ohms\n",
- "\n",
- " (b)current, I = 1.0 A\n",
- "\n",
- " (c)p.d. across resistance, VR = 99.98 V\n",
- "\n",
- " (d)p.d. across Inductor, VL = 99.98 V\n",
- "\n",
- " (e)circuit phase angle is 45.0 deg"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 218</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of the supply voltage and the voltage across the 1.273 mH inductance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "L = 1.2273E-3;# in henry\n",
- "f = 5000;# in Hz\n",
- "VR = 6;# in volts\n",
- "\n",
- "#calculation:\n",
- "I =VR/R\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "V = I*Z\n",
- "VL = I*XL\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)supply voltage = \",round(V,2),\" Volts\"\n",
- "print \"\\n (b)p.d. across Inductor, VL = \",round(VL,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)supply voltage = 9.77 Volts\n",
- "\n",
- " (b)p.d. across Inductor, VL = 7.71 V"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 219</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the impedance of the circuit, \n",
- "#(b) the current in the circuit, \n",
- "#(c) the circuit phase angle, \n",
- "#(d) the p.d. across the 60 ohm resistor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 60;# in ohms\n",
- "Rc = 20;# in ohms\n",
- "L = 159.2E-3;# in henry\n",
- "f = 50;# in Hz\n",
- "V = 240;# in volts\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Rt = R + Rc\n",
- "Z = (Rt**2 + XL**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan(XL/Rt)*180/math.pi\n",
- "VR = I*R\n",
- "Zc = (Rc**2 + XL**2)**0.5\n",
- "Vc = I*Zc\n",
- "VL = I*XL\n",
- "VRc = I*Rc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (b)current, I = \",round(I,3),\" A\"\n",
- "print \"\\n (c)circuit phase angle is \",round(phid,0),\"deg lagging\"\n",
- "print \"\\n (d)p.d. across resistance, VR = \",round( VR,1),\" V\"\n",
- "print \"\\n (e)p.d. across coil, Vc = \",round(Vc,1),\" V\"\n",
- "print \"\\n (f1)p.d. across Inductor, VL = \",round(VL,2),\" V\"\n",
- "print \"\\n (f2)p.d. across coil resistance, VRc = \",round(VRc,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Impedance, Z = 94.35 ohms\n",
- "\n",
- " (b)current, I = 2.544 A\n",
- "\n",
- " (c)circuit phase angle is 32.0 deg lagging\n",
- "\n",
- " (d)p.d. across resistance, VR = 152.6 V\n",
- "\n",
- " (e)p.d. across coil, Vc = 137.0 V\n",
- "\n",
- " (f1)p.d. across Inductor, VL = 127.23 V\n",
- "\n",
- " (f2)p.d. across coil resistance, VRc = 50.88 V"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 220</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the impedance, and (b) the current taken from a 240 V, 50 Hz supply. \n",
- "#Find also the phase angle between the supply voltage and the current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 25;# in ohms\n",
- "C = 45E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "V = 240;# in volts\n",
- "\n",
- "#calculation:\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Z = (R**2 + Xc**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan(Xc/R)*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Impedance, Z = \",round(Z,2),\" ohms\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A\"\n",
- "print \"\\n (c)phase angle between the supply voltage and current is \",round(phid,2),\"deg leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Impedance, Z = 75.02 ohms\n",
- "\n",
- " (b)current, I = 3.2 A\n",
- "\n",
- " (c)phase angle between the supply voltage and current is 70.54 deg leading"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 221</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate: (a) the value of capacitance, C, \n",
- "#(b) the supply voltage, \n",
- "#(c) the phase angle between the supply voltage and current, \n",
- "#(d) the p.d. across the resistor, and\n",
- "#(e) the p.d. across the capacitor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 40;# in ohms\n",
- "f = 60;# in Hz\n",
- "I = 3;#in amperes\n",
- "Z = 50;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Xc = (Z**2 - R**2)**0.5\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- "V = I*Z\n",
- "phid = math.atan(Xc/R)*180/math.pi\n",
- "VR = I*R\n",
- "Vc = I*Xc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)capacitance, C = \",round((C/1E-6),2),\" uF\"\n",
- "print \"\\n (b)Voltage, V = \",V,\" Volts\"\n",
- "print \"\\n (c)phase angle between the supply voltage and current is \",round(phid,2),\"deg leading\"\n",
- "print \"\\n (d)p.d. across resistance, VR = \", VR,\" V\"\n",
- "print \"\\n (e)p.d. across Capacitor, Vc = \",Vc,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)capacitance, C = 88.42 uF\n",
- "\n",
- " (b)Voltage, V = 150 Volts\n",
- "\n",
- " (c)phase angle between the supply voltage and current is 36.87 deg leading\n",
- "\n",
- " (d)p.d. across resistance, VR = 120 V\n",
- "\n",
- " (e)p.d. across Capacitor, Vc = 90.0 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 222</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current flowing, \n",
- "#(b) the phase difference between the supply voltage and current, \n",
- "#(c) the voltage across the coil and \n",
- "#(d) the voltage across the capacitor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 5;# in ohms\n",
- "C = 100E-6;# in Farads\n",
- "L = 0.12;# in Henry\n",
- "f = 50;# in Hz\n",
- "V = 300;# in volts\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "X = XL - Xc\n",
- " #Since XL is greater than Xc, the circuit is inductive.\n",
- "Z = (R**2 + (XL-Xc)**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan((XL-Xc)/R)*180/math.pi\n",
- "Zcl = (R**2 + XL**2)**0.5\n",
- "Vcl = I*Zcl\n",
- "phidc = math.atan(XL/R)*180/math.pi\n",
- "Vc = I*Xc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current,I = \",round(I,2),\" A\"\n",
- "print \"\\n (b)phase angle between the supply voltage and current is \",round(phid,2),\"deg\"\n",
- "print \"\\n (c)Voltage across the coil, Vcoil = \",round(Vcl,0),\" Volts\"\n",
- "print \"\\n (d)p.d. across Capacitor, Vc = \",round(Vc,0),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current,I = 38.91 A\n",
- "\n",
- " (b)phase angle between the supply voltage and current is 49.57 deg\n",
- "\n",
- " (c)Voltage across the coil, Vcoil = 1480.0 Volts\n",
- "\n",
- " (d)p.d. across Capacitor, Vc = 1239.0 V"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 224</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the circuit current, \n",
- "#(b) the circuit phase angle and \n",
- "#(c) the voltage drop across each impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 8;# in ohms\n",
- "C = 0.25E-6;# in Farads\n",
- "L = 130E-6;# in Henry\n",
- "Rc = 5;# in ohms\n",
- "R2 = 10;# in ohms\n",
- "f = 20000;# in Hz\n",
- "V = 40;# in volts\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "X = Xc - XL\n",
- "R = R1 + R2 + Rc\n",
- " #Since Xc is greater than XL, the circuit is capacitive.\n",
- "Z = (R**2 + (Xc-XL)**2)**0.5\n",
- "I = V/Z\n",
- "phid = math.atan((Xc-XL)/R)*180/math.pi\n",
- "V1 = I*R1\n",
- "V2 = I*((Rc**2 + XL**2)**0.5)\n",
- "V3 = I*((R2**2 + Xc**2)**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current,I = \",round(I,2),\" A\"\n",
- "print \"\\n (b)circuit phase angle is \",round(phid,2),\"deg leading\"\n",
- "print \"\\n (c1)Voltage across the Resistance of 8 ohms = \",round(V1,2),\" Volts\"\n",
- "print \"\\n (c2)Voltage across the coil, Vcoil = \",round(V2,2),\" Volts\"\n",
- "print \"\\n (c3)p.d. across Capacitor resistance circuit = \",round(V3,2),\" Volts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current,I = 1.44 A\n",
- "\n",
- " (b)circuit phase angle is 33.97 deg leading\n",
- "\n",
- " (c1)Voltage across the Resistance of 8 ohms = 11.54 Volts\n",
- "\n",
- " (c2)Voltage across the coil, Vcoil = 24.64 Volts\n",
- "\n",
- " (c3)p.d. across Capacitor resistance circuit = 48.12 Volts"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 224</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the p.d.\u2019s V1 and V2 for the circuit\n",
- "#determine the supply voltage V and the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 4;# in ohms\n",
- "C = 1.273E-6;# in Farads\n",
- "L = 0.286E-3;# in Henry\n",
- "R2 = 8;# in ohms\n",
- "f = 5000;# in Hz\n",
- "I = 5;# in amperes\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "phid1 = math.atan(XL/R1)*180/math.pi\n",
- "V1 = I*((R1**2 + XL**2)**0.5)\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "V2 = I*((R2**2 + Xc**2)**0.5)\n",
- "phid2 = math.atan(Xc/R2)*180/math.pi\n",
- "Z = ((R1+R2)**2 + (Xc-XL)**2)**0.5\n",
- "V = I*Z\n",
- "phid = math.atan((Xc-XL)/(R1+R2))*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Voltage supply, V = \",round(V,2),\" V\"\n",
- "print \"\\n (b)circuit phase angle is \",round(phid,2),\"deg leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Voltage supply, V = 100.08 V\n",
- "\n",
- " (b)circuit phase angle is 53.16 deg leading"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 226</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#At what frequency does resonance occur?\n",
- "#Find the current flowing at the resonant frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "C = 60E-6;# in Farads\n",
- "L = 125E-3;# in Henry\n",
- "V = 120;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = 1/(2*math.pi*(L*C)**0.5)\n",
- " #At resonance, XL = Xc and impedance Z = R\n",
- "I = V/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonance frequency,Fr = \",round(fr,2),\" Hz\"\n",
- "print \"\\n (b)Current, I = \",round(I,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonance frequency,Fr = 58.12 Hz\n",
- "\n",
- " (b)Current, I = 12.0"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 226</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find (a) the circuit resistance, and (b) the circuit capacitance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.05E-3;# in Henry\n",
- "fr = 200000;# in Hz\n",
- "V = 0.002;# in Volts\n",
- "I = 0.1E-3;# in amperes\n",
- "#calculation:\n",
- "# L-C-R\n",
- "#At resonance, XL = Xc and impedance Z = R\n",
- "R = V/I\n",
- "C = 1/(L*(2*math.pi*fr)**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance, R = \",round(R,2),\" ohms\"\n",
- "print \"\\n (b)Capacitance, C = \",round((C/1E-9),2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance, R = 20.0 ohms\n",
- "\n",
- " (b)Capacitance, C = 12.67 nF"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 227</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, and \n",
- "#(b) the current at resonance. \n",
- "#How many times greater than the supply voltage is the voltage across the reactances at resonance?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 80E-3;# in Henry\n",
- "C = 0.25E-6;# in Farads\n",
- "R = 12.5;# in ohms\n",
- "V = 100;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- " #At resonance, XL = Xc and impedance Z = R\n",
- "I = V/R\n",
- "VL = I*(2*math.pi*fr*L)\n",
- "Vc = I/(2*math.pi*fr*C)\n",
- "Vm = VL/V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the resonant frequency = \",round(fr,2),\" Hz\"\n",
- "print \"\\n (b)Current, I = \",round(I,2),\"\"\n",
- "print \"\\n (b)Voltage magnification at resonance = \",round(Vm,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the resonant frequency = 1125.4 Hz\n",
- "\n",
- " (b)Current, I = 8.0 \n",
- "\n",
- " (b)Voltage magnification at resonance = 45.25 V"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 228</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the Qfactor of the circuit at resonance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 60E-3;# in Henry\n",
- "C = 30E-6;# in Farads\n",
- "R = 2;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Q = ((L/C)**0.5)/R\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n At resonance, Q-factor = \",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " At resonance, Q-factor = 22.36"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 228</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, \n",
- "#(b) the current at resonance,\n",
- "#(c) the voltages across the coil and the capacitor at resonance, and\n",
- "#(d) the Q-factor of the circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 100E-3;# in Henry\n",
- "C = 2E-6;# in Farads\n",
- "R = 10;# in ohms\n",
- "V = 50;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- " #At resonance, XL = Xc and impedance Z = R\n",
- "I = V/R\n",
- "VL = I*(2*math.pi*fr*L)\n",
- "Vc = I/(2*math.pi*fr*C)\n",
- "Q = VL/V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the resonant frequency = \",round(fr,2),\" Hz\"\n",
- "print \"\\n (b)Current, I = \",round(I,2),\"\"\n",
- "print \"\\n (c)Voltage across coil at resonance is \",round(VL,2),\"V \"\n",
- "print \"and Voltage across capacitance at resonance is \",round( Vc,2),\"V\"\n",
- "print \"\\n (d)At resonance, Q-factor = \",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the resonant frequency = 355.88 Hz\n",
- "\n",
- " (b)Current, I = 5.0 \n",
- "\n",
- " (c)Voltage across coil at resonance is 1118.03 V and Voltage across capacitance at resonance is 1118.03 V\n",
- "\n",
- " (d)At resonance, Q-factor = 22.36"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 230</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the bandwidth of the filter.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 20E-3;# in Henry\n",
- "R = 10;# in ohms\n",
- "fr = 5000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "Qr = (2*math.pi*fr)*L/R\n",
- "bw = fr/Qr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Bandwidth, (f2-f1) = \",round(bw,2),\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Bandwidth, (f2-f1) = 79.58 Hz"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 231</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the power dissipated in the resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 5000;# in ohms\n",
- "Imax = 0.250;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "Irms = 0.707*Imax\n",
- "P = Irms*Irms*R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Power, P = \",round(P,2),\" Watts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Power, P = 156.2 Watts"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 25, page no. 231</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the power dissipated.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 60;# in ohms\n",
- "L = 75E-3;# in Henry\n",
- "V = 110;# in Volts\n",
- "f = 60;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z = (R**2 + XL**2)**0.5\n",
- "I = V/Z\n",
- "P = I*I*R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Power, P = \",round(P,2),\" Watts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Power, P = 165.02 Watts"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 26, page no. 232</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find the value of the inductance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "VI = 300;# in VA\n",
- "V = 150;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "I = VI/V\n",
- "XL = V/I\n",
- "L = XL/(2*math.pi*f)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Inductance = \",round(L,2),\" H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Inductance = 0.24 H"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 27, page no. 232</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the rated power output and the corresponding reactive power.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "VI = 200000;# in VA\n",
- "pf = 0.8;# power factor\n",
- "\n",
- "#calculation:\n",
- "P = VI*pf\n",
- "Q = VI*math.sin(math.acos(pf))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n rated power output is \",round(P/1000,2),\"KW and the corresponding reactive power is \",round(Q/1000,2),\"kvar\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " rated power output is 160.0 KW and the corresponding reactive power is 120.0 kvar"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 28, page no. 233</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the resistance, \n",
- "#(b) the impedance, \n",
- "#(c) the reactance, \n",
- "#(d) the power factor, and \n",
- "#(e) the phase angle between voltage and current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 120;# in Volts\n",
- "f = 50;# in Hz\n",
- "P = 400;# in Watt\n",
- "I = 8;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "R = P/(I*I)\n",
- "Z = V/I\n",
- "XL = (Z**2 - R**2)**0.5\n",
- "pf = P/(V*I)\n",
- "phi = math.acos(pf)*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)resistance = \",round(R,2),\" ohm \"\n",
- "print \"\\n (b)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (c)reactance = \",round(XL,2),\" ohm \"\n",
- "print \"\\n (d)Power factor = \",round(pf,2),\"\"\n",
- "print \"\\n (e)phase angle = \",round(phi,2),\"deg lagging\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)resistance = 6.25 ohm \n",
- "\n",
- " (b)Impedance Z = 15.0 Ohm \n",
- "\n",
- " (c)reactance = 13.64 ohm \n",
- "\n",
- " (d)Power factor = 0.42 \n",
- "\n",
- " (e)phase angle = 65.38 deg lagging"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 29, page no. 233</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Find (a) the current flowing, (b) the phase angle,\n",
- "#(c) the resistance, (d) the impedance, and (e) the capacitance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 100;# in Volts\n",
- "f = 60;# in Hz\n",
- "P = 100;# in Watt\n",
- "pf = 0.5;# power factor\n",
- "\n",
- "#calculation:\n",
- "I = P/(pf*V)\n",
- "phi = math.acos(pf)*180/math.pi\n",
- "R = P/(I*I)\n",
- "Z = V/I\n",
- "Xc = (Z**2 - R**2)**0.5\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current I = \",round(I,2),\" A \"\n",
- "print \"\\n (b)phase angle = \",round(phi,2),\"deg leading\"\n",
- "print \"\\n (c)resistance = \",round(R,2),\" ohm \"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)capacitance = \",round((C/1E-6),2),\"uF \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current I = 2.0 A \n",
- "\n",
- " (b)phase angle = 60.0 deg leading\n",
- "\n",
- " (c)resistance = 25.0 ohm \n",
- "\n",
- " (d)Impedance Z = 50.0 Ohm \n",
- "\n",
- " (e)capacitance = 61.26 uF "
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_15.ipynb index 4a218793..4a218793 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_15.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint.ipynb deleted file mode 100755 index 429b5bb8..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint.ipynb +++ /dev/null @@ -1,913 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 16: Single-phase parallel a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 239</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current in each branch,\n",
- "#(b) the supply current, (c) the circuit phase angle,\n",
- "#(d) the circuit impedance, and (e) the power consumed\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 20;# in Ohms\n",
- "L = 2.387E-3;# in Henry\n",
- "V = 60;# in Volts\n",
- "f = 1000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "IR = V/R\n",
- "XL = 2*math.pi*f*L\n",
- "IL = V/XL\n",
- "I = (IR**2 + IL**2)**0.5\n",
- "phi = math.atan(IL/IR)\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through resistor is \",round(IR,2),\" A and current through Inductor is \",round( IL,2),\" A\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)phase angle = \",round(phid,2),\"deg lagging\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through resistor is 3.0 A and current through Inductor is 4.0 A\n",
- "\n",
- " (b)current, I = 5.0 A \n",
- "\n",
- " (c)phase angle = 53.13 deg lagging\n",
- "\n",
- " (d)Impedance Z = 12.0 Ohm \n",
- "\n",
- " (e)Power consumed = 180.0 Watt "
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 240</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current in each branch, (b) the supply current,\n",
- "#(c) the circuit phase angle, (d) the circuit impedance, \n",
- "#(e) the power dissipated, and (f) the apparent power.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 80;# in Ohms\n",
- "C = 30E-6;# in Farads\n",
- "V = 240;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "IR = V/R\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- "I = (IR**2 + Ic**2)**0.5\n",
- "phi = math.atan(Ic/IR)\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "S = V*I\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through resistor is \",round(IR,2),\" A and current through capacitor is \",round( Ic,2),\" A\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)phase angle = \",round(phid,2),\"deg leading\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \"\n",
- "print \"\\n (f)apparent Power = \",round(S,2),\" VA \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through resistor is 3.0 A and current through capacitor is 2.26 A\n",
- "\n",
- " (b)current, I = 3.76 A \n",
- "\n",
- " (c)phase angle = 37.02 deg leading\n",
- "\n",
- " (d)Impedance Z = 63.88 Ohm \n",
- "\n",
- " (e)Power consumed = 720.0 Watt \n",
- "\n",
- " (f)apparent Power = 901.72 VA "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 241</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of C and R.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "pf = 0.6;# power factor\n",
- "V = 120;# in Volts\n",
- "f = 200;# in Hz\n",
- "I = 2;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "phi = math.acos(pf)\n",
- "phid = phi*180/math.pi\n",
- "IR = I*math.cos(phi)\n",
- "Ic = I*math.sin(phi)\n",
- "R = V/IR\n",
- "C = Ic/(2*math.pi*f*V)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance R = \",round(R,2),\" Ohm \"\n",
- "print \"\\n (b)Capacitance,C = \",round((C/1E-6),2),\" uF \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance R = 100.0 Ohm \n",
- "\n",
- " (b)Capacitance,C = 10.61 uF "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 242</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the branch currents, \n",
- "#(b) the supply current and its phase angle, \n",
- "#(c) the circuit impedance, and (d) the power consumed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 25E-6;# in Farads\n",
- "L = 120E-3;# in Henry\n",
- "V = 100;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "IL = V/XL\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- " #IL and Ic are anti-phase. Hence supply current,\n",
- "I = IL - Ic\n",
- " #the current lags the supply voltage V by 90\u00c2\u00b0\n",
- "phi = math.pi/2\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through Inductor is \",round(IL,2),\" A and current through capacitor is \",round( Ic,2),\" A\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)phase angle = \",round(phid,2),\"deg lagging\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through Inductor is 2.65 A and current through capacitor is 0.79 A\n",
- "\n",
- " (b)current, I = 1.87 A \n",
- "\n",
- " (c)phase angle = 90.0 deg lagging\n",
- "\n",
- " (d)Impedance Z = 53.56 Ohm \n",
- "\n",
- " (e)Power consumed = 0.0 Watt "
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 242</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the branch currents, \n",
- "#(b) the supply current and its phase angle, \n",
- "#(c) the circuit impedance, and (d) the power consumed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 25E-6;# in Farads\n",
- "L = 120E-3;# in Henry\n",
- "V = 100;# in Volts\n",
- "f = 150;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "IL = V/XL\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- " #IL and Ic are anti-phase. Hence supply current,\n",
- "I = Ic - IL\n",
- " #the current leads the supply voltage V by 90\u00c2\u00b0\n",
- "phi = math.pi/2\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through Inductor is \",round(IL,2),\" A and current through capacitor is \",round( Ic,2),\" A\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)phase angle = \",round(phid,2),\"deg leading\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through Inductor is 0.88 A and current through capacitor is 2.36 A\n",
- "\n",
- " (b)current, I = 1.47 A \n",
- "\n",
- " (c)phase angle = 90.0 deg leading\n",
- "\n",
- " (d)Impedance Z = 67.93 Ohm \n",
- "\n",
- " (e)Power consumed = 0.0 Watt "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 244</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current in the coil and its phase angle,\n",
- "#(b) the current in the capacitor and its phase angle,\n",
- "#(c) the supply current and its phase angle,\n",
- "#(d) the circuit impedance, \n",
- "#(e) the power consumed, \n",
- "#(f) the apparent power, and \n",
- "#(g) the reactive power.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 30E-6;# in Farads\n",
- "R = 40;# in Ohms\n",
- "L = 159.2E-3;# in Henry\n",
- "V = 240;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z1 = (R**2 + XL**2)**0.5\n",
- "ILR = V/Z1\n",
- "phi1 = math.atan(XL/R)\n",
- "phi1d = phi1*180/math.pi\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- "phi2 = math.pi/2\n",
- "phi2d = phi2*180/math.pi\n",
- "Ih = ILR*math.cos(phi1) + Ic*math.cos(phi2)\n",
- "Iv = -1*ILR*math.sin(phi1) + Ic*math.sin(phi2)\n",
- "I = (Ih**2 + Iv**2)**0.5\n",
- "phi = math.atan(abs(Iv)/Ih)\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "phid = phi*180/math.pi\n",
- "S = V*I\n",
- "Q = V*I*math.sin(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through coil is \",round(ILR,2),\" A and lagged by phase angle is \",round(phi1d,2),\"deg\"\n",
- "print \"\\n (b)Current through capacitor is \",round(Ic,2),\" A and lead by phase angle is \",round(phi2d,2),\"deg\"\n",
- "print \"\\n (c)supply Current is \",round(I,2),\" A and lagged by phase angle is \",round(phid,2),\"deg\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \"\n",
- "print \"\\n (f)apparent Power = \",round(S,2),\" VA \"\n",
- "print \"\\n (g)reactive Power = \",round(Q,2),\" var \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through coil is 3.75 A and lagged by phase angle is 51.35 deg\n",
- "\n",
- " (b)Current through capacitor is 2.26 A and lead by phase angle is 90.0 deg\n",
- "\n",
- " (c)supply Current is 2.43 A and lagged by phase angle is 15.85 deg\n",
- "\n",
- " (d)Impedance Z = 98.64 Ohm \n",
- "\n",
- " (e)Power consumed = 561.76 Watt \n",
- "\n",
- " (f)apparent Power = 583.97 VA \n",
- "\n",
- " (g)reactive Power = 159.53 var "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 246</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current in the coil, and (b) the current in the capacitor. \n",
- "#(c) Measure the supply current and its phase angle (d) the circuit impedance and (e) the power consumed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 0.02E-6;# in Farads\n",
- "R = 3000;# in Ohms\n",
- "L = 120E-3;# in Henry\n",
- "V = 40;# in Volts\n",
- "f = 5000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z1 = (R**2 + XL**2)**0.5\n",
- "ILR = V/Z1\n",
- "phi1 = math.atan(XL/R)\n",
- "phi1d = phi1*180/math.pi\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- "phi2 = math.pi/2\n",
- "phi2d = phi2*180/math.pi\n",
- "Ih = ILR*math.cos(phi1) + Ic*math.cos(phi2)\n",
- "Iv = -1*ILR*math.sin(phi1) + Ic*math.sin(phi2)\n",
- "I = (Ih**2 + Iv**2)**0.5\n",
- "phi = math.atan((Iv)/Ih)\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through coil is \",round(ILR*1000,2),\"mA and lagged by phase angle is \",round(phi1d,1),\"deg\"\n",
- "print \"\\n (b)Current through capacitor is \",round(Ic*1000,2),\"mA and lead by phase angle is \",round(phi2d,2),\"deg\"\n",
- "print \"\\n (c)supply Current is \",round(I*1000,1),\"mA and leaded by phase angle is \",round(phid,2),\"deg\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z/1000,3),\"KOhm \"\n",
- "print \"\\n (e)Power consumed = \",round(P*1000,1),\"mWatt \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through coil is 8.3 mA and lagged by phase angle is 51.5 deg\n",
- "\n",
- " (b)Current through capacitor is 25.13 mA and lead by phase angle is 90.0 deg\n",
- "\n",
- " (c)supply Current is 19.3 mA and leaded by phase angle is 74.5 deg\n",
- "\n",
- " (d)Impedance Z = 2.068 KOhm \n",
- "\n",
- " (e)Power consumed = 206.8 mWatt "
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 249</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency of the circuit and \n",
- "#(b) the current circulating in the capacitor and inductance at resonance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 40E-6;# in Farads\n",
- "R = 0;# in Ohms\n",
- "L = 150E-3;# in Henry\n",
- "V = 50;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = ((1/(L*C) - R*R/(L*L))**0.5)/(2*math.pi)\n",
- "Xc = 1/(2*math.pi*fr*C)\n",
- "Icirc = V/Xc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Parallel resonant frequency, fr = \",round(fr,2),\" Hz \"\n",
- "print \"\\n (b)Current circulating in L and C at resonance = \",round(Icirc,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Parallel resonant frequency, fr = 64.97 Hz \n",
- "\n",
- " (b)Current circulating in L and C at resonance = 0.82 A "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 250</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the resonant frequency, \n",
- "#(b) the dynamic resistance, \n",
- "#(c) the current at resonance and \n",
- "#(d) the circuit Q-factor at resonanc\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 20E-6;# in Farads\n",
- "R = 60;# in Ohms\n",
- "L = 200E-3;# in Henry\n",
- "V = 20;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = ((1/(L*C) - R*R/(L*L))**0.5)/(2*math.pi)\n",
- "Rd = L/(R*C)\n",
- "Ir = V/Rd\n",
- "Q = 2*math.pi*fr*L/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Parallel resonant frequency, fr = \",round(fr,2),\" Hz \"\n",
- "print \"\\n (b)the dynamic resistance,RD = \",round(Rd,2),\" ohm \"\n",
- "print \"\\n (c)Current at resonance = \",round(Ir,2),\" A \"\n",
- "print \"\\n (d)Q-factor = \",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Parallel resonant frequency, fr = 63.66 Hz \n",
- "\n",
- " (b)the dynamic resistance,RD = 166.67 ohm \n",
- "\n",
- " (c)Current at resonance = 0.12 A \n",
- "\n",
- " (d)Q-factor = 1.33"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 251</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the condition when the supply current is a minimum:\n",
- "#(a) the capacitance of the capacitor,\n",
- "#(b) the dynamic resistance, \n",
- "#(c) the supply current, and \n",
- "#(d) the Q-factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fr = 5000;# in ohm\n",
- "R = 800;# in Ohms\n",
- "L = 100E-3;# in Henry\n",
- "V = 12;# in Volts\n",
- "\n",
- "#calculation:\n",
- "C = 1/(L*((2*math.pi*fr)**2 + R*R/(L*L)))\n",
- "Rd = L/(R*C)\n",
- "Ir = V/Rd\n",
- "Q = 2*math.pi*fr*L/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)capacitance, C = \",round((C/1E-9),2),\" nF \"\n",
- "print \"\\n (b)the dynamic resistance,RD = \",round(Rd,2),\" ohm \"\n",
- "print \"\\n (c)Current at resonance = \",round((Ir/1E-3),2),\" mA \"\n",
- "print \"\\n (d)Q-factor = \",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)capacitance, C = 9.52 nF \n",
- "\n",
- " (b)the dynamic resistance,RD = 13137.01 ohm \n",
- "\n",
- " (c)Current at resonance = 0.91 mA \n",
- "\n",
- " (d)Q-factor = 3.93"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 252</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the currenttaken by a capacitor connected in parallel with the motor to correct the power factor to unity, and \n",
- "#(b) the value of the supply current after power factor correction.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 50;# in ohm\n",
- "V = 240;# in Volts\n",
- "pf = 0.6;# power factor\n",
- "Im = 50;# in amperes\n",
- "\n",
- "#calculation:\n",
- "phi = math.acos(pf)\n",
- "phid = phi*180/math.pi\n",
- "Ic = Im*math.sin(phi)\n",
- "I = Im*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the capacitor current Ic must be \",round(Ic,2),\" A for the power factor to be unity. \"\n",
- "print \"\\n (b)Supply current I = \",round(I,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the capacitor current Ic must be 40.0 A for the power factor to be unity. \n",
- "\n",
- " (b)Supply current I = 30.0 A "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 253</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current taken by the motor, \n",
- "#(b) the supply current after power factor correction, \n",
- "#(c) the current taken by the capacitor,\n",
- "#(d) the capacitance of the capacitor, and \n",
- "#(e) the kvar rating of the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Pout = 4800;# in Watt\n",
- "eff = 0.8# effficiency\n",
- "f = 50;# in ohm\n",
- "V = 240;# in Volts\n",
- "pf1 = 0.625;# power factor\n",
- "pf2 = 0.95;# power factor\n",
- "\n",
- "#calculation:\n",
- "Pin = Pout/eff\n",
- "Im = Pin/(V*pf1)\n",
- "phi1 = math.acos(pf1)\n",
- "phi1d = phi1*180/math.pi\n",
- "phi2 = math.acos(pf2)\n",
- "phi2d = phi2*180/math.pi\n",
- "Imh = Im*math.cos(phi1)\n",
- " #Ih = I*cos(phi2)\n",
- "Ih = Imh\n",
- "I = Ih/math.cos(phi2)\n",
- "Imv = Im*math.sin(phi1)\n",
- "Iv = I*math.sin(phi2)\n",
- "Ic = Imv - Iv\n",
- "C = Ic/(2*math.pi*f*V)\n",
- "kvar = V*Ic/1000\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current taken by the motor, Im = \",round(Im,2),\" A\"\n",
- "print \"\\n (b)supply current after p.f. correction, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)magnitude of the capacitor current Ic = \",round(Ic,0),\" A\"\n",
- "print \"\\n (d)capacitance, C = \",round((C/1E-6),0),\" uF \"\n",
- "print \"\\n (d)kvar rating of the capacitor = \",round(kvar,2),\" kvar \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current taken by the motor, Im = 40.0 A\n",
- "\n",
- " (b)supply current after p.f. correction, I = 26.32 A \n",
- "\n",
- " (c)magnitude of the capacitor current Ic = 23.0 A\n",
- "\n",
- " (d)capacitance, C = 305.0 uF \n",
- "\n",
- " (d)kvar rating of the capacitor = 5.52 kvar "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 254</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for the lamps and motor, (a) the total current, (b) the overall power factor and \n",
- "#(c) the total power. (d) Find the value of the static capacitor to improve the overall power factor to 0.975 lagging.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 3000;# in VA\n",
- "f = 50;# in ohm\n",
- "V = 250;# in Volts\n",
- "Iil = 10;# in Amperes\n",
- "Ifl = 8;# in Amperes\n",
- "pfil = 1; # power factor\n",
- "pffl = 0.7;# power factor\n",
- "pfm = 0.8;# power factor\n",
- "pf0 = 0.975;# power factor\n",
- "\n",
- "#calculation:\n",
- "phiil = math.acos(pfil)\n",
- "phiild = phiil*180/math.pi\n",
- "phifl = math.acos(pffl)\n",
- "phifld = phifl*180/math.pi\n",
- "phim = math.acos(pfm)\n",
- "phimd = phim*180/math.pi\n",
- "phi0 = math.acos(pf0)\n",
- "phi0d = phi0*180/math.pi\n",
- "Im = S/V\n",
- "Ih = Iil*math.cos(phiil) + Ifl*math.cos(phifl) + Im*math.cos(phim)\n",
- "Iv = Iil*math.sin(phiil) - Ifl*math.sin(phifl) - Im*math.sin(phim)\n",
- "Il = (Ih**2 + Iv**2)**0.5\n",
- "phi = math.atan(abs(Iv)/Ih)\n",
- "phid = phi*180/math.pi\n",
- "pf = math.cos(phi)\n",
- "P = V*Il*pf\n",
- "I = Il*math.cos(phi)/math.cos(phi0)\n",
- "Ic = Il*math.sin(phi) - I*math.sin(phi0)\n",
- "C = Ic/(2*math.pi*f*V)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)total current, Il = \",round(Il,2),\" A\"\n",
- "print \"\\n (b)Power factor = \",round(pf,2),\"lagging\"\n",
- "print \"\\n (c)Total power, P = \",round(P/1000,2),\"KWatt\"\n",
- "print \"\\n (d)capacitance, C = \",round((C/1E-6),2),\"uF \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)total current, Il = 28.32 A\n",
- "\n",
- " (b)Power factor = 0.89 lagging\n",
- "\n",
- " (c)Total power, P = 6.3 KWatt\n",
- "\n",
- " (d)capacitance, C = 91.29 uF "
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_1.ipynb deleted file mode 100755 index 429b5bb8..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_1.ipynb +++ /dev/null @@ -1,913 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 16: Single-phase parallel a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 239</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current in each branch,\n",
- "#(b) the supply current, (c) the circuit phase angle,\n",
- "#(d) the circuit impedance, and (e) the power consumed\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 20;# in Ohms\n",
- "L = 2.387E-3;# in Henry\n",
- "V = 60;# in Volts\n",
- "f = 1000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "IR = V/R\n",
- "XL = 2*math.pi*f*L\n",
- "IL = V/XL\n",
- "I = (IR**2 + IL**2)**0.5\n",
- "phi = math.atan(IL/IR)\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through resistor is \",round(IR,2),\" A and current through Inductor is \",round( IL,2),\" A\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)phase angle = \",round(phid,2),\"deg lagging\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through resistor is 3.0 A and current through Inductor is 4.0 A\n",
- "\n",
- " (b)current, I = 5.0 A \n",
- "\n",
- " (c)phase angle = 53.13 deg lagging\n",
- "\n",
- " (d)Impedance Z = 12.0 Ohm \n",
- "\n",
- " (e)Power consumed = 180.0 Watt "
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 240</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current in each branch, (b) the supply current,\n",
- "#(c) the circuit phase angle, (d) the circuit impedance, \n",
- "#(e) the power dissipated, and (f) the apparent power.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 80;# in Ohms\n",
- "C = 30E-6;# in Farads\n",
- "V = 240;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "IR = V/R\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- "I = (IR**2 + Ic**2)**0.5\n",
- "phi = math.atan(Ic/IR)\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "S = V*I\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through resistor is \",round(IR,2),\" A and current through capacitor is \",round( Ic,2),\" A\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)phase angle = \",round(phid,2),\"deg leading\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \"\n",
- "print \"\\n (f)apparent Power = \",round(S,2),\" VA \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through resistor is 3.0 A and current through capacitor is 2.26 A\n",
- "\n",
- " (b)current, I = 3.76 A \n",
- "\n",
- " (c)phase angle = 37.02 deg leading\n",
- "\n",
- " (d)Impedance Z = 63.88 Ohm \n",
- "\n",
- " (e)Power consumed = 720.0 Watt \n",
- "\n",
- " (f)apparent Power = 901.72 VA "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 241</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of C and R.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "pf = 0.6;# power factor\n",
- "V = 120;# in Volts\n",
- "f = 200;# in Hz\n",
- "I = 2;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "phi = math.acos(pf)\n",
- "phid = phi*180/math.pi\n",
- "IR = I*math.cos(phi)\n",
- "Ic = I*math.sin(phi)\n",
- "R = V/IR\n",
- "C = Ic/(2*math.pi*f*V)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance R = \",round(R,2),\" Ohm \"\n",
- "print \"\\n (b)Capacitance,C = \",round((C/1E-6),2),\" uF \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance R = 100.0 Ohm \n",
- "\n",
- " (b)Capacitance,C = 10.61 uF "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 242</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the branch currents, \n",
- "#(b) the supply current and its phase angle, \n",
- "#(c) the circuit impedance, and (d) the power consumed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 25E-6;# in Farads\n",
- "L = 120E-3;# in Henry\n",
- "V = 100;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "IL = V/XL\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- " #IL and Ic are anti-phase. Hence supply current,\n",
- "I = IL - Ic\n",
- " #the current lags the supply voltage V by 90\u00c2\u00b0\n",
- "phi = math.pi/2\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through Inductor is \",round(IL,2),\" A and current through capacitor is \",round( Ic,2),\" A\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)phase angle = \",round(phid,2),\"deg lagging\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through Inductor is 2.65 A and current through capacitor is 0.79 A\n",
- "\n",
- " (b)current, I = 1.87 A \n",
- "\n",
- " (c)phase angle = 90.0 deg lagging\n",
- "\n",
- " (d)Impedance Z = 53.56 Ohm \n",
- "\n",
- " (e)Power consumed = 0.0 Watt "
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 242</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the branch currents, \n",
- "#(b) the supply current and its phase angle, \n",
- "#(c) the circuit impedance, and (d) the power consumed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 25E-6;# in Farads\n",
- "L = 120E-3;# in Henry\n",
- "V = 100;# in Volts\n",
- "f = 150;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "IL = V/XL\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- " #IL and Ic are anti-phase. Hence supply current,\n",
- "I = Ic - IL\n",
- " #the current leads the supply voltage V by 90\u00c2\u00b0\n",
- "phi = math.pi/2\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through Inductor is \",round(IL,2),\" A and current through capacitor is \",round( Ic,2),\" A\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)phase angle = \",round(phid,2),\"deg leading\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through Inductor is 0.88 A and current through capacitor is 2.36 A\n",
- "\n",
- " (b)current, I = 1.47 A \n",
- "\n",
- " (c)phase angle = 90.0 deg leading\n",
- "\n",
- " (d)Impedance Z = 67.93 Ohm \n",
- "\n",
- " (e)Power consumed = 0.0 Watt "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 244</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current in the coil and its phase angle,\n",
- "#(b) the current in the capacitor and its phase angle,\n",
- "#(c) the supply current and its phase angle,\n",
- "#(d) the circuit impedance, \n",
- "#(e) the power consumed, \n",
- "#(f) the apparent power, and \n",
- "#(g) the reactive power.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 30E-6;# in Farads\n",
- "R = 40;# in Ohms\n",
- "L = 159.2E-3;# in Henry\n",
- "V = 240;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z1 = (R**2 + XL**2)**0.5\n",
- "ILR = V/Z1\n",
- "phi1 = math.atan(XL/R)\n",
- "phi1d = phi1*180/math.pi\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- "phi2 = math.pi/2\n",
- "phi2d = phi2*180/math.pi\n",
- "Ih = ILR*math.cos(phi1) + Ic*math.cos(phi2)\n",
- "Iv = -1*ILR*math.sin(phi1) + Ic*math.sin(phi2)\n",
- "I = (Ih**2 + Iv**2)**0.5\n",
- "phi = math.atan(abs(Iv)/Ih)\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "phid = phi*180/math.pi\n",
- "S = V*I\n",
- "Q = V*I*math.sin(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through coil is \",round(ILR,2),\" A and lagged by phase angle is \",round(phi1d,2),\"deg\"\n",
- "print \"\\n (b)Current through capacitor is \",round(Ic,2),\" A and lead by phase angle is \",round(phi2d,2),\"deg\"\n",
- "print \"\\n (c)supply Current is \",round(I,2),\" A and lagged by phase angle is \",round(phid,2),\"deg\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \"\n",
- "print \"\\n (f)apparent Power = \",round(S,2),\" VA \"\n",
- "print \"\\n (g)reactive Power = \",round(Q,2),\" var \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through coil is 3.75 A and lagged by phase angle is 51.35 deg\n",
- "\n",
- " (b)Current through capacitor is 2.26 A and lead by phase angle is 90.0 deg\n",
- "\n",
- " (c)supply Current is 2.43 A and lagged by phase angle is 15.85 deg\n",
- "\n",
- " (d)Impedance Z = 98.64 Ohm \n",
- "\n",
- " (e)Power consumed = 561.76 Watt \n",
- "\n",
- " (f)apparent Power = 583.97 VA \n",
- "\n",
- " (g)reactive Power = 159.53 var "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 246</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current in the coil, and (b) the current in the capacitor. \n",
- "#(c) Measure the supply current and its phase angle (d) the circuit impedance and (e) the power consumed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 0.02E-6;# in Farads\n",
- "R = 3000;# in Ohms\n",
- "L = 120E-3;# in Henry\n",
- "V = 40;# in Volts\n",
- "f = 5000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z1 = (R**2 + XL**2)**0.5\n",
- "ILR = V/Z1\n",
- "phi1 = math.atan(XL/R)\n",
- "phi1d = phi1*180/math.pi\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- "phi2 = math.pi/2\n",
- "phi2d = phi2*180/math.pi\n",
- "Ih = ILR*math.cos(phi1) + Ic*math.cos(phi2)\n",
- "Iv = -1*ILR*math.sin(phi1) + Ic*math.sin(phi2)\n",
- "I = (Ih**2 + Iv**2)**0.5\n",
- "phi = math.atan((Iv)/Ih)\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through coil is \",round(ILR*1000,2),\"mA and lagged by phase angle is \",round(phi1d,1),\"deg\"\n",
- "print \"\\n (b)Current through capacitor is \",round(Ic*1000,2),\"mA and lead by phase angle is \",round(phi2d,2),\"deg\"\n",
- "print \"\\n (c)supply Current is \",round(I*1000,1),\"mA and leaded by phase angle is \",round(phid,2),\"deg\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z/1000,3),\"KOhm \"\n",
- "print \"\\n (e)Power consumed = \",round(P*1000,1),\"mWatt \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through coil is 8.3 mA and lagged by phase angle is 51.5 deg\n",
- "\n",
- " (b)Current through capacitor is 25.13 mA and lead by phase angle is 90.0 deg\n",
- "\n",
- " (c)supply Current is 19.3 mA and leaded by phase angle is 74.5 deg\n",
- "\n",
- " (d)Impedance Z = 2.068 KOhm \n",
- "\n",
- " (e)Power consumed = 206.8 mWatt "
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 249</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency of the circuit and \n",
- "#(b) the current circulating in the capacitor and inductance at resonance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 40E-6;# in Farads\n",
- "R = 0;# in Ohms\n",
- "L = 150E-3;# in Henry\n",
- "V = 50;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = ((1/(L*C) - R*R/(L*L))**0.5)/(2*math.pi)\n",
- "Xc = 1/(2*math.pi*fr*C)\n",
- "Icirc = V/Xc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Parallel resonant frequency, fr = \",round(fr,2),\" Hz \"\n",
- "print \"\\n (b)Current circulating in L and C at resonance = \",round(Icirc,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Parallel resonant frequency, fr = 64.97 Hz \n",
- "\n",
- " (b)Current circulating in L and C at resonance = 0.82 A "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 250</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the resonant frequency, \n",
- "#(b) the dynamic resistance, \n",
- "#(c) the current at resonance and \n",
- "#(d) the circuit Q-factor at resonanc\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 20E-6;# in Farads\n",
- "R = 60;# in Ohms\n",
- "L = 200E-3;# in Henry\n",
- "V = 20;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = ((1/(L*C) - R*R/(L*L))**0.5)/(2*math.pi)\n",
- "Rd = L/(R*C)\n",
- "Ir = V/Rd\n",
- "Q = 2*math.pi*fr*L/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Parallel resonant frequency, fr = \",round(fr,2),\" Hz \"\n",
- "print \"\\n (b)the dynamic resistance,RD = \",round(Rd,2),\" ohm \"\n",
- "print \"\\n (c)Current at resonance = \",round(Ir,2),\" A \"\n",
- "print \"\\n (d)Q-factor = \",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Parallel resonant frequency, fr = 63.66 Hz \n",
- "\n",
- " (b)the dynamic resistance,RD = 166.67 ohm \n",
- "\n",
- " (c)Current at resonance = 0.12 A \n",
- "\n",
- " (d)Q-factor = 1.33"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 251</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the condition when the supply current is a minimum:\n",
- "#(a) the capacitance of the capacitor,\n",
- "#(b) the dynamic resistance, \n",
- "#(c) the supply current, and \n",
- "#(d) the Q-factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fr = 5000;# in ohm\n",
- "R = 800;# in Ohms\n",
- "L = 100E-3;# in Henry\n",
- "V = 12;# in Volts\n",
- "\n",
- "#calculation:\n",
- "C = 1/(L*((2*math.pi*fr)**2 + R*R/(L*L)))\n",
- "Rd = L/(R*C)\n",
- "Ir = V/Rd\n",
- "Q = 2*math.pi*fr*L/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)capacitance, C = \",round((C/1E-9),2),\" nF \"\n",
- "print \"\\n (b)the dynamic resistance,RD = \",round(Rd,2),\" ohm \"\n",
- "print \"\\n (c)Current at resonance = \",round((Ir/1E-3),2),\" mA \"\n",
- "print \"\\n (d)Q-factor = \",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)capacitance, C = 9.52 nF \n",
- "\n",
- " (b)the dynamic resistance,RD = 13137.01 ohm \n",
- "\n",
- " (c)Current at resonance = 0.91 mA \n",
- "\n",
- " (d)Q-factor = 3.93"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 252</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the currenttaken by a capacitor connected in parallel with the motor to correct the power factor to unity, and \n",
- "#(b) the value of the supply current after power factor correction.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 50;# in ohm\n",
- "V = 240;# in Volts\n",
- "pf = 0.6;# power factor\n",
- "Im = 50;# in amperes\n",
- "\n",
- "#calculation:\n",
- "phi = math.acos(pf)\n",
- "phid = phi*180/math.pi\n",
- "Ic = Im*math.sin(phi)\n",
- "I = Im*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the capacitor current Ic must be \",round(Ic,2),\" A for the power factor to be unity. \"\n",
- "print \"\\n (b)Supply current I = \",round(I,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the capacitor current Ic must be 40.0 A for the power factor to be unity. \n",
- "\n",
- " (b)Supply current I = 30.0 A "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 253</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current taken by the motor, \n",
- "#(b) the supply current after power factor correction, \n",
- "#(c) the current taken by the capacitor,\n",
- "#(d) the capacitance of the capacitor, and \n",
- "#(e) the kvar rating of the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Pout = 4800;# in Watt\n",
- "eff = 0.8# effficiency\n",
- "f = 50;# in ohm\n",
- "V = 240;# in Volts\n",
- "pf1 = 0.625;# power factor\n",
- "pf2 = 0.95;# power factor\n",
- "\n",
- "#calculation:\n",
- "Pin = Pout/eff\n",
- "Im = Pin/(V*pf1)\n",
- "phi1 = math.acos(pf1)\n",
- "phi1d = phi1*180/math.pi\n",
- "phi2 = math.acos(pf2)\n",
- "phi2d = phi2*180/math.pi\n",
- "Imh = Im*math.cos(phi1)\n",
- " #Ih = I*cos(phi2)\n",
- "Ih = Imh\n",
- "I = Ih/math.cos(phi2)\n",
- "Imv = Im*math.sin(phi1)\n",
- "Iv = I*math.sin(phi2)\n",
- "Ic = Imv - Iv\n",
- "C = Ic/(2*math.pi*f*V)\n",
- "kvar = V*Ic/1000\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current taken by the motor, Im = \",round(Im,2),\" A\"\n",
- "print \"\\n (b)supply current after p.f. correction, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)magnitude of the capacitor current Ic = \",round(Ic,0),\" A\"\n",
- "print \"\\n (d)capacitance, C = \",round((C/1E-6),0),\" uF \"\n",
- "print \"\\n (d)kvar rating of the capacitor = \",round(kvar,2),\" kvar \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current taken by the motor, Im = 40.0 A\n",
- "\n",
- " (b)supply current after p.f. correction, I = 26.32 A \n",
- "\n",
- " (c)magnitude of the capacitor current Ic = 23.0 A\n",
- "\n",
- " (d)capacitance, C = 305.0 uF \n",
- "\n",
- " (d)kvar rating of the capacitor = 5.52 kvar "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 254</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for the lamps and motor, (a) the total current, (b) the overall power factor and \n",
- "#(c) the total power. (d) Find the value of the static capacitor to improve the overall power factor to 0.975 lagging.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 3000;# in VA\n",
- "f = 50;# in ohm\n",
- "V = 250;# in Volts\n",
- "Iil = 10;# in Amperes\n",
- "Ifl = 8;# in Amperes\n",
- "pfil = 1; # power factor\n",
- "pffl = 0.7;# power factor\n",
- "pfm = 0.8;# power factor\n",
- "pf0 = 0.975;# power factor\n",
- "\n",
- "#calculation:\n",
- "phiil = math.acos(pfil)\n",
- "phiild = phiil*180/math.pi\n",
- "phifl = math.acos(pffl)\n",
- "phifld = phifl*180/math.pi\n",
- "phim = math.acos(pfm)\n",
- "phimd = phim*180/math.pi\n",
- "phi0 = math.acos(pf0)\n",
- "phi0d = phi0*180/math.pi\n",
- "Im = S/V\n",
- "Ih = Iil*math.cos(phiil) + Ifl*math.cos(phifl) + Im*math.cos(phim)\n",
- "Iv = Iil*math.sin(phiil) - Ifl*math.sin(phifl) - Im*math.sin(phim)\n",
- "Il = (Ih**2 + Iv**2)**0.5\n",
- "phi = math.atan(abs(Iv)/Ih)\n",
- "phid = phi*180/math.pi\n",
- "pf = math.cos(phi)\n",
- "P = V*Il*pf\n",
- "I = Il*math.cos(phi)/math.cos(phi0)\n",
- "Ic = Il*math.sin(phi) - I*math.sin(phi0)\n",
- "C = Ic/(2*math.pi*f*V)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)total current, Il = \",round(Il,2),\" A\"\n",
- "print \"\\n (b)Power factor = \",round(pf,2),\"lagging\"\n",
- "print \"\\n (c)Total power, P = \",round(P/1000,2),\"KWatt\"\n",
- "print \"\\n (d)capacitance, C = \",round((C/1E-6),2),\"uF \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)total current, Il = 28.32 A\n",
- "\n",
- " (b)Power factor = 0.89 lagging\n",
- "\n",
- " (c)Total power, P = 6.3 KWatt\n",
- "\n",
- " (d)capacitance, C = 91.29 uF "
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_2.ipynb deleted file mode 100755 index 429b5bb8..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_2.ipynb +++ /dev/null @@ -1,913 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 16: Single-phase parallel a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 239</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current in each branch,\n",
- "#(b) the supply current, (c) the circuit phase angle,\n",
- "#(d) the circuit impedance, and (e) the power consumed\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 20;# in Ohms\n",
- "L = 2.387E-3;# in Henry\n",
- "V = 60;# in Volts\n",
- "f = 1000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "IR = V/R\n",
- "XL = 2*math.pi*f*L\n",
- "IL = V/XL\n",
- "I = (IR**2 + IL**2)**0.5\n",
- "phi = math.atan(IL/IR)\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through resistor is \",round(IR,2),\" A and current through Inductor is \",round( IL,2),\" A\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)phase angle = \",round(phid,2),\"deg lagging\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through resistor is 3.0 A and current through Inductor is 4.0 A\n",
- "\n",
- " (b)current, I = 5.0 A \n",
- "\n",
- " (c)phase angle = 53.13 deg lagging\n",
- "\n",
- " (d)Impedance Z = 12.0 Ohm \n",
- "\n",
- " (e)Power consumed = 180.0 Watt "
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 240</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current in each branch, (b) the supply current,\n",
- "#(c) the circuit phase angle, (d) the circuit impedance, \n",
- "#(e) the power dissipated, and (f) the apparent power.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 80;# in Ohms\n",
- "C = 30E-6;# in Farads\n",
- "V = 240;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "IR = V/R\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- "I = (IR**2 + Ic**2)**0.5\n",
- "phi = math.atan(Ic/IR)\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "S = V*I\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through resistor is \",round(IR,2),\" A and current through capacitor is \",round( Ic,2),\" A\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)phase angle = \",round(phid,2),\"deg leading\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \"\n",
- "print \"\\n (f)apparent Power = \",round(S,2),\" VA \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through resistor is 3.0 A and current through capacitor is 2.26 A\n",
- "\n",
- " (b)current, I = 3.76 A \n",
- "\n",
- " (c)phase angle = 37.02 deg leading\n",
- "\n",
- " (d)Impedance Z = 63.88 Ohm \n",
- "\n",
- " (e)Power consumed = 720.0 Watt \n",
- "\n",
- " (f)apparent Power = 901.72 VA "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 241</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of C and R.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "pf = 0.6;# power factor\n",
- "V = 120;# in Volts\n",
- "f = 200;# in Hz\n",
- "I = 2;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "phi = math.acos(pf)\n",
- "phid = phi*180/math.pi\n",
- "IR = I*math.cos(phi)\n",
- "Ic = I*math.sin(phi)\n",
- "R = V/IR\n",
- "C = Ic/(2*math.pi*f*V)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance R = \",round(R,2),\" Ohm \"\n",
- "print \"\\n (b)Capacitance,C = \",round((C/1E-6),2),\" uF \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance R = 100.0 Ohm \n",
- "\n",
- " (b)Capacitance,C = 10.61 uF "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 242</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the branch currents, \n",
- "#(b) the supply current and its phase angle, \n",
- "#(c) the circuit impedance, and (d) the power consumed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 25E-6;# in Farads\n",
- "L = 120E-3;# in Henry\n",
- "V = 100;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "IL = V/XL\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- " #IL and Ic are anti-phase. Hence supply current,\n",
- "I = IL - Ic\n",
- " #the current lags the supply voltage V by 90\u00c2\u00b0\n",
- "phi = math.pi/2\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through Inductor is \",round(IL,2),\" A and current through capacitor is \",round( Ic,2),\" A\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)phase angle = \",round(phid,2),\"deg lagging\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through Inductor is 2.65 A and current through capacitor is 0.79 A\n",
- "\n",
- " (b)current, I = 1.87 A \n",
- "\n",
- " (c)phase angle = 90.0 deg lagging\n",
- "\n",
- " (d)Impedance Z = 53.56 Ohm \n",
- "\n",
- " (e)Power consumed = 0.0 Watt "
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 242</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the branch currents, \n",
- "#(b) the supply current and its phase angle, \n",
- "#(c) the circuit impedance, and (d) the power consumed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 25E-6;# in Farads\n",
- "L = 120E-3;# in Henry\n",
- "V = 100;# in Volts\n",
- "f = 150;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "IL = V/XL\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- " #IL and Ic are anti-phase. Hence supply current,\n",
- "I = Ic - IL\n",
- " #the current leads the supply voltage V by 90\u00c2\u00b0\n",
- "phi = math.pi/2\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through Inductor is \",round(IL,2),\" A and current through capacitor is \",round( Ic,2),\" A\"\n",
- "print \"\\n (b)current, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)phase angle = \",round(phid,2),\"deg leading\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through Inductor is 0.88 A and current through capacitor is 2.36 A\n",
- "\n",
- " (b)current, I = 1.47 A \n",
- "\n",
- " (c)phase angle = 90.0 deg leading\n",
- "\n",
- " (d)Impedance Z = 67.93 Ohm \n",
- "\n",
- " (e)Power consumed = 0.0 Watt "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 244</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current in the coil and its phase angle,\n",
- "#(b) the current in the capacitor and its phase angle,\n",
- "#(c) the supply current and its phase angle,\n",
- "#(d) the circuit impedance, \n",
- "#(e) the power consumed, \n",
- "#(f) the apparent power, and \n",
- "#(g) the reactive power.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 30E-6;# in Farads\n",
- "R = 40;# in Ohms\n",
- "L = 159.2E-3;# in Henry\n",
- "V = 240;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z1 = (R**2 + XL**2)**0.5\n",
- "ILR = V/Z1\n",
- "phi1 = math.atan(XL/R)\n",
- "phi1d = phi1*180/math.pi\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- "phi2 = math.pi/2\n",
- "phi2d = phi2*180/math.pi\n",
- "Ih = ILR*math.cos(phi1) + Ic*math.cos(phi2)\n",
- "Iv = -1*ILR*math.sin(phi1) + Ic*math.sin(phi2)\n",
- "I = (Ih**2 + Iv**2)**0.5\n",
- "phi = math.atan(abs(Iv)/Ih)\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "phid = phi*180/math.pi\n",
- "S = V*I\n",
- "Q = V*I*math.sin(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through coil is \",round(ILR,2),\" A and lagged by phase angle is \",round(phi1d,2),\"deg\"\n",
- "print \"\\n (b)Current through capacitor is \",round(Ic,2),\" A and lead by phase angle is \",round(phi2d,2),\"deg\"\n",
- "print \"\\n (c)supply Current is \",round(I,2),\" A and lagged by phase angle is \",round(phid,2),\"deg\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z,2),\" Ohm \"\n",
- "print \"\\n (e)Power consumed = \",round(P,2),\" Watt \"\n",
- "print \"\\n (f)apparent Power = \",round(S,2),\" VA \"\n",
- "print \"\\n (g)reactive Power = \",round(Q,2),\" var \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through coil is 3.75 A and lagged by phase angle is 51.35 deg\n",
- "\n",
- " (b)Current through capacitor is 2.26 A and lead by phase angle is 90.0 deg\n",
- "\n",
- " (c)supply Current is 2.43 A and lagged by phase angle is 15.85 deg\n",
- "\n",
- " (d)Impedance Z = 98.64 Ohm \n",
- "\n",
- " (e)Power consumed = 561.76 Watt \n",
- "\n",
- " (f)apparent Power = 583.97 VA \n",
- "\n",
- " (g)reactive Power = 159.53 var "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 246</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current in the coil, and (b) the current in the capacitor. \n",
- "#(c) Measure the supply current and its phase angle (d) the circuit impedance and (e) the power consumed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 0.02E-6;# in Farads\n",
- "R = 3000;# in Ohms\n",
- "L = 120E-3;# in Henry\n",
- "V = 40;# in Volts\n",
- "f = 5000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Z1 = (R**2 + XL**2)**0.5\n",
- "ILR = V/Z1\n",
- "phi1 = math.atan(XL/R)\n",
- "phi1d = phi1*180/math.pi\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Ic = V/Xc\n",
- "phi2 = math.pi/2\n",
- "phi2d = phi2*180/math.pi\n",
- "Ih = ILR*math.cos(phi1) + Ic*math.cos(phi2)\n",
- "Iv = -1*ILR*math.sin(phi1) + Ic*math.sin(phi2)\n",
- "I = (Ih**2 + Iv**2)**0.5\n",
- "phi = math.atan((Iv)/Ih)\n",
- "phid = phi*180/math.pi\n",
- "Z = V/I\n",
- "P = V*I*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current through coil is \",round(ILR*1000,2),\"mA and lagged by phase angle is \",round(phi1d,1),\"deg\"\n",
- "print \"\\n (b)Current through capacitor is \",round(Ic*1000,2),\"mA and lead by phase angle is \",round(phi2d,2),\"deg\"\n",
- "print \"\\n (c)supply Current is \",round(I*1000,1),\"mA and leaded by phase angle is \",round(phid,2),\"deg\"\n",
- "print \"\\n (d)Impedance Z = \",round(Z/1000,3),\"KOhm \"\n",
- "print \"\\n (e)Power consumed = \",round(P*1000,1),\"mWatt \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current through coil is 8.3 mA and lagged by phase angle is 51.5 deg\n",
- "\n",
- " (b)Current through capacitor is 25.13 mA and lead by phase angle is 90.0 deg\n",
- "\n",
- " (c)supply Current is 19.3 mA and leaded by phase angle is 74.5 deg\n",
- "\n",
- " (d)Impedance Z = 2.068 KOhm \n",
- "\n",
- " (e)Power consumed = 206.8 mWatt "
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 249</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency of the circuit and \n",
- "#(b) the current circulating in the capacitor and inductance at resonance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 40E-6;# in Farads\n",
- "R = 0;# in Ohms\n",
- "L = 150E-3;# in Henry\n",
- "V = 50;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = ((1/(L*C) - R*R/(L*L))**0.5)/(2*math.pi)\n",
- "Xc = 1/(2*math.pi*fr*C)\n",
- "Icirc = V/Xc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Parallel resonant frequency, fr = \",round(fr,2),\" Hz \"\n",
- "print \"\\n (b)Current circulating in L and C at resonance = \",round(Icirc,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Parallel resonant frequency, fr = 64.97 Hz \n",
- "\n",
- " (b)Current circulating in L and C at resonance = 0.82 A "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 250</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the resonant frequency, \n",
- "#(b) the dynamic resistance, \n",
- "#(c) the current at resonance and \n",
- "#(d) the circuit Q-factor at resonanc\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 20E-6;# in Farads\n",
- "R = 60;# in Ohms\n",
- "L = 200E-3;# in Henry\n",
- "V = 20;# in Volts\n",
- "\n",
- "#calculation:\n",
- "fr = ((1/(L*C) - R*R/(L*L))**0.5)/(2*math.pi)\n",
- "Rd = L/(R*C)\n",
- "Ir = V/Rd\n",
- "Q = 2*math.pi*fr*L/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Parallel resonant frequency, fr = \",round(fr,2),\" Hz \"\n",
- "print \"\\n (b)the dynamic resistance,RD = \",round(Rd,2),\" ohm \"\n",
- "print \"\\n (c)Current at resonance = \",round(Ir,2),\" A \"\n",
- "print \"\\n (d)Q-factor = \",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Parallel resonant frequency, fr = 63.66 Hz \n",
- "\n",
- " (b)the dynamic resistance,RD = 166.67 ohm \n",
- "\n",
- " (c)Current at resonance = 0.12 A \n",
- "\n",
- " (d)Q-factor = 1.33"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 251</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the condition when the supply current is a minimum:\n",
- "#(a) the capacitance of the capacitor,\n",
- "#(b) the dynamic resistance, \n",
- "#(c) the supply current, and \n",
- "#(d) the Q-factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fr = 5000;# in ohm\n",
- "R = 800;# in Ohms\n",
- "L = 100E-3;# in Henry\n",
- "V = 12;# in Volts\n",
- "\n",
- "#calculation:\n",
- "C = 1/(L*((2*math.pi*fr)**2 + R*R/(L*L)))\n",
- "Rd = L/(R*C)\n",
- "Ir = V/Rd\n",
- "Q = 2*math.pi*fr*L/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)capacitance, C = \",round((C/1E-9),2),\" nF \"\n",
- "print \"\\n (b)the dynamic resistance,RD = \",round(Rd,2),\" ohm \"\n",
- "print \"\\n (c)Current at resonance = \",round((Ir/1E-3),2),\" mA \"\n",
- "print \"\\n (d)Q-factor = \",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)capacitance, C = 9.52 nF \n",
- "\n",
- " (b)the dynamic resistance,RD = 13137.01 ohm \n",
- "\n",
- " (c)Current at resonance = 0.91 mA \n",
- "\n",
- " (d)Q-factor = 3.93"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 252</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the currenttaken by a capacitor connected in parallel with the motor to correct the power factor to unity, and \n",
- "#(b) the value of the supply current after power factor correction.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 50;# in ohm\n",
- "V = 240;# in Volts\n",
- "pf = 0.6;# power factor\n",
- "Im = 50;# in amperes\n",
- "\n",
- "#calculation:\n",
- "phi = math.acos(pf)\n",
- "phid = phi*180/math.pi\n",
- "Ic = Im*math.sin(phi)\n",
- "I = Im*math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the capacitor current Ic must be \",round(Ic,2),\" A for the power factor to be unity. \"\n",
- "print \"\\n (b)Supply current I = \",round(I,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the capacitor current Ic must be 40.0 A for the power factor to be unity. \n",
- "\n",
- " (b)Supply current I = 30.0 A "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 253</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current taken by the motor, \n",
- "#(b) the supply current after power factor correction, \n",
- "#(c) the current taken by the capacitor,\n",
- "#(d) the capacitance of the capacitor, and \n",
- "#(e) the kvar rating of the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Pout = 4800;# in Watt\n",
- "eff = 0.8# effficiency\n",
- "f = 50;# in ohm\n",
- "V = 240;# in Volts\n",
- "pf1 = 0.625;# power factor\n",
- "pf2 = 0.95;# power factor\n",
- "\n",
- "#calculation:\n",
- "Pin = Pout/eff\n",
- "Im = Pin/(V*pf1)\n",
- "phi1 = math.acos(pf1)\n",
- "phi1d = phi1*180/math.pi\n",
- "phi2 = math.acos(pf2)\n",
- "phi2d = phi2*180/math.pi\n",
- "Imh = Im*math.cos(phi1)\n",
- " #Ih = I*cos(phi2)\n",
- "Ih = Imh\n",
- "I = Ih/math.cos(phi2)\n",
- "Imv = Im*math.sin(phi1)\n",
- "Iv = I*math.sin(phi2)\n",
- "Ic = Imv - Iv\n",
- "C = Ic/(2*math.pi*f*V)\n",
- "kvar = V*Ic/1000\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current taken by the motor, Im = \",round(Im,2),\" A\"\n",
- "print \"\\n (b)supply current after p.f. correction, I = \",round(I,2),\" A \"\n",
- "print \"\\n (c)magnitude of the capacitor current Ic = \",round(Ic,0),\" A\"\n",
- "print \"\\n (d)capacitance, C = \",round((C/1E-6),0),\" uF \"\n",
- "print \"\\n (d)kvar rating of the capacitor = \",round(kvar,2),\" kvar \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current taken by the motor, Im = 40.0 A\n",
- "\n",
- " (b)supply current after p.f. correction, I = 26.32 A \n",
- "\n",
- " (c)magnitude of the capacitor current Ic = 23.0 A\n",
- "\n",
- " (d)capacitance, C = 305.0 uF \n",
- "\n",
- " (d)kvar rating of the capacitor = 5.52 kvar "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 254</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for the lamps and motor, (a) the total current, (b) the overall power factor and \n",
- "#(c) the total power. (d) Find the value of the static capacitor to improve the overall power factor to 0.975 lagging.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 3000;# in VA\n",
- "f = 50;# in ohm\n",
- "V = 250;# in Volts\n",
- "Iil = 10;# in Amperes\n",
- "Ifl = 8;# in Amperes\n",
- "pfil = 1; # power factor\n",
- "pffl = 0.7;# power factor\n",
- "pfm = 0.8;# power factor\n",
- "pf0 = 0.975;# power factor\n",
- "\n",
- "#calculation:\n",
- "phiil = math.acos(pfil)\n",
- "phiild = phiil*180/math.pi\n",
- "phifl = math.acos(pffl)\n",
- "phifld = phifl*180/math.pi\n",
- "phim = math.acos(pfm)\n",
- "phimd = phim*180/math.pi\n",
- "phi0 = math.acos(pf0)\n",
- "phi0d = phi0*180/math.pi\n",
- "Im = S/V\n",
- "Ih = Iil*math.cos(phiil) + Ifl*math.cos(phifl) + Im*math.cos(phim)\n",
- "Iv = Iil*math.sin(phiil) - Ifl*math.sin(phifl) - Im*math.sin(phim)\n",
- "Il = (Ih**2 + Iv**2)**0.5\n",
- "phi = math.atan(abs(Iv)/Ih)\n",
- "phid = phi*180/math.pi\n",
- "pf = math.cos(phi)\n",
- "P = V*Il*pf\n",
- "I = Il*math.cos(phi)/math.cos(phi0)\n",
- "Ic = Il*math.sin(phi) - I*math.sin(phi0)\n",
- "C = Ic/(2*math.pi*f*V)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)total current, Il = \",round(Il,2),\" A\"\n",
- "print \"\\n (b)Power factor = \",round(pf,2),\"lagging\"\n",
- "print \"\\n (c)Total power, P = \",round(P/1000,2),\"KWatt\"\n",
- "print \"\\n (d)capacitance, C = \",round((C/1E-6),2),\"uF \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)total current, Il = 28.32 A\n",
- "\n",
- " (b)Power factor = 0.89 lagging\n",
- "\n",
- " (c)Total power, P = 6.3 KWatt\n",
- "\n",
- " (d)capacitance, C = 91.29 uF "
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_16.ipynb index 691ebea9..691ebea9 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_16.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint.ipynb deleted file mode 100755 index 94d79ec6..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint.ipynb +++ /dev/null @@ -1,895 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:3c30db14cf5a722166d02bb2b24e5cc57f73fe8d238c68c893567c34f71054b0"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 17: D.c. transients</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 262</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitor voltage at a time equal to one time constant after being connected to the supply, \n",
- "#and also two seconds after being connected to the supply. \n",
- "#Also, find the time for the capacitor voltage to reach one half of its steady state value.\n",
- "from __future__ import division\n",
- "import math\n",
- "#from pylab import *\n",
- "%pylab inline\n",
- "#initializing the variables:\n",
- "C = 15E-6;# in Farads\n",
- "R = 47000;# in ohms\n",
- "V = 120;# in Volts\n",
- "\n",
- "#calculation:\n",
- "tou = R*C\n",
- "t1 = tou\n",
- "Vctou = V*(1-math.e**(-1*t1/tou))\n",
- "Vct = V/2\n",
- "t0 = -1*tou*math.log(1 - Vct/V)\n",
- "t=[]\n",
- "Vc=[]\n",
- "I = V/R\n",
- "for h in range(50):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " Vc.append(V*(1 - math.e**(-1*k/tou)))\n",
- "fig = plt.figure()\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(t,Vc,'-')\n",
- "#plot(t,Vc,'-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('Volts(V)')\n",
- "show()\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the capacitor voltage at a time equal to one time constant = \",round(Vctou,2),\" V\"\n",
- "print \"\\n (b)the time for the capacitor voltage to reach one half of its steady state value = \",round(t0,5),\" secs\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Populating the interactive namespace from numpy and matplotlib\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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kSBouSe24oRNb4VhCoEiSqtB5lPYGQvWEQWZJkiRJkiRJkiRJkiRJkiRJSrv/\nD8/iLwbt6hd+AAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x34660b8>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the capacitor voltage at a time equal to one time constant = 75.85 V\n",
- "\n",
- " (b)the time for the capacitor voltage to reach one half of its steady state value = 0.48867 secs\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 263</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#draw: (a) the capacitor voltage/time characteristic, \n",
- "#(b) the resistor voltage/time characteristic and \n",
- "#(c) the current/time characteristic,\n",
- "#From the characteristics determine the value of capacitor voltage, \n",
- "#resistor voltage and current one and a half seconds after discharge has started.\n",
- "from __future__ import division\n",
- "import math\n",
- "#from pylab import *\n",
- "%pylab inline\n",
- "#initializing the variables:\n",
- "C = 4E-6;# in Farads\n",
- "R = 220000;# in ohms\n",
- "V = 24;# in Volts\n",
- "t1 = 1.5;# in secs\n",
- "\n",
- "#calculation:\n",
- "tou = R*C\n",
- "I = V/R\n",
- "\n",
- "t=[]\n",
- "Vc=[]\n",
- "for h in range(50):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " Vc.append(V*math.e**(-1*k/tou))\n",
- "#plt.figsize(10,8)\n",
- "fig = plt.figure()\n",
- "#canvas = fc(fig)\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(t,Vc,'-')\n",
- "#plot(t,Vc,'-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('P.D across Capacitor(V)')\n",
- "show()\n",
- "\n",
- "t=[]\n",
- "VR=[]\n",
- "for h in range(50):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " VR.append(V*(1 - math.e**(-1*k/tou)))\n",
- "fig = plt.figure()\n",
- "#canvas = fc(fig)\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(t,VR,'-')\n",
- "#plot(t,VR,'-*')\n",
- "xlabel('time(sec)')\n",
- "ylabel('P.D across Resistor(V)')\n",
- "show()\n",
- "\n",
- "t=[]\n",
- "i=[]\n",
- "for h in range(50):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " i.append(I*math.e**(-1*k/tou))\n",
- "fig = plt.figure()\n",
- "#canvas = fc(fig)\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(t,i,'-')\n",
- "#plot(t,i,'*-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('current(A)')\n",
- "show()\n",
- "\n",
- "Vct1 = V*math.e**(-1*t1/tou)\n",
- "VRt1 = V*math.e**(-1*t1/tou)\n",
- "it1 = I*math.e**(-1*t1/tou)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the value of capacitor voltage is \",round(Vct1,1),\" V, resistor voltage is \",round(VRt1,1),\" V,\"\n",
- "print \"current is \",round(0.02,2),\" mA at one and a half seconds after discharge has started.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Populating the interactive namespace from numpy and matplotlib\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x77c6e10>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x7b9e4a8>"
- ]
- },
- {
- "metadata": {},
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1NbVBmbS0NCQkJAAAoqKiUFFRgdLS0hbr1q+TkJCA3bt3AwBSU1Mxa9Ys+Pr6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\nqer/AUdcZmZ6XUFsAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x7bac898>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the value of capacitor voltage is 4.4 V, resistor voltage is 4.4 V,\n",
- "current is 0.02 mA at one and a half seconds after discharge has started.\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 265</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the initial value of the current flowing,\n",
- "#(b) the time constant of the circuit, \n",
- "#(c) the value of the current one second after connection, \n",
- "#(d) the value of the capacitor voltage two seconds after connection, and \n",
- "#(e) the time after connection when the resistor voltage is 15 V.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 20E-6;# in Farads\n",
- "R = 50000;# in ohms\n",
- "V = 20;# in Volts\n",
- "t1 = 1;# in secs\n",
- "t2 = 2;# in secs\n",
- "VRt = 15;# in Volts\n",
- "\n",
- "#calculation:\n",
- "tou = R*C\n",
- "I = V/R\n",
- "Vct1 = V*(1-math.e**(-1*t2/tou))\n",
- "t3 = -1*tou*math.log(VRt/V)\n",
- "it1 = I*math.e**(-1*t1/tou)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)initial value of the current flowing is \",round(I*1000,1),\"mA\"\n",
- "print \"\\n (b)time constant of the circuit \",round(tou,2),\" Sec\"\n",
- "print \"\\n (c)the value of the current one second after connection, \",round((it1/1E-3),3),\" mA\"\n",
- "print \"\\n (d)the value of the capacitor voltage two seconds after connection \",round(Vct1,1),\" V\"\n",
- "print \"\\n (e)the time after connection when the resistor voltage is 15 V is \",round(t3,3),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)initial value of the current flowing is 0.4 mA\n",
- "\n",
- " (b)time constant of the circuit 1.0 Sec\n",
- "\n",
- " (c)the value of the current one second after connection, 0.147 mA\n",
- "\n",
- " (d)the value of the capacitor voltage two seconds after connection 17.3 V\n",
- "\n",
- " (e)the time after connection when the resistor voltage is 15 V is 0.288 sec\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 266</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of the resistor, and (b) the capacitor voltage 7 ms after connecting the circuit to a 10 V supply\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 0.5E-6;# in Farads\n",
- "V = 10;# in Volts\n",
- "tou = 0.012;# in secs\n",
- "t1 = 0.007;# in secs\n",
- "\n",
- "#calculation:\n",
- "R = tou/C\n",
- "Vc = V*(1-math.e**(-1*t1/tou))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)value of the resistor is \",R,\" ohm\"\n",
- "print \"\\n (b)capacitor voltage is \",round(Vc,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)value of the resistor is 24000.0 ohm\n",
- "\n",
- " (b)capacitor voltage is 4.42 V\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 267</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine: (a) the value of the capacitor, \n",
- "#(b) the time for the capacitor voltage to fall to 20 V, \n",
- "#(c) the current flowing when the capacitor has been discharging for 0.5 s, and \n",
- "#(d) the voltage drop across the resistor when the capacitor has been discharging for one second.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 50000;# in ohms\n",
- "V = 100;# in Volts\n",
- "Vc1 = 20;# in Volts\n",
- "tou = 0.8;# in secs\n",
- "t1 = 0.5;# in secs\n",
- "t2 = 1;# in secs\n",
- "\n",
- "#calculation:\n",
- "C = tou/R\n",
- "t = -1*tou*math.log(Vc1/V)\n",
- "I = V/R\n",
- "it1 = I*math.e**(-1*t1/tou)\n",
- "Vc = V*math.e**(-1*t2/tou)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the value of the capacitor is \",round((C/1E-6),2),\"uF\"\n",
- "print \"\\n (b)the time for the capacitor voltage to fall to 20 V is \",round(t,2),\" sec\"\n",
- "print \"\\n (c)the current flowing when the capacitor has been discharging for 0.5 s is \",round(it1*1000,2),\"mA\"\n",
- "print \"\\n (d)voltage drop across resistor when the capacitor has been discharging for one second is \",round(Vc,1),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the value of the capacitor is 16.0 uF\n",
- "\n",
- " (b)the time for the capacitor voltage to fall to 20 V is 1.29 sec\n",
- "\n",
- " (c)the current flowing when the capacitor has been discharging for 0.5 s is 1.07 mA\n",
- "\n",
- " (d)voltage drop across resistor when the capacitor has been discharging for one second is 28.7 V\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6 page no. 268</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the initial discharge current, \n",
- "#(b) the time constant of the circuit, and \n",
- "#(c) the minimum time required for the voltage across the capacitor to fall to less than 2 V\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 0.1E-6;# in Farads\n",
- "R = 4000;# in ohms\n",
- "V = 200;# in Volts\n",
- "Vc1 = 2;# in Volts\n",
- "\n",
- "#calculation:\n",
- "tou = R*C\n",
- "I = V/R\n",
- "t = -1*tou*math.log(Vc1/V)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \" \\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) initial discharge current is \",round(I,2),\" A\"\n",
- "print \"\\n (b)Time constant tou is \",round(tou,5),\" sec\"\n",
- "print \"\\n (c)min. time required for voltage across capacitor to fall to less than 2 V is \",round(t*1000,0),\" msec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " \n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) initial discharge current is 0.05 A\n",
- "\n",
- " (b)Time constant tou is 0.0004 sec\n",
- "\n",
- " (c)min. time required for voltage across capacitor to fall to less than 2 V is 2.0 msec\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 270</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#draw the current/time characteristic and \n",
- "#hence determine the value of current flowing at a time equal to two time constants and the time for the current to grow to 1.5 A\n",
- "from __future__ import division\n",
- "import math\n",
- "from pylab import *\n",
- "#initializing the variables:\n",
- "L = 0.1;# in Henry\n",
- "R = 20;# in ohms\n",
- "V = 60;# in Volts\n",
- "i2 = 1.5;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "tou = L/R\n",
- "t1 = 2*tou\n",
- "\n",
- "t=[]\n",
- "i=[]\n",
- "I = V/R\n",
- "for h in range(250):\n",
- " t.append((h-1)/10000)\n",
- " k=(h-1)/10000\n",
- " i.append(I*(1 - math.e**(-1*k/tou)))\n",
- "plot(t,i,'-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('current(A)')\n",
- "show()\n",
- "i1 = I*(1 - math.e**(-1*t1/tou))\n",
- "t2 = -1*tou*math.log(1 - i2/I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \" \\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the value of current flowing at a time equal to two time constants is \",round(i1,2),\" A\"\n",
- "print \"\\n (b)the time for the current to grow to 1.5 A is \",round(t2,5),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Pnw/AhAkTGDZsGFlZWURHR9O6dWsWLlxoV7k+t2WLFSrx8XZXIiLSOFrLy08d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\nUYpOSazjAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x7b9d748>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " \n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the value of current flowing at a time equal to two time constants is 2.59 A\n",
- "\n",
- " (b)the time for the current to grow to 1.5 A is 0.00347 sec\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 271</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the final value of current, \n",
- "#(b) the time constant of the circuit, \n",
- "#(c) the value of current after a time equal to the time constant from the instant the supply oltage is connected, \n",
- "#(d) the expected time for the current to rise to within 1% of its final value.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.04;# in Henry\n",
- "R = 10;# in ohms\n",
- "V = 120;# in Volts\n",
- "\n",
- "#calculation:\n",
- "tou = L/R\n",
- "t1 = tou\n",
- "I = V/R\n",
- "i1 = I*(1 - math.e**(-1*t1/tou))\n",
- "i2 = 0.01*I\n",
- "t2 = -1*tou*(-5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the final value of current is \",round(I,2),\" A\"\n",
- "print \"\\n (b)time constant of the circuit is \",round(tou*1000,2),\"msec\"\n",
- "print \"\\n (c) value of current after a time equal to the time constant is \",round(i1,2),\" A\"\n",
- "print \"\\n (d)expected time for current to rise to within 0.01 times of its final value is \",round(t2*1000,2),\"msec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the final value of current is 12.0 A\n",
- "\n",
- " (b)time constant of the circuit is 4.0 msec\n",
- "\n",
- " (c) value of current after a time equal to the time constant is 7.59 A\n",
- "\n",
- " (d)expected time for current to rise to within 0.01 times of its final value is 20.0 msec\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 271</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate:\n",
- "#(a) the steady state value of current flowing in the winding,\n",
- "#(b) the time constant of the circuit,\n",
- "#(c) the value of the induced e.m.f. after 0.1 s,\n",
- "#(d) the time for the current to rise to 85% of its final value, and\n",
- "#(e) the value of the current after 0.3 s\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 3;# in Henry\n",
- "R = 15;# in ohms\n",
- "V = 120;# in Volts\n",
- "t1 = 0.1;# in secs\n",
- "t3 = 0.3;# in secs\n",
- "\n",
- "#calculation:\n",
- "tou = L/R\n",
- "I = V/R\n",
- "i2 = 0.85*I;# in amperes\n",
- "VL = V*math.e**(-1*t1/tou)\n",
- "t2 = -1*tou*math.log(1 - (i2/I))\n",
- "i3 = I*(1 - math.e**(-1*t3/tou))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Results \\n\\n\"\n",
- "print \"\\n (a) steady state value of current is \",round(I,2),\" A\"\n",
- "print \"\\n (b)time constant of the circuit is \",round(tou,2),\" sec\"\n",
- "print \"\\n (c)value of the induced e.m.f. after 0.1 s is \",round(VL,2),\" V\"\n",
- "print \"\\n (d) time for the current to rise to 0.85 times of its final values is \",round(t2,2),\" A\"\n",
- "print \"\\n (e)value of the current after 0.3 s is \",round(i3,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Results \n",
- "\n",
- "\n",
- "\n",
- " (a) steady state value of current is 8.0 A\n",
- "\n",
- " (b)time constant of the circuit is 0.2 sec\n",
- "\n",
- " (c)value of the induced e.m.f. after 0.1 s is 72.78 V\n",
- "\n",
- " (d) time for the current to rise to 0.85 times of its final values is 0.38 A\n",
- "\n",
- " (e)value of the current after 0.3 s is 6.21 A\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 273</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#draw the current/time characteristic when the supply is removed and replaced by a shorting link\n",
- "#determine (a) the current flowing in the winding 3 s after being shorted-out and (b) the time for the current to decay to 5 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "from pylab import *\n",
- "#initializing the variables:\n",
- "R = 15;# in ohms\n",
- "V = 110;# in Volts\n",
- "tou = 2;# in secs\n",
- "t1 = 3;# in secs\n",
- "i2 = 5;# in amperes\n",
- "\n",
- "#calculation:\n",
- "L = tou*R\n",
- "I = V/R\n",
- "\n",
- "t=[]\n",
- "i=[]\n",
- "for h in range(100):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " i.append(I*math.e**(-1*k/tou))\n",
- "plot(t,i,'-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('current(A)')\n",
- "show()\n",
- "i1 = I*(math.e**(-1*t1/tou))\n",
- "t2 = -1*tou*log((i2/I))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \" \\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the current flowing in the winding 3 s after being shorted-out is \",round(i1,2),\" A\"\n",
- "print \"\\n (b)the time for the current to decay to 5 A is \",round(t2,2),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x77abf28>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " \n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the current flowing in the winding 3 s after being shorted-out is 1.64 A\n",
- "\n",
- " (b)the time for the current to decay to 5 A is 0.77 sec\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 273</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine: (a) the resistance of the coil,\n",
- "#(b) the current flowing in the circuit one second after the shorting link has been placed in the circuit, and \n",
- "#(c) the time taken for the current to fall to 10% of its initial value.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 6;# in Henry\n",
- "r = 10;# in ohms\n",
- "V = 120;# in Volts\n",
- "tou = 0.3;# in secs\n",
- "t1 = 1;# in secs\n",
- "\n",
- "#calculation:\n",
- "R = (L/tou) - r\n",
- "Rt = R + r\n",
- "I = V/Rt\n",
- "i2 = 0.1*I\n",
- "i1 = I*(math.e**(-1*t1/tou))\n",
- "t2 = -1*tou*math.log((i2/I))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \" \\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) resistance of the coil is \",round(R,2),\" ohm\"\n",
- "print \"\\n (b) current flowing in circuit one second after the shorting link has been placed is \",round(i1,2),\" A\"\n",
- "print \"\\n (c)the time for the current to decay to 0.1 times of initial value is \",round(t2,2),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " \n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) resistance of the coil is 10.0 ohm\n",
- "\n",
- " (b) current flowing in circuit one second after the shorting link has been placed is 0.21 A\n",
- "\n",
- " (c)the time for the current to decay to 0.1 times of initial value is 0.69 sec\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 274</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the time constant of the circuit and the steady-state value of the current flowing in the circuit. \n",
- "#Find (a) the current flowing in the circuit at a time equal to one time constant, \n",
- "#(b) the voltage drop across the inductor at a time equal to two time constants and \n",
- "#(c) the voltage drop across the resistor after a time equal to three time constants.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.2;# in Henry\n",
- "R = 1000;# in ohms\n",
- "V = 24;# in Volts\n",
- "t1 = 1*L/R;# in secs\n",
- "t2 = 2*L/R;# in secs\n",
- "t3 = 3*L/R;# in secs\n",
- "\n",
- "#calculation:\n",
- "tou = L/R\n",
- "I = V/R\n",
- "i1 = I*(1 - math.e**(-1*t1/tou))\n",
- "VL = V*(math.e**(-1*t2/tou))\n",
- "VR = V*(1 - math.e**(-1*t3/tou))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n time constant of circuit is \",round(tou*1000,6),\"msec, and steady-state value of current is \",round(I*1000,2),\"mA\"\n",
- "print \"\\n (a) urrent flowing in the circuit at a time equal to one time constant is \",round(i1*1000,2),\"mA\"\n",
- "print \"\\n (b) voltage drop across the inductor at a time equal to two time constants is \",round(VL,3),\" V\"\n",
- "print \"\\n (c)the voltage drop across the resistor after a time equal to three time constants is \",round(VR,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " time constant of circuit is 0.2 msec, and steady-state value of current is 24.0 mA\n",
- "\n",
- " (a) urrent flowing in the circuit at a time equal to one time constant is 15.17 mA\n",
- "\n",
- " (b) voltage drop across the inductor at a time equal to two time constants is 3.248 V\n",
- "\n",
- " (c)the voltage drop across the resistor after a time equal to three time constants is 22.81 V\n"
- ]
- }
- ],
- "prompt_number": 18
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_1.ipynb deleted file mode 100755 index 94d79ec6..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_1.ipynb +++ /dev/null @@ -1,895 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:3c30db14cf5a722166d02bb2b24e5cc57f73fe8d238c68c893567c34f71054b0"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 17: D.c. transients</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 262</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitor voltage at a time equal to one time constant after being connected to the supply, \n",
- "#and also two seconds after being connected to the supply. \n",
- "#Also, find the time for the capacitor voltage to reach one half of its steady state value.\n",
- "from __future__ import division\n",
- "import math\n",
- "#from pylab import *\n",
- "%pylab inline\n",
- "#initializing the variables:\n",
- "C = 15E-6;# in Farads\n",
- "R = 47000;# in ohms\n",
- "V = 120;# in Volts\n",
- "\n",
- "#calculation:\n",
- "tou = R*C\n",
- "t1 = tou\n",
- "Vctou = V*(1-math.e**(-1*t1/tou))\n",
- "Vct = V/2\n",
- "t0 = -1*tou*math.log(1 - Vct/V)\n",
- "t=[]\n",
- "Vc=[]\n",
- "I = V/R\n",
- "for h in range(50):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " Vc.append(V*(1 - math.e**(-1*k/tou)))\n",
- "fig = plt.figure()\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(t,Vc,'-')\n",
- "#plot(t,Vc,'-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('Volts(V)')\n",
- "show()\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the capacitor voltage at a time equal to one time constant = \",round(Vctou,2),\" V\"\n",
- "print \"\\n (b)the time for the capacitor voltage to reach one half of its steady state value = \",round(t0,5),\" secs\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Populating the interactive namespace from numpy and matplotlib\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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kSBouSe24oRNb4VhCoEiSqtB5lPYGQvWEQWZJkiRJkiRJkiRJkiRJkiRJSrv/\nD8/iLwbt6hd+AAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x34660b8>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the capacitor voltage at a time equal to one time constant = 75.85 V\n",
- "\n",
- " (b)the time for the capacitor voltage to reach one half of its steady state value = 0.48867 secs\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 263</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#draw: (a) the capacitor voltage/time characteristic, \n",
- "#(b) the resistor voltage/time characteristic and \n",
- "#(c) the current/time characteristic,\n",
- "#From the characteristics determine the value of capacitor voltage, \n",
- "#resistor voltage and current one and a half seconds after discharge has started.\n",
- "from __future__ import division\n",
- "import math\n",
- "#from pylab import *\n",
- "%pylab inline\n",
- "#initializing the variables:\n",
- "C = 4E-6;# in Farads\n",
- "R = 220000;# in ohms\n",
- "V = 24;# in Volts\n",
- "t1 = 1.5;# in secs\n",
- "\n",
- "#calculation:\n",
- "tou = R*C\n",
- "I = V/R\n",
- "\n",
- "t=[]\n",
- "Vc=[]\n",
- "for h in range(50):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " Vc.append(V*math.e**(-1*k/tou))\n",
- "#plt.figsize(10,8)\n",
- "fig = plt.figure()\n",
- "#canvas = fc(fig)\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(t,Vc,'-')\n",
- "#plot(t,Vc,'-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('P.D across Capacitor(V)')\n",
- "show()\n",
- "\n",
- "t=[]\n",
- "VR=[]\n",
- "for h in range(50):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " VR.append(V*(1 - math.e**(-1*k/tou)))\n",
- "fig = plt.figure()\n",
- "#canvas = fc(fig)\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(t,VR,'-')\n",
- "#plot(t,VR,'-*')\n",
- "xlabel('time(sec)')\n",
- "ylabel('P.D across Resistor(V)')\n",
- "show()\n",
- "\n",
- "t=[]\n",
- "i=[]\n",
- "for h in range(50):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " i.append(I*math.e**(-1*k/tou))\n",
- "fig = plt.figure()\n",
- "#canvas = fc(fig)\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(t,i,'-')\n",
- "#plot(t,i,'*-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('current(A)')\n",
- "show()\n",
- "\n",
- "Vct1 = V*math.e**(-1*t1/tou)\n",
- "VRt1 = V*math.e**(-1*t1/tou)\n",
- "it1 = I*math.e**(-1*t1/tou)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the value of capacitor voltage is \",round(Vct1,1),\" V, resistor voltage is \",round(VRt1,1),\" V,\"\n",
- "print \"current is \",round(0.02,2),\" mA at one and a half seconds after discharge has started.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Populating the interactive namespace from numpy and matplotlib\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x77c6e10>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x7b9e4a8>"
- ]
- },
- {
- "metadata": {},
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1NbVBmbS0NCQkJAAAoqKiUFFRgdLS0hbr1q+TkJCA3bt3AwBSU1Mxa9Ys+Pr6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\nqer/AUdcZmZ6XUFsAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x7bac898>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the value of capacitor voltage is 4.4 V, resistor voltage is 4.4 V,\n",
- "current is 0.02 mA at one and a half seconds after discharge has started.\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 265</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the initial value of the current flowing,\n",
- "#(b) the time constant of the circuit, \n",
- "#(c) the value of the current one second after connection, \n",
- "#(d) the value of the capacitor voltage two seconds after connection, and \n",
- "#(e) the time after connection when the resistor voltage is 15 V.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 20E-6;# in Farads\n",
- "R = 50000;# in ohms\n",
- "V = 20;# in Volts\n",
- "t1 = 1;# in secs\n",
- "t2 = 2;# in secs\n",
- "VRt = 15;# in Volts\n",
- "\n",
- "#calculation:\n",
- "tou = R*C\n",
- "I = V/R\n",
- "Vct1 = V*(1-math.e**(-1*t2/tou))\n",
- "t3 = -1*tou*math.log(VRt/V)\n",
- "it1 = I*math.e**(-1*t1/tou)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)initial value of the current flowing is \",round(I*1000,1),\"mA\"\n",
- "print \"\\n (b)time constant of the circuit \",round(tou,2),\" Sec\"\n",
- "print \"\\n (c)the value of the current one second after connection, \",round((it1/1E-3),3),\" mA\"\n",
- "print \"\\n (d)the value of the capacitor voltage two seconds after connection \",round(Vct1,1),\" V\"\n",
- "print \"\\n (e)the time after connection when the resistor voltage is 15 V is \",round(t3,3),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)initial value of the current flowing is 0.4 mA\n",
- "\n",
- " (b)time constant of the circuit 1.0 Sec\n",
- "\n",
- " (c)the value of the current one second after connection, 0.147 mA\n",
- "\n",
- " (d)the value of the capacitor voltage two seconds after connection 17.3 V\n",
- "\n",
- " (e)the time after connection when the resistor voltage is 15 V is 0.288 sec\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 266</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of the resistor, and (b) the capacitor voltage 7 ms after connecting the circuit to a 10 V supply\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 0.5E-6;# in Farads\n",
- "V = 10;# in Volts\n",
- "tou = 0.012;# in secs\n",
- "t1 = 0.007;# in secs\n",
- "\n",
- "#calculation:\n",
- "R = tou/C\n",
- "Vc = V*(1-math.e**(-1*t1/tou))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)value of the resistor is \",R,\" ohm\"\n",
- "print \"\\n (b)capacitor voltage is \",round(Vc,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)value of the resistor is 24000.0 ohm\n",
- "\n",
- " (b)capacitor voltage is 4.42 V\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 267</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine: (a) the value of the capacitor, \n",
- "#(b) the time for the capacitor voltage to fall to 20 V, \n",
- "#(c) the current flowing when the capacitor has been discharging for 0.5 s, and \n",
- "#(d) the voltage drop across the resistor when the capacitor has been discharging for one second.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 50000;# in ohms\n",
- "V = 100;# in Volts\n",
- "Vc1 = 20;# in Volts\n",
- "tou = 0.8;# in secs\n",
- "t1 = 0.5;# in secs\n",
- "t2 = 1;# in secs\n",
- "\n",
- "#calculation:\n",
- "C = tou/R\n",
- "t = -1*tou*math.log(Vc1/V)\n",
- "I = V/R\n",
- "it1 = I*math.e**(-1*t1/tou)\n",
- "Vc = V*math.e**(-1*t2/tou)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the value of the capacitor is \",round((C/1E-6),2),\"uF\"\n",
- "print \"\\n (b)the time for the capacitor voltage to fall to 20 V is \",round(t,2),\" sec\"\n",
- "print \"\\n (c)the current flowing when the capacitor has been discharging for 0.5 s is \",round(it1*1000,2),\"mA\"\n",
- "print \"\\n (d)voltage drop across resistor when the capacitor has been discharging for one second is \",round(Vc,1),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the value of the capacitor is 16.0 uF\n",
- "\n",
- " (b)the time for the capacitor voltage to fall to 20 V is 1.29 sec\n",
- "\n",
- " (c)the current flowing when the capacitor has been discharging for 0.5 s is 1.07 mA\n",
- "\n",
- " (d)voltage drop across resistor when the capacitor has been discharging for one second is 28.7 V\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6 page no. 268</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the initial discharge current, \n",
- "#(b) the time constant of the circuit, and \n",
- "#(c) the minimum time required for the voltage across the capacitor to fall to less than 2 V\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 0.1E-6;# in Farads\n",
- "R = 4000;# in ohms\n",
- "V = 200;# in Volts\n",
- "Vc1 = 2;# in Volts\n",
- "\n",
- "#calculation:\n",
- "tou = R*C\n",
- "I = V/R\n",
- "t = -1*tou*math.log(Vc1/V)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \" \\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) initial discharge current is \",round(I,2),\" A\"\n",
- "print \"\\n (b)Time constant tou is \",round(tou,5),\" sec\"\n",
- "print \"\\n (c)min. time required for voltage across capacitor to fall to less than 2 V is \",round(t*1000,0),\" msec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " \n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) initial discharge current is 0.05 A\n",
- "\n",
- " (b)Time constant tou is 0.0004 sec\n",
- "\n",
- " (c)min. time required for voltage across capacitor to fall to less than 2 V is 2.0 msec\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 270</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#draw the current/time characteristic and \n",
- "#hence determine the value of current flowing at a time equal to two time constants and the time for the current to grow to 1.5 A\n",
- "from __future__ import division\n",
- "import math\n",
- "from pylab import *\n",
- "#initializing the variables:\n",
- "L = 0.1;# in Henry\n",
- "R = 20;# in ohms\n",
- "V = 60;# in Volts\n",
- "i2 = 1.5;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "tou = L/R\n",
- "t1 = 2*tou\n",
- "\n",
- "t=[]\n",
- "i=[]\n",
- "I = V/R\n",
- "for h in range(250):\n",
- " t.append((h-1)/10000)\n",
- " k=(h-1)/10000\n",
- " i.append(I*(1 - math.e**(-1*k/tou)))\n",
- "plot(t,i,'-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('current(A)')\n",
- "show()\n",
- "i1 = I*(1 - math.e**(-1*t1/tou))\n",
- "t2 = -1*tou*math.log(1 - i2/I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \" \\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the value of current flowing at a time equal to two time constants is \",round(i1,2),\" A\"\n",
- "print \"\\n (b)the time for the current to grow to 1.5 A is \",round(t2,5),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Pnw/AhAkTGDZsGFlZWURHR9O6dWsWLlxoV7k+t2WLFSrx8XZXIiLSOFrLy08d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\nUYpOSazjAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x7b9d748>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " \n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the value of current flowing at a time equal to two time constants is 2.59 A\n",
- "\n",
- " (b)the time for the current to grow to 1.5 A is 0.00347 sec\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 271</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the final value of current, \n",
- "#(b) the time constant of the circuit, \n",
- "#(c) the value of current after a time equal to the time constant from the instant the supply oltage is connected, \n",
- "#(d) the expected time for the current to rise to within 1% of its final value.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.04;# in Henry\n",
- "R = 10;# in ohms\n",
- "V = 120;# in Volts\n",
- "\n",
- "#calculation:\n",
- "tou = L/R\n",
- "t1 = tou\n",
- "I = V/R\n",
- "i1 = I*(1 - math.e**(-1*t1/tou))\n",
- "i2 = 0.01*I\n",
- "t2 = -1*tou*(-5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the final value of current is \",round(I,2),\" A\"\n",
- "print \"\\n (b)time constant of the circuit is \",round(tou*1000,2),\"msec\"\n",
- "print \"\\n (c) value of current after a time equal to the time constant is \",round(i1,2),\" A\"\n",
- "print \"\\n (d)expected time for current to rise to within 0.01 times of its final value is \",round(t2*1000,2),\"msec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the final value of current is 12.0 A\n",
- "\n",
- " (b)time constant of the circuit is 4.0 msec\n",
- "\n",
- " (c) value of current after a time equal to the time constant is 7.59 A\n",
- "\n",
- " (d)expected time for current to rise to within 0.01 times of its final value is 20.0 msec\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 271</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate:\n",
- "#(a) the steady state value of current flowing in the winding,\n",
- "#(b) the time constant of the circuit,\n",
- "#(c) the value of the induced e.m.f. after 0.1 s,\n",
- "#(d) the time for the current to rise to 85% of its final value, and\n",
- "#(e) the value of the current after 0.3 s\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 3;# in Henry\n",
- "R = 15;# in ohms\n",
- "V = 120;# in Volts\n",
- "t1 = 0.1;# in secs\n",
- "t3 = 0.3;# in secs\n",
- "\n",
- "#calculation:\n",
- "tou = L/R\n",
- "I = V/R\n",
- "i2 = 0.85*I;# in amperes\n",
- "VL = V*math.e**(-1*t1/tou)\n",
- "t2 = -1*tou*math.log(1 - (i2/I))\n",
- "i3 = I*(1 - math.e**(-1*t3/tou))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Results \\n\\n\"\n",
- "print \"\\n (a) steady state value of current is \",round(I,2),\" A\"\n",
- "print \"\\n (b)time constant of the circuit is \",round(tou,2),\" sec\"\n",
- "print \"\\n (c)value of the induced e.m.f. after 0.1 s is \",round(VL,2),\" V\"\n",
- "print \"\\n (d) time for the current to rise to 0.85 times of its final values is \",round(t2,2),\" A\"\n",
- "print \"\\n (e)value of the current after 0.3 s is \",round(i3,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Results \n",
- "\n",
- "\n",
- "\n",
- " (a) steady state value of current is 8.0 A\n",
- "\n",
- " (b)time constant of the circuit is 0.2 sec\n",
- "\n",
- " (c)value of the induced e.m.f. after 0.1 s is 72.78 V\n",
- "\n",
- " (d) time for the current to rise to 0.85 times of its final values is 0.38 A\n",
- "\n",
- " (e)value of the current after 0.3 s is 6.21 A\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 273</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#draw the current/time characteristic when the supply is removed and replaced by a shorting link\n",
- "#determine (a) the current flowing in the winding 3 s after being shorted-out and (b) the time for the current to decay to 5 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "from pylab import *\n",
- "#initializing the variables:\n",
- "R = 15;# in ohms\n",
- "V = 110;# in Volts\n",
- "tou = 2;# in secs\n",
- "t1 = 3;# in secs\n",
- "i2 = 5;# in amperes\n",
- "\n",
- "#calculation:\n",
- "L = tou*R\n",
- "I = V/R\n",
- "\n",
- "t=[]\n",
- "i=[]\n",
- "for h in range(100):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " i.append(I*math.e**(-1*k/tou))\n",
- "plot(t,i,'-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('current(A)')\n",
- "show()\n",
- "i1 = I*(math.e**(-1*t1/tou))\n",
- "t2 = -1*tou*log((i2/I))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \" \\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the current flowing in the winding 3 s after being shorted-out is \",round(i1,2),\" A\"\n",
- "print \"\\n (b)the time for the current to decay to 5 A is \",round(t2,2),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x77abf28>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " \n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the current flowing in the winding 3 s after being shorted-out is 1.64 A\n",
- "\n",
- " (b)the time for the current to decay to 5 A is 0.77 sec\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 273</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine: (a) the resistance of the coil,\n",
- "#(b) the current flowing in the circuit one second after the shorting link has been placed in the circuit, and \n",
- "#(c) the time taken for the current to fall to 10% of its initial value.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 6;# in Henry\n",
- "r = 10;# in ohms\n",
- "V = 120;# in Volts\n",
- "tou = 0.3;# in secs\n",
- "t1 = 1;# in secs\n",
- "\n",
- "#calculation:\n",
- "R = (L/tou) - r\n",
- "Rt = R + r\n",
- "I = V/Rt\n",
- "i2 = 0.1*I\n",
- "i1 = I*(math.e**(-1*t1/tou))\n",
- "t2 = -1*tou*math.log((i2/I))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \" \\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) resistance of the coil is \",round(R,2),\" ohm\"\n",
- "print \"\\n (b) current flowing in circuit one second after the shorting link has been placed is \",round(i1,2),\" A\"\n",
- "print \"\\n (c)the time for the current to decay to 0.1 times of initial value is \",round(t2,2),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " \n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) resistance of the coil is 10.0 ohm\n",
- "\n",
- " (b) current flowing in circuit one second after the shorting link has been placed is 0.21 A\n",
- "\n",
- " (c)the time for the current to decay to 0.1 times of initial value is 0.69 sec\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 274</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the time constant of the circuit and the steady-state value of the current flowing in the circuit. \n",
- "#Find (a) the current flowing in the circuit at a time equal to one time constant, \n",
- "#(b) the voltage drop across the inductor at a time equal to two time constants and \n",
- "#(c) the voltage drop across the resistor after a time equal to three time constants.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.2;# in Henry\n",
- "R = 1000;# in ohms\n",
- "V = 24;# in Volts\n",
- "t1 = 1*L/R;# in secs\n",
- "t2 = 2*L/R;# in secs\n",
- "t3 = 3*L/R;# in secs\n",
- "\n",
- "#calculation:\n",
- "tou = L/R\n",
- "I = V/R\n",
- "i1 = I*(1 - math.e**(-1*t1/tou))\n",
- "VL = V*(math.e**(-1*t2/tou))\n",
- "VR = V*(1 - math.e**(-1*t3/tou))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n time constant of circuit is \",round(tou*1000,6),\"msec, and steady-state value of current is \",round(I*1000,2),\"mA\"\n",
- "print \"\\n (a) urrent flowing in the circuit at a time equal to one time constant is \",round(i1*1000,2),\"mA\"\n",
- "print \"\\n (b) voltage drop across the inductor at a time equal to two time constants is \",round(VL,3),\" V\"\n",
- "print \"\\n (c)the voltage drop across the resistor after a time equal to three time constants is \",round(VR,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " time constant of circuit is 0.2 msec, and steady-state value of current is 24.0 mA\n",
- "\n",
- " (a) urrent flowing in the circuit at a time equal to one time constant is 15.17 mA\n",
- "\n",
- " (b) voltage drop across the inductor at a time equal to two time constants is 3.248 V\n",
- "\n",
- " (c)the voltage drop across the resistor after a time equal to three time constants is 22.81 V\n"
- ]
- }
- ],
- "prompt_number": 18
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_2.ipynb deleted file mode 100755 index adc7f92a..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_2.ipynb +++ /dev/null @@ -1,895 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f205689956af649c31ad664e81a2e6ebba547e950f06ad3009a92a9bc3b8be80"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 17: D.c. transients</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 262</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitor voltage at a time equal to one time constant after being connected to the supply, \n",
- "#and also two seconds after being connected to the supply. \n",
- "#Also, find the time for the capacitor voltage to reach one half of its steady state value.\n",
- "from __future__ import division\n",
- "import math\n",
- "#from pylab import *\n",
- "%pylab inline\n",
- "#initializing the variables:\n",
- "C = 15E-6;# in Farads\n",
- "R = 47000;# in ohms\n",
- "V = 120;# in Volts\n",
- "\n",
- "#calculation:\n",
- "tou = R*C\n",
- "t1 = tou\n",
- "Vctou = V*(1-math.e**(-1*t1/tou))\n",
- "Vct = V/2\n",
- "t0 = -1*tou*math.log(1 - Vct/V)\n",
- "t=[]\n",
- "Vc=[]\n",
- "I = V/R\n",
- "for h in range(50):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " Vc.append(V*(1 - math.e**(-1*k/tou)))\n",
- "fig = plt.figure()\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(t,Vc,'-')\n",
- "#plot(t,Vc,'-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('Volts(V)')\n",
- "show()\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the capacitor voltage at a time equal to one time constant = \",round(Vctou,2),\" V\"\n",
- "print \"\\n (b)the time for the capacitor voltage to reach one half of its steady state value = \",round(t0,5),\" secs\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Populating the interactive namespace from numpy and matplotlib\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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kSBouSe24oRNb4VhCoEiSqtB5lPYGQvWEQWZJkiRJkiRJkiRJkiRJkiRJSrv/\nD8/iLwbt6hd+AAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x34660b8>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the capacitor voltage at a time equal to one time constant = 75.85 V\n",
- "\n",
- " (b)the time for the capacitor voltage to reach one half of its steady state value = 0.48867 secs\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 263</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#draw: (a) the capacitor voltage/time characteristic, \n",
- "#(b) the resistor voltage/time characteristic and \n",
- "#(c) the current/time characteristic,\n",
- "#From the characteristics determine the value of capacitor voltage, \n",
- "#resistor voltage and current one and a half seconds after discharge has started.\n",
- "from __future__ import division\n",
- "import math\n",
- "#from pylab import *\n",
- "%pylab inline\n",
- "#initializing the variables:\n",
- "C = 4E-6;# in Farads\n",
- "R = 220000;# in ohms\n",
- "V = 24;# in Volts\n",
- "t1 = 1.5;# in secs\n",
- "\n",
- "#calculation:\n",
- "tou = R*C\n",
- "I = V/R\n",
- "\n",
- "t=[]\n",
- "Vc=[]\n",
- "for h in range(50):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " Vc.append(V*math.e**(-1*k/tou))\n",
- "#plt.figsize(10,8)\n",
- "fig = plt.figure()\n",
- "#canvas = fc(fig)\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(t,Vc,'-')\n",
- "#plot(t,Vc,'-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('P.D across Capacitor(V)')\n",
- "show()\n",
- "\n",
- "t=[]\n",
- "VR=[]\n",
- "for h in range(50):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " VR.append(V*(1 - math.e**(-1*k/tou)))\n",
- "fig = plt.figure()\n",
- "#canvas = fc(fig)\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(t,VR,'-')\n",
- "#plot(t,VR,'-*')\n",
- "xlabel('time(sec)')\n",
- "ylabel('P.D across Resistor(V)')\n",
- "show()\n",
- "\n",
- "t=[]\n",
- "i=[]\n",
- "for h in range(50):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " i.append(I*math.e**(-1*k/tou))\n",
- "fig = plt.figure()\n",
- "#canvas = fc(fig)\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(t,i,'-')\n",
- "#plot(t,i,'*-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('current(A)')\n",
- "show()\n",
- "\n",
- "Vct1 = V*math.e**(-1*t1/tou)\n",
- "VRt1 = V*math.e**(-1*t1/tou)\n",
- "it1 = I*math.e**(-1*t1/tou)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the value of capacitor voltage is \",round(Vct1,1),\" V, resistor voltage is \",round(VRt1,1),\" V,\"\n",
- "print \"current is \",round(0.02,2),\" mA at one and a half seconds after discharge has started.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Populating the interactive namespace from numpy and matplotlib\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x77c6e10>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x7b9e4a8>"
- ]
- },
- {
- "metadata": {},
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1NbVBmbS0NCQkJAAAoqKiUFFRgdLS0hbr1q+TkJCA3bt3AwBSU1Mxa9Ys+Pr6\nwmQywWKxIDs7GwBw6dIlrF+/HsuXL+d1mA6mXz/lvpq5c4FTp7SOhohuhmqJpqioCP7+/s7XRqMR\nRUVFbpUpLi5utm5ZWRn0ej0AQK/Xo6ysDABQXFwMY73pf41GI4qLiwEAK1aswHPPPYcePXp4+Cyp\nLdx7L7BoETBrFlBVpXU0RNRaPmodWOfmDRDutDBEpMnj6XS6Fj9HRPDZZ5/h5MmTWL9+PQoKClr8\nnJUrVzqfR0dHIzo62mVs1DZ+8Qvgn/9UZg9o5nIfEbWBzMxMZGZmtqqOaonGYDDAbrc7X9vt9gYt\njqbKFBYWwmg0oqqqqtF+g8EAQGnFlJaWws/PDyUlJRg4cGCLx8rKysKhQ4cwdOhQVFdX4/Tp05gw\nYQLef//9RjHXTzTUvnTpAqSkAPfcA4werSyaRkRt78Y/wletWuW6kloXiKqqqmTYsGGSn58vlZWV\nLgcDfPLJJ87BAC3VXbRokSQlJYmIyNq1axsNBqisrJSTJ0/KsGHDpLa2tsHnFRQUSGhoaJPxqvhV\nkAcdOSJy550ihw9rHQkRibj326lai8bHxwcbN27E5MmTUVNTgzlz5iA4OBibN28GAMydOxdTpkxB\neno6LBYLevbsia1bt7ZYFwCWLl2KGTNmYMuWLTCZTHjrrbcAACEhIZgxYwZCQkLg4+ODl156qVG3\nmjTTBUcdR0QEsHkzMG0acPAgMGiQ1hERkSucGaAOZwboWF54AdizB8jMBG6/XetoiDovTqrZCkw0\nHYsIMHu2cu3m9dc5+SaRVjgFDXktnQ545RXgq6+ApCStoyGilqh2jYZIbbffDqSmAlFRQGAg8KMf\naR0RETWFiYY6tMGDlWTz4IPKLALjx2sdERHdiF1n1OGNGgW89ZYy0/Onn2odDRHdiImGvEJ0NLBl\nCxAfD+Tmah0NEdXHRENeIz4eWLcOmDwZcDHbEBG1IV6jIa/yk58A584BsbHAv/4F+PlpHRERMdGQ\n11mwQEk2kycD778PDBigdUREnRu7zsgrrVihjEQbPx44fVrraIg6N7ZoyCvpdMqNnD16AOPGAf/4\nB1A3ATgRtTEmGvJaOh3w/PPKjZ3Xko3JpHVURJ0PEw15vcWLr7ds3nsPCAjQOiKizoWJhjqF+fOV\nZDN+PPDuu0BoqNYREXUeTDTUaTz+uNKNFhMDbN8OTJigdUREnQNHnVGnMmsWsGOH8li3zh4RqYzr\n0dThejSlYKk+AAAS5ElEQVSdy5dfAlOnKvOjvfCCsq4NEbUeFz5rBSaazufMGWVJaKMRePVVrtRJ\ndDO48BlRC+66Sxny3KWLcr2mrEzriIi8ExMNdWrduwNvvKHMInDPPcAHH2gdEZH3YddZHXadUUYG\n8NhjwC9+ASxapNzwSUQt4zWaVmCiIQCw24EZM4CBA5XrNv36aR0RUfvGazREreTvDxw4AAwdqnSl\nHT6sdUREHZ/qiSYjIwNBQUGwWq1ITk5uskxiYiKsVivCw8ORk5Pjsm55eTliY2MREBCASZMmoaKi\nwvne2rVrYbVaERQUhH379gEArl69iqlTpyI4OBihoaFYtmyZSmdL3uC224Df/x5ITgbi4pTJOaur\ntY6KqAMTFVVXV4vZbJb8/HxxOBwSHh4uubm5Dcrs3btX4uLiREQkKytLoqKiXNZdtGiRJCcni4hI\nUlKSLFmyREREjh49KuHh4eJwOCQ/P1/MZrPU1tbKlStXJDMzU0REHA6HjB07Vt55550Gcaj8VVAH\ndeqUyIQJIvfeK3L8uNbRELU/7vx2qtqiyc7OhsVigclkgq+vL2bOnInU1NQGZdLS0pCQkAAAiIqK\nQkVFBUpLS1usW79OQkICdu/eDQBITU3FrFmz4OvrC5PJBIvFgoMHD+L222/HAw88AADw9fXFqFGj\nUFRUpOapk5e4+25g/37gxz8GxowB/vAHoLZW66iIOhZVE01RURH8/f2dr41GY6Mf+ObKFBcXN1u3\nrKwMer0eAKDX61FWdwNEcXExjEZji59XUVGBPXv2ICYmxkNnSd6uSxdlUs6PPgLefFNZJrqgQOuo\niDoOVSfV1Lk5PlTcGO0lIk0eT6fTtfg59d+rrq7GrFmzsHDhQpiaWJhk5cqVzufR0dGIjo52GRd1\nHgEBwIcfAuvWAaNHA7/8pbLddpvWkRG1nczMTGRmZraqjqqJxmAwwG63O1/b7fYGLY6myhQWFsJo\nNKKqqqrRfkPdEol6vR6lpaXw8/NDSUkJBg4c2OyxDPWWVXzqqacQGBiIxMTEJuOtn2iImtK1K7Bk\niTIEesECYNs24I9/5EzQ1Hnc+Ef4qlWrXNZRtets9OjRyMvLQ0FBARwOB3bu3AmbzdagjM1mw7Zt\n2wAAWVlZ6Nu3L/R6fYt1bTYbUlJSAAApKSmYNm2ac/+OHTvgcDiQn5+PvLw8REZGAgCWL1+OCxcu\nYP369WqeMnUSQ4cCe/YoI9J+9jPlGk5JidZREbVTao9ISE9Pl4CAADGbzbJmzRoREdm0aZNs2rTJ\nWebpp58Ws9ksYWFhcvjw4RbrioicPXtWYmJixGq1SmxsrJw7d8753urVq8VsNktgYKBkZGSIiIjd\nbhedTichISEycuRIGTlypGzZsqVBnG3wVZCXunRJZOlSkQEDRFavFrl8WeuIiNqOO7+dnBmgDmcG\noFuVlwf8938Dn3wCrFypTGfjw6UFyctxCppWYKIhTzl4EFi8GPj2W2DtWiA+nvOmkfdiomkFJhry\nJBFg715g6VKgTx9gxQpg8mQmHPI+TDStwERDaqipAf7yF2UVz+7dgeXLAZuNK3qS92CiaQUmGlJT\nbS2QmqoknKoq4Fe/Ah5+WBkuTdSRMdG0AhMNtQURZd2b1auBwkJlxoE5c7gcAXVcXCaAqJ3R6ZQZ\noT/8UOlS+/e/gWHDgHnzgGPHtI6OSB1MNEQa+d73gNdeA3JzlYXWxo9X5lHbuROorNQ6OiLPYddZ\nHXadkdYqK4Fdu4AtW5SWzuzZSrdaWJjWkRE1j9doWoGJhtqT/Hxg61Zl0+uBRx9V5lfz89M6MqKG\nmGhagYmG2qOaGmU9nDffVOZWGzUKmDUL+NGPgP79tY6OiImmVZhoqL27ehVITwd27AD27QPGjgWm\nTVNmHqhbnomozTHRtAITDXUkFy8qLZzUVODdd4GQEOAHP1BuBg0K4gwE1HaYaFqBiYY6qspK4MAB\nJemkpSkLsU2eDEyapIxk69NH6wjJmzHRtAITDXkDEeCLL5RWzr59ykzSI0cqSSc6GoiMBLp10zpK\n8iZMNK3AREPe6OpV4IMPlKRz4ADw5ZfK/TvjxgEPPADcey/Qo4fWUVJHxkTTCkw01BmcPw98/LGS\ndA4cUO7XCQlREs61zWzmNR5yHxNNKzDRUGd09SqQkwNkZV3frlwBRo9WhlJf24YOZfKhpjHRtAIT\nDZGiqAg4cuT6dvgwcPmycq1nxIjr2/DhQO/eWkdLWmOiaQUmGqLmlZUBn32mDDT4z3+Ux2PHlDna\nhg9XhlQHBiqPQUHAnXeyBdRZMNG0AhMNUevU1ABff60MMKi/HTumLOxmsSjXe+o/Dhum3FzKhd+8\nBxNNKzDREHmGCHDmjJKETpy4/njiBHDypHKzqb8/YDJd3/z9AaNReTQYgNtv1/gkyG1MNK3AREPU\nNq5cAU6dUraCAmWz25WF4Ox25RpR795K4hk0SNkGD77+3M9P6bLT64FevdhFpzXNE01GRgaeeeYZ\n1NTU4IknnsCSJUsalUlMTMQ777yDHj164NVXX0VERESLdcvLy/HII4/g1KlTMJlMeOutt9C3b18A\nwNq1a/HKK6+ga9euePHFFzFp0iQAwOHDh/HYY4/hu+++w5QpU7Bhw4bGXwQTDVG7UFsLfPutknhK\nSpStuPj687Ky65uIknQGDgTuuku5NnTt8c47gQEDlMlH629sLXmWpommpqYGgYGBeO+992AwGPC9\n730P27dvR3BwsLNMeno6Nm7ciPT0dBw8eBALFy5EVlZWi3UXL16MO++8E4sXL0ZycjLOnTuHpKQk\n5ObmYvbs2fj0009RVFSEiRMnIi8vDzqdDpGRkdi4cSMiIyMxZcoUJCYm4sEHH2z1l9WRZWZmIjo6\nWuswVMPz69hu9vwuX1YSzunTSnK6tp05o2zl5de3c+eAs2eVFlDfvo23Pn2U7Y47lO3a8969G2+9\negFdu6p7bh2FO7+dPmp9eHZ2NiwWC0wmEwBg5syZSE1NbZBo0tLSkJCQAACIiopCRUUFSktLkZ+f\n32zdtLQ0HDhwAACQkJCA6OhoJCUlITU1FbNmzYKvry9MJhMsFgsOHjyIIUOG4OLFi4iMjAQAPPro\no9i9e3ejROPtvP0/O8+vY7vZ8+vZUxlgMGyY+3WuXFFuXD13DqioaLhduKBsJSVKmfPngUuXlOtK\nFy9ef37pkjKnXM+e17devZTHHj0abp99lomJE6Nx++1otHXv3vTWrZuy1X/ekQdQqJZoioqK4O/v\n73xtNBpx8OBBl2WKiopQXFzcbN2ysjLo6+ZE1+v1KCsrAwAUFxfj3nvvbXQsX19fGI1G536DwYCi\noiIPnikRdSTXEsCgQTd/DBHgu++UFtWlS9cfr1xRtsuXrz8/cQLw9VUS1OnTyk2y17bvvmu8VVY2\nfqysBHx8lOTWrZvy2NTm63v98dpW/7WPT+PHG5/7+AAREcDEiZ77zlVLNDo3r9C5010lIk0eT6fT\nuf05RESeotNdb5XceWfLZU+fBn7961v7PBGgulpJOA7H9eRTVaW8vvZ4bauqur5de11d3fxjdbVS\n7soV5XlFxa3FeyPVEo3BYIDdbne+ttvtDVoWTZUpLCyE0WhEVVVVo/0GgwGA0oopLS2Fn58fSkpK\nMHDgwBaPZTAYUFhY2OSx6jObzV6ftFatWqV1CKri+XVs3nx+3nxuZrPZdSFRSVVVlQwbNkzy8/Ol\nsrJSwsPDJTc3t0GZvXv3SlxcnIiIfPLJJxIVFeWy7qJFiyQpKUlERNauXStLliwREZGjR49KeHi4\nVFZWysmTJ2XYsGFSW1srIiKRkZGSlZUltbW1EhcXJ++8845ap01ERDdQrUXj4+ODjRs3YvLkyaip\nqcGcOXMQHByMzZs3AwDmzp2LKVOmID09HRaLBT179sTWrVtbrAsAS5cuxYwZM7Blyxbn8GYACAkJ\nwYwZMxASEgIfHx+89NJLzhbKSy+9hMceewxXr17FlClTOt1AACIiLfGGTSIiUlUHHjDneX/5y18w\nfPhwdO3aFUeOHNE6HI/JyMhAUFAQrFYrkpOTtQ7Hox5//HHo9XqMGDFC61BUYbfbMX78eAwfPhyh\noaF48cUXtQ7JY7777jtERUVh5MiRCAkJwbJly7QOSRU1NTWIiIhAfHy81qF4nMlkQlhYGCIiIpy3\nkDRJ67679uTYsWNy/PhxiY6OlsOHD2sdjkdUV1eL2WyW/Px8cTgcTV4r68g++OADOXLkiISGhmod\niipKSkokJydHREQuXrwoAQEBXvXvd/nyZRFRrstGRUXJv/71L40j8rzf/e53Mnv2bImPj9c6FI8z\nmUxy9uxZl+XYoqknKCgIAQEBWofhUfVvnPX19XXe/Ootxo4di379+mkdhmr8/PwwcuRIAECvXr0Q\nHByM4uJijaPynB5160g7HA7U1NSgf//+GkfkWYWFhUhPT8cTTzzhtTOPuHNeTDRerrmbYqnjKSgo\nQE5ODqKiorQOxWNqa2sxcuRI6PV6jB8/HiEhIVqH5FHPPvssfvvb36JLR76tvwU6nQ4TJ07E6NGj\n8ac//anZcqqNOmuvYmNjUVpa2mj/mjVrvLIP1dvvDeosLl26hIcffhgbNmxAr169tA7HY7p06YLP\nPvsM58+fx+TJk71qqp23334bAwcOREREBDIzM7UORxUfffQRBg0ahDNnziA2NhZBQUEYO3Zso3Kd\nLtHs379f6xDalDs3zlL7VlVVhf/6r//CT37yE0ybNk3rcFTRp08fTJ06FYcOHfKaRPPxxx8jLS0N\n6enp+O6773DhwgU8+uij2LZtm9ahecygunl87rrrLvzwhz9EdnZ2k4nGO9tzHuAt/amjR49GXl4e\nCgoK4HA4sHPnTthsNq3DIjeJCObMmYOQkBA888wzWofjUd9++y0q6uY6uXr1Kvbv3+9cJsQbrFmz\nBna7Hfn5+dixYwcmTJjgVUnmypUruHjxIgDg8uXL2LdvX7OjP5lo6vn73/8Of39/ZGVlYerUqYiL\ni9M6pFtW/+bXkJAQPPLIIw1m0O7oZs2ahe9///v46quv4O/v77zp11t89NFHeP311/HPf/4TERER\niIiIQEZGhtZheURJSQkmTJiAkSNHIioqCvHx8YiJidE6LNV4Wzd2WVkZxo4d6/z3e+ihh5xrgN2I\nN2wSEZGq2KIhIiJVMdEQEZGqmGiIiEhVTDRERKQqJhoiIlIVEw0REamKiYbIQ86fP4//+7//A6Dc\nIzJ9+nSPHXvjxo149dVXPXa8GTNmID8/32PHI2oJ76Mh8pCCggLEx8fjP//5j0ePKyIYNWoUPv30\nU/j4eGbWqP3792PPnj1etb4NtV9s0RB5yNKlS/H1118jIiICM2bMcE7H8eqrr2LatGmYNGkShg4d\nio0bN2LdunUYNWoU7rvvPpw7dw4A8PXXXyMuLg6jR4/GuHHjcPz4cQDK7ABBQUHOJPPiiy9i+PDh\nCA8Px6xZswAoU4A8/vjjiIqKwqhRo5CWlgZAWXTrueeew4gRIxAeHo6NGzcCAKKjo5Gent6m3w91\nYuotiUPUuRQUFDgXYKv/fOvWrWKxWOTSpUty5swZueOOO2Tz5s0iIvLss8/K73//exERmTBhguTl\n5YmISFZWlkyYMEFERNauXSvr1q1zfs7gwYPF4XCIiMj58+dFRGTZsmXy+uuvi4jIuXPnJCAgQC5f\nviwvvfSSTJ8+XWpqakREpLy83HmccePGedUiatR+dbrZm4nUIvV6oeWGHunx48ejZ8+e6NmzJ/r2\n7etckmLEiBH4/PPPcfnyZXz88ccNrus4HA4AwDfffIP777/fuT8sLAyzZ8/GtGnTnLM579u3D3v2\n7MG6desAAJWVlfjmm2/wj3/8A/PmzXOuh1J/kbjBgwejoKDAq+a+o/aJiYaoDXTr1s35vEuXLs7X\nXbp0QXV1NWpra9GvXz/k5OQ0Wb9+4tq7dy8++OAD7NmzB6tXr3ZeE9q1axesVmuLdW/c760LclH7\nwv9lRB7Su3dv57Tp7rqWBHr37o2hQ4fir3/9q3P/559/DgAYMmSIc7E+EcE333yD6OhoJCUl4fz5\n87h06RImT57c4ML+tYQVGxuLzZs3o6amBgCc14MAZWTckCFDbvJsidzHREPkIQMGDMCYMWMwYsQI\nLF682DktvE6nazBF/I3Pr71+4403sGXLFowcORKhoaHOC/r3338/Dh06BACorq7GT3/6U4SFhWHU\nqFFYuHAh+vTpgxUrVqCqqgphYWEIDQ3F888/DwB44okncPfddyMsLAwjR47E9u3bASiLqRUWFiIo\nKEj9L4Y6PQ5vJmrnpG5488GDB3Hbbbd55Jj79u3D3r17sWHDBo8cj6glbNEQtXM6nQ5PPvkk3njj\nDY8d889//jOeffZZjx2PqCVs0RARkarYoiEiIlUx0RARkaqYaIiISFVMNEREpComGiIiUhUTDRER\nqer/AUdcZmZ6XUFsAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x7bac898>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the value of capacitor voltage is 4.4 V, resistor voltage is 4.4 V,\n",
- "current is 0.02 mA at one and a half seconds after discharge has started.\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 265</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the initial value of the current flowing,\n",
- "#(b) the time constant of the circuit, \n",
- "#(c) the value of the current one second after connection, \n",
- "#(d) the value of the capacitor voltage two seconds after connection, and \n",
- "#(e) the time after connection when the resistor voltage is 15 V.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 20E-6;# in Farads\n",
- "R = 50000;# in ohms\n",
- "V = 20;# in Volts\n",
- "t1 = 1;# in secs\n",
- "t2 = 2;# in secs\n",
- "VRt = 15;# in Volts\n",
- "\n",
- "#calculation:\n",
- "tou = R*C\n",
- "I = V/R\n",
- "Vct1 = V*(1-math.e**(-1*t2/tou))\n",
- "t3 = -1*tou*math.log(VRt/V)\n",
- "it1 = I*math.e**(-1*t1/tou)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)initial value of the current flowing is \",round(I*1000,1),\"mA\"\n",
- "print \"\\n (b)time constant of the circuit \",round(tou,2),\" Sec\"\n",
- "print \"\\n (c)the value of the current one second after connection, \",round((it1/1E-3),3),\" mA\"\n",
- "print \"\\n (d)the value of the capacitor voltage two seconds after connection \",round(Vct1,1),\" V\"\n",
- "print \"\\n (e)the time after connection when the resistor voltage is 15 V is \",round(t3,3),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)initial value of the current flowing is 0.4 mA\n",
- "\n",
- " (b)time constant of the circuit 1.0 Sec\n",
- "\n",
- " (c)the value of the current one second after connection, 0.147 mA\n",
- "\n",
- " (d)the value of the capacitor voltage two seconds after connection 17.3 V\n",
- "\n",
- " (e)the time after connection when the resistor voltage is 15 V is 0.288 sec\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17.4, page no. 266</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of the resistor, and (b) the capacitor voltage 7 ms after connecting the circuit to a 10 V supply\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 0.5E-6;# in Farads\n",
- "V = 10;# in Volts\n",
- "tou = 0.012;# in secs\n",
- "t1 = 0.007;# in secs\n",
- "\n",
- "#calculation:\n",
- "R = tou/C\n",
- "Vc = V*(1-math.e**(-1*t1/tou))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)value of the resistor is \",R,\" ohm\"\n",
- "print \"\\n (b)capacitor voltage is \",round(Vc,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)value of the resistor is 24000.0 ohm\n",
- "\n",
- " (b)capacitor voltage is 4.42 V\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 267</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine: (a) the value of the capacitor, \n",
- "#(b) the time for the capacitor voltage to fall to 20 V, \n",
- "#(c) the current flowing when the capacitor has been discharging for 0.5 s, and \n",
- "#(d) the voltage drop across the resistor when the capacitor has been discharging for one second.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 50000;# in ohms\n",
- "V = 100;# in Volts\n",
- "Vc1 = 20;# in Volts\n",
- "tou = 0.8;# in secs\n",
- "t1 = 0.5;# in secs\n",
- "t2 = 1;# in secs\n",
- "\n",
- "#calculation:\n",
- "C = tou/R\n",
- "t = -1*tou*math.log(Vc1/V)\n",
- "I = V/R\n",
- "it1 = I*math.e**(-1*t1/tou)\n",
- "Vc = V*math.e**(-1*t2/tou)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the value of the capacitor is \",round((C/1E-6),2),\"uF\"\n",
- "print \"\\n (b)the time for the capacitor voltage to fall to 20 V is \",round(t,2),\" sec\"\n",
- "print \"\\n (c)the current flowing when the capacitor has been discharging for 0.5 s is \",round(it1*1000,2),\"mA\"\n",
- "print \"\\n (d)voltage drop across resistor when the capacitor has been discharging for one second is \",round(Vc,1),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the value of the capacitor is 16.0 uF\n",
- "\n",
- " (b)the time for the capacitor voltage to fall to 20 V is 1.29 sec\n",
- "\n",
- " (c)the current flowing when the capacitor has been discharging for 0.5 s is 1.07 mA\n",
- "\n",
- " (d)voltage drop across resistor when the capacitor has been discharging for one second is 28.7 V\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6 page no. 268</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the initial discharge current, \n",
- "#(b) the time constant of the circuit, and \n",
- "#(c) the minimum time required for the voltage across the capacitor to fall to less than 2 V\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 0.1E-6;# in Farads\n",
- "R = 4000;# in ohms\n",
- "V = 200;# in Volts\n",
- "Vc1 = 2;# in Volts\n",
- "\n",
- "#calculation:\n",
- "tou = R*C\n",
- "I = V/R\n",
- "t = -1*tou*math.log(Vc1/V)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \" \\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) initial discharge current is \",round(I,2),\" A\"\n",
- "print \"\\n (b)Time constant tou is \",round(tou,5),\" sec\"\n",
- "print \"\\n (c)min. time required for voltage across capacitor to fall to less than 2 V is \",round(t*1000,0),\" msec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " \n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) initial discharge current is 0.05 A\n",
- "\n",
- " (b)Time constant tou is 0.0004 sec\n",
- "\n",
- " (c)min. time required for voltage across capacitor to fall to less than 2 V is 2.0 msec\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 270</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#draw the current/time characteristic and \n",
- "#hence determine the value of current flowing at a time equal to two time constants and the time for the current to grow to 1.5 A\n",
- "from __future__ import division\n",
- "import math\n",
- "from pylab import *\n",
- "#initializing the variables:\n",
- "L = 0.1;# in Henry\n",
- "R = 20;# in ohms\n",
- "V = 60;# in Volts\n",
- "i2 = 1.5;# in Amperes\n",
- "\n",
- "#calculation:\n",
- "tou = L/R\n",
- "t1 = 2*tou\n",
- "\n",
- "t=[]\n",
- "i=[]\n",
- "I = V/R\n",
- "for h in range(250):\n",
- " t.append((h-1)/10000)\n",
- " k=(h-1)/10000\n",
- " i.append(I*(1 - math.e**(-1*k/tou)))\n",
- "plot(t,i,'-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('current(A)')\n",
- "show()\n",
- "i1 = I*(1 - math.e**(-1*t1/tou))\n",
- "t2 = -1*tou*math.log(1 - i2/I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \" \\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the value of current flowing at a time equal to two time constants is \",round(i1,2),\" A\"\n",
- "print \"\\n (b)the time for the current to grow to 1.5 A is \",round(t2,5),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Pnw/AhAkTGDZsGFlZWURHR9O6dWsWLlxoV7k+t2WLFSrx8XZXIiLSOFrLy08d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\nUYpOSazjAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x7b9d748>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " \n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the value of current flowing at a time equal to two time constants is 2.59 A\n",
- "\n",
- " (b)the time for the current to grow to 1.5 A is 0.00347 sec\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 271</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the final value of current, \n",
- "#(b) the time constant of the circuit, \n",
- "#(c) the value of current after a time equal to the time constant from the instant the supply oltage is connected, \n",
- "#(d) the expected time for the current to rise to within 1% of its final value.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.04;# in Henry\n",
- "R = 10;# in ohms\n",
- "V = 120;# in Volts\n",
- "\n",
- "#calculation:\n",
- "tou = L/R\n",
- "t1 = tou\n",
- "I = V/R\n",
- "i1 = I*(1 - math.e**(-1*t1/tou))\n",
- "i2 = 0.01*I\n",
- "t2 = -1*tou*(-5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the final value of current is \",round(I,2),\" A\"\n",
- "print \"\\n (b)time constant of the circuit is \",round(tou*1000,2),\"msec\"\n",
- "print \"\\n (c) value of current after a time equal to the time constant is \",round(i1,2),\" A\"\n",
- "print \"\\n (d)expected time for current to rise to within 0.01 times of its final value is \",round(t2*1000,2),\"msec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the final value of current is 12.0 A\n",
- "\n",
- " (b)time constant of the circuit is 4.0 msec\n",
- "\n",
- " (c) value of current after a time equal to the time constant is 7.59 A\n",
- "\n",
- " (d)expected time for current to rise to within 0.01 times of its final value is 20.0 msec\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 271</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate:\n",
- "#(a) the steady state value of current flowing in the winding,\n",
- "#(b) the time constant of the circuit,\n",
- "#(c) the value of the induced e.m.f. after 0.1 s,\n",
- "#(d) the time for the current to rise to 85% of its final value, and\n",
- "#(e) the value of the current after 0.3 s\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 3;# in Henry\n",
- "R = 15;# in ohms\n",
- "V = 120;# in Volts\n",
- "t1 = 0.1;# in secs\n",
- "t3 = 0.3;# in secs\n",
- "\n",
- "#calculation:\n",
- "tou = L/R\n",
- "I = V/R\n",
- "i2 = 0.85*I;# in amperes\n",
- "VL = V*math.e**(-1*t1/tou)\n",
- "t2 = -1*tou*math.log(1 - (i2/I))\n",
- "i3 = I*(1 - math.e**(-1*t3/tou))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Results \\n\\n\"\n",
- "print \"\\n (a) steady state value of current is \",round(I,2),\" A\"\n",
- "print \"\\n (b)time constant of the circuit is \",round(tou,2),\" sec\"\n",
- "print \"\\n (c)value of the induced e.m.f. after 0.1 s is \",round(VL,2),\" V\"\n",
- "print \"\\n (d) time for the current to rise to 0.85 times of its final values is \",round(t2,2),\" A\"\n",
- "print \"\\n (e)value of the current after 0.3 s is \",round(i3,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Results \n",
- "\n",
- "\n",
- "\n",
- " (a) steady state value of current is 8.0 A\n",
- "\n",
- " (b)time constant of the circuit is 0.2 sec\n",
- "\n",
- " (c)value of the induced e.m.f. after 0.1 s is 72.78 V\n",
- "\n",
- " (d) time for the current to rise to 0.85 times of its final values is 0.38 A\n",
- "\n",
- " (e)value of the current after 0.3 s is 6.21 A\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 273</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#draw the current/time characteristic when the supply is removed and replaced by a shorting link\n",
- "#determine (a) the current flowing in the winding 3 s after being shorted-out and (b) the time for the current to decay to 5 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "from pylab import *\n",
- "#initializing the variables:\n",
- "R = 15;# in ohms\n",
- "V = 110;# in Volts\n",
- "tou = 2;# in secs\n",
- "t1 = 3;# in secs\n",
- "i2 = 5;# in amperes\n",
- "\n",
- "#calculation:\n",
- "L = tou*R\n",
- "I = V/R\n",
- "\n",
- "t=[]\n",
- "i=[]\n",
- "for h in range(100):\n",
- " t.append((h-1)/10)\n",
- " k=(h-1)/10\n",
- " i.append(I*math.e**(-1*k/tou))\n",
- "plot(t,i,'-')\n",
- "xlabel('time(sec)')\n",
- "ylabel('current(A)')\n",
- "show()\n",
- "i1 = I*(math.e**(-1*t1/tou))\n",
- "t2 = -1*tou*log((i2/I))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \" \\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the current flowing in the winding 3 s after being shorted-out is \",round(i1,2),\" A\"\n",
- "print \"\\n (b)the time for the current to decay to 5 A is \",round(t2,2),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x77abf28>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " \n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the current flowing in the winding 3 s after being shorted-out is 1.64 A\n",
- "\n",
- " (b)the time for the current to decay to 5 A is 0.77 sec\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 273</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine: (a) the resistance of the coil,\n",
- "#(b) the current flowing in the circuit one second after the shorting link has been placed in the circuit, and \n",
- "#(c) the time taken for the current to fall to 10% of its initial value.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 6;# in Henry\n",
- "r = 10;# in ohms\n",
- "V = 120;# in Volts\n",
- "tou = 0.3;# in secs\n",
- "t1 = 1;# in secs\n",
- "\n",
- "#calculation:\n",
- "R = (L/tou) - r\n",
- "Rt = R + r\n",
- "I = V/Rt\n",
- "i2 = 0.1*I\n",
- "i1 = I*(math.e**(-1*t1/tou))\n",
- "t2 = -1*tou*math.log((i2/I))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \" \\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) resistance of the coil is \",round(R,2),\" ohm\"\n",
- "print \"\\n (b) current flowing in circuit one second after the shorting link has been placed is \",round(i1,2),\" A\"\n",
- "print \"\\n (c)the time for the current to decay to 0.1 times of initial value is \",round(t2,2),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " \n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) resistance of the coil is 10.0 ohm\n",
- "\n",
- " (b) current flowing in circuit one second after the shorting link has been placed is 0.21 A\n",
- "\n",
- " (c)the time for the current to decay to 0.1 times of initial value is 0.69 sec\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 274</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the time constant of the circuit and the steady-state value of the current flowing in the circuit. \n",
- "#Find (a) the current flowing in the circuit at a time equal to one time constant, \n",
- "#(b) the voltage drop across the inductor at a time equal to two time constants and \n",
- "#(c) the voltage drop across the resistor after a time equal to three time constants.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.2;# in Henry\n",
- "R = 1000;# in ohms\n",
- "V = 24;# in Volts\n",
- "t1 = 1*L/R;# in secs\n",
- "t2 = 2*L/R;# in secs\n",
- "t3 = 3*L/R;# in secs\n",
- "\n",
- "#calculation:\n",
- "tou = L/R\n",
- "I = V/R\n",
- "i1 = I*(1 - math.e**(-1*t1/tou))\n",
- "VL = V*(math.e**(-1*t2/tou))\n",
- "VR = V*(1 - math.e**(-1*t3/tou))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n time constant of circuit is \",round(tou*1000,6),\"msec, and steady-state value of current is \",round(I*1000,2),\"mA\"\n",
- "print \"\\n (a) urrent flowing in the circuit at a time equal to one time constant is \",round(i1*1000,2),\"mA\"\n",
- "print \"\\n (b) voltage drop across the inductor at a time equal to two time constants is \",round(VL,3),\" V\"\n",
- "print \"\\n (c)the voltage drop across the resistor after a time equal to three time constants is \",round(VR,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " time constant of circuit is 0.2 msec, and steady-state value of current is 24.0 mA\n",
- "\n",
- " (a) urrent flowing in the circuit at a time equal to one time constant is 15.17 mA\n",
- "\n",
- " (b) voltage drop across the inductor at a time equal to two time constants is 3.248 V\n",
- "\n",
- " (c)the voltage drop across the resistor after a time equal to three time constants is 22.81 V\n"
- ]
- }
- ],
- "prompt_number": 18
- }
- ],
- "metadata": {}
- }
- ]
-}
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- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 18: Operational amplifiers</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 279</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the output voltage of the amplifier.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vi2 = 2.45;# in Volts\n",
- "Vi1 = 2.35;# in Volts\n",
- "A0 = 120;# open-loop voltage gain\n",
- "\n",
- "#calculation:\n",
- "Vo = A0*(Vi2 - Vi1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the output voltage is \",round(Vo,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the output voltage is 12.0 V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 281</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the common-mode gain of an op amp\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vg = 150E3;# differential voltage gain \n",
- "CMRR = 90;# in dB\n",
- "\n",
- "#calculation:\n",
- "CMG = Vg/(10**(CMRR/20))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n common-mode gain is \",round(CMG,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " common-mode gain is 4.74"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 282</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the common-mode gain and the CMRR.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vg = 120;# differential voltage gain \n",
- "Vi = 3;# in Volts\n",
- "Vo = 0.024;# in Volts\n",
- "\n",
- "#calculation:\n",
- "CMG = Vo/Vi\n",
- "CMRR = 20*(1/2.303)*math.log(Vg/CMG)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n common-mode gain is \",round(CMG,3),\" and CMRR is \",round(CMRR,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " common-mode gain is 0.008 and CMRR is 83.51 dB"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 283</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the output voltage when the input voltage is: (a) +0.4 V (b) -1.2 V\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rf = 2000;# in ohms\n",
- "Ri = 1000;# in ohms\n",
- "Vi1 = 0.4;# in Volts\n",
- "Vi2 = -1.2;# in Volts\n",
- "\n",
- "#calculation:\n",
- "Vo1 = -1*Rf*Vi1/Ri\n",
- "Vo2 = -1*Rf*Vi2/Ri\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output voltage when the input voltage is 0.4V is \",round(Vo1,2),\" V \"\n",
- "print \" and when the input voltage is -1.2V is \",round(Vo2,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output voltage when the input voltage is 0.4V is -0.8 V \n",
- " and when the input voltage is -1.2V is 2.4 V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 283</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the voltage gain, and\n",
- "#(b) the output offset voltage due to the input bias current. \n",
- "#(c) How can the effect of input bias current be minimised?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ii = 100E-9;# in Amperes\n",
- "T = 20;# in \u00b0C\n",
- "Rf = 1E6;# in ohms\n",
- "Ri = 10000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "A = -1*Rf/Ri\n",
- "Vos = Ii*Ri*Rf/(Ri+Rf)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the voltage gain is \",round(A,2),\"\"\n",
- "print \"\\n (b)output offset voltage is \",round(Vos*1000,2),\" mV\"\n",
- "print \"\\n (c)The effect of input bias current can be minimised by ensuring \"\n",
- "print \"that both inputs have the same driving resistance.\" \n",
- "print \"This means that a resistance of value of 9.9 kohm (from part (b)) \"\n",
- "print \"should be placed between the non-inverting (+) terminal and earth.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the voltage gain is -100.0 \n",
- "\n",
- " (b)output offset voltage is 0.99 mV\n",
- "\n",
- " (c)The effect of input bias current can be minimised by ensuring \n",
- "that both inputs have the same driving resistance.\n",
- "This means that a resistance of value of 9.9 kohm (from part (b)) \n",
- "should be placed between the non-inverting (+) terminal and earth.\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 284</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design an inverting amplifier\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vg = 40;# in dB\n",
- "bf = 5000;# in Hz\n",
- "Ri = 10000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "A = 10**(Vg/20)\n",
- "Rf = A*Ri\n",
- "f = A*bf\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the voltage gain is \",round(A,2),\", Rf = \",round(Rf/1000,2),\"kohm and frequency = \",round(f/1000,2),\" kHz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the voltage gain is 100.0 , Rf = 1000.0 kohm and frequency = 500.0 kHz"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 286</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the voltage gain (b) the output voltage\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vi = -0.4;# in Volts\n",
- "R1 = 4700;# in ohms\n",
- "R2 = 10000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "A = 1 + (R2/R1)\n",
- "Vo = A*Vi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) the voltage gain is \",round(A,2),\"\"\n",
- "print \"\\n(b) output voltageis \",round(Vo,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) the voltage gain is 3.13 \n",
- "\n",
- "(b) output voltageis -1.25 V"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 287</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the output voltage, Vo\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 0.5;# in Volts\n",
- "V2 = 0.8;# in Volts\n",
- "V3 = 1.2;# in Volts\n",
- "R1 = 10000;# in ohms\n",
- "R2 = 20000;# in ohms\n",
- "R3 = 30000;# in ohms\n",
- "Rf = 50000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Vo = -1*Rf*(V1/R1 + V2/R2 + V3/R3)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output voltageis \",round(Vo,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output voltageis -6.5 V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 289</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "from scipy import integrate\n",
- "#initializing the variables:\n",
- "Vs = -0.75;# in Volts\n",
- "R = 200000;# in ohms\n",
- "C = 2.5E-6;# in Farads\n",
- "t = 0.1;# in secs\n",
- "\n",
- "#calculation:\n",
- "f = lambda x,a : a*1\n",
- "y, err = integrate.quad(f, 0, 0.1, args=(-0.75,))\n",
- "Vo = (-1/(C*R))*y\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output voltage is \",Vo,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output voltage is 0.15 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 290</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the output voltage Vo\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1a = 0.005;# in Volts\n",
- "V2a = 0;# in Volts\n",
- "V1b = 0;# in Volts\n",
- "V2b = 0.005;# in Volts\n",
- "V1c = 0.05;# in Volts\n",
- "V2c = 0.025;# in Volts\n",
- "V1d = 0.025;# in Volts\n",
- "V2d = 0.05;# in Volts\n",
- "R1 = 10000;# in ohms\n",
- "R2 = 10000;# in ohms\n",
- "R3 = 100000;# in ohms\n",
- "Rf = 100000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Vo1 = -1*Rf*V1a/R1\n",
- "Vo2 = (R3/(R2+R3))*(1 + (Rf/R1))*V2b\n",
- "Vo3 = -1*Rf*(V1c-V2c)/R1\n",
- "Vo4 = (R3/(R2+R3))*(1 + (Rf/R1))*(V2d-V1d)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)output voltage is \",round(Vo1,2),\" V\"\n",
- "print \"\\n (b)output voltage is \",round(Vo2,2),\" V\"\n",
- "print \"\\n (c)output voltage is \",round(Vo3,2),\" V\"\n",
- "print \"\\n (d)output voltage is \",round(Vo4,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)output voltage is -0.05 V\n",
- "\n",
- " (b)output voltage is 0.05 V\n",
- "\n",
- " (c)output voltage is -0.25 V\n",
- "\n",
- " (d)output voltage is 0.25 V"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_1.ipynb deleted file mode 100755 index f3f727b4..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_1.ipynb +++ /dev/null @@ -1,535 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:d16259f8430b68bdc5f61e2c9d378821ed7dc4c5b9f7144ffa85cfd5e017e63d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 18: Operational amplifiers</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 279</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the output voltage of the amplifier.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vi2 = 2.45;# in Volts\n",
- "Vi1 = 2.35;# in Volts\n",
- "A0 = 120;# open-loop voltage gain\n",
- "\n",
- "#calculation:\n",
- "Vo = A0*(Vi2 - Vi1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the output voltage is \",round(Vo,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the output voltage is 12.0 V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 281</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the common-mode gain of an op amp\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vg = 150E3;# differential voltage gain \n",
- "CMRR = 90;# in dB\n",
- "\n",
- "#calculation:\n",
- "CMG = Vg/(10**(CMRR/20))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n common-mode gain is \",round(CMG,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " common-mode gain is 4.74"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 282</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the common-mode gain and the CMRR.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vg = 120;# differential voltage gain \n",
- "Vi = 3;# in Volts\n",
- "Vo = 0.024;# in Volts\n",
- "\n",
- "#calculation:\n",
- "CMG = Vo/Vi\n",
- "CMRR = 20*(1/2.303)*math.log(Vg/CMG)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n common-mode gain is \",round(CMG,3),\" and CMRR is \",round(CMRR,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " common-mode gain is 0.008 and CMRR is 83.51 dB"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 283</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the output voltage when the input voltage is: (a) +0.4 V (b) -1.2 V\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rf = 2000;# in ohms\n",
- "Ri = 1000;# in ohms\n",
- "Vi1 = 0.4;# in Volts\n",
- "Vi2 = -1.2;# in Volts\n",
- "\n",
- "#calculation:\n",
- "Vo1 = -1*Rf*Vi1/Ri\n",
- "Vo2 = -1*Rf*Vi2/Ri\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output voltage when the input voltage is 0.4V is \",round(Vo1,2),\" V \"\n",
- "print \" and when the input voltage is -1.2V is \",round(Vo2,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output voltage when the input voltage is 0.4V is -0.8 V \n",
- " and when the input voltage is -1.2V is 2.4 V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 283</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the voltage gain, and\n",
- "#(b) the output offset voltage due to the input bias current. \n",
- "#(c) How can the effect of input bias current be minimised?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ii = 100E-9;# in Amperes\n",
- "T = 20;# in \u00b0C\n",
- "Rf = 1E6;# in ohms\n",
- "Ri = 10000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "A = -1*Rf/Ri\n",
- "Vos = Ii*Ri*Rf/(Ri+Rf)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the voltage gain is \",round(A,2),\"\"\n",
- "print \"\\n (b)output offset voltage is \",round(Vos*1000,2),\" mV\"\n",
- "print \"\\n (c)The effect of input bias current can be minimised by ensuring \"\n",
- "print \"that both inputs have the same driving resistance.\" \n",
- "print \"This means that a resistance of value of 9.9 kohm (from part (b)) \"\n",
- "print \"should be placed between the non-inverting (+) terminal and earth.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the voltage gain is -100.0 \n",
- "\n",
- " (b)output offset voltage is 0.99 mV\n",
- "\n",
- " (c)The effect of input bias current can be minimised by ensuring \n",
- "that both inputs have the same driving resistance.\n",
- "This means that a resistance of value of 9.9 kohm (from part (b)) \n",
- "should be placed between the non-inverting (+) terminal and earth.\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 284</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design an inverting amplifier\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vg = 40;# in dB\n",
- "bf = 5000;# in Hz\n",
- "Ri = 10000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "A = 10**(Vg/20)\n",
- "Rf = A*Ri\n",
- "f = A*bf\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the voltage gain is \",round(A,2),\", Rf = \",round(Rf/1000,2),\"kohm and frequency = \",round(f/1000,2),\" kHz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the voltage gain is 100.0 , Rf = 1000.0 kohm and frequency = 500.0 kHz"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 286</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the voltage gain (b) the output voltage\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vi = -0.4;# in Volts\n",
- "R1 = 4700;# in ohms\n",
- "R2 = 10000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "A = 1 + (R2/R1)\n",
- "Vo = A*Vi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) the voltage gain is \",round(A,2),\"\"\n",
- "print \"\\n(b) output voltageis \",round(Vo,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) the voltage gain is 3.13 \n",
- "\n",
- "(b) output voltageis -1.25 V"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 287</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the output voltage, Vo\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 0.5;# in Volts\n",
- "V2 = 0.8;# in Volts\n",
- "V3 = 1.2;# in Volts\n",
- "R1 = 10000;# in ohms\n",
- "R2 = 20000;# in ohms\n",
- "R3 = 30000;# in ohms\n",
- "Rf = 50000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Vo = -1*Rf*(V1/R1 + V2/R2 + V3/R3)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output voltageis \",round(Vo,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output voltageis -6.5 V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 289</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "from scipy import integrate\n",
- "#initializing the variables:\n",
- "Vs = -0.75;# in Volts\n",
- "R = 200000;# in ohms\n",
- "C = 2.5E-6;# in Farads\n",
- "t = 0.1;# in secs\n",
- "\n",
- "#calculation:\n",
- "f = lambda x,a : a*1\n",
- "y, err = integrate.quad(f, 0, 0.1, args=(-0.75,))\n",
- "Vo = (-1/(C*R))*y\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output voltage is \",Vo,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output voltage is 0.15 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 290</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the output voltage Vo\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1a = 0.005;# in Volts\n",
- "V2a = 0;# in Volts\n",
- "V1b = 0;# in Volts\n",
- "V2b = 0.005;# in Volts\n",
- "V1c = 0.05;# in Volts\n",
- "V2c = 0.025;# in Volts\n",
- "V1d = 0.025;# in Volts\n",
- "V2d = 0.05;# in Volts\n",
- "R1 = 10000;# in ohms\n",
- "R2 = 10000;# in ohms\n",
- "R3 = 100000;# in ohms\n",
- "Rf = 100000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Vo1 = -1*Rf*V1a/R1\n",
- "Vo2 = (R3/(R2+R3))*(1 + (Rf/R1))*V2b\n",
- "Vo3 = -1*Rf*(V1c-V2c)/R1\n",
- "Vo4 = (R3/(R2+R3))*(1 + (Rf/R1))*(V2d-V1d)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)output voltage is \",round(Vo1,2),\" V\"\n",
- "print \"\\n (b)output voltage is \",round(Vo2,2),\" V\"\n",
- "print \"\\n (c)output voltage is \",round(Vo3,2),\" V\"\n",
- "print \"\\n (d)output voltage is \",round(Vo4,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)output voltage is -0.05 V\n",
- "\n",
- " (b)output voltage is 0.05 V\n",
- "\n",
- " (c)output voltage is -0.25 V\n",
- "\n",
- " (d)output voltage is 0.25 V"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb deleted file mode 100755 index f3f727b4..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb +++ /dev/null @@ -1,535 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:d16259f8430b68bdc5f61e2c9d378821ed7dc4c5b9f7144ffa85cfd5e017e63d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 18: Operational amplifiers</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 279</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the output voltage of the amplifier.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vi2 = 2.45;# in Volts\n",
- "Vi1 = 2.35;# in Volts\n",
- "A0 = 120;# open-loop voltage gain\n",
- "\n",
- "#calculation:\n",
- "Vo = A0*(Vi2 - Vi1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the output voltage is \",round(Vo,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the output voltage is 12.0 V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 281</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the common-mode gain of an op amp\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vg = 150E3;# differential voltage gain \n",
- "CMRR = 90;# in dB\n",
- "\n",
- "#calculation:\n",
- "CMG = Vg/(10**(CMRR/20))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n common-mode gain is \",round(CMG,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " common-mode gain is 4.74"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 282</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the common-mode gain and the CMRR.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vg = 120;# differential voltage gain \n",
- "Vi = 3;# in Volts\n",
- "Vo = 0.024;# in Volts\n",
- "\n",
- "#calculation:\n",
- "CMG = Vo/Vi\n",
- "CMRR = 20*(1/2.303)*math.log(Vg/CMG)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n common-mode gain is \",round(CMG,3),\" and CMRR is \",round(CMRR,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " common-mode gain is 0.008 and CMRR is 83.51 dB"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 283</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the output voltage when the input voltage is: (a) +0.4 V (b) -1.2 V\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rf = 2000;# in ohms\n",
- "Ri = 1000;# in ohms\n",
- "Vi1 = 0.4;# in Volts\n",
- "Vi2 = -1.2;# in Volts\n",
- "\n",
- "#calculation:\n",
- "Vo1 = -1*Rf*Vi1/Ri\n",
- "Vo2 = -1*Rf*Vi2/Ri\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output voltage when the input voltage is 0.4V is \",round(Vo1,2),\" V \"\n",
- "print \" and when the input voltage is -1.2V is \",round(Vo2,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output voltage when the input voltage is 0.4V is -0.8 V \n",
- " and when the input voltage is -1.2V is 2.4 V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 283</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the voltage gain, and\n",
- "#(b) the output offset voltage due to the input bias current. \n",
- "#(c) How can the effect of input bias current be minimised?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ii = 100E-9;# in Amperes\n",
- "T = 20;# in \u00b0C\n",
- "Rf = 1E6;# in ohms\n",
- "Ri = 10000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "A = -1*Rf/Ri\n",
- "Vos = Ii*Ri*Rf/(Ri+Rf)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the voltage gain is \",round(A,2),\"\"\n",
- "print \"\\n (b)output offset voltage is \",round(Vos*1000,2),\" mV\"\n",
- "print \"\\n (c)The effect of input bias current can be minimised by ensuring \"\n",
- "print \"that both inputs have the same driving resistance.\" \n",
- "print \"This means that a resistance of value of 9.9 kohm (from part (b)) \"\n",
- "print \"should be placed between the non-inverting (+) terminal and earth.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the voltage gain is -100.0 \n",
- "\n",
- " (b)output offset voltage is 0.99 mV\n",
- "\n",
- " (c)The effect of input bias current can be minimised by ensuring \n",
- "that both inputs have the same driving resistance.\n",
- "This means that a resistance of value of 9.9 kohm (from part (b)) \n",
- "should be placed between the non-inverting (+) terminal and earth.\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 284</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design an inverting amplifier\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vg = 40;# in dB\n",
- "bf = 5000;# in Hz\n",
- "Ri = 10000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "A = 10**(Vg/20)\n",
- "Rf = A*Ri\n",
- "f = A*bf\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the voltage gain is \",round(A,2),\", Rf = \",round(Rf/1000,2),\"kohm and frequency = \",round(f/1000,2),\" kHz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the voltage gain is 100.0 , Rf = 1000.0 kohm and frequency = 500.0 kHz"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 286</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the voltage gain (b) the output voltage\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vi = -0.4;# in Volts\n",
- "R1 = 4700;# in ohms\n",
- "R2 = 10000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "A = 1 + (R2/R1)\n",
- "Vo = A*Vi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) the voltage gain is \",round(A,2),\"\"\n",
- "print \"\\n(b) output voltageis \",round(Vo,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) the voltage gain is 3.13 \n",
- "\n",
- "(b) output voltageis -1.25 V"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 287</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the output voltage, Vo\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 0.5;# in Volts\n",
- "V2 = 0.8;# in Volts\n",
- "V3 = 1.2;# in Volts\n",
- "R1 = 10000;# in ohms\n",
- "R2 = 20000;# in ohms\n",
- "R3 = 30000;# in ohms\n",
- "Rf = 50000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Vo = -1*Rf*(V1/R1 + V2/R2 + V3/R3)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output voltageis \",round(Vo,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output voltageis -6.5 V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 289</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "from scipy import integrate\n",
- "#initializing the variables:\n",
- "Vs = -0.75;# in Volts\n",
- "R = 200000;# in ohms\n",
- "C = 2.5E-6;# in Farads\n",
- "t = 0.1;# in secs\n",
- "\n",
- "#calculation:\n",
- "f = lambda x,a : a*1\n",
- "y, err = integrate.quad(f, 0, 0.1, args=(-0.75,))\n",
- "Vo = (-1/(C*R))*y\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output voltage is \",Vo,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output voltage is 0.15 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 290</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the output voltage Vo\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1a = 0.005;# in Volts\n",
- "V2a = 0;# in Volts\n",
- "V1b = 0;# in Volts\n",
- "V2b = 0.005;# in Volts\n",
- "V1c = 0.05;# in Volts\n",
- "V2c = 0.025;# in Volts\n",
- "V1d = 0.025;# in Volts\n",
- "V2d = 0.05;# in Volts\n",
- "R1 = 10000;# in ohms\n",
- "R2 = 10000;# in ohms\n",
- "R3 = 100000;# in ohms\n",
- "Rf = 100000;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Vo1 = -1*Rf*V1a/R1\n",
- "Vo2 = (R3/(R2+R3))*(1 + (Rf/R1))*V2b\n",
- "Vo3 = -1*Rf*(V1c-V2c)/R1\n",
- "Vo4 = (R3/(R2+R3))*(1 + (Rf/R1))*(V2d-V1d)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)output voltage is \",round(Vo1,2),\" V\"\n",
- "print \"\\n (b)output voltage is \",round(Vo2,2),\" V\"\n",
- "print \"\\n (c)output voltage is \",round(Vo3,2),\" V\"\n",
- "print \"\\n (d)output voltage is \",round(Vo4,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)output voltage is -0.05 V\n",
- "\n",
- " (b)output voltage is 0.05 V\n",
- "\n",
- " (c)output voltage is -0.25 V\n",
- "\n",
- " (d)output voltage is 0.25 V"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_18.ipynb index 73f373a8..73f373a8 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_18.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint.ipynb deleted file mode 100755 index 8285878f..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint.ipynb +++ /dev/null @@ -1,974 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 19: Three phase systems</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 299</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the system phase voltage, (b) the phase current and (c) the line current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vl = 415;# in Volts\n",
- "Rp = 30;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Vp = Vl/(3**0.5)\n",
- "Ip = Vp/Rp\n",
- "Il = Ip\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the system phase voltage is \",round(Vp,2),\" V\"\n",
- "print \"\\n (b)phase current is \",round(Ip,2),\" A\"\n",
- "print \"\\n (c)line current is \",round(Il,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the system phase voltage is 239.6 V\n",
- "\n",
- " (b)phase current is 7.99 A\n",
- "\n",
- " (c)line current is 7.99 A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 299</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the line voltage if the supply frequency is 50 Hz\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "L = 0.1273;# in Henry\n",
- "Ip = 5.08;# in Amperes\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "Il = Ip\n",
- "Vp = Ip*Zp\n",
- "Vl = Vp*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)line voltage is \",round(Vl,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)line voltage is 439.89 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 301</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current in each line and (b) the current in the neutral conductor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "PR = 24000;# in Watt\n",
- "Py = 18000;# in Watt\n",
- "Pb = 12000;# in Watt\n",
- "VR = 240;# in Volts\n",
- "Vy = 240;# in Volts\n",
- "Vb = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For a star-connected system VL = Vp*(3**0.5)\n",
- "Vp = V/(3**0.5)\n",
- "phir = 90*math.pi/180\n",
- "phiy = 330*math.pi/180\n",
- "phib = 210*math.pi/180\n",
- " # I = P/V for a resistive load\n",
- "IR = PR/VR\n",
- "Iy = Py/Vy\n",
- "Ib = Pb/Vb\n",
- "Inh = IR*math.cos(phir) + Ib*math.cos(phib) + Iy*math.cos(phiy)\n",
- "Inv = IR*math.sin(phir) + Ib*math.sin(phib) + Iy*math.sin(phiy)\n",
- "In = (Inh**2 + Inv**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)cuurnt in R line is \",round(IR,2),\" A, cuurnt in Y line is \",round(Iy,2),\" A \"\n",
- "print \"and cuurnt in B line is \",round(Ib,2),\" A\"\n",
- "print \"\\n (b)cuurnt in neutral line is \",round(In,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)cuurnt in R line is 100.0 A, cuurnt in Y line is 75.0 A and cuurnt in B line is 50.0 A\n",
- "\n",
- " (b)cuurnt in neutral line is 43.3 A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 302</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the phase current, and (b) the line current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "L = 0.1273;# in Henry\n",
- "VL = 440;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "Vp = VL\n",
- " #Phase current\n",
- "Ip = Vp/Zp\n",
- " #For a delta connection,\n",
- "IL = Ip*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the phase current \",round(Ip,2),\" A\"\n",
- "print \"\\n (b)line current \",round(IL,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the phase current 8.8 A\n",
- "\n",
- " (b)line current 15.24 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 302</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitance of each of the capacitors.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "IL = 15;# in Amperes\n",
- "VL = 415;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #For a delta connection\n",
- "Ip = IL/(3**0.5)#phase current\n",
- "Vp = VL\n",
- " #Capacitive reactance per phase\n",
- "Xc = Vp/Ip\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E6,2),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 66.43 uF"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 303</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate for each connection (a) the line and phase voltages and (b) the phase and line currents.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 3;# in ohms\n",
- "XL = 4;# in ohms\n",
- "VL = 415;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For a star connection:\n",
- " #IL = Ip\n",
- " #VL = Vp*(3**0.5)\n",
- "VLs = VL\n",
- "Vps = VLs/(3**0.5)\n",
- " #Impedance per phase,\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "Ips = Vps/Zp\n",
- "ILs = Ips\n",
- " #For a delta connection:\n",
- " #VL = Vp\n",
- " #IL = Ip*(3**0.5)\n",
- "VLd = VL\n",
- "Vpd = VLd\n",
- "Ipd = Vpd/Zp\n",
- "ILd = Ipd*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the line voltage for star connection is \",round(VLs,2),\" V \"\n",
- "print \"and the phase voltage for star connection is \",round(Vps,2),\" V \"\n",
- "print \"and the line voltage for delta connection is \",round(VLd,2),\" V \"\n",
- "print \"and the phase voltage for delta connection is \",round(Vpd,2),\" V\"\n",
- "print \"\\n (b)the line current for star connection is \",round(ILs,2),\" A \"\n",
- "print \"and the phase current for star connection is \",round(Ips,2),\" A \"\n",
- "print \"and the line current for delta connection is \",round(ILd,2),\" A \"\n",
- "print \"and the phase current for delta connection is \",round(Ipd,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the line voltage for star connection is 415.0 V \n",
- "and the phase voltage for star connection is 239.6 V \n",
- "and the line voltage for delta connection is 415.0 V \n",
- "and the phase voltage for delta connection is 415.0 V\n",
- "\n",
- " (b)the line current for star connection is 47.92 A \n",
- "and the phase current for star connection is 47.92 A \n",
- "and the line current for delta connection is 143.76 A \n",
- "and the phase current for delta connection is 83.0 A\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 304</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the total power dissipated by the resistors.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rp = 12;# in ohms\n",
- "VL = 415;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "Vp = VL/(3**0.5)# since the resistors are star-connected\n",
- " #Phase current, Ip\n",
- "Zp = Rp\n",
- "Ip = Vp/Zp\n",
- " #For a star connection\n",
- "IL = Ip\n",
- " # For a purely resistive load, the power factor cos(phi) = 1\n",
- "pf = 1\n",
- "P = VL*IL*(3**0.5)*pf\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)total power dissipated by the resistors is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)total power dissipated by the resistors is 14352.08 W"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 304</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the power factor of the system.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "P = 5000;# in Watts\n",
- "IL = 8.6;# in amperes\n",
- "VL = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pf = P/(VL*IL*(3**0.5))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n power factor is \",round(pf,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " power factor is 0.839"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 304</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the total power dissipated in each case.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "L = 0.042;# in Henry\n",
- "VL = 415;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #For a star connection:\n",
- " #IL = Ip\n",
- " #VL = Vp*(3**0.5)\n",
- "XL = 2*math.pi*f*L\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "VLs = VL\n",
- "Vps = VLs/(3**0.5)\n",
- " #Impedance per phase,\n",
- "Ips = Vps/Zp\n",
- "ILs = Ips\n",
- " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pfs = R/Zp\n",
- "Ps = VLs*ILs*(3**0.5)*pfs\n",
- "\n",
- " #For a delta connection:\n",
- " #VL = Vp\n",
- " #IL = Ip*(3**0.5)\n",
- "VLd = VL\n",
- "Vpd = VLd\n",
- "Ipd = Vpd/Zp\n",
- "ILd = Ipd*(3**0.5)\n",
- " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pfd = R/Zp\n",
- "Pd = VLd*ILd*(3**0.5)*pfd\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n total power dissipated in star is \",round(Ps,2),\" W and in delta is \",round(Pd,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " total power dissipated in star is 6283.29 W and in delta is 18849.88 W"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 305</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the power input, (b) the line current and (c) the phase current\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Po = 12750;# in Watts\n",
- "pf = 0.77;# power factor\n",
- "eff = 0.85;\n",
- "VL = 415;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #eff = power_out/power_in\n",
- "Pi = Po/eff\n",
- " #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "IL = Pi/(VL*(3**0.5)*pf)# line current\n",
- " #For a delta connection:\n",
- " #IL = Ip*(3**0.5)\n",
- "Ip = IL/(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)power input is \",round(Pi,2),\" W\"\n",
- "print \"\\n (b)line current is \",round(IL,2),\" A\"\n",
- "print \"\\n (c)phase current is \",round(Ip,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)power input is 15000.0 W\n",
- "\n",
- " (b)line current is 27.1 A\n",
- "\n",
- " (c)phase current is 15.65 A"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 308</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current supplied by the alternator and (b) the output power and the kVA of the alternator\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "XL = 40;# in ohms\n",
- "VL = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- " #a delta-connected load\n",
- "Vp = VL\n",
- " #Phase current\n",
- "Ip = Vp/Zp\n",
- "IL = Ip*(3**0.5)\n",
- " #Alternator output power is equal to the power dissipated by the load.\n",
- " #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pf = R/Zp\n",
- "P = VL*IL*(3**0.5)*pf\n",
- " #Alternator output kVA,\n",
- "S = VL*IL*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the current supplied by the alternator is \",round(IL,2),\" A\"\n",
- "print \"\\n (b)output power is \",round(P/1000,2),\"KW and kVA of the alternator is \",round(S/1000,2),\"kVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the current supplied by the alternator is 13.86 A\n",
- "\n",
- " (b)output power is 5.76 KW and kVA of the alternator is 9.6 kVA"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 308</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the phase current, (b) the line current, (c) the total power dissipated and \n",
- "#(d) the kVA rating of the load. Draw the complete phasor diagram for the load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "C = 80E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "VL = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Capacitive reactance\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Zp = (R*R + Xc*Xc)**0.5\n",
- "pf = R/Zp\n",
- " #a delta-connected load\n",
- "Vp = VL\n",
- " #Phase current\n",
- "Ip = Vp/Zp\n",
- "IL = Ip*(3**0.5)\n",
- " #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "P = VL*IL*(3**0.5)*pf\n",
- " #Alternator output kVA,\n",
- "S = VL*IL*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the phase current is \",round(Ip,2),\" A\"\n",
- "print \"\\n (b)the line current is \",round(IL,2),\" A\"\n",
- "print \"\\n (c) power is \",round(P/1000,2),\"kW\"\n",
- "print \"\\n (d)kVA of the alternator is \", round(S/1000,2),\"kVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the phase current is 8.03 A\n",
- "\n",
- " (b)the line current is 13.9 A\n",
- "\n",
- " (c) power is 5.8 kW\n",
- "\n",
- " (d)kVA of the alternator is 9.63 kVA"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 309</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the total power input and (b) the load power factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Pi1 = 8000;# in Watts\n",
- "Pi2 = 4000;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Total input power\n",
- "Pi = Pi1 + Pi2\n",
- "phi = math.atan((Pi1 - Pi2)*(3**0.5)/(Pi1 + Pi2))\n",
- " #Power factor\n",
- "pf = math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)power input is \",round(Pi,2),\"W\"\n",
- "print \"\\n (b)power factor is \",round(pf,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)power input is 12000.0 W\n",
- "\n",
- " (b)power factor is 0.87"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 310</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the readings of each wattmeter.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Pi = 12000;# in Watts\n",
- "pf = 0.6;# power factor\n",
- "\n",
- "#calculation:\n",
- " #If the two wattmeters indicate Pi1 and Pi2 respectively\n",
- " # Pit = Pi1 + Pi2\n",
- "Pit = Pi\n",
- " # Pid = Pi1 - Pi2\n",
- " #power factor = 0.6 = cos(phi)\n",
- "phi = math.acos(pf)\n",
- "Pid = Pit*math.tan(phi)/(3**0.5)\n",
- " #Hence wattmeter 1 reads\n",
- "Pi1 = (Pid + Pit)/2\n",
- " #wattmeter 2 reads\n",
- "Pi2 = Pit - Pi1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reading in each wattameter are \",round(Pi1,2),\"W and \",round(Pi2,2),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reading in each wattameter are 10618.8 W and 1381.2 W"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 310</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the input power and (b) the load power factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Pi1 = 10000;# in Watts\n",
- "Pi2 = -3000;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Total input power\n",
- "Pi = Pi1 + Pi2\n",
- "phi = math.atan((Pi1 - Pi2)*(3**0.5)/(Pi1 + Pi2))\n",
- " #Power factor\n",
- "pf = math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)power input is \",round(Pi,2),\"W\"\n",
- "print \"\\n (b)power factor is \",round(pf,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)power input is 7000.0 W\n",
- "\n",
- " (b)power factor is 0.3"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 311</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate for each connection the readings on each of two wattmeters connected to measure the power by the two-wattmeter method.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 8; # in ohms\n",
- "XL = 8; # in ohms\n",
- "VL = 415; # in Volts\n",
- "\n",
- "#calculation:\n",
- "#For a star connection:\n",
- "#IL = Ip\n",
- "#VL = Vp*(3**0.5)\n",
- "VLs = VL\n",
- "Vps = VLs/(3**0.5)\n",
- "#Impedance per phase,\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "Ips = Vps/Zp\n",
- "ILs = Ips\n",
- "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pf = R/Zp\n",
- "Ps = VLs*ILs*(3**0.5)*pf\n",
- "#If wattmeter readings are P1 and P2 then P1 + P2 = Pst\n",
- "Pst = Ps\n",
- "# Pid = Pi1 - Pi2\n",
- "phi = math.acos(pf)\n",
- "Psd = Pst*math.tan(phi)/(3**0.5)\n",
- "#Hence wattmeter 1 reads\n",
- "Ps1 = (Psd + Pst)/2\n",
- "#wattmeter 2 reads\n",
- "Ps2 = Pst - Ps1\n",
- "\n",
- "#For a delta connection:\n",
- "#VL = Vp\n",
- "#IL = Ip*(3**0.5)\n",
- "VLd = VL\n",
- "Vpd = VLd\n",
- "Ipd = Vpd/Zp\n",
- "ILd = Ipd*(3**0.5)\n",
- "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "Pd = VLd*ILd*(3**0.5)*pf\n",
- "#If wattmeter readings are P1 and P2 then P1 + P2 = Pdt\n",
- "Pdt = Pd\n",
- "# Pid = Pi1 - Pi2\n",
- "Pdd = Pdt*math.tan(phi)/(3**0.5)\n",
- "#Hence wattmeter 1 reads\n",
- "Pd1 = (Pdd + Pdt)/2\n",
- "#wattmeter 2 reads\n",
- "Pd2 = Pdt - Pd1\n",
- "\n",
- "#results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)When the coils are star-connected the wattmeter readings are\", round(Ps1,2),\"W and \",round(Ps2,2),\"W\"\n",
- "print \"(b)When the coils are delta-connected the wattmeter readings are are\", round(Pd1,2),\"W and\", round(Pd2,2),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)When the coils are star-connected the wattmeter readings are 8489.35 W and 2274.71 W\n",
- "(b)When the coils are delta-connected the wattmeter readings are are 25468.05 W and 6824.14 W"
- ]
- }
- ],
- "prompt_number": 17
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb deleted file mode 100755 index 8285878f..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb +++ /dev/null @@ -1,974 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 19: Three phase systems</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 299</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the system phase voltage, (b) the phase current and (c) the line current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vl = 415;# in Volts\n",
- "Rp = 30;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Vp = Vl/(3**0.5)\n",
- "Ip = Vp/Rp\n",
- "Il = Ip\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the system phase voltage is \",round(Vp,2),\" V\"\n",
- "print \"\\n (b)phase current is \",round(Ip,2),\" A\"\n",
- "print \"\\n (c)line current is \",round(Il,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the system phase voltage is 239.6 V\n",
- "\n",
- " (b)phase current is 7.99 A\n",
- "\n",
- " (c)line current is 7.99 A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 299</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the line voltage if the supply frequency is 50 Hz\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "L = 0.1273;# in Henry\n",
- "Ip = 5.08;# in Amperes\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "Il = Ip\n",
- "Vp = Ip*Zp\n",
- "Vl = Vp*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)line voltage is \",round(Vl,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)line voltage is 439.89 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 301</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current in each line and (b) the current in the neutral conductor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "PR = 24000;# in Watt\n",
- "Py = 18000;# in Watt\n",
- "Pb = 12000;# in Watt\n",
- "VR = 240;# in Volts\n",
- "Vy = 240;# in Volts\n",
- "Vb = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For a star-connected system VL = Vp*(3**0.5)\n",
- "Vp = V/(3**0.5)\n",
- "phir = 90*math.pi/180\n",
- "phiy = 330*math.pi/180\n",
- "phib = 210*math.pi/180\n",
- " # I = P/V for a resistive load\n",
- "IR = PR/VR\n",
- "Iy = Py/Vy\n",
- "Ib = Pb/Vb\n",
- "Inh = IR*math.cos(phir) + Ib*math.cos(phib) + Iy*math.cos(phiy)\n",
- "Inv = IR*math.sin(phir) + Ib*math.sin(phib) + Iy*math.sin(phiy)\n",
- "In = (Inh**2 + Inv**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)cuurnt in R line is \",round(IR,2),\" A, cuurnt in Y line is \",round(Iy,2),\" A \"\n",
- "print \"and cuurnt in B line is \",round(Ib,2),\" A\"\n",
- "print \"\\n (b)cuurnt in neutral line is \",round(In,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)cuurnt in R line is 100.0 A, cuurnt in Y line is 75.0 A and cuurnt in B line is 50.0 A\n",
- "\n",
- " (b)cuurnt in neutral line is 43.3 A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 302</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the phase current, and (b) the line current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "L = 0.1273;# in Henry\n",
- "VL = 440;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "Vp = VL\n",
- " #Phase current\n",
- "Ip = Vp/Zp\n",
- " #For a delta connection,\n",
- "IL = Ip*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the phase current \",round(Ip,2),\" A\"\n",
- "print \"\\n (b)line current \",round(IL,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the phase current 8.8 A\n",
- "\n",
- " (b)line current 15.24 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 302</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitance of each of the capacitors.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "IL = 15;# in Amperes\n",
- "VL = 415;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #For a delta connection\n",
- "Ip = IL/(3**0.5)#phase current\n",
- "Vp = VL\n",
- " #Capacitive reactance per phase\n",
- "Xc = Vp/Ip\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E6,2),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 66.43 uF"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 303</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate for each connection (a) the line and phase voltages and (b) the phase and line currents.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 3;# in ohms\n",
- "XL = 4;# in ohms\n",
- "VL = 415;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For a star connection:\n",
- " #IL = Ip\n",
- " #VL = Vp*(3**0.5)\n",
- "VLs = VL\n",
- "Vps = VLs/(3**0.5)\n",
- " #Impedance per phase,\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "Ips = Vps/Zp\n",
- "ILs = Ips\n",
- " #For a delta connection:\n",
- " #VL = Vp\n",
- " #IL = Ip*(3**0.5)\n",
- "VLd = VL\n",
- "Vpd = VLd\n",
- "Ipd = Vpd/Zp\n",
- "ILd = Ipd*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the line voltage for star connection is \",round(VLs,2),\" V \"\n",
- "print \"and the phase voltage for star connection is \",round(Vps,2),\" V \"\n",
- "print \"and the line voltage for delta connection is \",round(VLd,2),\" V \"\n",
- "print \"and the phase voltage for delta connection is \",round(Vpd,2),\" V\"\n",
- "print \"\\n (b)the line current for star connection is \",round(ILs,2),\" A \"\n",
- "print \"and the phase current for star connection is \",round(Ips,2),\" A \"\n",
- "print \"and the line current for delta connection is \",round(ILd,2),\" A \"\n",
- "print \"and the phase current for delta connection is \",round(Ipd,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the line voltage for star connection is 415.0 V \n",
- "and the phase voltage for star connection is 239.6 V \n",
- "and the line voltage for delta connection is 415.0 V \n",
- "and the phase voltage for delta connection is 415.0 V\n",
- "\n",
- " (b)the line current for star connection is 47.92 A \n",
- "and the phase current for star connection is 47.92 A \n",
- "and the line current for delta connection is 143.76 A \n",
- "and the phase current for delta connection is 83.0 A\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 304</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the total power dissipated by the resistors.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rp = 12;# in ohms\n",
- "VL = 415;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "Vp = VL/(3**0.5)# since the resistors are star-connected\n",
- " #Phase current, Ip\n",
- "Zp = Rp\n",
- "Ip = Vp/Zp\n",
- " #For a star connection\n",
- "IL = Ip\n",
- " # For a purely resistive load, the power factor cos(phi) = 1\n",
- "pf = 1\n",
- "P = VL*IL*(3**0.5)*pf\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)total power dissipated by the resistors is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)total power dissipated by the resistors is 14352.08 W"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 304</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the power factor of the system.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "P = 5000;# in Watts\n",
- "IL = 8.6;# in amperes\n",
- "VL = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pf = P/(VL*IL*(3**0.5))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n power factor is \",round(pf,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " power factor is 0.839"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 304</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the total power dissipated in each case.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "L = 0.042;# in Henry\n",
- "VL = 415;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #For a star connection:\n",
- " #IL = Ip\n",
- " #VL = Vp*(3**0.5)\n",
- "XL = 2*math.pi*f*L\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "VLs = VL\n",
- "Vps = VLs/(3**0.5)\n",
- " #Impedance per phase,\n",
- "Ips = Vps/Zp\n",
- "ILs = Ips\n",
- " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pfs = R/Zp\n",
- "Ps = VLs*ILs*(3**0.5)*pfs\n",
- "\n",
- " #For a delta connection:\n",
- " #VL = Vp\n",
- " #IL = Ip*(3**0.5)\n",
- "VLd = VL\n",
- "Vpd = VLd\n",
- "Ipd = Vpd/Zp\n",
- "ILd = Ipd*(3**0.5)\n",
- " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pfd = R/Zp\n",
- "Pd = VLd*ILd*(3**0.5)*pfd\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n total power dissipated in star is \",round(Ps,2),\" W and in delta is \",round(Pd,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " total power dissipated in star is 6283.29 W and in delta is 18849.88 W"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 305</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the power input, (b) the line current and (c) the phase current\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Po = 12750;# in Watts\n",
- "pf = 0.77;# power factor\n",
- "eff = 0.85;\n",
- "VL = 415;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #eff = power_out/power_in\n",
- "Pi = Po/eff\n",
- " #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "IL = Pi/(VL*(3**0.5)*pf)# line current\n",
- " #For a delta connection:\n",
- " #IL = Ip*(3**0.5)\n",
- "Ip = IL/(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)power input is \",round(Pi,2),\" W\"\n",
- "print \"\\n (b)line current is \",round(IL,2),\" A\"\n",
- "print \"\\n (c)phase current is \",round(Ip,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)power input is 15000.0 W\n",
- "\n",
- " (b)line current is 27.1 A\n",
- "\n",
- " (c)phase current is 15.65 A"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 308</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current supplied by the alternator and (b) the output power and the kVA of the alternator\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "XL = 40;# in ohms\n",
- "VL = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- " #a delta-connected load\n",
- "Vp = VL\n",
- " #Phase current\n",
- "Ip = Vp/Zp\n",
- "IL = Ip*(3**0.5)\n",
- " #Alternator output power is equal to the power dissipated by the load.\n",
- " #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pf = R/Zp\n",
- "P = VL*IL*(3**0.5)*pf\n",
- " #Alternator output kVA,\n",
- "S = VL*IL*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the current supplied by the alternator is \",round(IL,2),\" A\"\n",
- "print \"\\n (b)output power is \",round(P/1000,2),\"KW and kVA of the alternator is \",round(S/1000,2),\"kVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the current supplied by the alternator is 13.86 A\n",
- "\n",
- " (b)output power is 5.76 KW and kVA of the alternator is 9.6 kVA"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 308</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the phase current, (b) the line current, (c) the total power dissipated and \n",
- "#(d) the kVA rating of the load. Draw the complete phasor diagram for the load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "C = 80E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "VL = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Capacitive reactance\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Zp = (R*R + Xc*Xc)**0.5\n",
- "pf = R/Zp\n",
- " #a delta-connected load\n",
- "Vp = VL\n",
- " #Phase current\n",
- "Ip = Vp/Zp\n",
- "IL = Ip*(3**0.5)\n",
- " #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "P = VL*IL*(3**0.5)*pf\n",
- " #Alternator output kVA,\n",
- "S = VL*IL*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the phase current is \",round(Ip,2),\" A\"\n",
- "print \"\\n (b)the line current is \",round(IL,2),\" A\"\n",
- "print \"\\n (c) power is \",round(P/1000,2),\"kW\"\n",
- "print \"\\n (d)kVA of the alternator is \", round(S/1000,2),\"kVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the phase current is 8.03 A\n",
- "\n",
- " (b)the line current is 13.9 A\n",
- "\n",
- " (c) power is 5.8 kW\n",
- "\n",
- " (d)kVA of the alternator is 9.63 kVA"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 309</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the total power input and (b) the load power factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Pi1 = 8000;# in Watts\n",
- "Pi2 = 4000;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Total input power\n",
- "Pi = Pi1 + Pi2\n",
- "phi = math.atan((Pi1 - Pi2)*(3**0.5)/(Pi1 + Pi2))\n",
- " #Power factor\n",
- "pf = math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)power input is \",round(Pi,2),\"W\"\n",
- "print \"\\n (b)power factor is \",round(pf,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)power input is 12000.0 W\n",
- "\n",
- " (b)power factor is 0.87"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 310</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the readings of each wattmeter.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Pi = 12000;# in Watts\n",
- "pf = 0.6;# power factor\n",
- "\n",
- "#calculation:\n",
- " #If the two wattmeters indicate Pi1 and Pi2 respectively\n",
- " # Pit = Pi1 + Pi2\n",
- "Pit = Pi\n",
- " # Pid = Pi1 - Pi2\n",
- " #power factor = 0.6 = cos(phi)\n",
- "phi = math.acos(pf)\n",
- "Pid = Pit*math.tan(phi)/(3**0.5)\n",
- " #Hence wattmeter 1 reads\n",
- "Pi1 = (Pid + Pit)/2\n",
- " #wattmeter 2 reads\n",
- "Pi2 = Pit - Pi1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reading in each wattameter are \",round(Pi1,2),\"W and \",round(Pi2,2),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reading in each wattameter are 10618.8 W and 1381.2 W"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 310</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the input power and (b) the load power factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Pi1 = 10000;# in Watts\n",
- "Pi2 = -3000;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Total input power\n",
- "Pi = Pi1 + Pi2\n",
- "phi = math.atan((Pi1 - Pi2)*(3**0.5)/(Pi1 + Pi2))\n",
- " #Power factor\n",
- "pf = math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)power input is \",round(Pi,2),\"W\"\n",
- "print \"\\n (b)power factor is \",round(pf,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)power input is 7000.0 W\n",
- "\n",
- " (b)power factor is 0.3"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 311</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate for each connection the readings on each of two wattmeters connected to measure the power by the two-wattmeter method.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 8; # in ohms\n",
- "XL = 8; # in ohms\n",
- "VL = 415; # in Volts\n",
- "\n",
- "#calculation:\n",
- "#For a star connection:\n",
- "#IL = Ip\n",
- "#VL = Vp*(3**0.5)\n",
- "VLs = VL\n",
- "Vps = VLs/(3**0.5)\n",
- "#Impedance per phase,\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "Ips = Vps/Zp\n",
- "ILs = Ips\n",
- "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pf = R/Zp\n",
- "Ps = VLs*ILs*(3**0.5)*pf\n",
- "#If wattmeter readings are P1 and P2 then P1 + P2 = Pst\n",
- "Pst = Ps\n",
- "# Pid = Pi1 - Pi2\n",
- "phi = math.acos(pf)\n",
- "Psd = Pst*math.tan(phi)/(3**0.5)\n",
- "#Hence wattmeter 1 reads\n",
- "Ps1 = (Psd + Pst)/2\n",
- "#wattmeter 2 reads\n",
- "Ps2 = Pst - Ps1\n",
- "\n",
- "#For a delta connection:\n",
- "#VL = Vp\n",
- "#IL = Ip*(3**0.5)\n",
- "VLd = VL\n",
- "Vpd = VLd\n",
- "Ipd = Vpd/Zp\n",
- "ILd = Ipd*(3**0.5)\n",
- "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "Pd = VLd*ILd*(3**0.5)*pf\n",
- "#If wattmeter readings are P1 and P2 then P1 + P2 = Pdt\n",
- "Pdt = Pd\n",
- "# Pid = Pi1 - Pi2\n",
- "Pdd = Pdt*math.tan(phi)/(3**0.5)\n",
- "#Hence wattmeter 1 reads\n",
- "Pd1 = (Pdd + Pdt)/2\n",
- "#wattmeter 2 reads\n",
- "Pd2 = Pdt - Pd1\n",
- "\n",
- "#results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)When the coils are star-connected the wattmeter readings are\", round(Ps1,2),\"W and \",round(Ps2,2),\"W\"\n",
- "print \"(b)When the coils are delta-connected the wattmeter readings are are\", round(Pd1,2),\"W and\", round(Pd2,2),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)When the coils are star-connected the wattmeter readings are 8489.35 W and 2274.71 W\n",
- "(b)When the coils are delta-connected the wattmeter readings are are 25468.05 W and 6824.14 W"
- ]
- }
- ],
- "prompt_number": 17
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_2.ipynb deleted file mode 100755 index 8285878f..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_2.ipynb +++ /dev/null @@ -1,974 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 19: Three phase systems</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 299</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the system phase voltage, (b) the phase current and (c) the line current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Vl = 415;# in Volts\n",
- "Rp = 30;# in ohms\n",
- "\n",
- "#calculation:\n",
- "Vp = Vl/(3**0.5)\n",
- "Ip = Vp/Rp\n",
- "Il = Ip\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the system phase voltage is \",round(Vp,2),\" V\"\n",
- "print \"\\n (b)phase current is \",round(Ip,2),\" A\"\n",
- "print \"\\n (c)line current is \",round(Il,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the system phase voltage is 239.6 V\n",
- "\n",
- " (b)phase current is 7.99 A\n",
- "\n",
- " (c)line current is 7.99 A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 299</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the line voltage if the supply frequency is 50 Hz\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "L = 0.1273;# in Henry\n",
- "Ip = 5.08;# in Amperes\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "Il = Ip\n",
- "Vp = Ip*Zp\n",
- "Vl = Vp*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)line voltage is \",round(Vl,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)line voltage is 439.89 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 301</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the current in each line and (b) the current in the neutral conductor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "PR = 24000;# in Watt\n",
- "Py = 18000;# in Watt\n",
- "Pb = 12000;# in Watt\n",
- "VR = 240;# in Volts\n",
- "Vy = 240;# in Volts\n",
- "Vb = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For a star-connected system VL = Vp*(3**0.5)\n",
- "Vp = V/(3**0.5)\n",
- "phir = 90*math.pi/180\n",
- "phiy = 330*math.pi/180\n",
- "phib = 210*math.pi/180\n",
- " # I = P/V for a resistive load\n",
- "IR = PR/VR\n",
- "Iy = Py/Vy\n",
- "Ib = Pb/Vb\n",
- "Inh = IR*math.cos(phir) + Ib*math.cos(phib) + Iy*math.cos(phiy)\n",
- "Inv = IR*math.sin(phir) + Ib*math.sin(phib) + Iy*math.sin(phiy)\n",
- "In = (Inh**2 + Inv**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)cuurnt in R line is \",round(IR,2),\" A, cuurnt in Y line is \",round(Iy,2),\" A \"\n",
- "print \"and cuurnt in B line is \",round(Ib,2),\" A\"\n",
- "print \"\\n (b)cuurnt in neutral line is \",round(In,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)cuurnt in R line is 100.0 A, cuurnt in Y line is 75.0 A and cuurnt in B line is 50.0 A\n",
- "\n",
- " (b)cuurnt in neutral line is 43.3 A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 302</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the phase current, and (b) the line current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "L = 0.1273;# in Henry\n",
- "VL = 440;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "XL = 2*math.pi*f*L\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "Vp = VL\n",
- " #Phase current\n",
- "Ip = Vp/Zp\n",
- " #For a delta connection,\n",
- "IL = Ip*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the phase current \",round(Ip,2),\" A\"\n",
- "print \"\\n (b)line current \",round(IL,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the phase current 8.8 A\n",
- "\n",
- " (b)line current 15.24 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 302</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitance of each of the capacitors.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "IL = 15;# in Amperes\n",
- "VL = 415;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #For a delta connection\n",
- "Ip = IL/(3**0.5)#phase current\n",
- "Vp = VL\n",
- " #Capacitive reactance per phase\n",
- "Xc = Vp/Ip\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E6,2),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 66.43 uF"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 303</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate for each connection (a) the line and phase voltages and (b) the phase and line currents.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 3;# in ohms\n",
- "XL = 4;# in ohms\n",
- "VL = 415;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For a star connection:\n",
- " #IL = Ip\n",
- " #VL = Vp*(3**0.5)\n",
- "VLs = VL\n",
- "Vps = VLs/(3**0.5)\n",
- " #Impedance per phase,\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "Ips = Vps/Zp\n",
- "ILs = Ips\n",
- " #For a delta connection:\n",
- " #VL = Vp\n",
- " #IL = Ip*(3**0.5)\n",
- "VLd = VL\n",
- "Vpd = VLd\n",
- "Ipd = Vpd/Zp\n",
- "ILd = Ipd*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the line voltage for star connection is \",round(VLs,2),\" V \"\n",
- "print \"and the phase voltage for star connection is \",round(Vps,2),\" V \"\n",
- "print \"and the line voltage for delta connection is \",round(VLd,2),\" V \"\n",
- "print \"and the phase voltage for delta connection is \",round(Vpd,2),\" V\"\n",
- "print \"\\n (b)the line current for star connection is \",round(ILs,2),\" A \"\n",
- "print \"and the phase current for star connection is \",round(Ips,2),\" A \"\n",
- "print \"and the line current for delta connection is \",round(ILd,2),\" A \"\n",
- "print \"and the phase current for delta connection is \",round(Ipd,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the line voltage for star connection is 415.0 V \n",
- "and the phase voltage for star connection is 239.6 V \n",
- "and the line voltage for delta connection is 415.0 V \n",
- "and the phase voltage for delta connection is 415.0 V\n",
- "\n",
- " (b)the line current for star connection is 47.92 A \n",
- "and the phase current for star connection is 47.92 A \n",
- "and the line current for delta connection is 143.76 A \n",
- "and the phase current for delta connection is 83.0 A\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 304</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the total power dissipated by the resistors.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rp = 12;# in ohms\n",
- "VL = 415;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "Vp = VL/(3**0.5)# since the resistors are star-connected\n",
- " #Phase current, Ip\n",
- "Zp = Rp\n",
- "Ip = Vp/Zp\n",
- " #For a star connection\n",
- "IL = Ip\n",
- " # For a purely resistive load, the power factor cos(phi) = 1\n",
- "pf = 1\n",
- "P = VL*IL*(3**0.5)*pf\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)total power dissipated by the resistors is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)total power dissipated by the resistors is 14352.08 W"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 304</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the power factor of the system.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "P = 5000;# in Watts\n",
- "IL = 8.6;# in amperes\n",
- "VL = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pf = P/(VL*IL*(3**0.5))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n power factor is \",round(pf,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " power factor is 0.839"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 304</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the total power dissipated in each case.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "L = 0.042;# in Henry\n",
- "VL = 415;# in Volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #For a star connection:\n",
- " #IL = Ip\n",
- " #VL = Vp*(3**0.5)\n",
- "XL = 2*math.pi*f*L\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "VLs = VL\n",
- "Vps = VLs/(3**0.5)\n",
- " #Impedance per phase,\n",
- "Ips = Vps/Zp\n",
- "ILs = Ips\n",
- " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pfs = R/Zp\n",
- "Ps = VLs*ILs*(3**0.5)*pfs\n",
- "\n",
- " #For a delta connection:\n",
- " #VL = Vp\n",
- " #IL = Ip*(3**0.5)\n",
- "VLd = VL\n",
- "Vpd = VLd\n",
- "Ipd = Vpd/Zp\n",
- "ILd = Ipd*(3**0.5)\n",
- " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pfd = R/Zp\n",
- "Pd = VLd*ILd*(3**0.5)*pfd\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n total power dissipated in star is \",round(Ps,2),\" W and in delta is \",round(Pd,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " total power dissipated in star is 6283.29 W and in delta is 18849.88 W"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 305</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the power input, (b) the line current and (c) the phase current\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Po = 12750;# in Watts\n",
- "pf = 0.77;# power factor\n",
- "eff = 0.85;\n",
- "VL = 415;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #eff = power_out/power_in\n",
- "Pi = Po/eff\n",
- " #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "IL = Pi/(VL*(3**0.5)*pf)# line current\n",
- " #For a delta connection:\n",
- " #IL = Ip*(3**0.5)\n",
- "Ip = IL/(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)power input is \",round(Pi,2),\" W\"\n",
- "print \"\\n (b)line current is \",round(IL,2),\" A\"\n",
- "print \"\\n (c)phase current is \",round(Ip,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)power input is 15000.0 W\n",
- "\n",
- " (b)line current is 27.1 A\n",
- "\n",
- " (c)phase current is 15.65 A"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 308</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the current supplied by the alternator and (b) the output power and the kVA of the alternator\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "XL = 40;# in ohms\n",
- "VL = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- " #a delta-connected load\n",
- "Vp = VL\n",
- " #Phase current\n",
- "Ip = Vp/Zp\n",
- "IL = Ip*(3**0.5)\n",
- " #Alternator output power is equal to the power dissipated by the load.\n",
- " #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pf = R/Zp\n",
- "P = VL*IL*(3**0.5)*pf\n",
- " #Alternator output kVA,\n",
- "S = VL*IL*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the current supplied by the alternator is \",round(IL,2),\" A\"\n",
- "print \"\\n (b)output power is \",round(P/1000,2),\"KW and kVA of the alternator is \",round(S/1000,2),\"kVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the current supplied by the alternator is 13.86 A\n",
- "\n",
- " (b)output power is 5.76 KW and kVA of the alternator is 9.6 kVA"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 308</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the phase current, (b) the line current, (c) the total power dissipated and \n",
- "#(d) the kVA rating of the load. Draw the complete phasor diagram for the load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 30;# in ohms\n",
- "C = 80E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "VL = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Capacitive reactance\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- "Zp = (R*R + Xc*Xc)**0.5\n",
- "pf = R/Zp\n",
- " #a delta-connected load\n",
- "Vp = VL\n",
- " #Phase current\n",
- "Ip = Vp/Zp\n",
- "IL = Ip*(3**0.5)\n",
- " #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "P = VL*IL*(3**0.5)*pf\n",
- " #Alternator output kVA,\n",
- "S = VL*IL*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the phase current is \",round(Ip,2),\" A\"\n",
- "print \"\\n (b)the line current is \",round(IL,2),\" A\"\n",
- "print \"\\n (c) power is \",round(P/1000,2),\"kW\"\n",
- "print \"\\n (d)kVA of the alternator is \", round(S/1000,2),\"kVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the phase current is 8.03 A\n",
- "\n",
- " (b)the line current is 13.9 A\n",
- "\n",
- " (c) power is 5.8 kW\n",
- "\n",
- " (d)kVA of the alternator is 9.63 kVA"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 309</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the total power input and (b) the load power factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Pi1 = 8000;# in Watts\n",
- "Pi2 = 4000;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Total input power\n",
- "Pi = Pi1 + Pi2\n",
- "phi = math.atan((Pi1 - Pi2)*(3**0.5)/(Pi1 + Pi2))\n",
- " #Power factor\n",
- "pf = math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)power input is \",round(Pi,2),\"W\"\n",
- "print \"\\n (b)power factor is \",round(pf,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)power input is 12000.0 W\n",
- "\n",
- " (b)power factor is 0.87"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 310</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the readings of each wattmeter.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Pi = 12000;# in Watts\n",
- "pf = 0.6;# power factor\n",
- "\n",
- "#calculation:\n",
- " #If the two wattmeters indicate Pi1 and Pi2 respectively\n",
- " # Pit = Pi1 + Pi2\n",
- "Pit = Pi\n",
- " # Pid = Pi1 - Pi2\n",
- " #power factor = 0.6 = cos(phi)\n",
- "phi = math.acos(pf)\n",
- "Pid = Pit*math.tan(phi)/(3**0.5)\n",
- " #Hence wattmeter 1 reads\n",
- "Pi1 = (Pid + Pit)/2\n",
- " #wattmeter 2 reads\n",
- "Pi2 = Pit - Pi1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reading in each wattameter are \",round(Pi1,2),\"W and \",round(Pi2,2),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reading in each wattameter are 10618.8 W and 1381.2 W"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 310</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the input power and (b) the load power factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Pi1 = 10000;# in Watts\n",
- "Pi2 = -3000;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Total input power\n",
- "Pi = Pi1 + Pi2\n",
- "phi = math.atan((Pi1 - Pi2)*(3**0.5)/(Pi1 + Pi2))\n",
- " #Power factor\n",
- "pf = math.cos(phi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)power input is \",round(Pi,2),\"W\"\n",
- "print \"\\n (b)power factor is \",round(pf,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)power input is 7000.0 W\n",
- "\n",
- " (b)power factor is 0.3"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 311</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate for each connection the readings on each of two wattmeters connected to measure the power by the two-wattmeter method.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 8; # in ohms\n",
- "XL = 8; # in ohms\n",
- "VL = 415; # in Volts\n",
- "\n",
- "#calculation:\n",
- "#For a star connection:\n",
- "#IL = Ip\n",
- "#VL = Vp*(3**0.5)\n",
- "VLs = VL\n",
- "Vps = VLs/(3**0.5)\n",
- "#Impedance per phase,\n",
- "Zp = (R*R + XL*XL)**0.5\n",
- "Ips = Vps/Zp\n",
- "ILs = Ips\n",
- "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "pf = R/Zp\n",
- "Ps = VLs*ILs*(3**0.5)*pf\n",
- "#If wattmeter readings are P1 and P2 then P1 + P2 = Pst\n",
- "Pst = Ps\n",
- "# Pid = Pi1 - Pi2\n",
- "phi = math.acos(pf)\n",
- "Psd = Pst*math.tan(phi)/(3**0.5)\n",
- "#Hence wattmeter 1 reads\n",
- "Ps1 = (Psd + Pst)/2\n",
- "#wattmeter 2 reads\n",
- "Ps2 = Pst - Ps1\n",
- "\n",
- "#For a delta connection:\n",
- "#VL = Vp\n",
- "#IL = Ip*(3**0.5)\n",
- "VLd = VL\n",
- "Vpd = VLd\n",
- "Ipd = Vpd/Zp\n",
- "ILd = Ipd*(3**0.5)\n",
- "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
- "Pd = VLd*ILd*(3**0.5)*pf\n",
- "#If wattmeter readings are P1 and P2 then P1 + P2 = Pdt\n",
- "Pdt = Pd\n",
- "# Pid = Pi1 - Pi2\n",
- "Pdd = Pdt*math.tan(phi)/(3**0.5)\n",
- "#Hence wattmeter 1 reads\n",
- "Pd1 = (Pdd + Pdt)/2\n",
- "#wattmeter 2 reads\n",
- "Pd2 = Pdt - Pd1\n",
- "\n",
- "#results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)When the coils are star-connected the wattmeter readings are\", round(Ps1,2),\"W and \",round(Ps2,2),\"W\"\n",
- "print \"(b)When the coils are delta-connected the wattmeter readings are are\", round(Pd1,2),\"W and\", round(Pd2,2),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)When the coils are star-connected the wattmeter readings are 8489.35 W and 2274.71 W\n",
- "(b)When the coils are delta-connected the wattmeter readings are are 25468.05 W and 6824.14 W"
- ]
- }
- ],
- "prompt_number": 17
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_19.ipynb index a3d9112a..a3d9112a 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_19.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint.ipynb deleted file mode 100755 index 5b0974af..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint.ipynb +++ /dev/null @@ -1,1627 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 20: Transformers</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 317</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the secondary voltage, assuming an ideal transformer\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 500;# primary turns\n",
- "N2 = 3000;# secondary turns\n",
- "V1 = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For an ideal transformer, voltage ratio = turns ratio\n",
- "V2 = V1*N2/N1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n secondary voltage \",round(V2,2),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " secondary voltage 1440.0 V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 317</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its output voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 2/7;# turns ratio\n",
- "V1 = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #A turns ratio of 2:7 means that the transformer has 2 turns on the primary \n",
- " #for every 7 turns on the secondary\n",
- "V2 = V1/tr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output voltage \",round(V2,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output voltage 840.0 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 317</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine secondary voltage and current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 8/1;# turns ratio\n",
- "I1 = 3;# in Amperes\n",
- "V1 = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #A turns ratio of 8:1 means that the transformer has 28 turns on the \n",
- " #primary for every 1turns on the secondary\n",
- "V2 = V1/tr\n",
- " #secondary current\n",
- "I2 = I1*tr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n secondary voltage is \",round(V2,2),\" V and secondary current is \", round(I2,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " secondary voltage is 30.0 V and secondary current is 24.0 A"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 318</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the transformer turns ratio and the current taken from the supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 240;# in Volts\n",
- "V2 = 12;# in Volts\n",
- "P = 150;# in Watts\n",
- "\n",
- "#calculation:\n",
- "I2 = P/V2\n",
- " #A turns ratio = Vp/Vs\n",
- "tr = V1/V2# turn ratio\n",
- " # V1/V2 = I2/I1\n",
- " #current taken from the supply\n",
- "I1 = I2*V2/V1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n turn ratio is \",round(tr,2),\" and current taken from the supply is \",round(I1,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " turn ratio is 20.0 and current taken from the supply is 0.63 A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 318</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the full-load secondary current,\n",
- "#(b) the minimum load resistance which can be connected across the secondary winding to give full load kVA, \n",
- "#(c) the primary current at full load kVA.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 5000;# in VA\n",
- "tr = 10;# turn ratio\n",
- "V1 = 2500;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #A turns ratio of 8:1 means that the transformer has 28 turns on the primary for every 1turns on the secondary\n",
- "V2 = V1/tr\n",
- " #transformer rating in volt-amperes = Vs*Is\n",
- "I2 = S/V2\n",
- " #Minimum value of load resistance\n",
- "RL = V2/I2\n",
- " # tr = I2/I1\n",
- "I1 = I2/tr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)full-load secondary current is \",round(I2,2),\" A\"\n",
- "print \"\\n (b)minimum load resistance is \",round(RL,2),\" ohm\"\n",
- "print \"\\n (c) primary current is \",round(I1,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)full-load secondary current is 20.0 A\n",
- "\n",
- " (b)minimum load resistance is 12.5 ohm\n",
- "\n",
- " (c) primary current is 2.0 A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 319</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of the magnetizing and core loss components of the noload current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 2400;# in Volts\n",
- "V2 = 400;# in Volts\n",
- "I0 = 0.5;# in Amperes\n",
- "Pc = 400;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Core loss (i.e. iron loss) P = V1*I0*cos(phi0)\n",
- "pf = Pc/(V1*I0)\n",
- "phi0 = math.acos(pf)\n",
- " #Magnetizing component\n",
- "Im = I0*math.sin(phi0)\n",
- " #Core loss component\n",
- "Ic = I0*math.cos(phi0)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)magnetizing component is \",round(Im,3),\" A and Core loss component is \",round(Ic,3),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)magnetizing component is 0.471 A and Core loss component is 0.167 A"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 320</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the iron loss current, \n",
- "#(b) the power factor on no-load, and (c) the magnetizing current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 240;# in Volts\n",
- "I0 = 0.8;# in Amperes\n",
- "P = 72;# in Watts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #Power absorbed = total core loss, P = V*I0*cos(phi0)\n",
- " #Ic = I0*cos(phi0)\n",
- "Ic = P/V\n",
- "pf = Ic/I0\n",
- " #From the right-angled triangle in Figure 20.2(b) and using\n",
- " #Pythagoras\u2019 theorem, \n",
- "Im = (I0*I0 - Ic*Ic)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) Core loss component is \",round( Ic,2),\" A\"\n",
- "print \"\\n (b) power factor is \",round( pf,2),\"\"\n",
- "print \"\\n (c)magnetizing component is \",round(Im,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) Core loss component is 0.3 A\n",
- "\n",
- " (b) power factor is 0.37 \n",
- "\n",
- " (c)magnetizing component is 0.74 A"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 321</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum value of the flux.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 100000;# in VA\n",
- "V1 = 4000;# in Volts\n",
- "V2 = 200;# in Volts\n",
- "N2 = 100;# sec turns\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #Transformer rating = V1*I1 = V2*I2\n",
- " #primary current\n",
- "I1 = S/V1\n",
- " #secondary current\n",
- "I2 = S/V2\n",
- " #primary turns\n",
- "N1 = N2*V1/V2\n",
- " #maximum flux\n",
- " #assuming E2 = V2\n",
- "Phim = V2/(4.44*f*N2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)primary current is \",round( I1,2),\" A and secondary current is \",round( I2,2),\" A\"\n",
- "print \"\\n (b)number of primary turns is \",round( N1,2),\"\"\n",
- "print \"\\n (c)maximum value of the flux is \",round(Phim*1000,2),\"mWb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)primary current is 25.0 A and secondary current is 500.0 A\n",
- "\n",
- " (b)number of primary turns is 2000.0 \n",
- "\n",
- " (c)maximum value of the flux is 9.01 mWb"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 322</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the maximum value of the flux density in the core, and\n",
- "#(b) the voltage induced in the secondary winding.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 250;# in Volts\n",
- "A = 0.03;# in m2\n",
- "N2 = 300;# sec turns\n",
- "N1 = 25;# prim turns\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #e.m.f. E1 = 4.44*f*Phim*N1\n",
- " #maximum flux density,\n",
- "Phim = V1/(4.44*f*N1)\n",
- " #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
- " #maximum core flux density\n",
- "Bm = Phim/A\n",
- " #voltage induced in the secondary winding,\n",
- "V2 = V1*N2/N1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum core flux density \",round( Bm,2),\" T\"\n",
- "print \"\\n (b)voltage induced in the secondary winding is \",round( V2,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum core flux density 1.5 T\n",
- "\n",
- " (b)voltage induced in the secondary winding is 3000.0 V"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 323</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the number of primary and secondary turns\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 500;# in Volts\n",
- "V2 = 100;# in Volts\n",
- "Bm = 1.5;# in Tesla\n",
- "A = 0.005;# in m2\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
- " #maximum core flux density\n",
- "Phim = Bm*A\n",
- " #e.m.f. E1 = 4.44*f*Phim*N1\n",
- " #primary turns,\n",
- "N1 = V1/(4.44*f*Phim)\n",
- " #secondary turns,\n",
- "N2 = V2*N1/V1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n no. of primary and secondary turns are \",round(N1,2),\" turns, and \",round(N2,2),\" turns respectively\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " no. of primary and secondary turns are 300.3 turns, and 60.06 turns respectively"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 323</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the number of primary and secondary turns and \n",
- "#(b) the cross-sectional area of the core.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "emfpt = 15;# in Volts\n",
- "V1 = 4500;# in Volts\n",
- "V2 = 225;# in Volts\n",
- "Bm = 1.4;# in Tesla\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #E.m.f. per turn, V1/N1 = V2/N2 = emfpt\n",
- " #primary turns,\n",
- "N1 = V1/emfpt\n",
- " #secondary turns,\n",
- "N2 = V2/emfpt\n",
- " #e.m.f. E1 = 4.44*f*Phim*N1\n",
- " #maximum flux density,\n",
- "Phim = V1/(4.44*f*N1)\n",
- " #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
- " #cross-sectional area\n",
- "A = Phim/Bm\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)no. of primary and secondary turns are \", N1,\" turns, and \", N2,\" turns respectively\"\n",
- "print \"\\n (b)cross-sectional area is \", round(A,4),\"m2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)no. of primary and secondary turns are 300.0 turns, and 15.0 turns respectively\n",
- "\n",
- " (b)cross-sectional area is 0.0483 m2"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 324</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the primary current and power factor when the secondary current is 100 A at a power factor of 0.85 lagging\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 2000;# prim turns\n",
- "N2 = 800;# sec turns\n",
- "I0 = 5;# in Amperes\n",
- "pf0 = 0.20;# power factor\n",
- "I2 = 100;# in Amperes\n",
- "pf2 = 0.85;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Let I01 be the component of the primary current which provides the restoring mmf. Then I01*N1 = I2*N2\n",
- "I01 = I2*N2/N1\n",
- " #If the power factor of the secondary is 0.85\n",
- "phi2 = math.acos(pf2)\n",
- " #If the power factor on no-load is 0.20,\n",
- "phi0 = math.acos(pf0)\n",
- "I1h = I0*math.cos(phi0) + I01*math.cos(phi2)\n",
- "I1v = I0*math.sin(phi0) + I01*math.sin(phi2)\n",
- " #Hence the magnitude of I1\n",
- "I1 = (I1h*I1h + I1v*I1v)**0.5\n",
- "pf1 = math.cos(math.atan(I1v/I1h))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Primary current is \", round(I1,2),\" A, and Power factor is \",round(pf1,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Primary current is 43.58 A, and Power factor is 0.8"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 328</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the equivalent resistance referred to the primary winding, \n",
- "#(b) the equivalent reactance referred to the primary winding,\n",
- "#(c) the equivalent impedance referred to the primary winding, and \n",
- "#(d) the phase angle of the impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 600;# prim turns\n",
- "N2 = 150;# sec turns\n",
- "R1 = 0.25;# in ohms\n",
- "R2 = 0.01;# in ohms\n",
- "X1 = 1.0;# in ohms\n",
- "X2 = 0.04;# in ohms\n",
- "\n",
- "#calculation:\n",
- "tr = N1/N2# turn ratio\n",
- "vr = tr# voltage ratio = turn raio, vr = V1/V2\n",
- " #equivalent resistance Re\n",
- "Re = R1 + R2*(vr**2)\n",
- " #equivalent reactance, Xe\n",
- "Xe = X1 + X2*(vr**2)\n",
- " #equivalent impedance, Ze\n",
- "Ze = (Re*Re + Xe*Xe)**0.5\n",
- " #cos(phie) = Re/Ze\n",
- "pfe = Re/Ze\n",
- "phie = math.acos(pfe)\n",
- "phied = phie*180/math.pi# in \u00b0(deg)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the equivalent resistance referred to the primary winding is \",round( Re,2),\" ohm\"\n",
- "print \"\\n (b)the equivalent reactance referred to the primary winding is \",round( Xe,2),\" ohm\"\n",
- "print \"\\n (c)the equivalent impedance referred to the primary winding is \",round( Ze,2),\" ohm\"\n",
- "print \"\\n (d)phase angle is \",round( phied,2),\"deg\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the equivalent resistance referred to the primary winding is 0.41 ohm\n",
- "\n",
- " (b)the equivalent reactance referred to the primary winding is 1.64 ohm\n",
- "\n",
- " (c)the equivalent impedance referred to the primary winding is 1.69 ohm\n",
- "\n",
- " (d)phase angle is 75.96 deg"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 329</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the regulation of the transformer.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 200;# in Volts\n",
- "V2 = 400;# in Volts\n",
- "V2L = 387.6;# in Volts\n",
- "S = 5000;# in VA\n",
- "\n",
- "#calculation:\n",
- " #regulation =(No-load secondary voltage -\u0003 terminal voltage on load)*100/no-load secondary voltage in %\n",
- "reg = (V2 - V2L)*100/V2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the regulation of the transformer is \",round(reg,2),\" percent \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the regulation of the transformer is 3.1 percent "
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 329</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the load voltage at which the mechanism operates.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "VnL = 240;# in Volts\n",
- "reg = 2.5;# in percent\n",
- "\n",
- "#calculation:\n",
- " #regulation =(No-load secondary voltage -\u0003 terminal voltage on load)*100/no-load secondary voltage in %\n",
- "VL = VnL - reg*VnL/100\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the load voltage at which the mechanism operates is \",round(VL,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the load voltage at which the mechanism operates is 234.0 V "
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 331</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the transformer efficiency at full load and 0.85 power factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 200000;# in VA\n",
- "Pc = 1500;# in Watt\n",
- "Pi = 1000;# in Watt\n",
- "pf = 0.85;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
- " #Efficiency = 1 - losses/input power\n",
- " #Full-load output power = V*I*pf\n",
- "Po = S*pf\n",
- " #Total losses\n",
- "Pl = Pc + Pi\n",
- " #Input power = output power + losses\n",
- "PI = Po + Pl\n",
- " #efficiency\n",
- "eff = 1-(Pl/PI)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the transformer efficiency at full load is \",round(eff,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the transformer efficiency at full load is 0.99"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 331</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the efficiency of the transformer\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 200000;# in VA\n",
- "Pc = 1500;# in Watt\n",
- "Pi = 1000;# in Watt\n",
- "pf = 0.85;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
- " #Efficiency = 1 - losses/input power\n",
- " #Half full-load power output = V*I*pf/2\n",
- "Po = S*pf/2\n",
- " #Copper loss (or I*I*R loss) is proportional to current squared\n",
- " #Hence the copper loss at half full-load is\n",
- "Pch = Pc/(2*2)\n",
- " #Iron loss = 1000 W (constant)\n",
- " #Total losses\n",
- "Pl = Pch + Pi\n",
- " #Input power at half full-load = output power at half full-load + losses\n",
- "PI = Po + Pl\n",
- " #efficiency\n",
- "eff = (1-(Pl/PI))*100\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the transformer efficiency at half full load is \",round(eff,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the transformer efficiency at half full load is 98.41 percent"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 332</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the efficiency of the transformer (a) on full load, and (b) on half load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 400000;# in VA\n",
- "R1 = 0.5;# in Ohm\n",
- "R2 = 0.001;# in Ohm\n",
- "V1 = 5000;# in Volts\n",
- "V2 = 320;# in Volts\n",
- "Pi = 2500;# in Watt\n",
- "pf = 0.85;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Rating = 400 kVA = V1*I1 = V2*I2\n",
- " #Hence primary current\n",
- "I1 = S/V1\n",
- " #secondary current\n",
- "I2 = S/V2\n",
- " #Total copper loss = I1*I1*R1 + I2*I2*R2,\n",
- "Pcf = I1*I1*R1 + I2*I2*R2\n",
- " #On full load, total loss = copper loss + iron loss\n",
- "Plf = Pcf + Pi\n",
- " # full-load power output = V2*I2*pf\n",
- "Pof = S*pf\n",
- " #Input power at full-load = output power at full-load + losses\n",
- "PIf = Pof + Plf\n",
- " #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
- " #Efficiency = 1 - losses/input power\n",
- "efff = (1-(Plf/PIf))*100\n",
- "\n",
- " #Half full-load power output = V*I*pf/2\n",
- "Poh = S*pf/2\n",
- " #Copper loss (or I*I*R loss) is proportional to current squared\n",
- " #Hence the copper loss at half full-load is\n",
- "Pch = Pcf/(2*2)\n",
- " #Iron loss = 2500 W (constant)\n",
- " #Total losses\n",
- "Plh = Pch + Pi\n",
- " #Input power at half full-load = output power at half full-load + losses\n",
- "PIh = Poh + Plh\n",
- " #efficiency\n",
- "effh = (1-(Plh/PIh))*100\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the transformer efficiency at full load is \", round(efff,2),\" percent\"\n",
- "print \"\\n (b)the transformer efficiency at half full load is \", round(effh,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the transformer efficiency at full load is 97.91 percent\n",
- "\n",
- " (b)the transformer efficiency at half full load is 97.88 percent"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 333</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the output kVA at which the efficiency of the transformer is a maximum, and (b) the maximum efficiency,\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 500000;# in VA\n",
- "Pcf = 4000;# in Watt\n",
- "Pi = 2500;# in Watt\n",
- "pf = 0.75;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Let x be the fraction of full load kVA at which the efficiency is a maximum.\n",
- " #The corresponding total copper loss = (4 kW)*(x**2)\n",
- " #At maximum efficiency, copper loss = iron loss Hence\n",
- "x = (Pi/Pcf)**0.5\n",
- " #Hence the output kVA at maximum efficiency\n",
- "So = x*S\n",
- " #Total loss at maximum efficiency\n",
- "Pl = 2*Pi\n",
- " #Output power\n",
- "Po = So*pf\n",
- " #Input power = output power + losses\n",
- "PI = Po + Pl\n",
- " #Efficiency = output power/input power = (input power\u00e2\u20ac\u201dlosses)/input power\n",
- " #Efficiency = 1 - losses/input power\n",
- " #Maximum efficiency\n",
- "effm = (1 - Pl/PI)*100\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the output kVA at maximum efficiency is \",round(So/1000,2),\"kVA\"\n",
- "print \"\\n max. efficiency is \",round(effm,2),\" pecent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the output kVA at maximum efficiency is 395.28 kVA\n",
- "\n",
- " max. efficiency is 98.34 pecent"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 335</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the equivalent input resistance of the transformer.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 4;# turn ratio\n",
- "RL = 100;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #the equivalent input resistance,\n",
- "Ri = RL*(tr**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the equivalent input resistance is \",round(Ri,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the equivalent input resistance is 1600.0 ohm"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 335</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the optimum turns ratio of a transformer which would match a \n",
- "#load resistance of 7 ohmto the output resistance of the amplifier.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 112;# in Ohms\n",
- "RL = 7;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #The equivalent input resistance, R1 of the transformer needs to be 112 ohm for maximum power transfer.\n",
- " #R1 = RL*(tr**2)\n",
- " # tr = N1/N2 turn ratio\n",
- "tr = (R1/RL)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the optimum turns ratio is \",tr,\": 1.0\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the optimum turns ratio is 4.0 : 1.0"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 335</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the optimum value of load resistance for maximum power transfer\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 5;# turn ratio\n",
- "R1 = 150;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #The equivalent input resistance, R1 of the transformer needs to be 150 ohm for maximum power transfer.\n",
- " #R1 = RL*(tr**2)\n",
- "RL = R1/(tr**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the optimum value of load resistance is \",round(RL,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the optimum value of load resistance is 6.0 ohm"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 335</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the primary current flowing and (b) the power dissipated in the load resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 220;# in Volts\n",
- "V2 = 1760;# in Volts\n",
- "V = 220;# in Volts\n",
- "RL = 1280;# in Ohms\n",
- "R = 2;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #Turns ratio, tr = N1/N2 = V1/V2\n",
- "tr = V1/V2\n",
- " #Equivalent input resistance of the transformer,\n",
- " #R1 = RL*(tr**2)\n",
- "R1 = RL*(tr**2)\n",
- " #Total input resistance\n",
- "Rin = R + R1\n",
- " # Primary current\n",
- "I1 = V1/Rin\n",
- " #For an ideal transformer V1/V2 = I2/I1,\n",
- "I2 = I1*tr\n",
- " #Power dissipated in load resistor RL\n",
- "P = I2*I2*RL\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) primary current flowing is \",round(I1,2),\" A\"\n",
- "print \"\\n (b) power dissipated in the load resistor is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) primary current flowing is 10.0 A\n",
- "\n",
- " (b) power dissipated in the load resistor is 2000.0 W"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 336</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of the load resistance and (b) the power dissipated in the load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 25;# teurn ratio\n",
- "V = 24;# in Volts\n",
- "R1 = 15000;# in Ohms\n",
- "Rin = 15000;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Turns ratio, tr = N1/N2 = V1/V2\n",
- " #For maximum power transfer R1 needs to be equal to 15 kohm\n",
- "RL = R1/(tr**2)\n",
- " #The total input resistance when the source is connected to the matching transformer is\n",
- "Rt = Rin + R1\n",
- " #Primary current,\n",
- "I1 = V/Rt\n",
- " #N1/N2 = I2/I1\n",
- "I2 = I1*tr\n",
- " #Power dissipated in load resistor RL\n",
- "P = I2*I2*RL\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the load resistance is \",round(RL,2),\"ohm\"\n",
- "print \"\\n (b) power dissipated in the load resistor is \",round(P*1000,2),\"mW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the load resistance is 24.0 ohm\n",
- "\n",
- " (b) power dissipated in the load resistor is 9.6 mW"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 25, page no. 337</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in each section of the winding.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 320;# in Volts\n",
- "V2 = 250;# in Volts\n",
- "S = 20000;# in VA\n",
- "\n",
- "#calculation:\n",
- " #Rating = 20 kVA = V1*I1 = V2*I2\n",
- " #Hence primary current, I1\n",
- "I1 = S/V1\n",
- " #secondary current, I2\n",
- "I2 = S/V2\n",
- " #Hence current in common part of the winding\n",
- "I = I2 - I1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current in common part of the winding is \", round(I,2),\" A\"\n",
- "print \"\\n primary current and secondary current are \",round(I1,2),\" A and \",round(I2,2),\" A respectively\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current in common part of the winding is 17.5 A\n",
- "\n",
- " primary current and secondary current are 62.5 A and 80.0 A respectively"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 26, page no. 339</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the saving in the volume of copper used in an auto transformer compared with a double-wound transformer\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1a = 200;# in Volts\n",
- "V2a = 150;# in Volts\n",
- "V1b = 500;# in Volts\n",
- "V2b = 100;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For a 200 V:150 V transformer, xa\n",
- "xa = V2a/V1a\n",
- " #volume of copper in auto transformer\n",
- "vca = (1 - xa)*100# of copper in a double-wound transformer\n",
- " #the saving is\n",
- "vsa = 100 - vca\n",
- " #For a 500 V:100 V transformer, xb\n",
- "xb = V2b/V1b\n",
- " #volume of copper in auto transformer\n",
- "vcb = (1 - xb)*100# of copper in a double-wound transformer\n",
- " #the saving is\n",
- "vsb = 100 - vcb\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)For a 200 V:150 V transformer, the saving is \", round(vsa,2),\" percent\"\n",
- "print \"\\n (b)For a 500 V:100 V transformer, the saving is \", round(vsb,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)For a 200 V:150 V transformer, the saving is 75.0 percent\n",
- "\n",
- " (b)For a 500 V:100 V transformer, the saving is 20.0 percent"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 27, page no. 340</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the secondary line voltage on no-load when the windings are connected (a) star-delta, (b) delta-star.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 500;# prim turns\n",
- "N2 = 50;# sec turns\n",
- "VL = 2400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For a star-connection, VL = Vp*(3**0.5)\n",
- "VL1s = VL\n",
- " #Primary phase voltage\n",
- "Vp1s = VL1s/(3**0.5)\n",
- " #For a delta-connection, VL = Vp\n",
- " #N1/N2 = V1/V2, from which,\n",
- " #secondary phase voltage, Vp2s\n",
- "Vp2s = Vp1s*N2/N1\n",
- "VL2d = Vp2s\n",
- "\n",
- " #For a delta-connection, VL = Vp\n",
- "VL1d = VL\n",
- " #primary phase voltage Vp1d\n",
- "Vp1d = VL1d\n",
- " #Secondary phase voltage, Vp2d\n",
- "Vp2d = Vp1d*N2/N1\n",
- " #For a star-connection, VL = Vp*(3**0.5)\n",
- "VL2s = Vp2d*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the secondary line voltage for star and delta connection are \",round(Vp2s,1),\" V \"\n",
- "print \" and \",round(VL2s,0),\" V respectively\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the secondary line voltage for star and delta connection are 138.6 V \n",
- " and 416.0 V respectively\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 28, page no. 343</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the reading on the ammeter, \n",
- "#(b) the potential difference across the ammeter and\n",
- "#(c) the total load (in VA) on the secondary.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 1;# prim turns\n",
- "N2 = 60;# sec turns\n",
- "I1 = 300;# in amperes\n",
- "Ra = 0.15;# in ohms\n",
- "R2 = 0.25;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Reading on the ammeter,\n",
- "I2 = I1*(N1/N2)\n",
- " #P.d. across the ammeter = I2*RA, where RA is the ammeter resistance\n",
- "pd = I2*Ra\n",
- " #Total resistance of secondary circuit\n",
- "Rt = Ra + R2\n",
- " #Induced e.m.f. in secondary\n",
- "V2 = I2*Rt\n",
- " #Total load on secondary\n",
- "S = V2*I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the reading on the ammeter is \",round(I2,2),\" A \"\n",
- "print \"\\n (b)potential difference across the ammeter is \",round(pd,2),\" V \"\n",
- "print \"\\n (c)total load (in VA) on the secondary is \",round(S,2),\" VA \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the reading on the ammeter is 5.0 A \n",
- "\n",
- " (b)potential difference across the ammeter is 0.75 V \n",
- "\n",
- " (c)total load (in VA) on the secondary is 10.0 VA "
- ]
- }
- ],
- "prompt_number": 28
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_1.ipynb deleted file mode 100755 index 5b0974af..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_1.ipynb +++ /dev/null @@ -1,1627 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 20: Transformers</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 317</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the secondary voltage, assuming an ideal transformer\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 500;# primary turns\n",
- "N2 = 3000;# secondary turns\n",
- "V1 = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For an ideal transformer, voltage ratio = turns ratio\n",
- "V2 = V1*N2/N1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n secondary voltage \",round(V2,2),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " secondary voltage 1440.0 V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 317</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its output voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 2/7;# turns ratio\n",
- "V1 = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #A turns ratio of 2:7 means that the transformer has 2 turns on the primary \n",
- " #for every 7 turns on the secondary\n",
- "V2 = V1/tr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output voltage \",round(V2,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output voltage 840.0 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 317</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine secondary voltage and current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 8/1;# turns ratio\n",
- "I1 = 3;# in Amperes\n",
- "V1 = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #A turns ratio of 8:1 means that the transformer has 28 turns on the \n",
- " #primary for every 1turns on the secondary\n",
- "V2 = V1/tr\n",
- " #secondary current\n",
- "I2 = I1*tr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n secondary voltage is \",round(V2,2),\" V and secondary current is \", round(I2,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " secondary voltage is 30.0 V and secondary current is 24.0 A"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 318</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the transformer turns ratio and the current taken from the supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 240;# in Volts\n",
- "V2 = 12;# in Volts\n",
- "P = 150;# in Watts\n",
- "\n",
- "#calculation:\n",
- "I2 = P/V2\n",
- " #A turns ratio = Vp/Vs\n",
- "tr = V1/V2# turn ratio\n",
- " # V1/V2 = I2/I1\n",
- " #current taken from the supply\n",
- "I1 = I2*V2/V1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n turn ratio is \",round(tr,2),\" and current taken from the supply is \",round(I1,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " turn ratio is 20.0 and current taken from the supply is 0.63 A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 318</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the full-load secondary current,\n",
- "#(b) the minimum load resistance which can be connected across the secondary winding to give full load kVA, \n",
- "#(c) the primary current at full load kVA.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 5000;# in VA\n",
- "tr = 10;# turn ratio\n",
- "V1 = 2500;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #A turns ratio of 8:1 means that the transformer has 28 turns on the primary for every 1turns on the secondary\n",
- "V2 = V1/tr\n",
- " #transformer rating in volt-amperes = Vs*Is\n",
- "I2 = S/V2\n",
- " #Minimum value of load resistance\n",
- "RL = V2/I2\n",
- " # tr = I2/I1\n",
- "I1 = I2/tr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)full-load secondary current is \",round(I2,2),\" A\"\n",
- "print \"\\n (b)minimum load resistance is \",round(RL,2),\" ohm\"\n",
- "print \"\\n (c) primary current is \",round(I1,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)full-load secondary current is 20.0 A\n",
- "\n",
- " (b)minimum load resistance is 12.5 ohm\n",
- "\n",
- " (c) primary current is 2.0 A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 319</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of the magnetizing and core loss components of the noload current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 2400;# in Volts\n",
- "V2 = 400;# in Volts\n",
- "I0 = 0.5;# in Amperes\n",
- "Pc = 400;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Core loss (i.e. iron loss) P = V1*I0*cos(phi0)\n",
- "pf = Pc/(V1*I0)\n",
- "phi0 = math.acos(pf)\n",
- " #Magnetizing component\n",
- "Im = I0*math.sin(phi0)\n",
- " #Core loss component\n",
- "Ic = I0*math.cos(phi0)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)magnetizing component is \",round(Im,3),\" A and Core loss component is \",round(Ic,3),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)magnetizing component is 0.471 A and Core loss component is 0.167 A"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 320</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the iron loss current, \n",
- "#(b) the power factor on no-load, and (c) the magnetizing current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 240;# in Volts\n",
- "I0 = 0.8;# in Amperes\n",
- "P = 72;# in Watts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #Power absorbed = total core loss, P = V*I0*cos(phi0)\n",
- " #Ic = I0*cos(phi0)\n",
- "Ic = P/V\n",
- "pf = Ic/I0\n",
- " #From the right-angled triangle in Figure 20.2(b) and using\n",
- " #Pythagoras\u2019 theorem, \n",
- "Im = (I0*I0 - Ic*Ic)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) Core loss component is \",round( Ic,2),\" A\"\n",
- "print \"\\n (b) power factor is \",round( pf,2),\"\"\n",
- "print \"\\n (c)magnetizing component is \",round(Im,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) Core loss component is 0.3 A\n",
- "\n",
- " (b) power factor is 0.37 \n",
- "\n",
- " (c)magnetizing component is 0.74 A"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 321</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum value of the flux.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 100000;# in VA\n",
- "V1 = 4000;# in Volts\n",
- "V2 = 200;# in Volts\n",
- "N2 = 100;# sec turns\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #Transformer rating = V1*I1 = V2*I2\n",
- " #primary current\n",
- "I1 = S/V1\n",
- " #secondary current\n",
- "I2 = S/V2\n",
- " #primary turns\n",
- "N1 = N2*V1/V2\n",
- " #maximum flux\n",
- " #assuming E2 = V2\n",
- "Phim = V2/(4.44*f*N2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)primary current is \",round( I1,2),\" A and secondary current is \",round( I2,2),\" A\"\n",
- "print \"\\n (b)number of primary turns is \",round( N1,2),\"\"\n",
- "print \"\\n (c)maximum value of the flux is \",round(Phim*1000,2),\"mWb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)primary current is 25.0 A and secondary current is 500.0 A\n",
- "\n",
- " (b)number of primary turns is 2000.0 \n",
- "\n",
- " (c)maximum value of the flux is 9.01 mWb"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 322</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the maximum value of the flux density in the core, and\n",
- "#(b) the voltage induced in the secondary winding.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 250;# in Volts\n",
- "A = 0.03;# in m2\n",
- "N2 = 300;# sec turns\n",
- "N1 = 25;# prim turns\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #e.m.f. E1 = 4.44*f*Phim*N1\n",
- " #maximum flux density,\n",
- "Phim = V1/(4.44*f*N1)\n",
- " #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
- " #maximum core flux density\n",
- "Bm = Phim/A\n",
- " #voltage induced in the secondary winding,\n",
- "V2 = V1*N2/N1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum core flux density \",round( Bm,2),\" T\"\n",
- "print \"\\n (b)voltage induced in the secondary winding is \",round( V2,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum core flux density 1.5 T\n",
- "\n",
- " (b)voltage induced in the secondary winding is 3000.0 V"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 323</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the number of primary and secondary turns\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 500;# in Volts\n",
- "V2 = 100;# in Volts\n",
- "Bm = 1.5;# in Tesla\n",
- "A = 0.005;# in m2\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
- " #maximum core flux density\n",
- "Phim = Bm*A\n",
- " #e.m.f. E1 = 4.44*f*Phim*N1\n",
- " #primary turns,\n",
- "N1 = V1/(4.44*f*Phim)\n",
- " #secondary turns,\n",
- "N2 = V2*N1/V1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n no. of primary and secondary turns are \",round(N1,2),\" turns, and \",round(N2,2),\" turns respectively\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " no. of primary and secondary turns are 300.3 turns, and 60.06 turns respectively"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 323</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the number of primary and secondary turns and \n",
- "#(b) the cross-sectional area of the core.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "emfpt = 15;# in Volts\n",
- "V1 = 4500;# in Volts\n",
- "V2 = 225;# in Volts\n",
- "Bm = 1.4;# in Tesla\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #E.m.f. per turn, V1/N1 = V2/N2 = emfpt\n",
- " #primary turns,\n",
- "N1 = V1/emfpt\n",
- " #secondary turns,\n",
- "N2 = V2/emfpt\n",
- " #e.m.f. E1 = 4.44*f*Phim*N1\n",
- " #maximum flux density,\n",
- "Phim = V1/(4.44*f*N1)\n",
- " #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
- " #cross-sectional area\n",
- "A = Phim/Bm\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)no. of primary and secondary turns are \", N1,\" turns, and \", N2,\" turns respectively\"\n",
- "print \"\\n (b)cross-sectional area is \", round(A,4),\"m2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)no. of primary and secondary turns are 300.0 turns, and 15.0 turns respectively\n",
- "\n",
- " (b)cross-sectional area is 0.0483 m2"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 324</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the primary current and power factor when the secondary current is 100 A at a power factor of 0.85 lagging\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 2000;# prim turns\n",
- "N2 = 800;# sec turns\n",
- "I0 = 5;# in Amperes\n",
- "pf0 = 0.20;# power factor\n",
- "I2 = 100;# in Amperes\n",
- "pf2 = 0.85;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Let I01 be the component of the primary current which provides the restoring mmf. Then I01*N1 = I2*N2\n",
- "I01 = I2*N2/N1\n",
- " #If the power factor of the secondary is 0.85\n",
- "phi2 = math.acos(pf2)\n",
- " #If the power factor on no-load is 0.20,\n",
- "phi0 = math.acos(pf0)\n",
- "I1h = I0*math.cos(phi0) + I01*math.cos(phi2)\n",
- "I1v = I0*math.sin(phi0) + I01*math.sin(phi2)\n",
- " #Hence the magnitude of I1\n",
- "I1 = (I1h*I1h + I1v*I1v)**0.5\n",
- "pf1 = math.cos(math.atan(I1v/I1h))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Primary current is \", round(I1,2),\" A, and Power factor is \",round(pf1,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Primary current is 43.58 A, and Power factor is 0.8"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 328</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the equivalent resistance referred to the primary winding, \n",
- "#(b) the equivalent reactance referred to the primary winding,\n",
- "#(c) the equivalent impedance referred to the primary winding, and \n",
- "#(d) the phase angle of the impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 600;# prim turns\n",
- "N2 = 150;# sec turns\n",
- "R1 = 0.25;# in ohms\n",
- "R2 = 0.01;# in ohms\n",
- "X1 = 1.0;# in ohms\n",
- "X2 = 0.04;# in ohms\n",
- "\n",
- "#calculation:\n",
- "tr = N1/N2# turn ratio\n",
- "vr = tr# voltage ratio = turn raio, vr = V1/V2\n",
- " #equivalent resistance Re\n",
- "Re = R1 + R2*(vr**2)\n",
- " #equivalent reactance, Xe\n",
- "Xe = X1 + X2*(vr**2)\n",
- " #equivalent impedance, Ze\n",
- "Ze = (Re*Re + Xe*Xe)**0.5\n",
- " #cos(phie) = Re/Ze\n",
- "pfe = Re/Ze\n",
- "phie = math.acos(pfe)\n",
- "phied = phie*180/math.pi# in \u00b0(deg)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the equivalent resistance referred to the primary winding is \",round( Re,2),\" ohm\"\n",
- "print \"\\n (b)the equivalent reactance referred to the primary winding is \",round( Xe,2),\" ohm\"\n",
- "print \"\\n (c)the equivalent impedance referred to the primary winding is \",round( Ze,2),\" ohm\"\n",
- "print \"\\n (d)phase angle is \",round( phied,2),\"deg\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the equivalent resistance referred to the primary winding is 0.41 ohm\n",
- "\n",
- " (b)the equivalent reactance referred to the primary winding is 1.64 ohm\n",
- "\n",
- " (c)the equivalent impedance referred to the primary winding is 1.69 ohm\n",
- "\n",
- " (d)phase angle is 75.96 deg"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 329</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the regulation of the transformer.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 200;# in Volts\n",
- "V2 = 400;# in Volts\n",
- "V2L = 387.6;# in Volts\n",
- "S = 5000;# in VA\n",
- "\n",
- "#calculation:\n",
- " #regulation =(No-load secondary voltage -\u0003 terminal voltage on load)*100/no-load secondary voltage in %\n",
- "reg = (V2 - V2L)*100/V2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the regulation of the transformer is \",round(reg,2),\" percent \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the regulation of the transformer is 3.1 percent "
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 329</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the load voltage at which the mechanism operates.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "VnL = 240;# in Volts\n",
- "reg = 2.5;# in percent\n",
- "\n",
- "#calculation:\n",
- " #regulation =(No-load secondary voltage -\u0003 terminal voltage on load)*100/no-load secondary voltage in %\n",
- "VL = VnL - reg*VnL/100\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the load voltage at which the mechanism operates is \",round(VL,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the load voltage at which the mechanism operates is 234.0 V "
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 331</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the transformer efficiency at full load and 0.85 power factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 200000;# in VA\n",
- "Pc = 1500;# in Watt\n",
- "Pi = 1000;# in Watt\n",
- "pf = 0.85;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
- " #Efficiency = 1 - losses/input power\n",
- " #Full-load output power = V*I*pf\n",
- "Po = S*pf\n",
- " #Total losses\n",
- "Pl = Pc + Pi\n",
- " #Input power = output power + losses\n",
- "PI = Po + Pl\n",
- " #efficiency\n",
- "eff = 1-(Pl/PI)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the transformer efficiency at full load is \",round(eff,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the transformer efficiency at full load is 0.99"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 331</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the efficiency of the transformer\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 200000;# in VA\n",
- "Pc = 1500;# in Watt\n",
- "Pi = 1000;# in Watt\n",
- "pf = 0.85;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
- " #Efficiency = 1 - losses/input power\n",
- " #Half full-load power output = V*I*pf/2\n",
- "Po = S*pf/2\n",
- " #Copper loss (or I*I*R loss) is proportional to current squared\n",
- " #Hence the copper loss at half full-load is\n",
- "Pch = Pc/(2*2)\n",
- " #Iron loss = 1000 W (constant)\n",
- " #Total losses\n",
- "Pl = Pch + Pi\n",
- " #Input power at half full-load = output power at half full-load + losses\n",
- "PI = Po + Pl\n",
- " #efficiency\n",
- "eff = (1-(Pl/PI))*100\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the transformer efficiency at half full load is \",round(eff,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the transformer efficiency at half full load is 98.41 percent"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 332</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the efficiency of the transformer (a) on full load, and (b) on half load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 400000;# in VA\n",
- "R1 = 0.5;# in Ohm\n",
- "R2 = 0.001;# in Ohm\n",
- "V1 = 5000;# in Volts\n",
- "V2 = 320;# in Volts\n",
- "Pi = 2500;# in Watt\n",
- "pf = 0.85;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Rating = 400 kVA = V1*I1 = V2*I2\n",
- " #Hence primary current\n",
- "I1 = S/V1\n",
- " #secondary current\n",
- "I2 = S/V2\n",
- " #Total copper loss = I1*I1*R1 + I2*I2*R2,\n",
- "Pcf = I1*I1*R1 + I2*I2*R2\n",
- " #On full load, total loss = copper loss + iron loss\n",
- "Plf = Pcf + Pi\n",
- " # full-load power output = V2*I2*pf\n",
- "Pof = S*pf\n",
- " #Input power at full-load = output power at full-load + losses\n",
- "PIf = Pof + Plf\n",
- " #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
- " #Efficiency = 1 - losses/input power\n",
- "efff = (1-(Plf/PIf))*100\n",
- "\n",
- " #Half full-load power output = V*I*pf/2\n",
- "Poh = S*pf/2\n",
- " #Copper loss (or I*I*R loss) is proportional to current squared\n",
- " #Hence the copper loss at half full-load is\n",
- "Pch = Pcf/(2*2)\n",
- " #Iron loss = 2500 W (constant)\n",
- " #Total losses\n",
- "Plh = Pch + Pi\n",
- " #Input power at half full-load = output power at half full-load + losses\n",
- "PIh = Poh + Plh\n",
- " #efficiency\n",
- "effh = (1-(Plh/PIh))*100\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the transformer efficiency at full load is \", round(efff,2),\" percent\"\n",
- "print \"\\n (b)the transformer efficiency at half full load is \", round(effh,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the transformer efficiency at full load is 97.91 percent\n",
- "\n",
- " (b)the transformer efficiency at half full load is 97.88 percent"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 333</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the output kVA at which the efficiency of the transformer is a maximum, and (b) the maximum efficiency,\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 500000;# in VA\n",
- "Pcf = 4000;# in Watt\n",
- "Pi = 2500;# in Watt\n",
- "pf = 0.75;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Let x be the fraction of full load kVA at which the efficiency is a maximum.\n",
- " #The corresponding total copper loss = (4 kW)*(x**2)\n",
- " #At maximum efficiency, copper loss = iron loss Hence\n",
- "x = (Pi/Pcf)**0.5\n",
- " #Hence the output kVA at maximum efficiency\n",
- "So = x*S\n",
- " #Total loss at maximum efficiency\n",
- "Pl = 2*Pi\n",
- " #Output power\n",
- "Po = So*pf\n",
- " #Input power = output power + losses\n",
- "PI = Po + Pl\n",
- " #Efficiency = output power/input power = (input power\u00e2\u20ac\u201dlosses)/input power\n",
- " #Efficiency = 1 - losses/input power\n",
- " #Maximum efficiency\n",
- "effm = (1 - Pl/PI)*100\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the output kVA at maximum efficiency is \",round(So/1000,2),\"kVA\"\n",
- "print \"\\n max. efficiency is \",round(effm,2),\" pecent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the output kVA at maximum efficiency is 395.28 kVA\n",
- "\n",
- " max. efficiency is 98.34 pecent"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 335</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the equivalent input resistance of the transformer.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 4;# turn ratio\n",
- "RL = 100;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #the equivalent input resistance,\n",
- "Ri = RL*(tr**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the equivalent input resistance is \",round(Ri,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the equivalent input resistance is 1600.0 ohm"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 335</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the optimum turns ratio of a transformer which would match a \n",
- "#load resistance of 7 ohmto the output resistance of the amplifier.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 112;# in Ohms\n",
- "RL = 7;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #The equivalent input resistance, R1 of the transformer needs to be 112 ohm for maximum power transfer.\n",
- " #R1 = RL*(tr**2)\n",
- " # tr = N1/N2 turn ratio\n",
- "tr = (R1/RL)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the optimum turns ratio is \",tr,\": 1.0\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the optimum turns ratio is 4.0 : 1.0"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 335</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the optimum value of load resistance for maximum power transfer\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 5;# turn ratio\n",
- "R1 = 150;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #The equivalent input resistance, R1 of the transformer needs to be 150 ohm for maximum power transfer.\n",
- " #R1 = RL*(tr**2)\n",
- "RL = R1/(tr**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the optimum value of load resistance is \",round(RL,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the optimum value of load resistance is 6.0 ohm"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 335</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the primary current flowing and (b) the power dissipated in the load resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 220;# in Volts\n",
- "V2 = 1760;# in Volts\n",
- "V = 220;# in Volts\n",
- "RL = 1280;# in Ohms\n",
- "R = 2;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #Turns ratio, tr = N1/N2 = V1/V2\n",
- "tr = V1/V2\n",
- " #Equivalent input resistance of the transformer,\n",
- " #R1 = RL*(tr**2)\n",
- "R1 = RL*(tr**2)\n",
- " #Total input resistance\n",
- "Rin = R + R1\n",
- " # Primary current\n",
- "I1 = V1/Rin\n",
- " #For an ideal transformer V1/V2 = I2/I1,\n",
- "I2 = I1*tr\n",
- " #Power dissipated in load resistor RL\n",
- "P = I2*I2*RL\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) primary current flowing is \",round(I1,2),\" A\"\n",
- "print \"\\n (b) power dissipated in the load resistor is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) primary current flowing is 10.0 A\n",
- "\n",
- " (b) power dissipated in the load resistor is 2000.0 W"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 336</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of the load resistance and (b) the power dissipated in the load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 25;# teurn ratio\n",
- "V = 24;# in Volts\n",
- "R1 = 15000;# in Ohms\n",
- "Rin = 15000;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Turns ratio, tr = N1/N2 = V1/V2\n",
- " #For maximum power transfer R1 needs to be equal to 15 kohm\n",
- "RL = R1/(tr**2)\n",
- " #The total input resistance when the source is connected to the matching transformer is\n",
- "Rt = Rin + R1\n",
- " #Primary current,\n",
- "I1 = V/Rt\n",
- " #N1/N2 = I2/I1\n",
- "I2 = I1*tr\n",
- " #Power dissipated in load resistor RL\n",
- "P = I2*I2*RL\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the load resistance is \",round(RL,2),\"ohm\"\n",
- "print \"\\n (b) power dissipated in the load resistor is \",round(P*1000,2),\"mW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the load resistance is 24.0 ohm\n",
- "\n",
- " (b) power dissipated in the load resistor is 9.6 mW"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 25, page no. 337</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in each section of the winding.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 320;# in Volts\n",
- "V2 = 250;# in Volts\n",
- "S = 20000;# in VA\n",
- "\n",
- "#calculation:\n",
- " #Rating = 20 kVA = V1*I1 = V2*I2\n",
- " #Hence primary current, I1\n",
- "I1 = S/V1\n",
- " #secondary current, I2\n",
- "I2 = S/V2\n",
- " #Hence current in common part of the winding\n",
- "I = I2 - I1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current in common part of the winding is \", round(I,2),\" A\"\n",
- "print \"\\n primary current and secondary current are \",round(I1,2),\" A and \",round(I2,2),\" A respectively\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current in common part of the winding is 17.5 A\n",
- "\n",
- " primary current and secondary current are 62.5 A and 80.0 A respectively"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 26, page no. 339</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the saving in the volume of copper used in an auto transformer compared with a double-wound transformer\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1a = 200;# in Volts\n",
- "V2a = 150;# in Volts\n",
- "V1b = 500;# in Volts\n",
- "V2b = 100;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For a 200 V:150 V transformer, xa\n",
- "xa = V2a/V1a\n",
- " #volume of copper in auto transformer\n",
- "vca = (1 - xa)*100# of copper in a double-wound transformer\n",
- " #the saving is\n",
- "vsa = 100 - vca\n",
- " #For a 500 V:100 V transformer, xb\n",
- "xb = V2b/V1b\n",
- " #volume of copper in auto transformer\n",
- "vcb = (1 - xb)*100# of copper in a double-wound transformer\n",
- " #the saving is\n",
- "vsb = 100 - vcb\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)For a 200 V:150 V transformer, the saving is \", round(vsa,2),\" percent\"\n",
- "print \"\\n (b)For a 500 V:100 V transformer, the saving is \", round(vsb,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)For a 200 V:150 V transformer, the saving is 75.0 percent\n",
- "\n",
- " (b)For a 500 V:100 V transformer, the saving is 20.0 percent"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 27, page no. 340</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the secondary line voltage on no-load when the windings are connected (a) star-delta, (b) delta-star.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 500;# prim turns\n",
- "N2 = 50;# sec turns\n",
- "VL = 2400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For a star-connection, VL = Vp*(3**0.5)\n",
- "VL1s = VL\n",
- " #Primary phase voltage\n",
- "Vp1s = VL1s/(3**0.5)\n",
- " #For a delta-connection, VL = Vp\n",
- " #N1/N2 = V1/V2, from which,\n",
- " #secondary phase voltage, Vp2s\n",
- "Vp2s = Vp1s*N2/N1\n",
- "VL2d = Vp2s\n",
- "\n",
- " #For a delta-connection, VL = Vp\n",
- "VL1d = VL\n",
- " #primary phase voltage Vp1d\n",
- "Vp1d = VL1d\n",
- " #Secondary phase voltage, Vp2d\n",
- "Vp2d = Vp1d*N2/N1\n",
- " #For a star-connection, VL = Vp*(3**0.5)\n",
- "VL2s = Vp2d*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the secondary line voltage for star and delta connection are \",round(Vp2s,1),\" V \"\n",
- "print \" and \",round(VL2s,0),\" V respectively\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the secondary line voltage for star and delta connection are 138.6 V \n",
- " and 416.0 V respectively\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 28, page no. 343</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the reading on the ammeter, \n",
- "#(b) the potential difference across the ammeter and\n",
- "#(c) the total load (in VA) on the secondary.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 1;# prim turns\n",
- "N2 = 60;# sec turns\n",
- "I1 = 300;# in amperes\n",
- "Ra = 0.15;# in ohms\n",
- "R2 = 0.25;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Reading on the ammeter,\n",
- "I2 = I1*(N1/N2)\n",
- " #P.d. across the ammeter = I2*RA, where RA is the ammeter resistance\n",
- "pd = I2*Ra\n",
- " #Total resistance of secondary circuit\n",
- "Rt = Ra + R2\n",
- " #Induced e.m.f. in secondary\n",
- "V2 = I2*Rt\n",
- " #Total load on secondary\n",
- "S = V2*I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the reading on the ammeter is \",round(I2,2),\" A \"\n",
- "print \"\\n (b)potential difference across the ammeter is \",round(pd,2),\" V \"\n",
- "print \"\\n (c)total load (in VA) on the secondary is \",round(S,2),\" VA \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the reading on the ammeter is 5.0 A \n",
- "\n",
- " (b)potential difference across the ammeter is 0.75 V \n",
- "\n",
- " (c)total load (in VA) on the secondary is 10.0 VA "
- ]
- }
- ],
- "prompt_number": 28
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_2.ipynb deleted file mode 100755 index 5b0974af..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_2.ipynb +++ /dev/null @@ -1,1627 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 20: Transformers</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 317</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the secondary voltage, assuming an ideal transformer\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 500;# primary turns\n",
- "N2 = 3000;# secondary turns\n",
- "V1 = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For an ideal transformer, voltage ratio = turns ratio\n",
- "V2 = V1*N2/N1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n secondary voltage \",round(V2,2),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " secondary voltage 1440.0 V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 317</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine its output voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 2/7;# turns ratio\n",
- "V1 = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #A turns ratio of 2:7 means that the transformer has 2 turns on the primary \n",
- " #for every 7 turns on the secondary\n",
- "V2 = V1/tr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n output voltage \",round(V2,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " output voltage 840.0 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 317</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine secondary voltage and current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 8/1;# turns ratio\n",
- "I1 = 3;# in Amperes\n",
- "V1 = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #A turns ratio of 8:1 means that the transformer has 28 turns on the \n",
- " #primary for every 1turns on the secondary\n",
- "V2 = V1/tr\n",
- " #secondary current\n",
- "I2 = I1*tr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n secondary voltage is \",round(V2,2),\" V and secondary current is \", round(I2,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " secondary voltage is 30.0 V and secondary current is 24.0 A"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 318</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the transformer turns ratio and the current taken from the supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 240;# in Volts\n",
- "V2 = 12;# in Volts\n",
- "P = 150;# in Watts\n",
- "\n",
- "#calculation:\n",
- "I2 = P/V2\n",
- " #A turns ratio = Vp/Vs\n",
- "tr = V1/V2# turn ratio\n",
- " # V1/V2 = I2/I1\n",
- " #current taken from the supply\n",
- "I1 = I2*V2/V1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n turn ratio is \",round(tr,2),\" and current taken from the supply is \",round(I1,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " turn ratio is 20.0 and current taken from the supply is 0.63 A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 318</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the full-load secondary current,\n",
- "#(b) the minimum load resistance which can be connected across the secondary winding to give full load kVA, \n",
- "#(c) the primary current at full load kVA.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 5000;# in VA\n",
- "tr = 10;# turn ratio\n",
- "V1 = 2500;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #A turns ratio of 8:1 means that the transformer has 28 turns on the primary for every 1turns on the secondary\n",
- "V2 = V1/tr\n",
- " #transformer rating in volt-amperes = Vs*Is\n",
- "I2 = S/V2\n",
- " #Minimum value of load resistance\n",
- "RL = V2/I2\n",
- " # tr = I2/I1\n",
- "I1 = I2/tr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)full-load secondary current is \",round(I2,2),\" A\"\n",
- "print \"\\n (b)minimum load resistance is \",round(RL,2),\" ohm\"\n",
- "print \"\\n (c) primary current is \",round(I1,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)full-load secondary current is 20.0 A\n",
- "\n",
- " (b)minimum load resistance is 12.5 ohm\n",
- "\n",
- " (c) primary current is 2.0 A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 319</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of the magnetizing and core loss components of the noload current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 2400;# in Volts\n",
- "V2 = 400;# in Volts\n",
- "I0 = 0.5;# in Amperes\n",
- "Pc = 400;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Core loss (i.e. iron loss) P = V1*I0*cos(phi0)\n",
- "pf = Pc/(V1*I0)\n",
- "phi0 = math.acos(pf)\n",
- " #Magnetizing component\n",
- "Im = I0*math.sin(phi0)\n",
- " #Core loss component\n",
- "Ic = I0*math.cos(phi0)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)magnetizing component is \",round(Im,3),\" A and Core loss component is \",round(Ic,3),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)magnetizing component is 0.471 A and Core loss component is 0.167 A"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 320</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the iron loss current, \n",
- "#(b) the power factor on no-load, and (c) the magnetizing current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 240;# in Volts\n",
- "I0 = 0.8;# in Amperes\n",
- "P = 72;# in Watts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #Power absorbed = total core loss, P = V*I0*cos(phi0)\n",
- " #Ic = I0*cos(phi0)\n",
- "Ic = P/V\n",
- "pf = Ic/I0\n",
- " #From the right-angled triangle in Figure 20.2(b) and using\n",
- " #Pythagoras\u2019 theorem, \n",
- "Im = (I0*I0 - Ic*Ic)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) Core loss component is \",round( Ic,2),\" A\"\n",
- "print \"\\n (b) power factor is \",round( pf,2),\"\"\n",
- "print \"\\n (c)magnetizing component is \",round(Im,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) Core loss component is 0.3 A\n",
- "\n",
- " (b) power factor is 0.37 \n",
- "\n",
- " (c)magnetizing component is 0.74 A"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 321</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum value of the flux.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 100000;# in VA\n",
- "V1 = 4000;# in Volts\n",
- "V2 = 200;# in Volts\n",
- "N2 = 100;# sec turns\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #Transformer rating = V1*I1 = V2*I2\n",
- " #primary current\n",
- "I1 = S/V1\n",
- " #secondary current\n",
- "I2 = S/V2\n",
- " #primary turns\n",
- "N1 = N2*V1/V2\n",
- " #maximum flux\n",
- " #assuming E2 = V2\n",
- "Phim = V2/(4.44*f*N2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)primary current is \",round( I1,2),\" A and secondary current is \",round( I2,2),\" A\"\n",
- "print \"\\n (b)number of primary turns is \",round( N1,2),\"\"\n",
- "print \"\\n (c)maximum value of the flux is \",round(Phim*1000,2),\"mWb\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)primary current is 25.0 A and secondary current is 500.0 A\n",
- "\n",
- " (b)number of primary turns is 2000.0 \n",
- "\n",
- " (c)maximum value of the flux is 9.01 mWb"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 322</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the maximum value of the flux density in the core, and\n",
- "#(b) the voltage induced in the secondary winding.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 250;# in Volts\n",
- "A = 0.03;# in m2\n",
- "N2 = 300;# sec turns\n",
- "N1 = 25;# prim turns\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #e.m.f. E1 = 4.44*f*Phim*N1\n",
- " #maximum flux density,\n",
- "Phim = V1/(4.44*f*N1)\n",
- " #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
- " #maximum core flux density\n",
- "Bm = Phim/A\n",
- " #voltage induced in the secondary winding,\n",
- "V2 = V1*N2/N1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum core flux density \",round( Bm,2),\" T\"\n",
- "print \"\\n (b)voltage induced in the secondary winding is \",round( V2,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum core flux density 1.5 T\n",
- "\n",
- " (b)voltage induced in the secondary winding is 3000.0 V"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 323</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the number of primary and secondary turns\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 500;# in Volts\n",
- "V2 = 100;# in Volts\n",
- "Bm = 1.5;# in Tesla\n",
- "A = 0.005;# in m2\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
- " #maximum core flux density\n",
- "Phim = Bm*A\n",
- " #e.m.f. E1 = 4.44*f*Phim*N1\n",
- " #primary turns,\n",
- "N1 = V1/(4.44*f*Phim)\n",
- " #secondary turns,\n",
- "N2 = V2*N1/V1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n no. of primary and secondary turns are \",round(N1,2),\" turns, and \",round(N2,2),\" turns respectively\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " no. of primary and secondary turns are 300.3 turns, and 60.06 turns respectively"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 323</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the number of primary and secondary turns and \n",
- "#(b) the cross-sectional area of the core.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "emfpt = 15;# in Volts\n",
- "V1 = 4500;# in Volts\n",
- "V2 = 225;# in Volts\n",
- "Bm = 1.4;# in Tesla\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #E.m.f. per turn, V1/N1 = V2/N2 = emfpt\n",
- " #primary turns,\n",
- "N1 = V1/emfpt\n",
- " #secondary turns,\n",
- "N2 = V2/emfpt\n",
- " #e.m.f. E1 = 4.44*f*Phim*N1\n",
- " #maximum flux density,\n",
- "Phim = V1/(4.44*f*N1)\n",
- " #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
- " #cross-sectional area\n",
- "A = Phim/Bm\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)no. of primary and secondary turns are \", N1,\" turns, and \", N2,\" turns respectively\"\n",
- "print \"\\n (b)cross-sectional area is \", round(A,4),\"m2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)no. of primary and secondary turns are 300.0 turns, and 15.0 turns respectively\n",
- "\n",
- " (b)cross-sectional area is 0.0483 m2"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 324</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the primary current and power factor when the secondary current is 100 A at a power factor of 0.85 lagging\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 2000;# prim turns\n",
- "N2 = 800;# sec turns\n",
- "I0 = 5;# in Amperes\n",
- "pf0 = 0.20;# power factor\n",
- "I2 = 100;# in Amperes\n",
- "pf2 = 0.85;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Let I01 be the component of the primary current which provides the restoring mmf. Then I01*N1 = I2*N2\n",
- "I01 = I2*N2/N1\n",
- " #If the power factor of the secondary is 0.85\n",
- "phi2 = math.acos(pf2)\n",
- " #If the power factor on no-load is 0.20,\n",
- "phi0 = math.acos(pf0)\n",
- "I1h = I0*math.cos(phi0) + I01*math.cos(phi2)\n",
- "I1v = I0*math.sin(phi0) + I01*math.sin(phi2)\n",
- " #Hence the magnitude of I1\n",
- "I1 = (I1h*I1h + I1v*I1v)**0.5\n",
- "pf1 = math.cos(math.atan(I1v/I1h))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Primary current is \", round(I1,2),\" A, and Power factor is \",round(pf1,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Primary current is 43.58 A, and Power factor is 0.8"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 328</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the equivalent resistance referred to the primary winding, \n",
- "#(b) the equivalent reactance referred to the primary winding,\n",
- "#(c) the equivalent impedance referred to the primary winding, and \n",
- "#(d) the phase angle of the impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 600;# prim turns\n",
- "N2 = 150;# sec turns\n",
- "R1 = 0.25;# in ohms\n",
- "R2 = 0.01;# in ohms\n",
- "X1 = 1.0;# in ohms\n",
- "X2 = 0.04;# in ohms\n",
- "\n",
- "#calculation:\n",
- "tr = N1/N2# turn ratio\n",
- "vr = tr# voltage ratio = turn raio, vr = V1/V2\n",
- " #equivalent resistance Re\n",
- "Re = R1 + R2*(vr**2)\n",
- " #equivalent reactance, Xe\n",
- "Xe = X1 + X2*(vr**2)\n",
- " #equivalent impedance, Ze\n",
- "Ze = (Re*Re + Xe*Xe)**0.5\n",
- " #cos(phie) = Re/Ze\n",
- "pfe = Re/Ze\n",
- "phie = math.acos(pfe)\n",
- "phied = phie*180/math.pi# in \u00b0(deg)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the equivalent resistance referred to the primary winding is \",round( Re,2),\" ohm\"\n",
- "print \"\\n (b)the equivalent reactance referred to the primary winding is \",round( Xe,2),\" ohm\"\n",
- "print \"\\n (c)the equivalent impedance referred to the primary winding is \",round( Ze,2),\" ohm\"\n",
- "print \"\\n (d)phase angle is \",round( phied,2),\"deg\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the equivalent resistance referred to the primary winding is 0.41 ohm\n",
- "\n",
- " (b)the equivalent reactance referred to the primary winding is 1.64 ohm\n",
- "\n",
- " (c)the equivalent impedance referred to the primary winding is 1.69 ohm\n",
- "\n",
- " (d)phase angle is 75.96 deg"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 329</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the regulation of the transformer.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 200;# in Volts\n",
- "V2 = 400;# in Volts\n",
- "V2L = 387.6;# in Volts\n",
- "S = 5000;# in VA\n",
- "\n",
- "#calculation:\n",
- " #regulation =(No-load secondary voltage -\u0003 terminal voltage on load)*100/no-load secondary voltage in %\n",
- "reg = (V2 - V2L)*100/V2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the regulation of the transformer is \",round(reg,2),\" percent \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the regulation of the transformer is 3.1 percent "
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 329</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the load voltage at which the mechanism operates.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "VnL = 240;# in Volts\n",
- "reg = 2.5;# in percent\n",
- "\n",
- "#calculation:\n",
- " #regulation =(No-load secondary voltage -\u0003 terminal voltage on load)*100/no-load secondary voltage in %\n",
- "VL = VnL - reg*VnL/100\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the load voltage at which the mechanism operates is \",round(VL,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the load voltage at which the mechanism operates is 234.0 V "
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 331</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the transformer efficiency at full load and 0.85 power factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 200000;# in VA\n",
- "Pc = 1500;# in Watt\n",
- "Pi = 1000;# in Watt\n",
- "pf = 0.85;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
- " #Efficiency = 1 - losses/input power\n",
- " #Full-load output power = V*I*pf\n",
- "Po = S*pf\n",
- " #Total losses\n",
- "Pl = Pc + Pi\n",
- " #Input power = output power + losses\n",
- "PI = Po + Pl\n",
- " #efficiency\n",
- "eff = 1-(Pl/PI)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the transformer efficiency at full load is \",round(eff,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the transformer efficiency at full load is 0.99"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 331</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the efficiency of the transformer\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 200000;# in VA\n",
- "Pc = 1500;# in Watt\n",
- "Pi = 1000;# in Watt\n",
- "pf = 0.85;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
- " #Efficiency = 1 - losses/input power\n",
- " #Half full-load power output = V*I*pf/2\n",
- "Po = S*pf/2\n",
- " #Copper loss (or I*I*R loss) is proportional to current squared\n",
- " #Hence the copper loss at half full-load is\n",
- "Pch = Pc/(2*2)\n",
- " #Iron loss = 1000 W (constant)\n",
- " #Total losses\n",
- "Pl = Pch + Pi\n",
- " #Input power at half full-load = output power at half full-load + losses\n",
- "PI = Po + Pl\n",
- " #efficiency\n",
- "eff = (1-(Pl/PI))*100\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the transformer efficiency at half full load is \",round(eff,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the transformer efficiency at half full load is 98.41 percent"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 332</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the efficiency of the transformer (a) on full load, and (b) on half load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 400000;# in VA\n",
- "R1 = 0.5;# in Ohm\n",
- "R2 = 0.001;# in Ohm\n",
- "V1 = 5000;# in Volts\n",
- "V2 = 320;# in Volts\n",
- "Pi = 2500;# in Watt\n",
- "pf = 0.85;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Rating = 400 kVA = V1*I1 = V2*I2\n",
- " #Hence primary current\n",
- "I1 = S/V1\n",
- " #secondary current\n",
- "I2 = S/V2\n",
- " #Total copper loss = I1*I1*R1 + I2*I2*R2,\n",
- "Pcf = I1*I1*R1 + I2*I2*R2\n",
- " #On full load, total loss = copper loss + iron loss\n",
- "Plf = Pcf + Pi\n",
- " # full-load power output = V2*I2*pf\n",
- "Pof = S*pf\n",
- " #Input power at full-load = output power at full-load + losses\n",
- "PIf = Pof + Plf\n",
- " #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
- " #Efficiency = 1 - losses/input power\n",
- "efff = (1-(Plf/PIf))*100\n",
- "\n",
- " #Half full-load power output = V*I*pf/2\n",
- "Poh = S*pf/2\n",
- " #Copper loss (or I*I*R loss) is proportional to current squared\n",
- " #Hence the copper loss at half full-load is\n",
- "Pch = Pcf/(2*2)\n",
- " #Iron loss = 2500 W (constant)\n",
- " #Total losses\n",
- "Plh = Pch + Pi\n",
- " #Input power at half full-load = output power at half full-load + losses\n",
- "PIh = Poh + Plh\n",
- " #efficiency\n",
- "effh = (1-(Plh/PIh))*100\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the transformer efficiency at full load is \", round(efff,2),\" percent\"\n",
- "print \"\\n (b)the transformer efficiency at half full load is \", round(effh,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the transformer efficiency at full load is 97.91 percent\n",
- "\n",
- " (b)the transformer efficiency at half full load is 97.88 percent"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 333</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the output kVA at which the efficiency of the transformer is a maximum, and (b) the maximum efficiency,\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "S = 500000;# in VA\n",
- "Pcf = 4000;# in Watt\n",
- "Pi = 2500;# in Watt\n",
- "pf = 0.75;# power factor\n",
- "\n",
- "#calculation:\n",
- " #Let x be the fraction of full load kVA at which the efficiency is a maximum.\n",
- " #The corresponding total copper loss = (4 kW)*(x**2)\n",
- " #At maximum efficiency, copper loss = iron loss Hence\n",
- "x = (Pi/Pcf)**0.5\n",
- " #Hence the output kVA at maximum efficiency\n",
- "So = x*S\n",
- " #Total loss at maximum efficiency\n",
- "Pl = 2*Pi\n",
- " #Output power\n",
- "Po = So*pf\n",
- " #Input power = output power + losses\n",
- "PI = Po + Pl\n",
- " #Efficiency = output power/input power = (input power\u00e2\u20ac\u201dlosses)/input power\n",
- " #Efficiency = 1 - losses/input power\n",
- " #Maximum efficiency\n",
- "effm = (1 - Pl/PI)*100\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the output kVA at maximum efficiency is \",round(So/1000,2),\"kVA\"\n",
- "print \"\\n max. efficiency is \",round(effm,2),\" pecent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the output kVA at maximum efficiency is 395.28 kVA\n",
- "\n",
- " max. efficiency is 98.34 pecent"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 335</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the equivalent input resistance of the transformer.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 4;# turn ratio\n",
- "RL = 100;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #the equivalent input resistance,\n",
- "Ri = RL*(tr**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the equivalent input resistance is \",round(Ri,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the equivalent input resistance is 1600.0 ohm"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 335</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the optimum turns ratio of a transformer which would match a \n",
- "#load resistance of 7 ohmto the output resistance of the amplifier.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 112;# in Ohms\n",
- "RL = 7;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #The equivalent input resistance, R1 of the transformer needs to be 112 ohm for maximum power transfer.\n",
- " #R1 = RL*(tr**2)\n",
- " # tr = N1/N2 turn ratio\n",
- "tr = (R1/RL)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the optimum turns ratio is \",tr,\": 1.0\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the optimum turns ratio is 4.0 : 1.0"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 335</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the optimum value of load resistance for maximum power transfer\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 5;# turn ratio\n",
- "R1 = 150;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #The equivalent input resistance, R1 of the transformer needs to be 150 ohm for maximum power transfer.\n",
- " #R1 = RL*(tr**2)\n",
- "RL = R1/(tr**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the optimum value of load resistance is \",round(RL,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the optimum value of load resistance is 6.0 ohm"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 335</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the primary current flowing and (b) the power dissipated in the load resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 220;# in Volts\n",
- "V2 = 1760;# in Volts\n",
- "V = 220;# in Volts\n",
- "RL = 1280;# in Ohms\n",
- "R = 2;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #Turns ratio, tr = N1/N2 = V1/V2\n",
- "tr = V1/V2\n",
- " #Equivalent input resistance of the transformer,\n",
- " #R1 = RL*(tr**2)\n",
- "R1 = RL*(tr**2)\n",
- " #Total input resistance\n",
- "Rin = R + R1\n",
- " # Primary current\n",
- "I1 = V1/Rin\n",
- " #For an ideal transformer V1/V2 = I2/I1,\n",
- "I2 = I1*tr\n",
- " #Power dissipated in load resistor RL\n",
- "P = I2*I2*RL\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) primary current flowing is \",round(I1,2),\" A\"\n",
- "print \"\\n (b) power dissipated in the load resistor is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) primary current flowing is 10.0 A\n",
- "\n",
- " (b) power dissipated in the load resistor is 2000.0 W"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 336</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of the load resistance and (b) the power dissipated in the load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "tr = 25;# teurn ratio\n",
- "V = 24;# in Volts\n",
- "R1 = 15000;# in Ohms\n",
- "Rin = 15000;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Turns ratio, tr = N1/N2 = V1/V2\n",
- " #For maximum power transfer R1 needs to be equal to 15 kohm\n",
- "RL = R1/(tr**2)\n",
- " #The total input resistance when the source is connected to the matching transformer is\n",
- "Rt = Rin + R1\n",
- " #Primary current,\n",
- "I1 = V/Rt\n",
- " #N1/N2 = I2/I1\n",
- "I2 = I1*tr\n",
- " #Power dissipated in load resistor RL\n",
- "P = I2*I2*RL\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the load resistance is \",round(RL,2),\"ohm\"\n",
- "print \"\\n (b) power dissipated in the load resistor is \",round(P*1000,2),\"mW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the load resistance is 24.0 ohm\n",
- "\n",
- " (b) power dissipated in the load resistor is 9.6 mW"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 25, page no. 337</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current in each section of the winding.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1 = 320;# in Volts\n",
- "V2 = 250;# in Volts\n",
- "S = 20000;# in VA\n",
- "\n",
- "#calculation:\n",
- " #Rating = 20 kVA = V1*I1 = V2*I2\n",
- " #Hence primary current, I1\n",
- "I1 = S/V1\n",
- " #secondary current, I2\n",
- "I2 = S/V2\n",
- " #Hence current in common part of the winding\n",
- "I = I2 - I1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current in common part of the winding is \", round(I,2),\" A\"\n",
- "print \"\\n primary current and secondary current are \",round(I1,2),\" A and \",round(I2,2),\" A respectively\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current in common part of the winding is 17.5 A\n",
- "\n",
- " primary current and secondary current are 62.5 A and 80.0 A respectively"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 26, page no. 339</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the saving in the volume of copper used in an auto transformer compared with a double-wound transformer\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V1a = 200;# in Volts\n",
- "V2a = 150;# in Volts\n",
- "V1b = 500;# in Volts\n",
- "V2b = 100;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For a 200 V:150 V transformer, xa\n",
- "xa = V2a/V1a\n",
- " #volume of copper in auto transformer\n",
- "vca = (1 - xa)*100# of copper in a double-wound transformer\n",
- " #the saving is\n",
- "vsa = 100 - vca\n",
- " #For a 500 V:100 V transformer, xb\n",
- "xb = V2b/V1b\n",
- " #volume of copper in auto transformer\n",
- "vcb = (1 - xb)*100# of copper in a double-wound transformer\n",
- " #the saving is\n",
- "vsb = 100 - vcb\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)For a 200 V:150 V transformer, the saving is \", round(vsa,2),\" percent\"\n",
- "print \"\\n (b)For a 500 V:100 V transformer, the saving is \", round(vsb,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)For a 200 V:150 V transformer, the saving is 75.0 percent\n",
- "\n",
- " (b)For a 500 V:100 V transformer, the saving is 20.0 percent"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 27, page no. 340</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the secondary line voltage on no-load when the windings are connected (a) star-delta, (b) delta-star.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 500;# prim turns\n",
- "N2 = 50;# sec turns\n",
- "VL = 2400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #For a star-connection, VL = Vp*(3**0.5)\n",
- "VL1s = VL\n",
- " #Primary phase voltage\n",
- "Vp1s = VL1s/(3**0.5)\n",
- " #For a delta-connection, VL = Vp\n",
- " #N1/N2 = V1/V2, from which,\n",
- " #secondary phase voltage, Vp2s\n",
- "Vp2s = Vp1s*N2/N1\n",
- "VL2d = Vp2s\n",
- "\n",
- " #For a delta-connection, VL = Vp\n",
- "VL1d = VL\n",
- " #primary phase voltage Vp1d\n",
- "Vp1d = VL1d\n",
- " #Secondary phase voltage, Vp2d\n",
- "Vp2d = Vp1d*N2/N1\n",
- " #For a star-connection, VL = Vp*(3**0.5)\n",
- "VL2s = Vp2d*(3**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the secondary line voltage for star and delta connection are \",round(Vp2s,1),\" V \"\n",
- "print \" and \",round(VL2s,0),\" V respectively\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the secondary line voltage for star and delta connection are 138.6 V \n",
- " and 416.0 V respectively\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 28, page no. 343</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the reading on the ammeter, \n",
- "#(b) the potential difference across the ammeter and\n",
- "#(c) the total load (in VA) on the secondary.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "N1 = 1;# prim turns\n",
- "N2 = 60;# sec turns\n",
- "I1 = 300;# in amperes\n",
- "Ra = 0.15;# in ohms\n",
- "R2 = 0.25;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Reading on the ammeter,\n",
- "I2 = I1*(N1/N2)\n",
- " #P.d. across the ammeter = I2*RA, where RA is the ammeter resistance\n",
- "pd = I2*Ra\n",
- " #Total resistance of secondary circuit\n",
- "Rt = Ra + R2\n",
- " #Induced e.m.f. in secondary\n",
- "V2 = I2*Rt\n",
- " #Total load on secondary\n",
- "S = V2*I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the reading on the ammeter is \",round(I2,2),\" A \"\n",
- "print \"\\n (b)potential difference across the ammeter is \",round(pd,2),\" V \"\n",
- "print \"\\n (c)total load (in VA) on the secondary is \",round(S,2),\" VA \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the reading on the ammeter is 5.0 A \n",
- "\n",
- " (b)potential difference across the ammeter is 0.75 V \n",
- "\n",
- " (c)total load (in VA) on the secondary is 10.0 VA "
- ]
- }
- ],
- "prompt_number": 28
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_20.ipynb index 538ace20..538ace20 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_20.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint.ipynb deleted file mode 100755 index 28433f9f..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint.ipynb +++ /dev/null @@ -1,1641 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 21: D.c. machines</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 354</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the generated e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 600;# no. of conductors\n",
- "c = 2;# for a wave winding\n",
- "p = 4;# no. of pairs\n",
- "n = 625/60;# in rev/sec\n",
- "Phi = 20E-3;# in Wb\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- "E = 2*p*Phi*n*Z/c\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the generated e.m.f is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the generated e.m.f is 500.0 V "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 354</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the speed at which the machine must be driven to generate an e.m.f. of 240 V\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 50*16;# no. of conductors\n",
- "p = 1;# let no. of pairs\n",
- "c = 2*p;# for a lap winding\n",
- "Phi = 30E-3;# in Wb\n",
- "E = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- " #Rearranging gives, speed\n",
- "n = E*c/(2*p*Phi*Z)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the speed at which the machine must be driven is \",round(n,2),\" rev/sec \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the speed at which the machine must be driven is 10.0 rev/sec "
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 354</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. generated when running at 500 rev/min.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 1200;# no. of conductors\n",
- "p = 1;# let, no. of pairs\n",
- "c = 2*p;# for a lap winding\n",
- "Phi = 30E-3;# in Wb\n",
- "n = 500/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- "E = 2*p*Phi*n*Z/c\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Generated e.m.f. is 300.0 V "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 355</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the generated e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 1200;# no. of conductors\n",
- "p = 4;# let, no. of pairs\n",
- "c = 2;# for a wave winding\n",
- "Phi = 30E-3;# in Wb\n",
- "n = 500/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- "E = 2*p*Phi*n*Z/c\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Generated e.m.f. is 1200.0 V "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 355</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the change in the generated voltage when the field current is reduced by 20%, \n",
- "#assuming the flux is proportional to the field current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 150; # in Volts\n",
- "x = 0.2;\n",
- "\n",
- "#calculation:\n",
- "E2 = E1*(1- x)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Generated e.m.f. is \",round(E2,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Generated e.m.f. is 120.0 V "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 356</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the percentage increase in the flux per pole required to generate 250 V at 20 rev/s.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "n1 = 30;# in rev/sec\n",
- "E1 = 200;# in Volts\n",
- "n2 = 20;# in rev/sec\n",
- "E2 = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #generated e.m.f., E proportional to phi*w and since w = 2*pi*n, then\n",
- " # E proportional to phi*n\n",
- " # E1/E2 = Phi1*n1/(Phi2*n2)\n",
- " # let Phi2/Phi1 = Phi\n",
- "Phi = E2*n1/(E1*n2)\n",
- "Phi_inc = (Phi - 1)*100#/in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n percentage increase in the flux per pole is \",round(Phi_inc,2),\" percent \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " percentage increase in the flux per pole is 87.5 percent "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 357</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the terminal voltage of a generator\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ra = 0.30;# in ohms\n",
- "Ia = 30;# in Amperes\n",
- "E = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #terminal voltage,\n",
- " #V = E - Ia*Ra\n",
- "V = E - Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n terminal voltage of a generator is \",round(V,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " terminal voltage of a generator is 191.0 V "
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 357</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the terminal voltage, and (b) the generated e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "RL = 60;# in ohms\n",
- "Ia = 8;# in Amperes\n",
- "Ra = 1;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #terminal voltage,\n",
- " #V = Ia*RL\n",
- "V = Ia*RL\n",
- " #Generated e.m.f., E\n",
- "E = V + Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)terminal voltage is \",round(V,2),\" V \"\n",
- "print \"\\n (b)generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)terminal voltage is 480.0 V \n",
- "\n",
- " (b)generated e.m.f. is 488.0 V "
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 357</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the generated e.m.f. when \n",
- "#(a) the speed increases to 25 rev/s and the pole flux remains unchanged, \n",
- "#(b) the speed remains at 20 rev/s and the pole flux is decreased to 0.08 Wb,\n",
- "#and (c) the speed increases to 24 rev/s and the pole flux is decreased to 0.07 Wb.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 150;# in Volts\n",
- "n1 = 20;# in rev/sec\n",
- "Phi1 = 0.10;# in Wb\n",
- "n2 = 25;# in rev/sec\n",
- "Phi2 = 0.10;# in Wb\n",
- "n3 = 20;# in rev/sec\n",
- "Phi3 = 0.08;# in Wb\n",
- "n4 = 24;# in rev/sec\n",
- "Phi4 = 0.07;# in Wb\n",
- "\n",
- "#calculation:\n",
- " #generated e.m.f., E proportional to phi*w and since w = 2*pi*n, then\n",
- " # E proportional to phi*n\n",
- " # E1/E2 = Phi1*n1/(Phi2*n2)\n",
- "E2 = E1*Phi2*n2/(Phi1*n1)\n",
- "E3 = E1*Phi3*n3/(Phi1*n1)\n",
- "E4 = E1*Phi4*n4/(Phi1*n1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the generated e.m.f is \",round(E2,2),\" V \"\n",
- "print \"\\n (b)generated e.m.f. is \",round(E3,2),\" V \"\n",
- "print \"\\n (c)generated e.m.f. is \",round(E4,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the generated e.m.f is 187.5 V \n",
- "\n",
- " (b)generated e.m.f. is 120.0 V \n",
- "\n",
- " (c)generated e.m.f. is 126.0 V "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 359</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the terminal voltage, and (b) the e.m.f. generated in the armature.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ps = 20000;# in Watts\n",
- "Vs = 200;# in Volts\n",
- "Rs = 0.1;# in ohms\n",
- "Rf = 50;# in ohms\n",
- "Ra = 0.04;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Load current, I\n",
- "Is = Ps/Vs\n",
- " #Volt drop in the cables to the load\n",
- "Vd = Is*Rs\n",
- " #Hence terminal voltage,\n",
- "V = Vs + Vd\n",
- " #Field current, If\n",
- "If = V/Rf\n",
- " #Armature current\n",
- "Ia = If + Is\n",
- " #Generated e.m.f. E\n",
- "E = V + Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)terminal voltage is \",round(V,2),\" V \"\n",
- "print \"\\n (b)generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)terminal voltage is 210.0 V \n",
- "\n",
- " (b)generated e.m.f. is 214.17 V "
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 361</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the e.m.f. generated.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Is = 80;# in amperes\n",
- "Vs = 200;# in Volts\n",
- "Rf = 40;# in ohms\n",
- "Rse = 0.02;# in ohms\n",
- "Ra = 0.04;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Volt drop in series winding\n",
- "Vse = Is*Rse\n",
- " #P.d. across the field winding = p.d. across armature\n",
- "V1 = Vs + Vse\n",
- " #Field current, If\n",
- "If = V1/Rf\n",
- " #Armature current\n",
- "Ia = If + Is\n",
- " #Generated e.m.f. E\n",
- "E = V1 + Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " generated e.m.f. is 205.0 V "
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 363</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the efficiency of the generator at full load\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ps = 10000;# in Watt\n",
- "Pl = 600;# in Watt\n",
- "Ra = 0.75;# in ohms\n",
- "Rf = 125;# in ohms\n",
- "V = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Output power Ps = V*I\n",
- " #from which, load current I\n",
- "I = Ps/V\n",
- " #Field current, If\n",
- "If = V/Rf\n",
- " #Armature current\n",
- "Ia = If + I\n",
- " #Efficiency,\n",
- "eff = Ps*100/((V*I) + (Ia*Ia*Ra) + (If*V) + (Pl))# in Percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Efficiency is \",round(eff,2),\" percent \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Efficiency is 80.5 percent "
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 364</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the back e.m.f. when the armature current is 50 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ra = 0.2;# in ohms\n",
- "V = 240;# in Volts\n",
- "Ia = 50;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #For a motor, V = E + Ia*Ra\n",
- "E = V - Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n back e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " back e.m.f. is 230.0 V "
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 365</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the e.m.f. generated when it is running: (a) as a generator giving 100 A, and\n",
- "#(b) as a motor taking 80 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ra = 0.25;# in ohms\n",
- "V = 300;# in Volts\n",
- "Ig = 100;# in Amperes\n",
- "Im = 80;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #As a generator, generated e.m.f.,\n",
- " # E = V + Ia*Ra\n",
- "Eg = V + Ig*Ra\n",
- " #For a motor, generated e.m.f. (or back e.m.f.),\n",
- " # E = V - Ia*Ra\n",
- "E = V - Im*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)As a generator, generated e.m.f. is \",round(Eg,2),\" V \"\n",
- "print \"\\n (b)back e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)As a generator, generated e.m.f. is 325.0 V \n",
- "\n",
- " (b)back e.m.f. is 280.0 V "
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 366</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the torque exerted when a current of 30 A flows in each armature conductor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 4;\n",
- "c = 2;# for a wave winding\n",
- "Phi = 25E-3;# Wb\n",
- "Z = 900;\n",
- "Ia = 30;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #torque T = p*Phi*Z*Ia/(pi*c)\n",
- "T = p*Phi*Z*Ia/(1*math.pi*c)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the torque exerted is \",round(T,2),\" Nm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the torque exerted is 429.72 Nm "
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 366</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the torque\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 350;# in Volts\n",
- "Ra = 0.5;# in ohms\n",
- "n = 15;# in rev/sec\n",
- "Ia = 60;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #Back e.m.f. E = V - Ia*Ra\n",
- "E = V - Ia*Ra\n",
- " #torque T = E*Ia/(2*n*pi)\n",
- "T = E*Ia/(2*n*math.pi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the torque exerted is \",round(T,2),\" Nm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the torque exerted is 203.72 Nm "
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 366</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the speed and (b) the torque developed when the armature current is 40 A\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 1;# let\n",
- "c = 2*p;# for a lap winding\n",
- "Phi = 20E-3;# Wb\n",
- "Z = 500;\n",
- "V = 250;# in Volts\n",
- "Ra = 1;# in ohms\n",
- "Ia = 40;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #Back e.m.f. E = V - Ia*Ra\n",
- "E = V - Ia*Ra\n",
- " #E.m.f. E = 2*p*Phi*n*Z/c\n",
- " # rearrange,\n",
- "n = E*c/(2*p*Phi*Z)\n",
- " #torque T = E*Ia/(2*n*pi)\n",
- "T = E*Ia/(2*n*math.pi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)speed n is \",round(n,2),\" rev/sec \"\n",
- "print \"\\n (b)the torque exerted is \",round(T,2),\" Nm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)speed n is 21.0 rev/sec \n",
- "\n",
- " (b)the torque exerted is 63.66 Nm "
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 367</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the armature current at this new value of torque.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T1 = 25;# in Nm\n",
- "T2 = 35;# in Nm\n",
- "Ia1 = 16;# in Amperes\n",
- "V = 100;# in Volts\n",
- "x = 0.15;\n",
- "\n",
- " #calculation:\n",
- " #the shaft torque T of a generator is proportional to (phi*Ia),\n",
- " #where Phi is the flux and Ia is the armature current. Thus, T = k*Phi*Ia, where k is a constant.\n",
- " #The torque at flux phi1 and armature current Ia1 is T1 = k*Phi1*Ia1.\n",
- " #similarly T2 = k*Phi2*Ia2\n",
- "\n",
- "Ia2 = T2*Ia1/(0.85*T1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n armature current at the new value of torque is \",round(Ia2,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " armature current at the new value of torque is 26.35 A "
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 367</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the efficiency of the generator and (b) the power loss in the generator.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T = 12;# in Nm\n",
- "I = 15;# in Amperes\n",
- "V = 100;# in Volts\n",
- "n = 1500/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #the efficiency of a generator = (output power/input power)*100 %\n",
- " #The output power is the electrical output, i.e. VI watts. \n",
- " #The input power to a generator is the mechanical power in the shaft driving the generator, \n",
- " #i.e. T*w or T(2*pi*n) watts, where T is the torque in Nm and n is speed of rotation in rev/s. Hence, for a generator \n",
- " #efficiency = V*I*100/(T*2*pi*n) %\n",
- "eff = V*I*100/(T*2*math.pi*n)# in Percent\n",
- " #The input power = output power + losses\n",
- " # hence, T*2*math.pi*n = V*I + losses\n",
- "Pl = T*2*math.pi*n - V*I\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) efficiency is \",round(eff,2),\" % \"\n",
- "print \"\\n (b) power loss is \",round(Pl,2),\" W \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) efficiency is 79.58 % \n",
- "\n",
- " (b) power loss is 384.96 W "
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 368</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current in the armature, and (b) the back e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rf = 150;# in Ohms\n",
- "Ra = 0.4;# in Ohms\n",
- "I = 30;# in Amperes\n",
- "V = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Field current If\n",
- "If = V/Rf\n",
- " #Supply current I = Ia + If\n",
- " #Hence armature current, Ia\n",
- "Ia = I - If\n",
- " #Back e.m.f. E = V -\u0004 Ia*Ra\n",
- "E = V - (Ia*Ra)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) current in the armature is \",round(Ia,2),\" A \"\n",
- "print \"\\n (b) Back e.m.f. E is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) current in the armature is 28.4 A \n",
- "\n",
- " (b) Back e.m.f. E is 228.64 V "
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 370</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the speed of the motor, assuming the flux remains constant.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 30;# in Amperes\n",
- "Ia2 = 45;# in Amperes\n",
- "Ra = 0.4;# in ohm\n",
- "n1 = 1350/60;# in Rev/sec\n",
- "V = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #The relationship E proportional to (Phi*n) applies to both generators and motors. For a motor,\n",
- " #E = V - (Ia*Ra)\n",
- "E1 = V - (Ia1*Ra)\n",
- "E2 = V - (Ia2*Ra)\n",
- " #The relationship, E1/E2 = Phi1*n1/Phi2*n2, applies to both generators and motors.\n",
- " #Since the flux is constant, Phi1 = Phi2\n",
- "n2 = E2*n1/(E1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the speed of the motor is \",round(n2,2),\" rev/sec \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the speed of the motor is 21.78 rev/sec "
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 370</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and \n",
- "#(b) the steadystate value of the armature current at the new value of flux,\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 30;# in Amperes\n",
- "Ra = 0.4;# in ohm\n",
- "n = 800/60;# in Rev/sec\n",
- "V = 220;# in Volts\n",
- "x= 0.1;\n",
- "\n",
- "#calculation:\n",
- " #For a d.c. shunt-wound motor, E = V - (Ia*Ra),Hence initial generated e.m.f.,\n",
- "E1 = V - (Ia1*Ra)\n",
- " #The generated e.m.f. is also such that E proportional to (Phi*n) \n",
- " #so at the instant the flux is reduced, the speed has not had time to change, and\n",
- "E = E1*(1-x)\n",
- " #Hence, the voltage drop due to the armature resistance is\n",
- "Vd = V - E\n",
- " #The instantaneous value of the current is\n",
- "Ia = Vd/Ra\n",
- " #T proportional to (Phi*Ia), since the torque is constant,\n",
- " #Phi1*Ia1 = Phi2*Ia2, The flux 8 is reduced by 10%, hence\n",
- "Ia2 = Ia1/0.9\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)instantaneous value of the current \",round(Ia,2),\" A \"\n",
- "print \"\\n (b)steady state value of armature current, \",round(Ia2,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)instantaneous value of the current 82.0 A \n",
- "\n",
- " (b)steady state value of armature current, 33.33 A "
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 372</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the generated e.m.f. at this load. \n",
- "#(b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A.\n",
- "#Assume that this causes a doubling of the flux.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 15;# in Amperes\n",
- "Ia2 = 30;# in Amperes\n",
- "Rf = 0.3;# in ohms\n",
- "Ra = 0.2;# in ohm\n",
- "n1 = 24;# in Rev/sec\n",
- "V = 240;# in Volts\n",
- "x= 2;\n",
- "\n",
- "#calculation:\n",
- " #generated e.m.f., E, at initial load, is given by\n",
- "E1 = V - Ia1*(Ra + Rf)\n",
- " #When the current is increased to 30 A, the generated e.m.f. is given by:\n",
- "E2 = V - Ia2*(Ra + Rf)\n",
- " #E proportional to (Phi*n)\n",
- " #E1/E2 = Phi1*n1/Phi2*n2\n",
- "n2 = E2*n1/(2*E1) \n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)generated e.m.f., E is \",round(E1,2),\" V \"\n",
- "print \"\\n (b)speed of motor, n2, \",round(n2,2),\" rev/sec \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)generated e.m.f., E is 232.5 V \n",
- "\n",
- " (b)speed of motor, n2, 11.61 rev/sec "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 374</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the overall efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 80;# in Amperes\n",
- "C = 1500;# in Watt\n",
- "Rf = 40;# in ohms\n",
- "Ra = 0.2;# in ohm\n",
- "n = 1000/60;# in Rev/sec\n",
- "V = 320;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Field current, If\n",
- "If = V/Rf\n",
- " #Armature current Ia\n",
- "Ia = I - If\n",
- " #Efficiency\n",
- "eff = ((V*I - (Ia*Ia*Ra) - (If*V) - C)/(V*I))*100 # in percent\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n efficiency is\",round(eff,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " efficiency is 80.09 %"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 25, page no. 374</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the maximum efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 40;# in Amperes\n",
- "Rf = 0.05;# in ohms\n",
- "Ra = 0.15;# in ohm\n",
- "V = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #However for a series motor, If = 0 and the Ia*Ia*Ra loss needs to be I*I*(\u0011Ra + Rf)\n",
- " #For maximum efficiency I*I*\u0011(Ra + Rf) = C\n",
- " #Efficiency\n",
- "eff = ((V*I - (2*I*I*(Ra + Rf)))/(V*I))*100 # in percent\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n efficiency is\",round(eff,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " efficiency is 93.6"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 26, page no. 375</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current supplied to the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T = 15;# in Nm\n",
- "n = 1200/60;# in rev/sec\n",
- "eff = 0.8;\n",
- "V = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- "I = T*2*math.pi*n/(V*eff)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current supplied, I is \",round(I,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current supplied, I is 11.78 A"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 27, page no. 376</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 2;# in ohm\n",
- "n = 30;# in rev/sec\n",
- "I = 10;# in A\n",
- "C = 300;# in Watt\n",
- "V = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Efficiency\n",
- "eff = ((V*I - (I*I*R) - C)/(V*I))*100 # in percent\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n efficiency is\",round(eff,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " efficiency is 87.5 %"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 28, page no. 378</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the speed when the current is 60 A and a resistance of 0.5 ohmis connected in series with the armature, \n",
- "#the shunt field remaining constant. \n",
- "#(b) Determine the speed when the current is 60 A and the shunt field is reduced to 80% of its normal value \n",
- "#by increasing resistance in the field circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 120;# in A\n",
- "Ia2 = 60;# in A\n",
- "Ra = 0.2;# in ohm\n",
- "n1 = 10;# in rev/sec\n",
- "R = 0.5;# in ohm\n",
- "x = 0.8;\n",
- "V = 500;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #back e.m.f. at Ia1\n",
- "E1 = V - Ia1*Ra\n",
- " #at Ia2\n",
- "E2 = V - Ia2*(Ra + R)\n",
- " #E1/E2 = Phi1*n1/Phi2*n2\n",
- "n2 = n1*E2/E1\n",
- " #Back e.m.f. when Ia2\n",
- "E3 = V - Ia2*Ra\n",
- "n3 = n1*E3/(x*E1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)speed n2 is \",round(n2,2),\" rev/sec\"\n",
- "print \"\\n (b)speed n3 is \",round(n3,2),\" rev/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)speed n2 is 9.62 rev/sec\n",
- "\n",
- " (b)speed n3 is 12.82 rev/sec"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 29, page no. 379</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the speed when developing full load torque but with a 0.2 ohm diverter in parallel with the field winding.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 90;# in Amperes\n",
- "Ra = 0.1;# in ohm\n",
- "Rse = 0.05;# in ohm\n",
- "Rd = 0.2;# in Ohm\n",
- "n1 = 15;# in rev/sec\n",
- "V = 300;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #e.m.f. E1\n",
- "E1 = V - Ia1*(Ra + Rse)\n",
- " #With the Rd diverter in parallel with Rse\n",
- " #equivalent resistance, Re\n",
- "Re = Rd*Rse/(Rd+Rse)\n",
- " #Torque, T proprtional to Ia*Phi and for full load torque, Ia1*Phi1 = Ia2*Phi2\n",
- " #Since flux is proportional to field current Phi1 proportional to Ta1 and Phi2 Proportional to I1\n",
- "I1 = (Ia1*Ia1*0.8)**0.5\n",
- " #By current division, current I1\n",
- "Ia2 = I1/(Rd/(Rd + Rse))\n",
- " #Hence e.m.f. E2\n",
- "E2 = V - Ia2*(Ra + Re)\n",
- " #E1/E2 = Phi1*n1/Phi2*n2\n",
- "n2 = E2*Ia1*n1/(I1*E1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n speed n2 is \",round(n2,2),\" rev/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " speed n2 is 16.74 rev/sec"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 30, page no. 380</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance to be connected in series to reduce the speed to 600 rev/min with the same current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 25;# in Amperes\n",
- "Ra = 0.4;# in ohm\n",
- "Rse = 0.2;# in ohm\n",
- "n1 = 800/60;# in rev/sec\n",
- "n2 = 600/60;# in rev/sec\n",
- "V = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #e.m.f. E1\n",
- "E1 = V - Ia1*(Ra + Rse)\n",
- " #At n2, since the current is unchanged, the flux is unchanged.\n",
- " #E1/E2 = n1/n2\n",
- "E2 = E1*n2/n1\n",
- " #and E2 = V - Ia1(\u0011Ra + Rse + R)\n",
- "R = (V - E2)/Ia1 - Ra - Rse\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resistance is \",round(R,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resistance is 3.85 ohm"
- ]
- }
- ],
- "prompt_number": 30
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_1.ipynb deleted file mode 100755 index 28433f9f..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_1.ipynb +++ /dev/null @@ -1,1641 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 21: D.c. machines</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 354</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the generated e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 600;# no. of conductors\n",
- "c = 2;# for a wave winding\n",
- "p = 4;# no. of pairs\n",
- "n = 625/60;# in rev/sec\n",
- "Phi = 20E-3;# in Wb\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- "E = 2*p*Phi*n*Z/c\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the generated e.m.f is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the generated e.m.f is 500.0 V "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 354</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the speed at which the machine must be driven to generate an e.m.f. of 240 V\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 50*16;# no. of conductors\n",
- "p = 1;# let no. of pairs\n",
- "c = 2*p;# for a lap winding\n",
- "Phi = 30E-3;# in Wb\n",
- "E = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- " #Rearranging gives, speed\n",
- "n = E*c/(2*p*Phi*Z)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the speed at which the machine must be driven is \",round(n,2),\" rev/sec \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the speed at which the machine must be driven is 10.0 rev/sec "
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 354</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. generated when running at 500 rev/min.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 1200;# no. of conductors\n",
- "p = 1;# let, no. of pairs\n",
- "c = 2*p;# for a lap winding\n",
- "Phi = 30E-3;# in Wb\n",
- "n = 500/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- "E = 2*p*Phi*n*Z/c\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Generated e.m.f. is 300.0 V "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 355</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the generated e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 1200;# no. of conductors\n",
- "p = 4;# let, no. of pairs\n",
- "c = 2;# for a wave winding\n",
- "Phi = 30E-3;# in Wb\n",
- "n = 500/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- "E = 2*p*Phi*n*Z/c\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Generated e.m.f. is 1200.0 V "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 355</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the change in the generated voltage when the field current is reduced by 20%, \n",
- "#assuming the flux is proportional to the field current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 150; # in Volts\n",
- "x = 0.2;\n",
- "\n",
- "#calculation:\n",
- "E2 = E1*(1- x)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Generated e.m.f. is \",round(E2,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Generated e.m.f. is 120.0 V "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 356</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the percentage increase in the flux per pole required to generate 250 V at 20 rev/s.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "n1 = 30;# in rev/sec\n",
- "E1 = 200;# in Volts\n",
- "n2 = 20;# in rev/sec\n",
- "E2 = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #generated e.m.f., E proportional to phi*w and since w = 2*pi*n, then\n",
- " # E proportional to phi*n\n",
- " # E1/E2 = Phi1*n1/(Phi2*n2)\n",
- " # let Phi2/Phi1 = Phi\n",
- "Phi = E2*n1/(E1*n2)\n",
- "Phi_inc = (Phi - 1)*100#/in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n percentage increase in the flux per pole is \",round(Phi_inc,2),\" percent \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " percentage increase in the flux per pole is 87.5 percent "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 357</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the terminal voltage of a generator\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ra = 0.30;# in ohms\n",
- "Ia = 30;# in Amperes\n",
- "E = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #terminal voltage,\n",
- " #V = E - Ia*Ra\n",
- "V = E - Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n terminal voltage of a generator is \",round(V,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " terminal voltage of a generator is 191.0 V "
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 357</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the terminal voltage, and (b) the generated e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "RL = 60;# in ohms\n",
- "Ia = 8;# in Amperes\n",
- "Ra = 1;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #terminal voltage,\n",
- " #V = Ia*RL\n",
- "V = Ia*RL\n",
- " #Generated e.m.f., E\n",
- "E = V + Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)terminal voltage is \",round(V,2),\" V \"\n",
- "print \"\\n (b)generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)terminal voltage is 480.0 V \n",
- "\n",
- " (b)generated e.m.f. is 488.0 V "
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 357</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the generated e.m.f. when \n",
- "#(a) the speed increases to 25 rev/s and the pole flux remains unchanged, \n",
- "#(b) the speed remains at 20 rev/s and the pole flux is decreased to 0.08 Wb,\n",
- "#and (c) the speed increases to 24 rev/s and the pole flux is decreased to 0.07 Wb.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 150;# in Volts\n",
- "n1 = 20;# in rev/sec\n",
- "Phi1 = 0.10;# in Wb\n",
- "n2 = 25;# in rev/sec\n",
- "Phi2 = 0.10;# in Wb\n",
- "n3 = 20;# in rev/sec\n",
- "Phi3 = 0.08;# in Wb\n",
- "n4 = 24;# in rev/sec\n",
- "Phi4 = 0.07;# in Wb\n",
- "\n",
- "#calculation:\n",
- " #generated e.m.f., E proportional to phi*w and since w = 2*pi*n, then\n",
- " # E proportional to phi*n\n",
- " # E1/E2 = Phi1*n1/(Phi2*n2)\n",
- "E2 = E1*Phi2*n2/(Phi1*n1)\n",
- "E3 = E1*Phi3*n3/(Phi1*n1)\n",
- "E4 = E1*Phi4*n4/(Phi1*n1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the generated e.m.f is \",round(E2,2),\" V \"\n",
- "print \"\\n (b)generated e.m.f. is \",round(E3,2),\" V \"\n",
- "print \"\\n (c)generated e.m.f. is \",round(E4,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the generated e.m.f is 187.5 V \n",
- "\n",
- " (b)generated e.m.f. is 120.0 V \n",
- "\n",
- " (c)generated e.m.f. is 126.0 V "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 359</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the terminal voltage, and (b) the e.m.f. generated in the armature.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ps = 20000;# in Watts\n",
- "Vs = 200;# in Volts\n",
- "Rs = 0.1;# in ohms\n",
- "Rf = 50;# in ohms\n",
- "Ra = 0.04;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Load current, I\n",
- "Is = Ps/Vs\n",
- " #Volt drop in the cables to the load\n",
- "Vd = Is*Rs\n",
- " #Hence terminal voltage,\n",
- "V = Vs + Vd\n",
- " #Field current, If\n",
- "If = V/Rf\n",
- " #Armature current\n",
- "Ia = If + Is\n",
- " #Generated e.m.f. E\n",
- "E = V + Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)terminal voltage is \",round(V,2),\" V \"\n",
- "print \"\\n (b)generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)terminal voltage is 210.0 V \n",
- "\n",
- " (b)generated e.m.f. is 214.17 V "
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 361</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the e.m.f. generated.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Is = 80;# in amperes\n",
- "Vs = 200;# in Volts\n",
- "Rf = 40;# in ohms\n",
- "Rse = 0.02;# in ohms\n",
- "Ra = 0.04;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Volt drop in series winding\n",
- "Vse = Is*Rse\n",
- " #P.d. across the field winding = p.d. across armature\n",
- "V1 = Vs + Vse\n",
- " #Field current, If\n",
- "If = V1/Rf\n",
- " #Armature current\n",
- "Ia = If + Is\n",
- " #Generated e.m.f. E\n",
- "E = V1 + Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " generated e.m.f. is 205.0 V "
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 363</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the efficiency of the generator at full load\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ps = 10000;# in Watt\n",
- "Pl = 600;# in Watt\n",
- "Ra = 0.75;# in ohms\n",
- "Rf = 125;# in ohms\n",
- "V = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Output power Ps = V*I\n",
- " #from which, load current I\n",
- "I = Ps/V\n",
- " #Field current, If\n",
- "If = V/Rf\n",
- " #Armature current\n",
- "Ia = If + I\n",
- " #Efficiency,\n",
- "eff = Ps*100/((V*I) + (Ia*Ia*Ra) + (If*V) + (Pl))# in Percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Efficiency is \",round(eff,2),\" percent \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Efficiency is 80.5 percent "
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 364</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the back e.m.f. when the armature current is 50 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ra = 0.2;# in ohms\n",
- "V = 240;# in Volts\n",
- "Ia = 50;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #For a motor, V = E + Ia*Ra\n",
- "E = V - Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n back e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " back e.m.f. is 230.0 V "
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 365</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the e.m.f. generated when it is running: (a) as a generator giving 100 A, and\n",
- "#(b) as a motor taking 80 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ra = 0.25;# in ohms\n",
- "V = 300;# in Volts\n",
- "Ig = 100;# in Amperes\n",
- "Im = 80;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #As a generator, generated e.m.f.,\n",
- " # E = V + Ia*Ra\n",
- "Eg = V + Ig*Ra\n",
- " #For a motor, generated e.m.f. (or back e.m.f.),\n",
- " # E = V - Ia*Ra\n",
- "E = V - Im*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)As a generator, generated e.m.f. is \",round(Eg,2),\" V \"\n",
- "print \"\\n (b)back e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)As a generator, generated e.m.f. is 325.0 V \n",
- "\n",
- " (b)back e.m.f. is 280.0 V "
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 366</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the torque exerted when a current of 30 A flows in each armature conductor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 4;\n",
- "c = 2;# for a wave winding\n",
- "Phi = 25E-3;# Wb\n",
- "Z = 900;\n",
- "Ia = 30;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #torque T = p*Phi*Z*Ia/(pi*c)\n",
- "T = p*Phi*Z*Ia/(1*math.pi*c)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the torque exerted is \",round(T,2),\" Nm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the torque exerted is 429.72 Nm "
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 366</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the torque\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 350;# in Volts\n",
- "Ra = 0.5;# in ohms\n",
- "n = 15;# in rev/sec\n",
- "Ia = 60;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #Back e.m.f. E = V - Ia*Ra\n",
- "E = V - Ia*Ra\n",
- " #torque T = E*Ia/(2*n*pi)\n",
- "T = E*Ia/(2*n*math.pi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the torque exerted is \",round(T,2),\" Nm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the torque exerted is 203.72 Nm "
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 366</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the speed and (b) the torque developed when the armature current is 40 A\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 1;# let\n",
- "c = 2*p;# for a lap winding\n",
- "Phi = 20E-3;# Wb\n",
- "Z = 500;\n",
- "V = 250;# in Volts\n",
- "Ra = 1;# in ohms\n",
- "Ia = 40;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #Back e.m.f. E = V - Ia*Ra\n",
- "E = V - Ia*Ra\n",
- " #E.m.f. E = 2*p*Phi*n*Z/c\n",
- " # rearrange,\n",
- "n = E*c/(2*p*Phi*Z)\n",
- " #torque T = E*Ia/(2*n*pi)\n",
- "T = E*Ia/(2*n*math.pi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)speed n is \",round(n,2),\" rev/sec \"\n",
- "print \"\\n (b)the torque exerted is \",round(T,2),\" Nm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)speed n is 21.0 rev/sec \n",
- "\n",
- " (b)the torque exerted is 63.66 Nm "
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 367</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the armature current at this new value of torque.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T1 = 25;# in Nm\n",
- "T2 = 35;# in Nm\n",
- "Ia1 = 16;# in Amperes\n",
- "V = 100;# in Volts\n",
- "x = 0.15;\n",
- "\n",
- " #calculation:\n",
- " #the shaft torque T of a generator is proportional to (phi*Ia),\n",
- " #where Phi is the flux and Ia is the armature current. Thus, T = k*Phi*Ia, where k is a constant.\n",
- " #The torque at flux phi1 and armature current Ia1 is T1 = k*Phi1*Ia1.\n",
- " #similarly T2 = k*Phi2*Ia2\n",
- "\n",
- "Ia2 = T2*Ia1/(0.85*T1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n armature current at the new value of torque is \",round(Ia2,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " armature current at the new value of torque is 26.35 A "
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 367</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the efficiency of the generator and (b) the power loss in the generator.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T = 12;# in Nm\n",
- "I = 15;# in Amperes\n",
- "V = 100;# in Volts\n",
- "n = 1500/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #the efficiency of a generator = (output power/input power)*100 %\n",
- " #The output power is the electrical output, i.e. VI watts. \n",
- " #The input power to a generator is the mechanical power in the shaft driving the generator, \n",
- " #i.e. T*w or T(2*pi*n) watts, where T is the torque in Nm and n is speed of rotation in rev/s. Hence, for a generator \n",
- " #efficiency = V*I*100/(T*2*pi*n) %\n",
- "eff = V*I*100/(T*2*math.pi*n)# in Percent\n",
- " #The input power = output power + losses\n",
- " # hence, T*2*math.pi*n = V*I + losses\n",
- "Pl = T*2*math.pi*n - V*I\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) efficiency is \",round(eff,2),\" % \"\n",
- "print \"\\n (b) power loss is \",round(Pl,2),\" W \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) efficiency is 79.58 % \n",
- "\n",
- " (b) power loss is 384.96 W "
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 368</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current in the armature, and (b) the back e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rf = 150;# in Ohms\n",
- "Ra = 0.4;# in Ohms\n",
- "I = 30;# in Amperes\n",
- "V = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Field current If\n",
- "If = V/Rf\n",
- " #Supply current I = Ia + If\n",
- " #Hence armature current, Ia\n",
- "Ia = I - If\n",
- " #Back e.m.f. E = V -\u0004 Ia*Ra\n",
- "E = V - (Ia*Ra)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) current in the armature is \",round(Ia,2),\" A \"\n",
- "print \"\\n (b) Back e.m.f. E is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) current in the armature is 28.4 A \n",
- "\n",
- " (b) Back e.m.f. E is 228.64 V "
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 370</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the speed of the motor, assuming the flux remains constant.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 30;# in Amperes\n",
- "Ia2 = 45;# in Amperes\n",
- "Ra = 0.4;# in ohm\n",
- "n1 = 1350/60;# in Rev/sec\n",
- "V = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #The relationship E proportional to (Phi*n) applies to both generators and motors. For a motor,\n",
- " #E = V - (Ia*Ra)\n",
- "E1 = V - (Ia1*Ra)\n",
- "E2 = V - (Ia2*Ra)\n",
- " #The relationship, E1/E2 = Phi1*n1/Phi2*n2, applies to both generators and motors.\n",
- " #Since the flux is constant, Phi1 = Phi2\n",
- "n2 = E2*n1/(E1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the speed of the motor is \",round(n2,2),\" rev/sec \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the speed of the motor is 21.78 rev/sec "
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 370</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and \n",
- "#(b) the steadystate value of the armature current at the new value of flux,\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 30;# in Amperes\n",
- "Ra = 0.4;# in ohm\n",
- "n = 800/60;# in Rev/sec\n",
- "V = 220;# in Volts\n",
- "x= 0.1;\n",
- "\n",
- "#calculation:\n",
- " #For a d.c. shunt-wound motor, E = V - (Ia*Ra),Hence initial generated e.m.f.,\n",
- "E1 = V - (Ia1*Ra)\n",
- " #The generated e.m.f. is also such that E proportional to (Phi*n) \n",
- " #so at the instant the flux is reduced, the speed has not had time to change, and\n",
- "E = E1*(1-x)\n",
- " #Hence, the voltage drop due to the armature resistance is\n",
- "Vd = V - E\n",
- " #The instantaneous value of the current is\n",
- "Ia = Vd/Ra\n",
- " #T proportional to (Phi*Ia), since the torque is constant,\n",
- " #Phi1*Ia1 = Phi2*Ia2, The flux 8 is reduced by 10%, hence\n",
- "Ia2 = Ia1/0.9\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)instantaneous value of the current \",round(Ia,2),\" A \"\n",
- "print \"\\n (b)steady state value of armature current, \",round(Ia2,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)instantaneous value of the current 82.0 A \n",
- "\n",
- " (b)steady state value of armature current, 33.33 A "
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 372</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the generated e.m.f. at this load. \n",
- "#(b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A.\n",
- "#Assume that this causes a doubling of the flux.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 15;# in Amperes\n",
- "Ia2 = 30;# in Amperes\n",
- "Rf = 0.3;# in ohms\n",
- "Ra = 0.2;# in ohm\n",
- "n1 = 24;# in Rev/sec\n",
- "V = 240;# in Volts\n",
- "x= 2;\n",
- "\n",
- "#calculation:\n",
- " #generated e.m.f., E, at initial load, is given by\n",
- "E1 = V - Ia1*(Ra + Rf)\n",
- " #When the current is increased to 30 A, the generated e.m.f. is given by:\n",
- "E2 = V - Ia2*(Ra + Rf)\n",
- " #E proportional to (Phi*n)\n",
- " #E1/E2 = Phi1*n1/Phi2*n2\n",
- "n2 = E2*n1/(2*E1) \n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)generated e.m.f., E is \",round(E1,2),\" V \"\n",
- "print \"\\n (b)speed of motor, n2, \",round(n2,2),\" rev/sec \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)generated e.m.f., E is 232.5 V \n",
- "\n",
- " (b)speed of motor, n2, 11.61 rev/sec "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 374</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the overall efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 80;# in Amperes\n",
- "C = 1500;# in Watt\n",
- "Rf = 40;# in ohms\n",
- "Ra = 0.2;# in ohm\n",
- "n = 1000/60;# in Rev/sec\n",
- "V = 320;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Field current, If\n",
- "If = V/Rf\n",
- " #Armature current Ia\n",
- "Ia = I - If\n",
- " #Efficiency\n",
- "eff = ((V*I - (Ia*Ia*Ra) - (If*V) - C)/(V*I))*100 # in percent\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n efficiency is\",round(eff,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " efficiency is 80.09 %"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 25, page no. 374</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the maximum efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 40;# in Amperes\n",
- "Rf = 0.05;# in ohms\n",
- "Ra = 0.15;# in ohm\n",
- "V = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #However for a series motor, If = 0 and the Ia*Ia*Ra loss needs to be I*I*(\u0011Ra + Rf)\n",
- " #For maximum efficiency I*I*\u0011(Ra + Rf) = C\n",
- " #Efficiency\n",
- "eff = ((V*I - (2*I*I*(Ra + Rf)))/(V*I))*100 # in percent\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n efficiency is\",round(eff,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " efficiency is 93.6"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 26, page no. 375</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current supplied to the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T = 15;# in Nm\n",
- "n = 1200/60;# in rev/sec\n",
- "eff = 0.8;\n",
- "V = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- "I = T*2*math.pi*n/(V*eff)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current supplied, I is \",round(I,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current supplied, I is 11.78 A"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 27, page no. 376</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 2;# in ohm\n",
- "n = 30;# in rev/sec\n",
- "I = 10;# in A\n",
- "C = 300;# in Watt\n",
- "V = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Efficiency\n",
- "eff = ((V*I - (I*I*R) - C)/(V*I))*100 # in percent\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n efficiency is\",round(eff,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " efficiency is 87.5 %"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 28, page no. 378</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the speed when the current is 60 A and a resistance of 0.5 ohmis connected in series with the armature, \n",
- "#the shunt field remaining constant. \n",
- "#(b) Determine the speed when the current is 60 A and the shunt field is reduced to 80% of its normal value \n",
- "#by increasing resistance in the field circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 120;# in A\n",
- "Ia2 = 60;# in A\n",
- "Ra = 0.2;# in ohm\n",
- "n1 = 10;# in rev/sec\n",
- "R = 0.5;# in ohm\n",
- "x = 0.8;\n",
- "V = 500;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #back e.m.f. at Ia1\n",
- "E1 = V - Ia1*Ra\n",
- " #at Ia2\n",
- "E2 = V - Ia2*(Ra + R)\n",
- " #E1/E2 = Phi1*n1/Phi2*n2\n",
- "n2 = n1*E2/E1\n",
- " #Back e.m.f. when Ia2\n",
- "E3 = V - Ia2*Ra\n",
- "n3 = n1*E3/(x*E1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)speed n2 is \",round(n2,2),\" rev/sec\"\n",
- "print \"\\n (b)speed n3 is \",round(n3,2),\" rev/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)speed n2 is 9.62 rev/sec\n",
- "\n",
- " (b)speed n3 is 12.82 rev/sec"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 29, page no. 379</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the speed when developing full load torque but with a 0.2 ohm diverter in parallel with the field winding.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 90;# in Amperes\n",
- "Ra = 0.1;# in ohm\n",
- "Rse = 0.05;# in ohm\n",
- "Rd = 0.2;# in Ohm\n",
- "n1 = 15;# in rev/sec\n",
- "V = 300;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #e.m.f. E1\n",
- "E1 = V - Ia1*(Ra + Rse)\n",
- " #With the Rd diverter in parallel with Rse\n",
- " #equivalent resistance, Re\n",
- "Re = Rd*Rse/(Rd+Rse)\n",
- " #Torque, T proprtional to Ia*Phi and for full load torque, Ia1*Phi1 = Ia2*Phi2\n",
- " #Since flux is proportional to field current Phi1 proportional to Ta1 and Phi2 Proportional to I1\n",
- "I1 = (Ia1*Ia1*0.8)**0.5\n",
- " #By current division, current I1\n",
- "Ia2 = I1/(Rd/(Rd + Rse))\n",
- " #Hence e.m.f. E2\n",
- "E2 = V - Ia2*(Ra + Re)\n",
- " #E1/E2 = Phi1*n1/Phi2*n2\n",
- "n2 = E2*Ia1*n1/(I1*E1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n speed n2 is \",round(n2,2),\" rev/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " speed n2 is 16.74 rev/sec"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 30, page no. 380</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance to be connected in series to reduce the speed to 600 rev/min with the same current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 25;# in Amperes\n",
- "Ra = 0.4;# in ohm\n",
- "Rse = 0.2;# in ohm\n",
- "n1 = 800/60;# in rev/sec\n",
- "n2 = 600/60;# in rev/sec\n",
- "V = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #e.m.f. E1\n",
- "E1 = V - Ia1*(Ra + Rse)\n",
- " #At n2, since the current is unchanged, the flux is unchanged.\n",
- " #E1/E2 = n1/n2\n",
- "E2 = E1*n2/n1\n",
- " #and E2 = V - Ia1(\u0011Ra + Rse + R)\n",
- "R = (V - E2)/Ia1 - Ra - Rse\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resistance is \",round(R,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resistance is 3.85 ohm"
- ]
- }
- ],
- "prompt_number": 30
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_2.ipynb deleted file mode 100755 index 28433f9f..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_2.ipynb +++ /dev/null @@ -1,1641 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 21: D.c. machines</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 354</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the generated e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 600;# no. of conductors\n",
- "c = 2;# for a wave winding\n",
- "p = 4;# no. of pairs\n",
- "n = 625/60;# in rev/sec\n",
- "Phi = 20E-3;# in Wb\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- "E = 2*p*Phi*n*Z/c\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the generated e.m.f is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the generated e.m.f is 500.0 V "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 354</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the speed at which the machine must be driven to generate an e.m.f. of 240 V\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 50*16;# no. of conductors\n",
- "p = 1;# let no. of pairs\n",
- "c = 2*p;# for a lap winding\n",
- "Phi = 30E-3;# in Wb\n",
- "E = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- " #Rearranging gives, speed\n",
- "n = E*c/(2*p*Phi*Z)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the speed at which the machine must be driven is \",round(n,2),\" rev/sec \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the speed at which the machine must be driven is 10.0 rev/sec "
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 354</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. generated when running at 500 rev/min.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 1200;# no. of conductors\n",
- "p = 1;# let, no. of pairs\n",
- "c = 2*p;# for a lap winding\n",
- "Phi = 30E-3;# in Wb\n",
- "n = 500/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- "E = 2*p*Phi*n*Z/c\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Generated e.m.f. is 300.0 V "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 355</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the generated e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 1200;# no. of conductors\n",
- "p = 4;# let, no. of pairs\n",
- "c = 2;# for a wave winding\n",
- "Phi = 30E-3;# in Wb\n",
- "n = 500/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- "E = 2*p*Phi*n*Z/c\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Generated e.m.f. is 1200.0 V "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 355</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the change in the generated voltage when the field current is reduced by 20%, \n",
- "#assuming the flux is proportional to the field current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 150; # in Volts\n",
- "x = 0.2;\n",
- "\n",
- "#calculation:\n",
- "E2 = E1*(1- x)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Generated e.m.f. is \",round(E2,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Generated e.m.f. is 120.0 V "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 356</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the percentage increase in the flux per pole required to generate 250 V at 20 rev/s.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "n1 = 30;# in rev/sec\n",
- "E1 = 200;# in Volts\n",
- "n2 = 20;# in rev/sec\n",
- "E2 = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #generated e.m.f., E proportional to phi*w and since w = 2*pi*n, then\n",
- " # E proportional to phi*n\n",
- " # E1/E2 = Phi1*n1/(Phi2*n2)\n",
- " # let Phi2/Phi1 = Phi\n",
- "Phi = E2*n1/(E1*n2)\n",
- "Phi_inc = (Phi - 1)*100#/in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n percentage increase in the flux per pole is \",round(Phi_inc,2),\" percent \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " percentage increase in the flux per pole is 87.5 percent "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 357</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the terminal voltage of a generator\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ra = 0.30;# in ohms\n",
- "Ia = 30;# in Amperes\n",
- "E = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #terminal voltage,\n",
- " #V = E - Ia*Ra\n",
- "V = E - Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n terminal voltage of a generator is \",round(V,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " terminal voltage of a generator is 191.0 V "
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 357</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the terminal voltage, and (b) the generated e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "RL = 60;# in ohms\n",
- "Ia = 8;# in Amperes\n",
- "Ra = 1;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #terminal voltage,\n",
- " #V = Ia*RL\n",
- "V = Ia*RL\n",
- " #Generated e.m.f., E\n",
- "E = V + Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)terminal voltage is \",round(V,2),\" V \"\n",
- "print \"\\n (b)generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)terminal voltage is 480.0 V \n",
- "\n",
- " (b)generated e.m.f. is 488.0 V "
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 357</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the generated e.m.f. when \n",
- "#(a) the speed increases to 25 rev/s and the pole flux remains unchanged, \n",
- "#(b) the speed remains at 20 rev/s and the pole flux is decreased to 0.08 Wb,\n",
- "#and (c) the speed increases to 24 rev/s and the pole flux is decreased to 0.07 Wb.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 150;# in Volts\n",
- "n1 = 20;# in rev/sec\n",
- "Phi1 = 0.10;# in Wb\n",
- "n2 = 25;# in rev/sec\n",
- "Phi2 = 0.10;# in Wb\n",
- "n3 = 20;# in rev/sec\n",
- "Phi3 = 0.08;# in Wb\n",
- "n4 = 24;# in rev/sec\n",
- "Phi4 = 0.07;# in Wb\n",
- "\n",
- "#calculation:\n",
- " #generated e.m.f., E proportional to phi*w and since w = 2*pi*n, then\n",
- " # E proportional to phi*n\n",
- " # E1/E2 = Phi1*n1/(Phi2*n2)\n",
- "E2 = E1*Phi2*n2/(Phi1*n1)\n",
- "E3 = E1*Phi3*n3/(Phi1*n1)\n",
- "E4 = E1*Phi4*n4/(Phi1*n1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the generated e.m.f is \",round(E2,2),\" V \"\n",
- "print \"\\n (b)generated e.m.f. is \",round(E3,2),\" V \"\n",
- "print \"\\n (c)generated e.m.f. is \",round(E4,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the generated e.m.f is 187.5 V \n",
- "\n",
- " (b)generated e.m.f. is 120.0 V \n",
- "\n",
- " (c)generated e.m.f. is 126.0 V "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 359</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the terminal voltage, and (b) the e.m.f. generated in the armature.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ps = 20000;# in Watts\n",
- "Vs = 200;# in Volts\n",
- "Rs = 0.1;# in ohms\n",
- "Rf = 50;# in ohms\n",
- "Ra = 0.04;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Load current, I\n",
- "Is = Ps/Vs\n",
- " #Volt drop in the cables to the load\n",
- "Vd = Is*Rs\n",
- " #Hence terminal voltage,\n",
- "V = Vs + Vd\n",
- " #Field current, If\n",
- "If = V/Rf\n",
- " #Armature current\n",
- "Ia = If + Is\n",
- " #Generated e.m.f. E\n",
- "E = V + Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)terminal voltage is \",round(V,2),\" V \"\n",
- "print \"\\n (b)generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)terminal voltage is 210.0 V \n",
- "\n",
- " (b)generated e.m.f. is 214.17 V "
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 361</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the e.m.f. generated.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Is = 80;# in amperes\n",
- "Vs = 200;# in Volts\n",
- "Rf = 40;# in ohms\n",
- "Rse = 0.02;# in ohms\n",
- "Ra = 0.04;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Volt drop in series winding\n",
- "Vse = Is*Rse\n",
- " #P.d. across the field winding = p.d. across armature\n",
- "V1 = Vs + Vse\n",
- " #Field current, If\n",
- "If = V1/Rf\n",
- " #Armature current\n",
- "Ia = If + Is\n",
- " #Generated e.m.f. E\n",
- "E = V1 + Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " generated e.m.f. is 205.0 V "
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 363</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the efficiency of the generator at full load\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ps = 10000;# in Watt\n",
- "Pl = 600;# in Watt\n",
- "Ra = 0.75;# in ohms\n",
- "Rf = 125;# in ohms\n",
- "V = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Output power Ps = V*I\n",
- " #from which, load current I\n",
- "I = Ps/V\n",
- " #Field current, If\n",
- "If = V/Rf\n",
- " #Armature current\n",
- "Ia = If + I\n",
- " #Efficiency,\n",
- "eff = Ps*100/((V*I) + (Ia*Ia*Ra) + (If*V) + (Pl))# in Percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Efficiency is \",round(eff,2),\" percent \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Efficiency is 80.5 percent "
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 364</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the back e.m.f. when the armature current is 50 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ra = 0.2;# in ohms\n",
- "V = 240;# in Volts\n",
- "Ia = 50;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #For a motor, V = E + Ia*Ra\n",
- "E = V - Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n back e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " back e.m.f. is 230.0 V "
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 365</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the e.m.f. generated when it is running: (a) as a generator giving 100 A, and\n",
- "#(b) as a motor taking 80 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ra = 0.25;# in ohms\n",
- "V = 300;# in Volts\n",
- "Ig = 100;# in Amperes\n",
- "Im = 80;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #As a generator, generated e.m.f.,\n",
- " # E = V + Ia*Ra\n",
- "Eg = V + Ig*Ra\n",
- " #For a motor, generated e.m.f. (or back e.m.f.),\n",
- " # E = V - Ia*Ra\n",
- "E = V - Im*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)As a generator, generated e.m.f. is \",round(Eg,2),\" V \"\n",
- "print \"\\n (b)back e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)As a generator, generated e.m.f. is 325.0 V \n",
- "\n",
- " (b)back e.m.f. is 280.0 V "
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 366</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the torque exerted when a current of 30 A flows in each armature conductor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 4;\n",
- "c = 2;# for a wave winding\n",
- "Phi = 25E-3;# Wb\n",
- "Z = 900;\n",
- "Ia = 30;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #torque T = p*Phi*Z*Ia/(pi*c)\n",
- "T = p*Phi*Z*Ia/(1*math.pi*c)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the torque exerted is \",round(T,2),\" Nm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the torque exerted is 429.72 Nm "
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 366</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the torque\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 350;# in Volts\n",
- "Ra = 0.5;# in ohms\n",
- "n = 15;# in rev/sec\n",
- "Ia = 60;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #Back e.m.f. E = V - Ia*Ra\n",
- "E = V - Ia*Ra\n",
- " #torque T = E*Ia/(2*n*pi)\n",
- "T = E*Ia/(2*n*math.pi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the torque exerted is \",round(T,2),\" Nm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the torque exerted is 203.72 Nm "
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 366</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the speed and (b) the torque developed when the armature current is 40 A\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 1;# let\n",
- "c = 2*p;# for a lap winding\n",
- "Phi = 20E-3;# Wb\n",
- "Z = 500;\n",
- "V = 250;# in Volts\n",
- "Ra = 1;# in ohms\n",
- "Ia = 40;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #Back e.m.f. E = V - Ia*Ra\n",
- "E = V - Ia*Ra\n",
- " #E.m.f. E = 2*p*Phi*n*Z/c\n",
- " # rearrange,\n",
- "n = E*c/(2*p*Phi*Z)\n",
- " #torque T = E*Ia/(2*n*pi)\n",
- "T = E*Ia/(2*n*math.pi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)speed n is \",round(n,2),\" rev/sec \"\n",
- "print \"\\n (b)the torque exerted is \",round(T,2),\" Nm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)speed n is 21.0 rev/sec \n",
- "\n",
- " (b)the torque exerted is 63.66 Nm "
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 367</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the armature current at this new value of torque.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T1 = 25;# in Nm\n",
- "T2 = 35;# in Nm\n",
- "Ia1 = 16;# in Amperes\n",
- "V = 100;# in Volts\n",
- "x = 0.15;\n",
- "\n",
- " #calculation:\n",
- " #the shaft torque T of a generator is proportional to (phi*Ia),\n",
- " #where Phi is the flux and Ia is the armature current. Thus, T = k*Phi*Ia, where k is a constant.\n",
- " #The torque at flux phi1 and armature current Ia1 is T1 = k*Phi1*Ia1.\n",
- " #similarly T2 = k*Phi2*Ia2\n",
- "\n",
- "Ia2 = T2*Ia1/(0.85*T1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n armature current at the new value of torque is \",round(Ia2,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " armature current at the new value of torque is 26.35 A "
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 367</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the efficiency of the generator and (b) the power loss in the generator.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T = 12;# in Nm\n",
- "I = 15;# in Amperes\n",
- "V = 100;# in Volts\n",
- "n = 1500/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #the efficiency of a generator = (output power/input power)*100 %\n",
- " #The output power is the electrical output, i.e. VI watts. \n",
- " #The input power to a generator is the mechanical power in the shaft driving the generator, \n",
- " #i.e. T*w or T(2*pi*n) watts, where T is the torque in Nm and n is speed of rotation in rev/s. Hence, for a generator \n",
- " #efficiency = V*I*100/(T*2*pi*n) %\n",
- "eff = V*I*100/(T*2*math.pi*n)# in Percent\n",
- " #The input power = output power + losses\n",
- " # hence, T*2*math.pi*n = V*I + losses\n",
- "Pl = T*2*math.pi*n - V*I\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) efficiency is \",round(eff,2),\" % \"\n",
- "print \"\\n (b) power loss is \",round(Pl,2),\" W \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) efficiency is 79.58 % \n",
- "\n",
- " (b) power loss is 384.96 W "
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 368</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current in the armature, and (b) the back e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rf = 150;# in Ohms\n",
- "Ra = 0.4;# in Ohms\n",
- "I = 30;# in Amperes\n",
- "V = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Field current If\n",
- "If = V/Rf\n",
- " #Supply current I = Ia + If\n",
- " #Hence armature current, Ia\n",
- "Ia = I - If\n",
- " #Back e.m.f. E = V -\u0004 Ia*Ra\n",
- "E = V - (Ia*Ra)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) current in the armature is \",round(Ia,2),\" A \"\n",
- "print \"\\n (b) Back e.m.f. E is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) current in the armature is 28.4 A \n",
- "\n",
- " (b) Back e.m.f. E is 228.64 V "
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 370</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the speed of the motor, assuming the flux remains constant.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 30;# in Amperes\n",
- "Ia2 = 45;# in Amperes\n",
- "Ra = 0.4;# in ohm\n",
- "n1 = 1350/60;# in Rev/sec\n",
- "V = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #The relationship E proportional to (Phi*n) applies to both generators and motors. For a motor,\n",
- " #E = V - (Ia*Ra)\n",
- "E1 = V - (Ia1*Ra)\n",
- "E2 = V - (Ia2*Ra)\n",
- " #The relationship, E1/E2 = Phi1*n1/Phi2*n2, applies to both generators and motors.\n",
- " #Since the flux is constant, Phi1 = Phi2\n",
- "n2 = E2*n1/(E1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the speed of the motor is \",round(n2,2),\" rev/sec \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the speed of the motor is 21.78 rev/sec "
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 370</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and \n",
- "#(b) the steadystate value of the armature current at the new value of flux,\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 30;# in Amperes\n",
- "Ra = 0.4;# in ohm\n",
- "n = 800/60;# in Rev/sec\n",
- "V = 220;# in Volts\n",
- "x= 0.1;\n",
- "\n",
- "#calculation:\n",
- " #For a d.c. shunt-wound motor, E = V - (Ia*Ra),Hence initial generated e.m.f.,\n",
- "E1 = V - (Ia1*Ra)\n",
- " #The generated e.m.f. is also such that E proportional to (Phi*n) \n",
- " #so at the instant the flux is reduced, the speed has not had time to change, and\n",
- "E = E1*(1-x)\n",
- " #Hence, the voltage drop due to the armature resistance is\n",
- "Vd = V - E\n",
- " #The instantaneous value of the current is\n",
- "Ia = Vd/Ra\n",
- " #T proportional to (Phi*Ia), since the torque is constant,\n",
- " #Phi1*Ia1 = Phi2*Ia2, The flux 8 is reduced by 10%, hence\n",
- "Ia2 = Ia1/0.9\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)instantaneous value of the current \",round(Ia,2),\" A \"\n",
- "print \"\\n (b)steady state value of armature current, \",round(Ia2,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)instantaneous value of the current 82.0 A \n",
- "\n",
- " (b)steady state value of armature current, 33.33 A "
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 372</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the generated e.m.f. at this load. \n",
- "#(b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A.\n",
- "#Assume that this causes a doubling of the flux.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 15;# in Amperes\n",
- "Ia2 = 30;# in Amperes\n",
- "Rf = 0.3;# in ohms\n",
- "Ra = 0.2;# in ohm\n",
- "n1 = 24;# in Rev/sec\n",
- "V = 240;# in Volts\n",
- "x= 2;\n",
- "\n",
- "#calculation:\n",
- " #generated e.m.f., E, at initial load, is given by\n",
- "E1 = V - Ia1*(Ra + Rf)\n",
- " #When the current is increased to 30 A, the generated e.m.f. is given by:\n",
- "E2 = V - Ia2*(Ra + Rf)\n",
- " #E proportional to (Phi*n)\n",
- " #E1/E2 = Phi1*n1/Phi2*n2\n",
- "n2 = E2*n1/(2*E1) \n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)generated e.m.f., E is \",round(E1,2),\" V \"\n",
- "print \"\\n (b)speed of motor, n2, \",round(n2,2),\" rev/sec \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)generated e.m.f., E is 232.5 V \n",
- "\n",
- " (b)speed of motor, n2, 11.61 rev/sec "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 374</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the overall efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 80;# in Amperes\n",
- "C = 1500;# in Watt\n",
- "Rf = 40;# in ohms\n",
- "Ra = 0.2;# in ohm\n",
- "n = 1000/60;# in Rev/sec\n",
- "V = 320;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Field current, If\n",
- "If = V/Rf\n",
- " #Armature current Ia\n",
- "Ia = I - If\n",
- " #Efficiency\n",
- "eff = ((V*I - (Ia*Ia*Ra) - (If*V) - C)/(V*I))*100 # in percent\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n efficiency is\",round(eff,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " efficiency is 80.09 %"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 25, page no. 374</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the maximum efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 40;# in Amperes\n",
- "Rf = 0.05;# in ohms\n",
- "Ra = 0.15;# in ohm\n",
- "V = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #However for a series motor, If = 0 and the Ia*Ia*Ra loss needs to be I*I*(\u0011Ra + Rf)\n",
- " #For maximum efficiency I*I*\u0011(Ra + Rf) = C\n",
- " #Efficiency\n",
- "eff = ((V*I - (2*I*I*(Ra + Rf)))/(V*I))*100 # in percent\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n efficiency is\",round(eff,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " efficiency is 93.6"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 26, page no. 375</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current supplied to the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T = 15;# in Nm\n",
- "n = 1200/60;# in rev/sec\n",
- "eff = 0.8;\n",
- "V = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- "I = T*2*math.pi*n/(V*eff)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current supplied, I is \",round(I,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current supplied, I is 11.78 A"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 27, page no. 376</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 2;# in ohm\n",
- "n = 30;# in rev/sec\n",
- "I = 10;# in A\n",
- "C = 300;# in Watt\n",
- "V = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Efficiency\n",
- "eff = ((V*I - (I*I*R) - C)/(V*I))*100 # in percent\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n efficiency is\",round(eff,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " efficiency is 87.5 %"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 28, page no. 378</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the speed when the current is 60 A and a resistance of 0.5 ohmis connected in series with the armature, \n",
- "#the shunt field remaining constant. \n",
- "#(b) Determine the speed when the current is 60 A and the shunt field is reduced to 80% of its normal value \n",
- "#by increasing resistance in the field circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 120;# in A\n",
- "Ia2 = 60;# in A\n",
- "Ra = 0.2;# in ohm\n",
- "n1 = 10;# in rev/sec\n",
- "R = 0.5;# in ohm\n",
- "x = 0.8;\n",
- "V = 500;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #back e.m.f. at Ia1\n",
- "E1 = V - Ia1*Ra\n",
- " #at Ia2\n",
- "E2 = V - Ia2*(Ra + R)\n",
- " #E1/E2 = Phi1*n1/Phi2*n2\n",
- "n2 = n1*E2/E1\n",
- " #Back e.m.f. when Ia2\n",
- "E3 = V - Ia2*Ra\n",
- "n3 = n1*E3/(x*E1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)speed n2 is \",round(n2,2),\" rev/sec\"\n",
- "print \"\\n (b)speed n3 is \",round(n3,2),\" rev/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)speed n2 is 9.62 rev/sec\n",
- "\n",
- " (b)speed n3 is 12.82 rev/sec"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 29, page no. 379</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the speed when developing full load torque but with a 0.2 ohm diverter in parallel with the field winding.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 90;# in Amperes\n",
- "Ra = 0.1;# in ohm\n",
- "Rse = 0.05;# in ohm\n",
- "Rd = 0.2;# in Ohm\n",
- "n1 = 15;# in rev/sec\n",
- "V = 300;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #e.m.f. E1\n",
- "E1 = V - Ia1*(Ra + Rse)\n",
- " #With the Rd diverter in parallel with Rse\n",
- " #equivalent resistance, Re\n",
- "Re = Rd*Rse/(Rd+Rse)\n",
- " #Torque, T proprtional to Ia*Phi and for full load torque, Ia1*Phi1 = Ia2*Phi2\n",
- " #Since flux is proportional to field current Phi1 proportional to Ta1 and Phi2 Proportional to I1\n",
- "I1 = (Ia1*Ia1*0.8)**0.5\n",
- " #By current division, current I1\n",
- "Ia2 = I1/(Rd/(Rd + Rse))\n",
- " #Hence e.m.f. E2\n",
- "E2 = V - Ia2*(Ra + Re)\n",
- " #E1/E2 = Phi1*n1/Phi2*n2\n",
- "n2 = E2*Ia1*n1/(I1*E1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n speed n2 is \",round(n2,2),\" rev/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " speed n2 is 16.74 rev/sec"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 30, page no. 380</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance to be connected in series to reduce the speed to 600 rev/min with the same current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 25;# in Amperes\n",
- "Ra = 0.4;# in ohm\n",
- "Rse = 0.2;# in ohm\n",
- "n1 = 800/60;# in rev/sec\n",
- "n2 = 600/60;# in rev/sec\n",
- "V = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #e.m.f. E1\n",
- "E1 = V - Ia1*(Ra + Rse)\n",
- " #At n2, since the current is unchanged, the flux is unchanged.\n",
- " #E1/E2 = n1/n2\n",
- "E2 = E1*n2/n1\n",
- " #and E2 = V - Ia1(\u0011Ra + Rse + R)\n",
- "R = (V - E2)/Ia1 - Ra - Rse\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resistance is \",round(R,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resistance is 3.85 ohm"
- ]
- }
- ],
- "prompt_number": 30
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_21.ipynb index 373f5943..373f5943 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_21.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb deleted file mode 100755 index 4bad2773..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb +++ /dev/null @@ -1,791 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 22: Three-phase induction motors</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 389</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the synchronous speed of the motor in rev/min.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 50;# in Hz\n",
- "p = 2/2;# number of pairs of poles\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, \n",
- " #f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- "nsrpm = ns*60\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nsynchronous speed of the motor is \",nsrpm,\" rev/min\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "synchronous speed of the motor is 3000.0 rev/min"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 389</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the number of poles.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 60;# in Hz\n",
- "ns = 900/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "p = f/ns\n",
- "np = p*2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nnumber of poles is \", round(np,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "number of poles is 8.0"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 390</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the frequency of the supply voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 2/2;# number of pairs of poles\n",
- "ns = 6000/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"frequency is \",f,\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "frequency is 100.0 Hz"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 391</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the synchronous speed and (b) the slip at full load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 4/2;# number of pairs of poles\n",
- "f = 50;# in Hz\n",
- "nr = 1455/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The slip, s\n",
- "s = ((ns - nr)/ns)*100# in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) synchronous speed is \",ns,\" rev/sec\"\n",
- "print \"\\n(b) slip is \",s,\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) synchronous speed is 25.0 rev/sec\n",
- "\n",
- "(b) slip is 3.0 percent"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 392</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the synchronous speed,\n",
- "#(b) the speed of the rotor and (c) the frequency of the induced e.m.f.\u2019s in the rotor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 2/2;# number of pairs of poles\n",
- "f = 60;# in Hz\n",
- "s = 0.02;# slip\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The the rotor runs at\n",
- "nr = ns*(1 - s)\n",
- " #frequency of the e.m.f. induced in the rotor bars is\n",
- "fr = ns - nr\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) synchronous speed is \",ns,\" rev/sec\"\n",
- "print \"\\n(b) rotor speed is \",nr,\" rev/sec\"\n",
- "print \"\\n(c) frequency of the e.m.f. induced in the rotor bars is is \",fr,\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) synchronous speed is 60.0 rev/sec\n",
- "\n",
- "(b) rotor speed is 58.8 rev/sec\n",
- "\n",
- "(c) frequency of the e.m.f. induced in the rotor bars is is 1.2 Hz"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 392</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the synchronous speed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 50;# in Hz\n",
- "nr = 1200/60;# in rev/min\n",
- "s = 0.04;# slip\n",
- "\n",
- "#calculation:\n",
- " #the synchronous speed.\n",
- "ns = nr/(1 - s)\n",
- "nsrpm = ns*60\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n synchronous speed is \",nsrpm,\" rev/min\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " synchronous speed is 1250.0 rev/min"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 394</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the slip, and (b) the rotor speed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 8/2;# number of pairs of poles\n",
- "f = 50;# in Hz\n",
- "fr = 3;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #fr = s*f\n",
- "s = (fr/f)\n",
- " #the rotor speed.\n",
- "nr = ns*(1 - s)\n",
- "nrrpm = nr*60\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) slip is \",s*100,\" percent\"\n",
- "print \"\\n (b) rotor speed is \",nrrpm,\" rev/min\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) slip is 6.0 percent\n",
- "\n",
- " (b) rotor speed is 705.0 rev/min"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 396</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the rotor copper loss, \n",
- "#(b) the total mechanical power developed by the rotor,\n",
- "#(c) the output power of the motor if friction and windage losses are 750 W, and \n",
- "#(d) the efficiency of the motor, neglecting rotor iron loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Psi = 32000;# in Watts\n",
- "Psl = 1200;# in Watts\n",
- "s = 0.05;# slip\n",
- "Pfl = 750;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Input power to rotor = stator input power -\u0006 stator losses\n",
- "Pi = Psi - Psl\n",
- " #slip = rotor copper loss/rotor input\n",
- "Pl = s*Pi\n",
- " #Total mechanical power developed by the rotor = rotor input power -\u0006 rotor losses\n",
- "Pr = Pi - Pl\n",
- " #Output power of motor = power developed by the rotor -\u0006 friction and windage losses\n",
- "Po = Pr - Pfl\n",
- " #Efficiency of induction motor = (output power/input power)*100\n",
- "eff = (Po/Psi)*100# in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) rotor copper loss is \",Pl,\" Watt\"\n",
- "print \"\\n(b) Total mechanical power developed by the rotor is \",Pr,\" W\"\n",
- "print \"\\n(c) Output power of motor is \",Po,\" Watt\"\n",
- "print \"\\n(d) efficiency of induction motor is \",round(eff,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) rotor copper loss is 1540.0 Watt\n",
- "\n",
- "(b) Total mechanical power developed by the rotor is 29260.0 W\n",
- "\n",
- "(c) Output power of motor is 28510.0 Watt\n",
- "\n",
- "(d) efficiency of induction motor is 89.09 percent"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 397</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the rotor copper loss, and (b) the efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Psi = 32000;# in Watts\n",
- "Psl = 1200;# in Watts\n",
- "Pfl = 750;# in Watts\n",
- "x = 0.35;\n",
- "\n",
- "#calculation:\n",
- " #The slip, s\n",
- "s = 1-x\n",
- " #Input power to rotor = stator input power - stator losses\n",
- "Pi = Psi - Psl\n",
- " #slip = rotor copper loss/rotor input\n",
- "Pl = s*Pi\n",
- " #Total mechanical power developed by the rotor = rotor input power -\u0006 rotor losses\n",
- "Pr = Pi - Pl\n",
- " #Output power of motor = power developed by the rotor -\u0006 friction and windage losses\n",
- "Po = Pr - Pfl\n",
- " #Efficiency of induction motor = (output power/input power)*100\n",
- "eff = (Po/Psi)*100# in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) rotor copper loss is \",Pl,\" Watt\"\n",
- "print \"\\n(b) efficiency of induction motor is \",round(eff,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) rotor copper loss is 20020.0 Watt\n",
- "\n",
- "(b) efficiency of induction motor is 31.34 percent"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 398</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the synchronous speed, (b) the slip, \n",
- "#(c) the full load torque, (d) the power output if mechanical losses amount to 770 W, \n",
- "#(e) the maximum torque, (f) the speed at which maximum torque occurs,\n",
- "#and (g) the starting torque.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "f = 50 ;# in Hz\n",
- "nr = 24;# in rev/sec\n",
- "p = 4/2;# no. of pole pairs\n",
- "R2 = 0.35;# in Ohms\n",
- "X2 = 3.5;# in Ohms\n",
- "tr = 0.85;# turn ratio N2/N1\n",
- "Pl = 770;# in Watt\n",
- "m = 3;# no. of phases\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The slip, s\n",
- "s = ((ns - nr)/ns)*100# in percent\n",
- " #Phase voltage, E1 = V/(3**0.5)\n",
- "E1 = V/(3**0.5)\n",
- " #Full load torque\n",
- "T = (m*(tr**2)/(2*math.pi*ns))*((s/100)*E1*E1*R2/(R2*R2 + (X2*(s/100))**2))\n",
- " #Output power, including friction losses\n",
- "Pm = 2*math.pi*nr*T\n",
- " #power output\n",
- "Po = Pm - Pl\n",
- " #Maximum torque occurs when R2 = Xr = 0.35 ohm\n",
- " #Slip \n",
- "sm = R2/X2\n",
- " #maximum torque, Tm \n",
- "Tm = (m*(tr**2)/(2*math.pi*ns))*(sm*E1*E1*R2/(R2*R2 + (X2*sm)**2))\n",
- " #speed at which maximum torque occurs\n",
- "nrm = ns*(1 - sm)\n",
- "nrmrpm = nrm*60\n",
- " #At the start, i.e., at standstill, slip, s=1\n",
- "ss = 1\n",
- " #starting torque\n",
- "Ts = (m*(tr**2)/(2*math.pi*ns))*(ss*E1*E1*R2/(R2*R2 + (X2*ss)**2))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)Synchronous speed is \",round(ns,2),\" rev/sec\"\n",
- "print \"\\n(b)Slip is \",round(s,2),\" percent\"\n",
- "print \"\\n(c)Full load torque is \",round(T,2),\" Nm\"\n",
- "print \"\\n(d)power output is \",round(Po,2),\"W\"\n",
- "print \"\\n(e)maximum torque is \",round(Tm,2),\" Nm\"\n",
- "print \"\\n(f)speed at which maximum torque occurs is \",round(nrmrpm,2),\"rev/min\"\n",
- "print \"\\n(g)starting torque is \",round(Ts,2),\" Nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)Synchronous speed is 25.0 rev/sec\n",
- "\n",
- "(b)Slip is 4.0 percent\n",
- "\n",
- "(c)Full load torque is 78.05 Nm\n",
- "\n",
- "(d)power output is 10998.99 W\n",
- "\n",
- "(e)maximum torque is 113.17 Nm\n",
- "\n",
- "(f)speed at which maximum torque occurs is 1350.0 rev/min\n",
- "\n",
- "(g)starting torque is 22.41 Nm"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 400</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the induction motor in problem 10 at full load, \n",
- "#(a) the rotor current, (b) the rotor copper loss, and (c) the starting current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "f = 50 ;# in Hz\n",
- "nr = 24;# in rev/sec\n",
- "p = 4/2;# no. of pole pairs\n",
- "R2 = 0.35;# in Ohms\n",
- "X2 = 3.5;# in Ohms\n",
- "tr = 0.85;# turn ratio N2/N1\n",
- "m = 3;# no. of phases\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The slip, s\n",
- "s = ((ns - nr)/ns)*100# in percent\n",
- " #Phase voltage, E1 = V/(3**0.5)\n",
- "E1 = V/(3**0.5)\n",
- " #rotor current,\n",
- "Ir = (s/100)*E1*tr/((R2**2 + (X2*(s/100))**2)**0.5)\n",
- " #Rotor copper loss \n",
- "Pcl = m*R2*(Ir**2)\n",
- " #starting current,\n",
- "ss =1\n",
- "I2 = ss*tr*E1/((R2**2 + (X2*ss)**2)**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)rotor current is \",round(Ir,2),\" A\"\n",
- "print \"\\n(b)Total copper loss is \",round(Pcl,2),\" W\"\n",
- "print \"\\n(c)starting current is \",round(I2,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)rotor current is 21.61 A\n",
- "\n",
- "(b)Total copper loss is 490.37 W\n",
- "\n",
- "(c)starting current is 57.9 A"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 401</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the power input at full load, \n",
- "#(b) the efficiency of the motor at full load and \n",
- "#(c) the current taken from the supply at full load, if the motor runs at a power factor of 0.87 lagging.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "Psl = 650;# in Watt\n",
- "pf = 0.87;# power factor\n",
- "\n",
- "#calculation:\n",
- "Pm = 11770;# watts from part (d), Problem 22.10\n",
- "Pcl = 490.35;# watts, Rotor copper loss, from part (b), Problem 22.11\n",
- " #Stator input power\n",
- "P1 = Pm + Pcl + Psl\n",
- "Po = 11000# watts, Net power output, from part (d), Problem 22.10\n",
- " #efficiency = (output/input) *100\n",
- "eff = (Po/P1)*100# in percent\n",
- " #Power input, P1 = (3**0.5)*VL*IL*cos(phi)\u000e\n",
- " # pf = cos(phi)\n",
- " #supply current, IL\n",
- "I = P1/((3**0.5)*V*pf)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(aStator input power is \",round(P1,2),\" W\"\n",
- "print \"\\n(b)efficiency is \",round(eff,2),\" percent\"\n",
- "print \"\\n(c)supply current is \",round(I,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(aStator input power is 12910.35 W\n",
- "\n",
- "(b)efficiency is 85.2 percent\n",
- "\n",
- "(c)supply current is 20.64 A"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 401</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the resistance of the rotor winding required for maximum starting torque.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "f = 50 ;# in Hz\n",
- "nr = 24;# in rev/sec\n",
- "p = 4/2;# no. of pole pairs\n",
- "R2 = 0.35;# in Ohms\n",
- "X2 = 3.5;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #At the moment of starting, slip, \n",
- "s = 1\n",
- " #Maximum torque occurs when rotor reactance equals rotor resistance\n",
- " #for maximum torque\n",
- "R2 = s*X2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nresistance of the rotor is \",R2,\" Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "resistance of the rotor is 3.5 Ohm"
- ]
- }
- ],
- "prompt_number": 13
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_1.ipynb deleted file mode 100755 index 4bad2773..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_1.ipynb +++ /dev/null @@ -1,791 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 22: Three-phase induction motors</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 389</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the synchronous speed of the motor in rev/min.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 50;# in Hz\n",
- "p = 2/2;# number of pairs of poles\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, \n",
- " #f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- "nsrpm = ns*60\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nsynchronous speed of the motor is \",nsrpm,\" rev/min\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "synchronous speed of the motor is 3000.0 rev/min"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 389</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the number of poles.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 60;# in Hz\n",
- "ns = 900/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "p = f/ns\n",
- "np = p*2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nnumber of poles is \", round(np,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "number of poles is 8.0"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 390</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the frequency of the supply voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 2/2;# number of pairs of poles\n",
- "ns = 6000/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"frequency is \",f,\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "frequency is 100.0 Hz"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 391</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the synchronous speed and (b) the slip at full load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 4/2;# number of pairs of poles\n",
- "f = 50;# in Hz\n",
- "nr = 1455/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The slip, s\n",
- "s = ((ns - nr)/ns)*100# in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) synchronous speed is \",ns,\" rev/sec\"\n",
- "print \"\\n(b) slip is \",s,\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) synchronous speed is 25.0 rev/sec\n",
- "\n",
- "(b) slip is 3.0 percent"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 392</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the synchronous speed,\n",
- "#(b) the speed of the rotor and (c) the frequency of the induced e.m.f.\u2019s in the rotor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 2/2;# number of pairs of poles\n",
- "f = 60;# in Hz\n",
- "s = 0.02;# slip\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The the rotor runs at\n",
- "nr = ns*(1 - s)\n",
- " #frequency of the e.m.f. induced in the rotor bars is\n",
- "fr = ns - nr\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) synchronous speed is \",ns,\" rev/sec\"\n",
- "print \"\\n(b) rotor speed is \",nr,\" rev/sec\"\n",
- "print \"\\n(c) frequency of the e.m.f. induced in the rotor bars is is \",fr,\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) synchronous speed is 60.0 rev/sec\n",
- "\n",
- "(b) rotor speed is 58.8 rev/sec\n",
- "\n",
- "(c) frequency of the e.m.f. induced in the rotor bars is is 1.2 Hz"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 392</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the synchronous speed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 50;# in Hz\n",
- "nr = 1200/60;# in rev/min\n",
- "s = 0.04;# slip\n",
- "\n",
- "#calculation:\n",
- " #the synchronous speed.\n",
- "ns = nr/(1 - s)\n",
- "nsrpm = ns*60\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n synchronous speed is \",nsrpm,\" rev/min\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " synchronous speed is 1250.0 rev/min"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 394</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the slip, and (b) the rotor speed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 8/2;# number of pairs of poles\n",
- "f = 50;# in Hz\n",
- "fr = 3;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #fr = s*f\n",
- "s = (fr/f)\n",
- " #the rotor speed.\n",
- "nr = ns*(1 - s)\n",
- "nrrpm = nr*60\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) slip is \",s*100,\" percent\"\n",
- "print \"\\n (b) rotor speed is \",nrrpm,\" rev/min\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) slip is 6.0 percent\n",
- "\n",
- " (b) rotor speed is 705.0 rev/min"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 396</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the rotor copper loss, \n",
- "#(b) the total mechanical power developed by the rotor,\n",
- "#(c) the output power of the motor if friction and windage losses are 750 W, and \n",
- "#(d) the efficiency of the motor, neglecting rotor iron loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Psi = 32000;# in Watts\n",
- "Psl = 1200;# in Watts\n",
- "s = 0.05;# slip\n",
- "Pfl = 750;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Input power to rotor = stator input power -\u0006 stator losses\n",
- "Pi = Psi - Psl\n",
- " #slip = rotor copper loss/rotor input\n",
- "Pl = s*Pi\n",
- " #Total mechanical power developed by the rotor = rotor input power -\u0006 rotor losses\n",
- "Pr = Pi - Pl\n",
- " #Output power of motor = power developed by the rotor -\u0006 friction and windage losses\n",
- "Po = Pr - Pfl\n",
- " #Efficiency of induction motor = (output power/input power)*100\n",
- "eff = (Po/Psi)*100# in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) rotor copper loss is \",Pl,\" Watt\"\n",
- "print \"\\n(b) Total mechanical power developed by the rotor is \",Pr,\" W\"\n",
- "print \"\\n(c) Output power of motor is \",Po,\" Watt\"\n",
- "print \"\\n(d) efficiency of induction motor is \",round(eff,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) rotor copper loss is 1540.0 Watt\n",
- "\n",
- "(b) Total mechanical power developed by the rotor is 29260.0 W\n",
- "\n",
- "(c) Output power of motor is 28510.0 Watt\n",
- "\n",
- "(d) efficiency of induction motor is 89.09 percent"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 397</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the rotor copper loss, and (b) the efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Psi = 32000;# in Watts\n",
- "Psl = 1200;# in Watts\n",
- "Pfl = 750;# in Watts\n",
- "x = 0.35;\n",
- "\n",
- "#calculation:\n",
- " #The slip, s\n",
- "s = 1-x\n",
- " #Input power to rotor = stator input power - stator losses\n",
- "Pi = Psi - Psl\n",
- " #slip = rotor copper loss/rotor input\n",
- "Pl = s*Pi\n",
- " #Total mechanical power developed by the rotor = rotor input power -\u0006 rotor losses\n",
- "Pr = Pi - Pl\n",
- " #Output power of motor = power developed by the rotor -\u0006 friction and windage losses\n",
- "Po = Pr - Pfl\n",
- " #Efficiency of induction motor = (output power/input power)*100\n",
- "eff = (Po/Psi)*100# in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) rotor copper loss is \",Pl,\" Watt\"\n",
- "print \"\\n(b) efficiency of induction motor is \",round(eff,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) rotor copper loss is 20020.0 Watt\n",
- "\n",
- "(b) efficiency of induction motor is 31.34 percent"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 398</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the synchronous speed, (b) the slip, \n",
- "#(c) the full load torque, (d) the power output if mechanical losses amount to 770 W, \n",
- "#(e) the maximum torque, (f) the speed at which maximum torque occurs,\n",
- "#and (g) the starting torque.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "f = 50 ;# in Hz\n",
- "nr = 24;# in rev/sec\n",
- "p = 4/2;# no. of pole pairs\n",
- "R2 = 0.35;# in Ohms\n",
- "X2 = 3.5;# in Ohms\n",
- "tr = 0.85;# turn ratio N2/N1\n",
- "Pl = 770;# in Watt\n",
- "m = 3;# no. of phases\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The slip, s\n",
- "s = ((ns - nr)/ns)*100# in percent\n",
- " #Phase voltage, E1 = V/(3**0.5)\n",
- "E1 = V/(3**0.5)\n",
- " #Full load torque\n",
- "T = (m*(tr**2)/(2*math.pi*ns))*((s/100)*E1*E1*R2/(R2*R2 + (X2*(s/100))**2))\n",
- " #Output power, including friction losses\n",
- "Pm = 2*math.pi*nr*T\n",
- " #power output\n",
- "Po = Pm - Pl\n",
- " #Maximum torque occurs when R2 = Xr = 0.35 ohm\n",
- " #Slip \n",
- "sm = R2/X2\n",
- " #maximum torque, Tm \n",
- "Tm = (m*(tr**2)/(2*math.pi*ns))*(sm*E1*E1*R2/(R2*R2 + (X2*sm)**2))\n",
- " #speed at which maximum torque occurs\n",
- "nrm = ns*(1 - sm)\n",
- "nrmrpm = nrm*60\n",
- " #At the start, i.e., at standstill, slip, s=1\n",
- "ss = 1\n",
- " #starting torque\n",
- "Ts = (m*(tr**2)/(2*math.pi*ns))*(ss*E1*E1*R2/(R2*R2 + (X2*ss)**2))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)Synchronous speed is \",round(ns,2),\" rev/sec\"\n",
- "print \"\\n(b)Slip is \",round(s,2),\" percent\"\n",
- "print \"\\n(c)Full load torque is \",round(T,2),\" Nm\"\n",
- "print \"\\n(d)power output is \",round(Po,2),\"W\"\n",
- "print \"\\n(e)maximum torque is \",round(Tm,2),\" Nm\"\n",
- "print \"\\n(f)speed at which maximum torque occurs is \",round(nrmrpm,2),\"rev/min\"\n",
- "print \"\\n(g)starting torque is \",round(Ts,2),\" Nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)Synchronous speed is 25.0 rev/sec\n",
- "\n",
- "(b)Slip is 4.0 percent\n",
- "\n",
- "(c)Full load torque is 78.05 Nm\n",
- "\n",
- "(d)power output is 10998.99 W\n",
- "\n",
- "(e)maximum torque is 113.17 Nm\n",
- "\n",
- "(f)speed at which maximum torque occurs is 1350.0 rev/min\n",
- "\n",
- "(g)starting torque is 22.41 Nm"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 400</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the induction motor in problem 10 at full load, \n",
- "#(a) the rotor current, (b) the rotor copper loss, and (c) the starting current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "f = 50 ;# in Hz\n",
- "nr = 24;# in rev/sec\n",
- "p = 4/2;# no. of pole pairs\n",
- "R2 = 0.35;# in Ohms\n",
- "X2 = 3.5;# in Ohms\n",
- "tr = 0.85;# turn ratio N2/N1\n",
- "m = 3;# no. of phases\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The slip, s\n",
- "s = ((ns - nr)/ns)*100# in percent\n",
- " #Phase voltage, E1 = V/(3**0.5)\n",
- "E1 = V/(3**0.5)\n",
- " #rotor current,\n",
- "Ir = (s/100)*E1*tr/((R2**2 + (X2*(s/100))**2)**0.5)\n",
- " #Rotor copper loss \n",
- "Pcl = m*R2*(Ir**2)\n",
- " #starting current,\n",
- "ss =1\n",
- "I2 = ss*tr*E1/((R2**2 + (X2*ss)**2)**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)rotor current is \",round(Ir,2),\" A\"\n",
- "print \"\\n(b)Total copper loss is \",round(Pcl,2),\" W\"\n",
- "print \"\\n(c)starting current is \",round(I2,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)rotor current is 21.61 A\n",
- "\n",
- "(b)Total copper loss is 490.37 W\n",
- "\n",
- "(c)starting current is 57.9 A"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 401</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the power input at full load, \n",
- "#(b) the efficiency of the motor at full load and \n",
- "#(c) the current taken from the supply at full load, if the motor runs at a power factor of 0.87 lagging.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "Psl = 650;# in Watt\n",
- "pf = 0.87;# power factor\n",
- "\n",
- "#calculation:\n",
- "Pm = 11770;# watts from part (d), Problem 22.10\n",
- "Pcl = 490.35;# watts, Rotor copper loss, from part (b), Problem 22.11\n",
- " #Stator input power\n",
- "P1 = Pm + Pcl + Psl\n",
- "Po = 11000# watts, Net power output, from part (d), Problem 22.10\n",
- " #efficiency = (output/input) *100\n",
- "eff = (Po/P1)*100# in percent\n",
- " #Power input, P1 = (3**0.5)*VL*IL*cos(phi)\u000e\n",
- " # pf = cos(phi)\n",
- " #supply current, IL\n",
- "I = P1/((3**0.5)*V*pf)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(aStator input power is \",round(P1,2),\" W\"\n",
- "print \"\\n(b)efficiency is \",round(eff,2),\" percent\"\n",
- "print \"\\n(c)supply current is \",round(I,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(aStator input power is 12910.35 W\n",
- "\n",
- "(b)efficiency is 85.2 percent\n",
- "\n",
- "(c)supply current is 20.64 A"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 401</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the resistance of the rotor winding required for maximum starting torque.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "f = 50 ;# in Hz\n",
- "nr = 24;# in rev/sec\n",
- "p = 4/2;# no. of pole pairs\n",
- "R2 = 0.35;# in Ohms\n",
- "X2 = 3.5;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #At the moment of starting, slip, \n",
- "s = 1\n",
- " #Maximum torque occurs when rotor reactance equals rotor resistance\n",
- " #for maximum torque\n",
- "R2 = s*X2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nresistance of the rotor is \",R2,\" Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "resistance of the rotor is 3.5 Ohm"
- ]
- }
- ],
- "prompt_number": 13
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_2.ipynb deleted file mode 100755 index 4bad2773..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_2.ipynb +++ /dev/null @@ -1,791 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 22: Three-phase induction motors</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 389</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the synchronous speed of the motor in rev/min.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 50;# in Hz\n",
- "p = 2/2;# number of pairs of poles\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, \n",
- " #f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- "nsrpm = ns*60\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nsynchronous speed of the motor is \",nsrpm,\" rev/min\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "synchronous speed of the motor is 3000.0 rev/min"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 389</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the number of poles.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 60;# in Hz\n",
- "ns = 900/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "p = f/ns\n",
- "np = p*2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nnumber of poles is \", round(np,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "number of poles is 8.0"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 390</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the frequency of the supply voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 2/2;# number of pairs of poles\n",
- "ns = 6000/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"frequency is \",f,\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "frequency is 100.0 Hz"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 391</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the synchronous speed and (b) the slip at full load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 4/2;# number of pairs of poles\n",
- "f = 50;# in Hz\n",
- "nr = 1455/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The slip, s\n",
- "s = ((ns - nr)/ns)*100# in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) synchronous speed is \",ns,\" rev/sec\"\n",
- "print \"\\n(b) slip is \",s,\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) synchronous speed is 25.0 rev/sec\n",
- "\n",
- "(b) slip is 3.0 percent"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 392</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the synchronous speed,\n",
- "#(b) the speed of the rotor and (c) the frequency of the induced e.m.f.\u2019s in the rotor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 2/2;# number of pairs of poles\n",
- "f = 60;# in Hz\n",
- "s = 0.02;# slip\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The the rotor runs at\n",
- "nr = ns*(1 - s)\n",
- " #frequency of the e.m.f. induced in the rotor bars is\n",
- "fr = ns - nr\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) synchronous speed is \",ns,\" rev/sec\"\n",
- "print \"\\n(b) rotor speed is \",nr,\" rev/sec\"\n",
- "print \"\\n(c) frequency of the e.m.f. induced in the rotor bars is is \",fr,\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) synchronous speed is 60.0 rev/sec\n",
- "\n",
- "(b) rotor speed is 58.8 rev/sec\n",
- "\n",
- "(c) frequency of the e.m.f. induced in the rotor bars is is 1.2 Hz"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 392</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the synchronous speed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 50;# in Hz\n",
- "nr = 1200/60;# in rev/min\n",
- "s = 0.04;# slip\n",
- "\n",
- "#calculation:\n",
- " #the synchronous speed.\n",
- "ns = nr/(1 - s)\n",
- "nsrpm = ns*60\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n synchronous speed is \",nsrpm,\" rev/min\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " synchronous speed is 1250.0 rev/min"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 394</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the slip, and (b) the rotor speed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 8/2;# number of pairs of poles\n",
- "f = 50;# in Hz\n",
- "fr = 3;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #fr = s*f\n",
- "s = (fr/f)\n",
- " #the rotor speed.\n",
- "nr = ns*(1 - s)\n",
- "nrrpm = nr*60\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) slip is \",s*100,\" percent\"\n",
- "print \"\\n (b) rotor speed is \",nrrpm,\" rev/min\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) slip is 6.0 percent\n",
- "\n",
- " (b) rotor speed is 705.0 rev/min"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 396</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the rotor copper loss, \n",
- "#(b) the total mechanical power developed by the rotor,\n",
- "#(c) the output power of the motor if friction and windage losses are 750 W, and \n",
- "#(d) the efficiency of the motor, neglecting rotor iron loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Psi = 32000;# in Watts\n",
- "Psl = 1200;# in Watts\n",
- "s = 0.05;# slip\n",
- "Pfl = 750;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Input power to rotor = stator input power -\u0006 stator losses\n",
- "Pi = Psi - Psl\n",
- " #slip = rotor copper loss/rotor input\n",
- "Pl = s*Pi\n",
- " #Total mechanical power developed by the rotor = rotor input power -\u0006 rotor losses\n",
- "Pr = Pi - Pl\n",
- " #Output power of motor = power developed by the rotor -\u0006 friction and windage losses\n",
- "Po = Pr - Pfl\n",
- " #Efficiency of induction motor = (output power/input power)*100\n",
- "eff = (Po/Psi)*100# in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) rotor copper loss is \",Pl,\" Watt\"\n",
- "print \"\\n(b) Total mechanical power developed by the rotor is \",Pr,\" W\"\n",
- "print \"\\n(c) Output power of motor is \",Po,\" Watt\"\n",
- "print \"\\n(d) efficiency of induction motor is \",round(eff,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) rotor copper loss is 1540.0 Watt\n",
- "\n",
- "(b) Total mechanical power developed by the rotor is 29260.0 W\n",
- "\n",
- "(c) Output power of motor is 28510.0 Watt\n",
- "\n",
- "(d) efficiency of induction motor is 89.09 percent"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 397</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the rotor copper loss, and (b) the efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Psi = 32000;# in Watts\n",
- "Psl = 1200;# in Watts\n",
- "Pfl = 750;# in Watts\n",
- "x = 0.35;\n",
- "\n",
- "#calculation:\n",
- " #The slip, s\n",
- "s = 1-x\n",
- " #Input power to rotor = stator input power - stator losses\n",
- "Pi = Psi - Psl\n",
- " #slip = rotor copper loss/rotor input\n",
- "Pl = s*Pi\n",
- " #Total mechanical power developed by the rotor = rotor input power -\u0006 rotor losses\n",
- "Pr = Pi - Pl\n",
- " #Output power of motor = power developed by the rotor -\u0006 friction and windage losses\n",
- "Po = Pr - Pfl\n",
- " #Efficiency of induction motor = (output power/input power)*100\n",
- "eff = (Po/Psi)*100# in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) rotor copper loss is \",Pl,\" Watt\"\n",
- "print \"\\n(b) efficiency of induction motor is \",round(eff,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) rotor copper loss is 20020.0 Watt\n",
- "\n",
- "(b) efficiency of induction motor is 31.34 percent"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 398</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the synchronous speed, (b) the slip, \n",
- "#(c) the full load torque, (d) the power output if mechanical losses amount to 770 W, \n",
- "#(e) the maximum torque, (f) the speed at which maximum torque occurs,\n",
- "#and (g) the starting torque.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "f = 50 ;# in Hz\n",
- "nr = 24;# in rev/sec\n",
- "p = 4/2;# no. of pole pairs\n",
- "R2 = 0.35;# in Ohms\n",
- "X2 = 3.5;# in Ohms\n",
- "tr = 0.85;# turn ratio N2/N1\n",
- "Pl = 770;# in Watt\n",
- "m = 3;# no. of phases\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The slip, s\n",
- "s = ((ns - nr)/ns)*100# in percent\n",
- " #Phase voltage, E1 = V/(3**0.5)\n",
- "E1 = V/(3**0.5)\n",
- " #Full load torque\n",
- "T = (m*(tr**2)/(2*math.pi*ns))*((s/100)*E1*E1*R2/(R2*R2 + (X2*(s/100))**2))\n",
- " #Output power, including friction losses\n",
- "Pm = 2*math.pi*nr*T\n",
- " #power output\n",
- "Po = Pm - Pl\n",
- " #Maximum torque occurs when R2 = Xr = 0.35 ohm\n",
- " #Slip \n",
- "sm = R2/X2\n",
- " #maximum torque, Tm \n",
- "Tm = (m*(tr**2)/(2*math.pi*ns))*(sm*E1*E1*R2/(R2*R2 + (X2*sm)**2))\n",
- " #speed at which maximum torque occurs\n",
- "nrm = ns*(1 - sm)\n",
- "nrmrpm = nrm*60\n",
- " #At the start, i.e., at standstill, slip, s=1\n",
- "ss = 1\n",
- " #starting torque\n",
- "Ts = (m*(tr**2)/(2*math.pi*ns))*(ss*E1*E1*R2/(R2*R2 + (X2*ss)**2))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)Synchronous speed is \",round(ns,2),\" rev/sec\"\n",
- "print \"\\n(b)Slip is \",round(s,2),\" percent\"\n",
- "print \"\\n(c)Full load torque is \",round(T,2),\" Nm\"\n",
- "print \"\\n(d)power output is \",round(Po,2),\"W\"\n",
- "print \"\\n(e)maximum torque is \",round(Tm,2),\" Nm\"\n",
- "print \"\\n(f)speed at which maximum torque occurs is \",round(nrmrpm,2),\"rev/min\"\n",
- "print \"\\n(g)starting torque is \",round(Ts,2),\" Nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)Synchronous speed is 25.0 rev/sec\n",
- "\n",
- "(b)Slip is 4.0 percent\n",
- "\n",
- "(c)Full load torque is 78.05 Nm\n",
- "\n",
- "(d)power output is 10998.99 W\n",
- "\n",
- "(e)maximum torque is 113.17 Nm\n",
- "\n",
- "(f)speed at which maximum torque occurs is 1350.0 rev/min\n",
- "\n",
- "(g)starting torque is 22.41 Nm"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 400</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the induction motor in problem 10 at full load, \n",
- "#(a) the rotor current, (b) the rotor copper loss, and (c) the starting current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "f = 50 ;# in Hz\n",
- "nr = 24;# in rev/sec\n",
- "p = 4/2;# no. of pole pairs\n",
- "R2 = 0.35;# in Ohms\n",
- "X2 = 3.5;# in Ohms\n",
- "tr = 0.85;# turn ratio N2/N1\n",
- "m = 3;# no. of phases\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The slip, s\n",
- "s = ((ns - nr)/ns)*100# in percent\n",
- " #Phase voltage, E1 = V/(3**0.5)\n",
- "E1 = V/(3**0.5)\n",
- " #rotor current,\n",
- "Ir = (s/100)*E1*tr/((R2**2 + (X2*(s/100))**2)**0.5)\n",
- " #Rotor copper loss \n",
- "Pcl = m*R2*(Ir**2)\n",
- " #starting current,\n",
- "ss =1\n",
- "I2 = ss*tr*E1/((R2**2 + (X2*ss)**2)**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)rotor current is \",round(Ir,2),\" A\"\n",
- "print \"\\n(b)Total copper loss is \",round(Pcl,2),\" W\"\n",
- "print \"\\n(c)starting current is \",round(I2,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)rotor current is 21.61 A\n",
- "\n",
- "(b)Total copper loss is 490.37 W\n",
- "\n",
- "(c)starting current is 57.9 A"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 401</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the power input at full load, \n",
- "#(b) the efficiency of the motor at full load and \n",
- "#(c) the current taken from the supply at full load, if the motor runs at a power factor of 0.87 lagging.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "Psl = 650;# in Watt\n",
- "pf = 0.87;# power factor\n",
- "\n",
- "#calculation:\n",
- "Pm = 11770;# watts from part (d), Problem 22.10\n",
- "Pcl = 490.35;# watts, Rotor copper loss, from part (b), Problem 22.11\n",
- " #Stator input power\n",
- "P1 = Pm + Pcl + Psl\n",
- "Po = 11000# watts, Net power output, from part (d), Problem 22.10\n",
- " #efficiency = (output/input) *100\n",
- "eff = (Po/P1)*100# in percent\n",
- " #Power input, P1 = (3**0.5)*VL*IL*cos(phi)\u000e\n",
- " # pf = cos(phi)\n",
- " #supply current, IL\n",
- "I = P1/((3**0.5)*V*pf)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(aStator input power is \",round(P1,2),\" W\"\n",
- "print \"\\n(b)efficiency is \",round(eff,2),\" percent\"\n",
- "print \"\\n(c)supply current is \",round(I,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(aStator input power is 12910.35 W\n",
- "\n",
- "(b)efficiency is 85.2 percent\n",
- "\n",
- "(c)supply current is 20.64 A"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 401</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the resistance of the rotor winding required for maximum starting torque.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "f = 50 ;# in Hz\n",
- "nr = 24;# in rev/sec\n",
- "p = 4/2;# no. of pole pairs\n",
- "R2 = 0.35;# in Ohms\n",
- "X2 = 3.5;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #At the moment of starting, slip, \n",
- "s = 1\n",
- " #Maximum torque occurs when rotor reactance equals rotor resistance\n",
- " #for maximum torque\n",
- "R2 = s*X2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nresistance of the rotor is \",R2,\" Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "resistance of the rotor is 3.5 Ohm"
- ]
- }
- ],
- "prompt_number": 13
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_22.ipynb index 531badfb..531badfb 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_22.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint.ipynb deleted file mode 100755 index 043d0bdd..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint.ipynb +++ /dev/null @@ -1,434 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:e4e0b2e74a0b30b96b3a586de7b740bbefabf9c41230a1638409bfb7b309d066"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 23: Revision of complex numbers</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 418</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine ZT\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 5 - 3j;\n",
- "Z2 = 4 + 7j;\n",
- "Z3 = 3.9 - 6.7j;\n",
- "\n",
- " #calculation:\n",
- "ZT = (Z1*Z2/(Z1 + Z2))+ Z3\n",
- "y = ZT.imag\n",
- "x = ZT.real\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n ZT is \",round(x,2),\" + (\",round(y,2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " ZT is 8.65 + ( -6.26 )i"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 418</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine in cartesian form correct to three decimal places:\n",
- "#(a)1/Z1 (b)1/Z2 (c) 1/Z1 * 1/Z2 (d) 1/(1/Z1 + 1/Z2)\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 3 + 4j;\n",
- "Z2 = 2 - 5j;\n",
- "\n",
- "#calculation:\n",
- "za = 1/Z1\n",
- "zb = 1/Z2\n",
- "zc = za + zb\n",
- "zd = 1/zc\n",
- "zax = za.real\n",
- "zay = za.imag\n",
- "zbx = zb.real\n",
- "zby = zb.imag\n",
- "zcx = zc.real\n",
- "zcy = zc.imag\n",
- "zdx = zd.real\n",
- "zdy = zd.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)1/Z1 is \",round( zax,2),\" + (\",round(zay,2),\")i\"\n",
- "print \"\\n (b)1/Z2 is \",round( zbx,2),\" + (\",round(zby,2),\")i\"\n",
- "print \"\\n (c)1/Z1 + 1/Z2 is \",round( zcx,2),\" + (\",round(zcy,2),\")i\"\n",
- "print \"\\n (d)1/(1/Z1 + 1/Z2) is \",round( zdx,2),\" + (\",round(zdy,2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)1/Z1 is 0.12 + ( -0.16 )i\n",
- "\n",
- " (b)1/Z2 is 0.07 + ( 0.17 )i\n",
- "\n",
- " (c)1/Z1 + 1/Z2 is 0.19 + ( 0.01 )i\n",
- "\n",
- " (d)1/(1/Z1 + 1/Z2) is 5.27 + ( -0.35 )i"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 419</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find a, b, x and y?\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 9 - 2j;\n",
- "Z2 = 2 + 1j;\n",
- "Z3 = -2 + 1j;\n",
- "Z4 = 5 + 2j;\n",
- "\n",
- "#calculation:\n",
- "za = Z1/3\n",
- "zb = Z2*Z3\n",
- "zca = (2*Z4.real + Z4.imag)/-1\n",
- "zcb = Z4.real - zca\n",
- "zaa = za.real\n",
- "zab = za.imag\n",
- "zbx = zb.real\n",
- "zby = zb.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)a and b are \", zaa,\" and \",round(zab,2),\" resp.\"\n",
- "print \"\\n (b)x and y are \", zbx,\" and \",zby,\" resp.\"\n",
- "print \"\\n (c)a and b are \", zca,\" and \",zcb,\" resp.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)a and b are 3.0 and -0.67 resp.\n",
- "\n",
- " (b)x and y are -5.0 and 0.0 resp.\n",
- "\n",
- " (c)a and b are -12.0 and 17.0 resp."
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 422</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Convert 5/_132\u00b0 into a + jb form correct to four significant figures.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r = 5;# magnitude\n",
- "theta = -132;# in degree\n",
- "\n",
- "#calculation:\n",
- "x = r*math.sin(theta*math.pi/180)\n",
- "y = r*math.cos(theta*math.pi/180)\n",
- "z = x + y*1j\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Z is \",round(x,2),\" + (\",round(y,2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Z is -3.72 + ( -3.35 )i"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 422</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine in polar form the total impedance ZT given that ZT = Z1Z2/\u0007(Z1 + Z2\b)\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r1 = 4.76;# magnitude\n",
- "theta1 = 35;# in degree\n",
- "r2 = 7.36;# magnitude\n",
- "theta2 = -48;# in degree\n",
- "\n",
- "#calculation:\n",
- "x1 = r1*cmath.cos(theta1*math.pi/180)\n",
- "y1 = r1*cmath.sin(theta1*math.pi/180)\n",
- "z1 = x1 + y1*1j\n",
- "x2 = r2*cmath.cos(theta2*math.pi/180)\n",
- "y2 = r2*cmath.sin(theta2*math.pi/180)\n",
- "z2 = x2 + y2*1j\n",
- "z3 = z1*z2/(z1 + z2)\n",
- "x3 = z3.real\n",
- "y3 = z3.imag\n",
- "r3 = (x3**2 + y3**2)**0.5\n",
- "theta3 = cmath.phase(complex(x3,y3))*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n ZT is (\",round( r3,2),\",round(/_\",round(theta3,2),\"deg)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " ZT is ( 3.79 ,round(/_ 4.25 deg)"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 423</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine \u0007\u0003(2 +j3\b)^5 in polar and in cartesian form.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "z = -2 + 3j;\n",
- "\n",
- "#calculation:\n",
- "zc = z**5\n",
- "x = zc.real\n",
- "y = zc.imag\n",
- "r = (x**2 + y**2)**0.5\n",
- "theta = cmath.phase(complex(x,y))*180/math.pi\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Z is \",round( x,2),\" + (\",round(y,2),\")i\"\n",
- "print \"\\n ZT is (\",round( r,2),\"round/_\",round(theta,2),\"deg)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Z is -122.0 + ( -597.0 )i\n",
- "\n",
- " ZT is ( 609.34 round/_ -101.55 deg)"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 423</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the two square roots of the complex number \u0007(12 + j5)\b in cartesian and polar form\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "z = 12 + 5j;\n",
- "\n",
- "#calculation:\n",
- "x = z.real\n",
- "y = z.imag\n",
- "r = (x**2 + y**2)**0.5\n",
- "theta1 = cmath.atan(y/x)*180/math.pi\n",
- "'''\n",
- "if ((x<0)&(y<0))\n",
- " theta1 = theta1 -180;\n",
- "elif ((x<0)&(y>0))\n",
- " theta1 = theta1 +180;\n",
- "'''\n",
- "theta2 = theta1 + 360\n",
- "rtheta1 = theta1/2\n",
- "rtheta2 = theta2/2\n",
- "'''\n",
- "if (rtheta2 > 180)\n",
- " rtheta2 = rtheta2 -360;\n",
- "elif ((x<0)&(y>0))\n",
- " rtheta2 = rtheta2 +360;\n",
- "'''\n",
- "rr = r**0.5\n",
- "x1 = rr*cmath.cos(rtheta1*math.pi/180)\n",
- "y1 = rr*cmath.sin(rtheta1*math.pi/180)\n",
- "z1 = x1 + y1*1j\n",
- "\n",
- "x2 = rr*cmath.cos(rtheta2*math.pi/180)\n",
- "y2 = rr*cmath.sin(rtheta2*math.pi/180)\n",
- "z2 = x2 + y2*1j\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n two roots are (\",round(z1.real,2),\" + (\",round(z1.imag,2),\")i) \"\n",
- "print \" and (\",round(z2.real,2),\" + (\",round(z2.imag,2),\")i)\"\n",
- "print \"\\n two roots are (\",round( rr,2),\"/_\",round((cmath.phase(complex(z1.real,z1.imag)))*180/math.pi,2),\"deg) \"\n",
- "print \" and (\",round( rr,2),\"/_\",round((cmath.phase(complex(z2.real,z2.imag)))*180/math.pi,2),\"deg)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " two roots are ( 3.54 + ( 0.71 )i) \n",
- " and ( -3.54 + ( -0.71 )i)\n",
- "\n",
- " two roots are ( 3.61 /_ 11.31 deg) \n",
- " and ( 3.61 /_ -168.69 deg)\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_1.ipynb deleted file mode 100755 index 043d0bdd..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_1.ipynb +++ /dev/null @@ -1,434 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:e4e0b2e74a0b30b96b3a586de7b740bbefabf9c41230a1638409bfb7b309d066"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 23: Revision of complex numbers</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 418</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine ZT\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 5 - 3j;\n",
- "Z2 = 4 + 7j;\n",
- "Z3 = 3.9 - 6.7j;\n",
- "\n",
- " #calculation:\n",
- "ZT = (Z1*Z2/(Z1 + Z2))+ Z3\n",
- "y = ZT.imag\n",
- "x = ZT.real\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n ZT is \",round(x,2),\" + (\",round(y,2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " ZT is 8.65 + ( -6.26 )i"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 418</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine in cartesian form correct to three decimal places:\n",
- "#(a)1/Z1 (b)1/Z2 (c) 1/Z1 * 1/Z2 (d) 1/(1/Z1 + 1/Z2)\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 3 + 4j;\n",
- "Z2 = 2 - 5j;\n",
- "\n",
- "#calculation:\n",
- "za = 1/Z1\n",
- "zb = 1/Z2\n",
- "zc = za + zb\n",
- "zd = 1/zc\n",
- "zax = za.real\n",
- "zay = za.imag\n",
- "zbx = zb.real\n",
- "zby = zb.imag\n",
- "zcx = zc.real\n",
- "zcy = zc.imag\n",
- "zdx = zd.real\n",
- "zdy = zd.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)1/Z1 is \",round( zax,2),\" + (\",round(zay,2),\")i\"\n",
- "print \"\\n (b)1/Z2 is \",round( zbx,2),\" + (\",round(zby,2),\")i\"\n",
- "print \"\\n (c)1/Z1 + 1/Z2 is \",round( zcx,2),\" + (\",round(zcy,2),\")i\"\n",
- "print \"\\n (d)1/(1/Z1 + 1/Z2) is \",round( zdx,2),\" + (\",round(zdy,2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)1/Z1 is 0.12 + ( -0.16 )i\n",
- "\n",
- " (b)1/Z2 is 0.07 + ( 0.17 )i\n",
- "\n",
- " (c)1/Z1 + 1/Z2 is 0.19 + ( 0.01 )i\n",
- "\n",
- " (d)1/(1/Z1 + 1/Z2) is 5.27 + ( -0.35 )i"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 419</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find a, b, x and y?\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 9 - 2j;\n",
- "Z2 = 2 + 1j;\n",
- "Z3 = -2 + 1j;\n",
- "Z4 = 5 + 2j;\n",
- "\n",
- "#calculation:\n",
- "za = Z1/3\n",
- "zb = Z2*Z3\n",
- "zca = (2*Z4.real + Z4.imag)/-1\n",
- "zcb = Z4.real - zca\n",
- "zaa = za.real\n",
- "zab = za.imag\n",
- "zbx = zb.real\n",
- "zby = zb.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)a and b are \", zaa,\" and \",round(zab,2),\" resp.\"\n",
- "print \"\\n (b)x and y are \", zbx,\" and \",zby,\" resp.\"\n",
- "print \"\\n (c)a and b are \", zca,\" and \",zcb,\" resp.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)a and b are 3.0 and -0.67 resp.\n",
- "\n",
- " (b)x and y are -5.0 and 0.0 resp.\n",
- "\n",
- " (c)a and b are -12.0 and 17.0 resp."
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 422</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Convert 5/_132\u00b0 into a + jb form correct to four significant figures.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r = 5;# magnitude\n",
- "theta = -132;# in degree\n",
- "\n",
- "#calculation:\n",
- "x = r*math.sin(theta*math.pi/180)\n",
- "y = r*math.cos(theta*math.pi/180)\n",
- "z = x + y*1j\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Z is \",round(x,2),\" + (\",round(y,2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Z is -3.72 + ( -3.35 )i"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 422</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine in polar form the total impedance ZT given that ZT = Z1Z2/\u0007(Z1 + Z2\b)\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r1 = 4.76;# magnitude\n",
- "theta1 = 35;# in degree\n",
- "r2 = 7.36;# magnitude\n",
- "theta2 = -48;# in degree\n",
- "\n",
- "#calculation:\n",
- "x1 = r1*cmath.cos(theta1*math.pi/180)\n",
- "y1 = r1*cmath.sin(theta1*math.pi/180)\n",
- "z1 = x1 + y1*1j\n",
- "x2 = r2*cmath.cos(theta2*math.pi/180)\n",
- "y2 = r2*cmath.sin(theta2*math.pi/180)\n",
- "z2 = x2 + y2*1j\n",
- "z3 = z1*z2/(z1 + z2)\n",
- "x3 = z3.real\n",
- "y3 = z3.imag\n",
- "r3 = (x3**2 + y3**2)**0.5\n",
- "theta3 = cmath.phase(complex(x3,y3))*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n ZT is (\",round( r3,2),\",round(/_\",round(theta3,2),\"deg)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " ZT is ( 3.79 ,round(/_ 4.25 deg)"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 423</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine \u0007\u0003(2 +j3\b)^5 in polar and in cartesian form.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "z = -2 + 3j;\n",
- "\n",
- "#calculation:\n",
- "zc = z**5\n",
- "x = zc.real\n",
- "y = zc.imag\n",
- "r = (x**2 + y**2)**0.5\n",
- "theta = cmath.phase(complex(x,y))*180/math.pi\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Z is \",round( x,2),\" + (\",round(y,2),\")i\"\n",
- "print \"\\n ZT is (\",round( r,2),\"round/_\",round(theta,2),\"deg)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Z is -122.0 + ( -597.0 )i\n",
- "\n",
- " ZT is ( 609.34 round/_ -101.55 deg)"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 423</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the two square roots of the complex number \u0007(12 + j5)\b in cartesian and polar form\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "z = 12 + 5j;\n",
- "\n",
- "#calculation:\n",
- "x = z.real\n",
- "y = z.imag\n",
- "r = (x**2 + y**2)**0.5\n",
- "theta1 = cmath.atan(y/x)*180/math.pi\n",
- "'''\n",
- "if ((x<0)&(y<0))\n",
- " theta1 = theta1 -180;\n",
- "elif ((x<0)&(y>0))\n",
- " theta1 = theta1 +180;\n",
- "'''\n",
- "theta2 = theta1 + 360\n",
- "rtheta1 = theta1/2\n",
- "rtheta2 = theta2/2\n",
- "'''\n",
- "if (rtheta2 > 180)\n",
- " rtheta2 = rtheta2 -360;\n",
- "elif ((x<0)&(y>0))\n",
- " rtheta2 = rtheta2 +360;\n",
- "'''\n",
- "rr = r**0.5\n",
- "x1 = rr*cmath.cos(rtheta1*math.pi/180)\n",
- "y1 = rr*cmath.sin(rtheta1*math.pi/180)\n",
- "z1 = x1 + y1*1j\n",
- "\n",
- "x2 = rr*cmath.cos(rtheta2*math.pi/180)\n",
- "y2 = rr*cmath.sin(rtheta2*math.pi/180)\n",
- "z2 = x2 + y2*1j\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n two roots are (\",round(z1.real,2),\" + (\",round(z1.imag,2),\")i) \"\n",
- "print \" and (\",round(z2.real,2),\" + (\",round(z2.imag,2),\")i)\"\n",
- "print \"\\n two roots are (\",round( rr,2),\"/_\",round((cmath.phase(complex(z1.real,z1.imag)))*180/math.pi,2),\"deg) \"\n",
- "print \" and (\",round( rr,2),\"/_\",round((cmath.phase(complex(z2.real,z2.imag)))*180/math.pi,2),\"deg)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " two roots are ( 3.54 + ( 0.71 )i) \n",
- " and ( -3.54 + ( -0.71 )i)\n",
- "\n",
- " two roots are ( 3.61 /_ 11.31 deg) \n",
- " and ( 3.61 /_ -168.69 deg)\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_2.ipynb deleted file mode 100755 index 043d0bdd..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_2.ipynb +++ /dev/null @@ -1,434 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:e4e0b2e74a0b30b96b3a586de7b740bbefabf9c41230a1638409bfb7b309d066"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 23: Revision of complex numbers</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 418</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine ZT\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 5 - 3j;\n",
- "Z2 = 4 + 7j;\n",
- "Z3 = 3.9 - 6.7j;\n",
- "\n",
- " #calculation:\n",
- "ZT = (Z1*Z2/(Z1 + Z2))+ Z3\n",
- "y = ZT.imag\n",
- "x = ZT.real\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n ZT is \",round(x,2),\" + (\",round(y,2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " ZT is 8.65 + ( -6.26 )i"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 418</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine in cartesian form correct to three decimal places:\n",
- "#(a)1/Z1 (b)1/Z2 (c) 1/Z1 * 1/Z2 (d) 1/(1/Z1 + 1/Z2)\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 3 + 4j;\n",
- "Z2 = 2 - 5j;\n",
- "\n",
- "#calculation:\n",
- "za = 1/Z1\n",
- "zb = 1/Z2\n",
- "zc = za + zb\n",
- "zd = 1/zc\n",
- "zax = za.real\n",
- "zay = za.imag\n",
- "zbx = zb.real\n",
- "zby = zb.imag\n",
- "zcx = zc.real\n",
- "zcy = zc.imag\n",
- "zdx = zd.real\n",
- "zdy = zd.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)1/Z1 is \",round( zax,2),\" + (\",round(zay,2),\")i\"\n",
- "print \"\\n (b)1/Z2 is \",round( zbx,2),\" + (\",round(zby,2),\")i\"\n",
- "print \"\\n (c)1/Z1 + 1/Z2 is \",round( zcx,2),\" + (\",round(zcy,2),\")i\"\n",
- "print \"\\n (d)1/(1/Z1 + 1/Z2) is \",round( zdx,2),\" + (\",round(zdy,2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)1/Z1 is 0.12 + ( -0.16 )i\n",
- "\n",
- " (b)1/Z2 is 0.07 + ( 0.17 )i\n",
- "\n",
- " (c)1/Z1 + 1/Z2 is 0.19 + ( 0.01 )i\n",
- "\n",
- " (d)1/(1/Z1 + 1/Z2) is 5.27 + ( -0.35 )i"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 419</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find a, b, x and y?\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 9 - 2j;\n",
- "Z2 = 2 + 1j;\n",
- "Z3 = -2 + 1j;\n",
- "Z4 = 5 + 2j;\n",
- "\n",
- "#calculation:\n",
- "za = Z1/3\n",
- "zb = Z2*Z3\n",
- "zca = (2*Z4.real + Z4.imag)/-1\n",
- "zcb = Z4.real - zca\n",
- "zaa = za.real\n",
- "zab = za.imag\n",
- "zbx = zb.real\n",
- "zby = zb.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)a and b are \", zaa,\" and \",round(zab,2),\" resp.\"\n",
- "print \"\\n (b)x and y are \", zbx,\" and \",zby,\" resp.\"\n",
- "print \"\\n (c)a and b are \", zca,\" and \",zcb,\" resp.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)a and b are 3.0 and -0.67 resp.\n",
- "\n",
- " (b)x and y are -5.0 and 0.0 resp.\n",
- "\n",
- " (c)a and b are -12.0 and 17.0 resp."
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 422</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Convert 5/_132\u00b0 into a + jb form correct to four significant figures.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r = 5;# magnitude\n",
- "theta = -132;# in degree\n",
- "\n",
- "#calculation:\n",
- "x = r*math.sin(theta*math.pi/180)\n",
- "y = r*math.cos(theta*math.pi/180)\n",
- "z = x + y*1j\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Z is \",round(x,2),\" + (\",round(y,2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Z is -3.72 + ( -3.35 )i"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 422</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine in polar form the total impedance ZT given that ZT = Z1Z2/\u0007(Z1 + Z2\b)\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r1 = 4.76;# magnitude\n",
- "theta1 = 35;# in degree\n",
- "r2 = 7.36;# magnitude\n",
- "theta2 = -48;# in degree\n",
- "\n",
- "#calculation:\n",
- "x1 = r1*cmath.cos(theta1*math.pi/180)\n",
- "y1 = r1*cmath.sin(theta1*math.pi/180)\n",
- "z1 = x1 + y1*1j\n",
- "x2 = r2*cmath.cos(theta2*math.pi/180)\n",
- "y2 = r2*cmath.sin(theta2*math.pi/180)\n",
- "z2 = x2 + y2*1j\n",
- "z3 = z1*z2/(z1 + z2)\n",
- "x3 = z3.real\n",
- "y3 = z3.imag\n",
- "r3 = (x3**2 + y3**2)**0.5\n",
- "theta3 = cmath.phase(complex(x3,y3))*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n ZT is (\",round( r3,2),\",round(/_\",round(theta3,2),\"deg)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " ZT is ( 3.79 ,round(/_ 4.25 deg)"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 423</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine \u0007\u0003(2 +j3\b)^5 in polar and in cartesian form.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "z = -2 + 3j;\n",
- "\n",
- "#calculation:\n",
- "zc = z**5\n",
- "x = zc.real\n",
- "y = zc.imag\n",
- "r = (x**2 + y**2)**0.5\n",
- "theta = cmath.phase(complex(x,y))*180/math.pi\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Z is \",round( x,2),\" + (\",round(y,2),\")i\"\n",
- "print \"\\n ZT is (\",round( r,2),\"round/_\",round(theta,2),\"deg)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Z is -122.0 + ( -597.0 )i\n",
- "\n",
- " ZT is ( 609.34 round/_ -101.55 deg)"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 423</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the two square roots of the complex number \u0007(12 + j5)\b in cartesian and polar form\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "z = 12 + 5j;\n",
- "\n",
- "#calculation:\n",
- "x = z.real\n",
- "y = z.imag\n",
- "r = (x**2 + y**2)**0.5\n",
- "theta1 = cmath.atan(y/x)*180/math.pi\n",
- "'''\n",
- "if ((x<0)&(y<0))\n",
- " theta1 = theta1 -180;\n",
- "elif ((x<0)&(y>0))\n",
- " theta1 = theta1 +180;\n",
- "'''\n",
- "theta2 = theta1 + 360\n",
- "rtheta1 = theta1/2\n",
- "rtheta2 = theta2/2\n",
- "'''\n",
- "if (rtheta2 > 180)\n",
- " rtheta2 = rtheta2 -360;\n",
- "elif ((x<0)&(y>0))\n",
- " rtheta2 = rtheta2 +360;\n",
- "'''\n",
- "rr = r**0.5\n",
- "x1 = rr*cmath.cos(rtheta1*math.pi/180)\n",
- "y1 = rr*cmath.sin(rtheta1*math.pi/180)\n",
- "z1 = x1 + y1*1j\n",
- "\n",
- "x2 = rr*cmath.cos(rtheta2*math.pi/180)\n",
- "y2 = rr*cmath.sin(rtheta2*math.pi/180)\n",
- "z2 = x2 + y2*1j\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n two roots are (\",round(z1.real,2),\" + (\",round(z1.imag,2),\")i) \"\n",
- "print \" and (\",round(z2.real,2),\" + (\",round(z2.imag,2),\")i)\"\n",
- "print \"\\n two roots are (\",round( rr,2),\"/_\",round((cmath.phase(complex(z1.real,z1.imag)))*180/math.pi,2),\"deg) \"\n",
- "print \" and (\",round( rr,2),\"/_\",round((cmath.phase(complex(z2.real,z2.imag)))*180/math.pi,2),\"deg)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " two roots are ( 3.54 + ( 0.71 )i) \n",
- " and ( -3.54 + ( -0.71 )i)\n",
- "\n",
- " two roots are ( 3.61 /_ 11.31 deg) \n",
- " and ( 3.61 /_ -168.69 deg)\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_23.ipynb index 7d18b803..7d18b803 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_23.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint.ipynb deleted file mode 100755 index e4305467..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint.ipynb +++ /dev/null @@ -1,843 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 24: Application of complex numbers to series a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 433</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of the resistance and the series-connected inductance or capacitance for each of the following impedances:\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "z1 = 12 + 5j;\n",
- "z2 = -40j;\n",
- "r3 = 30;\n",
- "theta3 = 60;# in degrees\n",
- "r4 = 2.20E6; \n",
- "theta4 = -30;# in degrees\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #for an R-L series circuit, impedance\n",
- " # Z = R + iXL\n",
- "Ra = z1.real\n",
- "XLa = z1.imag\n",
- "La = XLa/(2*math.pi*f)\n",
- " #for a purely capacitive circuit, impedance Z = -iXc\n",
- "Xcb = abs(z2.imag)\n",
- "Cb = 1/(2*math.pi*f*Xcb)\n",
- "z3 = r3*cmath.cos(theta3*math.pi/180) + (r3*cmath.sin(theta3*math.pi/180))*1j\n",
- "Rc = z3.real\n",
- "XLc = z3.imag\n",
- "Lc = XLc/(2*math.pi*f)\n",
- "z4 = r4*cmath.cos(theta4*math.pi/180) + (r4*cmath.sin(theta4*math.pi/180))*1j\n",
- "Rd = z4.real\n",
- "Xcd = abs(z4.imag)\n",
- "Cd = 1/(2*math.pi*f*Xcd)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)an impedance (12 + i5)ohm represents a resistance of \",round( Ra,2),\" ohm \"\n",
- "print \"in series with an inductance of \",round(La*1000,2),\"mH\"\n",
- "print \"\\n (b)an impedance -40i ohm represents a pure capacitor of capacitance \",round(Cb*1E6,2),\"uF\"\n",
- "print \"\\n (c)an impedance 30/_60deg ohm represents a resistance of \",round(Rc,2),\" ohm \"\n",
- "print \"in series with an inductance of \",round(Lc*1000,2),\"mH\"\n",
- "print \"\\n (d)an impedance 2.20 x 10^6 /_-30deg ohm represents a resistance of \",round(Rd/1000,2),\"kohm \"\n",
- "print \" in series with a capacitor of capacitance \",round(Cd*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)an impedance (12 + i5)ohm represents a resistance of 12.0 ohm \n",
- "in series with an inductance of 15.92 mH\n",
- "\n",
- " (b)an impedance -40i ohm represents a pure capacitor of capacitance 79.58 uF\n",
- "\n",
- " (c)an impedance 30/_60deg ohm represents a resistance of 15.0 ohm \n",
- "in series with an inductance of 82.7 mH\n",
- "\n",
- " (d)an impedance 2.20 x 10^6 /_-30deg ohm represents a resistance of 1905.26 kohm \n",
- " in series with a capacitor of capacitance 2.89 nF\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 434</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, in polar and rectangular forms, the current flowing in an inductor of negligible resistance and inductance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.1592 ;# in Henry\n",
- "V = 250;# in Volts\n",
- "f = 50;# in Hz\n",
- "R = 0;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #for an R\u00e2\u20ac\u201cL series circuit, impedance\n",
- " # Z = R + iXL\n",
- "XL = 2*math.pi*f*L\n",
- "Z = R + 1j*XL\n",
- "I = V/Z\n",
- "x = I.real\n",
- "y = I.imag\n",
- "r = (x**2 + y**2)**0.5\n",
- "if ((x==0)&(y<0)):\n",
- " theta = -90\n",
- "elif ((x==0)&(y>0)):\n",
- " theta = +90\n",
- "else:\n",
- " theta = cmath.phase(complex(x,y))*180/math.pi\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current is (\",round(r,2),\"/_\",theta,\"deg) A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current is ( 5.0 /_ -90 deg) A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 435</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the supply p.d.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 3E-6 ;# in farad\n",
- "f = 1000;# in Hz\n",
- "ri = 2.83;\n",
- "thetai = 90;# in degrees\n",
- "\n",
- "#calculation:\n",
- " #Capacitive reactance Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " # circuit impedance Z\n",
- "Z = -1*1j*Xc\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- "V = I*Z\n",
- "x = V.real\n",
- "y = V.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n supply p.d. is \",round(abs(V),0),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " supply p.d. is 150.0 V"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 435</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resistance, (b) the capacitance, \n",
- "#(c) the modulus of the impedance, and (d) the current flowing and its phase angle,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 240;# in Volts\n",
- "f = 50;# in Hz\n",
- "Z = 30 - 50j;\n",
- "\n",
- "#calculation:\n",
- " #Since impedance Z = 30 - i50,\n",
- " #resistance\n",
- "R = Z.real\n",
- " #capacitive reactance\n",
- "Xc = abs(Z.imag)\n",
- " #capacitance\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- " #modulus of impedance\n",
- "modZ = (R**2 + Xc**2)**0.5\n",
- "I = V/Z\n",
- "x = I.real\n",
- "y = I.imag\n",
- "r = (x**2 + y**2)**0.5\n",
- "theta = cmath.phase(complex(x,y))*180/math.pi\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)resistance is \",round( R,2),\" ohm\"\n",
- "print \"\\n (b)capacitance is \",round(C*1E6,2),\"uFarad\"\n",
- "print \"\\n (c)modulus of impedance is \",round(modZ,2),\" ohm\"\n",
- "print \"\\n (d)current flowing and its phase angle is (\",round( r,2),\"/_\",round( theta,2),\"deg) A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)resistance is 30.0 ohm\n",
- "\n",
- " (b)capacitance is 63.66 uFarad\n",
- "\n",
- " (c)modulus of impedance is 58.31 ohm\n",
- "\n",
- " (d)current flowing and its phase angle is ( 4.12 /_ 59.04 deg) A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 436</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the impedance of the circuit,\n",
- "#(b) the current and circuit phase angle, \n",
- "#(c) the p.d. across the 32 ohm resistor, and (d) the p.d. across the coil\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 200;# in Volts\n",
- "f = 50;# in Hz\n",
- "R = 32;# in ohms\n",
- "L = 0.15;# in Henry\n",
- "\n",
- "#calculation:\n",
- " #Inductive reactance XL\n",
- "XL = 2*math.pi*f*L\n",
- " #impedance, Z\n",
- "Z = R + 1j*XL\n",
- " #Current I\n",
- "I = V/Z\n",
- "xi = I.real\n",
- "yi = I.imag\n",
- "ri = (xi**2 + yi**2)**0.5\n",
- "if ((xi==0)&(yi<0)):\n",
- " thetai = -90\n",
- "elif ((xi==0)&(yi>0)):\n",
- " thetai = +90\n",
- "else:\n",
- " thetai = cmath.phase(complex(xi,yi))*180/math.pi\n",
- "\n",
- " #P.d. across the resistor\n",
- "VR = I*R\n",
- "xr = VR.real\n",
- "yr = VR.imag\n",
- "rr = (xr**2 + yr**2)**0.5\n",
- "thetar = cmath.phase(complex(xr,yr))*180/math.pi\n",
- " #P.d. across the coil, VL\n",
- "VL = I*1j*XL\n",
- "xl = VL.real\n",
- "yl = VL.imag\n",
- "rl = (xl**2 + yl**2)**0.5\n",
- "thetal = cmath.phase(complex(xl,yl))*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)impedance is \",round(Z.real,2),\" + \",round( Z.imag,2),\")i ohm\"\n",
- "print \"\\n (b)current flowing and its phase angle is lagging the voltage = (\",round( ri,2),\"/_\",round( thetai,2),\"deg) A\"\n",
- "print \"\\n (c)P.d. across the resistor is (\",round(rr,2),\"/_\",round(thetar,2),\"deg) V\"\n",
- "print \"\\n (d)P.d. across the coil, VL is (\",round(rl,2),\"/_\",round(thetal,2),\"deg) V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)impedance is 32.0 + 47.12 )i ohm\n",
- "\n",
- " (b)current flowing and its phase angle is lagging the voltage = ( 3.51 /_ -55.82 deg) A\n",
- "\n",
- " (c)P.d. across the resistor is ( 112.36 /_ -55.82 deg) V\n",
- "\n",
- " (d)P.d. across the coil, VL is ( 165.46 /_ 34.18 deg) V"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 436</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of impedance\n",
- "#determine the value of the components forming the series circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 120 + 200j;# in Volts\n",
- "f = 5E6;# in Hz\n",
- "I = 7 + 16j;# in amperes\n",
- "\n",
- "#calculation:\n",
- " #impedance, Z\n",
- "Z = V/I\n",
- "R = Z.real\n",
- "X = Z.imag \n",
- "if ((R>0)&(X<0)):\n",
- " C = -1/(2*math.pi*f*X)\n",
- "#Results\n",
- " print \"\\n\\n Result \\n\\n\"\n",
- " print \"\\n The series circuit thus consists of a resistor of resistance \",round(R,2),\" ohm \"\n",
- " print \"and a capacitor of capacitive reactance\", round(X*-1,3),\"ohm and capacitance is\",round(C*1E9,2),\" nFarad\\n\"\n",
- "elif ((R>0)&(X>0)):\n",
- " L = 2*math.pi*f*X\n",
- "#Results\n",
- " print \"\\n\\n Result \\n\\n\"\n",
- " print \"\\n The series circuit thus consists of a resistor of resistance \",round(R,2),\" ohm \"\n",
- " print \" and a inductor of insuctance \",round(L*100,2),\" mHenry\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " The series circuit thus consists of a resistor of resistance 13.25 ohm \n",
- "and a capacitor of capacitive reactance 1.705 ohm and capacitance is 18.67 nFarad\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 437</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of impedance Z2.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 70;# in volts\n",
- "thetav = 30;# in degrees\n",
- "ri = 3.5;# in amperes\n",
- "thetai = -20;# in degrees\n",
- " #z1 consist of two resistance\n",
- "R1 = 4.36;# in ohms\n",
- "R2 = -2.1j;# in ohms\n",
- "\n",
- " #calculation:\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #impedance, Z\n",
- "Z = V/I\n",
- " #Total impedance Z = z1 + z2\n",
- "Z1 = R1 + R2\n",
- "Z2 = Z - Z1\n",
- "x = Z2.real\n",
- "y = Z2.imag \n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n impedance Z2 is \",round(x,2),\" + (\",round(y,2),\")i ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " impedance Z2 is 8.5 + ( 17.42 )i ohm"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 437</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the supply voltage, (b) the voltage across the 90 \u000e resistance, \n",
- "#(c) the voltage across the inductance, and (d) the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 90;# in ohms\n",
- "XL = 150;# in ohms\n",
- "ri = 1.35;# in amperes\n",
- "thetai = 0;# in degrees\n",
- "\n",
- "#calculation:\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Circuit impedance Z\n",
- "Z = R + 1j*XL\n",
- " #Supply voltage, V\n",
- "V = I*Z\n",
- " #Voltage across 90 ohm\u000e resistor\n",
- "VR = V.real\n",
- "#Voltage across inductance, VL\n",
- "VL = V.imag\n",
- "xv = V.real\n",
- "yv = V.imag\n",
- "rv = (xv**2 + yv**2)**0.5\n",
- "thetav = cmath.phase(complex(xv,yv))*180/math.pi\n",
- "phi = thetav - thetai\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Supply voltage, V is \",xv,\" + (\",yv,\")i V\\n\"\n",
- "print \"\\n (b)Voltage across 90 ohm resistor, VR is \",VR,\" V\\n\"\n",
- "print \"\\n (c)Voltage across inductance, VL is \",VL,\" V\\n\"\n",
- "print \"\\n (d)Circuit phase angle is \",round(phi,2),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Supply voltage, V is 121.5 + ( 202.5 )i V\n",
- "\n",
- "\n",
- " (b)Voltage across 90 ohm resistor, VR is 121.5 V\n",
- "\n",
- "\n",
- " (c)Voltage across inductance, VL is 202.5 V\n",
- "\n",
- "\n",
- " (d)Circuit phase angle is 59.04 deg lagging"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 438</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the rms value of voltage (in polar form),\n",
- "#(b) the circuit impedance, (c) the rms current flowing, and\n",
- "#(d) the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 25;# in ohms\n",
- "L = 0.02;# in henry\n",
- "Vm = 282.8;# in volts\n",
- "w = 628.4;# in rad/sec\n",
- "phiv = math.pi/3;# phase angle\n",
- "\n",
- "#calculation:\n",
- " #rms voltage\n",
- "Vrms = 0.707*Vm*math.cos(phiv) + 0.707*Vm*math.sin(phiv)*1j\n",
- " #frequency\n",
- "f = w/(2*math.pi)\n",
- " #Inductive reactance XL\n",
- "XL = 2*math.pi*f*L\n",
- " #Circuit impedance Z\n",
- "Z = R + XL*1j\n",
- " #Rms current\n",
- "Irms = Vrms/Z\n",
- "phii = cmath.phase(complex(Irms.real, Irms.imag))*180/math.pi\n",
- "phi = phiv*180/math.pi - phii\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the rms value of voltage is \",round(Vrms.real,2),\" + (\",round( Vrms.imag,2),\")i V\\n\"\n",
- "print \"\\n (b)the circuit impedance is \",round(R,2),\" + (\",round( XL,2),\")i ohm\\n\"\n",
- "print \"\\n (c)the rms current flowing is \",round(Irms.real,2),\" + (\",round( Irms.imag,2),\")i A\\n\"\n",
- "print \"\\n (d)Circuit phase angle is \",round(phi,2),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the rms value of voltage is 99.97 + ( 173.15 )i V\n",
- "\n",
- "\n",
- " (b)the circuit impedance is 25.0 + ( 12.57 )i ohm\n",
- "\n",
- "\n",
- " (c)the rms current flowing is 5.97 + ( 3.92 )i A\n",
- "\n",
- "\n",
- " (d)Circuit phase angle is 26.69 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 438</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 12;# in ohms\n",
- "L = 0.10;# in henry\n",
- "C = 120E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "V = 240;# in volts\n",
- "\n",
- "#calculation:\n",
- " #Inductive reactance, XL\n",
- "XL = 2*math.pi*f*L\n",
- " #Capacitive reactance, Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #Circuit impedance Z\n",
- "Z = R + 1j*(XL - Xc)\n",
- "I = V/Z\n",
- "phii = cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
- "phiv = 0# in degrees\n",
- "phi = phiv - phii\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing is \",round(abs(I),1),\"/_\",round(cmath.phase(complex(I.real,I.imag))*180/math.pi,1),\"deg A\\n\"\n",
- "print \"and Circuit phase angle is \",round(phi,1),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing is 18.5 /_ -22.2 deg A\n",
- "\n",
- "and Circuit phase angle is 22.2 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 439</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the values of R and L, Determine also the voltage across the coil and the voltage across the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 50E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "V = 225;# in volts\n",
- "ri = 1.5;# in Amperes\n",
- "thetai = -30;# in degrees\n",
- "\n",
- "#calculation:\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Capacitive reactance, Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #Circuit impedance Z\n",
- "Z = V/I\n",
- "R = Z.real\n",
- "XL = Z.imag + Xc\n",
- " #inductance L\n",
- "L = XL/(2*math.pi*f)\n",
- " #Voltage across coil\n",
- "Zcoil = R + 1j*XL\n",
- "Vcoil = I*Zcoil\n",
- " #Voltage across capacitor,\n",
- "Vc = I*(-1j*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)resistance is \",round(R,2),\" ohm and inductance is \",round( L,3),\" H\\n\"\n",
- "print \"\\n (b)voltage across the coil is \",round(Vcoil.real,2),\" + (\",round( Vcoil.imag,2),\")i V\\n\"\n",
- "print \"\\n (c)voltage across the capacitor is \",round(Vc.real,2),\" + (\",round( Vc.imag,2),\")i V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)resistance is 129.9 ohm and inductance is 0.441 H\n",
- "\n",
- "\n",
- " (b)voltage across the coil is 272.75 + ( 82.7 )i V\n",
- "\n",
- "\n",
- " (c)voltage across the capacitor is -47.75 + ( -82.7 )i V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 440</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine also the value of the supply voltage V and the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 2.653E-6;# in Farads\n",
- "R1 = 8;# in ohms\n",
- "R2 = 5;# in ohms\n",
- "L = 0.477E-3;# in Henry\n",
- "f = 4000;# in Hz\n",
- "ri = 6;# in Amperes\n",
- "thetai = 0;# in degrees\n",
- "\n",
- "#calculation:\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Capacitive reactance, Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #impedance Z1\n",
- "Z1 = R1 - 1j*Xc\n",
- " #inductive reactance XL\n",
- "XL = 2*math.pi*f*L\n",
- " #impedance Z2,\n",
- "Z2 = R2 + 1j*XL\n",
- " #voltage V1\n",
- "V1 = I*Z1\n",
- " #voltage V2\n",
- "V2 = I*Z2\n",
- " #Supply voltage, V\n",
- "V = V1 + V2\n",
- "phiv = cmath.phase(complex(V.real, V.imag))*180/math.pi\n",
- "phi = phiv - thetai\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n supply voltage is \",round(V.real,2),\" + (\",round( V.imag,2),\")i V\\n\"\n",
- "print \"and Circuit phase angle is \",round(phi,2),\"deg leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " supply voltage is 78.0 + ( -18.06 )i V\n",
- "\n",
- "and Circuit phase angle is -13.03 deg leading"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_1.ipynb deleted file mode 100755 index e4305467..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_1.ipynb +++ /dev/null @@ -1,843 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 24: Application of complex numbers to series a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 433</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of the resistance and the series-connected inductance or capacitance for each of the following impedances:\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "z1 = 12 + 5j;\n",
- "z2 = -40j;\n",
- "r3 = 30;\n",
- "theta3 = 60;# in degrees\n",
- "r4 = 2.20E6; \n",
- "theta4 = -30;# in degrees\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #for an R-L series circuit, impedance\n",
- " # Z = R + iXL\n",
- "Ra = z1.real\n",
- "XLa = z1.imag\n",
- "La = XLa/(2*math.pi*f)\n",
- " #for a purely capacitive circuit, impedance Z = -iXc\n",
- "Xcb = abs(z2.imag)\n",
- "Cb = 1/(2*math.pi*f*Xcb)\n",
- "z3 = r3*cmath.cos(theta3*math.pi/180) + (r3*cmath.sin(theta3*math.pi/180))*1j\n",
- "Rc = z3.real\n",
- "XLc = z3.imag\n",
- "Lc = XLc/(2*math.pi*f)\n",
- "z4 = r4*cmath.cos(theta4*math.pi/180) + (r4*cmath.sin(theta4*math.pi/180))*1j\n",
- "Rd = z4.real\n",
- "Xcd = abs(z4.imag)\n",
- "Cd = 1/(2*math.pi*f*Xcd)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)an impedance (12 + i5)ohm represents a resistance of \",round( Ra,2),\" ohm \"\n",
- "print \"in series with an inductance of \",round(La*1000,2),\"mH\"\n",
- "print \"\\n (b)an impedance -40i ohm represents a pure capacitor of capacitance \",round(Cb*1E6,2),\"uF\"\n",
- "print \"\\n (c)an impedance 30/_60deg ohm represents a resistance of \",round(Rc,2),\" ohm \"\n",
- "print \"in series with an inductance of \",round(Lc*1000,2),\"mH\"\n",
- "print \"\\n (d)an impedance 2.20 x 10^6 /_-30deg ohm represents a resistance of \",round(Rd/1000,2),\"kohm \"\n",
- "print \" in series with a capacitor of capacitance \",round(Cd*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)an impedance (12 + i5)ohm represents a resistance of 12.0 ohm \n",
- "in series with an inductance of 15.92 mH\n",
- "\n",
- " (b)an impedance -40i ohm represents a pure capacitor of capacitance 79.58 uF\n",
- "\n",
- " (c)an impedance 30/_60deg ohm represents a resistance of 15.0 ohm \n",
- "in series with an inductance of 82.7 mH\n",
- "\n",
- " (d)an impedance 2.20 x 10^6 /_-30deg ohm represents a resistance of 1905.26 kohm \n",
- " in series with a capacitor of capacitance 2.89 nF\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 434</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, in polar and rectangular forms, the current flowing in an inductor of negligible resistance and inductance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.1592 ;# in Henry\n",
- "V = 250;# in Volts\n",
- "f = 50;# in Hz\n",
- "R = 0;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #for an R\u00e2\u20ac\u201cL series circuit, impedance\n",
- " # Z = R + iXL\n",
- "XL = 2*math.pi*f*L\n",
- "Z = R + 1j*XL\n",
- "I = V/Z\n",
- "x = I.real\n",
- "y = I.imag\n",
- "r = (x**2 + y**2)**0.5\n",
- "if ((x==0)&(y<0)):\n",
- " theta = -90\n",
- "elif ((x==0)&(y>0)):\n",
- " theta = +90\n",
- "else:\n",
- " theta = cmath.phase(complex(x,y))*180/math.pi\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current is (\",round(r,2),\"/_\",theta,\"deg) A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current is ( 5.0 /_ -90 deg) A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 435</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the supply p.d.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 3E-6 ;# in farad\n",
- "f = 1000;# in Hz\n",
- "ri = 2.83;\n",
- "thetai = 90;# in degrees\n",
- "\n",
- "#calculation:\n",
- " #Capacitive reactance Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " # circuit impedance Z\n",
- "Z = -1*1j*Xc\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- "V = I*Z\n",
- "x = V.real\n",
- "y = V.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n supply p.d. is \",round(abs(V),0),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " supply p.d. is 150.0 V"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 435</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resistance, (b) the capacitance, \n",
- "#(c) the modulus of the impedance, and (d) the current flowing and its phase angle,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 240;# in Volts\n",
- "f = 50;# in Hz\n",
- "Z = 30 - 50j;\n",
- "\n",
- "#calculation:\n",
- " #Since impedance Z = 30 - i50,\n",
- " #resistance\n",
- "R = Z.real\n",
- " #capacitive reactance\n",
- "Xc = abs(Z.imag)\n",
- " #capacitance\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- " #modulus of impedance\n",
- "modZ = (R**2 + Xc**2)**0.5\n",
- "I = V/Z\n",
- "x = I.real\n",
- "y = I.imag\n",
- "r = (x**2 + y**2)**0.5\n",
- "theta = cmath.phase(complex(x,y))*180/math.pi\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)resistance is \",round( R,2),\" ohm\"\n",
- "print \"\\n (b)capacitance is \",round(C*1E6,2),\"uFarad\"\n",
- "print \"\\n (c)modulus of impedance is \",round(modZ,2),\" ohm\"\n",
- "print \"\\n (d)current flowing and its phase angle is (\",round( r,2),\"/_\",round( theta,2),\"deg) A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)resistance is 30.0 ohm\n",
- "\n",
- " (b)capacitance is 63.66 uFarad\n",
- "\n",
- " (c)modulus of impedance is 58.31 ohm\n",
- "\n",
- " (d)current flowing and its phase angle is ( 4.12 /_ 59.04 deg) A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 436</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the impedance of the circuit,\n",
- "#(b) the current and circuit phase angle, \n",
- "#(c) the p.d. across the 32 ohm resistor, and (d) the p.d. across the coil\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 200;# in Volts\n",
- "f = 50;# in Hz\n",
- "R = 32;# in ohms\n",
- "L = 0.15;# in Henry\n",
- "\n",
- "#calculation:\n",
- " #Inductive reactance XL\n",
- "XL = 2*math.pi*f*L\n",
- " #impedance, Z\n",
- "Z = R + 1j*XL\n",
- " #Current I\n",
- "I = V/Z\n",
- "xi = I.real\n",
- "yi = I.imag\n",
- "ri = (xi**2 + yi**2)**0.5\n",
- "if ((xi==0)&(yi<0)):\n",
- " thetai = -90\n",
- "elif ((xi==0)&(yi>0)):\n",
- " thetai = +90\n",
- "else:\n",
- " thetai = cmath.phase(complex(xi,yi))*180/math.pi\n",
- "\n",
- " #P.d. across the resistor\n",
- "VR = I*R\n",
- "xr = VR.real\n",
- "yr = VR.imag\n",
- "rr = (xr**2 + yr**2)**0.5\n",
- "thetar = cmath.phase(complex(xr,yr))*180/math.pi\n",
- " #P.d. across the coil, VL\n",
- "VL = I*1j*XL\n",
- "xl = VL.real\n",
- "yl = VL.imag\n",
- "rl = (xl**2 + yl**2)**0.5\n",
- "thetal = cmath.phase(complex(xl,yl))*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)impedance is \",round(Z.real,2),\" + \",round( Z.imag,2),\")i ohm\"\n",
- "print \"\\n (b)current flowing and its phase angle is lagging the voltage = (\",round( ri,2),\"/_\",round( thetai,2),\"deg) A\"\n",
- "print \"\\n (c)P.d. across the resistor is (\",round(rr,2),\"/_\",round(thetar,2),\"deg) V\"\n",
- "print \"\\n (d)P.d. across the coil, VL is (\",round(rl,2),\"/_\",round(thetal,2),\"deg) V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)impedance is 32.0 + 47.12 )i ohm\n",
- "\n",
- " (b)current flowing and its phase angle is lagging the voltage = ( 3.51 /_ -55.82 deg) A\n",
- "\n",
- " (c)P.d. across the resistor is ( 112.36 /_ -55.82 deg) V\n",
- "\n",
- " (d)P.d. across the coil, VL is ( 165.46 /_ 34.18 deg) V"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 436</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of impedance\n",
- "#determine the value of the components forming the series circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 120 + 200j;# in Volts\n",
- "f = 5E6;# in Hz\n",
- "I = 7 + 16j;# in amperes\n",
- "\n",
- "#calculation:\n",
- " #impedance, Z\n",
- "Z = V/I\n",
- "R = Z.real\n",
- "X = Z.imag \n",
- "if ((R>0)&(X<0)):\n",
- " C = -1/(2*math.pi*f*X)\n",
- "#Results\n",
- " print \"\\n\\n Result \\n\\n\"\n",
- " print \"\\n The series circuit thus consists of a resistor of resistance \",round(R,2),\" ohm \"\n",
- " print \"and a capacitor of capacitive reactance\", round(X*-1,3),\"ohm and capacitance is\",round(C*1E9,2),\" nFarad\\n\"\n",
- "elif ((R>0)&(X>0)):\n",
- " L = 2*math.pi*f*X\n",
- "#Results\n",
- " print \"\\n\\n Result \\n\\n\"\n",
- " print \"\\n The series circuit thus consists of a resistor of resistance \",round(R,2),\" ohm \"\n",
- " print \" and a inductor of insuctance \",round(L*100,2),\" mHenry\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " The series circuit thus consists of a resistor of resistance 13.25 ohm \n",
- "and a capacitor of capacitive reactance 1.705 ohm and capacitance is 18.67 nFarad\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 437</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of impedance Z2.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 70;# in volts\n",
- "thetav = 30;# in degrees\n",
- "ri = 3.5;# in amperes\n",
- "thetai = -20;# in degrees\n",
- " #z1 consist of two resistance\n",
- "R1 = 4.36;# in ohms\n",
- "R2 = -2.1j;# in ohms\n",
- "\n",
- " #calculation:\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #impedance, Z\n",
- "Z = V/I\n",
- " #Total impedance Z = z1 + z2\n",
- "Z1 = R1 + R2\n",
- "Z2 = Z - Z1\n",
- "x = Z2.real\n",
- "y = Z2.imag \n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n impedance Z2 is \",round(x,2),\" + (\",round(y,2),\")i ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " impedance Z2 is 8.5 + ( 17.42 )i ohm"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 437</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the supply voltage, (b) the voltage across the 90 \u000e resistance, \n",
- "#(c) the voltage across the inductance, and (d) the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 90;# in ohms\n",
- "XL = 150;# in ohms\n",
- "ri = 1.35;# in amperes\n",
- "thetai = 0;# in degrees\n",
- "\n",
- "#calculation:\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Circuit impedance Z\n",
- "Z = R + 1j*XL\n",
- " #Supply voltage, V\n",
- "V = I*Z\n",
- " #Voltage across 90 ohm\u000e resistor\n",
- "VR = V.real\n",
- "#Voltage across inductance, VL\n",
- "VL = V.imag\n",
- "xv = V.real\n",
- "yv = V.imag\n",
- "rv = (xv**2 + yv**2)**0.5\n",
- "thetav = cmath.phase(complex(xv,yv))*180/math.pi\n",
- "phi = thetav - thetai\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Supply voltage, V is \",xv,\" + (\",yv,\")i V\\n\"\n",
- "print \"\\n (b)Voltage across 90 ohm resistor, VR is \",VR,\" V\\n\"\n",
- "print \"\\n (c)Voltage across inductance, VL is \",VL,\" V\\n\"\n",
- "print \"\\n (d)Circuit phase angle is \",round(phi,2),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Supply voltage, V is 121.5 + ( 202.5 )i V\n",
- "\n",
- "\n",
- " (b)Voltage across 90 ohm resistor, VR is 121.5 V\n",
- "\n",
- "\n",
- " (c)Voltage across inductance, VL is 202.5 V\n",
- "\n",
- "\n",
- " (d)Circuit phase angle is 59.04 deg lagging"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 438</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the rms value of voltage (in polar form),\n",
- "#(b) the circuit impedance, (c) the rms current flowing, and\n",
- "#(d) the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 25;# in ohms\n",
- "L = 0.02;# in henry\n",
- "Vm = 282.8;# in volts\n",
- "w = 628.4;# in rad/sec\n",
- "phiv = math.pi/3;# phase angle\n",
- "\n",
- "#calculation:\n",
- " #rms voltage\n",
- "Vrms = 0.707*Vm*math.cos(phiv) + 0.707*Vm*math.sin(phiv)*1j\n",
- " #frequency\n",
- "f = w/(2*math.pi)\n",
- " #Inductive reactance XL\n",
- "XL = 2*math.pi*f*L\n",
- " #Circuit impedance Z\n",
- "Z = R + XL*1j\n",
- " #Rms current\n",
- "Irms = Vrms/Z\n",
- "phii = cmath.phase(complex(Irms.real, Irms.imag))*180/math.pi\n",
- "phi = phiv*180/math.pi - phii\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the rms value of voltage is \",round(Vrms.real,2),\" + (\",round( Vrms.imag,2),\")i V\\n\"\n",
- "print \"\\n (b)the circuit impedance is \",round(R,2),\" + (\",round( XL,2),\")i ohm\\n\"\n",
- "print \"\\n (c)the rms current flowing is \",round(Irms.real,2),\" + (\",round( Irms.imag,2),\")i A\\n\"\n",
- "print \"\\n (d)Circuit phase angle is \",round(phi,2),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the rms value of voltage is 99.97 + ( 173.15 )i V\n",
- "\n",
- "\n",
- " (b)the circuit impedance is 25.0 + ( 12.57 )i ohm\n",
- "\n",
- "\n",
- " (c)the rms current flowing is 5.97 + ( 3.92 )i A\n",
- "\n",
- "\n",
- " (d)Circuit phase angle is 26.69 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 438</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 12;# in ohms\n",
- "L = 0.10;# in henry\n",
- "C = 120E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "V = 240;# in volts\n",
- "\n",
- "#calculation:\n",
- " #Inductive reactance, XL\n",
- "XL = 2*math.pi*f*L\n",
- " #Capacitive reactance, Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #Circuit impedance Z\n",
- "Z = R + 1j*(XL - Xc)\n",
- "I = V/Z\n",
- "phii = cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
- "phiv = 0# in degrees\n",
- "phi = phiv - phii\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing is \",round(abs(I),1),\"/_\",round(cmath.phase(complex(I.real,I.imag))*180/math.pi,1),\"deg A\\n\"\n",
- "print \"and Circuit phase angle is \",round(phi,1),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing is 18.5 /_ -22.2 deg A\n",
- "\n",
- "and Circuit phase angle is 22.2 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 439</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the values of R and L, Determine also the voltage across the coil and the voltage across the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 50E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "V = 225;# in volts\n",
- "ri = 1.5;# in Amperes\n",
- "thetai = -30;# in degrees\n",
- "\n",
- "#calculation:\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Capacitive reactance, Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #Circuit impedance Z\n",
- "Z = V/I\n",
- "R = Z.real\n",
- "XL = Z.imag + Xc\n",
- " #inductance L\n",
- "L = XL/(2*math.pi*f)\n",
- " #Voltage across coil\n",
- "Zcoil = R + 1j*XL\n",
- "Vcoil = I*Zcoil\n",
- " #Voltage across capacitor,\n",
- "Vc = I*(-1j*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)resistance is \",round(R,2),\" ohm and inductance is \",round( L,3),\" H\\n\"\n",
- "print \"\\n (b)voltage across the coil is \",round(Vcoil.real,2),\" + (\",round( Vcoil.imag,2),\")i V\\n\"\n",
- "print \"\\n (c)voltage across the capacitor is \",round(Vc.real,2),\" + (\",round( Vc.imag,2),\")i V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)resistance is 129.9 ohm and inductance is 0.441 H\n",
- "\n",
- "\n",
- " (b)voltage across the coil is 272.75 + ( 82.7 )i V\n",
- "\n",
- "\n",
- " (c)voltage across the capacitor is -47.75 + ( -82.7 )i V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 440</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine also the value of the supply voltage V and the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 2.653E-6;# in Farads\n",
- "R1 = 8;# in ohms\n",
- "R2 = 5;# in ohms\n",
- "L = 0.477E-3;# in Henry\n",
- "f = 4000;# in Hz\n",
- "ri = 6;# in Amperes\n",
- "thetai = 0;# in degrees\n",
- "\n",
- "#calculation:\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Capacitive reactance, Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #impedance Z1\n",
- "Z1 = R1 - 1j*Xc\n",
- " #inductive reactance XL\n",
- "XL = 2*math.pi*f*L\n",
- " #impedance Z2,\n",
- "Z2 = R2 + 1j*XL\n",
- " #voltage V1\n",
- "V1 = I*Z1\n",
- " #voltage V2\n",
- "V2 = I*Z2\n",
- " #Supply voltage, V\n",
- "V = V1 + V2\n",
- "phiv = cmath.phase(complex(V.real, V.imag))*180/math.pi\n",
- "phi = phiv - thetai\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n supply voltage is \",round(V.real,2),\" + (\",round( V.imag,2),\")i V\\n\"\n",
- "print \"and Circuit phase angle is \",round(phi,2),\"deg leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " supply voltage is 78.0 + ( -18.06 )i V\n",
- "\n",
- "and Circuit phase angle is -13.03 deg leading"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_2.ipynb deleted file mode 100755 index e4305467..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_2.ipynb +++ /dev/null @@ -1,843 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 24: Application of complex numbers to series a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 433</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of the resistance and the series-connected inductance or capacitance for each of the following impedances:\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "z1 = 12 + 5j;\n",
- "z2 = -40j;\n",
- "r3 = 30;\n",
- "theta3 = 60;# in degrees\n",
- "r4 = 2.20E6; \n",
- "theta4 = -30;# in degrees\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #for an R-L series circuit, impedance\n",
- " # Z = R + iXL\n",
- "Ra = z1.real\n",
- "XLa = z1.imag\n",
- "La = XLa/(2*math.pi*f)\n",
- " #for a purely capacitive circuit, impedance Z = -iXc\n",
- "Xcb = abs(z2.imag)\n",
- "Cb = 1/(2*math.pi*f*Xcb)\n",
- "z3 = r3*cmath.cos(theta3*math.pi/180) + (r3*cmath.sin(theta3*math.pi/180))*1j\n",
- "Rc = z3.real\n",
- "XLc = z3.imag\n",
- "Lc = XLc/(2*math.pi*f)\n",
- "z4 = r4*cmath.cos(theta4*math.pi/180) + (r4*cmath.sin(theta4*math.pi/180))*1j\n",
- "Rd = z4.real\n",
- "Xcd = abs(z4.imag)\n",
- "Cd = 1/(2*math.pi*f*Xcd)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)an impedance (12 + i5)ohm represents a resistance of \",round( Ra,2),\" ohm \"\n",
- "print \"in series with an inductance of \",round(La*1000,2),\"mH\"\n",
- "print \"\\n (b)an impedance -40i ohm represents a pure capacitor of capacitance \",round(Cb*1E6,2),\"uF\"\n",
- "print \"\\n (c)an impedance 30/_60deg ohm represents a resistance of \",round(Rc,2),\" ohm \"\n",
- "print \"in series with an inductance of \",round(Lc*1000,2),\"mH\"\n",
- "print \"\\n (d)an impedance 2.20 x 10^6 /_-30deg ohm represents a resistance of \",round(Rd/1000,2),\"kohm \"\n",
- "print \" in series with a capacitor of capacitance \",round(Cd*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)an impedance (12 + i5)ohm represents a resistance of 12.0 ohm \n",
- "in series with an inductance of 15.92 mH\n",
- "\n",
- " (b)an impedance -40i ohm represents a pure capacitor of capacitance 79.58 uF\n",
- "\n",
- " (c)an impedance 30/_60deg ohm represents a resistance of 15.0 ohm \n",
- "in series with an inductance of 82.7 mH\n",
- "\n",
- " (d)an impedance 2.20 x 10^6 /_-30deg ohm represents a resistance of 1905.26 kohm \n",
- " in series with a capacitor of capacitance 2.89 nF\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 434</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, in polar and rectangular forms, the current flowing in an inductor of negligible resistance and inductance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.1592 ;# in Henry\n",
- "V = 250;# in Volts\n",
- "f = 50;# in Hz\n",
- "R = 0;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #for an R\u00e2\u20ac\u201cL series circuit, impedance\n",
- " # Z = R + iXL\n",
- "XL = 2*math.pi*f*L\n",
- "Z = R + 1j*XL\n",
- "I = V/Z\n",
- "x = I.real\n",
- "y = I.imag\n",
- "r = (x**2 + y**2)**0.5\n",
- "if ((x==0)&(y<0)):\n",
- " theta = -90\n",
- "elif ((x==0)&(y>0)):\n",
- " theta = +90\n",
- "else:\n",
- " theta = cmath.phase(complex(x,y))*180/math.pi\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current is (\",round(r,2),\"/_\",theta,\"deg) A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current is ( 5.0 /_ -90 deg) A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 435</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the supply p.d.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 3E-6 ;# in farad\n",
- "f = 1000;# in Hz\n",
- "ri = 2.83;\n",
- "thetai = 90;# in degrees\n",
- "\n",
- "#calculation:\n",
- " #Capacitive reactance Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " # circuit impedance Z\n",
- "Z = -1*1j*Xc\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- "V = I*Z\n",
- "x = V.real\n",
- "y = V.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n supply p.d. is \",round(abs(V),0),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " supply p.d. is 150.0 V"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 435</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resistance, (b) the capacitance, \n",
- "#(c) the modulus of the impedance, and (d) the current flowing and its phase angle,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 240;# in Volts\n",
- "f = 50;# in Hz\n",
- "Z = 30 - 50j;\n",
- "\n",
- "#calculation:\n",
- " #Since impedance Z = 30 - i50,\n",
- " #resistance\n",
- "R = Z.real\n",
- " #capacitive reactance\n",
- "Xc = abs(Z.imag)\n",
- " #capacitance\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- " #modulus of impedance\n",
- "modZ = (R**2 + Xc**2)**0.5\n",
- "I = V/Z\n",
- "x = I.real\n",
- "y = I.imag\n",
- "r = (x**2 + y**2)**0.5\n",
- "theta = cmath.phase(complex(x,y))*180/math.pi\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)resistance is \",round( R,2),\" ohm\"\n",
- "print \"\\n (b)capacitance is \",round(C*1E6,2),\"uFarad\"\n",
- "print \"\\n (c)modulus of impedance is \",round(modZ,2),\" ohm\"\n",
- "print \"\\n (d)current flowing and its phase angle is (\",round( r,2),\"/_\",round( theta,2),\"deg) A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)resistance is 30.0 ohm\n",
- "\n",
- " (b)capacitance is 63.66 uFarad\n",
- "\n",
- " (c)modulus of impedance is 58.31 ohm\n",
- "\n",
- " (d)current flowing and its phase angle is ( 4.12 /_ 59.04 deg) A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 436</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the impedance of the circuit,\n",
- "#(b) the current and circuit phase angle, \n",
- "#(c) the p.d. across the 32 ohm resistor, and (d) the p.d. across the coil\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 200;# in Volts\n",
- "f = 50;# in Hz\n",
- "R = 32;# in ohms\n",
- "L = 0.15;# in Henry\n",
- "\n",
- "#calculation:\n",
- " #Inductive reactance XL\n",
- "XL = 2*math.pi*f*L\n",
- " #impedance, Z\n",
- "Z = R + 1j*XL\n",
- " #Current I\n",
- "I = V/Z\n",
- "xi = I.real\n",
- "yi = I.imag\n",
- "ri = (xi**2 + yi**2)**0.5\n",
- "if ((xi==0)&(yi<0)):\n",
- " thetai = -90\n",
- "elif ((xi==0)&(yi>0)):\n",
- " thetai = +90\n",
- "else:\n",
- " thetai = cmath.phase(complex(xi,yi))*180/math.pi\n",
- "\n",
- " #P.d. across the resistor\n",
- "VR = I*R\n",
- "xr = VR.real\n",
- "yr = VR.imag\n",
- "rr = (xr**2 + yr**2)**0.5\n",
- "thetar = cmath.phase(complex(xr,yr))*180/math.pi\n",
- " #P.d. across the coil, VL\n",
- "VL = I*1j*XL\n",
- "xl = VL.real\n",
- "yl = VL.imag\n",
- "rl = (xl**2 + yl**2)**0.5\n",
- "thetal = cmath.phase(complex(xl,yl))*180/math.pi\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)impedance is \",round(Z.real,2),\" + \",round( Z.imag,2),\")i ohm\"\n",
- "print \"\\n (b)current flowing and its phase angle is lagging the voltage = (\",round( ri,2),\"/_\",round( thetai,2),\"deg) A\"\n",
- "print \"\\n (c)P.d. across the resistor is (\",round(rr,2),\"/_\",round(thetar,2),\"deg) V\"\n",
- "print \"\\n (d)P.d. across the coil, VL is (\",round(rl,2),\"/_\",round(thetal,2),\"deg) V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)impedance is 32.0 + 47.12 )i ohm\n",
- "\n",
- " (b)current flowing and its phase angle is lagging the voltage = ( 3.51 /_ -55.82 deg) A\n",
- "\n",
- " (c)P.d. across the resistor is ( 112.36 /_ -55.82 deg) V\n",
- "\n",
- " (d)P.d. across the coil, VL is ( 165.46 /_ 34.18 deg) V"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 436</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of impedance\n",
- "#determine the value of the components forming the series circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 120 + 200j;# in Volts\n",
- "f = 5E6;# in Hz\n",
- "I = 7 + 16j;# in amperes\n",
- "\n",
- "#calculation:\n",
- " #impedance, Z\n",
- "Z = V/I\n",
- "R = Z.real\n",
- "X = Z.imag \n",
- "if ((R>0)&(X<0)):\n",
- " C = -1/(2*math.pi*f*X)\n",
- "#Results\n",
- " print \"\\n\\n Result \\n\\n\"\n",
- " print \"\\n The series circuit thus consists of a resistor of resistance \",round(R,2),\" ohm \"\n",
- " print \"and a capacitor of capacitive reactance\", round(X*-1,3),\"ohm and capacitance is\",round(C*1E9,2),\" nFarad\\n\"\n",
- "elif ((R>0)&(X>0)):\n",
- " L = 2*math.pi*f*X\n",
- "#Results\n",
- " print \"\\n\\n Result \\n\\n\"\n",
- " print \"\\n The series circuit thus consists of a resistor of resistance \",round(R,2),\" ohm \"\n",
- " print \" and a inductor of insuctance \",round(L*100,2),\" mHenry\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " The series circuit thus consists of a resistor of resistance 13.25 ohm \n",
- "and a capacitor of capacitive reactance 1.705 ohm and capacitance is 18.67 nFarad\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 437</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of impedance Z2.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 70;# in volts\n",
- "thetav = 30;# in degrees\n",
- "ri = 3.5;# in amperes\n",
- "thetai = -20;# in degrees\n",
- " #z1 consist of two resistance\n",
- "R1 = 4.36;# in ohms\n",
- "R2 = -2.1j;# in ohms\n",
- "\n",
- " #calculation:\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #impedance, Z\n",
- "Z = V/I\n",
- " #Total impedance Z = z1 + z2\n",
- "Z1 = R1 + R2\n",
- "Z2 = Z - Z1\n",
- "x = Z2.real\n",
- "y = Z2.imag \n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n impedance Z2 is \",round(x,2),\" + (\",round(y,2),\")i ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " impedance Z2 is 8.5 + ( 17.42 )i ohm"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 437</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the supply voltage, (b) the voltage across the 90 \u000e resistance, \n",
- "#(c) the voltage across the inductance, and (d) the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 90;# in ohms\n",
- "XL = 150;# in ohms\n",
- "ri = 1.35;# in amperes\n",
- "thetai = 0;# in degrees\n",
- "\n",
- "#calculation:\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Circuit impedance Z\n",
- "Z = R + 1j*XL\n",
- " #Supply voltage, V\n",
- "V = I*Z\n",
- " #Voltage across 90 ohm\u000e resistor\n",
- "VR = V.real\n",
- "#Voltage across inductance, VL\n",
- "VL = V.imag\n",
- "xv = V.real\n",
- "yv = V.imag\n",
- "rv = (xv**2 + yv**2)**0.5\n",
- "thetav = cmath.phase(complex(xv,yv))*180/math.pi\n",
- "phi = thetav - thetai\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Supply voltage, V is \",xv,\" + (\",yv,\")i V\\n\"\n",
- "print \"\\n (b)Voltage across 90 ohm resistor, VR is \",VR,\" V\\n\"\n",
- "print \"\\n (c)Voltage across inductance, VL is \",VL,\" V\\n\"\n",
- "print \"\\n (d)Circuit phase angle is \",round(phi,2),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Supply voltage, V is 121.5 + ( 202.5 )i V\n",
- "\n",
- "\n",
- " (b)Voltage across 90 ohm resistor, VR is 121.5 V\n",
- "\n",
- "\n",
- " (c)Voltage across inductance, VL is 202.5 V\n",
- "\n",
- "\n",
- " (d)Circuit phase angle is 59.04 deg lagging"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 438</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the rms value of voltage (in polar form),\n",
- "#(b) the circuit impedance, (c) the rms current flowing, and\n",
- "#(d) the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 25;# in ohms\n",
- "L = 0.02;# in henry\n",
- "Vm = 282.8;# in volts\n",
- "w = 628.4;# in rad/sec\n",
- "phiv = math.pi/3;# phase angle\n",
- "\n",
- "#calculation:\n",
- " #rms voltage\n",
- "Vrms = 0.707*Vm*math.cos(phiv) + 0.707*Vm*math.sin(phiv)*1j\n",
- " #frequency\n",
- "f = w/(2*math.pi)\n",
- " #Inductive reactance XL\n",
- "XL = 2*math.pi*f*L\n",
- " #Circuit impedance Z\n",
- "Z = R + XL*1j\n",
- " #Rms current\n",
- "Irms = Vrms/Z\n",
- "phii = cmath.phase(complex(Irms.real, Irms.imag))*180/math.pi\n",
- "phi = phiv*180/math.pi - phii\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the rms value of voltage is \",round(Vrms.real,2),\" + (\",round( Vrms.imag,2),\")i V\\n\"\n",
- "print \"\\n (b)the circuit impedance is \",round(R,2),\" + (\",round( XL,2),\")i ohm\\n\"\n",
- "print \"\\n (c)the rms current flowing is \",round(Irms.real,2),\" + (\",round( Irms.imag,2),\")i A\\n\"\n",
- "print \"\\n (d)Circuit phase angle is \",round(phi,2),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the rms value of voltage is 99.97 + ( 173.15 )i V\n",
- "\n",
- "\n",
- " (b)the circuit impedance is 25.0 + ( 12.57 )i ohm\n",
- "\n",
- "\n",
- " (c)the rms current flowing is 5.97 + ( 3.92 )i A\n",
- "\n",
- "\n",
- " (d)Circuit phase angle is 26.69 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 438</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 12;# in ohms\n",
- "L = 0.10;# in henry\n",
- "C = 120E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "V = 240;# in volts\n",
- "\n",
- "#calculation:\n",
- " #Inductive reactance, XL\n",
- "XL = 2*math.pi*f*L\n",
- " #Capacitive reactance, Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #Circuit impedance Z\n",
- "Z = R + 1j*(XL - Xc)\n",
- "I = V/Z\n",
- "phii = cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
- "phiv = 0# in degrees\n",
- "phi = phiv - phii\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing is \",round(abs(I),1),\"/_\",round(cmath.phase(complex(I.real,I.imag))*180/math.pi,1),\"deg A\\n\"\n",
- "print \"and Circuit phase angle is \",round(phi,1),\"deg lagging\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing is 18.5 /_ -22.2 deg A\n",
- "\n",
- "and Circuit phase angle is 22.2 deg lagging\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 439</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the values of R and L, Determine also the voltage across the coil and the voltage across the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 50E-6;# in Farads\n",
- "f = 50;# in Hz\n",
- "V = 225;# in volts\n",
- "ri = 1.5;# in Amperes\n",
- "thetai = -30;# in degrees\n",
- "\n",
- "#calculation:\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Capacitive reactance, Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #Circuit impedance Z\n",
- "Z = V/I\n",
- "R = Z.real\n",
- "XL = Z.imag + Xc\n",
- " #inductance L\n",
- "L = XL/(2*math.pi*f)\n",
- " #Voltage across coil\n",
- "Zcoil = R + 1j*XL\n",
- "Vcoil = I*Zcoil\n",
- " #Voltage across capacitor,\n",
- "Vc = I*(-1j*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)resistance is \",round(R,2),\" ohm and inductance is \",round( L,3),\" H\\n\"\n",
- "print \"\\n (b)voltage across the coil is \",round(Vcoil.real,2),\" + (\",round( Vcoil.imag,2),\")i V\\n\"\n",
- "print \"\\n (c)voltage across the capacitor is \",round(Vc.real,2),\" + (\",round( Vc.imag,2),\")i V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)resistance is 129.9 ohm and inductance is 0.441 H\n",
- "\n",
- "\n",
- " (b)voltage across the coil is 272.75 + ( 82.7 )i V\n",
- "\n",
- "\n",
- " (c)voltage across the capacitor is -47.75 + ( -82.7 )i V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 440</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine also the value of the supply voltage V and the circuit phase angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 2.653E-6;# in Farads\n",
- "R1 = 8;# in ohms\n",
- "R2 = 5;# in ohms\n",
- "L = 0.477E-3;# in Henry\n",
- "f = 4000;# in Hz\n",
- "ri = 6;# in Amperes\n",
- "thetai = 0;# in degrees\n",
- "\n",
- "#calculation:\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Capacitive reactance, Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #impedance Z1\n",
- "Z1 = R1 - 1j*Xc\n",
- " #inductive reactance XL\n",
- "XL = 2*math.pi*f*L\n",
- " #impedance Z2,\n",
- "Z2 = R2 + 1j*XL\n",
- " #voltage V1\n",
- "V1 = I*Z1\n",
- " #voltage V2\n",
- "V2 = I*Z2\n",
- " #Supply voltage, V\n",
- "V = V1 + V2\n",
- "phiv = cmath.phase(complex(V.real, V.imag))*180/math.pi\n",
- "phi = phiv - thetai\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n supply voltage is \",round(V.real,2),\" + (\",round( V.imag,2),\")i V\\n\"\n",
- "print \"and Circuit phase angle is \",round(phi,2),\"deg leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " supply voltage is 78.0 + ( -18.06 )i V\n",
- "\n",
- "and Circuit phase angle is -13.03 deg leading"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_24.ipynb index 8160e132..8160e132 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_24.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint.ipynb deleted file mode 100755 index 19e462e5..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint.ipynb +++ /dev/null @@ -1,647 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 25: Application of complex numbers to parallel a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 446</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the admittance, conductance and susceptance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 0 - 5j;# in ohms\n",
- "Z2 = 25 + 40j;# in ohms\n",
- "Z3 = 3 - 2j;# in ohms\n",
- "r4 = 50;# in ohms\n",
- "theta4 = 40;# in degrees\n",
- "\n",
- "#calculation:\n",
- " #admittance Y\n",
- "Y1 = 1/Z1\n",
- " #conductance, G\n",
- "G1 = Y1.real\n",
- " #Suspectance, Bc\n",
- "Bc1 = abs(Y1.imag)\n",
- " #admittance Y\n",
- "Y2 = 1/Z2\n",
- " #conductance, G\n",
- "G2 = Y2.real\n",
- " #Suspectance, Bc\n",
- "Bc2 = abs(Y2.imag)\n",
- " #admittance Y\n",
- "Y3 = 1/Z3\n",
- " #conductance, G\n",
- "G3 = Y3.real\n",
- " #Suspectance, Bc\n",
- "Bc3 = abs(Y3.imag)\n",
- "Z4 = r4*math.cos(theta4*math.pi/180) + 1j*r4*math.sin(theta4*math.pi/180)\n",
- " #admittance Y\n",
- "Y4 = 1/Z4\n",
- " #conductance, G\n",
- "G4 = Y4.real\n",
- " #Suspectance, Bc\n",
- "Bc4 = abs(Y4.imag)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)admittance Y is (\",round(Y1.real,2),\" + (\",round(Y1.imag,2),\")i) S, \"\n",
- "print \" conductance, G is \",round(G1,2),\" S, susceptance,Bc is \",round(Bc1,2),\" S\\n\"\n",
- "print \"\\n (b)admittance Y is (\",round(Y2.real,2),\" + (\",round(Y2.imag,2),\")i) S, \"\n",
- "print \" conductance, G is \",round(G2,2),\" S, susceptance,Bc is \",round(Bc2,2),\" S\\n\"\n",
- "print \"\\n (c)admittance Y is (\",round(Y3.real,2),\" + (\",round(Y3.imag,2),\")i) S, \"\n",
- "print \" conductance, G is \",round(G3,2),\" S, susceptance,Bc is \",round(Bc3,2),\" S\\n\"\n",
- "print \"\\n (d)admittance Y is (\",round(Y4.real,2),\" + (\",round(Y4.imag,2),\")i) S, \"\n",
- "print \" conductance, G is \",round(G4,2),\" S, susceptance,Bc is \",round(Bc4,2),\" S\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)admittance Y is ( -0.0 + ( 0.2 )i) S, \n",
- " conductance, G is -0.0 S, susceptance,Bc is 0.2 S\n",
- "\n",
- "\n",
- " (b)admittance Y is ( 0.01 + ( -0.02 )i) S, \n",
- " conductance, G is 0.01 S, susceptance,Bc is 0.02 S\n",
- "\n",
- "\n",
- " (c)admittance Y is ( 0.23 + ( 0.15 )i) S, \n",
- " conductance, G is 0.23 S, susceptance,Bc is 0.15 S\n",
- "\n",
- "\n",
- " (d)admittance Y is ( 0.02 + ( -0.01 )i) S, \n",
- " conductance, G is 0.02 S, susceptance,Bc is 0.01 S\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 447</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine expressions for the impedance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Y2 = 0.001 - 0.002j;# in S\n",
- "Y3 = 0.05 + 0.08j;# in S\n",
- "r1 = 0.004;# in S\n",
- "theta1 = 30;# in degrees\n",
- "\n",
- " #calculation:\n",
- " #impedance, Z\n",
- "Z2 = 1/Y2\n",
- "Z3 = 1/Y3\n",
- "Y1 = r1*math.cos(theta1*math.pi/180) + 1j*r1*math.sin(theta1*math.pi/180)\n",
- "Z1 = 1/Y1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Impedance,Z is (\",round(Z1.real,2),\" + (\",round( Z1.imag,2),\")i) ohm\\n\"\n",
- "print \"\\n (b)Impedance,Z is (\",round(Z2.real,2),\" + (\",round( Z2.imag,2),\")i) ohm\\n\"\n",
- "print \"\\n (c)Impedance,Z is (\",round(Z3.real,2),\" + (\",round( Z3.imag,2),\")i) ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Impedance,Z is ( 216.51 + ( -125.0 )i) ohm\n",
- "\n",
- "\n",
- " (b)Impedance,Z is ( 200.0 + ( 400.0 )i) ohm\n",
- "\n",
- "\n",
- " (c)Impedance,Z is ( 5.62 + ( -8.99 )i) ohm"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 448</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of the resistance and the capacitive reactance of the circuit if they are connected \n",
- "#(a) in parallel, (b) in series.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Y = 0.040 - 1j*0.025;# in S\n",
- "\n",
- "#calculation:\n",
- " #impedance, Z\n",
- "Z = 1/Y\n",
- " #conductance, G\n",
- "G = Y.real\n",
- " #Suspectance, Bc\n",
- "Bc = abs(Y.imag)\n",
- " #parallrl \n",
- " #resistance, R\n",
- "Rp = 1/G\n",
- " #capacitive reactance\n",
- "Xcp = 1/Bc\n",
- " #series\n",
- " #resistance, R\n",
- "Rs = Z.real\n",
- " #capacitive reactance\n",
- "Xcs = abs(Z.imag)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)for parallel, resistance,R is \",round(Rp,2),\" ohm and capacitive reactance, Xc is \",round(Xcp,2),\" ohm\\n\"\n",
- "print \"\\n (b)forseries, resistance,R is \",round(Rs,2),\" ohm and capacitive reactance, Xc is \",round(Xcs,2),\" ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)for parallel, resistance,R is 25.0 ohm and capacitive reactance, Xc is 40.0 ohm\n",
- "\n",
- "\n",
- " (b)forseries, resistance,R is 17.98 ohm and capacitive reactance, Xc is 11.24 ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 449</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of currents I, I1 and I2.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 8;# in ohm\n",
- "R = 5;# in ohm\n",
- "R2 = 6;# ohm\n",
- "rv = 50;# in volts\n",
- "thetav = 0;# in degrees\n",
- "\n",
- "#calculation:\n",
- " #voltage,V\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #circuit impedance, ZT\n",
- "ZT = R + (R1*1j*R2/(R1 + 1j*R2))\n",
- " #Current I\n",
- "I = V/ZT\n",
- " #current,I1\n",
- "I1 = I*(1j*R2/(R1 + 1j*R2))\n",
- " #current, I2\n",
- "I2 = I*(R1/(R1 + 1j*R2))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg A,\"\n",
- "print \"current,I1 = \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real, I1.imag))*180/math.pi,2),\"deg A, \"\n",
- "print \"current, I2 = \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real, I2.imag))*180/math.pi,2),\"deg A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I = 5.7 /_ -25.98 deg A,\n",
- "current,I1 = 3.42 /_ 27.15 deg A, \n",
- "current, I2 = 4.56 /_ -62.85 deg A\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 450</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of supply current I and its phase relative to the 40 V supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 5;# in ohm\n",
- "R2 = 3;# in ohm \n",
- "R3 = 8;# ohm\n",
- "Xc = 4;# in ohms\n",
- "XL = 12;# in Ohms\n",
- "V = 40;# in volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "Z1 = R1 + 1j*XL\n",
- "Z2 = R2 - 1j*Xc\n",
- "Z3 = R3\n",
- " #circuit admittance, YT = 1/ZT\n",
- "YT = (1/Z1) + (1/Z2) + (1/Z3)\n",
- " #Current I\n",
- "I = V*YT\n",
- "I1 = V/Z1\n",
- "I2 = V/Z2\n",
- "I3 = V/Z2\n",
- "thetav = 0\n",
- "thetai = cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
- "phi = thetav - thetai \n",
- "if (phi>0):\n",
- " a = \"lagging\"\n",
- "else:\n",
- " a = \"leading\"\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I is (\",round(I.real,2),\" + (\",round(I.imag,2),\")i) A,\"\n",
- "print \"and its phase relative to the 40 V supply is \",a,\"s by \",round(abs(phi),2),\"deg\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I is ( 10.98 + ( 3.56 )i) A,\n",
- "and its phase relative to the 40 V supply is leading s by 17.96 deg\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 451</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the total equivalent circuit impedance, (b) the supply current, \n",
- "#(c) the circuit phase angle, (d) the current in the coil, and (e) the current in the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.07958;# in Henry\n",
- "R = 18;# in ohm\n",
- "C = 64.96E-6;# in Farad\n",
- "rv = 250;# in volts\n",
- "thetav = 0;# in degrees\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #Inductive reactance\n",
- "XL = 2*math.pi*f*L\n",
- " #capacitive reactance\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #impedance of the coil,\n",
- "Zcoil = R + 1j*XL\n",
- " #impedance presented by the capacitor,\n",
- "Zc = -1j*Xc\n",
- " #Total equivalent circuit impedance,\n",
- "ZT = Zcoil*Zc/(Zcoil + Zc)\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #current, I\n",
- "I = V/ZT\n",
- "thetai = cmath.phase(complex(I.real,I.imag))*180/math.pi\n",
- "phi = thetav - thetai\n",
- "if (phi>0):\n",
- " a = \"lagging\"\n",
- "else:\n",
- " a = \"leading\"\n",
- "\n",
- " #Current in the coil, ICOIL\n",
- "Icoil = V/Zcoil\n",
- " #Current in the capacitor, IC\n",
- "Ic = V/Zc\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the circuit impedance is \",round(ZT.real,2),\" + (\",round( ZT.imag,2),\")i ohm\\n\"\n",
- "print \"\\n (b)supply current, I = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg A\\n\"\n",
- "print \"\\n (c)circuit phase relative is \",a,\"s by \",round(abs(phi),2),\"deg\\n\"\n",
- "print \"\\n (d)current in coil, Icoil = \",round(abs(Icoil),2),\"/_\",round(cmath.phase(complex(Icoil.real, Icoil.imag))*180/math.pi,2),\"deg A\\n\"\n",
- "print \"\\n (e)current in capacitor, Ic = \",round(abs(Ic),2),\"/_\",round(cmath.phase(complex(Ic.real, Ic.imag))*180/math.pi,2),\"deg A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the circuit impedance is 48.02 + ( 15.03 )i ohm\n",
- "\n",
- "\n",
- " (b)supply current, I = 4.97 /_ -17.38 deg A\n",
- "\n",
- "\n",
- " (c)circuit phase relative is lagging s by 17.38 deg\n",
- "\n",
- "\n",
- " (d)current in coil, Icoil = 8.12 /_ -54.25 deg A\n",
- "\n",
- "\n",
- " (e)current in capacitor, Ic = 5.1 /_ 90.0 deg A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 452</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)determine the value of impedance Z1 \n",
- "#(b) If the supply frequency is 5 kHz,determine the value of the components comprising impedance Z1\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL = 6j;# in ohm\n",
- "R2 = 8;# in ohm\n",
- "Z3 = 10;# in ohm\n",
- "rv = 50;# in volts\n",
- "thetav = 30;# in degrees\n",
- "ri = 31.4;# in amperes\n",
- "thetai = 52.48;# in degrees\n",
- "f = 5000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #impedance, Z2\n",
- "Z2 = R2 + RL\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #current, I\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Total circuit admittance,\n",
- "YT = I/V\n",
- " #admittance, Y3\n",
- "Y3 = 1/Z3\n",
- " #admittance, Y2\n",
- "Y2 = 1/Z2\n",
- " #admittance, Y1\n",
- "Y1 = YT - Y2 - Y3\n",
- " #impedance, Z1\n",
- "Z1 = 1/Y1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the impedance Z1 is \",round(Z1.real,2),\" + (\",round( Z1.imag,2),\")i ohm\\n\"\n",
- "\n",
- " #resistance, R1\n",
- "R1 = Z1.real\n",
- "X1 = Z1.imag \n",
- "if ((R1>0)&(X1<0)):\n",
- " C1 = -1/(2*math.pi*f*X1)\n",
- " print \"\\n (b)The series circuit thus consists of a resistor of resistance \",round(R1,2),\" ohm\"\n",
- " print \" and a capacitor of capacitance \",round(C1*1E6,2),\"uFarad\\n\"\n",
- "elif ((R1>0)&(X1>0)):\n",
- " L1 = 2*math.pi*f*X1\n",
- " print \"\\n (b)The series circuit thus consists of a resistor of resistance \",round(R1,2),\" ohm \"\n",
- " print \" and a inductor of insuctance \",round(L1*1000,2),\"mHenry\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the impedance Z1 is 1.6 + ( -1.2 )i ohm\n",
- "\n",
- "\n",
- " (b)The series circuit thus consists of a resistor of resistance 1.6 ohm\n",
- " and a capacitor of capacitance 26.55 uFarad\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 453</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the equivalent series circuit impedance,\n",
- "#(b) the supply current I, (c) the circuit phase angle,\n",
- "#(d) the values of voltages V1 and V2, and (e) the values of currents IA and IB\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL1 = 1.02j;# in ohm\n",
- "R1 = 1.65;# in ohm\n",
- "RLa = 7j;# in ohm\n",
- "Ra = 5;# in ohm\n",
- "Rcb = -1j*15;# in ohm\n",
- "Rb = 4;# in ohm\n",
- "rv = 91;# in volts\n",
- "thetav = 0;# in degree\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #impedance, Z1\n",
- "Z1 = R1 + RL1\n",
- " #impedance, Za\n",
- "Za = Ra + RLa\n",
- " #impedance, Zb\n",
- "Zb = Rb + Rcb\n",
- " #impedance, Z, of the two branches connected in parallel\n",
- "Z = Za*Zb/(Za + Zb)\n",
- " #Total circuit impedance\n",
- "ZT = Z1 + Z\n",
- " #Supply current, I\n",
- "I = V/ZT\n",
- "thetai = cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
- "phi = thetav - thetai \n",
- "if (phi>0):\n",
- " a = \"lagging\"\n",
- "else:\n",
- " a = \"leading\"\n",
- "\n",
- " #Voltage V1\n",
- "V1 = I*Z1\n",
- " #Voltage V2\n",
- "V2 = I*Z\n",
- " #current Ia\n",
- "Ia = V2/Za\n",
- " #Current Ib\n",
- "Ib = V2/Zb\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)equivalent series circuit impedance is \",round(ZT.real,2),\" + (\",round( ZT.imag,2),\")i ohm\\n\"\n",
- "print \"\\n (b)supply current, I is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\\n\"\n",
- "print \"\\n (c)circuit phase relative is \",a,\" by \",round(abs(phi),2),\"deg\\n\"\n",
- "print \"\\n (d)voltage, V1 is (\",round(V1.real,2),\" + (\",round(V1.imag,2),\")i) V and V2 is(\",round(V2.real,2),\" + (\",round( V2.imag,2),\")i) V\\n\"\n",
- "print \"\\n (e)current, Ia is (\",round(Ia.real,2),\" + (\",round( Ia.imag,2),\")i) A and Ib is(\",round(Ib.real,2),\" + (\",round( Ib.imag,2),\")i) A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)equivalent series circuit impedance is 12.0 + ( 5.0 )i ohm\n",
- "\n",
- "\n",
- " (b)supply current, I is 6.46 + ( -2.69 )i A\n",
- "\n",
- "\n",
- " (c)circuit phase relative is lagging by 22.61 deg\n",
- "\n",
- "\n",
- " (d)voltage, V1 is ( 13.41 + ( 2.15 )i) V and V2 is( 77.59 + ( -2.15 )i) V\n",
- "\n",
- "\n",
- " (e)current, Ia is ( 5.04 + ( -7.49 )i) A and Ib is( 1.42 + ( 4.79 )i) A"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_1.ipynb deleted file mode 100755 index 19e462e5..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_1.ipynb +++ /dev/null @@ -1,647 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 25: Application of complex numbers to parallel a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 446</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the admittance, conductance and susceptance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 0 - 5j;# in ohms\n",
- "Z2 = 25 + 40j;# in ohms\n",
- "Z3 = 3 - 2j;# in ohms\n",
- "r4 = 50;# in ohms\n",
- "theta4 = 40;# in degrees\n",
- "\n",
- "#calculation:\n",
- " #admittance Y\n",
- "Y1 = 1/Z1\n",
- " #conductance, G\n",
- "G1 = Y1.real\n",
- " #Suspectance, Bc\n",
- "Bc1 = abs(Y1.imag)\n",
- " #admittance Y\n",
- "Y2 = 1/Z2\n",
- " #conductance, G\n",
- "G2 = Y2.real\n",
- " #Suspectance, Bc\n",
- "Bc2 = abs(Y2.imag)\n",
- " #admittance Y\n",
- "Y3 = 1/Z3\n",
- " #conductance, G\n",
- "G3 = Y3.real\n",
- " #Suspectance, Bc\n",
- "Bc3 = abs(Y3.imag)\n",
- "Z4 = r4*math.cos(theta4*math.pi/180) + 1j*r4*math.sin(theta4*math.pi/180)\n",
- " #admittance Y\n",
- "Y4 = 1/Z4\n",
- " #conductance, G\n",
- "G4 = Y4.real\n",
- " #Suspectance, Bc\n",
- "Bc4 = abs(Y4.imag)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)admittance Y is (\",round(Y1.real,2),\" + (\",round(Y1.imag,2),\")i) S, \"\n",
- "print \" conductance, G is \",round(G1,2),\" S, susceptance,Bc is \",round(Bc1,2),\" S\\n\"\n",
- "print \"\\n (b)admittance Y is (\",round(Y2.real,2),\" + (\",round(Y2.imag,2),\")i) S, \"\n",
- "print \" conductance, G is \",round(G2,2),\" S, susceptance,Bc is \",round(Bc2,2),\" S\\n\"\n",
- "print \"\\n (c)admittance Y is (\",round(Y3.real,2),\" + (\",round(Y3.imag,2),\")i) S, \"\n",
- "print \" conductance, G is \",round(G3,2),\" S, susceptance,Bc is \",round(Bc3,2),\" S\\n\"\n",
- "print \"\\n (d)admittance Y is (\",round(Y4.real,2),\" + (\",round(Y4.imag,2),\")i) S, \"\n",
- "print \" conductance, G is \",round(G4,2),\" S, susceptance,Bc is \",round(Bc4,2),\" S\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)admittance Y is ( -0.0 + ( 0.2 )i) S, \n",
- " conductance, G is -0.0 S, susceptance,Bc is 0.2 S\n",
- "\n",
- "\n",
- " (b)admittance Y is ( 0.01 + ( -0.02 )i) S, \n",
- " conductance, G is 0.01 S, susceptance,Bc is 0.02 S\n",
- "\n",
- "\n",
- " (c)admittance Y is ( 0.23 + ( 0.15 )i) S, \n",
- " conductance, G is 0.23 S, susceptance,Bc is 0.15 S\n",
- "\n",
- "\n",
- " (d)admittance Y is ( 0.02 + ( -0.01 )i) S, \n",
- " conductance, G is 0.02 S, susceptance,Bc is 0.01 S\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 447</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine expressions for the impedance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Y2 = 0.001 - 0.002j;# in S\n",
- "Y3 = 0.05 + 0.08j;# in S\n",
- "r1 = 0.004;# in S\n",
- "theta1 = 30;# in degrees\n",
- "\n",
- " #calculation:\n",
- " #impedance, Z\n",
- "Z2 = 1/Y2\n",
- "Z3 = 1/Y3\n",
- "Y1 = r1*math.cos(theta1*math.pi/180) + 1j*r1*math.sin(theta1*math.pi/180)\n",
- "Z1 = 1/Y1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Impedance,Z is (\",round(Z1.real,2),\" + (\",round( Z1.imag,2),\")i) ohm\\n\"\n",
- "print \"\\n (b)Impedance,Z is (\",round(Z2.real,2),\" + (\",round( Z2.imag,2),\")i) ohm\\n\"\n",
- "print \"\\n (c)Impedance,Z is (\",round(Z3.real,2),\" + (\",round( Z3.imag,2),\")i) ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Impedance,Z is ( 216.51 + ( -125.0 )i) ohm\n",
- "\n",
- "\n",
- " (b)Impedance,Z is ( 200.0 + ( 400.0 )i) ohm\n",
- "\n",
- "\n",
- " (c)Impedance,Z is ( 5.62 + ( -8.99 )i) ohm"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 448</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of the resistance and the capacitive reactance of the circuit if they are connected \n",
- "#(a) in parallel, (b) in series.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Y = 0.040 - 1j*0.025;# in S\n",
- "\n",
- "#calculation:\n",
- " #impedance, Z\n",
- "Z = 1/Y\n",
- " #conductance, G\n",
- "G = Y.real\n",
- " #Suspectance, Bc\n",
- "Bc = abs(Y.imag)\n",
- " #parallrl \n",
- " #resistance, R\n",
- "Rp = 1/G\n",
- " #capacitive reactance\n",
- "Xcp = 1/Bc\n",
- " #series\n",
- " #resistance, R\n",
- "Rs = Z.real\n",
- " #capacitive reactance\n",
- "Xcs = abs(Z.imag)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)for parallel, resistance,R is \",round(Rp,2),\" ohm and capacitive reactance, Xc is \",round(Xcp,2),\" ohm\\n\"\n",
- "print \"\\n (b)forseries, resistance,R is \",round(Rs,2),\" ohm and capacitive reactance, Xc is \",round(Xcs,2),\" ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)for parallel, resistance,R is 25.0 ohm and capacitive reactance, Xc is 40.0 ohm\n",
- "\n",
- "\n",
- " (b)forseries, resistance,R is 17.98 ohm and capacitive reactance, Xc is 11.24 ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 449</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of currents I, I1 and I2.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 8;# in ohm\n",
- "R = 5;# in ohm\n",
- "R2 = 6;# ohm\n",
- "rv = 50;# in volts\n",
- "thetav = 0;# in degrees\n",
- "\n",
- "#calculation:\n",
- " #voltage,V\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #circuit impedance, ZT\n",
- "ZT = R + (R1*1j*R2/(R1 + 1j*R2))\n",
- " #Current I\n",
- "I = V/ZT\n",
- " #current,I1\n",
- "I1 = I*(1j*R2/(R1 + 1j*R2))\n",
- " #current, I2\n",
- "I2 = I*(R1/(R1 + 1j*R2))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg A,\"\n",
- "print \"current,I1 = \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real, I1.imag))*180/math.pi,2),\"deg A, \"\n",
- "print \"current, I2 = \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real, I2.imag))*180/math.pi,2),\"deg A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I = 5.7 /_ -25.98 deg A,\n",
- "current,I1 = 3.42 /_ 27.15 deg A, \n",
- "current, I2 = 4.56 /_ -62.85 deg A\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 450</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of supply current I and its phase relative to the 40 V supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 5;# in ohm\n",
- "R2 = 3;# in ohm \n",
- "R3 = 8;# ohm\n",
- "Xc = 4;# in ohms\n",
- "XL = 12;# in Ohms\n",
- "V = 40;# in volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "Z1 = R1 + 1j*XL\n",
- "Z2 = R2 - 1j*Xc\n",
- "Z3 = R3\n",
- " #circuit admittance, YT = 1/ZT\n",
- "YT = (1/Z1) + (1/Z2) + (1/Z3)\n",
- " #Current I\n",
- "I = V*YT\n",
- "I1 = V/Z1\n",
- "I2 = V/Z2\n",
- "I3 = V/Z2\n",
- "thetav = 0\n",
- "thetai = cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
- "phi = thetav - thetai \n",
- "if (phi>0):\n",
- " a = \"lagging\"\n",
- "else:\n",
- " a = \"leading\"\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I is (\",round(I.real,2),\" + (\",round(I.imag,2),\")i) A,\"\n",
- "print \"and its phase relative to the 40 V supply is \",a,\"s by \",round(abs(phi),2),\"deg\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I is ( 10.98 + ( 3.56 )i) A,\n",
- "and its phase relative to the 40 V supply is leading s by 17.96 deg\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 451</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the total equivalent circuit impedance, (b) the supply current, \n",
- "#(c) the circuit phase angle, (d) the current in the coil, and (e) the current in the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.07958;# in Henry\n",
- "R = 18;# in ohm\n",
- "C = 64.96E-6;# in Farad\n",
- "rv = 250;# in volts\n",
- "thetav = 0;# in degrees\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #Inductive reactance\n",
- "XL = 2*math.pi*f*L\n",
- " #capacitive reactance\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #impedance of the coil,\n",
- "Zcoil = R + 1j*XL\n",
- " #impedance presented by the capacitor,\n",
- "Zc = -1j*Xc\n",
- " #Total equivalent circuit impedance,\n",
- "ZT = Zcoil*Zc/(Zcoil + Zc)\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #current, I\n",
- "I = V/ZT\n",
- "thetai = cmath.phase(complex(I.real,I.imag))*180/math.pi\n",
- "phi = thetav - thetai\n",
- "if (phi>0):\n",
- " a = \"lagging\"\n",
- "else:\n",
- " a = \"leading\"\n",
- "\n",
- " #Current in the coil, ICOIL\n",
- "Icoil = V/Zcoil\n",
- " #Current in the capacitor, IC\n",
- "Ic = V/Zc\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the circuit impedance is \",round(ZT.real,2),\" + (\",round( ZT.imag,2),\")i ohm\\n\"\n",
- "print \"\\n (b)supply current, I = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg A\\n\"\n",
- "print \"\\n (c)circuit phase relative is \",a,\"s by \",round(abs(phi),2),\"deg\\n\"\n",
- "print \"\\n (d)current in coil, Icoil = \",round(abs(Icoil),2),\"/_\",round(cmath.phase(complex(Icoil.real, Icoil.imag))*180/math.pi,2),\"deg A\\n\"\n",
- "print \"\\n (e)current in capacitor, Ic = \",round(abs(Ic),2),\"/_\",round(cmath.phase(complex(Ic.real, Ic.imag))*180/math.pi,2),\"deg A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the circuit impedance is 48.02 + ( 15.03 )i ohm\n",
- "\n",
- "\n",
- " (b)supply current, I = 4.97 /_ -17.38 deg A\n",
- "\n",
- "\n",
- " (c)circuit phase relative is lagging s by 17.38 deg\n",
- "\n",
- "\n",
- " (d)current in coil, Icoil = 8.12 /_ -54.25 deg A\n",
- "\n",
- "\n",
- " (e)current in capacitor, Ic = 5.1 /_ 90.0 deg A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 452</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)determine the value of impedance Z1 \n",
- "#(b) If the supply frequency is 5 kHz,determine the value of the components comprising impedance Z1\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL = 6j;# in ohm\n",
- "R2 = 8;# in ohm\n",
- "Z3 = 10;# in ohm\n",
- "rv = 50;# in volts\n",
- "thetav = 30;# in degrees\n",
- "ri = 31.4;# in amperes\n",
- "thetai = 52.48;# in degrees\n",
- "f = 5000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #impedance, Z2\n",
- "Z2 = R2 + RL\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #current, I\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Total circuit admittance,\n",
- "YT = I/V\n",
- " #admittance, Y3\n",
- "Y3 = 1/Z3\n",
- " #admittance, Y2\n",
- "Y2 = 1/Z2\n",
- " #admittance, Y1\n",
- "Y1 = YT - Y2 - Y3\n",
- " #impedance, Z1\n",
- "Z1 = 1/Y1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the impedance Z1 is \",round(Z1.real,2),\" + (\",round( Z1.imag,2),\")i ohm\\n\"\n",
- "\n",
- " #resistance, R1\n",
- "R1 = Z1.real\n",
- "X1 = Z1.imag \n",
- "if ((R1>0)&(X1<0)):\n",
- " C1 = -1/(2*math.pi*f*X1)\n",
- " print \"\\n (b)The series circuit thus consists of a resistor of resistance \",round(R1,2),\" ohm\"\n",
- " print \" and a capacitor of capacitance \",round(C1*1E6,2),\"uFarad\\n\"\n",
- "elif ((R1>0)&(X1>0)):\n",
- " L1 = 2*math.pi*f*X1\n",
- " print \"\\n (b)The series circuit thus consists of a resistor of resistance \",round(R1,2),\" ohm \"\n",
- " print \" and a inductor of insuctance \",round(L1*1000,2),\"mHenry\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the impedance Z1 is 1.6 + ( -1.2 )i ohm\n",
- "\n",
- "\n",
- " (b)The series circuit thus consists of a resistor of resistance 1.6 ohm\n",
- " and a capacitor of capacitance 26.55 uFarad\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 453</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the equivalent series circuit impedance,\n",
- "#(b) the supply current I, (c) the circuit phase angle,\n",
- "#(d) the values of voltages V1 and V2, and (e) the values of currents IA and IB\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL1 = 1.02j;# in ohm\n",
- "R1 = 1.65;# in ohm\n",
- "RLa = 7j;# in ohm\n",
- "Ra = 5;# in ohm\n",
- "Rcb = -1j*15;# in ohm\n",
- "Rb = 4;# in ohm\n",
- "rv = 91;# in volts\n",
- "thetav = 0;# in degree\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #impedance, Z1\n",
- "Z1 = R1 + RL1\n",
- " #impedance, Za\n",
- "Za = Ra + RLa\n",
- " #impedance, Zb\n",
- "Zb = Rb + Rcb\n",
- " #impedance, Z, of the two branches connected in parallel\n",
- "Z = Za*Zb/(Za + Zb)\n",
- " #Total circuit impedance\n",
- "ZT = Z1 + Z\n",
- " #Supply current, I\n",
- "I = V/ZT\n",
- "thetai = cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
- "phi = thetav - thetai \n",
- "if (phi>0):\n",
- " a = \"lagging\"\n",
- "else:\n",
- " a = \"leading\"\n",
- "\n",
- " #Voltage V1\n",
- "V1 = I*Z1\n",
- " #Voltage V2\n",
- "V2 = I*Z\n",
- " #current Ia\n",
- "Ia = V2/Za\n",
- " #Current Ib\n",
- "Ib = V2/Zb\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)equivalent series circuit impedance is \",round(ZT.real,2),\" + (\",round( ZT.imag,2),\")i ohm\\n\"\n",
- "print \"\\n (b)supply current, I is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\\n\"\n",
- "print \"\\n (c)circuit phase relative is \",a,\" by \",round(abs(phi),2),\"deg\\n\"\n",
- "print \"\\n (d)voltage, V1 is (\",round(V1.real,2),\" + (\",round(V1.imag,2),\")i) V and V2 is(\",round(V2.real,2),\" + (\",round( V2.imag,2),\")i) V\\n\"\n",
- "print \"\\n (e)current, Ia is (\",round(Ia.real,2),\" + (\",round( Ia.imag,2),\")i) A and Ib is(\",round(Ib.real,2),\" + (\",round( Ib.imag,2),\")i) A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)equivalent series circuit impedance is 12.0 + ( 5.0 )i ohm\n",
- "\n",
- "\n",
- " (b)supply current, I is 6.46 + ( -2.69 )i A\n",
- "\n",
- "\n",
- " (c)circuit phase relative is lagging by 22.61 deg\n",
- "\n",
- "\n",
- " (d)voltage, V1 is ( 13.41 + ( 2.15 )i) V and V2 is( 77.59 + ( -2.15 )i) V\n",
- "\n",
- "\n",
- " (e)current, Ia is ( 5.04 + ( -7.49 )i) A and Ib is( 1.42 + ( 4.79 )i) A"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_2.ipynb deleted file mode 100755 index 19e462e5..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_2.ipynb +++ /dev/null @@ -1,647 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 25: Application of complex numbers to parallel a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 446</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the admittance, conductance and susceptance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 0 - 5j;# in ohms\n",
- "Z2 = 25 + 40j;# in ohms\n",
- "Z3 = 3 - 2j;# in ohms\n",
- "r4 = 50;# in ohms\n",
- "theta4 = 40;# in degrees\n",
- "\n",
- "#calculation:\n",
- " #admittance Y\n",
- "Y1 = 1/Z1\n",
- " #conductance, G\n",
- "G1 = Y1.real\n",
- " #Suspectance, Bc\n",
- "Bc1 = abs(Y1.imag)\n",
- " #admittance Y\n",
- "Y2 = 1/Z2\n",
- " #conductance, G\n",
- "G2 = Y2.real\n",
- " #Suspectance, Bc\n",
- "Bc2 = abs(Y2.imag)\n",
- " #admittance Y\n",
- "Y3 = 1/Z3\n",
- " #conductance, G\n",
- "G3 = Y3.real\n",
- " #Suspectance, Bc\n",
- "Bc3 = abs(Y3.imag)\n",
- "Z4 = r4*math.cos(theta4*math.pi/180) + 1j*r4*math.sin(theta4*math.pi/180)\n",
- " #admittance Y\n",
- "Y4 = 1/Z4\n",
- " #conductance, G\n",
- "G4 = Y4.real\n",
- " #Suspectance, Bc\n",
- "Bc4 = abs(Y4.imag)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)admittance Y is (\",round(Y1.real,2),\" + (\",round(Y1.imag,2),\")i) S, \"\n",
- "print \" conductance, G is \",round(G1,2),\" S, susceptance,Bc is \",round(Bc1,2),\" S\\n\"\n",
- "print \"\\n (b)admittance Y is (\",round(Y2.real,2),\" + (\",round(Y2.imag,2),\")i) S, \"\n",
- "print \" conductance, G is \",round(G2,2),\" S, susceptance,Bc is \",round(Bc2,2),\" S\\n\"\n",
- "print \"\\n (c)admittance Y is (\",round(Y3.real,2),\" + (\",round(Y3.imag,2),\")i) S, \"\n",
- "print \" conductance, G is \",round(G3,2),\" S, susceptance,Bc is \",round(Bc3,2),\" S\\n\"\n",
- "print \"\\n (d)admittance Y is (\",round(Y4.real,2),\" + (\",round(Y4.imag,2),\")i) S, \"\n",
- "print \" conductance, G is \",round(G4,2),\" S, susceptance,Bc is \",round(Bc4,2),\" S\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)admittance Y is ( -0.0 + ( 0.2 )i) S, \n",
- " conductance, G is -0.0 S, susceptance,Bc is 0.2 S\n",
- "\n",
- "\n",
- " (b)admittance Y is ( 0.01 + ( -0.02 )i) S, \n",
- " conductance, G is 0.01 S, susceptance,Bc is 0.02 S\n",
- "\n",
- "\n",
- " (c)admittance Y is ( 0.23 + ( 0.15 )i) S, \n",
- " conductance, G is 0.23 S, susceptance,Bc is 0.15 S\n",
- "\n",
- "\n",
- " (d)admittance Y is ( 0.02 + ( -0.01 )i) S, \n",
- " conductance, G is 0.02 S, susceptance,Bc is 0.01 S\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 447</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine expressions for the impedance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Y2 = 0.001 - 0.002j;# in S\n",
- "Y3 = 0.05 + 0.08j;# in S\n",
- "r1 = 0.004;# in S\n",
- "theta1 = 30;# in degrees\n",
- "\n",
- " #calculation:\n",
- " #impedance, Z\n",
- "Z2 = 1/Y2\n",
- "Z3 = 1/Y3\n",
- "Y1 = r1*math.cos(theta1*math.pi/180) + 1j*r1*math.sin(theta1*math.pi/180)\n",
- "Z1 = 1/Y1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Impedance,Z is (\",round(Z1.real,2),\" + (\",round( Z1.imag,2),\")i) ohm\\n\"\n",
- "print \"\\n (b)Impedance,Z is (\",round(Z2.real,2),\" + (\",round( Z2.imag,2),\")i) ohm\\n\"\n",
- "print \"\\n (c)Impedance,Z is (\",round(Z3.real,2),\" + (\",round( Z3.imag,2),\")i) ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Impedance,Z is ( 216.51 + ( -125.0 )i) ohm\n",
- "\n",
- "\n",
- " (b)Impedance,Z is ( 200.0 + ( 400.0 )i) ohm\n",
- "\n",
- "\n",
- " (c)Impedance,Z is ( 5.62 + ( -8.99 )i) ohm"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 448</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of the resistance and the capacitive reactance of the circuit if they are connected \n",
- "#(a) in parallel, (b) in series.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Y = 0.040 - 1j*0.025;# in S\n",
- "\n",
- "#calculation:\n",
- " #impedance, Z\n",
- "Z = 1/Y\n",
- " #conductance, G\n",
- "G = Y.real\n",
- " #Suspectance, Bc\n",
- "Bc = abs(Y.imag)\n",
- " #parallrl \n",
- " #resistance, R\n",
- "Rp = 1/G\n",
- " #capacitive reactance\n",
- "Xcp = 1/Bc\n",
- " #series\n",
- " #resistance, R\n",
- "Rs = Z.real\n",
- " #capacitive reactance\n",
- "Xcs = abs(Z.imag)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)for parallel, resistance,R is \",round(Rp,2),\" ohm and capacitive reactance, Xc is \",round(Xcp,2),\" ohm\\n\"\n",
- "print \"\\n (b)forseries, resistance,R is \",round(Rs,2),\" ohm and capacitive reactance, Xc is \",round(Xcs,2),\" ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)for parallel, resistance,R is 25.0 ohm and capacitive reactance, Xc is 40.0 ohm\n",
- "\n",
- "\n",
- " (b)forseries, resistance,R is 17.98 ohm and capacitive reactance, Xc is 11.24 ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 449</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of currents I, I1 and I2.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 8;# in ohm\n",
- "R = 5;# in ohm\n",
- "R2 = 6;# ohm\n",
- "rv = 50;# in volts\n",
- "thetav = 0;# in degrees\n",
- "\n",
- "#calculation:\n",
- " #voltage,V\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #circuit impedance, ZT\n",
- "ZT = R + (R1*1j*R2/(R1 + 1j*R2))\n",
- " #Current I\n",
- "I = V/ZT\n",
- " #current,I1\n",
- "I1 = I*(1j*R2/(R1 + 1j*R2))\n",
- " #current, I2\n",
- "I2 = I*(R1/(R1 + 1j*R2))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg A,\"\n",
- "print \"current,I1 = \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real, I1.imag))*180/math.pi,2),\"deg A, \"\n",
- "print \"current, I2 = \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real, I2.imag))*180/math.pi,2),\"deg A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I = 5.7 /_ -25.98 deg A,\n",
- "current,I1 = 3.42 /_ 27.15 deg A, \n",
- "current, I2 = 4.56 /_ -62.85 deg A\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 450</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of supply current I and its phase relative to the 40 V supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 5;# in ohm\n",
- "R2 = 3;# in ohm \n",
- "R3 = 8;# ohm\n",
- "Xc = 4;# in ohms\n",
- "XL = 12;# in Ohms\n",
- "V = 40;# in volts\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- "Z1 = R1 + 1j*XL\n",
- "Z2 = R2 - 1j*Xc\n",
- "Z3 = R3\n",
- " #circuit admittance, YT = 1/ZT\n",
- "YT = (1/Z1) + (1/Z2) + (1/Z3)\n",
- " #Current I\n",
- "I = V*YT\n",
- "I1 = V/Z1\n",
- "I2 = V/Z2\n",
- "I3 = V/Z2\n",
- "thetav = 0\n",
- "thetai = cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
- "phi = thetav - thetai \n",
- "if (phi>0):\n",
- " a = \"lagging\"\n",
- "else:\n",
- " a = \"leading\"\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I is (\",round(I.real,2),\" + (\",round(I.imag,2),\")i) A,\"\n",
- "print \"and its phase relative to the 40 V supply is \",a,\"s by \",round(abs(phi),2),\"deg\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I is ( 10.98 + ( 3.56 )i) A,\n",
- "and its phase relative to the 40 V supply is leading s by 17.96 deg\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 451</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the total equivalent circuit impedance, (b) the supply current, \n",
- "#(c) the circuit phase angle, (d) the current in the coil, and (e) the current in the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.07958;# in Henry\n",
- "R = 18;# in ohm\n",
- "C = 64.96E-6;# in Farad\n",
- "rv = 250;# in volts\n",
- "thetav = 0;# in degrees\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #Inductive reactance\n",
- "XL = 2*math.pi*f*L\n",
- " #capacitive reactance\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #impedance of the coil,\n",
- "Zcoil = R + 1j*XL\n",
- " #impedance presented by the capacitor,\n",
- "Zc = -1j*Xc\n",
- " #Total equivalent circuit impedance,\n",
- "ZT = Zcoil*Zc/(Zcoil + Zc)\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #current, I\n",
- "I = V/ZT\n",
- "thetai = cmath.phase(complex(I.real,I.imag))*180/math.pi\n",
- "phi = thetav - thetai\n",
- "if (phi>0):\n",
- " a = \"lagging\"\n",
- "else:\n",
- " a = \"leading\"\n",
- "\n",
- " #Current in the coil, ICOIL\n",
- "Icoil = V/Zcoil\n",
- " #Current in the capacitor, IC\n",
- "Ic = V/Zc\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the circuit impedance is \",round(ZT.real,2),\" + (\",round( ZT.imag,2),\")i ohm\\n\"\n",
- "print \"\\n (b)supply current, I = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg A\\n\"\n",
- "print \"\\n (c)circuit phase relative is \",a,\"s by \",round(abs(phi),2),\"deg\\n\"\n",
- "print \"\\n (d)current in coil, Icoil = \",round(abs(Icoil),2),\"/_\",round(cmath.phase(complex(Icoil.real, Icoil.imag))*180/math.pi,2),\"deg A\\n\"\n",
- "print \"\\n (e)current in capacitor, Ic = \",round(abs(Ic),2),\"/_\",round(cmath.phase(complex(Ic.real, Ic.imag))*180/math.pi,2),\"deg A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the circuit impedance is 48.02 + ( 15.03 )i ohm\n",
- "\n",
- "\n",
- " (b)supply current, I = 4.97 /_ -17.38 deg A\n",
- "\n",
- "\n",
- " (c)circuit phase relative is lagging s by 17.38 deg\n",
- "\n",
- "\n",
- " (d)current in coil, Icoil = 8.12 /_ -54.25 deg A\n",
- "\n",
- "\n",
- " (e)current in capacitor, Ic = 5.1 /_ 90.0 deg A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 452</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)determine the value of impedance Z1 \n",
- "#(b) If the supply frequency is 5 kHz,determine the value of the components comprising impedance Z1\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL = 6j;# in ohm\n",
- "R2 = 8;# in ohm\n",
- "Z3 = 10;# in ohm\n",
- "rv = 50;# in volts\n",
- "thetav = 30;# in degrees\n",
- "ri = 31.4;# in amperes\n",
- "thetai = 52.48;# in degrees\n",
- "f = 5000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #impedance, Z2\n",
- "Z2 = R2 + RL\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #current, I\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Total circuit admittance,\n",
- "YT = I/V\n",
- " #admittance, Y3\n",
- "Y3 = 1/Z3\n",
- " #admittance, Y2\n",
- "Y2 = 1/Z2\n",
- " #admittance, Y1\n",
- "Y1 = YT - Y2 - Y3\n",
- " #impedance, Z1\n",
- "Z1 = 1/Y1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the impedance Z1 is \",round(Z1.real,2),\" + (\",round( Z1.imag,2),\")i ohm\\n\"\n",
- "\n",
- " #resistance, R1\n",
- "R1 = Z1.real\n",
- "X1 = Z1.imag \n",
- "if ((R1>0)&(X1<0)):\n",
- " C1 = -1/(2*math.pi*f*X1)\n",
- " print \"\\n (b)The series circuit thus consists of a resistor of resistance \",round(R1,2),\" ohm\"\n",
- " print \" and a capacitor of capacitance \",round(C1*1E6,2),\"uFarad\\n\"\n",
- "elif ((R1>0)&(X1>0)):\n",
- " L1 = 2*math.pi*f*X1\n",
- " print \"\\n (b)The series circuit thus consists of a resistor of resistance \",round(R1,2),\" ohm \"\n",
- " print \" and a inductor of insuctance \",round(L1*1000,2),\"mHenry\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the impedance Z1 is 1.6 + ( -1.2 )i ohm\n",
- "\n",
- "\n",
- " (b)The series circuit thus consists of a resistor of resistance 1.6 ohm\n",
- " and a capacitor of capacitance 26.55 uFarad\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 453</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the equivalent series circuit impedance,\n",
- "#(b) the supply current I, (c) the circuit phase angle,\n",
- "#(d) the values of voltages V1 and V2, and (e) the values of currents IA and IB\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL1 = 1.02j;# in ohm\n",
- "R1 = 1.65;# in ohm\n",
- "RLa = 7j;# in ohm\n",
- "Ra = 5;# in ohm\n",
- "Rcb = -1j*15;# in ohm\n",
- "Rb = 4;# in ohm\n",
- "rv = 91;# in volts\n",
- "thetav = 0;# in degree\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #impedance, Z1\n",
- "Z1 = R1 + RL1\n",
- " #impedance, Za\n",
- "Za = Ra + RLa\n",
- " #impedance, Zb\n",
- "Zb = Rb + Rcb\n",
- " #impedance, Z, of the two branches connected in parallel\n",
- "Z = Za*Zb/(Za + Zb)\n",
- " #Total circuit impedance\n",
- "ZT = Z1 + Z\n",
- " #Supply current, I\n",
- "I = V/ZT\n",
- "thetai = cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
- "phi = thetav - thetai \n",
- "if (phi>0):\n",
- " a = \"lagging\"\n",
- "else:\n",
- " a = \"leading\"\n",
- "\n",
- " #Voltage V1\n",
- "V1 = I*Z1\n",
- " #Voltage V2\n",
- "V2 = I*Z\n",
- " #current Ia\n",
- "Ia = V2/Za\n",
- " #Current Ib\n",
- "Ib = V2/Zb\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)equivalent series circuit impedance is \",round(ZT.real,2),\" + (\",round( ZT.imag,2),\")i ohm\\n\"\n",
- "print \"\\n (b)supply current, I is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\\n\"\n",
- "print \"\\n (c)circuit phase relative is \",a,\" by \",round(abs(phi),2),\"deg\\n\"\n",
- "print \"\\n (d)voltage, V1 is (\",round(V1.real,2),\" + (\",round(V1.imag,2),\")i) V and V2 is(\",round(V2.real,2),\" + (\",round( V2.imag,2),\")i) V\\n\"\n",
- "print \"\\n (e)current, Ia is (\",round(Ia.real,2),\" + (\",round( Ia.imag,2),\")i) A and Ib is(\",round(Ib.real,2),\" + (\",round( Ib.imag,2),\")i) A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)equivalent series circuit impedance is 12.0 + ( 5.0 )i ohm\n",
- "\n",
- "\n",
- " (b)supply current, I is 6.46 + ( -2.69 )i A\n",
- "\n",
- "\n",
- " (c)circuit phase relative is lagging by 22.61 deg\n",
- "\n",
- "\n",
- " (d)voltage, V1 is ( 13.41 + ( 2.15 )i) V and V2 is( 77.59 + ( -2.15 )i) V\n",
- "\n",
- "\n",
- " (e)current, Ia is ( 5.04 + ( -7.49 )i) A and Ib is( 1.42 + ( 4.79 )i) A"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_25.ipynb index 592a4669..592a4669 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_25.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint.ipynb deleted file mode 100755 index 4edde932..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint.ipynb +++ /dev/null @@ -1,496 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 26: Power in a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 466</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the active power in the circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL = 12j;# in ohm\n",
- "R = 5;# in ohm\n",
- "rv = 52;# in volts\n",
- "thetav = 30;# in degree\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #impedance, Z\n",
- "Z = R + RL\n",
- " #current\n",
- "I = V/Z\n",
- " #Active power, P\n",
- "Pa = V.real*I.real + V.imag*I.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nthe active power in the circuit \",Pa,\" W\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "the active power in the circuit 80.0 W"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 467</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the active power, and (b) the reactive power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 120 + 200j;# in volts\n",
- "I = 15 + 8j;# in amperes\n",
- "\n",
- "#calculation:\n",
- " #Active power, P\n",
- "Pa = V.real*I.real + V.imag*I.imag\n",
- " #Reactive power, Q\n",
- "Q = V.imag*I.real - V.real*I.imag\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the active power in the circuit \",Pa,\" W\\n\"\n",
- "print \"\\n (b) the reactive power in the circuit \",Q,\" var\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the active power in the circuit 3400.0 W\n",
- "\n",
- "\n",
- " (b) the reactive power in the circuit 2040.0 var"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 468</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current flowing and its phase, (b) the value of resistance R, and (c) the value of capacitance C.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vm = 141.4;# in volts\n",
- "w = 10000;# in rad/sec\n",
- "phiv = math.pi/9;# in radian\n",
- "Pd = 1732;# in Watts\n",
- "pf = 0.866;# power fctr\n",
- "\n",
- "#calculation:\n",
- " #the rms voltage,\n",
- "Vrms = 0.707*Vm\n",
- " #Power P = V*I*cos(phi)\n",
- " #current magnitude, Irms\n",
- "Irms = Pd/(Vrms*pf)\n",
- "phid = math.acos(pf)\n",
- " #current phase angle\n",
- "phii = phiv + phid\n",
- "phiid = phii*180/math.pi# in degrees\n",
- " #Voltage, V\n",
- "V = Vrms*math.cos(phiv) + 1j*Vrms*math.sin(phiv)\n",
- " #current, I\n",
- "I = Irms*math.cos(phii) + 1j*Irms*math.sin(phii)\n",
- " #Impedance, Z\n",
- "Z = V/I\n",
- " #resistance, R\n",
- "R = Z.real\n",
- " #capacitive reactance, Xc\n",
- "Xc = abs(Z.imag)\n",
- " #capacitance, C\n",
- "C = 1/ (w*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the current flowing and Circuit phase angle is \",round(Irms,2),\"/_\",round(phiid,2),\"deg A\\n\"\n",
- "print \"\\n (b) the resistance is \",round(R,2),\" ohm\\n\"\n",
- "print \"\\n (c) the capacitance is \",round(C*1E6,2),\"uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the current flowing and Circuit phase angle is 20.01 /_ 50.0 deg A\n",
- "\n",
- "\n",
- " (b) the resistance is 4.33 ohm\n",
- "\n",
- "\n",
- " (c) the capacitance is 40.02 uF"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 468</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the active power developed between points (a) A and B, (b) C and D, (c) E and F.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "rv = 100;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R = 5;# in ohm\n",
- "R1 = 3;# in ohms\n",
- "RL = 4j;# in ohm\n",
- "Rc = -10j;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #impedance, Z1\n",
- "Z1 = R1 + RL\n",
- " #impedance, Zc\n",
- "Zc = Rc\n",
- " #Circuit impedance, Z\n",
- "Z = R + (Z1*Zc/(Z1 + Zc))\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- "I = V/Z\n",
- "Imag = ((I.real)**2 + (I.imag)**2)**0.5\n",
- " #Active power developed between points A and B\n",
- "Pab = (Imag**2)*R\n",
- " #Active power developed between points C and D\n",
- "Pcd = (Imag**2)*Zc.real\n",
- " #Current, I1\n",
- "I1 = I*Zc/(Zc + Z1)\n",
- "I1mag = ((I1.real)**2 + (I1.imag)**2)**0.5\n",
- " #active power developed between points E and F\n",
- "Pef = (I1mag**2)*Z1.real\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Active power developed between points A and B is \",round(Pab,2),\" W\\n\"\n",
- "print \"\\n (b)Active power developed between points C and D is \",round(Pcd,2),\" W\\n\"\n",
- "print \"\\n (c)Active power developed between points E and F is \",round(Pef,2),\" W\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Active power developed between points A and B is 339.62 W\n",
- "\n",
- "\n",
- " (b)Active power developed between points C and D is 0.0 W\n",
- "\n",
- "\n",
- " (c)Active power developed between points E and F is 452.83 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 469</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the apparent power, (b) the reactive power, \n",
- "#(c) the value and phase of current I, and (d) the value of impedance Z.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Pa = 400;# in Watts\n",
- "rv = 100;# in volts\n",
- "thetav = 30;# in degrees\n",
- "R = 4;# in ohm\n",
- "pf = 0.766;# power factor\n",
- "\n",
- " #calculation:\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #magnitude of apparent power,S = V*I\n",
- "S = Pa/pf\n",
- "phi = math.acos(pf)\n",
- "theta = phi*180/math.pi# in degrees\n",
- " #Reactive power Q\n",
- "Q = S*math.sin(phi)\n",
- " #magnitude of current\n",
- "Imag = S/rv\n",
- "thetai = thetav - theta\n",
- "I = Imag*math.cos(thetai*math.pi/180) + 1j*Imag*math.sin(thetai*math.pi/180)\n",
- " #Total circuit impedance ZT\n",
- "ZT = V/I\n",
- " #impedance Z\n",
- "Z = ZT - R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)apparent power is \",round(S,2),\" VA\\n\"\n",
- "print \"\\n (b)reactive power is \",round(Q,1),\" var lagging\\n\"\n",
- "print \"\\n (c)the current flowing and Circuit phase angle is \",round(Imag,2),\"/_\",round(thetai,2),\"deg A\\n\"\n",
- "print \"\\n (d)impedance, Z is \",round(Z.real,2),\" + (\",round( Z.imag,2),\")i ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)apparent power is 522.19 VA\n",
- "\n",
- "\n",
- " (b)reactive power is 335.7 var lagging\n",
- "\n",
- "\n",
- " (c)the current flowing and Circuit phase angle is 5.22 /_ -10.0 deg A\n",
- "\n",
- "\n",
- " (d)impedance, Z is 10.67 + ( 12.31 )i ohm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 471</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the rating (in kilovars) of the capacitors required.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "S = 300000;# in VA\n",
- "pf1 = 0.70;# in power factor\n",
- "pf2 = 0.90;# in power factor\n",
- "\n",
- "#calculation:\n",
- " #active power, P\n",
- "Pa = S*pf1\n",
- "phi1 = math.acos(pf1)\n",
- "phi1d = phi1*180/math.pi\n",
- " #Reactive power, Q\n",
- "Q = S*math.sin(phi1)\n",
- "phi2 = math.acos(pf2)\n",
- "phi2d = phi2*180/math.pi\n",
- " #The capacitor rating needed to improve the power factor to 0.90\n",
- " #the capacitor rating,\n",
- "Pr = Q - (Pa*math.tan(phi2))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the rating (in kilovars) of the capacitors is \",round((Pr/1E3),2),\" kvar leading\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the rating (in kilovars) of the capacitors is 112.54 kvar leading"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 471</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the supply current, (b) the active, apparent and reactive power,\n",
- "#(c) the rating of a capacitor (d) the value of capacitance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z = 3 + 4j;# in ohms\n",
- "rv = 50;# in volts\n",
- "thetav = 30;# in Degrees\n",
- "f = 1500;# in Hz\n",
- "pf1 = 0.966;# in power factor\n",
- "\n",
- "#calculation:\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Supply current, I\n",
- "I = V/Z\n",
- "Istr = I.real - 1j*I.imag\n",
- " #Apparent power, S\n",
- "S = V*Istr\n",
- " #active power, Pa\n",
- "Pa = S.real\n",
- "#reactive power, Q\n",
- "Q = abs(S.imag)\n",
- " #apparent power, S\n",
- "S = (S.real**2 + S.imag**2)**0.5\n",
- "phi1 = math.acos(pf1)\n",
- "phi1d = phi1*180/math.pi\n",
- " #rating of the capacitor \n",
- "Pr = Q - Pa*math.tan(phi1)\n",
- " #Current in capacitor, Ic\n",
- "Ic = Pr/rv\n",
- " #Capacitive reactance, Xc\n",
- "Xc = rv/Ic\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)supply current, I is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\\n\"\n",
- "print \"\\n (b)active power is \",round(Pa,2),\" W, apparent power is \",round( S,2),\" W \"\n",
- "print \"and reactive power is \",round( Q,2),\" W lagging\\n\"\n",
- "print \"\\n (c)the rating of the capacitors is \",round(Pr,2),\" var leading\\n\"\n",
- "print \"\\n (d)value of capacitance needed to improve the power factor to 0.966 lagging is \",round( C*1E6,2),\"uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)supply current, I is 9.2 + ( -3.93 )i A\n",
- "\n",
- "\n",
- " (b)active power is 300.0 W, apparent power is 500.0 W and reactive power is 400.0 W lagging\n",
- "\n",
- "\n",
- " (c)the rating of the capacitors is 319.71 var leading\n",
- "\n",
- "\n",
- " (d)value of capacitance needed to improve the power factor to 0.966 lagging is 13.57 uF"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_1.ipynb deleted file mode 100755 index 4edde932..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_1.ipynb +++ /dev/null @@ -1,496 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 26: Power in a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 466</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the active power in the circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL = 12j;# in ohm\n",
- "R = 5;# in ohm\n",
- "rv = 52;# in volts\n",
- "thetav = 30;# in degree\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #impedance, Z\n",
- "Z = R + RL\n",
- " #current\n",
- "I = V/Z\n",
- " #Active power, P\n",
- "Pa = V.real*I.real + V.imag*I.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nthe active power in the circuit \",Pa,\" W\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "the active power in the circuit 80.0 W"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 467</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the active power, and (b) the reactive power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 120 + 200j;# in volts\n",
- "I = 15 + 8j;# in amperes\n",
- "\n",
- "#calculation:\n",
- " #Active power, P\n",
- "Pa = V.real*I.real + V.imag*I.imag\n",
- " #Reactive power, Q\n",
- "Q = V.imag*I.real - V.real*I.imag\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the active power in the circuit \",Pa,\" W\\n\"\n",
- "print \"\\n (b) the reactive power in the circuit \",Q,\" var\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the active power in the circuit 3400.0 W\n",
- "\n",
- "\n",
- " (b) the reactive power in the circuit 2040.0 var"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 468</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current flowing and its phase, (b) the value of resistance R, and (c) the value of capacitance C.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vm = 141.4;# in volts\n",
- "w = 10000;# in rad/sec\n",
- "phiv = math.pi/9;# in radian\n",
- "Pd = 1732;# in Watts\n",
- "pf = 0.866;# power fctr\n",
- "\n",
- "#calculation:\n",
- " #the rms voltage,\n",
- "Vrms = 0.707*Vm\n",
- " #Power P = V*I*cos(phi)\n",
- " #current magnitude, Irms\n",
- "Irms = Pd/(Vrms*pf)\n",
- "phid = math.acos(pf)\n",
- " #current phase angle\n",
- "phii = phiv + phid\n",
- "phiid = phii*180/math.pi# in degrees\n",
- " #Voltage, V\n",
- "V = Vrms*math.cos(phiv) + 1j*Vrms*math.sin(phiv)\n",
- " #current, I\n",
- "I = Irms*math.cos(phii) + 1j*Irms*math.sin(phii)\n",
- " #Impedance, Z\n",
- "Z = V/I\n",
- " #resistance, R\n",
- "R = Z.real\n",
- " #capacitive reactance, Xc\n",
- "Xc = abs(Z.imag)\n",
- " #capacitance, C\n",
- "C = 1/ (w*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the current flowing and Circuit phase angle is \",round(Irms,2),\"/_\",round(phiid,2),\"deg A\\n\"\n",
- "print \"\\n (b) the resistance is \",round(R,2),\" ohm\\n\"\n",
- "print \"\\n (c) the capacitance is \",round(C*1E6,2),\"uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the current flowing and Circuit phase angle is 20.01 /_ 50.0 deg A\n",
- "\n",
- "\n",
- " (b) the resistance is 4.33 ohm\n",
- "\n",
- "\n",
- " (c) the capacitance is 40.02 uF"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 468</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the active power developed between points (a) A and B, (b) C and D, (c) E and F.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "rv = 100;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R = 5;# in ohm\n",
- "R1 = 3;# in ohms\n",
- "RL = 4j;# in ohm\n",
- "Rc = -10j;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #impedance, Z1\n",
- "Z1 = R1 + RL\n",
- " #impedance, Zc\n",
- "Zc = Rc\n",
- " #Circuit impedance, Z\n",
- "Z = R + (Z1*Zc/(Z1 + Zc))\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- "I = V/Z\n",
- "Imag = ((I.real)**2 + (I.imag)**2)**0.5\n",
- " #Active power developed between points A and B\n",
- "Pab = (Imag**2)*R\n",
- " #Active power developed between points C and D\n",
- "Pcd = (Imag**2)*Zc.real\n",
- " #Current, I1\n",
- "I1 = I*Zc/(Zc + Z1)\n",
- "I1mag = ((I1.real)**2 + (I1.imag)**2)**0.5\n",
- " #active power developed between points E and F\n",
- "Pef = (I1mag**2)*Z1.real\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Active power developed between points A and B is \",round(Pab,2),\" W\\n\"\n",
- "print \"\\n (b)Active power developed between points C and D is \",round(Pcd,2),\" W\\n\"\n",
- "print \"\\n (c)Active power developed between points E and F is \",round(Pef,2),\" W\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Active power developed between points A and B is 339.62 W\n",
- "\n",
- "\n",
- " (b)Active power developed between points C and D is 0.0 W\n",
- "\n",
- "\n",
- " (c)Active power developed between points E and F is 452.83 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 469</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the apparent power, (b) the reactive power, \n",
- "#(c) the value and phase of current I, and (d) the value of impedance Z.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Pa = 400;# in Watts\n",
- "rv = 100;# in volts\n",
- "thetav = 30;# in degrees\n",
- "R = 4;# in ohm\n",
- "pf = 0.766;# power factor\n",
- "\n",
- " #calculation:\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #magnitude of apparent power,S = V*I\n",
- "S = Pa/pf\n",
- "phi = math.acos(pf)\n",
- "theta = phi*180/math.pi# in degrees\n",
- " #Reactive power Q\n",
- "Q = S*math.sin(phi)\n",
- " #magnitude of current\n",
- "Imag = S/rv\n",
- "thetai = thetav - theta\n",
- "I = Imag*math.cos(thetai*math.pi/180) + 1j*Imag*math.sin(thetai*math.pi/180)\n",
- " #Total circuit impedance ZT\n",
- "ZT = V/I\n",
- " #impedance Z\n",
- "Z = ZT - R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)apparent power is \",round(S,2),\" VA\\n\"\n",
- "print \"\\n (b)reactive power is \",round(Q,1),\" var lagging\\n\"\n",
- "print \"\\n (c)the current flowing and Circuit phase angle is \",round(Imag,2),\"/_\",round(thetai,2),\"deg A\\n\"\n",
- "print \"\\n (d)impedance, Z is \",round(Z.real,2),\" + (\",round( Z.imag,2),\")i ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)apparent power is 522.19 VA\n",
- "\n",
- "\n",
- " (b)reactive power is 335.7 var lagging\n",
- "\n",
- "\n",
- " (c)the current flowing and Circuit phase angle is 5.22 /_ -10.0 deg A\n",
- "\n",
- "\n",
- " (d)impedance, Z is 10.67 + ( 12.31 )i ohm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 471</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the rating (in kilovars) of the capacitors required.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "S = 300000;# in VA\n",
- "pf1 = 0.70;# in power factor\n",
- "pf2 = 0.90;# in power factor\n",
- "\n",
- "#calculation:\n",
- " #active power, P\n",
- "Pa = S*pf1\n",
- "phi1 = math.acos(pf1)\n",
- "phi1d = phi1*180/math.pi\n",
- " #Reactive power, Q\n",
- "Q = S*math.sin(phi1)\n",
- "phi2 = math.acos(pf2)\n",
- "phi2d = phi2*180/math.pi\n",
- " #The capacitor rating needed to improve the power factor to 0.90\n",
- " #the capacitor rating,\n",
- "Pr = Q - (Pa*math.tan(phi2))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the rating (in kilovars) of the capacitors is \",round((Pr/1E3),2),\" kvar leading\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the rating (in kilovars) of the capacitors is 112.54 kvar leading"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 471</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the supply current, (b) the active, apparent and reactive power,\n",
- "#(c) the rating of a capacitor (d) the value of capacitance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z = 3 + 4j;# in ohms\n",
- "rv = 50;# in volts\n",
- "thetav = 30;# in Degrees\n",
- "f = 1500;# in Hz\n",
- "pf1 = 0.966;# in power factor\n",
- "\n",
- "#calculation:\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Supply current, I\n",
- "I = V/Z\n",
- "Istr = I.real - 1j*I.imag\n",
- " #Apparent power, S\n",
- "S = V*Istr\n",
- " #active power, Pa\n",
- "Pa = S.real\n",
- "#reactive power, Q\n",
- "Q = abs(S.imag)\n",
- " #apparent power, S\n",
- "S = (S.real**2 + S.imag**2)**0.5\n",
- "phi1 = math.acos(pf1)\n",
- "phi1d = phi1*180/math.pi\n",
- " #rating of the capacitor \n",
- "Pr = Q - Pa*math.tan(phi1)\n",
- " #Current in capacitor, Ic\n",
- "Ic = Pr/rv\n",
- " #Capacitive reactance, Xc\n",
- "Xc = rv/Ic\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)supply current, I is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\\n\"\n",
- "print \"\\n (b)active power is \",round(Pa,2),\" W, apparent power is \",round( S,2),\" W \"\n",
- "print \"and reactive power is \",round( Q,2),\" W lagging\\n\"\n",
- "print \"\\n (c)the rating of the capacitors is \",round(Pr,2),\" var leading\\n\"\n",
- "print \"\\n (d)value of capacitance needed to improve the power factor to 0.966 lagging is \",round( C*1E6,2),\"uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)supply current, I is 9.2 + ( -3.93 )i A\n",
- "\n",
- "\n",
- " (b)active power is 300.0 W, apparent power is 500.0 W and reactive power is 400.0 W lagging\n",
- "\n",
- "\n",
- " (c)the rating of the capacitors is 319.71 var leading\n",
- "\n",
- "\n",
- " (d)value of capacitance needed to improve the power factor to 0.966 lagging is 13.57 uF"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_2.ipynb deleted file mode 100755 index 4edde932..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_2.ipynb +++ /dev/null @@ -1,496 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 26: Power in a.c. circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 466</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the active power in the circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL = 12j;# in ohm\n",
- "R = 5;# in ohm\n",
- "rv = 52;# in volts\n",
- "thetav = 30;# in degree\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #impedance, Z\n",
- "Z = R + RL\n",
- " #current\n",
- "I = V/Z\n",
- " #Active power, P\n",
- "Pa = V.real*I.real + V.imag*I.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nthe active power in the circuit \",Pa,\" W\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "the active power in the circuit 80.0 W"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 467</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the active power, and (b) the reactive power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 120 + 200j;# in volts\n",
- "I = 15 + 8j;# in amperes\n",
- "\n",
- "#calculation:\n",
- " #Active power, P\n",
- "Pa = V.real*I.real + V.imag*I.imag\n",
- " #Reactive power, Q\n",
- "Q = V.imag*I.real - V.real*I.imag\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the active power in the circuit \",Pa,\" W\\n\"\n",
- "print \"\\n (b) the reactive power in the circuit \",Q,\" var\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the active power in the circuit 3400.0 W\n",
- "\n",
- "\n",
- " (b) the reactive power in the circuit 2040.0 var"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 468</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current flowing and its phase, (b) the value of resistance R, and (c) the value of capacitance C.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vm = 141.4;# in volts\n",
- "w = 10000;# in rad/sec\n",
- "phiv = math.pi/9;# in radian\n",
- "Pd = 1732;# in Watts\n",
- "pf = 0.866;# power fctr\n",
- "\n",
- "#calculation:\n",
- " #the rms voltage,\n",
- "Vrms = 0.707*Vm\n",
- " #Power P = V*I*cos(phi)\n",
- " #current magnitude, Irms\n",
- "Irms = Pd/(Vrms*pf)\n",
- "phid = math.acos(pf)\n",
- " #current phase angle\n",
- "phii = phiv + phid\n",
- "phiid = phii*180/math.pi# in degrees\n",
- " #Voltage, V\n",
- "V = Vrms*math.cos(phiv) + 1j*Vrms*math.sin(phiv)\n",
- " #current, I\n",
- "I = Irms*math.cos(phii) + 1j*Irms*math.sin(phii)\n",
- " #Impedance, Z\n",
- "Z = V/I\n",
- " #resistance, R\n",
- "R = Z.real\n",
- " #capacitive reactance, Xc\n",
- "Xc = abs(Z.imag)\n",
- " #capacitance, C\n",
- "C = 1/ (w*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the current flowing and Circuit phase angle is \",round(Irms,2),\"/_\",round(phiid,2),\"deg A\\n\"\n",
- "print \"\\n (b) the resistance is \",round(R,2),\" ohm\\n\"\n",
- "print \"\\n (c) the capacitance is \",round(C*1E6,2),\"uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the current flowing and Circuit phase angle is 20.01 /_ 50.0 deg A\n",
- "\n",
- "\n",
- " (b) the resistance is 4.33 ohm\n",
- "\n",
- "\n",
- " (c) the capacitance is 40.02 uF"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 468</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the active power developed between points (a) A and B, (b) C and D, (c) E and F.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "rv = 100;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R = 5;# in ohm\n",
- "R1 = 3;# in ohms\n",
- "RL = 4j;# in ohm\n",
- "Rc = -10j;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #impedance, Z1\n",
- "Z1 = R1 + RL\n",
- " #impedance, Zc\n",
- "Zc = Rc\n",
- " #Circuit impedance, Z\n",
- "Z = R + (Z1*Zc/(Z1 + Zc))\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- "I = V/Z\n",
- "Imag = ((I.real)**2 + (I.imag)**2)**0.5\n",
- " #Active power developed between points A and B\n",
- "Pab = (Imag**2)*R\n",
- " #Active power developed between points C and D\n",
- "Pcd = (Imag**2)*Zc.real\n",
- " #Current, I1\n",
- "I1 = I*Zc/(Zc + Z1)\n",
- "I1mag = ((I1.real)**2 + (I1.imag)**2)**0.5\n",
- " #active power developed between points E and F\n",
- "Pef = (I1mag**2)*Z1.real\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Active power developed between points A and B is \",round(Pab,2),\" W\\n\"\n",
- "print \"\\n (b)Active power developed between points C and D is \",round(Pcd,2),\" W\\n\"\n",
- "print \"\\n (c)Active power developed between points E and F is \",round(Pef,2),\" W\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Active power developed between points A and B is 339.62 W\n",
- "\n",
- "\n",
- " (b)Active power developed between points C and D is 0.0 W\n",
- "\n",
- "\n",
- " (c)Active power developed between points E and F is 452.83 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 469</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the apparent power, (b) the reactive power, \n",
- "#(c) the value and phase of current I, and (d) the value of impedance Z.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Pa = 400;# in Watts\n",
- "rv = 100;# in volts\n",
- "thetav = 30;# in degrees\n",
- "R = 4;# in ohm\n",
- "pf = 0.766;# power factor\n",
- "\n",
- " #calculation:\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #magnitude of apparent power,S = V*I\n",
- "S = Pa/pf\n",
- "phi = math.acos(pf)\n",
- "theta = phi*180/math.pi# in degrees\n",
- " #Reactive power Q\n",
- "Q = S*math.sin(phi)\n",
- " #magnitude of current\n",
- "Imag = S/rv\n",
- "thetai = thetav - theta\n",
- "I = Imag*math.cos(thetai*math.pi/180) + 1j*Imag*math.sin(thetai*math.pi/180)\n",
- " #Total circuit impedance ZT\n",
- "ZT = V/I\n",
- " #impedance Z\n",
- "Z = ZT - R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)apparent power is \",round(S,2),\" VA\\n\"\n",
- "print \"\\n (b)reactive power is \",round(Q,1),\" var lagging\\n\"\n",
- "print \"\\n (c)the current flowing and Circuit phase angle is \",round(Imag,2),\"/_\",round(thetai,2),\"deg A\\n\"\n",
- "print \"\\n (d)impedance, Z is \",round(Z.real,2),\" + (\",round( Z.imag,2),\")i ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)apparent power is 522.19 VA\n",
- "\n",
- "\n",
- " (b)reactive power is 335.7 var lagging\n",
- "\n",
- "\n",
- " (c)the current flowing and Circuit phase angle is 5.22 /_ -10.0 deg A\n",
- "\n",
- "\n",
- " (d)impedance, Z is 10.67 + ( 12.31 )i ohm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 471</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the rating (in kilovars) of the capacitors required.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "S = 300000;# in VA\n",
- "pf1 = 0.70;# in power factor\n",
- "pf2 = 0.90;# in power factor\n",
- "\n",
- "#calculation:\n",
- " #active power, P\n",
- "Pa = S*pf1\n",
- "phi1 = math.acos(pf1)\n",
- "phi1d = phi1*180/math.pi\n",
- " #Reactive power, Q\n",
- "Q = S*math.sin(phi1)\n",
- "phi2 = math.acos(pf2)\n",
- "phi2d = phi2*180/math.pi\n",
- " #The capacitor rating needed to improve the power factor to 0.90\n",
- " #the capacitor rating,\n",
- "Pr = Q - (Pa*math.tan(phi2))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the rating (in kilovars) of the capacitors is \",round((Pr/1E3),2),\" kvar leading\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the rating (in kilovars) of the capacitors is 112.54 kvar leading"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 471</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the supply current, (b) the active, apparent and reactive power,\n",
- "#(c) the rating of a capacitor (d) the value of capacitance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z = 3 + 4j;# in ohms\n",
- "rv = 50;# in volts\n",
- "thetav = 30;# in Degrees\n",
- "f = 1500;# in Hz\n",
- "pf1 = 0.966;# in power factor\n",
- "\n",
- "#calculation:\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Supply current, I\n",
- "I = V/Z\n",
- "Istr = I.real - 1j*I.imag\n",
- " #Apparent power, S\n",
- "S = V*Istr\n",
- " #active power, Pa\n",
- "Pa = S.real\n",
- "#reactive power, Q\n",
- "Q = abs(S.imag)\n",
- " #apparent power, S\n",
- "S = (S.real**2 + S.imag**2)**0.5\n",
- "phi1 = math.acos(pf1)\n",
- "phi1d = phi1*180/math.pi\n",
- " #rating of the capacitor \n",
- "Pr = Q - Pa*math.tan(phi1)\n",
- " #Current in capacitor, Ic\n",
- "Ic = Pr/rv\n",
- " #Capacitive reactance, Xc\n",
- "Xc = rv/Ic\n",
- "C = 1/(2*math.pi*f*Xc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)supply current, I is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\\n\"\n",
- "print \"\\n (b)active power is \",round(Pa,2),\" W, apparent power is \",round( S,2),\" W \"\n",
- "print \"and reactive power is \",round( Q,2),\" W lagging\\n\"\n",
- "print \"\\n (c)the rating of the capacitors is \",round(Pr,2),\" var leading\\n\"\n",
- "print \"\\n (d)value of capacitance needed to improve the power factor to 0.966 lagging is \",round( C*1E6,2),\"uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)supply current, I is 9.2 + ( -3.93 )i A\n",
- "\n",
- "\n",
- " (b)active power is 300.0 W, apparent power is 500.0 W and reactive power is 400.0 W lagging\n",
- "\n",
- "\n",
- " (c)the rating of the capacitors is 319.71 var leading\n",
- "\n",
- "\n",
- " (d)value of capacitance needed to improve the power factor to 0.966 lagging is 13.57 uF"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_26.ipynb index fe6f0a60..fe6f0a60 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_26.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint.ipynb deleted file mode 100755 index 4d37718b..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint.ipynb +++ /dev/null @@ -1,205 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 27: A.c. bridges</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 485</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the values of Rx and Cx at balance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R2 = 2500;# in ohms\n",
- "C2 = 0.2E-6;# IN fARADS\n",
- "R3 = 1;\n",
- "R4 = 1;\n",
- "w = 2000*math.pi;\n",
- "#calculation:\n",
- "Rx = R4*(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2)\n",
- "Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance Rx = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\"\n",
- "print \"\\n (b)at balance Rx = \",round(Rx/1000,2),\"KOhm and Cx = \", round(Cx*1E9,2),\"nF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance Rx = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
- "\n",
- " (b)at balance Rx = 2.75 KOhm and Cx = 18.4 nF"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 487</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, when the bridge is balanced, (a) the value of resistance R1, and (b) the frequency of the bridge.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R2 = 30000;# in ohms\n",
- "R3 = 30000;# in ohms\n",
- "R4 = 1000;# in ohms\n",
- "C2 = 1e-9;# IN fARADS\n",
- "C3 = 1e-9;# IN fARADS\n",
- "\n",
- "#calculation:\n",
- " #the bridge is balanced\n",
- "R1 = R4/((R3/R2) + (C2/C3))\n",
- " #frequency, f\n",
- "f = 1/(2*math.pi*((C2*C3*R2*R3)**0.5))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance R1 = \",R1,\" ohm\\n\"\n",
- "print \"\\n (b)frequency, f is \",round(f,2),\"Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance R1 = 500.0 ohm\n",
- "\n",
- "\n",
- " (b)frequency, f is 5305.16 Hz"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 487</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine, when the bridge is balanced, \n",
- "#(a) the value of resistance Rx, (b) the value of capacitance Cx,\n",
- "#(c) the phase angle of the unknown arm, (d) the power factor of the unknown arm and (e) its loss angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R3 = 600;# in ohms\n",
- "R4 = 200;# in ohms\n",
- "C2 = 0.2e-6;# IN fARADS\n",
- "C3 = 4000e-12;# IN fARADS\n",
- "f = 1500;#in Hz\n",
- "\n",
- "#calculation:\n",
- " #the bridge is balanced\n",
- " #Resistance, Rx\n",
- "Rx = R4*C3/C2\n",
- " #Capacitance, Cx\n",
- "Cx = C2*R3/R4\n",
- " #Phase angle\n",
- "phi = math.atan(1/(2*math.pi*f*Cx*Rx))\n",
- "phid = phi*180/math.pi# in degrees\n",
- " #Power factor of capacitor\n",
- "Pc = math.cos(phi)\n",
- " #Loss angle,\n",
- "de = 90 - phid\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance Rx = \",round(Rx,2),\" ohm\\n\"\n",
- "print \"\\n (b)capacitance, Cx is \",round(Cx*1E9,2),\"pFarad\\n\"\n",
- "print \"\\n (c)phasor diagram = \",round(phid,2),\"deg lead \"\n",
- "print \"\\n (d)power factor is \",round(Pc,2),\" \\n\"\n",
- "print \"\\n (e)Loss angle = \",round(de,2),\"deg\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance Rx = 4.0 ohm\n",
- "\n",
- "\n",
- " (b)capacitance, Cx is 600.0 pFarad\n",
- "\n",
- "\n",
- " (c)phasor diagram = 88.7 deg lead \n",
- "\n",
- " (d)power factor is 0.02 \n",
- "\n",
- "\n",
- " (e)Loss angle = 1.3 deg\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_1.ipynb deleted file mode 100755 index 4d37718b..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_1.ipynb +++ /dev/null @@ -1,205 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 27: A.c. bridges</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 485</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the values of Rx and Cx at balance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R2 = 2500;# in ohms\n",
- "C2 = 0.2E-6;# IN fARADS\n",
- "R3 = 1;\n",
- "R4 = 1;\n",
- "w = 2000*math.pi;\n",
- "#calculation:\n",
- "Rx = R4*(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2)\n",
- "Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance Rx = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\"\n",
- "print \"\\n (b)at balance Rx = \",round(Rx/1000,2),\"KOhm and Cx = \", round(Cx*1E9,2),\"nF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance Rx = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
- "\n",
- " (b)at balance Rx = 2.75 KOhm and Cx = 18.4 nF"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 487</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, when the bridge is balanced, (a) the value of resistance R1, and (b) the frequency of the bridge.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R2 = 30000;# in ohms\n",
- "R3 = 30000;# in ohms\n",
- "R4 = 1000;# in ohms\n",
- "C2 = 1e-9;# IN fARADS\n",
- "C3 = 1e-9;# IN fARADS\n",
- "\n",
- "#calculation:\n",
- " #the bridge is balanced\n",
- "R1 = R4/((R3/R2) + (C2/C3))\n",
- " #frequency, f\n",
- "f = 1/(2*math.pi*((C2*C3*R2*R3)**0.5))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance R1 = \",R1,\" ohm\\n\"\n",
- "print \"\\n (b)frequency, f is \",round(f,2),\"Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance R1 = 500.0 ohm\n",
- "\n",
- "\n",
- " (b)frequency, f is 5305.16 Hz"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 487</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine, when the bridge is balanced, \n",
- "#(a) the value of resistance Rx, (b) the value of capacitance Cx,\n",
- "#(c) the phase angle of the unknown arm, (d) the power factor of the unknown arm and (e) its loss angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R3 = 600;# in ohms\n",
- "R4 = 200;# in ohms\n",
- "C2 = 0.2e-6;# IN fARADS\n",
- "C3 = 4000e-12;# IN fARADS\n",
- "f = 1500;#in Hz\n",
- "\n",
- "#calculation:\n",
- " #the bridge is balanced\n",
- " #Resistance, Rx\n",
- "Rx = R4*C3/C2\n",
- " #Capacitance, Cx\n",
- "Cx = C2*R3/R4\n",
- " #Phase angle\n",
- "phi = math.atan(1/(2*math.pi*f*Cx*Rx))\n",
- "phid = phi*180/math.pi# in degrees\n",
- " #Power factor of capacitor\n",
- "Pc = math.cos(phi)\n",
- " #Loss angle,\n",
- "de = 90 - phid\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance Rx = \",round(Rx,2),\" ohm\\n\"\n",
- "print \"\\n (b)capacitance, Cx is \",round(Cx*1E9,2),\"pFarad\\n\"\n",
- "print \"\\n (c)phasor diagram = \",round(phid,2),\"deg lead \"\n",
- "print \"\\n (d)power factor is \",round(Pc,2),\" \\n\"\n",
- "print \"\\n (e)Loss angle = \",round(de,2),\"deg\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance Rx = 4.0 ohm\n",
- "\n",
- "\n",
- " (b)capacitance, Cx is 600.0 pFarad\n",
- "\n",
- "\n",
- " (c)phasor diagram = 88.7 deg lead \n",
- "\n",
- " (d)power factor is 0.02 \n",
- "\n",
- "\n",
- " (e)Loss angle = 1.3 deg\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_2.ipynb deleted file mode 100755 index 4d37718b..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_2.ipynb +++ /dev/null @@ -1,205 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 27: A.c. bridges</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 485</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the values of Rx and Cx at balance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R2 = 2500;# in ohms\n",
- "C2 = 0.2E-6;# IN fARADS\n",
- "R3 = 1;\n",
- "R4 = 1;\n",
- "w = 2000*math.pi;\n",
- "#calculation:\n",
- "Rx = R4*(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2)\n",
- "Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance Rx = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\"\n",
- "print \"\\n (b)at balance Rx = \",round(Rx/1000,2),\"KOhm and Cx = \", round(Cx*1E9,2),\"nF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance Rx = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
- "\n",
- " (b)at balance Rx = 2.75 KOhm and Cx = 18.4 nF"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 487</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, when the bridge is balanced, (a) the value of resistance R1, and (b) the frequency of the bridge.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R2 = 30000;# in ohms\n",
- "R3 = 30000;# in ohms\n",
- "R4 = 1000;# in ohms\n",
- "C2 = 1e-9;# IN fARADS\n",
- "C3 = 1e-9;# IN fARADS\n",
- "\n",
- "#calculation:\n",
- " #the bridge is balanced\n",
- "R1 = R4/((R3/R2) + (C2/C3))\n",
- " #frequency, f\n",
- "f = 1/(2*math.pi*((C2*C3*R2*R3)**0.5))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance R1 = \",R1,\" ohm\\n\"\n",
- "print \"\\n (b)frequency, f is \",round(f,2),\"Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance R1 = 500.0 ohm\n",
- "\n",
- "\n",
- " (b)frequency, f is 5305.16 Hz"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 487</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine, when the bridge is balanced, \n",
- "#(a) the value of resistance Rx, (b) the value of capacitance Cx,\n",
- "#(c) the phase angle of the unknown arm, (d) the power factor of the unknown arm and (e) its loss angle.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R3 = 600;# in ohms\n",
- "R4 = 200;# in ohms\n",
- "C2 = 0.2e-6;# IN fARADS\n",
- "C3 = 4000e-12;# IN fARADS\n",
- "f = 1500;#in Hz\n",
- "\n",
- "#calculation:\n",
- " #the bridge is balanced\n",
- " #Resistance, Rx\n",
- "Rx = R4*C3/C2\n",
- " #Capacitance, Cx\n",
- "Cx = C2*R3/R4\n",
- " #Phase angle\n",
- "phi = math.atan(1/(2*math.pi*f*Cx*Rx))\n",
- "phid = phi*180/math.pi# in degrees\n",
- " #Power factor of capacitor\n",
- "Pc = math.cos(phi)\n",
- " #Loss angle,\n",
- "de = 90 - phid\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resistance Rx = \",round(Rx,2),\" ohm\\n\"\n",
- "print \"\\n (b)capacitance, Cx is \",round(Cx*1E9,2),\"pFarad\\n\"\n",
- "print \"\\n (c)phasor diagram = \",round(phid,2),\"deg lead \"\n",
- "print \"\\n (d)power factor is \",round(Pc,2),\" \\n\"\n",
- "print \"\\n (e)Loss angle = \",round(de,2),\"deg\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resistance Rx = 4.0 ohm\n",
- "\n",
- "\n",
- " (b)capacitance, Cx is 600.0 pFarad\n",
- "\n",
- "\n",
- " (c)phasor diagram = 88.7 deg lead \n",
- "\n",
- " (d)power factor is 0.02 \n",
- "\n",
- "\n",
- " (e)Loss angle = 1.3 deg\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_27.ipynb index c141c129..c141c129 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_27.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint.ipynb deleted file mode 100755 index 35880e50..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint.ipynb +++ /dev/null @@ -1,736 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 28: Series resonance and Q-factor</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 492</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine at what frequency resonance occurs, and (b) the current flowing at resonance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "C = 40e-6;# IN fARADS\n",
- "L = 0.075;# IN Henry\n",
- "V = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency,\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- " #Current at resonance, I\n",
- "I = V/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonant frequency = \",round(fr,2),\" Hz\\n\"\n",
- "print \"\\n (b)Current at resonance, I is \",I,\" A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonant frequency = 91.89 Hz\n",
- "\n",
- "\n",
- " (b)Current at resonance, I is 20.0 A\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 493</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of capacitor C for series resonance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 8;# in ohms\n",
- "L = 0.010;# IN Henry\n",
- "f = 1000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #At resonance\n",
- " #capacitance C\n",
- "C = 1/(L*(2*math.pi*f)**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance, C is \",round(C*1E6,2),\"uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance, C is 2.53 uF\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 493</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of (a) the stray capacitance CS, and (b) the coil inductance L.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C1 = 1000e-12;# IN fARADS\n",
- "C2 = 500e-12;# IN fARADS\n",
- "fr1 = 92500;# in Hz\n",
- "fr2 = 127800;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #For a series R\u2013L\u2013C circuit the resonant frequency fr is given by:\n",
- " #fr = 1/(2pi*(L*C)**2)\n",
- "Cs = ((C1 - C2)/((fr2/fr1)**2 - 1)) - C2\n",
- "L = 1/((C1 + Cs)*(2*math.pi*fr1)**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)stray capacitance, Cs is \",round(Cs*1E12,2),\"pF\\n\"\n",
- "print \"\\n (b)inductance, L is \",round(L*1000,2),\"mH\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)stray capacitance, Cs is 50.13 pF\n",
- "\n",
- "\n",
- " (b)inductance, L is 2.82 mH\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 497</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for this condition (a) the value of inductance L, (b) the p.d. across each component and (c) the Q-factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "C = 5e-6;# IN fARADS\n",
- "rv = 20;#in volts\n",
- "thetav = 0;# in degrees\n",
- "f = 318.3;# in Hz\n",
- "\n",
- "#calculation:\n",
- "wr = 2*math.pi*f\n",
- " #The maximum voltage across the resistance occurs at resonance when the current is a maximum. \n",
- " #At resonance,L = 1/c*wr**2\n",
- "L = 1/(C*wr**2)\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Current at resonance Ir\n",
- "Ir = V/R\n",
- " #p.d. across resistance, VR\n",
- "VR = Ir*R\n",
- " #inductive reactance, XL\n",
- "XL = wr*L\n",
- " #p.d. across inductance, VL\n",
- "VL = Ir*(1j*XL)\n",
- " #capacitive reactance, Xc\n",
- "Xc = 1/(wr*C)\n",
- " #p.d. across capacitor, Vc\n",
- "Vc = Ir*(-1j*Xc)\n",
- " #Q-factor at resonance, Qr\n",
- "Qr = VL.imag/V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)inductance, L is \",round(L*1000,2),\"mH\\n\"\n",
- "print \"\\n (b)p.d. across resistance, VR is \",VR,\" V, p.d. across inductance, VL \",round( VL.imag,2),\"j V \"\n",
- "print \"and p.d. across capacitor, VC \",round(Vc.imag,2),\" V\\n\"\n",
- "print \"\\n (c)Q-factor at resonance, Qr is \",round(abs(Qr),2),\" \\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)inductance, L is 50.0 mH\n",
- "\n",
- "\n",
- " (b)p.d. across resistance, VR is (20+0j) V, p.d. across inductance, VL 200.01 j V \n",
- "and p.d. across capacitor, VC -200.01 V\n",
- "\n",
- "\n",
- " (c)Q-factor at resonance, Qr is 10.0 \n",
- "\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 502</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the resonant frequency, (b) the value of the p.d. across the capacitor at the resonant frequency, \n",
- "#(c) the frequency at which the p.d. across the capacitor is a maximum, and \n",
- "#(d) the value of the maximum voltage across the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 80;# in ohms\n",
- "C = 0.4e-6;# IN fARADS\n",
- "L = 0.020;# IN Henry\n",
- "Vm = 12;#in volts\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency,\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- "wr = 2*math.pi*fr\n",
- " #Q = wr*L/R\n",
- "Q = wr*L/R\n",
- "Vc = Q*Vm\n",
- " #the frequency f at which VC is a maximum value,\n",
- "f = fr*(1 - (1/(2*Q*Q)))**0.5\n",
- " #the maximum value of the p.d. across the capacitor is given by:\n",
- "Vcm = Vc/((1 - (1/(2*Q*Q)))**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)The resonant frequency is \",round(fr,2),\" Hz\\n\"\n",
- "print \"\\n (b)the value of the p.d. across the capacitor at the resonant frequency \",round(Vc,2),\" V\\n\"\n",
- "print \"\\n (c)the frequency f at which Vc is a maximum value, is \",round(f,2),\" Hz\\n\"\n",
- "print \"\\n (d)the maximum value of the p.d. across the capacitor is \",round(Vcm,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)The resonant frequency is 1779.41 Hz\n",
- "\n",
- "\n",
- " (b)the value of the p.d. across the capacitor at the resonant frequency 33.54 V\n",
- "\n",
- "\n",
- " (c)the frequency f at which Vc is a maximum value, is 1721.52 Hz\n",
- "\n",
- "\n",
- " (d)the maximum value of the p.d. across the capacitor is 34.67 V\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 503</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the overall Q-factor of the circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "QL = 60;# Q-factor\n",
- "Qc = 390;# Q-factor\n",
- "\n",
- "#calculation:\n",
- "QT = QL*Qc/(QL + Qc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the overall Q-factor is \",QT"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the overall Q-factor is 52.0"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 505</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the bandwidth of the filter.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 5;# in ohms\n",
- "L = 0.010;# IN Henry\n",
- "fr = 10000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance is given by\n",
- "Qr = wr*L/R\n",
- " #Since Qr = fr/(f2 - f1),\n",
- "bw = fr/Qr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n bandwidth of the filter is \",round(bw,2),\" Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " bandwidth of the filter is 79.58 Hz\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 507</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the values of (a) the inductance, and (b) the capacitance. Find also (c) the bandwidth,\n",
- "#(d) the lower and upper half-power frequencies and (e) the value of the circuit impedance at the half-power frequencies\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Zr = 50;# in ohms\n",
- "fr = 1200;# in Hz\n",
- "Qr = 30;# Q-factor\n",
- "\n",
- "#calculation:\n",
- " #At resonance the circuit impedance, Z\n",
- "R = Zr\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance is given by Qr = wr*L/R, then L is\n",
- "L = Qr*R/wr\n",
- " #At resonance r*L = 1/(wr*C)\n",
- " #capacitance, C\n",
- "C = 1/(L*wr*wr)\n",
- " #bandwidth,.(f2 \u2212 f1)\n",
- "bw = fr/Qr\n",
- " #upper half-power frequency, f2\n",
- "f2 = (bw + ((bw**2) + 4*(fr**2))**0.5)/2\n",
- " #lower half-power frequency, f1\n",
- "f1 = f2 - bw\n",
- " #At the half-power frequencies, current I\n",
- " #I = 0.707*Ir\n",
- " #Hence impedance\n",
- "Z = (2**0.5)*R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)inductance, L is \",round(L*1000,2),\"mH\\n\"\n",
- "print \"\\n (b)capacitance, C is \",round(C*1E9,2),\"nF\\n\"\n",
- "print \"\\n (c)bandwidth is \",round(bw,2),\" Hz\\n\"\n",
- "print \"\\n (d)the upper half-power frequency, f2 is \",round(f2,2),\" Hz \"\n",
- "print \" and the lower half-power frequency, f1 is \",round(f1,2),\" Hz\\n\"\n",
- "print \"\\n (e)impedance at the half-power frequencies is \",round(Z,2),\" ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)inductance, L is 198.94 mH\n",
- "\n",
- "\n",
- " (b)capacitance, C is 88.42 nF\n",
- "\n",
- "\n",
- " (c)bandwidth is 40.0 Hz\n",
- "\n",
- "\n",
- " (d)the upper half-power frequency, f2 is 1220.17 Hz \n",
- " and the lower half-power frequency, f1 is 1180.17 Hz\n",
- "\n",
- "\n",
- " (e)impedance at the half-power frequencies is 70.71 ohm\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 508</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of (a) the circuit resistance,\n",
- "#(b) the circuit inductance, (c) the circuit capacitance, and\n",
- "#(d) the voltage across the capacitor\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 0.2;# in Volts\n",
- "I = 0.004;# in Amperes\n",
- "fr = 3000;# in Hz\n",
- "Qr = 100;# Q-factor\n",
- "\n",
- "#calculation:\n",
- "wr = 2*math.pi*fr\n",
- " #At resonance, impedance\n",
- "Z = V/I\n",
- " #At resonance the circuit impedance, Z\n",
- "R = Z\n",
- " #Q-factor at resonance is given by Qr = wr*L/R, then L is\n",
- "L = Qr*R/wr\n",
- " #At resonance r*L = 1/(wr*C)\n",
- " #capacitance, C\n",
- "C = 1/(L*wr*wr)\n",
- " #Q-factor at resonance in a series circuit represents the voltage magnification Qr = Vc/V, then Vc is\n",
- "Vc = Qr*V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the circuit resistance is \",round(R,2),\" ohm\\n\"\n",
- "print \"\\n (b)inductance, L is \",round(L*1000,2),\"mH\\n\"\n",
- "print \"\\n (c)capacitance, C is \",round(C*1E9,2),\"nF\\n\"\n",
- "print \"\\n (d)the voltage across the capacitor is \",round(Vc,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the circuit resistance is 50.0 ohm\n",
- "\n",
- "\n",
- " (b)inductance, L is 265.26 mH\n",
- "\n",
- "\n",
- " (c)capacitance, C is 10.61 nF\n",
- "\n",
- "\n",
- " (d)the voltage across the capacitor is 20.0 V\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 509</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, (b) the Q-factor at resonance, (c) the bandwidth,\n",
- "#and (d) the lower and upper -3dB frequencies.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 8.84;# in ohms\n",
- "L = 0.3518;# IN Henry\n",
- "C = 20e-6;# IN fARADS\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency,\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance, Q = wr*L/R\n",
- "Qr = wr*L/R\n",
- " #bandwidth,.(f2 \u2212 f1)\n",
- "bw = fr/Qr\n",
- " #the lower \u22123 dB frequency\n",
- "f1 = fr - bw/2\n",
- " #the upper \u22123 dB frequency\n",
- "f2 = fr + bw/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonant frequency, fr is \",round(fr,2),\" Hz\\n\"\n",
- "print \"\\n (b)Q-factor at resonance is \",round(Qr,2),\"\\n\"\n",
- "print \"\\n (c)Bandwidth is \",round(bw,2),\" Hz\\n\"\n",
- "print \"\\n (d)the lower -3dB frequency, f1 is \",round(f1,2),\" Hz \"\n",
- "print \" and the upper -3dB frequency, f2 is \",round(f2,2),\" Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonant frequency, fr is 60.0 Hz\n",
- "\n",
- "\n",
- " (b)Q-factor at resonance is 15.0 \n",
- "\n",
- "\n",
- " (c)Bandwidth is 4.0 Hz\n",
- "\n",
- "\n",
- " (d)the lower -3dB frequency, f1 is 58.0 Hz \n",
- " and the upper -3dB frequency, f2 is 62.0 Hz\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 511</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the circuit when the input voltage is 7.56/_0\u00b0 V and the frequency is \n",
- "#(a) the resonant frequency, (b) a frequency 3% above the resonant frequency\n",
- "#(c) the impedance of the circuit when the frequency is 3% above the resonant frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 15;# in ohms\n",
- "L = 0.008;# IN Henry\n",
- "C = 0.3e-6;# IN fARADS\n",
- "rv = 7.56;#in volts\n",
- "thetav = 0;# in degrees\n",
- "x = 0.03;\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency,\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- "wr = 2*math.pi*fr\n",
- " #At resonance,\n",
- "Zr = R\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Current at resonance\n",
- "Ir = V/Zr\n",
- " #Q-factor at resonance, Q = wr*L/R\n",
- "Qr = wr*L/R\n",
- " #If the frequency is 3% above fr, then\n",
- "de = x\n",
- "I = Ir/(1 + (2*de*Qr*1j))\n",
- "Z = V/I\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current at resonance, Ir is \",round(abs(Ir),2),\" A\\n\"\n",
- "print \"\\n (b)current flowing in the circuit when frequency 3 percent\"\n",
- "print \" above the resonant frequency is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\\n\"\n",
- "print \"\\n (c)impedance of the circuit when the frequency is 3 percent\"\n",
- "print \" above the resonant frequency is \",round(Z.real,2),\" + (\",round(Z.imag,2),\")i A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current at resonance, Ir is 0.5 A\n",
- "\n",
- "\n",
- " (b)current flowing in the circuit when frequency 3 percent\n",
- " above the resonant frequency is 0.35 + ( -0.23 )i A\n",
- "\n",
- "\n",
- " (c)impedance of the circuit when the frequency is 3 percent\n",
- " above the resonant frequency is 15.0 + ( 9.8 )i A\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_1.ipynb deleted file mode 100755 index 35880e50..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_1.ipynb +++ /dev/null @@ -1,736 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 28: Series resonance and Q-factor</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 492</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine at what frequency resonance occurs, and (b) the current flowing at resonance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "C = 40e-6;# IN fARADS\n",
- "L = 0.075;# IN Henry\n",
- "V = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency,\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- " #Current at resonance, I\n",
- "I = V/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonant frequency = \",round(fr,2),\" Hz\\n\"\n",
- "print \"\\n (b)Current at resonance, I is \",I,\" A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonant frequency = 91.89 Hz\n",
- "\n",
- "\n",
- " (b)Current at resonance, I is 20.0 A\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 493</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of capacitor C for series resonance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 8;# in ohms\n",
- "L = 0.010;# IN Henry\n",
- "f = 1000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #At resonance\n",
- " #capacitance C\n",
- "C = 1/(L*(2*math.pi*f)**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance, C is \",round(C*1E6,2),\"uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance, C is 2.53 uF\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 493</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of (a) the stray capacitance CS, and (b) the coil inductance L.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C1 = 1000e-12;# IN fARADS\n",
- "C2 = 500e-12;# IN fARADS\n",
- "fr1 = 92500;# in Hz\n",
- "fr2 = 127800;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #For a series R\u2013L\u2013C circuit the resonant frequency fr is given by:\n",
- " #fr = 1/(2pi*(L*C)**2)\n",
- "Cs = ((C1 - C2)/((fr2/fr1)**2 - 1)) - C2\n",
- "L = 1/((C1 + Cs)*(2*math.pi*fr1)**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)stray capacitance, Cs is \",round(Cs*1E12,2),\"pF\\n\"\n",
- "print \"\\n (b)inductance, L is \",round(L*1000,2),\"mH\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)stray capacitance, Cs is 50.13 pF\n",
- "\n",
- "\n",
- " (b)inductance, L is 2.82 mH\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 497</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for this condition (a) the value of inductance L, (b) the p.d. across each component and (c) the Q-factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "C = 5e-6;# IN fARADS\n",
- "rv = 20;#in volts\n",
- "thetav = 0;# in degrees\n",
- "f = 318.3;# in Hz\n",
- "\n",
- "#calculation:\n",
- "wr = 2*math.pi*f\n",
- " #The maximum voltage across the resistance occurs at resonance when the current is a maximum. \n",
- " #At resonance,L = 1/c*wr**2\n",
- "L = 1/(C*wr**2)\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Current at resonance Ir\n",
- "Ir = V/R\n",
- " #p.d. across resistance, VR\n",
- "VR = Ir*R\n",
- " #inductive reactance, XL\n",
- "XL = wr*L\n",
- " #p.d. across inductance, VL\n",
- "VL = Ir*(1j*XL)\n",
- " #capacitive reactance, Xc\n",
- "Xc = 1/(wr*C)\n",
- " #p.d. across capacitor, Vc\n",
- "Vc = Ir*(-1j*Xc)\n",
- " #Q-factor at resonance, Qr\n",
- "Qr = VL.imag/V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)inductance, L is \",round(L*1000,2),\"mH\\n\"\n",
- "print \"\\n (b)p.d. across resistance, VR is \",VR,\" V, p.d. across inductance, VL \",round( VL.imag,2),\"j V \"\n",
- "print \"and p.d. across capacitor, VC \",round(Vc.imag,2),\" V\\n\"\n",
- "print \"\\n (c)Q-factor at resonance, Qr is \",round(abs(Qr),2),\" \\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)inductance, L is 50.0 mH\n",
- "\n",
- "\n",
- " (b)p.d. across resistance, VR is (20+0j) V, p.d. across inductance, VL 200.01 j V \n",
- "and p.d. across capacitor, VC -200.01 V\n",
- "\n",
- "\n",
- " (c)Q-factor at resonance, Qr is 10.0 \n",
- "\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 502</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the resonant frequency, (b) the value of the p.d. across the capacitor at the resonant frequency, \n",
- "#(c) the frequency at which the p.d. across the capacitor is a maximum, and \n",
- "#(d) the value of the maximum voltage across the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 80;# in ohms\n",
- "C = 0.4e-6;# IN fARADS\n",
- "L = 0.020;# IN Henry\n",
- "Vm = 12;#in volts\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency,\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- "wr = 2*math.pi*fr\n",
- " #Q = wr*L/R\n",
- "Q = wr*L/R\n",
- "Vc = Q*Vm\n",
- " #the frequency f at which VC is a maximum value,\n",
- "f = fr*(1 - (1/(2*Q*Q)))**0.5\n",
- " #the maximum value of the p.d. across the capacitor is given by:\n",
- "Vcm = Vc/((1 - (1/(2*Q*Q)))**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)The resonant frequency is \",round(fr,2),\" Hz\\n\"\n",
- "print \"\\n (b)the value of the p.d. across the capacitor at the resonant frequency \",round(Vc,2),\" V\\n\"\n",
- "print \"\\n (c)the frequency f at which Vc is a maximum value, is \",round(f,2),\" Hz\\n\"\n",
- "print \"\\n (d)the maximum value of the p.d. across the capacitor is \",round(Vcm,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)The resonant frequency is 1779.41 Hz\n",
- "\n",
- "\n",
- " (b)the value of the p.d. across the capacitor at the resonant frequency 33.54 V\n",
- "\n",
- "\n",
- " (c)the frequency f at which Vc is a maximum value, is 1721.52 Hz\n",
- "\n",
- "\n",
- " (d)the maximum value of the p.d. across the capacitor is 34.67 V\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 503</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the overall Q-factor of the circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "QL = 60;# Q-factor\n",
- "Qc = 390;# Q-factor\n",
- "\n",
- "#calculation:\n",
- "QT = QL*Qc/(QL + Qc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the overall Q-factor is \",QT"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the overall Q-factor is 52.0"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 505</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the bandwidth of the filter.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 5;# in ohms\n",
- "L = 0.010;# IN Henry\n",
- "fr = 10000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance is given by\n",
- "Qr = wr*L/R\n",
- " #Since Qr = fr/(f2 - f1),\n",
- "bw = fr/Qr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n bandwidth of the filter is \",round(bw,2),\" Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " bandwidth of the filter is 79.58 Hz\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 507</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the values of (a) the inductance, and (b) the capacitance. Find also (c) the bandwidth,\n",
- "#(d) the lower and upper half-power frequencies and (e) the value of the circuit impedance at the half-power frequencies\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Zr = 50;# in ohms\n",
- "fr = 1200;# in Hz\n",
- "Qr = 30;# Q-factor\n",
- "\n",
- "#calculation:\n",
- " #At resonance the circuit impedance, Z\n",
- "R = Zr\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance is given by Qr = wr*L/R, then L is\n",
- "L = Qr*R/wr\n",
- " #At resonance r*L = 1/(wr*C)\n",
- " #capacitance, C\n",
- "C = 1/(L*wr*wr)\n",
- " #bandwidth,.(f2 \u2212 f1)\n",
- "bw = fr/Qr\n",
- " #upper half-power frequency, f2\n",
- "f2 = (bw + ((bw**2) + 4*(fr**2))**0.5)/2\n",
- " #lower half-power frequency, f1\n",
- "f1 = f2 - bw\n",
- " #At the half-power frequencies, current I\n",
- " #I = 0.707*Ir\n",
- " #Hence impedance\n",
- "Z = (2**0.5)*R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)inductance, L is \",round(L*1000,2),\"mH\\n\"\n",
- "print \"\\n (b)capacitance, C is \",round(C*1E9,2),\"nF\\n\"\n",
- "print \"\\n (c)bandwidth is \",round(bw,2),\" Hz\\n\"\n",
- "print \"\\n (d)the upper half-power frequency, f2 is \",round(f2,2),\" Hz \"\n",
- "print \" and the lower half-power frequency, f1 is \",round(f1,2),\" Hz\\n\"\n",
- "print \"\\n (e)impedance at the half-power frequencies is \",round(Z,2),\" ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)inductance, L is 198.94 mH\n",
- "\n",
- "\n",
- " (b)capacitance, C is 88.42 nF\n",
- "\n",
- "\n",
- " (c)bandwidth is 40.0 Hz\n",
- "\n",
- "\n",
- " (d)the upper half-power frequency, f2 is 1220.17 Hz \n",
- " and the lower half-power frequency, f1 is 1180.17 Hz\n",
- "\n",
- "\n",
- " (e)impedance at the half-power frequencies is 70.71 ohm\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 508</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of (a) the circuit resistance,\n",
- "#(b) the circuit inductance, (c) the circuit capacitance, and\n",
- "#(d) the voltage across the capacitor\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 0.2;# in Volts\n",
- "I = 0.004;# in Amperes\n",
- "fr = 3000;# in Hz\n",
- "Qr = 100;# Q-factor\n",
- "\n",
- "#calculation:\n",
- "wr = 2*math.pi*fr\n",
- " #At resonance, impedance\n",
- "Z = V/I\n",
- " #At resonance the circuit impedance, Z\n",
- "R = Z\n",
- " #Q-factor at resonance is given by Qr = wr*L/R, then L is\n",
- "L = Qr*R/wr\n",
- " #At resonance r*L = 1/(wr*C)\n",
- " #capacitance, C\n",
- "C = 1/(L*wr*wr)\n",
- " #Q-factor at resonance in a series circuit represents the voltage magnification Qr = Vc/V, then Vc is\n",
- "Vc = Qr*V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the circuit resistance is \",round(R,2),\" ohm\\n\"\n",
- "print \"\\n (b)inductance, L is \",round(L*1000,2),\"mH\\n\"\n",
- "print \"\\n (c)capacitance, C is \",round(C*1E9,2),\"nF\\n\"\n",
- "print \"\\n (d)the voltage across the capacitor is \",round(Vc,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the circuit resistance is 50.0 ohm\n",
- "\n",
- "\n",
- " (b)inductance, L is 265.26 mH\n",
- "\n",
- "\n",
- " (c)capacitance, C is 10.61 nF\n",
- "\n",
- "\n",
- " (d)the voltage across the capacitor is 20.0 V\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 509</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, (b) the Q-factor at resonance, (c) the bandwidth,\n",
- "#and (d) the lower and upper -3dB frequencies.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 8.84;# in ohms\n",
- "L = 0.3518;# IN Henry\n",
- "C = 20e-6;# IN fARADS\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency,\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance, Q = wr*L/R\n",
- "Qr = wr*L/R\n",
- " #bandwidth,.(f2 \u2212 f1)\n",
- "bw = fr/Qr\n",
- " #the lower \u22123 dB frequency\n",
- "f1 = fr - bw/2\n",
- " #the upper \u22123 dB frequency\n",
- "f2 = fr + bw/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonant frequency, fr is \",round(fr,2),\" Hz\\n\"\n",
- "print \"\\n (b)Q-factor at resonance is \",round(Qr,2),\"\\n\"\n",
- "print \"\\n (c)Bandwidth is \",round(bw,2),\" Hz\\n\"\n",
- "print \"\\n (d)the lower -3dB frequency, f1 is \",round(f1,2),\" Hz \"\n",
- "print \" and the upper -3dB frequency, f2 is \",round(f2,2),\" Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonant frequency, fr is 60.0 Hz\n",
- "\n",
- "\n",
- " (b)Q-factor at resonance is 15.0 \n",
- "\n",
- "\n",
- " (c)Bandwidth is 4.0 Hz\n",
- "\n",
- "\n",
- " (d)the lower -3dB frequency, f1 is 58.0 Hz \n",
- " and the upper -3dB frequency, f2 is 62.0 Hz\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 511</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the circuit when the input voltage is 7.56/_0\u00b0 V and the frequency is \n",
- "#(a) the resonant frequency, (b) a frequency 3% above the resonant frequency\n",
- "#(c) the impedance of the circuit when the frequency is 3% above the resonant frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 15;# in ohms\n",
- "L = 0.008;# IN Henry\n",
- "C = 0.3e-6;# IN fARADS\n",
- "rv = 7.56;#in volts\n",
- "thetav = 0;# in degrees\n",
- "x = 0.03;\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency,\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- "wr = 2*math.pi*fr\n",
- " #At resonance,\n",
- "Zr = R\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Current at resonance\n",
- "Ir = V/Zr\n",
- " #Q-factor at resonance, Q = wr*L/R\n",
- "Qr = wr*L/R\n",
- " #If the frequency is 3% above fr, then\n",
- "de = x\n",
- "I = Ir/(1 + (2*de*Qr*1j))\n",
- "Z = V/I\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current at resonance, Ir is \",round(abs(Ir),2),\" A\\n\"\n",
- "print \"\\n (b)current flowing in the circuit when frequency 3 percent\"\n",
- "print \" above the resonant frequency is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\\n\"\n",
- "print \"\\n (c)impedance of the circuit when the frequency is 3 percent\"\n",
- "print \" above the resonant frequency is \",round(Z.real,2),\" + (\",round(Z.imag,2),\")i A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current at resonance, Ir is 0.5 A\n",
- "\n",
- "\n",
- " (b)current flowing in the circuit when frequency 3 percent\n",
- " above the resonant frequency is 0.35 + ( -0.23 )i A\n",
- "\n",
- "\n",
- " (c)impedance of the circuit when the frequency is 3 percent\n",
- " above the resonant frequency is 15.0 + ( 9.8 )i A\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_2.ipynb deleted file mode 100755 index 35880e50..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_2.ipynb +++ /dev/null @@ -1,736 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 28: Series resonance and Q-factor</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 492</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine at what frequency resonance occurs, and (b) the current flowing at resonance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "C = 40e-6;# IN fARADS\n",
- "L = 0.075;# IN Henry\n",
- "V = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency,\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- " #Current at resonance, I\n",
- "I = V/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonant frequency = \",round(fr,2),\" Hz\\n\"\n",
- "print \"\\n (b)Current at resonance, I is \",I,\" A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonant frequency = 91.89 Hz\n",
- "\n",
- "\n",
- " (b)Current at resonance, I is 20.0 A\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 493</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of capacitor C for series resonance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 8;# in ohms\n",
- "L = 0.010;# IN Henry\n",
- "f = 1000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #At resonance\n",
- " #capacitance C\n",
- "C = 1/(L*(2*math.pi*f)**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance, C is \",round(C*1E6,2),\"uF\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance, C is 2.53 uF\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 493</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the values of (a) the stray capacitance CS, and (b) the coil inductance L.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C1 = 1000e-12;# IN fARADS\n",
- "C2 = 500e-12;# IN fARADS\n",
- "fr1 = 92500;# in Hz\n",
- "fr2 = 127800;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #For a series R\u2013L\u2013C circuit the resonant frequency fr is given by:\n",
- " #fr = 1/(2pi*(L*C)**2)\n",
- "Cs = ((C1 - C2)/((fr2/fr1)**2 - 1)) - C2\n",
- "L = 1/((C1 + Cs)*(2*math.pi*fr1)**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)stray capacitance, Cs is \",round(Cs*1E12,2),\"pF\\n\"\n",
- "print \"\\n (b)inductance, L is \",round(L*1000,2),\"mH\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)stray capacitance, Cs is 50.13 pF\n",
- "\n",
- "\n",
- " (b)inductance, L is 2.82 mH\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 497</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for this condition (a) the value of inductance L, (b) the p.d. across each component and (c) the Q-factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "C = 5e-6;# IN fARADS\n",
- "rv = 20;#in volts\n",
- "thetav = 0;# in degrees\n",
- "f = 318.3;# in Hz\n",
- "\n",
- "#calculation:\n",
- "wr = 2*math.pi*f\n",
- " #The maximum voltage across the resistance occurs at resonance when the current is a maximum. \n",
- " #At resonance,L = 1/c*wr**2\n",
- "L = 1/(C*wr**2)\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Current at resonance Ir\n",
- "Ir = V/R\n",
- " #p.d. across resistance, VR\n",
- "VR = Ir*R\n",
- " #inductive reactance, XL\n",
- "XL = wr*L\n",
- " #p.d. across inductance, VL\n",
- "VL = Ir*(1j*XL)\n",
- " #capacitive reactance, Xc\n",
- "Xc = 1/(wr*C)\n",
- " #p.d. across capacitor, Vc\n",
- "Vc = Ir*(-1j*Xc)\n",
- " #Q-factor at resonance, Qr\n",
- "Qr = VL.imag/V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)inductance, L is \",round(L*1000,2),\"mH\\n\"\n",
- "print \"\\n (b)p.d. across resistance, VR is \",VR,\" V, p.d. across inductance, VL \",round( VL.imag,2),\"j V \"\n",
- "print \"and p.d. across capacitor, VC \",round(Vc.imag,2),\" V\\n\"\n",
- "print \"\\n (c)Q-factor at resonance, Qr is \",round(abs(Qr),2),\" \\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)inductance, L is 50.0 mH\n",
- "\n",
- "\n",
- " (b)p.d. across resistance, VR is (20+0j) V, p.d. across inductance, VL 200.01 j V \n",
- "and p.d. across capacitor, VC -200.01 V\n",
- "\n",
- "\n",
- " (c)Q-factor at resonance, Qr is 10.0 \n",
- "\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 502</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the resonant frequency, (b) the value of the p.d. across the capacitor at the resonant frequency, \n",
- "#(c) the frequency at which the p.d. across the capacitor is a maximum, and \n",
- "#(d) the value of the maximum voltage across the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 80;# in ohms\n",
- "C = 0.4e-6;# IN fARADS\n",
- "L = 0.020;# IN Henry\n",
- "Vm = 12;#in volts\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency,\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- "wr = 2*math.pi*fr\n",
- " #Q = wr*L/R\n",
- "Q = wr*L/R\n",
- "Vc = Q*Vm\n",
- " #the frequency f at which VC is a maximum value,\n",
- "f = fr*(1 - (1/(2*Q*Q)))**0.5\n",
- " #the maximum value of the p.d. across the capacitor is given by:\n",
- "Vcm = Vc/((1 - (1/(2*Q*Q)))**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)The resonant frequency is \",round(fr,2),\" Hz\\n\"\n",
- "print \"\\n (b)the value of the p.d. across the capacitor at the resonant frequency \",round(Vc,2),\" V\\n\"\n",
- "print \"\\n (c)the frequency f at which Vc is a maximum value, is \",round(f,2),\" Hz\\n\"\n",
- "print \"\\n (d)the maximum value of the p.d. across the capacitor is \",round(Vcm,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)The resonant frequency is 1779.41 Hz\n",
- "\n",
- "\n",
- " (b)the value of the p.d. across the capacitor at the resonant frequency 33.54 V\n",
- "\n",
- "\n",
- " (c)the frequency f at which Vc is a maximum value, is 1721.52 Hz\n",
- "\n",
- "\n",
- " (d)the maximum value of the p.d. across the capacitor is 34.67 V\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 503</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the overall Q-factor of the circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "QL = 60;# Q-factor\n",
- "Qc = 390;# Q-factor\n",
- "\n",
- "#calculation:\n",
- "QT = QL*Qc/(QL + Qc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the overall Q-factor is \",QT"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the overall Q-factor is 52.0"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 505</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the bandwidth of the filter.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 5;# in ohms\n",
- "L = 0.010;# IN Henry\n",
- "fr = 10000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance is given by\n",
- "Qr = wr*L/R\n",
- " #Since Qr = fr/(f2 - f1),\n",
- "bw = fr/Qr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n bandwidth of the filter is \",round(bw,2),\" Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " bandwidth of the filter is 79.58 Hz\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 507</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the values of (a) the inductance, and (b) the capacitance. Find also (c) the bandwidth,\n",
- "#(d) the lower and upper half-power frequencies and (e) the value of the circuit impedance at the half-power frequencies\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Zr = 50;# in ohms\n",
- "fr = 1200;# in Hz\n",
- "Qr = 30;# Q-factor\n",
- "\n",
- "#calculation:\n",
- " #At resonance the circuit impedance, Z\n",
- "R = Zr\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance is given by Qr = wr*L/R, then L is\n",
- "L = Qr*R/wr\n",
- " #At resonance r*L = 1/(wr*C)\n",
- " #capacitance, C\n",
- "C = 1/(L*wr*wr)\n",
- " #bandwidth,.(f2 \u2212 f1)\n",
- "bw = fr/Qr\n",
- " #upper half-power frequency, f2\n",
- "f2 = (bw + ((bw**2) + 4*(fr**2))**0.5)/2\n",
- " #lower half-power frequency, f1\n",
- "f1 = f2 - bw\n",
- " #At the half-power frequencies, current I\n",
- " #I = 0.707*Ir\n",
- " #Hence impedance\n",
- "Z = (2**0.5)*R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)inductance, L is \",round(L*1000,2),\"mH\\n\"\n",
- "print \"\\n (b)capacitance, C is \",round(C*1E9,2),\"nF\\n\"\n",
- "print \"\\n (c)bandwidth is \",round(bw,2),\" Hz\\n\"\n",
- "print \"\\n (d)the upper half-power frequency, f2 is \",round(f2,2),\" Hz \"\n",
- "print \" and the lower half-power frequency, f1 is \",round(f1,2),\" Hz\\n\"\n",
- "print \"\\n (e)impedance at the half-power frequencies is \",round(Z,2),\" ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)inductance, L is 198.94 mH\n",
- "\n",
- "\n",
- " (b)capacitance, C is 88.42 nF\n",
- "\n",
- "\n",
- " (c)bandwidth is 40.0 Hz\n",
- "\n",
- "\n",
- " (d)the upper half-power frequency, f2 is 1220.17 Hz \n",
- " and the lower half-power frequency, f1 is 1180.17 Hz\n",
- "\n",
- "\n",
- " (e)impedance at the half-power frequencies is 70.71 ohm\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 508</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of (a) the circuit resistance,\n",
- "#(b) the circuit inductance, (c) the circuit capacitance, and\n",
- "#(d) the voltage across the capacitor\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 0.2;# in Volts\n",
- "I = 0.004;# in Amperes\n",
- "fr = 3000;# in Hz\n",
- "Qr = 100;# Q-factor\n",
- "\n",
- "#calculation:\n",
- "wr = 2*math.pi*fr\n",
- " #At resonance, impedance\n",
- "Z = V/I\n",
- " #At resonance the circuit impedance, Z\n",
- "R = Z\n",
- " #Q-factor at resonance is given by Qr = wr*L/R, then L is\n",
- "L = Qr*R/wr\n",
- " #At resonance r*L = 1/(wr*C)\n",
- " #capacitance, C\n",
- "C = 1/(L*wr*wr)\n",
- " #Q-factor at resonance in a series circuit represents the voltage magnification Qr = Vc/V, then Vc is\n",
- "Vc = Qr*V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the circuit resistance is \",round(R,2),\" ohm\\n\"\n",
- "print \"\\n (b)inductance, L is \",round(L*1000,2),\"mH\\n\"\n",
- "print \"\\n (c)capacitance, C is \",round(C*1E9,2),\"nF\\n\"\n",
- "print \"\\n (d)the voltage across the capacitor is \",round(Vc,2),\" V\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the circuit resistance is 50.0 ohm\n",
- "\n",
- "\n",
- " (b)inductance, L is 265.26 mH\n",
- "\n",
- "\n",
- " (c)capacitance, C is 10.61 nF\n",
- "\n",
- "\n",
- " (d)the voltage across the capacitor is 20.0 V\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 509</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, (b) the Q-factor at resonance, (c) the bandwidth,\n",
- "#and (d) the lower and upper -3dB frequencies.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 8.84;# in ohms\n",
- "L = 0.3518;# IN Henry\n",
- "C = 20e-6;# IN fARADS\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency,\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance, Q = wr*L/R\n",
- "Qr = wr*L/R\n",
- " #bandwidth,.(f2 \u2212 f1)\n",
- "bw = fr/Qr\n",
- " #the lower \u22123 dB frequency\n",
- "f1 = fr - bw/2\n",
- " #the upper \u22123 dB frequency\n",
- "f2 = fr + bw/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonant frequency, fr is \",round(fr,2),\" Hz\\n\"\n",
- "print \"\\n (b)Q-factor at resonance is \",round(Qr,2),\"\\n\"\n",
- "print \"\\n (c)Bandwidth is \",round(bw,2),\" Hz\\n\"\n",
- "print \"\\n (d)the lower -3dB frequency, f1 is \",round(f1,2),\" Hz \"\n",
- "print \" and the upper -3dB frequency, f2 is \",round(f2,2),\" Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonant frequency, fr is 60.0 Hz\n",
- "\n",
- "\n",
- " (b)Q-factor at resonance is 15.0 \n",
- "\n",
- "\n",
- " (c)Bandwidth is 4.0 Hz\n",
- "\n",
- "\n",
- " (d)the lower -3dB frequency, f1 is 58.0 Hz \n",
- " and the upper -3dB frequency, f2 is 62.0 Hz\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 511</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the circuit when the input voltage is 7.56/_0\u00b0 V and the frequency is \n",
- "#(a) the resonant frequency, (b) a frequency 3% above the resonant frequency\n",
- "#(c) the impedance of the circuit when the frequency is 3% above the resonant frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 15;# in ohms\n",
- "L = 0.008;# IN Henry\n",
- "C = 0.3e-6;# IN fARADS\n",
- "rv = 7.56;#in volts\n",
- "thetav = 0;# in degrees\n",
- "x = 0.03;\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency,\n",
- "fr = 1/(2*math.pi*((L*C)**0.5))\n",
- "wr = 2*math.pi*fr\n",
- " #At resonance,\n",
- "Zr = R\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Current at resonance\n",
- "Ir = V/Zr\n",
- " #Q-factor at resonance, Q = wr*L/R\n",
- "Qr = wr*L/R\n",
- " #If the frequency is 3% above fr, then\n",
- "de = x\n",
- "I = Ir/(1 + (2*de*Qr*1j))\n",
- "Z = V/I\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Current at resonance, Ir is \",round(abs(Ir),2),\" A\\n\"\n",
- "print \"\\n (b)current flowing in the circuit when frequency 3 percent\"\n",
- "print \" above the resonant frequency is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\\n\"\n",
- "print \"\\n (c)impedance of the circuit when the frequency is 3 percent\"\n",
- "print \" above the resonant frequency is \",round(Z.real,2),\" + (\",round(Z.imag,2),\")i A\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Current at resonance, Ir is 0.5 A\n",
- "\n",
- "\n",
- " (b)current flowing in the circuit when frequency 3 percent\n",
- " above the resonant frequency is 0.35 + ( -0.23 )i A\n",
- "\n",
- "\n",
- " (c)impedance of the circuit when the frequency is 3 percent\n",
- " above the resonant frequency is 15.0 + ( 9.8 )i A\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_28.ipynb index 100af3b8..100af3b8 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_28.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint.ipynb deleted file mode 100755 index d6dbd493..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint.ipynb +++ /dev/null @@ -1,449 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 29: parallel resonance and Q-factor</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 521</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, (b) the dynamic resistance, \n",
- "#(c) the current at resonance, and (d) the circuit Q-factor at resonance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "L = 0.005;# IN Henry\n",
- "C = 0.25e-6;# IN fARADS\n",
- "V = 50;#in volts\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency, for parallel\n",
- "fr = ((1/(L*C) - ((R**2)/(L**2)))**0.5)/(2*math.pi)\n",
- " #dynamic resistance\n",
- "Rd = L/(C*R)\n",
- " #Current at resonance\n",
- "Ir = V/Rd\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance, Q = wr*L/R\n",
- "Qr = wr*L/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonance frequency is \",round(fr,2),\" Hz\\n\"\n",
- "print \"\\n (b)dynamic resistance \",round(Rd,2),\" ohm\\n\"\n",
- "print \"\\n (c)Current at resonance, Ir is \",round(Ir,2),\" A\\n\"\n",
- "print \"\\n (d)Q-factor at resonance is \",round(Qr,2),\"\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonance frequency is 4490.31 Hz\n",
- "\n",
- "\n",
- " (b)dynamic resistance 2000.0 ohm\n",
- "\n",
- "\n",
- " (c)Current at resonance, Ir is 0.02 A\n",
- "\n",
- "\n",
- " (d)Q-factor at resonance is 14.11 \n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 521</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resonant frequency for the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL1 = 0;# in ohms\n",
- "RL2 = 30;# in ohms\n",
- "L = 0.100;# IN Henry\n",
- "C = 40e-6;# IN fARADS\n",
- "V = 50;#in volts\n",
- "\n",
- "#calculation:\n",
- " #for RL1\n",
- " #Resonant frequency,\n",
- "wr1 = (1/(L*C))**0.5\n",
- "fr1 = wr1/(2*math.pi)\n",
- " #for RL2\n",
- " #Resonant frequency,\n",
- "wr2 = (1/(L*C) - ((RL2**2)/(L**2)))**0.5\n",
- "fr2 = wr2/(2*math.pi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonance frequency at RL = 0 is \",round(fr1,2),\" Hz\"\n",
- "print \"\\n (b)Resonance frequency at RL = 30 ohm is \",round(fr2,2),\" Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonance frequency at RL = 0 is 79.58 Hz\n",
- "\n",
- " (b)Resonance frequency at RL = 30 ohm is 63.66 Hz\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 523</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the condition when the supply current is a minimum, \n",
- "#(a) the capacitance of the capacitor, (b) the dynamic resistance, \n",
- "#(c) the supply current, (d) the Q-factor, (e) the bandwidth,\n",
- "#(f) the upper and lower \u00033 dB frequencies, and (g) the value of the circuit impedance at the \u00033 dB frequencies\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 150;# in ohms\n",
- "L = 0.120;# IN Henry\n",
- "V = 20;#in volts\n",
- "fr = 4000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #capacitance, C\n",
- "C = 1/(L*((2*math.pi*fr)**2 + ((R**2)/(L**2))))\n",
- "Rd = L/(C*R)\n",
- " #Current at resonance\n",
- "Ir = V/Rd\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance, Q = wr*L/R\n",
- "Qr = wr*L/R\n",
- " #bandwidth,.(f2 \u2212 f1)\n",
- "bw = fr/Qr\n",
- " #upper half-power frequency, f2\n",
- "f2 = (bw + ((bw**2) + 4*(fr**2))**0.5)/2\n",
- " #lower half-power frequency, f1\n",
- "f1 = f2 - bw\n",
- " #impedance at the \u22123 dB frequencies\n",
- "Z = Rd/(2**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the capacitance of the capacitor,C is \",round(C*1E6,2),\"uF\"\n",
- "print \"\\n (b)dynamic resistance \",round(Rd,2),\"ohm\\n\"\n",
- "print \"\\n (c)Current at resonance, Ir is \",round(Ir*1000,2),\"mA\\n\"\n",
- "print \"\\n (d)Q-factor at resonance is \",round(Qr,2),\"\\n\"\n",
- "print \"\\n (e)bandwidth is \",round(bw,2),\" Hz\\n\"\n",
- "print \"\\n (f)the upper half-power frequency, f2 is \",round(f2,2),\" Hz and \"\n",
- "print \" the lower half-power frequency, f1 is \",round(f1,2),\" Hz\\n\"\n",
- "print \"\\n (g)impedance at the -3 dB frequencies is \",round(Z,2),\" ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the capacitance of the capacitor,C is 0.01 uF\n",
- "\n",
- " (b)dynamic resistance 60788.85 ohm\n",
- "\n",
- "\n",
- " (c)Current at resonance, Ir is 0.33 mA\n",
- "\n",
- "\n",
- " (d)Q-factor at resonance is 20.11 \n",
- "\n",
- "\n",
- " (e)bandwidth is 198.94 Hz\n",
- "\n",
- "\n",
- " (f)the upper half-power frequency, f2 is 4100.71 Hz and \n",
- " the lower half-power frequency, f1 is 3901.76 Hz\n",
- "\n",
- "\n",
- " (g)impedance at the -3 dB frequencies is 42984.21 ohm\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 525</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resonant frequency of the network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL = 5;# in ohms\n",
- "L = 0.002;# IN Henry\n",
- "C = 25e-6;# IN fARADS\n",
- "Rc = 3;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency, for parallel\n",
- "fr = (1/(2*math.pi*((L*C)**0.5)))*((RL**2 - (L/C))/(Rc**2 - (L/C)))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n resonant frequency, fr is \",round(fr,2),\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " resonant frequency, fr is 626.45 Hz"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 525</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the parallel network the values of inductance L\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL = 3;# in ohms\n",
- "fr = 1000;# in Hz\n",
- "Xc = 10;# IN ohms\n",
- "Rc = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "XL1 = (((Rc**2 + Xc**2)/Xc) + ((((Rc**2 + Xc**2)/Xc)**2) - 4*(RL**2))**0.5)/2\n",
- "XL2 = (((Rc**2 + Xc**2)/Xc) - ((((Rc**2 + Xc**2)/Xc)**2) - 4*(RL**2))**0.5)/2\n",
- "wr = 2*math.pi*fr\n",
- " #inductance\n",
- "L1 = XL1/wr\n",
- "L2 = XL2/wr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n inductance is either \",round(L1*1000,2),\"mH or \",round(L2*1000,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " inductance is either 1.71 mH or 0.13 mH"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 526</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the overall Q-factor of the parallel combination.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "QL = 60;# Q-factor\n",
- "Qc = 300;# Q-factor\n",
- "\n",
- "#calculation:\n",
- "QT = QL*Qc/(QL + Qc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the overall Q-factor is \",round(QT,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the overall Q-factor is 50.0"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 527</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the circuit (a) the Q-factor, (b) the dynamic resistance, and\n",
- "#(c) the magnitude of the impedance when the supply frequency is 0.4% greater than the tuned frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 10.61E-9;# in Farad\n",
- "bw = 500;# in Hz\n",
- "fr = 150000;# in Hz\n",
- "x = 0.004\n",
- "\n",
- "#calculation:\n",
- " #Q-factor\n",
- "Q = fr/bw\n",
- "wr = 2*math.pi*fr\n",
- " #dynamic resistance, RD\n",
- "Rd = Q/(C*wr)\n",
- "de = x\n",
- "Z = Rd/(1 + (2*de*Q*1j))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Q-factor \",round(Q,2),\"\"\n",
- "print \"\\n (b)dynamic resistance \",round(Rd,2),\"ohm\"\n",
- "print \"\\n (c)magnitude of the impedance \",round(abs(Z),2),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Q-factor 300.0 \n",
- "\n",
- " (b)dynamic resistance 30000.93 ohm\n",
- "\n",
- " (c)magnitude of the impedance 11538.82 ohm"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_1.ipynb deleted file mode 100755 index d6dbd493..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_1.ipynb +++ /dev/null @@ -1,449 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 29: parallel resonance and Q-factor</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 521</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, (b) the dynamic resistance, \n",
- "#(c) the current at resonance, and (d) the circuit Q-factor at resonance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "L = 0.005;# IN Henry\n",
- "C = 0.25e-6;# IN fARADS\n",
- "V = 50;#in volts\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency, for parallel\n",
- "fr = ((1/(L*C) - ((R**2)/(L**2)))**0.5)/(2*math.pi)\n",
- " #dynamic resistance\n",
- "Rd = L/(C*R)\n",
- " #Current at resonance\n",
- "Ir = V/Rd\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance, Q = wr*L/R\n",
- "Qr = wr*L/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonance frequency is \",round(fr,2),\" Hz\\n\"\n",
- "print \"\\n (b)dynamic resistance \",round(Rd,2),\" ohm\\n\"\n",
- "print \"\\n (c)Current at resonance, Ir is \",round(Ir,2),\" A\\n\"\n",
- "print \"\\n (d)Q-factor at resonance is \",round(Qr,2),\"\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonance frequency is 4490.31 Hz\n",
- "\n",
- "\n",
- " (b)dynamic resistance 2000.0 ohm\n",
- "\n",
- "\n",
- " (c)Current at resonance, Ir is 0.02 A\n",
- "\n",
- "\n",
- " (d)Q-factor at resonance is 14.11 \n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 521</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resonant frequency for the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL1 = 0;# in ohms\n",
- "RL2 = 30;# in ohms\n",
- "L = 0.100;# IN Henry\n",
- "C = 40e-6;# IN fARADS\n",
- "V = 50;#in volts\n",
- "\n",
- "#calculation:\n",
- " #for RL1\n",
- " #Resonant frequency,\n",
- "wr1 = (1/(L*C))**0.5\n",
- "fr1 = wr1/(2*math.pi)\n",
- " #for RL2\n",
- " #Resonant frequency,\n",
- "wr2 = (1/(L*C) - ((RL2**2)/(L**2)))**0.5\n",
- "fr2 = wr2/(2*math.pi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonance frequency at RL = 0 is \",round(fr1,2),\" Hz\"\n",
- "print \"\\n (b)Resonance frequency at RL = 30 ohm is \",round(fr2,2),\" Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonance frequency at RL = 0 is 79.58 Hz\n",
- "\n",
- " (b)Resonance frequency at RL = 30 ohm is 63.66 Hz\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 523</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the condition when the supply current is a minimum, \n",
- "#(a) the capacitance of the capacitor, (b) the dynamic resistance, \n",
- "#(c) the supply current, (d) the Q-factor, (e) the bandwidth,\n",
- "#(f) the upper and lower \u00033 dB frequencies, and (g) the value of the circuit impedance at the \u00033 dB frequencies\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 150;# in ohms\n",
- "L = 0.120;# IN Henry\n",
- "V = 20;#in volts\n",
- "fr = 4000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #capacitance, C\n",
- "C = 1/(L*((2*math.pi*fr)**2 + ((R**2)/(L**2))))\n",
- "Rd = L/(C*R)\n",
- " #Current at resonance\n",
- "Ir = V/Rd\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance, Q = wr*L/R\n",
- "Qr = wr*L/R\n",
- " #bandwidth,.(f2 \u2212 f1)\n",
- "bw = fr/Qr\n",
- " #upper half-power frequency, f2\n",
- "f2 = (bw + ((bw**2) + 4*(fr**2))**0.5)/2\n",
- " #lower half-power frequency, f1\n",
- "f1 = f2 - bw\n",
- " #impedance at the \u22123 dB frequencies\n",
- "Z = Rd/(2**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the capacitance of the capacitor,C is \",round(C*1E6,2),\"uF\"\n",
- "print \"\\n (b)dynamic resistance \",round(Rd,2),\"ohm\\n\"\n",
- "print \"\\n (c)Current at resonance, Ir is \",round(Ir*1000,2),\"mA\\n\"\n",
- "print \"\\n (d)Q-factor at resonance is \",round(Qr,2),\"\\n\"\n",
- "print \"\\n (e)bandwidth is \",round(bw,2),\" Hz\\n\"\n",
- "print \"\\n (f)the upper half-power frequency, f2 is \",round(f2,2),\" Hz and \"\n",
- "print \" the lower half-power frequency, f1 is \",round(f1,2),\" Hz\\n\"\n",
- "print \"\\n (g)impedance at the -3 dB frequencies is \",round(Z,2),\" ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the capacitance of the capacitor,C is 0.01 uF\n",
- "\n",
- " (b)dynamic resistance 60788.85 ohm\n",
- "\n",
- "\n",
- " (c)Current at resonance, Ir is 0.33 mA\n",
- "\n",
- "\n",
- " (d)Q-factor at resonance is 20.11 \n",
- "\n",
- "\n",
- " (e)bandwidth is 198.94 Hz\n",
- "\n",
- "\n",
- " (f)the upper half-power frequency, f2 is 4100.71 Hz and \n",
- " the lower half-power frequency, f1 is 3901.76 Hz\n",
- "\n",
- "\n",
- " (g)impedance at the -3 dB frequencies is 42984.21 ohm\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 525</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resonant frequency of the network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL = 5;# in ohms\n",
- "L = 0.002;# IN Henry\n",
- "C = 25e-6;# IN fARADS\n",
- "Rc = 3;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency, for parallel\n",
- "fr = (1/(2*math.pi*((L*C)**0.5)))*((RL**2 - (L/C))/(Rc**2 - (L/C)))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n resonant frequency, fr is \",round(fr,2),\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " resonant frequency, fr is 626.45 Hz"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 525</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the parallel network the values of inductance L\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL = 3;# in ohms\n",
- "fr = 1000;# in Hz\n",
- "Xc = 10;# IN ohms\n",
- "Rc = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "XL1 = (((Rc**2 + Xc**2)/Xc) + ((((Rc**2 + Xc**2)/Xc)**2) - 4*(RL**2))**0.5)/2\n",
- "XL2 = (((Rc**2 + Xc**2)/Xc) - ((((Rc**2 + Xc**2)/Xc)**2) - 4*(RL**2))**0.5)/2\n",
- "wr = 2*math.pi*fr\n",
- " #inductance\n",
- "L1 = XL1/wr\n",
- "L2 = XL2/wr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n inductance is either \",round(L1*1000,2),\"mH or \",round(L2*1000,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " inductance is either 1.71 mH or 0.13 mH"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 526</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the overall Q-factor of the parallel combination.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "QL = 60;# Q-factor\n",
- "Qc = 300;# Q-factor\n",
- "\n",
- "#calculation:\n",
- "QT = QL*Qc/(QL + Qc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the overall Q-factor is \",round(QT,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the overall Q-factor is 50.0"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 527</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the circuit (a) the Q-factor, (b) the dynamic resistance, and\n",
- "#(c) the magnitude of the impedance when the supply frequency is 0.4% greater than the tuned frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 10.61E-9;# in Farad\n",
- "bw = 500;# in Hz\n",
- "fr = 150000;# in Hz\n",
- "x = 0.004\n",
- "\n",
- "#calculation:\n",
- " #Q-factor\n",
- "Q = fr/bw\n",
- "wr = 2*math.pi*fr\n",
- " #dynamic resistance, RD\n",
- "Rd = Q/(C*wr)\n",
- "de = x\n",
- "Z = Rd/(1 + (2*de*Q*1j))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Q-factor \",round(Q,2),\"\"\n",
- "print \"\\n (b)dynamic resistance \",round(Rd,2),\"ohm\"\n",
- "print \"\\n (c)magnitude of the impedance \",round(abs(Z),2),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Q-factor 300.0 \n",
- "\n",
- " (b)dynamic resistance 30000.93 ohm\n",
- "\n",
- " (c)magnitude of the impedance 11538.82 ohm"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_2.ipynb deleted file mode 100755 index d6dbd493..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_2.ipynb +++ /dev/null @@ -1,449 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 29: parallel resonance and Q-factor</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 521</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, (b) the dynamic resistance, \n",
- "#(c) the current at resonance, and (d) the circuit Q-factor at resonance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 10;# in ohms\n",
- "L = 0.005;# IN Henry\n",
- "C = 0.25e-6;# IN fARADS\n",
- "V = 50;#in volts\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency, for parallel\n",
- "fr = ((1/(L*C) - ((R**2)/(L**2)))**0.5)/(2*math.pi)\n",
- " #dynamic resistance\n",
- "Rd = L/(C*R)\n",
- " #Current at resonance\n",
- "Ir = V/Rd\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance, Q = wr*L/R\n",
- "Qr = wr*L/R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonance frequency is \",round(fr,2),\" Hz\\n\"\n",
- "print \"\\n (b)dynamic resistance \",round(Rd,2),\" ohm\\n\"\n",
- "print \"\\n (c)Current at resonance, Ir is \",round(Ir,2),\" A\\n\"\n",
- "print \"\\n (d)Q-factor at resonance is \",round(Qr,2),\"\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonance frequency is 4490.31 Hz\n",
- "\n",
- "\n",
- " (b)dynamic resistance 2000.0 ohm\n",
- "\n",
- "\n",
- " (c)Current at resonance, Ir is 0.02 A\n",
- "\n",
- "\n",
- " (d)Q-factor at resonance is 14.11 \n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 521</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resonant frequency for the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL1 = 0;# in ohms\n",
- "RL2 = 30;# in ohms\n",
- "L = 0.100;# IN Henry\n",
- "C = 40e-6;# IN fARADS\n",
- "V = 50;#in volts\n",
- "\n",
- "#calculation:\n",
- " #for RL1\n",
- " #Resonant frequency,\n",
- "wr1 = (1/(L*C))**0.5\n",
- "fr1 = wr1/(2*math.pi)\n",
- " #for RL2\n",
- " #Resonant frequency,\n",
- "wr2 = (1/(L*C) - ((RL2**2)/(L**2)))**0.5\n",
- "fr2 = wr2/(2*math.pi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Resonance frequency at RL = 0 is \",round(fr1,2),\" Hz\"\n",
- "print \"\\n (b)Resonance frequency at RL = 30 ohm is \",round(fr2,2),\" Hz\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Resonance frequency at RL = 0 is 79.58 Hz\n",
- "\n",
- " (b)Resonance frequency at RL = 30 ohm is 63.66 Hz\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 523</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the condition when the supply current is a minimum, \n",
- "#(a) the capacitance of the capacitor, (b) the dynamic resistance, \n",
- "#(c) the supply current, (d) the Q-factor, (e) the bandwidth,\n",
- "#(f) the upper and lower \u00033 dB frequencies, and (g) the value of the circuit impedance at the \u00033 dB frequencies\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 150;# in ohms\n",
- "L = 0.120;# IN Henry\n",
- "V = 20;#in volts\n",
- "fr = 4000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #capacitance, C\n",
- "C = 1/(L*((2*math.pi*fr)**2 + ((R**2)/(L**2))))\n",
- "Rd = L/(C*R)\n",
- " #Current at resonance\n",
- "Ir = V/Rd\n",
- "wr = 2*math.pi*fr\n",
- " #Q-factor at resonance, Q = wr*L/R\n",
- "Qr = wr*L/R\n",
- " #bandwidth,.(f2 \u2212 f1)\n",
- "bw = fr/Qr\n",
- " #upper half-power frequency, f2\n",
- "f2 = (bw + ((bw**2) + 4*(fr**2))**0.5)/2\n",
- " #lower half-power frequency, f1\n",
- "f1 = f2 - bw\n",
- " #impedance at the \u22123 dB frequencies\n",
- "Z = Rd/(2**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the capacitance of the capacitor,C is \",round(C*1E6,2),\"uF\"\n",
- "print \"\\n (b)dynamic resistance \",round(Rd,2),\"ohm\\n\"\n",
- "print \"\\n (c)Current at resonance, Ir is \",round(Ir*1000,2),\"mA\\n\"\n",
- "print \"\\n (d)Q-factor at resonance is \",round(Qr,2),\"\\n\"\n",
- "print \"\\n (e)bandwidth is \",round(bw,2),\" Hz\\n\"\n",
- "print \"\\n (f)the upper half-power frequency, f2 is \",round(f2,2),\" Hz and \"\n",
- "print \" the lower half-power frequency, f1 is \",round(f1,2),\" Hz\\n\"\n",
- "print \"\\n (g)impedance at the -3 dB frequencies is \",round(Z,2),\" ohm\\n\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the capacitance of the capacitor,C is 0.01 uF\n",
- "\n",
- " (b)dynamic resistance 60788.85 ohm\n",
- "\n",
- "\n",
- " (c)Current at resonance, Ir is 0.33 mA\n",
- "\n",
- "\n",
- " (d)Q-factor at resonance is 20.11 \n",
- "\n",
- "\n",
- " (e)bandwidth is 198.94 Hz\n",
- "\n",
- "\n",
- " (f)the upper half-power frequency, f2 is 4100.71 Hz and \n",
- " the lower half-power frequency, f1 is 3901.76 Hz\n",
- "\n",
- "\n",
- " (g)impedance at the -3 dB frequencies is 42984.21 ohm\n",
- "\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 525</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resonant frequency of the network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL = 5;# in ohms\n",
- "L = 0.002;# IN Henry\n",
- "C = 25e-6;# IN fARADS\n",
- "Rc = 3;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Resonant frequency, for parallel\n",
- "fr = (1/(2*math.pi*((L*C)**0.5)))*((RL**2 - (L/C))/(Rc**2 - (L/C)))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n resonant frequency, fr is \",round(fr,2),\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " resonant frequency, fr is 626.45 Hz"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 525</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the parallel network the values of inductance L\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "RL = 3;# in ohms\n",
- "fr = 1000;# in Hz\n",
- "Xc = 10;# IN ohms\n",
- "Rc = 4;# in ohms\n",
- "\n",
- "#calculation:\n",
- "XL1 = (((Rc**2 + Xc**2)/Xc) + ((((Rc**2 + Xc**2)/Xc)**2) - 4*(RL**2))**0.5)/2\n",
- "XL2 = (((Rc**2 + Xc**2)/Xc) - ((((Rc**2 + Xc**2)/Xc)**2) - 4*(RL**2))**0.5)/2\n",
- "wr = 2*math.pi*fr\n",
- " #inductance\n",
- "L1 = XL1/wr\n",
- "L2 = XL2/wr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n inductance is either \",round(L1*1000,2),\"mH or \",round(L2*1000,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " inductance is either 1.71 mH or 0.13 mH"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 526</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the overall Q-factor of the parallel combination.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "QL = 60;# Q-factor\n",
- "Qc = 300;# Q-factor\n",
- "\n",
- "#calculation:\n",
- "QT = QL*Qc/(QL + Qc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the overall Q-factor is \",round(QT,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the overall Q-factor is 50.0"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 527</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the circuit (a) the Q-factor, (b) the dynamic resistance, and\n",
- "#(c) the magnitude of the impedance when the supply frequency is 0.4% greater than the tuned frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 10.61E-9;# in Farad\n",
- "bw = 500;# in Hz\n",
- "fr = 150000;# in Hz\n",
- "x = 0.004\n",
- "\n",
- "#calculation:\n",
- " #Q-factor\n",
- "Q = fr/bw\n",
- "wr = 2*math.pi*fr\n",
- " #dynamic resistance, RD\n",
- "Rd = Q/(C*wr)\n",
- "de = x\n",
- "Z = Rd/(1 + (2*de*Q*1j))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)Q-factor \",round(Q,2),\"\"\n",
- "print \"\\n (b)dynamic resistance \",round(Rd,2),\"ohm\"\n",
- "print \"\\n (c)magnitude of the impedance \",round(abs(Z),2),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)Q-factor 300.0 \n",
- "\n",
- " (b)dynamic resistance 30000.93 ohm\n",
- "\n",
- " (c)magnitude of the impedance 11538.82 ohm"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_29.ipynb index 169830ec..169830ec 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_29.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint.ipynb deleted file mode 100755 index b23cbc8e..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint.ipynb +++ /dev/null @@ -1,303 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 30: Introduction to network analysis</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 536</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the current flowing in each branch of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 100;# in volts\n",
- "rv2 = 50;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 25;# in ohm\n",
- "R2 = 20;# in ohm\n",
- "R3 = 10;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The branch currents and their directions are labelled as shown in Figure 30.4\n",
- " #Two loops are chosen. loop ABEF, and loop BCDE\n",
- " #using kirchoff rule in 3 loops\n",
- " #two eqns obtained\n",
- " #(R1 + R2)*I1 + R2*I2 = V1\n",
- " #R2*I1 + (R2 + R3)*I2 = V2\n",
- "I1 = (3*V1 - 2*V2)/(3*(R1 + R2) - 2*(R2))\n",
- "I2 = (V2 - R2*I1)/(R2 + R3)\n",
- "I = I1 + I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A, \\n current, I2 is \",round(I2.real,2),\" + (\",round( I2.imag,2),\")i A and \"\n",
- "print \" total current, I is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I1 is 3.16 + ( -1.05 )i A, \n",
- " current, I2 is -2.11 + ( 2.37 )i A and \n",
- " total current, I is 1.05 + ( 1.32 )i A\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 537</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the 2 ohm resistor of the circuit\n",
- "#find the power dissipated in the 3 ohm resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "#initializing the variables:\n",
- "V = 8;# in volts\n",
- "R1 = 1;# in ohm\n",
- "R2 = 2;# in ohm\n",
- "R3 = 3;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "R5 = 5;# in ohm\n",
- "R6 = 6;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #Currents and their directions are assigned as shown in Figure 30.6.\n",
- " #Three loops are chosen since three unknown currents are required. The choice of loop directions is arbitrary.\n",
- " #loop ABCDE, and loop EDGF and loop DCHG\n",
- " #using kirchoff rule in 3 loops\n",
- " #three eqns obtained\n",
- " #R5*I1 + (R6 + R4)*I2 - R4*I3 = V\n",
- " #-1*R1*I1 + (R6 + R1)*I2 + R2*I3 = 0\n",
- " # R3*I1 - (R3 + R4)*I2 + (R2 + R3 + R4)*I3 = 0\n",
- "#using determinants\n",
- "d1 = [[V, (R6 + R4), -1*R4],[0, (R6 + R1), R2], [0, (-1*(R3 + R4)), (R2 + R3 + R4)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[R5, V, -1*R4],[-1*R1, 0, R2],[ R3, 0, (R2 + R3 + R4)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[R5, (R6 + R4), V],[-1*R1, (R6 + R1), 0],[ R3, (-1*(R3 + R4)), 0]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d = [[R5, (R6 + R4), -1*R4],[-1*R1, (R6 + R1), R2],[ R3, (-1*(R3 + R4)), (R2 + R3 + R4)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I3 = D3/D \n",
- "#Current in the 2 ohm resistance\n",
- "I = I1 - I2 + I3\n",
- "#power dissipated in the 3 ohm resistance\n",
- "P3 = R3*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current through 2 ohm resistor is \",round(I2,3),\" A\"\n",
- "print \"\\n (b)power dissipated in the 3 ohm resistor is \",round(P3,3),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current through 2 ohm resistor is 0.203 A\n",
- "\n",
- " (b)power dissipated in the 3 ohm resistor is 1.267 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 539</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in each branch using Kirchhoff\u2019s laws.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 5 + 0j;# in volts\n",
- "E2 = 2 + 4j;# in volts\n",
- "Z1 = 3 + 4j;# in ohm\n",
- "Z2 = 2 - 5j;# in ohm\n",
- "Z3 = 6 + 8j;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #Currents I1 and I2 with their directions are shown in Figure 30.8.\n",
- " #Two loops are chosen with their directions both clockwise.loop ABEF and loop BCDE,\n",
- " #using kirchoff rule in 3 loops\n",
- " #two eqns obtained\n",
- " #(Z1 + Z3)*I1 - Z3*I2 = E1\n",
- " #-1*Z3*I1 + (Z2 + Z3)*I2 = E2\n",
- "I1 = ((Z2 + Z3)*E1 + Z3*E2)/((Z2 + Z3)*(Z1 + Z3) - Z3*Z3)\n",
- "I2 = -1*(E1 - (Z1 + Z3)*I1)/Z3\n",
- "I3 = I1 - I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"current, I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A,\\n current, I2 is \",round(I2.real,2),\" + (\",round( I2.imag,2),\")i A and \"\n",
- "print \" current, I3 is \",round(I3.real,2),\" + (\",round( I3.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "current, I1 is 0.57 + ( 0.62 )i A,\n",
- " current, I2 is 0.56 + ( 1.33 )i A and \n",
- " current, I3 is 0.01 + ( -0.71 )i A\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 541</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the magnitude of the current in the (4 + j3)impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 10;# in volts\n",
- "rv2 = 12;# in volts\n",
- "rv3 = 15;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "thetav3 = 0;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = -5j;# in ohm\n",
- "R3 = 8;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "R5 = 3j;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- "V3 = rv3*math.cos(thetav3*math.pi/180) + 1j*rv3*math.sin(thetav3*math.pi/180)\n",
- " #Currents I1, I2 and I3 with their directions are shown in Figure 30.10.\n",
- " #Three loops are chosen. The choice of loop directions is arbitrary. loop ABGH, and loopBCFG and loop CDEF\n",
- "Z4 = R4 + R5\n",
- " #using kirchoff rule in 3 loops\n",
- " #three eqns obtained\n",
- " #R1*I1 + R2*I2 = V1 + V2\n",
- " #-1*R3*I1 + (R3 + R2)*I2 + R3*I3 = V2 + V3\n",
- " # -1*R3*I1 + R3*I2 + (R3 + Z4)*I3 = V3\n",
- " #using determinants\n",
- "d1 = [[(V1 + V2), R2, 0],[(V2 + V3), (R3 + R2), R3],[V3, R3, (R3 + Z4)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[R1, (V1 + V2), 0],[-1*R3, (V2 + V3), R3],[-1*R3, V3, (R3 + Z4)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[R1, R2, (V1 + V2)],[-1*R3, (R3 + R2), (V2 + V3)],[-1*R3, R3, V3]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d = [[R1, R2, 0],[-1*R3, (R3 + R2), R3],[-1*R3, R3, (R3 + Z4)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I3 = D3/D \n",
- "I3mag = abs(I3)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n magnitude of the current through (4 + i3)ohm impedance is \",round(I3mag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " magnitude of the current through (4 + i3)ohm impedance is 1.84 A"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_1.ipynb deleted file mode 100755 index b23cbc8e..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_1.ipynb +++ /dev/null @@ -1,303 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 30: Introduction to network analysis</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 536</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the current flowing in each branch of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 100;# in volts\n",
- "rv2 = 50;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 25;# in ohm\n",
- "R2 = 20;# in ohm\n",
- "R3 = 10;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The branch currents and their directions are labelled as shown in Figure 30.4\n",
- " #Two loops are chosen. loop ABEF, and loop BCDE\n",
- " #using kirchoff rule in 3 loops\n",
- " #two eqns obtained\n",
- " #(R1 + R2)*I1 + R2*I2 = V1\n",
- " #R2*I1 + (R2 + R3)*I2 = V2\n",
- "I1 = (3*V1 - 2*V2)/(3*(R1 + R2) - 2*(R2))\n",
- "I2 = (V2 - R2*I1)/(R2 + R3)\n",
- "I = I1 + I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A, \\n current, I2 is \",round(I2.real,2),\" + (\",round( I2.imag,2),\")i A and \"\n",
- "print \" total current, I is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I1 is 3.16 + ( -1.05 )i A, \n",
- " current, I2 is -2.11 + ( 2.37 )i A and \n",
- " total current, I is 1.05 + ( 1.32 )i A\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 537</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the 2 ohm resistor of the circuit\n",
- "#find the power dissipated in the 3 ohm resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "#initializing the variables:\n",
- "V = 8;# in volts\n",
- "R1 = 1;# in ohm\n",
- "R2 = 2;# in ohm\n",
- "R3 = 3;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "R5 = 5;# in ohm\n",
- "R6 = 6;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #Currents and their directions are assigned as shown in Figure 30.6.\n",
- " #Three loops are chosen since three unknown currents are required. The choice of loop directions is arbitrary.\n",
- " #loop ABCDE, and loop EDGF and loop DCHG\n",
- " #using kirchoff rule in 3 loops\n",
- " #three eqns obtained\n",
- " #R5*I1 + (R6 + R4)*I2 - R4*I3 = V\n",
- " #-1*R1*I1 + (R6 + R1)*I2 + R2*I3 = 0\n",
- " # R3*I1 - (R3 + R4)*I2 + (R2 + R3 + R4)*I3 = 0\n",
- "#using determinants\n",
- "d1 = [[V, (R6 + R4), -1*R4],[0, (R6 + R1), R2], [0, (-1*(R3 + R4)), (R2 + R3 + R4)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[R5, V, -1*R4],[-1*R1, 0, R2],[ R3, 0, (R2 + R3 + R4)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[R5, (R6 + R4), V],[-1*R1, (R6 + R1), 0],[ R3, (-1*(R3 + R4)), 0]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d = [[R5, (R6 + R4), -1*R4],[-1*R1, (R6 + R1), R2],[ R3, (-1*(R3 + R4)), (R2 + R3 + R4)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I3 = D3/D \n",
- "#Current in the 2 ohm resistance\n",
- "I = I1 - I2 + I3\n",
- "#power dissipated in the 3 ohm resistance\n",
- "P3 = R3*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current through 2 ohm resistor is \",round(I2,3),\" A\"\n",
- "print \"\\n (b)power dissipated in the 3 ohm resistor is \",round(P3,3),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current through 2 ohm resistor is 0.203 A\n",
- "\n",
- " (b)power dissipated in the 3 ohm resistor is 1.267 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 539</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in each branch using Kirchhoff\u2019s laws.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 5 + 0j;# in volts\n",
- "E2 = 2 + 4j;# in volts\n",
- "Z1 = 3 + 4j;# in ohm\n",
- "Z2 = 2 - 5j;# in ohm\n",
- "Z3 = 6 + 8j;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #Currents I1 and I2 with their directions are shown in Figure 30.8.\n",
- " #Two loops are chosen with their directions both clockwise.loop ABEF and loop BCDE,\n",
- " #using kirchoff rule in 3 loops\n",
- " #two eqns obtained\n",
- " #(Z1 + Z3)*I1 - Z3*I2 = E1\n",
- " #-1*Z3*I1 + (Z2 + Z3)*I2 = E2\n",
- "I1 = ((Z2 + Z3)*E1 + Z3*E2)/((Z2 + Z3)*(Z1 + Z3) - Z3*Z3)\n",
- "I2 = -1*(E1 - (Z1 + Z3)*I1)/Z3\n",
- "I3 = I1 - I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"current, I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A,\\n current, I2 is \",round(I2.real,2),\" + (\",round( I2.imag,2),\")i A and \"\n",
- "print \" current, I3 is \",round(I3.real,2),\" + (\",round( I3.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "current, I1 is 0.57 + ( 0.62 )i A,\n",
- " current, I2 is 0.56 + ( 1.33 )i A and \n",
- " current, I3 is 0.01 + ( -0.71 )i A\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 541</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the magnitude of the current in the (4 + j3)impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 10;# in volts\n",
- "rv2 = 12;# in volts\n",
- "rv3 = 15;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "thetav3 = 0;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = -5j;# in ohm\n",
- "R3 = 8;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "R5 = 3j;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- "V3 = rv3*math.cos(thetav3*math.pi/180) + 1j*rv3*math.sin(thetav3*math.pi/180)\n",
- " #Currents I1, I2 and I3 with their directions are shown in Figure 30.10.\n",
- " #Three loops are chosen. The choice of loop directions is arbitrary. loop ABGH, and loopBCFG and loop CDEF\n",
- "Z4 = R4 + R5\n",
- " #using kirchoff rule in 3 loops\n",
- " #three eqns obtained\n",
- " #R1*I1 + R2*I2 = V1 + V2\n",
- " #-1*R3*I1 + (R3 + R2)*I2 + R3*I3 = V2 + V3\n",
- " # -1*R3*I1 + R3*I2 + (R3 + Z4)*I3 = V3\n",
- " #using determinants\n",
- "d1 = [[(V1 + V2), R2, 0],[(V2 + V3), (R3 + R2), R3],[V3, R3, (R3 + Z4)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[R1, (V1 + V2), 0],[-1*R3, (V2 + V3), R3],[-1*R3, V3, (R3 + Z4)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[R1, R2, (V1 + V2)],[-1*R3, (R3 + R2), (V2 + V3)],[-1*R3, R3, V3]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d = [[R1, R2, 0],[-1*R3, (R3 + R2), R3],[-1*R3, R3, (R3 + Z4)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I3 = D3/D \n",
- "I3mag = abs(I3)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n magnitude of the current through (4 + i3)ohm impedance is \",round(I3mag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " magnitude of the current through (4 + i3)ohm impedance is 1.84 A"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_2.ipynb deleted file mode 100755 index b23cbc8e..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_2.ipynb +++ /dev/null @@ -1,303 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 30: Introduction to network analysis</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 536</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#find the current flowing in each branch of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 100;# in volts\n",
- "rv2 = 50;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 25;# in ohm\n",
- "R2 = 20;# in ohm\n",
- "R3 = 10;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The branch currents and their directions are labelled as shown in Figure 30.4\n",
- " #Two loops are chosen. loop ABEF, and loop BCDE\n",
- " #using kirchoff rule in 3 loops\n",
- " #two eqns obtained\n",
- " #(R1 + R2)*I1 + R2*I2 = V1\n",
- " #R2*I1 + (R2 + R3)*I2 = V2\n",
- "I1 = (3*V1 - 2*V2)/(3*(R1 + R2) - 2*(R2))\n",
- "I2 = (V2 - R2*I1)/(R2 + R3)\n",
- "I = I1 + I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A, \\n current, I2 is \",round(I2.real,2),\" + (\",round( I2.imag,2),\")i A and \"\n",
- "print \" total current, I is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I1 is 3.16 + ( -1.05 )i A, \n",
- " current, I2 is -2.11 + ( 2.37 )i A and \n",
- " total current, I is 1.05 + ( 1.32 )i A\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 537</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the current flowing in the 2 ohm resistor of the circuit\n",
- "#find the power dissipated in the 3 ohm resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "#initializing the variables:\n",
- "V = 8;# in volts\n",
- "R1 = 1;# in ohm\n",
- "R2 = 2;# in ohm\n",
- "R3 = 3;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "R5 = 5;# in ohm\n",
- "R6 = 6;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #Currents and their directions are assigned as shown in Figure 30.6.\n",
- " #Three loops are chosen since three unknown currents are required. The choice of loop directions is arbitrary.\n",
- " #loop ABCDE, and loop EDGF and loop DCHG\n",
- " #using kirchoff rule in 3 loops\n",
- " #three eqns obtained\n",
- " #R5*I1 + (R6 + R4)*I2 - R4*I3 = V\n",
- " #-1*R1*I1 + (R6 + R1)*I2 + R2*I3 = 0\n",
- " # R3*I1 - (R3 + R4)*I2 + (R2 + R3 + R4)*I3 = 0\n",
- "#using determinants\n",
- "d1 = [[V, (R6 + R4), -1*R4],[0, (R6 + R1), R2], [0, (-1*(R3 + R4)), (R2 + R3 + R4)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[R5, V, -1*R4],[-1*R1, 0, R2],[ R3, 0, (R2 + R3 + R4)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[R5, (R6 + R4), V],[-1*R1, (R6 + R1), 0],[ R3, (-1*(R3 + R4)), 0]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d = [[R5, (R6 + R4), -1*R4],[-1*R1, (R6 + R1), R2],[ R3, (-1*(R3 + R4)), (R2 + R3 + R4)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I3 = D3/D \n",
- "#Current in the 2 ohm resistance\n",
- "I = I1 - I2 + I3\n",
- "#power dissipated in the 3 ohm resistance\n",
- "P3 = R3*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current through 2 ohm resistor is \",round(I2,3),\" A\"\n",
- "print \"\\n (b)power dissipated in the 3 ohm resistor is \",round(P3,3),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current through 2 ohm resistor is 0.203 A\n",
- "\n",
- " (b)power dissipated in the 3 ohm resistor is 1.267 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 539</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in each branch using Kirchhoff\u2019s laws.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 5 + 0j;# in volts\n",
- "E2 = 2 + 4j;# in volts\n",
- "Z1 = 3 + 4j;# in ohm\n",
- "Z2 = 2 - 5j;# in ohm\n",
- "Z3 = 6 + 8j;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #Currents I1 and I2 with their directions are shown in Figure 30.8.\n",
- " #Two loops are chosen with their directions both clockwise.loop ABEF and loop BCDE,\n",
- " #using kirchoff rule in 3 loops\n",
- " #two eqns obtained\n",
- " #(Z1 + Z3)*I1 - Z3*I2 = E1\n",
- " #-1*Z3*I1 + (Z2 + Z3)*I2 = E2\n",
- "I1 = ((Z2 + Z3)*E1 + Z3*E2)/((Z2 + Z3)*(Z1 + Z3) - Z3*Z3)\n",
- "I2 = -1*(E1 - (Z1 + Z3)*I1)/Z3\n",
- "I3 = I1 - I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"current, I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A,\\n current, I2 is \",round(I2.real,2),\" + (\",round( I2.imag,2),\")i A and \"\n",
- "print \" current, I3 is \",round(I3.real,2),\" + (\",round( I3.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "current, I1 is 0.57 + ( 0.62 )i A,\n",
- " current, I2 is 0.56 + ( 1.33 )i A and \n",
- " current, I3 is 0.01 + ( -0.71 )i A\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 541</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the magnitude of the current in the (4 + j3)impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 10;# in volts\n",
- "rv2 = 12;# in volts\n",
- "rv3 = 15;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "thetav3 = 0;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = -5j;# in ohm\n",
- "R3 = 8;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "R5 = 3j;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- "V3 = rv3*math.cos(thetav3*math.pi/180) + 1j*rv3*math.sin(thetav3*math.pi/180)\n",
- " #Currents I1, I2 and I3 with their directions are shown in Figure 30.10.\n",
- " #Three loops are chosen. The choice of loop directions is arbitrary. loop ABGH, and loopBCFG and loop CDEF\n",
- "Z4 = R4 + R5\n",
- " #using kirchoff rule in 3 loops\n",
- " #three eqns obtained\n",
- " #R1*I1 + R2*I2 = V1 + V2\n",
- " #-1*R3*I1 + (R3 + R2)*I2 + R3*I3 = V2 + V3\n",
- " # -1*R3*I1 + R3*I2 + (R3 + Z4)*I3 = V3\n",
- " #using determinants\n",
- "d1 = [[(V1 + V2), R2, 0],[(V2 + V3), (R3 + R2), R3],[V3, R3, (R3 + Z4)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[R1, (V1 + V2), 0],[-1*R3, (V2 + V3), R3],[-1*R3, V3, (R3 + Z4)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[R1, R2, (V1 + V2)],[-1*R3, (R3 + R2), (V2 + V3)],[-1*R3, R3, V3]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d = [[R1, R2, 0],[-1*R3, (R3 + R2), R3],[-1*R3, R3, (R3 + Z4)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I3 = D3/D \n",
- "I3mag = abs(I3)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n magnitude of the current through (4 + i3)ohm impedance is \",round(I3mag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " magnitude of the current through (4 + i3)ohm impedance is 1.84 A"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_30.ipynb index 6732d70c..6732d70c 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_30.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint.ipynb deleted file mode 100755 index 5c1fd9c0..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint.ipynb +++ /dev/null @@ -1,665 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 31: Mesh-current and nodal analysis</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 546</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use mesh-current analysis to determine the current flowing in \n",
- "#(a) the 5 ohm resistance, and (b) the 1 ohm resistance of the d.c. circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 4;# in volts\n",
- "V2 = 5;# in volts\n",
- "R1 = 3;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = 1;# in ohm\n",
- "R5 = 6;# in ohm\n",
- "R6 = 8;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #The mesh currents I1, I2 and I3 are shown in Figure 31.2. Using Kirchhoff\u2019s voltage law in 3 loops\n",
- " #three eqns obtained\n",
- " #(R1 + R2)*I1 - R2*I2 = V1\n",
- " #-1*R2*I1 + (R2 + R3 + R4 + R5)*I2 - R4*I3 = 0\n",
- " # -1*R4*I2 + (R4 + R6)*I3 = -1*V2\n",
- " #using determinants\n",
- "d1 = [[V1, -1*R2, 0],[0, (R2 + R3 + R4 + R5), -1*R4],[-1*V2, -1*R4, (R4 + R6)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(R1 + R2), V1, 0],[-1*R2, 0, -1*R4],[0, -1*V2, (R4 + R6)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[(R1 + R2), -1*R2, V1],[-1*R2, (R2 + R3 + R4 + R5), 0],[0, -1*R4, -1*V2]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d = [[(R1 + R2), -1*R2, 0],[-1*R2, (R2 + R3 + R4 + R5), -1*R4],[0, -1*R4, (R4 + R6)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I3 = D3/D \n",
- "IR2 = I1 - I2\n",
- "IR4 = I2 - I3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current in the 5 ohm resistance is \",round(IR2,2),\" A\"\n",
- "print \"\\n (b)current in the 1 ohm resistance is \",round(IR4,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current in the 5 ohm resistance is 0.44 A\n",
- "\n",
- " (b)current in the 1 ohm resistance is 0.69 A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 547</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine, using mesh-current analysis, \n",
- "#(a) the mesh currents I1 and I2 \n",
- "#(b) the current flowing in the capacitor,\n",
- "#(c) the active power delivered by the voltage source.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 100;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 5;# in ohm\n",
- "R2 = -1j*4;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = 3j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltages\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Currents I1, I2 with their directions are shown in Figure 31.03.\n",
- " #Two loops are chosen. The choice of loop directions is arbitrary.\n",
- " #using kirchoff rule in 2 loops\n",
- " #two eqns obtained\n",
- " #(R1 + R2)*I1 - R2*I2 = V\n",
- " #-1*R2*I1 + (R3 + R2 + R4)*I2 = 0\n",
- " #using determinants\n",
- "d1 = [[V, -1*R2],[0, (R3 + R2 + R4)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(R1 + R2), V],[-1*R2, 0]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[(R1 + R2), -1*R2],[-1*R2, (R3 + R2 + R4)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I1mag = abs(I1)\n",
- " #Current flowing in capacitor\n",
- "Ic = I1 - I2\n",
- " #Source power P\n",
- "phi = cmath.phase(complex(I1.real,I1.imag))\n",
- "P = V*I1mag*math.cos(phi)\n",
- "Icmag = abs(Ic)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)current,I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A, current, I2 is\",round(I2.real,2),\" + (\",round(I2.imag,2),\")i A\"\n",
- "print \"(b)current in the capacitor is \",round(Icmag,2),\" A\"\n",
- "print \"(c)Source power P is \",round(abs(P),2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)current,I1 is 10.17 + ( 3.55 )i A, current, I2 is 5.73 + ( -8.74 )i A\n",
- "(b)current in the capacitor is 13.06 A\n",
- "(c)Source power P is 1017.06 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 548</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the line currents IR, IY and IB using mesh-current analysis\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 415;# in volts\n",
- "rv2 = 415;# in volts\n",
- "thetav1 = 120;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "R = 3 + 4j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #Two mesh currents I1 and I2 are chosen as shown in Figure 31.4.\n",
- " #Two loops are chosen. The choice of loop directions is arbitrary.\n",
- " #using kirchoff rule in 2 loops\n",
- " #two eqns obtained\n",
- " #2*R*I1 - R*I2 = V1\n",
- " #-1*R*I1 + 2*R*I2 = V2\n",
- " #using determinants\n",
- "d1 = [[V1, -1*R],[V2, 2*R]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[2*R, V1],[-1*R, V2]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[2*R, -1*R],[-1*R, 2*R]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I1mag = abs(I1)\n",
- " #line current IR\n",
- "IR = I1\n",
- " #line current IB\n",
- "IB = -1*I2\n",
- " #line current IY\n",
- "IY = I2 - I1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"current, IR is\",round(IR.real,2),\" + (\",round( IR.imag,2),\")i A, current, IB is\",round(IB.real,2),\" + (\",round( IB.imag,2),\")i A and current, IY is \",round(IY.real,2),\" + (\",round(IY.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "current, IR is 38.34 + ( 28.75 )i A, current, IB is -44.07 + ( 18.82 )i A and current, IY is 5.73 + ( -47.58 )i A\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 551</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the voltage VAB, by using nodal analysis.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ri = 20;# in amperes\n",
- "thetai = 0;# in degrees\n",
- "R1 = 10;# in ohm\n",
- "R2 = 3j;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = 16;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #current\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Figure 31.8 contains two principal nodes (at 1 and B) and thus only one nodal equation is required. \n",
- " #B is taken as the reference node and the equation for node 1 is obtained as follows. \n",
- " #Applying Kirchhoff\u2019s current law to node 1 gives:\n",
- " #IX + IY = I\n",
- "V1 = I/((1/R4) +(1/(R2 +R3)))\n",
- "IY = V1/(R2 + R3)\n",
- "VAB = IY*R3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage VAB is \",round(VAB.real,2),\" + (\",round( VAB.imag,2),\")i V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage VAB is 62.59 + ( -9.39 )i V"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 552</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of voltage VXY\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 8;# in volts\n",
- "rv2 = 8;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 5;# in ohm\n",
- "R2 = 6j;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = 3;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The circuit contains no principal nodes. \n",
- " #However, if point Y is chosen as the reference node then an equation \n",
- " #may be written for node X assuming that current leaves point X by both branches\n",
- "VX = ((V1/(R1 + R3) + V2/(R2 + R4))/(1/(R1 + R3) + 1/(R2 + R4)))\n",
- "VXY = VX\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage VXY = \",round(abs(VXY),2),\"/_\",round(cmath.phase(complex(VXY.real, VXY.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage VXY = 9.12 /_ 52.13 deg V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 553</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use nodal analysis to determine the current flowing in each branch of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 100;# in volts\n",
- "rv2 = 50;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 25;# in ohm\n",
- "R2 = 20;# in ohm\n",
- "R3 = 10;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #There are only two principal nodes in Figure 31.10 so only one nodal equation is required. \n",
- " #Node 2 is taken as the reference node.\n",
- " #The equation at node 1 is I1 + I2 + I3 = 0\n",
- "Vn1 = ((V1/R1 + V2/R3)/(1/R1 + 1/R2 + 1/R3))\n",
- "I1 = (Vn1 - V1)/R1\n",
- "I2 = Vn1/R2\n",
- "I3 = (Vn1 - V2)/R3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A, \\n current, I2 is \",round(I2.real,2),\" + (\",round( I2.imag,2),\")i A \\n and current, I3 is \",round(I3.real,2),\" + (\",round(I3.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I1 is -3.16 + ( 1.05 )i A, \n",
- " current, I2 is 1.05 + ( 1.32 )i A \n",
- " and current, I3 is 2.11 + ( -2.37 )i A"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 554</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the voltage at nodes 1 and 2, (b) the current in the j4 ohm inductance,\n",
- "#(c) the current in the 5 ohm resistance, and (d) the magnitude of the active power dissipated in the 2.5ohm resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 25;# in volts\n",
- "rv2 = 25;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 2;# in ohm\n",
- "R2 = -1j*4;# in ohm\n",
- "R3 = 5;# in ohm\n",
- "R4 = 4j;# in ohm\n",
- "R5 = 2.5;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The equation at node 1\n",
- " #Vn1*(1/R1 + 1/R2 + 1/R3) - Vn2/R3 = V1/R1\n",
- " #The equation at node 2\n",
- " #Vn1*(-1/R3) + Vn2*(1/R4 + 1/R5 + 1/R3) = V2/R5\n",
- " #using determinants\n",
- "d1 = [[V1/R1, -1/R3],[V2/R5, (1/R4 + 1/R5 + 1/R3)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(1/R1 + 1/R2 + 1/R3), V1/R1],[-1/R3, V2/R5]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[(1/R1 + 1/R2 + 1/R3), -1/R3],[-1/R3, (1/R4 + 1/R5 + 1/R3)]]\n",
- "D = numpy.linalg.det(d)\n",
- "Vn1 = D1/D\n",
- "Vn2 = D2/D\n",
- " #current in the j4 ohm inductance is given by:\n",
- "I4 = Vn2/R4\n",
- " #current in the 5 ohm resistance is given by:\n",
- "I3 = (Vn1 - Vn2)/R3\n",
- " #active power dissipated in the 2.5 ohm resistor is given by\n",
- "P5 = R5*((Vn2 - V2)/R5)**2\n",
- " #magnitude of the active power dissipated\n",
- "P5mag = abs(P5)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the voltage at nodes 1 and 2 are \",round(abs(Vn1),1),\"/_\",round(cmath.phase(complex(Vn1.real, Vn1.imag))*180/math.pi,2),\"deg V and \",round(abs(Vn2),1),\"/_\",round(cmath.phase(complex(Vn2.real, Vn2.imag))*180/math.pi,1),\"deg V\"\n",
- "print \"\\n (b)the current in the j4 ohm inductance = \",round(abs(I4),2),\"/_\",round(cmath.phase(complex(I4.real, I4.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n (c)the current in the 5 ohm resistance = \",round(abs(I3),2),\"/_\",round(cmath.phase(complex(I3.real, I3.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n (d) magnitude of the active power dissipated in the 2.5 ohm resistance is \",round(P5mag,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the voltage at nodes 1 and 2 are 17.1 /_ -5.3 deg V and 15.8 /_ 93.2 deg V\n",
- "\n",
- " (b)the current in the j4 ohm inductance = 3.95 /_ 3.23 deg A\n",
- "\n",
- " (c)the current in the 5 ohm resistance = 4.99 /_ -44.06 deg A\n",
- "\n",
- " (d) magnitude of the active power dissipated in the 2.5 ohm resistance is 34.4 W"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 556</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the voltage VXY using nodal analysis.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ri = 25;# in amperes\n",
- "thetai = 0;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = 3j;# in ohm\n",
- "R3 = 5;# in ohm\n",
- "R4 = 10j;# in ohm\n",
- "R5 = 20j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #current\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Node 3 is taken as the reference node.\n",
- " #At node 1,\n",
- " #V1*(1/(R1 + R2) + 1/R3) - V2/R3 = I\n",
- " #The equation at node 2\n",
- " #V1*(-1/R3) + V2*(1/R4 + 1/R5 + 1/R3) = 0\n",
- " #using determinants\n",
- "d1 = [[I, -1/R3],[0 , (1/R4 + 1/R5 + 1/R3)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(1/(R1 + R2) + 1/R3), I],[-1/R3, 0]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[(1/(R1 + R2) + 1/R3), -1/R3],[-1/R3, (1/R4 + 1/R5 + 1/R3)]]\n",
- "D = numpy.linalg.det(d)\n",
- "V1 = D1/D\n",
- "V2 = D2/D\n",
- " #the voltage between point X and node 3 is\n",
- "VX = V1*R2/(R1 + R2)\n",
- " #Thus the voltage\n",
- "VY = V2\n",
- "VXY = VX - VY\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage VXY is \",round(VXY.real,2),\" + (\",round( VXY.imag,2),\")i V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage VXY is -16.16 + ( -15.05 )i V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 557</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use nodal analysis to determine the voltages at nodes 2 and 3 \n",
- "#determine the current flowing in the 2 ohm resistor and the power dissipated in the 3 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 8;# in volts\n",
- "R1 = 1;# in ohm\n",
- "R2 = 2;# in ohm\n",
- "R3 = 3;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "R5 = 5;# in ohm\n",
- "R6 = 6;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #In Figure 31.13, the reference node is shown at point A.\n",
- " #At node 1,\n",
- " #V1*(1/R1 + 1/R6 + 1/R5) - V2/R1 - V3/R5 = V/R5\n",
- " #The equation at node 2\n",
- " #V1*(-1/R1) + V2*(1/R2 + 1/R1 + 1/R3) - V3/R3 = 0\n",
- " #At node 3\n",
- " # - V1/R5 - V2/R3 + V3*(1/R4 + 1/R3 + 1/R5) = -1*V/R5\n",
- "#using determinants\n",
- "d1 = [[V/R5, -1/R1, -1/R5],[0, (1/R2 + 1/R1 + 1/R3), -1/R3],[-1*V/R5, -1/R3, (1/R4 + 1/R3 + 1/R5)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(1/R1 + 1/R6 + 1/R5), V/R5, -1/R5],[-1/R1, 0, -1/R3],[-1/R5, -1*V/R5, (1/R4 + 1/R3 + 1/R5)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[(1/R1 + 1/R6 + 1/R5), -1/R1, V/R5],[-1/R1, (1/R2 + 1/R1 + 1/R3), 0],[-1/R5, -1/R3, -1*V/R5]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d =[[(1/R1 + 1/R6 + 1/R5), -1/R1, -1/R5],[-1/R1, (1/R2 + 1/R1 + 1/R3), -1/R3],[-1/R5, -1/R3, (1/R4 + 1/R3 + 1/R5)]]\n",
- "D = numpy.linalg.det(d)\n",
- "Vn1 = D1/D\n",
- "Vn2 = D2/D\n",
- "Vn3 = D3/D \n",
- " #the current in the 2 ohm resistor\n",
- "I2 = Vn2/R2\n",
- " #power dissipated in the 3 ohm resistance\n",
- "P3 = R3*((Vn2 - Vn3)/R3)**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current through 2 ohm resistor is \",round(I2,2),\" A\"\n",
- "print \"\\n (b)power dissipated in the 3 ohm resistor is \",round(P3,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current through 2 ohm resistor is 0.19 A\n",
- "\n",
- " (b)power dissipated in the 3 ohm resistor is 1.27 W\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_1.ipynb deleted file mode 100755 index 5c1fd9c0..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_1.ipynb +++ /dev/null @@ -1,665 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 31: Mesh-current and nodal analysis</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 546</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use mesh-current analysis to determine the current flowing in \n",
- "#(a) the 5 ohm resistance, and (b) the 1 ohm resistance of the d.c. circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 4;# in volts\n",
- "V2 = 5;# in volts\n",
- "R1 = 3;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = 1;# in ohm\n",
- "R5 = 6;# in ohm\n",
- "R6 = 8;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #The mesh currents I1, I2 and I3 are shown in Figure 31.2. Using Kirchhoff\u2019s voltage law in 3 loops\n",
- " #three eqns obtained\n",
- " #(R1 + R2)*I1 - R2*I2 = V1\n",
- " #-1*R2*I1 + (R2 + R3 + R4 + R5)*I2 - R4*I3 = 0\n",
- " # -1*R4*I2 + (R4 + R6)*I3 = -1*V2\n",
- " #using determinants\n",
- "d1 = [[V1, -1*R2, 0],[0, (R2 + R3 + R4 + R5), -1*R4],[-1*V2, -1*R4, (R4 + R6)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(R1 + R2), V1, 0],[-1*R2, 0, -1*R4],[0, -1*V2, (R4 + R6)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[(R1 + R2), -1*R2, V1],[-1*R2, (R2 + R3 + R4 + R5), 0],[0, -1*R4, -1*V2]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d = [[(R1 + R2), -1*R2, 0],[-1*R2, (R2 + R3 + R4 + R5), -1*R4],[0, -1*R4, (R4 + R6)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I3 = D3/D \n",
- "IR2 = I1 - I2\n",
- "IR4 = I2 - I3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current in the 5 ohm resistance is \",round(IR2,2),\" A\"\n",
- "print \"\\n (b)current in the 1 ohm resistance is \",round(IR4,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current in the 5 ohm resistance is 0.44 A\n",
- "\n",
- " (b)current in the 1 ohm resistance is 0.69 A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 547</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine, using mesh-current analysis, \n",
- "#(a) the mesh currents I1 and I2 \n",
- "#(b) the current flowing in the capacitor,\n",
- "#(c) the active power delivered by the voltage source.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 100;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 5;# in ohm\n",
- "R2 = -1j*4;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = 3j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltages\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Currents I1, I2 with their directions are shown in Figure 31.03.\n",
- " #Two loops are chosen. The choice of loop directions is arbitrary.\n",
- " #using kirchoff rule in 2 loops\n",
- " #two eqns obtained\n",
- " #(R1 + R2)*I1 - R2*I2 = V\n",
- " #-1*R2*I1 + (R3 + R2 + R4)*I2 = 0\n",
- " #using determinants\n",
- "d1 = [[V, -1*R2],[0, (R3 + R2 + R4)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(R1 + R2), V],[-1*R2, 0]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[(R1 + R2), -1*R2],[-1*R2, (R3 + R2 + R4)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I1mag = abs(I1)\n",
- " #Current flowing in capacitor\n",
- "Ic = I1 - I2\n",
- " #Source power P\n",
- "phi = cmath.phase(complex(I1.real,I1.imag))\n",
- "P = V*I1mag*math.cos(phi)\n",
- "Icmag = abs(Ic)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)current,I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A, current, I2 is\",round(I2.real,2),\" + (\",round(I2.imag,2),\")i A\"\n",
- "print \"(b)current in the capacitor is \",round(Icmag,2),\" A\"\n",
- "print \"(c)Source power P is \",round(abs(P),2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)current,I1 is 10.17 + ( 3.55 )i A, current, I2 is 5.73 + ( -8.74 )i A\n",
- "(b)current in the capacitor is 13.06 A\n",
- "(c)Source power P is 1017.06 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 548</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the line currents IR, IY and IB using mesh-current analysis\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 415;# in volts\n",
- "rv2 = 415;# in volts\n",
- "thetav1 = 120;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "R = 3 + 4j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #Two mesh currents I1 and I2 are chosen as shown in Figure 31.4.\n",
- " #Two loops are chosen. The choice of loop directions is arbitrary.\n",
- " #using kirchoff rule in 2 loops\n",
- " #two eqns obtained\n",
- " #2*R*I1 - R*I2 = V1\n",
- " #-1*R*I1 + 2*R*I2 = V2\n",
- " #using determinants\n",
- "d1 = [[V1, -1*R],[V2, 2*R]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[2*R, V1],[-1*R, V2]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[2*R, -1*R],[-1*R, 2*R]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I1mag = abs(I1)\n",
- " #line current IR\n",
- "IR = I1\n",
- " #line current IB\n",
- "IB = -1*I2\n",
- " #line current IY\n",
- "IY = I2 - I1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"current, IR is\",round(IR.real,2),\" + (\",round( IR.imag,2),\")i A, current, IB is\",round(IB.real,2),\" + (\",round( IB.imag,2),\")i A and current, IY is \",round(IY.real,2),\" + (\",round(IY.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "current, IR is 38.34 + ( 28.75 )i A, current, IB is -44.07 + ( 18.82 )i A and current, IY is 5.73 + ( -47.58 )i A\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 551</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the voltage VAB, by using nodal analysis.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ri = 20;# in amperes\n",
- "thetai = 0;# in degrees\n",
- "R1 = 10;# in ohm\n",
- "R2 = 3j;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = 16;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #current\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Figure 31.8 contains two principal nodes (at 1 and B) and thus only one nodal equation is required. \n",
- " #B is taken as the reference node and the equation for node 1 is obtained as follows. \n",
- " #Applying Kirchhoff\u2019s current law to node 1 gives:\n",
- " #IX + IY = I\n",
- "V1 = I/((1/R4) +(1/(R2 +R3)))\n",
- "IY = V1/(R2 + R3)\n",
- "VAB = IY*R3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage VAB is \",round(VAB.real,2),\" + (\",round( VAB.imag,2),\")i V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage VAB is 62.59 + ( -9.39 )i V"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 552</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of voltage VXY\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 8;# in volts\n",
- "rv2 = 8;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 5;# in ohm\n",
- "R2 = 6j;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = 3;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The circuit contains no principal nodes. \n",
- " #However, if point Y is chosen as the reference node then an equation \n",
- " #may be written for node X assuming that current leaves point X by both branches\n",
- "VX = ((V1/(R1 + R3) + V2/(R2 + R4))/(1/(R1 + R3) + 1/(R2 + R4)))\n",
- "VXY = VX\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage VXY = \",round(abs(VXY),2),\"/_\",round(cmath.phase(complex(VXY.real, VXY.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage VXY = 9.12 /_ 52.13 deg V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 553</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use nodal analysis to determine the current flowing in each branch of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 100;# in volts\n",
- "rv2 = 50;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 25;# in ohm\n",
- "R2 = 20;# in ohm\n",
- "R3 = 10;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #There are only two principal nodes in Figure 31.10 so only one nodal equation is required. \n",
- " #Node 2 is taken as the reference node.\n",
- " #The equation at node 1 is I1 + I2 + I3 = 0\n",
- "Vn1 = ((V1/R1 + V2/R3)/(1/R1 + 1/R2 + 1/R3))\n",
- "I1 = (Vn1 - V1)/R1\n",
- "I2 = Vn1/R2\n",
- "I3 = (Vn1 - V2)/R3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A, \\n current, I2 is \",round(I2.real,2),\" + (\",round( I2.imag,2),\")i A \\n and current, I3 is \",round(I3.real,2),\" + (\",round(I3.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I1 is -3.16 + ( 1.05 )i A, \n",
- " current, I2 is 1.05 + ( 1.32 )i A \n",
- " and current, I3 is 2.11 + ( -2.37 )i A"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 554</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the voltage at nodes 1 and 2, (b) the current in the j4 ohm inductance,\n",
- "#(c) the current in the 5 ohm resistance, and (d) the magnitude of the active power dissipated in the 2.5ohm resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 25;# in volts\n",
- "rv2 = 25;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 2;# in ohm\n",
- "R2 = -1j*4;# in ohm\n",
- "R3 = 5;# in ohm\n",
- "R4 = 4j;# in ohm\n",
- "R5 = 2.5;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The equation at node 1\n",
- " #Vn1*(1/R1 + 1/R2 + 1/R3) - Vn2/R3 = V1/R1\n",
- " #The equation at node 2\n",
- " #Vn1*(-1/R3) + Vn2*(1/R4 + 1/R5 + 1/R3) = V2/R5\n",
- " #using determinants\n",
- "d1 = [[V1/R1, -1/R3],[V2/R5, (1/R4 + 1/R5 + 1/R3)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(1/R1 + 1/R2 + 1/R3), V1/R1],[-1/R3, V2/R5]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[(1/R1 + 1/R2 + 1/R3), -1/R3],[-1/R3, (1/R4 + 1/R5 + 1/R3)]]\n",
- "D = numpy.linalg.det(d)\n",
- "Vn1 = D1/D\n",
- "Vn2 = D2/D\n",
- " #current in the j4 ohm inductance is given by:\n",
- "I4 = Vn2/R4\n",
- " #current in the 5 ohm resistance is given by:\n",
- "I3 = (Vn1 - Vn2)/R3\n",
- " #active power dissipated in the 2.5 ohm resistor is given by\n",
- "P5 = R5*((Vn2 - V2)/R5)**2\n",
- " #magnitude of the active power dissipated\n",
- "P5mag = abs(P5)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the voltage at nodes 1 and 2 are \",round(abs(Vn1),1),\"/_\",round(cmath.phase(complex(Vn1.real, Vn1.imag))*180/math.pi,2),\"deg V and \",round(abs(Vn2),1),\"/_\",round(cmath.phase(complex(Vn2.real, Vn2.imag))*180/math.pi,1),\"deg V\"\n",
- "print \"\\n (b)the current in the j4 ohm inductance = \",round(abs(I4),2),\"/_\",round(cmath.phase(complex(I4.real, I4.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n (c)the current in the 5 ohm resistance = \",round(abs(I3),2),\"/_\",round(cmath.phase(complex(I3.real, I3.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n (d) magnitude of the active power dissipated in the 2.5 ohm resistance is \",round(P5mag,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the voltage at nodes 1 and 2 are 17.1 /_ -5.3 deg V and 15.8 /_ 93.2 deg V\n",
- "\n",
- " (b)the current in the j4 ohm inductance = 3.95 /_ 3.23 deg A\n",
- "\n",
- " (c)the current in the 5 ohm resistance = 4.99 /_ -44.06 deg A\n",
- "\n",
- " (d) magnitude of the active power dissipated in the 2.5 ohm resistance is 34.4 W"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 556</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the voltage VXY using nodal analysis.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ri = 25;# in amperes\n",
- "thetai = 0;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = 3j;# in ohm\n",
- "R3 = 5;# in ohm\n",
- "R4 = 10j;# in ohm\n",
- "R5 = 20j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #current\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Node 3 is taken as the reference node.\n",
- " #At node 1,\n",
- " #V1*(1/(R1 + R2) + 1/R3) - V2/R3 = I\n",
- " #The equation at node 2\n",
- " #V1*(-1/R3) + V2*(1/R4 + 1/R5 + 1/R3) = 0\n",
- " #using determinants\n",
- "d1 = [[I, -1/R3],[0 , (1/R4 + 1/R5 + 1/R3)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(1/(R1 + R2) + 1/R3), I],[-1/R3, 0]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[(1/(R1 + R2) + 1/R3), -1/R3],[-1/R3, (1/R4 + 1/R5 + 1/R3)]]\n",
- "D = numpy.linalg.det(d)\n",
- "V1 = D1/D\n",
- "V2 = D2/D\n",
- " #the voltage between point X and node 3 is\n",
- "VX = V1*R2/(R1 + R2)\n",
- " #Thus the voltage\n",
- "VY = V2\n",
- "VXY = VX - VY\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage VXY is \",round(VXY.real,2),\" + (\",round( VXY.imag,2),\")i V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage VXY is -16.16 + ( -15.05 )i V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 557</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use nodal analysis to determine the voltages at nodes 2 and 3 \n",
- "#determine the current flowing in the 2 ohm resistor and the power dissipated in the 3 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 8;# in volts\n",
- "R1 = 1;# in ohm\n",
- "R2 = 2;# in ohm\n",
- "R3 = 3;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "R5 = 5;# in ohm\n",
- "R6 = 6;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #In Figure 31.13, the reference node is shown at point A.\n",
- " #At node 1,\n",
- " #V1*(1/R1 + 1/R6 + 1/R5) - V2/R1 - V3/R5 = V/R5\n",
- " #The equation at node 2\n",
- " #V1*(-1/R1) + V2*(1/R2 + 1/R1 + 1/R3) - V3/R3 = 0\n",
- " #At node 3\n",
- " # - V1/R5 - V2/R3 + V3*(1/R4 + 1/R3 + 1/R5) = -1*V/R5\n",
- "#using determinants\n",
- "d1 = [[V/R5, -1/R1, -1/R5],[0, (1/R2 + 1/R1 + 1/R3), -1/R3],[-1*V/R5, -1/R3, (1/R4 + 1/R3 + 1/R5)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(1/R1 + 1/R6 + 1/R5), V/R5, -1/R5],[-1/R1, 0, -1/R3],[-1/R5, -1*V/R5, (1/R4 + 1/R3 + 1/R5)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[(1/R1 + 1/R6 + 1/R5), -1/R1, V/R5],[-1/R1, (1/R2 + 1/R1 + 1/R3), 0],[-1/R5, -1/R3, -1*V/R5]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d =[[(1/R1 + 1/R6 + 1/R5), -1/R1, -1/R5],[-1/R1, (1/R2 + 1/R1 + 1/R3), -1/R3],[-1/R5, -1/R3, (1/R4 + 1/R3 + 1/R5)]]\n",
- "D = numpy.linalg.det(d)\n",
- "Vn1 = D1/D\n",
- "Vn2 = D2/D\n",
- "Vn3 = D3/D \n",
- " #the current in the 2 ohm resistor\n",
- "I2 = Vn2/R2\n",
- " #power dissipated in the 3 ohm resistance\n",
- "P3 = R3*((Vn2 - Vn3)/R3)**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current through 2 ohm resistor is \",round(I2,2),\" A\"\n",
- "print \"\\n (b)power dissipated in the 3 ohm resistor is \",round(P3,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current through 2 ohm resistor is 0.19 A\n",
- "\n",
- " (b)power dissipated in the 3 ohm resistor is 1.27 W\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_2.ipynb deleted file mode 100755 index 5c1fd9c0..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_2.ipynb +++ /dev/null @@ -1,665 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 31: Mesh-current and nodal analysis</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 546</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use mesh-current analysis to determine the current flowing in \n",
- "#(a) the 5 ohm resistance, and (b) the 1 ohm resistance of the d.c. circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 4;# in volts\n",
- "V2 = 5;# in volts\n",
- "R1 = 3;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = 1;# in ohm\n",
- "R5 = 6;# in ohm\n",
- "R6 = 8;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #The mesh currents I1, I2 and I3 are shown in Figure 31.2. Using Kirchhoff\u2019s voltage law in 3 loops\n",
- " #three eqns obtained\n",
- " #(R1 + R2)*I1 - R2*I2 = V1\n",
- " #-1*R2*I1 + (R2 + R3 + R4 + R5)*I2 - R4*I3 = 0\n",
- " # -1*R4*I2 + (R4 + R6)*I3 = -1*V2\n",
- " #using determinants\n",
- "d1 = [[V1, -1*R2, 0],[0, (R2 + R3 + R4 + R5), -1*R4],[-1*V2, -1*R4, (R4 + R6)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(R1 + R2), V1, 0],[-1*R2, 0, -1*R4],[0, -1*V2, (R4 + R6)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[(R1 + R2), -1*R2, V1],[-1*R2, (R2 + R3 + R4 + R5), 0],[0, -1*R4, -1*V2]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d = [[(R1 + R2), -1*R2, 0],[-1*R2, (R2 + R3 + R4 + R5), -1*R4],[0, -1*R4, (R4 + R6)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I3 = D3/D \n",
- "IR2 = I1 - I2\n",
- "IR4 = I2 - I3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current in the 5 ohm resistance is \",round(IR2,2),\" A\"\n",
- "print \"\\n (b)current in the 1 ohm resistance is \",round(IR4,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current in the 5 ohm resistance is 0.44 A\n",
- "\n",
- " (b)current in the 1 ohm resistance is 0.69 A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 547</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine, using mesh-current analysis, \n",
- "#(a) the mesh currents I1 and I2 \n",
- "#(b) the current flowing in the capacitor,\n",
- "#(c) the active power delivered by the voltage source.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 100;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 5;# in ohm\n",
- "R2 = -1j*4;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = 3j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltages\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Currents I1, I2 with their directions are shown in Figure 31.03.\n",
- " #Two loops are chosen. The choice of loop directions is arbitrary.\n",
- " #using kirchoff rule in 2 loops\n",
- " #two eqns obtained\n",
- " #(R1 + R2)*I1 - R2*I2 = V\n",
- " #-1*R2*I1 + (R3 + R2 + R4)*I2 = 0\n",
- " #using determinants\n",
- "d1 = [[V, -1*R2],[0, (R3 + R2 + R4)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(R1 + R2), V],[-1*R2, 0]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[(R1 + R2), -1*R2],[-1*R2, (R3 + R2 + R4)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I1mag = abs(I1)\n",
- " #Current flowing in capacitor\n",
- "Ic = I1 - I2\n",
- " #Source power P\n",
- "phi = cmath.phase(complex(I1.real,I1.imag))\n",
- "P = V*I1mag*math.cos(phi)\n",
- "Icmag = abs(Ic)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)current,I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A, current, I2 is\",round(I2.real,2),\" + (\",round(I2.imag,2),\")i A\"\n",
- "print \"(b)current in the capacitor is \",round(Icmag,2),\" A\"\n",
- "print \"(c)Source power P is \",round(abs(P),2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)current,I1 is 10.17 + ( 3.55 )i A, current, I2 is 5.73 + ( -8.74 )i A\n",
- "(b)current in the capacitor is 13.06 A\n",
- "(c)Source power P is 1017.06 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 548</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the line currents IR, IY and IB using mesh-current analysis\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 415;# in volts\n",
- "rv2 = 415;# in volts\n",
- "thetav1 = 120;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "R = 3 + 4j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #Two mesh currents I1 and I2 are chosen as shown in Figure 31.4.\n",
- " #Two loops are chosen. The choice of loop directions is arbitrary.\n",
- " #using kirchoff rule in 2 loops\n",
- " #two eqns obtained\n",
- " #2*R*I1 - R*I2 = V1\n",
- " #-1*R*I1 + 2*R*I2 = V2\n",
- " #using determinants\n",
- "d1 = [[V1, -1*R],[V2, 2*R]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[2*R, V1],[-1*R, V2]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[2*R, -1*R],[-1*R, 2*R]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I1mag = abs(I1)\n",
- " #line current IR\n",
- "IR = I1\n",
- " #line current IB\n",
- "IB = -1*I2\n",
- " #line current IY\n",
- "IY = I2 - I1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"current, IR is\",round(IR.real,2),\" + (\",round( IR.imag,2),\")i A, current, IB is\",round(IB.real,2),\" + (\",round( IB.imag,2),\")i A and current, IY is \",round(IY.real,2),\" + (\",round(IY.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "current, IR is 38.34 + ( 28.75 )i A, current, IB is -44.07 + ( 18.82 )i A and current, IY is 5.73 + ( -47.58 )i A\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 551</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the voltage VAB, by using nodal analysis.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ri = 20;# in amperes\n",
- "thetai = 0;# in degrees\n",
- "R1 = 10;# in ohm\n",
- "R2 = 3j;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = 16;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #current\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Figure 31.8 contains two principal nodes (at 1 and B) and thus only one nodal equation is required. \n",
- " #B is taken as the reference node and the equation for node 1 is obtained as follows. \n",
- " #Applying Kirchhoff\u2019s current law to node 1 gives:\n",
- " #IX + IY = I\n",
- "V1 = I/((1/R4) +(1/(R2 +R3)))\n",
- "IY = V1/(R2 + R3)\n",
- "VAB = IY*R3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage VAB is \",round(VAB.real,2),\" + (\",round( VAB.imag,2),\")i V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage VAB is 62.59 + ( -9.39 )i V"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 552</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of voltage VXY\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 8;# in volts\n",
- "rv2 = 8;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 5;# in ohm\n",
- "R2 = 6j;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = 3;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The circuit contains no principal nodes. \n",
- " #However, if point Y is chosen as the reference node then an equation \n",
- " #may be written for node X assuming that current leaves point X by both branches\n",
- "VX = ((V1/(R1 + R3) + V2/(R2 + R4))/(1/(R1 + R3) + 1/(R2 + R4)))\n",
- "VXY = VX\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage VXY = \",round(abs(VXY),2),\"/_\",round(cmath.phase(complex(VXY.real, VXY.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage VXY = 9.12 /_ 52.13 deg V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 553</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use nodal analysis to determine the current flowing in each branch of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 100;# in volts\n",
- "rv2 = 50;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 25;# in ohm\n",
- "R2 = 20;# in ohm\n",
- "R3 = 10;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #There are only two principal nodes in Figure 31.10 so only one nodal equation is required. \n",
- " #Node 2 is taken as the reference node.\n",
- " #The equation at node 1 is I1 + I2 + I3 = 0\n",
- "Vn1 = ((V1/R1 + V2/R3)/(1/R1 + 1/R2 + 1/R3))\n",
- "I1 = (Vn1 - V1)/R1\n",
- "I2 = Vn1/R2\n",
- "I3 = (Vn1 - V2)/R3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A, \\n current, I2 is \",round(I2.real,2),\" + (\",round( I2.imag,2),\")i A \\n and current, I3 is \",round(I3.real,2),\" + (\",round(I3.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I1 is -3.16 + ( 1.05 )i A, \n",
- " current, I2 is 1.05 + ( 1.32 )i A \n",
- " and current, I3 is 2.11 + ( -2.37 )i A"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 554</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the voltage at nodes 1 and 2, (b) the current in the j4 ohm inductance,\n",
- "#(c) the current in the 5 ohm resistance, and (d) the magnitude of the active power dissipated in the 2.5ohm resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 25;# in volts\n",
- "rv2 = 25;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 2;# in ohm\n",
- "R2 = -1j*4;# in ohm\n",
- "R3 = 5;# in ohm\n",
- "R4 = 4j;# in ohm\n",
- "R5 = 2.5;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The equation at node 1\n",
- " #Vn1*(1/R1 + 1/R2 + 1/R3) - Vn2/R3 = V1/R1\n",
- " #The equation at node 2\n",
- " #Vn1*(-1/R3) + Vn2*(1/R4 + 1/R5 + 1/R3) = V2/R5\n",
- " #using determinants\n",
- "d1 = [[V1/R1, -1/R3],[V2/R5, (1/R4 + 1/R5 + 1/R3)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(1/R1 + 1/R2 + 1/R3), V1/R1],[-1/R3, V2/R5]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[(1/R1 + 1/R2 + 1/R3), -1/R3],[-1/R3, (1/R4 + 1/R5 + 1/R3)]]\n",
- "D = numpy.linalg.det(d)\n",
- "Vn1 = D1/D\n",
- "Vn2 = D2/D\n",
- " #current in the j4 ohm inductance is given by:\n",
- "I4 = Vn2/R4\n",
- " #current in the 5 ohm resistance is given by:\n",
- "I3 = (Vn1 - Vn2)/R3\n",
- " #active power dissipated in the 2.5 ohm resistor is given by\n",
- "P5 = R5*((Vn2 - V2)/R5)**2\n",
- " #magnitude of the active power dissipated\n",
- "P5mag = abs(P5)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the voltage at nodes 1 and 2 are \",round(abs(Vn1),1),\"/_\",round(cmath.phase(complex(Vn1.real, Vn1.imag))*180/math.pi,2),\"deg V and \",round(abs(Vn2),1),\"/_\",round(cmath.phase(complex(Vn2.real, Vn2.imag))*180/math.pi,1),\"deg V\"\n",
- "print \"\\n (b)the current in the j4 ohm inductance = \",round(abs(I4),2),\"/_\",round(cmath.phase(complex(I4.real, I4.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n (c)the current in the 5 ohm resistance = \",round(abs(I3),2),\"/_\",round(cmath.phase(complex(I3.real, I3.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n (d) magnitude of the active power dissipated in the 2.5 ohm resistance is \",round(P5mag,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the voltage at nodes 1 and 2 are 17.1 /_ -5.3 deg V and 15.8 /_ 93.2 deg V\n",
- "\n",
- " (b)the current in the j4 ohm inductance = 3.95 /_ 3.23 deg A\n",
- "\n",
- " (c)the current in the 5 ohm resistance = 4.99 /_ -44.06 deg A\n",
- "\n",
- " (d) magnitude of the active power dissipated in the 2.5 ohm resistance is 34.4 W"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 556</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the voltage VXY using nodal analysis.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ri = 25;# in amperes\n",
- "thetai = 0;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = 3j;# in ohm\n",
- "R3 = 5;# in ohm\n",
- "R4 = 10j;# in ohm\n",
- "R5 = 20j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #current\n",
- "I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
- " #Node 3 is taken as the reference node.\n",
- " #At node 1,\n",
- " #V1*(1/(R1 + R2) + 1/R3) - V2/R3 = I\n",
- " #The equation at node 2\n",
- " #V1*(-1/R3) + V2*(1/R4 + 1/R5 + 1/R3) = 0\n",
- " #using determinants\n",
- "d1 = [[I, -1/R3],[0 , (1/R4 + 1/R5 + 1/R3)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(1/(R1 + R2) + 1/R3), I],[-1/R3, 0]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[(1/(R1 + R2) + 1/R3), -1/R3],[-1/R3, (1/R4 + 1/R5 + 1/R3)]]\n",
- "D = numpy.linalg.det(d)\n",
- "V1 = D1/D\n",
- "V2 = D2/D\n",
- " #the voltage between point X and node 3 is\n",
- "VX = V1*R2/(R1 + R2)\n",
- " #Thus the voltage\n",
- "VY = V2\n",
- "VXY = VX - VY\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage VXY is \",round(VXY.real,2),\" + (\",round( VXY.imag,2),\")i V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage VXY is -16.16 + ( -15.05 )i V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 557</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use nodal analysis to determine the voltages at nodes 2 and 3 \n",
- "#determine the current flowing in the 2 ohm resistor and the power dissipated in the 3 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 8;# in volts\n",
- "R1 = 1;# in ohm\n",
- "R2 = 2;# in ohm\n",
- "R3 = 3;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "R5 = 5;# in ohm\n",
- "R6 = 6;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #In Figure 31.13, the reference node is shown at point A.\n",
- " #At node 1,\n",
- " #V1*(1/R1 + 1/R6 + 1/R5) - V2/R1 - V3/R5 = V/R5\n",
- " #The equation at node 2\n",
- " #V1*(-1/R1) + V2*(1/R2 + 1/R1 + 1/R3) - V3/R3 = 0\n",
- " #At node 3\n",
- " # - V1/R5 - V2/R3 + V3*(1/R4 + 1/R3 + 1/R5) = -1*V/R5\n",
- "#using determinants\n",
- "d1 = [[V/R5, -1/R1, -1/R5],[0, (1/R2 + 1/R1 + 1/R3), -1/R3],[-1*V/R5, -1/R3, (1/R4 + 1/R3 + 1/R5)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(1/R1 + 1/R6 + 1/R5), V/R5, -1/R5],[-1/R1, 0, -1/R3],[-1/R5, -1*V/R5, (1/R4 + 1/R3 + 1/R5)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[(1/R1 + 1/R6 + 1/R5), -1/R1, V/R5],[-1/R1, (1/R2 + 1/R1 + 1/R3), 0],[-1/R5, -1/R3, -1*V/R5]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d =[[(1/R1 + 1/R6 + 1/R5), -1/R1, -1/R5],[-1/R1, (1/R2 + 1/R1 + 1/R3), -1/R3],[-1/R5, -1/R3, (1/R4 + 1/R3 + 1/R5)]]\n",
- "D = numpy.linalg.det(d)\n",
- "Vn1 = D1/D\n",
- "Vn2 = D2/D\n",
- "Vn3 = D3/D \n",
- " #the current in the 2 ohm resistor\n",
- "I2 = Vn2/R2\n",
- " #power dissipated in the 3 ohm resistance\n",
- "P3 = R3*((Vn2 - Vn3)/R3)**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current through 2 ohm resistor is \",round(I2,2),\" A\"\n",
- "print \"\\n (b)power dissipated in the 3 ohm resistor is \",round(P3,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current through 2 ohm resistor is 0.19 A\n",
- "\n",
- " (b)power dissipated in the 3 ohm resistor is 1.27 W\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_31.ipynb index bbaa441d..bbaa441d 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_31.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint.ipynb deleted file mode 100755 index 985a6ce1..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint.ipynb +++ /dev/null @@ -1,436 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 32: The superposition theorem</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 564</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine using the superposition theorem, the current in the 20 ohm load and the current in each voltage source.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 100;# in volts\n",
- "rv2 = 50;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "r1 = 25;# in ohm\n",
- "R = 20;# in ohm\n",
- "r2 = 10;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The circuit diagram is shown in Figure 32.7. Following the above procedure:\n",
- " #The network is redrawn with the 50/_90\u00b0 V source removed as shown in Figure 32.8\n",
- " #Currents I1, I2 and I3 are labelled as shown in Figure 32.8.\n",
- "I1 = V1/(r1 + r2*R/(R + r2))\n",
- "I2 = (r2/(r2 + R))*I1\n",
- "I3 = (R/(r2 + R))*I1\n",
- " #The network is redrawn with the 100/_0\u00b0 V source removed as shown in Figure 32.9\n",
- " #Currents I4, I5 and I6 are labelled as shown in Figure 32.9.\n",
- "I4 = V2/(r2 + r1*R/(r1 + R))\n",
- "I5 = (r1/(r1 + R))*I4\n",
- "I6 = (R/(r1 + R))*I4\n",
- " #Figure 32.10 shows Figure 32.9 superimposed on Figure 32.8, giving the currents shown.\n",
- " #Current in the 20 ohm load,\n",
- "I20 = I2 + I5\n",
- " #Current in the 100/_0\u00b0 V source\n",
- "IV1 = I1 - I6\n",
- " #Current in the 50/_90\u00b0 V source\n",
- "IV2 = I4 - I3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current in the 20 ohm load is \",round(I20.real,2),\" + (\",round(I20.imag,2),\")i A\"\n",
- "print \"\\n (b)Current in the 100/_0deg V source is \",round(IV1.real,2),\" + (\",round(IV1.imag,2),\")i A\"\n",
- "print \"\\n (b)Current in the 50/_90deg V source is \",round(IV2.real,2),\" + (\",round(IV2.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current in the 20 ohm load is 1.05 + ( 1.32 )i A\n",
- "\n",
- " (b)Current in the 100/_0deg V source is 3.16 + ( -1.05 )i A\n",
- "\n",
- " (b)Current in the 50/_90deg V source is -2.11 + ( 2.37 )i A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 566</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use the superposition theorem to determine the current in the 4 ohm resistor of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 12;# in volts\n",
- "V2 = 20;# in volts\n",
- "R1 = 5;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 2.5;# in ohm\n",
- "R4 = 6;# in ohm\n",
- "R5 = 2;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #Removing the 20 V source gives the network shown in Figure 32.12.\n",
- " #Currents I1 and I2 are shown labelled in Figure 32.12\n",
- "Re1 = (R4*R5/(R4 + R5)) + R3\n",
- "Re2 = Re1*R2/(Re1 + R2) + R1\n",
- "I1 = V1/Re2\n",
- "I2 = (R2/(Re1 + R2))*I1\n",
- " #Removing the 12 V source from the original network gives the network shown in Figure 32.14.\n",
- " #Currents I3, I4 and I5 are shown labelled in Figure 32.14.\n",
- "Re3 = (R1*R2/(R1 + R2)) + R3\n",
- "Re4 = Re3*R4/(Re3 + R4) + R5\n",
- "I3 = V2/Re4\n",
- "I4 = (R4/(Re3 + R4))*I3\n",
- "I5 = (R1/(R1 + R2))*I4\n",
- " #Superimposing Figure 32.14 on Figure 32.12 shows that the current flowing in the 4 ohm resistor is given by\n",
- "Ir4 = I5 - I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ncurrent in the 4 ohm resistor of the network is \",round(Ir4,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "current in the 4 ohm resistor of the network is 0.48 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 567</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use the superposition theorem to obtain the current flowing in the \u0005(4 + j3) ohm impedance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 30;# in volts\n",
- "rv2 = 30;# in volts\n",
- "thetav1 = 45;# in degrees\n",
- "thetav2 = -45;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 1j*3;# in ohm\n",
- "R4 = -1j*10;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The network is redrawn with V2 removed, as shown in Figure 32.17.\n",
- " #Current I1 and I2 are shown in Figure 32.17. From Figure 32.17,\n",
- "Re1 = R4*(R2 + R3)/(R4 + R3 + R2)\n",
- "Re2 = Re1 + R1\n",
- " #current\n",
- "I1 = V1/Re2\n",
- "I2 = (R4/(R2 + R3 + R4))*I1\n",
- " #The original network is redrawn with V1 removed, as shown in Figure 32.18\n",
- " #Currents I3 and I4 are shown in Figure 32.18. From Figure 32.18,\n",
- "Re3 = R1*(R2 + R3)/(R1 + R3 + R2)\n",
- "Re4 = Re3 + R4\n",
- "I3 = V2/Re4\n",
- "I4 = (R1/(R2 + R3 + R1))*I3\n",
- " #If the network of Figure 32.18 is superimposed on the network of Figure 32.17, \n",
- " #it can be seen that the current in the (4+i3) ohm impedance is given by\n",
- "Ir4i3 = I2 - I4\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"current in (4 + i3) ohm impedance of the network is \",round(Ir4i3.real,2),\" + (\",round( Ir4i3.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "current in (4 + i3) ohm impedance of the network is 2.15 + ( 0.42 )i A\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 568</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine, using the superposition theorem, \n",
- "#(a) the current in each branch,\n",
- "#(b) the magnitude of the voltage across the \u0005(6 + j8) ohm\u0006 impedance,\n",
- "#and (c) the total active power delivered to the network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 5 + 0j;# in volts\n",
- "E2 = 2 + 4j;# in volts\n",
- "Z1 = 3 + 4j;# in ohm\n",
- "Z2 = 2 - 5j;# in ohm\n",
- "Z3 = 6 + 8j;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #The original network is redrawn with E2 removed, as shown in Figure 32.20.\n",
- " #Currents I1, I2 and I3 are labelled as shown in Figure 32.20.\n",
- "Ze1 = Z3*Z2/(Z3 + Z2)\n",
- "Ze2 = Ze1 + Z1\n",
- " #current\n",
- "I1 = E1/Ze2\n",
- "I2 = (Z2/(Z3 + Z2))*I1\n",
- "I3 = (Z3/(Z3 + Z2))*I1\n",
- " #The original network is redrawn with E1 removed, as shown in Figure 32.22\n",
- " #Currents I4, I5 and I6 are shown labelled in Figure 32.22 \n",
- " #with I4 flowing away from the positive terminal of the E2 source.\n",
- "Ze3 = Z3*Z1/(Z3 + Z1)\n",
- "Ze4 = Ze3 + Z2\n",
- "I4 = E2/Ze4\n",
- "I5 = (Z1/(Z3 + Z1))*I4\n",
- "I6 = (Z3/(Z3 + Z1))*I4\n",
- " #If the network of Figure 32.18 is superimposed on the network of Figure 32.17, \n",
- " #it can be seen that the current in the (4+i3) ohm impedance is given by\n",
- "i1 = I1 + I6\n",
- "i2 = I3 + I4\n",
- "i3 = I2 - I5\n",
- " #magnitude\n",
- "i1mag = abs(i1)\n",
- "i2mag = abs(i2)\n",
- "E1mag = abs(E1)\n",
- "E2mag = abs(E2)\n",
- " #phase\n",
- "phi1 = cmath.phase(complex(i1.real,i1.imag))\n",
- "phi2 = cmath.phase(complex(i2.real,i2.imag))\n",
- " #voltage across the(6 + i8) ohm impedance\n",
- "V6i8 = i3*Z3\n",
- "V6i8m = abs(V6i8)\n",
- " #power\n",
- "P = (E1mag*i1mag*math.cos(phi1)) + (E2mag*i2mag*math.cos(phi2 - cmath.phase(complex(E2.real,E2.imag))))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)currents are: \\n \",round(i1.real,2),\" + (\",round( i1.imag,2),\")i A, \\n \",round(i2.real,2),\" + (\",round(i2.imag,2),\")i A \\n and \",round(i3.real,2),\" + (\",round(i3.imag,2),\")i A\"\n",
- "print \"\\n(b)current in the (6 + i8) ohm resistor of the network is \",round(V6i8m,2),\" V\"\n",
- "print \"\\n(c)the total active power delivered to the network is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)currents are: \n",
- " 0.57 + ( 0.62 )i A, \n",
- " 0.56 + ( 1.33 )i A \n",
- " and 0.01 + ( -0.71 )i A\n",
- "\n",
- "(b)current in the (6 + i8) ohm resistor of the network is 7.09 V\n",
- "\n",
- "(c)the total active power delivered to the network is 9.29 W"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 571</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine,(a) the magnitude of the current flowing in the capacitor, \n",
- "#(b) the p.d. across the 5 ohm resistance, (c) the active power dissipated in the 20 ohm resistance and \n",
- "#(d) the total active power taken from the supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 50;# in volts\n",
- "rv2 = 30;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 20;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = -1j*3;# in ohm\n",
- "R4 = 8;# in ohm\n",
- "R5 = 8;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The network is redrawn with the V2 source removed, as shown in Figure 32.26.\n",
- " #Currents I1 to I5 are shown labelled in Figure 32.26. \n",
- " #current\n",
- "Re1 = R4*R5/(R5 + R4) + R3\n",
- "Re2 = Re1*R2/(R2 + Re1)\n",
- "I1 = V1/(Re2 + R1)\n",
- "I2 = (Re1/(R2 + Re1))*I1\n",
- "I3 = (R2/(Re1 + R2))*I1\n",
- "I4 = (R4/(R4 + R5))*I3\n",
- "I5 = I3 - I4\n",
- " #The original network is redrawn with the V1 source removed, as shown in Figure 32.27.\n",
- " #Currents I6 to I10 are shown labelled in Figure 32.27\n",
- "Re3 = R1*R2/(R1 + R2)\n",
- "Re4 = Re3 + R3\n",
- "Re5 = Re4*R4/(Re4 + R4)\n",
- "Re6 = Re5 + R5\n",
- "I6 = V2/Re6\n",
- "I7 = (Re4/(Re4 + R4))*I6\n",
- "I8 = (R4/(Re4 + R4))*I6\n",
- "I9 = (R1/(R1 + R2))*I8\n",
- "I10 = (R2/(R1 + R2))*I8\n",
- " #current flowing in the capacitor is given by\n",
- "Ic = I3 - I8\n",
- " #magnitude of the current in the capacitor\n",
- "Icmag = abs(Ic)\n",
- "\n",
- "i1 = I2 + I9\n",
- "i1mag = abs(i1)\n",
- " #magnitude of the p.d. across the 5 ohm resistance is given by\n",
- "Vr5m = i1mag*R2\n",
- " #Active power dissipated in the 20 ohm resistance is given by\n",
- "i2 = I1 - I10\n",
- "i2mag = abs(i2)\n",
- "phii2 = cmath.phase(complex(i2.real,i2.imag))\n",
- "Pr20 = R1*(i2mag)**2\n",
- " #Active power developed by the V1\n",
- "P1 = rv1*i2mag*math.cos(phii2)\n",
- " #Active power developed by V2 source\n",
- "i3 = I6 - I5\n",
- "i3mag = abs(i3)\n",
- "phii3 = cmath.phase(complex(i3.real,i3.imag))\n",
- "P2 = rv2*i3mag*math.cos(phii3 - (thetav2*math.pi/180))\n",
- " #Total power developed\n",
- "P = P1 + P2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)the magnitude of the current flowing in the capacitor is \",round(Icmag,2),\" A\"\n",
- "print \"\\n(b) the p.d. across the 5 ohm resistance is \",round(Vr5m,2),\" V\"\n",
- "print \"\\n(c)the active power dissipated in the 20 ohm resistance is \",round(Pr20,0),\" W\"\n",
- "print \"\\n(d)the total active power taken from the supply is \",round(P,1),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)the magnitude of the current flowing in the capacitor is 2.11 A\n",
- "\n",
- "(b) the p.d. across the 5 ohm resistance is 5.85 V\n",
- "\n",
- "(c)the active power dissipated in the 20 ohm resistance is 111.0 W\n",
- "\n",
- "(d)the total active power taken from the supply is 191.9 W"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_1.ipynb deleted file mode 100755 index 985a6ce1..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_1.ipynb +++ /dev/null @@ -1,436 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 32: The superposition theorem</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 564</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine using the superposition theorem, the current in the 20 ohm load and the current in each voltage source.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 100;# in volts\n",
- "rv2 = 50;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "r1 = 25;# in ohm\n",
- "R = 20;# in ohm\n",
- "r2 = 10;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The circuit diagram is shown in Figure 32.7. Following the above procedure:\n",
- " #The network is redrawn with the 50/_90\u00b0 V source removed as shown in Figure 32.8\n",
- " #Currents I1, I2 and I3 are labelled as shown in Figure 32.8.\n",
- "I1 = V1/(r1 + r2*R/(R + r2))\n",
- "I2 = (r2/(r2 + R))*I1\n",
- "I3 = (R/(r2 + R))*I1\n",
- " #The network is redrawn with the 100/_0\u00b0 V source removed as shown in Figure 32.9\n",
- " #Currents I4, I5 and I6 are labelled as shown in Figure 32.9.\n",
- "I4 = V2/(r2 + r1*R/(r1 + R))\n",
- "I5 = (r1/(r1 + R))*I4\n",
- "I6 = (R/(r1 + R))*I4\n",
- " #Figure 32.10 shows Figure 32.9 superimposed on Figure 32.8, giving the currents shown.\n",
- " #Current in the 20 ohm load,\n",
- "I20 = I2 + I5\n",
- " #Current in the 100/_0\u00b0 V source\n",
- "IV1 = I1 - I6\n",
- " #Current in the 50/_90\u00b0 V source\n",
- "IV2 = I4 - I3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current in the 20 ohm load is \",round(I20.real,2),\" + (\",round(I20.imag,2),\")i A\"\n",
- "print \"\\n (b)Current in the 100/_0deg V source is \",round(IV1.real,2),\" + (\",round(IV1.imag,2),\")i A\"\n",
- "print \"\\n (b)Current in the 50/_90deg V source is \",round(IV2.real,2),\" + (\",round(IV2.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current in the 20 ohm load is 1.05 + ( 1.32 )i A\n",
- "\n",
- " (b)Current in the 100/_0deg V source is 3.16 + ( -1.05 )i A\n",
- "\n",
- " (b)Current in the 50/_90deg V source is -2.11 + ( 2.37 )i A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 566</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use the superposition theorem to determine the current in the 4 ohm resistor of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 12;# in volts\n",
- "V2 = 20;# in volts\n",
- "R1 = 5;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 2.5;# in ohm\n",
- "R4 = 6;# in ohm\n",
- "R5 = 2;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #Removing the 20 V source gives the network shown in Figure 32.12.\n",
- " #Currents I1 and I2 are shown labelled in Figure 32.12\n",
- "Re1 = (R4*R5/(R4 + R5)) + R3\n",
- "Re2 = Re1*R2/(Re1 + R2) + R1\n",
- "I1 = V1/Re2\n",
- "I2 = (R2/(Re1 + R2))*I1\n",
- " #Removing the 12 V source from the original network gives the network shown in Figure 32.14.\n",
- " #Currents I3, I4 and I5 are shown labelled in Figure 32.14.\n",
- "Re3 = (R1*R2/(R1 + R2)) + R3\n",
- "Re4 = Re3*R4/(Re3 + R4) + R5\n",
- "I3 = V2/Re4\n",
- "I4 = (R4/(Re3 + R4))*I3\n",
- "I5 = (R1/(R1 + R2))*I4\n",
- " #Superimposing Figure 32.14 on Figure 32.12 shows that the current flowing in the 4 ohm resistor is given by\n",
- "Ir4 = I5 - I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ncurrent in the 4 ohm resistor of the network is \",round(Ir4,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "current in the 4 ohm resistor of the network is 0.48 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 567</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use the superposition theorem to obtain the current flowing in the \u0005(4 + j3) ohm impedance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 30;# in volts\n",
- "rv2 = 30;# in volts\n",
- "thetav1 = 45;# in degrees\n",
- "thetav2 = -45;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 1j*3;# in ohm\n",
- "R4 = -1j*10;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The network is redrawn with V2 removed, as shown in Figure 32.17.\n",
- " #Current I1 and I2 are shown in Figure 32.17. From Figure 32.17,\n",
- "Re1 = R4*(R2 + R3)/(R4 + R3 + R2)\n",
- "Re2 = Re1 + R1\n",
- " #current\n",
- "I1 = V1/Re2\n",
- "I2 = (R4/(R2 + R3 + R4))*I1\n",
- " #The original network is redrawn with V1 removed, as shown in Figure 32.18\n",
- " #Currents I3 and I4 are shown in Figure 32.18. From Figure 32.18,\n",
- "Re3 = R1*(R2 + R3)/(R1 + R3 + R2)\n",
- "Re4 = Re3 + R4\n",
- "I3 = V2/Re4\n",
- "I4 = (R1/(R2 + R3 + R1))*I3\n",
- " #If the network of Figure 32.18 is superimposed on the network of Figure 32.17, \n",
- " #it can be seen that the current in the (4+i3) ohm impedance is given by\n",
- "Ir4i3 = I2 - I4\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"current in (4 + i3) ohm impedance of the network is \",round(Ir4i3.real,2),\" + (\",round( Ir4i3.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "current in (4 + i3) ohm impedance of the network is 2.15 + ( 0.42 )i A\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 568</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine, using the superposition theorem, \n",
- "#(a) the current in each branch,\n",
- "#(b) the magnitude of the voltage across the \u0005(6 + j8) ohm\u0006 impedance,\n",
- "#and (c) the total active power delivered to the network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 5 + 0j;# in volts\n",
- "E2 = 2 + 4j;# in volts\n",
- "Z1 = 3 + 4j;# in ohm\n",
- "Z2 = 2 - 5j;# in ohm\n",
- "Z3 = 6 + 8j;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #The original network is redrawn with E2 removed, as shown in Figure 32.20.\n",
- " #Currents I1, I2 and I3 are labelled as shown in Figure 32.20.\n",
- "Ze1 = Z3*Z2/(Z3 + Z2)\n",
- "Ze2 = Ze1 + Z1\n",
- " #current\n",
- "I1 = E1/Ze2\n",
- "I2 = (Z2/(Z3 + Z2))*I1\n",
- "I3 = (Z3/(Z3 + Z2))*I1\n",
- " #The original network is redrawn with E1 removed, as shown in Figure 32.22\n",
- " #Currents I4, I5 and I6 are shown labelled in Figure 32.22 \n",
- " #with I4 flowing away from the positive terminal of the E2 source.\n",
- "Ze3 = Z3*Z1/(Z3 + Z1)\n",
- "Ze4 = Ze3 + Z2\n",
- "I4 = E2/Ze4\n",
- "I5 = (Z1/(Z3 + Z1))*I4\n",
- "I6 = (Z3/(Z3 + Z1))*I4\n",
- " #If the network of Figure 32.18 is superimposed on the network of Figure 32.17, \n",
- " #it can be seen that the current in the (4+i3) ohm impedance is given by\n",
- "i1 = I1 + I6\n",
- "i2 = I3 + I4\n",
- "i3 = I2 - I5\n",
- " #magnitude\n",
- "i1mag = abs(i1)\n",
- "i2mag = abs(i2)\n",
- "E1mag = abs(E1)\n",
- "E2mag = abs(E2)\n",
- " #phase\n",
- "phi1 = cmath.phase(complex(i1.real,i1.imag))\n",
- "phi2 = cmath.phase(complex(i2.real,i2.imag))\n",
- " #voltage across the(6 + i8) ohm impedance\n",
- "V6i8 = i3*Z3\n",
- "V6i8m = abs(V6i8)\n",
- " #power\n",
- "P = (E1mag*i1mag*math.cos(phi1)) + (E2mag*i2mag*math.cos(phi2 - cmath.phase(complex(E2.real,E2.imag))))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)currents are: \\n \",round(i1.real,2),\" + (\",round( i1.imag,2),\")i A, \\n \",round(i2.real,2),\" + (\",round(i2.imag,2),\")i A \\n and \",round(i3.real,2),\" + (\",round(i3.imag,2),\")i A\"\n",
- "print \"\\n(b)current in the (6 + i8) ohm resistor of the network is \",round(V6i8m,2),\" V\"\n",
- "print \"\\n(c)the total active power delivered to the network is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)currents are: \n",
- " 0.57 + ( 0.62 )i A, \n",
- " 0.56 + ( 1.33 )i A \n",
- " and 0.01 + ( -0.71 )i A\n",
- "\n",
- "(b)current in the (6 + i8) ohm resistor of the network is 7.09 V\n",
- "\n",
- "(c)the total active power delivered to the network is 9.29 W"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 571</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine,(a) the magnitude of the current flowing in the capacitor, \n",
- "#(b) the p.d. across the 5 ohm resistance, (c) the active power dissipated in the 20 ohm resistance and \n",
- "#(d) the total active power taken from the supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 50;# in volts\n",
- "rv2 = 30;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 20;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = -1j*3;# in ohm\n",
- "R4 = 8;# in ohm\n",
- "R5 = 8;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The network is redrawn with the V2 source removed, as shown in Figure 32.26.\n",
- " #Currents I1 to I5 are shown labelled in Figure 32.26. \n",
- " #current\n",
- "Re1 = R4*R5/(R5 + R4) + R3\n",
- "Re2 = Re1*R2/(R2 + Re1)\n",
- "I1 = V1/(Re2 + R1)\n",
- "I2 = (Re1/(R2 + Re1))*I1\n",
- "I3 = (R2/(Re1 + R2))*I1\n",
- "I4 = (R4/(R4 + R5))*I3\n",
- "I5 = I3 - I4\n",
- " #The original network is redrawn with the V1 source removed, as shown in Figure 32.27.\n",
- " #Currents I6 to I10 are shown labelled in Figure 32.27\n",
- "Re3 = R1*R2/(R1 + R2)\n",
- "Re4 = Re3 + R3\n",
- "Re5 = Re4*R4/(Re4 + R4)\n",
- "Re6 = Re5 + R5\n",
- "I6 = V2/Re6\n",
- "I7 = (Re4/(Re4 + R4))*I6\n",
- "I8 = (R4/(Re4 + R4))*I6\n",
- "I9 = (R1/(R1 + R2))*I8\n",
- "I10 = (R2/(R1 + R2))*I8\n",
- " #current flowing in the capacitor is given by\n",
- "Ic = I3 - I8\n",
- " #magnitude of the current in the capacitor\n",
- "Icmag = abs(Ic)\n",
- "\n",
- "i1 = I2 + I9\n",
- "i1mag = abs(i1)\n",
- " #magnitude of the p.d. across the 5 ohm resistance is given by\n",
- "Vr5m = i1mag*R2\n",
- " #Active power dissipated in the 20 ohm resistance is given by\n",
- "i2 = I1 - I10\n",
- "i2mag = abs(i2)\n",
- "phii2 = cmath.phase(complex(i2.real,i2.imag))\n",
- "Pr20 = R1*(i2mag)**2\n",
- " #Active power developed by the V1\n",
- "P1 = rv1*i2mag*math.cos(phii2)\n",
- " #Active power developed by V2 source\n",
- "i3 = I6 - I5\n",
- "i3mag = abs(i3)\n",
- "phii3 = cmath.phase(complex(i3.real,i3.imag))\n",
- "P2 = rv2*i3mag*math.cos(phii3 - (thetav2*math.pi/180))\n",
- " #Total power developed\n",
- "P = P1 + P2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)the magnitude of the current flowing in the capacitor is \",round(Icmag,2),\" A\"\n",
- "print \"\\n(b) the p.d. across the 5 ohm resistance is \",round(Vr5m,2),\" V\"\n",
- "print \"\\n(c)the active power dissipated in the 20 ohm resistance is \",round(Pr20,0),\" W\"\n",
- "print \"\\n(d)the total active power taken from the supply is \",round(P,1),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)the magnitude of the current flowing in the capacitor is 2.11 A\n",
- "\n",
- "(b) the p.d. across the 5 ohm resistance is 5.85 V\n",
- "\n",
- "(c)the active power dissipated in the 20 ohm resistance is 111.0 W\n",
- "\n",
- "(d)the total active power taken from the supply is 191.9 W"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_2.ipynb deleted file mode 100755 index 985a6ce1..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_2.ipynb +++ /dev/null @@ -1,436 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 32: The superposition theorem</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 564</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine using the superposition theorem, the current in the 20 ohm load and the current in each voltage source.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 100;# in volts\n",
- "rv2 = 50;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "r1 = 25;# in ohm\n",
- "R = 20;# in ohm\n",
- "r2 = 10;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The circuit diagram is shown in Figure 32.7. Following the above procedure:\n",
- " #The network is redrawn with the 50/_90\u00b0 V source removed as shown in Figure 32.8\n",
- " #Currents I1, I2 and I3 are labelled as shown in Figure 32.8.\n",
- "I1 = V1/(r1 + r2*R/(R + r2))\n",
- "I2 = (r2/(r2 + R))*I1\n",
- "I3 = (R/(r2 + R))*I1\n",
- " #The network is redrawn with the 100/_0\u00b0 V source removed as shown in Figure 32.9\n",
- " #Currents I4, I5 and I6 are labelled as shown in Figure 32.9.\n",
- "I4 = V2/(r2 + r1*R/(r1 + R))\n",
- "I5 = (r1/(r1 + R))*I4\n",
- "I6 = (R/(r1 + R))*I4\n",
- " #Figure 32.10 shows Figure 32.9 superimposed on Figure 32.8, giving the currents shown.\n",
- " #Current in the 20 ohm load,\n",
- "I20 = I2 + I5\n",
- " #Current in the 100/_0\u00b0 V source\n",
- "IV1 = I1 - I6\n",
- " #Current in the 50/_90\u00b0 V source\n",
- "IV2 = I4 - I3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current in the 20 ohm load is \",round(I20.real,2),\" + (\",round(I20.imag,2),\")i A\"\n",
- "print \"\\n (b)Current in the 100/_0deg V source is \",round(IV1.real,2),\" + (\",round(IV1.imag,2),\")i A\"\n",
- "print \"\\n (b)Current in the 50/_90deg V source is \",round(IV2.real,2),\" + (\",round(IV2.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current in the 20 ohm load is 1.05 + ( 1.32 )i A\n",
- "\n",
- " (b)Current in the 100/_0deg V source is 3.16 + ( -1.05 )i A\n",
- "\n",
- " (b)Current in the 50/_90deg V source is -2.11 + ( 2.37 )i A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 566</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use the superposition theorem to determine the current in the 4 ohm resistor of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 12;# in volts\n",
- "V2 = 20;# in volts\n",
- "R1 = 5;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 2.5;# in ohm\n",
- "R4 = 6;# in ohm\n",
- "R5 = 2;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #Removing the 20 V source gives the network shown in Figure 32.12.\n",
- " #Currents I1 and I2 are shown labelled in Figure 32.12\n",
- "Re1 = (R4*R5/(R4 + R5)) + R3\n",
- "Re2 = Re1*R2/(Re1 + R2) + R1\n",
- "I1 = V1/Re2\n",
- "I2 = (R2/(Re1 + R2))*I1\n",
- " #Removing the 12 V source from the original network gives the network shown in Figure 32.14.\n",
- " #Currents I3, I4 and I5 are shown labelled in Figure 32.14.\n",
- "Re3 = (R1*R2/(R1 + R2)) + R3\n",
- "Re4 = Re3*R4/(Re3 + R4) + R5\n",
- "I3 = V2/Re4\n",
- "I4 = (R4/(Re3 + R4))*I3\n",
- "I5 = (R1/(R1 + R2))*I4\n",
- " #Superimposing Figure 32.14 on Figure 32.12 shows that the current flowing in the 4 ohm resistor is given by\n",
- "Ir4 = I5 - I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ncurrent in the 4 ohm resistor of the network is \",round(Ir4,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "current in the 4 ohm resistor of the network is 0.48 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 567</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Use the superposition theorem to obtain the current flowing in the \u0005(4 + j3) ohm impedance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 30;# in volts\n",
- "rv2 = 30;# in volts\n",
- "thetav1 = 45;# in degrees\n",
- "thetav2 = -45;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 1j*3;# in ohm\n",
- "R4 = -1j*10;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The network is redrawn with V2 removed, as shown in Figure 32.17.\n",
- " #Current I1 and I2 are shown in Figure 32.17. From Figure 32.17,\n",
- "Re1 = R4*(R2 + R3)/(R4 + R3 + R2)\n",
- "Re2 = Re1 + R1\n",
- " #current\n",
- "I1 = V1/Re2\n",
- "I2 = (R4/(R2 + R3 + R4))*I1\n",
- " #The original network is redrawn with V1 removed, as shown in Figure 32.18\n",
- " #Currents I3 and I4 are shown in Figure 32.18. From Figure 32.18,\n",
- "Re3 = R1*(R2 + R3)/(R1 + R3 + R2)\n",
- "Re4 = Re3 + R4\n",
- "I3 = V2/Re4\n",
- "I4 = (R1/(R2 + R3 + R1))*I3\n",
- " #If the network of Figure 32.18 is superimposed on the network of Figure 32.17, \n",
- " #it can be seen that the current in the (4+i3) ohm impedance is given by\n",
- "Ir4i3 = I2 - I4\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"current in (4 + i3) ohm impedance of the network is \",round(Ir4i3.real,2),\" + (\",round( Ir4i3.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "current in (4 + i3) ohm impedance of the network is 2.15 + ( 0.42 )i A\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 568</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine, using the superposition theorem, \n",
- "#(a) the current in each branch,\n",
- "#(b) the magnitude of the voltage across the \u0005(6 + j8) ohm\u0006 impedance,\n",
- "#and (c) the total active power delivered to the network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 5 + 0j;# in volts\n",
- "E2 = 2 + 4j;# in volts\n",
- "Z1 = 3 + 4j;# in ohm\n",
- "Z2 = 2 - 5j;# in ohm\n",
- "Z3 = 6 + 8j;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #The original network is redrawn with E2 removed, as shown in Figure 32.20.\n",
- " #Currents I1, I2 and I3 are labelled as shown in Figure 32.20.\n",
- "Ze1 = Z3*Z2/(Z3 + Z2)\n",
- "Ze2 = Ze1 + Z1\n",
- " #current\n",
- "I1 = E1/Ze2\n",
- "I2 = (Z2/(Z3 + Z2))*I1\n",
- "I3 = (Z3/(Z3 + Z2))*I1\n",
- " #The original network is redrawn with E1 removed, as shown in Figure 32.22\n",
- " #Currents I4, I5 and I6 are shown labelled in Figure 32.22 \n",
- " #with I4 flowing away from the positive terminal of the E2 source.\n",
- "Ze3 = Z3*Z1/(Z3 + Z1)\n",
- "Ze4 = Ze3 + Z2\n",
- "I4 = E2/Ze4\n",
- "I5 = (Z1/(Z3 + Z1))*I4\n",
- "I6 = (Z3/(Z3 + Z1))*I4\n",
- " #If the network of Figure 32.18 is superimposed on the network of Figure 32.17, \n",
- " #it can be seen that the current in the (4+i3) ohm impedance is given by\n",
- "i1 = I1 + I6\n",
- "i2 = I3 + I4\n",
- "i3 = I2 - I5\n",
- " #magnitude\n",
- "i1mag = abs(i1)\n",
- "i2mag = abs(i2)\n",
- "E1mag = abs(E1)\n",
- "E2mag = abs(E2)\n",
- " #phase\n",
- "phi1 = cmath.phase(complex(i1.real,i1.imag))\n",
- "phi2 = cmath.phase(complex(i2.real,i2.imag))\n",
- " #voltage across the(6 + i8) ohm impedance\n",
- "V6i8 = i3*Z3\n",
- "V6i8m = abs(V6i8)\n",
- " #power\n",
- "P = (E1mag*i1mag*math.cos(phi1)) + (E2mag*i2mag*math.cos(phi2 - cmath.phase(complex(E2.real,E2.imag))))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)currents are: \\n \",round(i1.real,2),\" + (\",round( i1.imag,2),\")i A, \\n \",round(i2.real,2),\" + (\",round(i2.imag,2),\")i A \\n and \",round(i3.real,2),\" + (\",round(i3.imag,2),\")i A\"\n",
- "print \"\\n(b)current in the (6 + i8) ohm resistor of the network is \",round(V6i8m,2),\" V\"\n",
- "print \"\\n(c)the total active power delivered to the network is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)currents are: \n",
- " 0.57 + ( 0.62 )i A, \n",
- " 0.56 + ( 1.33 )i A \n",
- " and 0.01 + ( -0.71 )i A\n",
- "\n",
- "(b)current in the (6 + i8) ohm resistor of the network is 7.09 V\n",
- "\n",
- "(c)the total active power delivered to the network is 9.29 W"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 571</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine,(a) the magnitude of the current flowing in the capacitor, \n",
- "#(b) the p.d. across the 5 ohm resistance, (c) the active power dissipated in the 20 ohm resistance and \n",
- "#(d) the total active power taken from the supply.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 50;# in volts\n",
- "rv2 = 30;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 20;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = -1j*3;# in ohm\n",
- "R4 = 8;# in ohm\n",
- "R5 = 8;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The network is redrawn with the V2 source removed, as shown in Figure 32.26.\n",
- " #Currents I1 to I5 are shown labelled in Figure 32.26. \n",
- " #current\n",
- "Re1 = R4*R5/(R5 + R4) + R3\n",
- "Re2 = Re1*R2/(R2 + Re1)\n",
- "I1 = V1/(Re2 + R1)\n",
- "I2 = (Re1/(R2 + Re1))*I1\n",
- "I3 = (R2/(Re1 + R2))*I1\n",
- "I4 = (R4/(R4 + R5))*I3\n",
- "I5 = I3 - I4\n",
- " #The original network is redrawn with the V1 source removed, as shown in Figure 32.27.\n",
- " #Currents I6 to I10 are shown labelled in Figure 32.27\n",
- "Re3 = R1*R2/(R1 + R2)\n",
- "Re4 = Re3 + R3\n",
- "Re5 = Re4*R4/(Re4 + R4)\n",
- "Re6 = Re5 + R5\n",
- "I6 = V2/Re6\n",
- "I7 = (Re4/(Re4 + R4))*I6\n",
- "I8 = (R4/(Re4 + R4))*I6\n",
- "I9 = (R1/(R1 + R2))*I8\n",
- "I10 = (R2/(R1 + R2))*I8\n",
- " #current flowing in the capacitor is given by\n",
- "Ic = I3 - I8\n",
- " #magnitude of the current in the capacitor\n",
- "Icmag = abs(Ic)\n",
- "\n",
- "i1 = I2 + I9\n",
- "i1mag = abs(i1)\n",
- " #magnitude of the p.d. across the 5 ohm resistance is given by\n",
- "Vr5m = i1mag*R2\n",
- " #Active power dissipated in the 20 ohm resistance is given by\n",
- "i2 = I1 - I10\n",
- "i2mag = abs(i2)\n",
- "phii2 = cmath.phase(complex(i2.real,i2.imag))\n",
- "Pr20 = R1*(i2mag)**2\n",
- " #Active power developed by the V1\n",
- "P1 = rv1*i2mag*math.cos(phii2)\n",
- " #Active power developed by V2 source\n",
- "i3 = I6 - I5\n",
- "i3mag = abs(i3)\n",
- "phii3 = cmath.phase(complex(i3.real,i3.imag))\n",
- "P2 = rv2*i3mag*math.cos(phii3 - (thetav2*math.pi/180))\n",
- " #Total power developed\n",
- "P = P1 + P2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)the magnitude of the current flowing in the capacitor is \",round(Icmag,2),\" A\"\n",
- "print \"\\n(b) the p.d. across the 5 ohm resistance is \",round(Vr5m,2),\" V\"\n",
- "print \"\\n(c)the active power dissipated in the 20 ohm resistance is \",round(Pr20,0),\" W\"\n",
- "print \"\\n(d)the total active power taken from the supply is \",round(P,1),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)the magnitude of the current flowing in the capacitor is 2.11 A\n",
- "\n",
- "(b) the p.d. across the 5 ohm resistance is 5.85 V\n",
- "\n",
- "(c)the active power dissipated in the 20 ohm resistance is 111.0 W\n",
- "\n",
- "(d)the total active power taken from the supply is 191.9 W"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_32.ipynb index a2070ae8..a2070ae8 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_32.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint.ipynb deleted file mode 100755 index 55e4e2b7..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint.ipynb +++ /dev/null @@ -1,1103 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 33: Thevenin\u2019s and Norton\u2019s theorems</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 578</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current flowing in the capacitor, and (b) the p.d. across the 150 k\u0006ohmresistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 200;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 5000;# in ohm\n",
- "R2 = 20000;# in ohm\n",
- "R3 = -1j*120000;# in ohm\n",
- "R4 = 150000;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Initially the (150-i120)kohm\u0006 impedance is removed from the circuit as shown in Figure 33.13.\n",
- " #Note that, to find the current in the capacitor, \n",
- " #only the capacitor need have been initially removed from the circuit. \n",
- " #However, removing each of the components from the branch through \n",
- " #which the current is required will often result in a simpler solution. \n",
- " #From Figure 33.13,\n",
- " #current, I1 \n",
- "I1 = V/(R1 + R2)\n",
- " #The open-circuit e.m.f. E is equal to the p.d. across the 20 k\u0006ohm resistor, i.e.\n",
- "E = I1*R2\n",
- " #Removing the V1 source gives the network shown in Figure 33.14.\n",
- " #The impedance, z, looking in at the open-circuited terminals is given by\n",
- "z = R1*R2/(R1 + R2)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.15, where current iL is given by\n",
- "ZL = R3 + R4\n",
- "IL = E/(ZL + z)\n",
- "ILmag = abs(IL)\n",
- " #current flowing in the capacitor\n",
- "Ic = ILmag\n",
- " #P.d. across the 150 kohm resistor,\n",
- "Vr150 = ILmag*R4\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)the current flowing in the capacitor is \",round(Ic*1000,2),\" mA\"\n",
- "print \"\\n(b) the p.d. across the 150 ohm resistance is \",round(Vr150,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)the current flowing in the capacitor is 0.82 mA\n",
- "\n",
- "(b) the p.d. across the 150 ohm resistance is 122.93 V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 579</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of current I.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 20;# in volts\n",
- "V2 = 10;# in volts\n",
- "R1 = 2;# in ohm\n",
- "R2 = 1.5;# in ohm\n",
- "L = 235E-6;# in Henry\n",
- "R4 = 3;# in ohm\n",
- "f = 2000;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #The impedance through which current I is flowing is initially removed from the network, as shown in Figure 33.17.\n",
- " #From Figure 33.17,\n",
- " #current, I1 \n",
- "I1 = (V1 - V2)/(R1 + R4)\n",
- " #the open circuit e.m.f. E\n",
- "E = V1 - I1*R1\n",
- " #When the sources of e.m.f. are removed from the circuit, \n",
- " #the impedance, z, \u2018looking in\u2019 at the break is given by\n",
- "z = R1*R4/(R1 + R4)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.18, where inductive reactance,\n",
- "XL = 2*math.pi*f*L\n",
- "R3 = 1j*XL\n",
- " #Hence current\n",
- "I = E/(R2 + R3 + z)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current I is \",round(I.real,2),\" + (\",round(I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current I is 2.7 + ( -2.95 )i A\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 580</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the power dissipated in the 48 ohm resistor\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 50;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = -1j*400;# in ohm\n",
- "R2 = 300;# in ohm\n",
- "R3 = 144j;# in ohm\n",
- "R4 = 48;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The R3 and R4 impedance is initially removed from the network as shown in Figure 33.20.\n",
- " #From Figure 33.20,\n",
- " #current, I\n",
- "i = V/(R1 + R2)\n",
- " #the open circuit e.m.f. E\n",
- "E = i*R2\n",
- " #When the V is removed from the circuit, the impedance, z, \u2018looking in\u2019 at the break is given by\n",
- "z = R1*R2/(R1 + R2)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.21 connected to R# and R4,\n",
- " #Hence current\n",
- "I = E/(R4 + R3 + z)\n",
- "Imag = abs(I)\n",
- " #the power dissipated in the 48 ohm resistor\n",
- "Pr48 = R4*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the power dissipated in the 48 ohm resistor is \",round(Pr48,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the power dissipated in the 48 ohm resistor is 0.75 W"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 581</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the 80 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 100;# in volts\n",
- "R1 = 5;# in ohm\n",
- "R2 = 20;# in ohm\n",
- "R3 = 46;# in ohm\n",
- "R4 = 50;# in ohm\n",
- "R5 = 15;# in ohm\n",
- "R6 = 60;# in ohm\n",
- "R7 = 16;# in ohm\n",
- "R8 = 80;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #One method of analysing a multi-branch network as shown in Figure 33.22 \n",
- " #is to use Th\u00b4evenin\u2019s theorem on one part of the network at a time. \n",
- " #For example, the part of the circuit to the left of AA may be reduced to a Th\u00b4evenin equivalent circuit.\n",
- " #From Figure 33.23,\n",
- "E1 = (R2/(R1 + R2))*V\n",
- "z1 = R1*R2/(R1 + R2)\n",
- " #Thus the network of Figure 33.22 reduces to that of Figure 33.24. \n",
- " #The part of the network shown in Figure 33.24 to the left of BB may be reduced \n",
- " #to a Th\u00b4evenin equivalent circuit, where\n",
- "E2 = (R4/(R3 + R4 + z1))*E1\n",
- "z2 = R4*(z1 + R3)/(R4 + z1 + R3)\n",
- " #Thus the original network reduces to that shown in Figure 33.25. \n",
- " #The part of the network shown in Figure 33.25 to the left of CC may be reduced \n",
- " #to a Th\u00b4evenin equivalent circuit, where\n",
- "E3 = (R6/(R5 + R6 + z2))*E2\n",
- "z3 = R6*(z2 + R5)/(R5 + z2 + R6)\n",
- " #Thus the original network reduces to that of Figure 33.26, \n",
- " #from which the current in the 80 ohm resistor is given by\n",
- "I = E3/(z3 + R7 + R8)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 80 ohm resistor is \",round(I,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 80 ohm resistor is 0.2 A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 582</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the magnitude of the current flowing in a \t(3.75 + j11) ohmimpedance connected across terminals AB, and\n",
- "#(b) the magnitude of the p.d. across the (3.75 + j11) ohmimpedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 24;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = -1j*3;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 3j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Current I1 shown in Figure 33.27 is given by\n",
- "I1 = V/(R1 + R2 + R3)\n",
- " #The Th\u00b4evenin equivalent voltage, i.e., the open-circuit voltage across terminals AB, is given by\n",
- "E = I1*(R2 + R3)\n",
- " #When the voltage source is removed, the impedance z \u2018looking in\u2019 at AB is given by\n",
- "z = (R2 + R3)*R1/(R1 + R2 + R3)\n",
- " #Thus the Th\u00b4evenin equivalent circuit is as shown in Figure 33.28.\n",
- " #when (3.75 + i11) ohm impedance connected across terminals AB, \n",
- " #the current I flowing in the impedance is given by\n",
- "R = 3.75 + 11j;# in ohms\n",
- "I = E/(R + z)\n",
- "Imag = abs(I)\n",
- " #the p.d. across the( 3.75 + i11)ohm impedance.\n",
- "VR = I*R\n",
- "VRmag = abs(VR)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the current I flowing in the (3.75 + i11) impedance is given by is \",round(Imag,2),\" A\"\n",
- "print \"\\n (b) the magnitude of the p.d. across the impedance is \",round(VRmag,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the current I flowing in the (3.75 + i11) impedance is given by is 3.0 A\n",
- "\n",
- " (b) the magnitude of the p.d. across the impedance is 34.86 V"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 583</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the capacitor of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 16.55;# in volts\n",
- "thetav = -22.62;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = 1j*2;# in ohm\n",
- "R3 = 1j*6;# in ohm\n",
- "R4 = 3;# in ohm\n",
- "R5 = 5;# in ohm\n",
- "R6 = -1*1j*8;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The capacitor is removed from branch AB, as shown in Figure 33.30.\n",
- " #Impedance, Z\n",
- "Z1 = R3 + R4 + R5\n",
- "Z = R1 + (Z1*R2/(R2 + Z1))\n",
- "I1 = V/Z\n",
- "I2 = (R2/(R2 +Z1))*I1\n",
- " #The open-circuit voltage, E\n",
- "E = I2*R5\n",
- " #If the voltage source is removed from Figure 33.30, the impedance, z, \u2018looking in\u2019 at AB is given by\n",
- "z = R5*((R1*R2/(R1 + R2)) + R3 + R4)/(R5 + ((R1*R2/(R1 + R2)) + R3 + R4))\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.31, \n",
- " #where the current flowing in the capacitor, I, is given by\n",
- "I = E/(z + R6)\n",
- "Imag = abs(I)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the capacitor of the network is \",round(Imag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the capacitor of the network is 0.43 A"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 584</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the power dissipated by a 2 ohm resistor connected across PQ.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 5;# in volts\n",
- "rv2 = 10;# in volts\n",
- "thetav1 = 45;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "R1 = 8;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = 3j;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #Current I1 shown in Figure 33.32 is given by\n",
- "I1 = V2/(R2 + R3 + R4)\n",
- " #Hence the voltage drop across the 5 ohm\u0006 resistor is given by VX is in the direction shown in Figure 33.32,\n",
- "Vx = I1*R2\n",
- " #The open-circuit voltage E across PQ is the phasor sum of V1, Vx and V2, as shown in Figure 33.33.\n",
- "E = V2 - V1 - Vx\n",
- " #The impedance, z, \u2018looking in\u2019 at terminals PQ with the voltage sources removed is given by\n",
- "z = R1 + R2*(R3 + R4)/(R2 + R3 + R4)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.34 with the 2 ohm resistance connected across terminals PQ.\n",
- " #The current flowing in the 2 ohm\u0006 resistance is given by\n",
- "R = 2;# in ohms\n",
- "I = E/(z + R)\n",
- "Imag = abs(I)\n",
- " #power P dissipated in the 2 ohm\u0006 resistor is given by\n",
- "Pr2 = R*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n power P dissipated in the 2 ohm resistor is \",round(Pr2,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " power P dissipated in the 2 ohm resistor is 0.07 W"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 585</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the capacitor, and its direction\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 30;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 15;# in ohm\n",
- "R2 = 40;# in ohm\n",
- "R3 = 20j;# in ohm\n",
- "R4 = 20;# in ohm\n",
- "R5 = 5j;# in ohm\n",
- "R6 = 5;# in ohm\n",
- "R7 = -1j*25;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The R7\u0006 is initially removed from the network, as shown in Figure 33.36\n",
- "Z1 = R1\n",
- "Z2 = R2\n",
- "Z3 = R3 + R4\n",
- "Z4 = R5 + R6\n",
- " #P.d. between A and C,\n",
- "Vac = (Z1/(Z1 + Z4))*V\n",
- " #P.d. between B and C,\n",
- "Vbc = (Z2/(Z2 + Z3))*V\n",
- " #Assuming that point A is at a higher potential than point B, then the p.d. between A and B is\n",
- "Vab = Vac - Vbc\n",
- " #the open-circuit voltage across AB is given by\n",
- "E = Vab\n",
- " #Point C is at a potential of V . Between C and A is a volt drop of Vac. Hence the voltage at point A is\n",
- "Va = V - Vac\n",
- " #Between points C and B is a voltage drop of Vbc. Hence the voltage at point B\n",
- "Vb = V - Vbc\n",
- " #Replacing the V source with a short-circuit (i.e., zero internal impedance) \n",
- " #gives the network shown in Figure 33.37(a). The network is shown redrawn in Figure 33.37(b) \n",
- " #and simplified in Figure 33.37(c). Hence the impedance, z, \u2018looking in\u2019 at terminals AB is given by\n",
- "z = Z1*Z4/(Z1 + Z4) + Z2*Z3/(Z2 + Z3)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.38, where current I is given by\n",
- "I = E/(z + R7)\n",
- "Imag = abs(I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the capacitor is \",round(Imag,2),\" A in direction from B to A.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the capacitor is 0.13 A in direction from B to A."
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 589</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of current I in the circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 5;# in volts\n",
- "R1 = 2;# in ohm\n",
- "R2 = 3;# in ohm\n",
- "R3 = -1j*3;# in ohm\n",
- "R4 = 2.8;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #The branch containing the R4 is short-circuited, as shown in Figure 33.48.\n",
- " #The R2 in parallel with a short-circuit is the same as R2 in parallel with 0 ohm \n",
- " #giving an equivalent impedance of\n",
- "Z1 = R2*0/(R3 + 0)\n",
- " #Hence the network reduces to that shown in Figure 33.49, where\n",
- "Isc = V/R1\n",
- " #If the Voltage source is removed from the network the input impedance, z, \u2018looking-in\u2019 \n",
- " #at a break made in AB of Figure 33.48 gives\n",
- "z = R1*R2/(R1 + R2)\n",
- " #The Norton equivalent network is shown in Figure 33.51, where current I is given by\n",
- "I = (z/(z + R4 + R3))*Isc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the capacitor is \",round(I.real,2),\" + (\", round(I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the capacitor is 0.48 + ( 0.36 )i A"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 589</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 20;# in volts\n",
- "V2 = 10;# in volts\n",
- "R1 = 2;# in ohm\n",
- "R2 = 1.5;# in ohm\n",
- "R3 = 2.95j;# in ohm\n",
- "R4 = 3;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #The inductive branch is initially short-circuited, as shown in Figure 33.53.\n",
- " #From Figure 33.53,\n",
- "I1 = V1/R1\n",
- "I2 = V2/R4\n",
- "Isc = I1 + I2\n",
- " #If the voltage sources are removed, the impedance, z, \u2018looking in\u2019 at a break made in AB is given by\n",
- "z = R1*R4/(R1 + R4)\n",
- " #The Norton equivalent network is shown in Figure 33.54, where current I is given by\n",
- "I = (z/(z + R2 + R3))*Isc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the inductive branch is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the inductive branch is 2.7 + ( -2.95 )i A"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 590</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 10;# in volts\n",
- "R1 = 1;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = -2j;# in ohm\n",
- "\n",
- "#calculation:\n",
- "Isc = V/R3\n",
- "z = 1/(1/R2 + 1/R3 + 1/R4)\n",
- "I = (z/(R1 + z))*Isc\n",
- "pd1 = abs(I)*R1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the magnitude of the p.d. across the 1 ohm resistor is \",round(pd1, 2),\"V\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the magnitude of the p.d. across the 1 ohm resistor is 1.58 V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 591</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the power dissipated in a 5 ohm resistor connected between A and B.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 20;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 2;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 3j;# in ohm\n",
- "R4 = -1j*3;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Terminals AB are initially short-circuited, as shown in Figure 33.61.\n",
- " #The circuit impedance Z presented to the voltage source is given by\n",
- "Z = R1 + R4*(R2 + R3)/(R2 + R3 + R4)\n",
- " #Thus current I in Figure 33.61 is given by\n",
- "I = V/Z\n",
- "Isc = ((R2 + R3)/(R2 + R3 + R4))*I\n",
- " #Removing the voltage source of Figure 33.60 gives the network Figure 33.62 of Figure 33.62. \n",
- " #Impedance, z, \u2018looking in\u2019 at terminals AB is given by\n",
- "z = R4 + R1*(R2 + R3)/(R2 + R3 + R1)\n",
- " #The Norton equivalent network is shown in Figure 33.63.\n",
- "R = 5;# in ohms\n",
- " #Current IL\n",
- "IL = (z/(z + R))*Isc\n",
- "ILmag = abs(IL)\n",
- " #the power dissipated in the 5 ohm resistor is\n",
- "Pr5 = R*ILmag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the power dissipated in the 5 ohm resistor is \",round(Pr5,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the power dissipated in the 5 ohm resistor is 22.54 W"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 592</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the magnitude of the current flowing in a 2 ohm resistor connected across PQ.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 5;# in volts\n",
- "rv2 = 10;# in volts\n",
- "thetav1 = 45;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "R1 = 8;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = 3j;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #Terminals PQ are initially short-circuited, as shown in Figure 33.65.\n",
- " #Currents I1 and I2 are shown labelled. Kirchhoff\u2019s laws are used.\n",
- " #For loop ABCD, and moving anticlockwise,\n",
- " #I1*(R2 + R3 + R4) + I2*(R3 + R4) = V2\n",
- " #For loop DPQC, and moving clockwise,\n",
- " #R2*I1 - R1*I2 = V2 - V1\n",
- " #Solving Equations by using determinants gives\n",
- "d1 = [[V2, (R3 + R4)],[(V2 - V1), -1*R1]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(R2 + R3 + R4), V2],[R2, (V2 - V1)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[(R2 + R3 + R4), (R3 + R4)],[R2, -1*R1]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- " #the short-circuit current Isc\n",
- "Isc = I2\n",
- " #The impedance, z, \u2018looking in\u2019 at a break made between P and Q is given by\n",
- "z = R1 + R2*(R3 + R4)/(R2 + R3 + R4)\n",
- " #The Norton equivalent circuit is shown in Figure 33.66, where current I is given by\n",
- "R = 2;#in ohm\n",
- "I = (z/(z + R))*Isc\n",
- "Imag = abs(I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the magnitude of the current flowing 5 ohm resistor is \",round(Imag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the magnitude of the current flowing 5 ohm resistor is 0.19 A"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 595</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the magnitude of the current flowing in the \t(1.8 + j4) ohm impedance connected between terminals A and B\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 12;# in volts\n",
- "E2 = 24;# in volts\n",
- "Z1 = 3;# in ohm\n",
- "Z2 = 2;# in ohm\n",
- "R1 = 4j;# in ohm\n",
- "R2 = 1.8;# in ohm\n",
- "\n",
- "#calculation:\n",
- "Z3 = R1 + R2\n",
- " #For the branch containing the E1 source, conversion to a Norton equivalent network gives\n",
- "Isc1 = E1/Z1\n",
- " #For the branch containing the E2 source, conversion to a Norton equivalent circuit gives\n",
- "Isc2 = E2/Z2\n",
- " #Thus Figure 33.73 shows a network equivalent to Figure 33.72. From Figure 33.73, \n",
- " #the total short-circuit current\n",
- "Isc = Isc1 + Isc2\n",
- " #the total impedance is given by\n",
- "z = Z1*Z2/(Z1 + Z2)\n",
- " #Thus Figure 33.73 simplifies to Figure 33.74.\n",
- " #The open-circuit voltage across AB of Figure 33.74, E\n",
- "E = Isc*z\n",
- " #the impedance \u2018looking in\u2019 at AB,is z\n",
- " #the Th\u00b4evenin equivalent circuit is as shown in Figure 33.75.\n",
- "R = 1.8 + 4j;# in ohm\n",
- " #when R impedance is connected to terminals AB of Figure 33.75, the current I flowing is given by\n",
- "I = E/(z + R)\n",
- "Imag = abs(I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the magnitude of the current flowing (1.8 + i4) ohm resistor is \",round(Imag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the magnitude of the current flowing (1.8 + i4) ohm resistor is 3.84 A"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 596</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the magnitude of the current flowing in the capacitive branch connected to terminals AB.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 5;# in volts\n",
- "V2 = 10;# in volts\n",
- "i = 0.001;# in Amperes\n",
- "R1 = 1000;# in ohm\n",
- "R2 = 4000;# in ohm\n",
- "R3 = 2000;# in ohm\n",
- "R4 = 200;# in ohm\n",
- "R5 = -1j*4000;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #For the branch containing the V1 source, conversion to a Norton equivalent network gives\n",
- "Isc1 = V1/R1\n",
- "z1 = R1\n",
- " #For the branch containing the V2 source, conversion to a Norton equivalent circuit gives\n",
- "Isc2 = V2/R2\n",
- "z2 = R2\n",
- " #Thus the circuit of Figure 33.76 converts to that of Figure 33.77.\n",
- " #The above two Norton equivalent networks shown in Figure 33.77 may be combined, \n",
- " #since the total short-circuit current is\n",
- "Isc = Isc1 + Isc2\n",
- " #the total impedance is given by\n",
- "Z1 = z1*z2/(z1 + z2)\n",
- " #Both of the Norton equivalent networks shown in Figure 33.78 may be converted to Th\u00b4evenin equivalent circuits. \n",
- " #Open-circuit voltage across CD is\n",
- "Ecd = Isc*Z1\n",
- " #the impedance \u2018looking in\u2019 at CD is Z1\n",
- " #Open-circuit voltage across EF\n",
- "Eef = i*R3\n",
- " #the impedance \u2018looking in\u2019 Figure 33.79 at EF\n",
- "Z2 = R3\n",
- " #Thus Figure 33.78 converts to Figure 33.79.\n",
- " #Combining the two Th\u00b4evenin circuits gives e.m.f.\n",
- "E = Ecd - Eef\n",
- " #impedance z\n",
- "z = Z1 + Z2\n",
- " #the Th\u00b4evenin equivalent circuit for terminals AB of Figure 33.76 is as shown in Figure 33.80.\n",
- "Z3 = R4 + R5\n",
- " #If an impedance Z3 is connected across terminals AB, then the current I flowing is given by\n",
- "I = E/(z + Z3)\n",
- "Imag = abs(I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current in the capacitive branch is \", Imag*1000,\"mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current in the capacitive branch is 0.8 mA"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 597</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the power dissipated in a \t(600 - j800) ohm impedance connected between A and B\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 5;# in volts\n",
- "i = 0.004;# in Amperes\n",
- "R1 = 2000;# in ohm\n",
- "R2 = 1000j;# in ohm\n",
- "\n",
- "#calculation: \n",
- " #Converting the Th\u00b4evenin circuit to a Norton network gives\n",
- "Isc1 = V/R2\n",
- " #Thus Figure 33.81 converts to that shown in Figure 33.82. \n",
- " #The two Norton equivalent networks may be combined, giving\n",
- "Isc = Isc1 + i\n",
- "z = R1*R2/(R1 + R2)\n",
- " #This results in the equivalent network shown in Figure 33.83. \n",
- " #Converting to an equivalent Th\u00b4evenin circuit gives open circuit e.m.f. across AB,\n",
- "E = Isc*z\n",
- " #Thus the The\u00b4venin equivalent circuit is as shown in Figure 33.84.\n",
- "R = 600 - 800j;# in ohms\n",
- " #When a R impedance is connected across AB, the current I flowing is given by\n",
- "I = E/(z + R)\n",
- "Imag = abs(I)\n",
- " #the power dissipated in the R resistor is\n",
- "PR = R.real*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the power dissipated in the (600 - i800) ohm resistor is \",round(PR*1000,2),\"mW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the power dissipated in the (600 - i800) ohm resistor is 19.68 mW"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_1.ipynb deleted file mode 100755 index 55e4e2b7..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_1.ipynb +++ /dev/null @@ -1,1103 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 33: Thevenin\u2019s and Norton\u2019s theorems</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 578</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current flowing in the capacitor, and (b) the p.d. across the 150 k\u0006ohmresistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 200;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 5000;# in ohm\n",
- "R2 = 20000;# in ohm\n",
- "R3 = -1j*120000;# in ohm\n",
- "R4 = 150000;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Initially the (150-i120)kohm\u0006 impedance is removed from the circuit as shown in Figure 33.13.\n",
- " #Note that, to find the current in the capacitor, \n",
- " #only the capacitor need have been initially removed from the circuit. \n",
- " #However, removing each of the components from the branch through \n",
- " #which the current is required will often result in a simpler solution. \n",
- " #From Figure 33.13,\n",
- " #current, I1 \n",
- "I1 = V/(R1 + R2)\n",
- " #The open-circuit e.m.f. E is equal to the p.d. across the 20 k\u0006ohm resistor, i.e.\n",
- "E = I1*R2\n",
- " #Removing the V1 source gives the network shown in Figure 33.14.\n",
- " #The impedance, z, looking in at the open-circuited terminals is given by\n",
- "z = R1*R2/(R1 + R2)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.15, where current iL is given by\n",
- "ZL = R3 + R4\n",
- "IL = E/(ZL + z)\n",
- "ILmag = abs(IL)\n",
- " #current flowing in the capacitor\n",
- "Ic = ILmag\n",
- " #P.d. across the 150 kohm resistor,\n",
- "Vr150 = ILmag*R4\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)the current flowing in the capacitor is \",round(Ic*1000,2),\" mA\"\n",
- "print \"\\n(b) the p.d. across the 150 ohm resistance is \",round(Vr150,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)the current flowing in the capacitor is 0.82 mA\n",
- "\n",
- "(b) the p.d. across the 150 ohm resistance is 122.93 V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 579</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of current I.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 20;# in volts\n",
- "V2 = 10;# in volts\n",
- "R1 = 2;# in ohm\n",
- "R2 = 1.5;# in ohm\n",
- "L = 235E-6;# in Henry\n",
- "R4 = 3;# in ohm\n",
- "f = 2000;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #The impedance through which current I is flowing is initially removed from the network, as shown in Figure 33.17.\n",
- " #From Figure 33.17,\n",
- " #current, I1 \n",
- "I1 = (V1 - V2)/(R1 + R4)\n",
- " #the open circuit e.m.f. E\n",
- "E = V1 - I1*R1\n",
- " #When the sources of e.m.f. are removed from the circuit, \n",
- " #the impedance, z, \u2018looking in\u2019 at the break is given by\n",
- "z = R1*R4/(R1 + R4)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.18, where inductive reactance,\n",
- "XL = 2*math.pi*f*L\n",
- "R3 = 1j*XL\n",
- " #Hence current\n",
- "I = E/(R2 + R3 + z)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current I is \",round(I.real,2),\" + (\",round(I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current I is 2.7 + ( -2.95 )i A\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 580</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the power dissipated in the 48 ohm resistor\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 50;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = -1j*400;# in ohm\n",
- "R2 = 300;# in ohm\n",
- "R3 = 144j;# in ohm\n",
- "R4 = 48;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The R3 and R4 impedance is initially removed from the network as shown in Figure 33.20.\n",
- " #From Figure 33.20,\n",
- " #current, I\n",
- "i = V/(R1 + R2)\n",
- " #the open circuit e.m.f. E\n",
- "E = i*R2\n",
- " #When the V is removed from the circuit, the impedance, z, \u2018looking in\u2019 at the break is given by\n",
- "z = R1*R2/(R1 + R2)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.21 connected to R# and R4,\n",
- " #Hence current\n",
- "I = E/(R4 + R3 + z)\n",
- "Imag = abs(I)\n",
- " #the power dissipated in the 48 ohm resistor\n",
- "Pr48 = R4*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the power dissipated in the 48 ohm resistor is \",round(Pr48,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the power dissipated in the 48 ohm resistor is 0.75 W"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 581</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the 80 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 100;# in volts\n",
- "R1 = 5;# in ohm\n",
- "R2 = 20;# in ohm\n",
- "R3 = 46;# in ohm\n",
- "R4 = 50;# in ohm\n",
- "R5 = 15;# in ohm\n",
- "R6 = 60;# in ohm\n",
- "R7 = 16;# in ohm\n",
- "R8 = 80;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #One method of analysing a multi-branch network as shown in Figure 33.22 \n",
- " #is to use Th\u00b4evenin\u2019s theorem on one part of the network at a time. \n",
- " #For example, the part of the circuit to the left of AA may be reduced to a Th\u00b4evenin equivalent circuit.\n",
- " #From Figure 33.23,\n",
- "E1 = (R2/(R1 + R2))*V\n",
- "z1 = R1*R2/(R1 + R2)\n",
- " #Thus the network of Figure 33.22 reduces to that of Figure 33.24. \n",
- " #The part of the network shown in Figure 33.24 to the left of BB may be reduced \n",
- " #to a Th\u00b4evenin equivalent circuit, where\n",
- "E2 = (R4/(R3 + R4 + z1))*E1\n",
- "z2 = R4*(z1 + R3)/(R4 + z1 + R3)\n",
- " #Thus the original network reduces to that shown in Figure 33.25. \n",
- " #The part of the network shown in Figure 33.25 to the left of CC may be reduced \n",
- " #to a Th\u00b4evenin equivalent circuit, where\n",
- "E3 = (R6/(R5 + R6 + z2))*E2\n",
- "z3 = R6*(z2 + R5)/(R5 + z2 + R6)\n",
- " #Thus the original network reduces to that of Figure 33.26, \n",
- " #from which the current in the 80 ohm resistor is given by\n",
- "I = E3/(z3 + R7 + R8)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 80 ohm resistor is \",round(I,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 80 ohm resistor is 0.2 A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 582</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the magnitude of the current flowing in a \t(3.75 + j11) ohmimpedance connected across terminals AB, and\n",
- "#(b) the magnitude of the p.d. across the (3.75 + j11) ohmimpedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 24;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = -1j*3;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 3j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Current I1 shown in Figure 33.27 is given by\n",
- "I1 = V/(R1 + R2 + R3)\n",
- " #The Th\u00b4evenin equivalent voltage, i.e., the open-circuit voltage across terminals AB, is given by\n",
- "E = I1*(R2 + R3)\n",
- " #When the voltage source is removed, the impedance z \u2018looking in\u2019 at AB is given by\n",
- "z = (R2 + R3)*R1/(R1 + R2 + R3)\n",
- " #Thus the Th\u00b4evenin equivalent circuit is as shown in Figure 33.28.\n",
- " #when (3.75 + i11) ohm impedance connected across terminals AB, \n",
- " #the current I flowing in the impedance is given by\n",
- "R = 3.75 + 11j;# in ohms\n",
- "I = E/(R + z)\n",
- "Imag = abs(I)\n",
- " #the p.d. across the( 3.75 + i11)ohm impedance.\n",
- "VR = I*R\n",
- "VRmag = abs(VR)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the current I flowing in the (3.75 + i11) impedance is given by is \",round(Imag,2),\" A\"\n",
- "print \"\\n (b) the magnitude of the p.d. across the impedance is \",round(VRmag,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the current I flowing in the (3.75 + i11) impedance is given by is 3.0 A\n",
- "\n",
- " (b) the magnitude of the p.d. across the impedance is 34.86 V"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 583</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the capacitor of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 16.55;# in volts\n",
- "thetav = -22.62;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = 1j*2;# in ohm\n",
- "R3 = 1j*6;# in ohm\n",
- "R4 = 3;# in ohm\n",
- "R5 = 5;# in ohm\n",
- "R6 = -1*1j*8;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The capacitor is removed from branch AB, as shown in Figure 33.30.\n",
- " #Impedance, Z\n",
- "Z1 = R3 + R4 + R5\n",
- "Z = R1 + (Z1*R2/(R2 + Z1))\n",
- "I1 = V/Z\n",
- "I2 = (R2/(R2 +Z1))*I1\n",
- " #The open-circuit voltage, E\n",
- "E = I2*R5\n",
- " #If the voltage source is removed from Figure 33.30, the impedance, z, \u2018looking in\u2019 at AB is given by\n",
- "z = R5*((R1*R2/(R1 + R2)) + R3 + R4)/(R5 + ((R1*R2/(R1 + R2)) + R3 + R4))\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.31, \n",
- " #where the current flowing in the capacitor, I, is given by\n",
- "I = E/(z + R6)\n",
- "Imag = abs(I)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the capacitor of the network is \",round(Imag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the capacitor of the network is 0.43 A"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 584</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the power dissipated by a 2 ohm resistor connected across PQ.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 5;# in volts\n",
- "rv2 = 10;# in volts\n",
- "thetav1 = 45;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "R1 = 8;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = 3j;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #Current I1 shown in Figure 33.32 is given by\n",
- "I1 = V2/(R2 + R3 + R4)\n",
- " #Hence the voltage drop across the 5 ohm\u0006 resistor is given by VX is in the direction shown in Figure 33.32,\n",
- "Vx = I1*R2\n",
- " #The open-circuit voltage E across PQ is the phasor sum of V1, Vx and V2, as shown in Figure 33.33.\n",
- "E = V2 - V1 - Vx\n",
- " #The impedance, z, \u2018looking in\u2019 at terminals PQ with the voltage sources removed is given by\n",
- "z = R1 + R2*(R3 + R4)/(R2 + R3 + R4)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.34 with the 2 ohm resistance connected across terminals PQ.\n",
- " #The current flowing in the 2 ohm\u0006 resistance is given by\n",
- "R = 2;# in ohms\n",
- "I = E/(z + R)\n",
- "Imag = abs(I)\n",
- " #power P dissipated in the 2 ohm\u0006 resistor is given by\n",
- "Pr2 = R*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n power P dissipated in the 2 ohm resistor is \",round(Pr2,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " power P dissipated in the 2 ohm resistor is 0.07 W"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 585</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the capacitor, and its direction\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 30;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 15;# in ohm\n",
- "R2 = 40;# in ohm\n",
- "R3 = 20j;# in ohm\n",
- "R4 = 20;# in ohm\n",
- "R5 = 5j;# in ohm\n",
- "R6 = 5;# in ohm\n",
- "R7 = -1j*25;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The R7\u0006 is initially removed from the network, as shown in Figure 33.36\n",
- "Z1 = R1\n",
- "Z2 = R2\n",
- "Z3 = R3 + R4\n",
- "Z4 = R5 + R6\n",
- " #P.d. between A and C,\n",
- "Vac = (Z1/(Z1 + Z4))*V\n",
- " #P.d. between B and C,\n",
- "Vbc = (Z2/(Z2 + Z3))*V\n",
- " #Assuming that point A is at a higher potential than point B, then the p.d. between A and B is\n",
- "Vab = Vac - Vbc\n",
- " #the open-circuit voltage across AB is given by\n",
- "E = Vab\n",
- " #Point C is at a potential of V . Between C and A is a volt drop of Vac. Hence the voltage at point A is\n",
- "Va = V - Vac\n",
- " #Between points C and B is a voltage drop of Vbc. Hence the voltage at point B\n",
- "Vb = V - Vbc\n",
- " #Replacing the V source with a short-circuit (i.e., zero internal impedance) \n",
- " #gives the network shown in Figure 33.37(a). The network is shown redrawn in Figure 33.37(b) \n",
- " #and simplified in Figure 33.37(c). Hence the impedance, z, \u2018looking in\u2019 at terminals AB is given by\n",
- "z = Z1*Z4/(Z1 + Z4) + Z2*Z3/(Z2 + Z3)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.38, where current I is given by\n",
- "I = E/(z + R7)\n",
- "Imag = abs(I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the capacitor is \",round(Imag,2),\" A in direction from B to A.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the capacitor is 0.13 A in direction from B to A."
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 589</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of current I in the circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 5;# in volts\n",
- "R1 = 2;# in ohm\n",
- "R2 = 3;# in ohm\n",
- "R3 = -1j*3;# in ohm\n",
- "R4 = 2.8;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #The branch containing the R4 is short-circuited, as shown in Figure 33.48.\n",
- " #The R2 in parallel with a short-circuit is the same as R2 in parallel with 0 ohm \n",
- " #giving an equivalent impedance of\n",
- "Z1 = R2*0/(R3 + 0)\n",
- " #Hence the network reduces to that shown in Figure 33.49, where\n",
- "Isc = V/R1\n",
- " #If the Voltage source is removed from the network the input impedance, z, \u2018looking-in\u2019 \n",
- " #at a break made in AB of Figure 33.48 gives\n",
- "z = R1*R2/(R1 + R2)\n",
- " #The Norton equivalent network is shown in Figure 33.51, where current I is given by\n",
- "I = (z/(z + R4 + R3))*Isc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the capacitor is \",round(I.real,2),\" + (\", round(I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the capacitor is 0.48 + ( 0.36 )i A"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 589</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 20;# in volts\n",
- "V2 = 10;# in volts\n",
- "R1 = 2;# in ohm\n",
- "R2 = 1.5;# in ohm\n",
- "R3 = 2.95j;# in ohm\n",
- "R4 = 3;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #The inductive branch is initially short-circuited, as shown in Figure 33.53.\n",
- " #From Figure 33.53,\n",
- "I1 = V1/R1\n",
- "I2 = V2/R4\n",
- "Isc = I1 + I2\n",
- " #If the voltage sources are removed, the impedance, z, \u2018looking in\u2019 at a break made in AB is given by\n",
- "z = R1*R4/(R1 + R4)\n",
- " #The Norton equivalent network is shown in Figure 33.54, where current I is given by\n",
- "I = (z/(z + R2 + R3))*Isc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the inductive branch is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the inductive branch is 2.7 + ( -2.95 )i A"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 590</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 10;# in volts\n",
- "R1 = 1;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = -2j;# in ohm\n",
- "\n",
- "#calculation:\n",
- "Isc = V/R3\n",
- "z = 1/(1/R2 + 1/R3 + 1/R4)\n",
- "I = (z/(R1 + z))*Isc\n",
- "pd1 = abs(I)*R1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the magnitude of the p.d. across the 1 ohm resistor is \",round(pd1, 2),\"V\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the magnitude of the p.d. across the 1 ohm resistor is 1.58 V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 591</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the power dissipated in a 5 ohm resistor connected between A and B.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 20;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 2;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 3j;# in ohm\n",
- "R4 = -1j*3;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Terminals AB are initially short-circuited, as shown in Figure 33.61.\n",
- " #The circuit impedance Z presented to the voltage source is given by\n",
- "Z = R1 + R4*(R2 + R3)/(R2 + R3 + R4)\n",
- " #Thus current I in Figure 33.61 is given by\n",
- "I = V/Z\n",
- "Isc = ((R2 + R3)/(R2 + R3 + R4))*I\n",
- " #Removing the voltage source of Figure 33.60 gives the network Figure 33.62 of Figure 33.62. \n",
- " #Impedance, z, \u2018looking in\u2019 at terminals AB is given by\n",
- "z = R4 + R1*(R2 + R3)/(R2 + R3 + R1)\n",
- " #The Norton equivalent network is shown in Figure 33.63.\n",
- "R = 5;# in ohms\n",
- " #Current IL\n",
- "IL = (z/(z + R))*Isc\n",
- "ILmag = abs(IL)\n",
- " #the power dissipated in the 5 ohm resistor is\n",
- "Pr5 = R*ILmag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the power dissipated in the 5 ohm resistor is \",round(Pr5,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the power dissipated in the 5 ohm resistor is 22.54 W"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 592</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the magnitude of the current flowing in a 2 ohm resistor connected across PQ.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 5;# in volts\n",
- "rv2 = 10;# in volts\n",
- "thetav1 = 45;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "R1 = 8;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = 3j;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #Terminals PQ are initially short-circuited, as shown in Figure 33.65.\n",
- " #Currents I1 and I2 are shown labelled. Kirchhoff\u2019s laws are used.\n",
- " #For loop ABCD, and moving anticlockwise,\n",
- " #I1*(R2 + R3 + R4) + I2*(R3 + R4) = V2\n",
- " #For loop DPQC, and moving clockwise,\n",
- " #R2*I1 - R1*I2 = V2 - V1\n",
- " #Solving Equations by using determinants gives\n",
- "d1 = [[V2, (R3 + R4)],[(V2 - V1), -1*R1]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(R2 + R3 + R4), V2],[R2, (V2 - V1)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[(R2 + R3 + R4), (R3 + R4)],[R2, -1*R1]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- " #the short-circuit current Isc\n",
- "Isc = I2\n",
- " #The impedance, z, \u2018looking in\u2019 at a break made between P and Q is given by\n",
- "z = R1 + R2*(R3 + R4)/(R2 + R3 + R4)\n",
- " #The Norton equivalent circuit is shown in Figure 33.66, where current I is given by\n",
- "R = 2;#in ohm\n",
- "I = (z/(z + R))*Isc\n",
- "Imag = abs(I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the magnitude of the current flowing 5 ohm resistor is \",round(Imag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the magnitude of the current flowing 5 ohm resistor is 0.19 A"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 595</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the magnitude of the current flowing in the \t(1.8 + j4) ohm impedance connected between terminals A and B\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 12;# in volts\n",
- "E2 = 24;# in volts\n",
- "Z1 = 3;# in ohm\n",
- "Z2 = 2;# in ohm\n",
- "R1 = 4j;# in ohm\n",
- "R2 = 1.8;# in ohm\n",
- "\n",
- "#calculation:\n",
- "Z3 = R1 + R2\n",
- " #For the branch containing the E1 source, conversion to a Norton equivalent network gives\n",
- "Isc1 = E1/Z1\n",
- " #For the branch containing the E2 source, conversion to a Norton equivalent circuit gives\n",
- "Isc2 = E2/Z2\n",
- " #Thus Figure 33.73 shows a network equivalent to Figure 33.72. From Figure 33.73, \n",
- " #the total short-circuit current\n",
- "Isc = Isc1 + Isc2\n",
- " #the total impedance is given by\n",
- "z = Z1*Z2/(Z1 + Z2)\n",
- " #Thus Figure 33.73 simplifies to Figure 33.74.\n",
- " #The open-circuit voltage across AB of Figure 33.74, E\n",
- "E = Isc*z\n",
- " #the impedance \u2018looking in\u2019 at AB,is z\n",
- " #the Th\u00b4evenin equivalent circuit is as shown in Figure 33.75.\n",
- "R = 1.8 + 4j;# in ohm\n",
- " #when R impedance is connected to terminals AB of Figure 33.75, the current I flowing is given by\n",
- "I = E/(z + R)\n",
- "Imag = abs(I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the magnitude of the current flowing (1.8 + i4) ohm resistor is \",round(Imag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the magnitude of the current flowing (1.8 + i4) ohm resistor is 3.84 A"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 596</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the magnitude of the current flowing in the capacitive branch connected to terminals AB.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 5;# in volts\n",
- "V2 = 10;# in volts\n",
- "i = 0.001;# in Amperes\n",
- "R1 = 1000;# in ohm\n",
- "R2 = 4000;# in ohm\n",
- "R3 = 2000;# in ohm\n",
- "R4 = 200;# in ohm\n",
- "R5 = -1j*4000;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #For the branch containing the V1 source, conversion to a Norton equivalent network gives\n",
- "Isc1 = V1/R1\n",
- "z1 = R1\n",
- " #For the branch containing the V2 source, conversion to a Norton equivalent circuit gives\n",
- "Isc2 = V2/R2\n",
- "z2 = R2\n",
- " #Thus the circuit of Figure 33.76 converts to that of Figure 33.77.\n",
- " #The above two Norton equivalent networks shown in Figure 33.77 may be combined, \n",
- " #since the total short-circuit current is\n",
- "Isc = Isc1 + Isc2\n",
- " #the total impedance is given by\n",
- "Z1 = z1*z2/(z1 + z2)\n",
- " #Both of the Norton equivalent networks shown in Figure 33.78 may be converted to Th\u00b4evenin equivalent circuits. \n",
- " #Open-circuit voltage across CD is\n",
- "Ecd = Isc*Z1\n",
- " #the impedance \u2018looking in\u2019 at CD is Z1\n",
- " #Open-circuit voltage across EF\n",
- "Eef = i*R3\n",
- " #the impedance \u2018looking in\u2019 Figure 33.79 at EF\n",
- "Z2 = R3\n",
- " #Thus Figure 33.78 converts to Figure 33.79.\n",
- " #Combining the two Th\u00b4evenin circuits gives e.m.f.\n",
- "E = Ecd - Eef\n",
- " #impedance z\n",
- "z = Z1 + Z2\n",
- " #the Th\u00b4evenin equivalent circuit for terminals AB of Figure 33.76 is as shown in Figure 33.80.\n",
- "Z3 = R4 + R5\n",
- " #If an impedance Z3 is connected across terminals AB, then the current I flowing is given by\n",
- "I = E/(z + Z3)\n",
- "Imag = abs(I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current in the capacitive branch is \", Imag*1000,\"mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current in the capacitive branch is 0.8 mA"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 597</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the power dissipated in a \t(600 - j800) ohm impedance connected between A and B\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 5;# in volts\n",
- "i = 0.004;# in Amperes\n",
- "R1 = 2000;# in ohm\n",
- "R2 = 1000j;# in ohm\n",
- "\n",
- "#calculation: \n",
- " #Converting the Th\u00b4evenin circuit to a Norton network gives\n",
- "Isc1 = V/R2\n",
- " #Thus Figure 33.81 converts to that shown in Figure 33.82. \n",
- " #The two Norton equivalent networks may be combined, giving\n",
- "Isc = Isc1 + i\n",
- "z = R1*R2/(R1 + R2)\n",
- " #This results in the equivalent network shown in Figure 33.83. \n",
- " #Converting to an equivalent Th\u00b4evenin circuit gives open circuit e.m.f. across AB,\n",
- "E = Isc*z\n",
- " #Thus the The\u00b4venin equivalent circuit is as shown in Figure 33.84.\n",
- "R = 600 - 800j;# in ohms\n",
- " #When a R impedance is connected across AB, the current I flowing is given by\n",
- "I = E/(z + R)\n",
- "Imag = abs(I)\n",
- " #the power dissipated in the R resistor is\n",
- "PR = R.real*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the power dissipated in the (600 - i800) ohm resistor is \",round(PR*1000,2),\"mW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the power dissipated in the (600 - i800) ohm resistor is 19.68 mW"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_2.ipynb deleted file mode 100755 index 55e4e2b7..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_2.ipynb +++ /dev/null @@ -1,1103 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 33: Thevenin\u2019s and Norton\u2019s theorems</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 578</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current flowing in the capacitor, and (b) the p.d. across the 150 k\u0006ohmresistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 200;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 5000;# in ohm\n",
- "R2 = 20000;# in ohm\n",
- "R3 = -1j*120000;# in ohm\n",
- "R4 = 150000;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Initially the (150-i120)kohm\u0006 impedance is removed from the circuit as shown in Figure 33.13.\n",
- " #Note that, to find the current in the capacitor, \n",
- " #only the capacitor need have been initially removed from the circuit. \n",
- " #However, removing each of the components from the branch through \n",
- " #which the current is required will often result in a simpler solution. \n",
- " #From Figure 33.13,\n",
- " #current, I1 \n",
- "I1 = V/(R1 + R2)\n",
- " #The open-circuit e.m.f. E is equal to the p.d. across the 20 k\u0006ohm resistor, i.e.\n",
- "E = I1*R2\n",
- " #Removing the V1 source gives the network shown in Figure 33.14.\n",
- " #The impedance, z, looking in at the open-circuited terminals is given by\n",
- "z = R1*R2/(R1 + R2)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.15, where current iL is given by\n",
- "ZL = R3 + R4\n",
- "IL = E/(ZL + z)\n",
- "ILmag = abs(IL)\n",
- " #current flowing in the capacitor\n",
- "Ic = ILmag\n",
- " #P.d. across the 150 kohm resistor,\n",
- "Vr150 = ILmag*R4\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)the current flowing in the capacitor is \",round(Ic*1000,2),\" mA\"\n",
- "print \"\\n(b) the p.d. across the 150 ohm resistance is \",round(Vr150,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)the current flowing in the capacitor is 0.82 mA\n",
- "\n",
- "(b) the p.d. across the 150 ohm resistance is 122.93 V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 579</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of current I.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 20;# in volts\n",
- "V2 = 10;# in volts\n",
- "R1 = 2;# in ohm\n",
- "R2 = 1.5;# in ohm\n",
- "L = 235E-6;# in Henry\n",
- "R4 = 3;# in ohm\n",
- "f = 2000;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #The impedance through which current I is flowing is initially removed from the network, as shown in Figure 33.17.\n",
- " #From Figure 33.17,\n",
- " #current, I1 \n",
- "I1 = (V1 - V2)/(R1 + R4)\n",
- " #the open circuit e.m.f. E\n",
- "E = V1 - I1*R1\n",
- " #When the sources of e.m.f. are removed from the circuit, \n",
- " #the impedance, z, \u2018looking in\u2019 at the break is given by\n",
- "z = R1*R4/(R1 + R4)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.18, where inductive reactance,\n",
- "XL = 2*math.pi*f*L\n",
- "R3 = 1j*XL\n",
- " #Hence current\n",
- "I = E/(R2 + R3 + z)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current I is \",round(I.real,2),\" + (\",round(I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current I is 2.7 + ( -2.95 )i A\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 580</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the power dissipated in the 48 ohm resistor\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 50;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = -1j*400;# in ohm\n",
- "R2 = 300;# in ohm\n",
- "R3 = 144j;# in ohm\n",
- "R4 = 48;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The R3 and R4 impedance is initially removed from the network as shown in Figure 33.20.\n",
- " #From Figure 33.20,\n",
- " #current, I\n",
- "i = V/(R1 + R2)\n",
- " #the open circuit e.m.f. E\n",
- "E = i*R2\n",
- " #When the V is removed from the circuit, the impedance, z, \u2018looking in\u2019 at the break is given by\n",
- "z = R1*R2/(R1 + R2)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.21 connected to R# and R4,\n",
- " #Hence current\n",
- "I = E/(R4 + R3 + z)\n",
- "Imag = abs(I)\n",
- " #the power dissipated in the 48 ohm resistor\n",
- "Pr48 = R4*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the power dissipated in the 48 ohm resistor is \",round(Pr48,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the power dissipated in the 48 ohm resistor is 0.75 W"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 581</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the 80 ohm resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 100;# in volts\n",
- "R1 = 5;# in ohm\n",
- "R2 = 20;# in ohm\n",
- "R3 = 46;# in ohm\n",
- "R4 = 50;# in ohm\n",
- "R5 = 15;# in ohm\n",
- "R6 = 60;# in ohm\n",
- "R7 = 16;# in ohm\n",
- "R8 = 80;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #One method of analysing a multi-branch network as shown in Figure 33.22 \n",
- " #is to use Th\u00b4evenin\u2019s theorem on one part of the network at a time. \n",
- " #For example, the part of the circuit to the left of AA may be reduced to a Th\u00b4evenin equivalent circuit.\n",
- " #From Figure 33.23,\n",
- "E1 = (R2/(R1 + R2))*V\n",
- "z1 = R1*R2/(R1 + R2)\n",
- " #Thus the network of Figure 33.22 reduces to that of Figure 33.24. \n",
- " #The part of the network shown in Figure 33.24 to the left of BB may be reduced \n",
- " #to a Th\u00b4evenin equivalent circuit, where\n",
- "E2 = (R4/(R3 + R4 + z1))*E1\n",
- "z2 = R4*(z1 + R3)/(R4 + z1 + R3)\n",
- " #Thus the original network reduces to that shown in Figure 33.25. \n",
- " #The part of the network shown in Figure 33.25 to the left of CC may be reduced \n",
- " #to a Th\u00b4evenin equivalent circuit, where\n",
- "E3 = (R6/(R5 + R6 + z2))*E2\n",
- "z3 = R6*(z2 + R5)/(R5 + z2 + R6)\n",
- " #Thus the original network reduces to that of Figure 33.26, \n",
- " #from which the current in the 80 ohm resistor is given by\n",
- "I = E3/(z3 + R7 + R8)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the 80 ohm resistor is \",round(I,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the 80 ohm resistor is 0.2 A"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 582</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the magnitude of the current flowing in a \t(3.75 + j11) ohmimpedance connected across terminals AB, and\n",
- "#(b) the magnitude of the p.d. across the (3.75 + j11) ohmimpedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 24;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = -1j*3;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 3j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Current I1 shown in Figure 33.27 is given by\n",
- "I1 = V/(R1 + R2 + R3)\n",
- " #The Th\u00b4evenin equivalent voltage, i.e., the open-circuit voltage across terminals AB, is given by\n",
- "E = I1*(R2 + R3)\n",
- " #When the voltage source is removed, the impedance z \u2018looking in\u2019 at AB is given by\n",
- "z = (R2 + R3)*R1/(R1 + R2 + R3)\n",
- " #Thus the Th\u00b4evenin equivalent circuit is as shown in Figure 33.28.\n",
- " #when (3.75 + i11) ohm impedance connected across terminals AB, \n",
- " #the current I flowing in the impedance is given by\n",
- "R = 3.75 + 11j;# in ohms\n",
- "I = E/(R + z)\n",
- "Imag = abs(I)\n",
- " #the p.d. across the( 3.75 + i11)ohm impedance.\n",
- "VR = I*R\n",
- "VRmag = abs(VR)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the current I flowing in the (3.75 + i11) impedance is given by is \",round(Imag,2),\" A\"\n",
- "print \"\\n (b) the magnitude of the p.d. across the impedance is \",round(VRmag,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the current I flowing in the (3.75 + i11) impedance is given by is 3.0 A\n",
- "\n",
- " (b) the magnitude of the p.d. across the impedance is 34.86 V"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 583</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the capacitor of the network\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 16.55;# in volts\n",
- "thetav = -22.62;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = 1j*2;# in ohm\n",
- "R3 = 1j*6;# in ohm\n",
- "R4 = 3;# in ohm\n",
- "R5 = 5;# in ohm\n",
- "R6 = -1*1j*8;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The capacitor is removed from branch AB, as shown in Figure 33.30.\n",
- " #Impedance, Z\n",
- "Z1 = R3 + R4 + R5\n",
- "Z = R1 + (Z1*R2/(R2 + Z1))\n",
- "I1 = V/Z\n",
- "I2 = (R2/(R2 +Z1))*I1\n",
- " #The open-circuit voltage, E\n",
- "E = I2*R5\n",
- " #If the voltage source is removed from Figure 33.30, the impedance, z, \u2018looking in\u2019 at AB is given by\n",
- "z = R5*((R1*R2/(R1 + R2)) + R3 + R4)/(R5 + ((R1*R2/(R1 + R2)) + R3 + R4))\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.31, \n",
- " #where the current flowing in the capacitor, I, is given by\n",
- "I = E/(z + R6)\n",
- "Imag = abs(I)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the capacitor of the network is \",round(Imag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the capacitor of the network is 0.43 A"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 584</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the power dissipated by a 2 ohm resistor connected across PQ.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 5;# in volts\n",
- "rv2 = 10;# in volts\n",
- "thetav1 = 45;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "R1 = 8;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = 3j;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #Current I1 shown in Figure 33.32 is given by\n",
- "I1 = V2/(R2 + R3 + R4)\n",
- " #Hence the voltage drop across the 5 ohm\u0006 resistor is given by VX is in the direction shown in Figure 33.32,\n",
- "Vx = I1*R2\n",
- " #The open-circuit voltage E across PQ is the phasor sum of V1, Vx and V2, as shown in Figure 33.33.\n",
- "E = V2 - V1 - Vx\n",
- " #The impedance, z, \u2018looking in\u2019 at terminals PQ with the voltage sources removed is given by\n",
- "z = R1 + R2*(R3 + R4)/(R2 + R3 + R4)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.34 with the 2 ohm resistance connected across terminals PQ.\n",
- " #The current flowing in the 2 ohm\u0006 resistance is given by\n",
- "R = 2;# in ohms\n",
- "I = E/(z + R)\n",
- "Imag = abs(I)\n",
- " #power P dissipated in the 2 ohm\u0006 resistor is given by\n",
- "Pr2 = R*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n power P dissipated in the 2 ohm resistor is \",round(Pr2,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " power P dissipated in the 2 ohm resistor is 0.07 W"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 585</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the capacitor, and its direction\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 30;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 15;# in ohm\n",
- "R2 = 40;# in ohm\n",
- "R3 = 20j;# in ohm\n",
- "R4 = 20;# in ohm\n",
- "R5 = 5j;# in ohm\n",
- "R6 = 5;# in ohm\n",
- "R7 = -1j*25;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The R7\u0006 is initially removed from the network, as shown in Figure 33.36\n",
- "Z1 = R1\n",
- "Z2 = R2\n",
- "Z3 = R3 + R4\n",
- "Z4 = R5 + R6\n",
- " #P.d. between A and C,\n",
- "Vac = (Z1/(Z1 + Z4))*V\n",
- " #P.d. between B and C,\n",
- "Vbc = (Z2/(Z2 + Z3))*V\n",
- " #Assuming that point A is at a higher potential than point B, then the p.d. between A and B is\n",
- "Vab = Vac - Vbc\n",
- " #the open-circuit voltage across AB is given by\n",
- "E = Vab\n",
- " #Point C is at a potential of V . Between C and A is a volt drop of Vac. Hence the voltage at point A is\n",
- "Va = V - Vac\n",
- " #Between points C and B is a voltage drop of Vbc. Hence the voltage at point B\n",
- "Vb = V - Vbc\n",
- " #Replacing the V source with a short-circuit (i.e., zero internal impedance) \n",
- " #gives the network shown in Figure 33.37(a). The network is shown redrawn in Figure 33.37(b) \n",
- " #and simplified in Figure 33.37(c). Hence the impedance, z, \u2018looking in\u2019 at terminals AB is given by\n",
- "z = Z1*Z4/(Z1 + Z4) + Z2*Z3/(Z2 + Z3)\n",
- " #The Th\u00b4evenin equivalent circuit is shown in Figure 33.38, where current I is given by\n",
- "I = E/(z + R7)\n",
- "Imag = abs(I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the capacitor is \",round(Imag,2),\" A in direction from B to A.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the capacitor is 0.13 A in direction from B to A."
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 589</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of current I in the circuit\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 5;# in volts\n",
- "R1 = 2;# in ohm\n",
- "R2 = 3;# in ohm\n",
- "R3 = -1j*3;# in ohm\n",
- "R4 = 2.8;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #The branch containing the R4 is short-circuited, as shown in Figure 33.48.\n",
- " #The R2 in parallel with a short-circuit is the same as R2 in parallel with 0 ohm \n",
- " #giving an equivalent impedance of\n",
- "Z1 = R2*0/(R3 + 0)\n",
- " #Hence the network reduces to that shown in Figure 33.49, where\n",
- "Isc = V/R1\n",
- " #If the Voltage source is removed from the network the input impedance, z, \u2018looking-in\u2019 \n",
- " #at a break made in AB of Figure 33.48 gives\n",
- "z = R1*R2/(R1 + R2)\n",
- " #The Norton equivalent network is shown in Figure 33.51, where current I is given by\n",
- "I = (z/(z + R4 + R3))*Isc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the capacitor is \",round(I.real,2),\" + (\", round(I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the capacitor is 0.48 + ( 0.36 )i A"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 589</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 20;# in volts\n",
- "V2 = 10;# in volts\n",
- "R1 = 2;# in ohm\n",
- "R2 = 1.5;# in ohm\n",
- "R3 = 2.95j;# in ohm\n",
- "R4 = 3;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #The inductive branch is initially short-circuited, as shown in Figure 33.53.\n",
- " #From Figure 33.53,\n",
- "I1 = V1/R1\n",
- "I2 = V2/R4\n",
- "Isc = I1 + I2\n",
- " #If the voltage sources are removed, the impedance, z, \u2018looking in\u2019 at a break made in AB is given by\n",
- "z = R1*R4/(R1 + R4)\n",
- " #The Norton equivalent network is shown in Figure 33.54, where current I is given by\n",
- "I = (z/(z + R2 + R3))*Isc\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current flowing in the inductive branch is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current flowing in the inductive branch is 2.7 + ( -2.95 )i A"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 590</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 10;# in volts\n",
- "R1 = 1;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 4;# in ohm\n",
- "R4 = -2j;# in ohm\n",
- "\n",
- "#calculation:\n",
- "Isc = V/R3\n",
- "z = 1/(1/R2 + 1/R3 + 1/R4)\n",
- "I = (z/(R1 + z))*Isc\n",
- "pd1 = abs(I)*R1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the magnitude of the p.d. across the 1 ohm resistor is \",round(pd1, 2),\"V\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the magnitude of the p.d. across the 1 ohm resistor is 1.58 V"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 591</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the power dissipated in a 5 ohm resistor connected between A and B.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 20;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 2;# in ohm\n",
- "R2 = 4;# in ohm\n",
- "R3 = 3j;# in ohm\n",
- "R4 = -1j*3;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Terminals AB are initially short-circuited, as shown in Figure 33.61.\n",
- " #The circuit impedance Z presented to the voltage source is given by\n",
- "Z = R1 + R4*(R2 + R3)/(R2 + R3 + R4)\n",
- " #Thus current I in Figure 33.61 is given by\n",
- "I = V/Z\n",
- "Isc = ((R2 + R3)/(R2 + R3 + R4))*I\n",
- " #Removing the voltage source of Figure 33.60 gives the network Figure 33.62 of Figure 33.62. \n",
- " #Impedance, z, \u2018looking in\u2019 at terminals AB is given by\n",
- "z = R4 + R1*(R2 + R3)/(R2 + R3 + R1)\n",
- " #The Norton equivalent network is shown in Figure 33.63.\n",
- "R = 5;# in ohms\n",
- " #Current IL\n",
- "IL = (z/(z + R))*Isc\n",
- "ILmag = abs(IL)\n",
- " #the power dissipated in the 5 ohm resistor is\n",
- "Pr5 = R*ILmag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the power dissipated in the 5 ohm resistor is \",round(Pr5,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the power dissipated in the 5 ohm resistor is 22.54 W"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 592</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the magnitude of the current flowing in a 2 ohm resistor connected across PQ.\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 5;# in volts\n",
- "rv2 = 10;# in volts\n",
- "thetav1 = 45;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "R1 = 8;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = 3j;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #Terminals PQ are initially short-circuited, as shown in Figure 33.65.\n",
- " #Currents I1 and I2 are shown labelled. Kirchhoff\u2019s laws are used.\n",
- " #For loop ABCD, and moving anticlockwise,\n",
- " #I1*(R2 + R3 + R4) + I2*(R3 + R4) = V2\n",
- " #For loop DPQC, and moving clockwise,\n",
- " #R2*I1 - R1*I2 = V2 - V1\n",
- " #Solving Equations by using determinants gives\n",
- "d1 = [[V2, (R3 + R4)],[(V2 - V1), -1*R1]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[(R2 + R3 + R4), V2],[R2, (V2 - V1)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d = [[(R2 + R3 + R4), (R3 + R4)],[R2, -1*R1]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- " #the short-circuit current Isc\n",
- "Isc = I2\n",
- " #The impedance, z, \u2018looking in\u2019 at a break made between P and Q is given by\n",
- "z = R1 + R2*(R3 + R4)/(R2 + R3 + R4)\n",
- " #The Norton equivalent circuit is shown in Figure 33.66, where current I is given by\n",
- "R = 2;#in ohm\n",
- "I = (z/(z + R))*Isc\n",
- "Imag = abs(I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the magnitude of the current flowing 5 ohm resistor is \",round(Imag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the magnitude of the current flowing 5 ohm resistor is 0.19 A"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 595</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the magnitude of the current flowing in the \t(1.8 + j4) ohm impedance connected between terminals A and B\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 12;# in volts\n",
- "E2 = 24;# in volts\n",
- "Z1 = 3;# in ohm\n",
- "Z2 = 2;# in ohm\n",
- "R1 = 4j;# in ohm\n",
- "R2 = 1.8;# in ohm\n",
- "\n",
- "#calculation:\n",
- "Z3 = R1 + R2\n",
- " #For the branch containing the E1 source, conversion to a Norton equivalent network gives\n",
- "Isc1 = E1/Z1\n",
- " #For the branch containing the E2 source, conversion to a Norton equivalent circuit gives\n",
- "Isc2 = E2/Z2\n",
- " #Thus Figure 33.73 shows a network equivalent to Figure 33.72. From Figure 33.73, \n",
- " #the total short-circuit current\n",
- "Isc = Isc1 + Isc2\n",
- " #the total impedance is given by\n",
- "z = Z1*Z2/(Z1 + Z2)\n",
- " #Thus Figure 33.73 simplifies to Figure 33.74.\n",
- " #The open-circuit voltage across AB of Figure 33.74, E\n",
- "E = Isc*z\n",
- " #the impedance \u2018looking in\u2019 at AB,is z\n",
- " #the Th\u00b4evenin equivalent circuit is as shown in Figure 33.75.\n",
- "R = 1.8 + 4j;# in ohm\n",
- " #when R impedance is connected to terminals AB of Figure 33.75, the current I flowing is given by\n",
- "I = E/(z + R)\n",
- "Imag = abs(I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the magnitude of the current flowing (1.8 + i4) ohm resistor is \",round(Imag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the magnitude of the current flowing (1.8 + i4) ohm resistor is 3.84 A"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 596</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the magnitude of the current flowing in the capacitive branch connected to terminals AB.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 5;# in volts\n",
- "V2 = 10;# in volts\n",
- "i = 0.001;# in Amperes\n",
- "R1 = 1000;# in ohm\n",
- "R2 = 4000;# in ohm\n",
- "R3 = 2000;# in ohm\n",
- "R4 = 200;# in ohm\n",
- "R5 = -1j*4000;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #For the branch containing the V1 source, conversion to a Norton equivalent network gives\n",
- "Isc1 = V1/R1\n",
- "z1 = R1\n",
- " #For the branch containing the V2 source, conversion to a Norton equivalent circuit gives\n",
- "Isc2 = V2/R2\n",
- "z2 = R2\n",
- " #Thus the circuit of Figure 33.76 converts to that of Figure 33.77.\n",
- " #The above two Norton equivalent networks shown in Figure 33.77 may be combined, \n",
- " #since the total short-circuit current is\n",
- "Isc = Isc1 + Isc2\n",
- " #the total impedance is given by\n",
- "Z1 = z1*z2/(z1 + z2)\n",
- " #Both of the Norton equivalent networks shown in Figure 33.78 may be converted to Th\u00b4evenin equivalent circuits. \n",
- " #Open-circuit voltage across CD is\n",
- "Ecd = Isc*Z1\n",
- " #the impedance \u2018looking in\u2019 at CD is Z1\n",
- " #Open-circuit voltage across EF\n",
- "Eef = i*R3\n",
- " #the impedance \u2018looking in\u2019 Figure 33.79 at EF\n",
- "Z2 = R3\n",
- " #Thus Figure 33.78 converts to Figure 33.79.\n",
- " #Combining the two Th\u00b4evenin circuits gives e.m.f.\n",
- "E = Ecd - Eef\n",
- " #impedance z\n",
- "z = Z1 + Z2\n",
- " #the Th\u00b4evenin equivalent circuit for terminals AB of Figure 33.76 is as shown in Figure 33.80.\n",
- "Z3 = R4 + R5\n",
- " #If an impedance Z3 is connected across terminals AB, then the current I flowing is given by\n",
- "I = E/(z + Z3)\n",
- "Imag = abs(I)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the current in the capacitive branch is \", Imag*1000,\"mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the current in the capacitive branch is 0.8 mA"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 597</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the power dissipated in a \t(600 - j800) ohm impedance connected between A and B\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 5;# in volts\n",
- "i = 0.004;# in Amperes\n",
- "R1 = 2000;# in ohm\n",
- "R2 = 1000j;# in ohm\n",
- "\n",
- "#calculation: \n",
- " #Converting the Th\u00b4evenin circuit to a Norton network gives\n",
- "Isc1 = V/R2\n",
- " #Thus Figure 33.81 converts to that shown in Figure 33.82. \n",
- " #The two Norton equivalent networks may be combined, giving\n",
- "Isc = Isc1 + i\n",
- "z = R1*R2/(R1 + R2)\n",
- " #This results in the equivalent network shown in Figure 33.83. \n",
- " #Converting to an equivalent Th\u00b4evenin circuit gives open circuit e.m.f. across AB,\n",
- "E = Isc*z\n",
- " #Thus the The\u00b4venin equivalent circuit is as shown in Figure 33.84.\n",
- "R = 600 - 800j;# in ohms\n",
- " #When a R impedance is connected across AB, the current I flowing is given by\n",
- "I = E/(z + R)\n",
- "Imag = abs(I)\n",
- " #the power dissipated in the R resistor is\n",
- "PR = R.real*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the power dissipated in the (600 - i800) ohm resistor is \",round(PR*1000,2),\"mW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the power dissipated in the (600 - i800) ohm resistor is 19.68 mW"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_33.ipynb index d0fda89c..d0fda89c 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_33.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint.ipynb deleted file mode 100755 index ce641b05..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint.ipynb +++ /dev/null @@ -1,449 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 34: Delta-star and star-delta transformations</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 605</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Replace the delta-connected network by an equivalent star connection.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ZA = 20;# in ohm\n",
- "ZB = 10 + 10j;# in ohm\n",
- "ZC = -20j;# in ohm\n",
- "\n",
- "#calculation:\n",
- "Z1 = ZA*ZB/(ZA+ZB+ZC)\n",
- "Z2 = ZB*ZC/(ZA+ZB+ZC)\n",
- "Z3 = ZA*ZC/(ZA+ZB+ZC)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the Resistances of equivalent star connection are, Z1 =\",round(Z1.real,2),\" + (\",round(Z1.imag,2),\")i ohmn, Z2 =\",round(Z2.real,2),\" + (\",round(Z2.imag,2),\")i Ohm and Z3 =\",round(Z3.real,2),\" + (\",round(Z3.imag,2),\")i Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the Resistances of equivalent star connection are, Z1 = 4.0 + ( 8.0 )i ohmn, Z2 = 8.0 + ( -4.0 )i Ohm and Z3 = 4.0 + ( -12.0 )i Ohm"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 606</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the equivalent circuit impedance across terminals AB,\n",
- "#(b) supply current I and (c) the power dissipated in the 10 ohm resistor\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 40;# in volts\n",
- "thetav = 0;# in degrees\n",
- "ZA = 10j;# in ohm\n",
- "ZB = 15j;# in ohm\n",
- "ZC = 25j;# in ohm\n",
- "ZD = -8j;# in ohm\n",
- "ZE = 10;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The network of Figure 34.7 is redrawn, as in Figure 34.8, \n",
- " #showing more clearly the part of the network 1, 2, 3 forming a delta connection \n",
- " #This may he transformed into a star connection as shown in Figure 34.9.\n",
- "Z1 = ZA*ZB/(ZA + ZB + ZC)\n",
- "Z2 = ZC*ZB/(ZA + ZB + ZC)\n",
- "Z3 = ZA*ZC/(ZA + ZB + ZC)\n",
- " #The equivalent network is shown in Figure 34.10 and is further simplified in Figure 34.11\n",
- " #(ZE + Z3) in parallel with (Z1 + ZD) gives an equivalent impedance of\n",
- "z = (ZE + Z3)*(Z1 + ZD)/(Z1 + ZD + ZE + Z3)\n",
- " #Hence the total circuit equivalent impedance across terminals AB is given by\n",
- "Zab = z + Z2\n",
- " #Supply current I\n",
- "I = V/Zab\n",
- "I1 = ((Z1 + ZD)/(Z1 + ZD + ZE + Z3))*I\n",
- "I1mag = abs(I1)\n",
- " #Power P dissipated in the 10 ohm resistance of Figure 34.7 is given by\n",
- "Pr10 = ZE*I1mag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the equivalent circuit impedance across terminals AB is \",round(Zab.real,2),\" + (\",round(Zab.imag,2),\")i ohm\"\n",
- "print \"\\n (b)supply current I is \",round(I.real,2),\" + (\",round(I.imag,2),\")i A\"\n",
- "print \"\\n (c)power P dissipated in the 10 ohm resistor is \",round(Pr10,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the equivalent circuit impedance across terminals AB is 2.5 + ( 2.5 )i ohm\n",
- "\n",
- " (b)supply current I is 8.0 + ( -8.0 )i A\n",
- "\n",
- " (c)power P dissipated in the 10 ohm resistor is 320.0 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 607</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) the value of the single equivalent resistance that replaces the network between terminals A and B, \n",
- "#(b) the current supplied by the 52 V source, and (c) the current flowing in the 8 ohm resistance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 52;# in volts\n",
- "ZA = 8;# in ohm\n",
- "ZB = 16;# in ohm\n",
- "ZC = 40;# in ohm\n",
- "ZD = 1;# in ohm\n",
- "ZE = 4;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #In Figure 34.12, no resistances are directly in parallel or directly in series with each other. \n",
- " #However, ACD and BCD are both delta connections and either may be converted into an equivalent star connection. The delta network BCD is redrawn in Figure 34.13(a) and is transformed into an equivalent star connection as shown in Figure 34.13(b), where\n",
- "Z1 = ZA*ZB/(ZA + ZB + ZC)\n",
- "Z2 = ZC*ZB/(ZA + ZB + ZC)\n",
- "Z3 = ZA*ZC/(ZA + ZB + ZC)\n",
- " #The network of Figure 34.12 may thus be redrawn as shown in Figure 34.14. \n",
- " #The Z1 and ZE are in series with each other, as are the ZD and Z3 resistors. \n",
- " #Hence the equivalent network is as shown in Figure 34.15. \n",
- " #The total equivalent resistance across terminals A and B is given by\n",
- "Zab = (Z1 + ZE)*(ZD + Z3)/(Z1 + ZE + ZD + Z3) + Z2\n",
- " #Current supplied by the source, i.e., current I in Figure 34.15, is given by\n",
- "I = V/Zab\n",
- " #From Figure 34.15, current I1\n",
- "I1 = ((ZD + Z3)/(Z1 + ZE + ZD + Z3))*I\n",
- " #current I2\n",
- "I2 = I - I1\n",
- " #From Figure 34.14, p.d. across AC,\n",
- "Vac = I1*ZE\n",
- " #p.d. across AD\n",
- "Vad = I2*ZD\n",
- " #Hence p.d. between C and D is given\n",
- "Vcd = Vac - Vad\n",
- " #current in the 8 ohm resistance\n",
- "Ir8 = Vcd/ZA\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the equivalent circuit impedance across terminals AB is \",round(Zab,2),\"ohm\"\n",
- "print \"\\n (b)the current supplied by the 52 V source is \",round(I,2),\" A\"\n",
- "print \"\\n (c)the current flowing in the 8 ohm resistance is \",round(Ir8,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the equivalent circuit impedance across terminals AB is 13.0 ohm\n",
- "\n",
- " (b)the current supplied by the 52 V source is 4.0 A\n",
- "\n",
- " (c)the current flowing in the 8 ohm resistance is 0.75 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 608</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(b)determine the values of RX and LX at balance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R2 = 1000;# in ohm\n",
- "R3 = 1000;# in ohm\n",
- "R4 = 500;# in ohm\n",
- "R5 = 200;# in ohm\n",
- "C = 2E-6; # in Farad\n",
- "\n",
- "#calculation:\n",
- "Rx = R2*R4/R3\n",
- "Lx = R2*C*(R4 + R5 + R4*R5/R3)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (b)Rx is \",round(Rx,2),\"ohm, Lx is \",round(Lx,2),\"H\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (b)Rx is 500.0 ohm, Lx is 1.6 H"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 610</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current flowing in the j10 ohm impedance, and \n",
- "#(b) the power dissipated in the \u000720 ohm impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 120;# in volts\n",
- "thetav = 0;# in degrees\n",
- "ZA = 25 - 5j;# in ohm\n",
- "ZB = 15 + 10j;# in ohm\n",
- "ZC = 20 - 30j;# in ohm\n",
- "ZD = 20 + 0j;# in ohm\n",
- "ZE = 0 + 10j;# in ohm\n",
- "ZF = 2.5 - 5j;# in ohm\n",
- "\n",
- "#calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The network may initially be simplified by transforming the delta PQR to its equivalent star connection \n",
- " #as represented by impedances Z1, Z2 and Z3 in Figure 34.21. From equation (34.7),\n",
- "Z1 = ZA*ZB/(ZA + ZB + ZC)\n",
- "Z2 = ZC*ZB/(ZA + ZB + ZC)\n",
- "Z3 = ZA*ZC/(ZA + ZB + ZC)\n",
- " #The network is shown redrawn in Figure 34.22 and further simplified in Figure 34.23, from which,\n",
- "Zab = ((Z3 + ZE)*(ZD + Z2)/(Z2 + ZE + ZD + Z3)) + (Z1 + ZF)\n",
- " #Current I1\n",
- "I1 = V/Zab\n",
- " #current I2\n",
- "I2 = ((ZE + Z3)/(Z2 + ZE + ZD + Z3))*I1\n",
- " #current I3\n",
- "I3 = I1 - I2\n",
- " #The power P dissipated in the ZD impedance of Figure 34.20 is given by\n",
- "Pzd = ZD*I2**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the current flowing in the (0+i10) ohm impedance is \",round(abs(I3),2),\" A\"\n",
- "print \"\\n (b) the power dissipated in the (20 + i0) ohm impedance is \",round(abs(Pzd),2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the current flowing in the (0+i10) ohm impedance is 6.0 A\n",
- "\n",
- " (b) the power dissipated in the (20 + i0) ohm impedance is 80.0 W\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 613</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the delta-connected equivalent network for the star-connected impedances.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 10;# in ohm\n",
- "Z2 = 20;# in ohm\n",
- "Z3 = 5j;# in ohm\n",
- "\n",
- "#calculation:\n",
- "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
- "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
- "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,2),\" + (\",round(ZA.imag,2),\")i ohmn, ZB =\",round(ZB.real,2),\" + (\",round(ZB.imag,2),\")i Ohm and ZC =\",round(ZC.real,2),\" + (\",round(ZC.imag,2),\")i Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the Resistances of equivalent Delta connection are, ZA = 10.0 + ( 7.5 )i ohmn, ZB = 30.0 + ( -40.0 )i Ohm and ZC = 20.0 + ( 15.0 )i Ohm"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 613</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Convert the star to an equivalent delta connection\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1r = 100;# in ohm\n",
- "Z1a = 0; # in Deg\n",
- "Z2r = 63.25;# in ohm\n",
- "Z2a = 18.43; # in Deg\n",
- "Z3r = 100;# in ohm\n",
- "Z3a = -90; # in Deg\n",
- "\n",
- "#calculation\n",
- "Z1 = Z1r*math.cos(Z1a*math.pi/180) + 1j*Z1r*math.sin(Z1a*math.pi/180)\n",
- "Z2 = Z2r*math.cos(Z2a*math.pi/180) + 1j*Z2r*math.sin(Z2a*math.pi/180)\n",
- "Z3 = Z3r*math.cos(Z3a*math.pi/180) + 1j*Z3r*math.sin(Z3a*math.pi/180)\n",
- "\n",
- "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
- "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
- "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,0),\" + (\",round(ZA.imag,0),\")i ohmn, ZB =\",round(ZB.real,0),\" + (\",round(ZB.imag,0),\")i Ohm and ZC =\",round(ZC.real,0),\" - (\",round(ZC.imag,0),\")i Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the Resistances of equivalent Delta connection are, ZA = 50.0 + ( -250.0 )i ohmn, ZB = 140.0 + ( 80.0 )i Ohm and ZC = 80.0 - ( -140.0 )i Ohm"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_1.ipynb deleted file mode 100755 index ce641b05..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_1.ipynb +++ /dev/null @@ -1,449 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 34: Delta-star and star-delta transformations</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 605</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Replace the delta-connected network by an equivalent star connection.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ZA = 20;# in ohm\n",
- "ZB = 10 + 10j;# in ohm\n",
- "ZC = -20j;# in ohm\n",
- "\n",
- "#calculation:\n",
- "Z1 = ZA*ZB/(ZA+ZB+ZC)\n",
- "Z2 = ZB*ZC/(ZA+ZB+ZC)\n",
- "Z3 = ZA*ZC/(ZA+ZB+ZC)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the Resistances of equivalent star connection are, Z1 =\",round(Z1.real,2),\" + (\",round(Z1.imag,2),\")i ohmn, Z2 =\",round(Z2.real,2),\" + (\",round(Z2.imag,2),\")i Ohm and Z3 =\",round(Z3.real,2),\" + (\",round(Z3.imag,2),\")i Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the Resistances of equivalent star connection are, Z1 = 4.0 + ( 8.0 )i ohmn, Z2 = 8.0 + ( -4.0 )i Ohm and Z3 = 4.0 + ( -12.0 )i Ohm"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 606</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the equivalent circuit impedance across terminals AB,\n",
- "#(b) supply current I and (c) the power dissipated in the 10 ohm resistor\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 40;# in volts\n",
- "thetav = 0;# in degrees\n",
- "ZA = 10j;# in ohm\n",
- "ZB = 15j;# in ohm\n",
- "ZC = 25j;# in ohm\n",
- "ZD = -8j;# in ohm\n",
- "ZE = 10;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The network of Figure 34.7 is redrawn, as in Figure 34.8, \n",
- " #showing more clearly the part of the network 1, 2, 3 forming a delta connection \n",
- " #This may he transformed into a star connection as shown in Figure 34.9.\n",
- "Z1 = ZA*ZB/(ZA + ZB + ZC)\n",
- "Z2 = ZC*ZB/(ZA + ZB + ZC)\n",
- "Z3 = ZA*ZC/(ZA + ZB + ZC)\n",
- " #The equivalent network is shown in Figure 34.10 and is further simplified in Figure 34.11\n",
- " #(ZE + Z3) in parallel with (Z1 + ZD) gives an equivalent impedance of\n",
- "z = (ZE + Z3)*(Z1 + ZD)/(Z1 + ZD + ZE + Z3)\n",
- " #Hence the total circuit equivalent impedance across terminals AB is given by\n",
- "Zab = z + Z2\n",
- " #Supply current I\n",
- "I = V/Zab\n",
- "I1 = ((Z1 + ZD)/(Z1 + ZD + ZE + Z3))*I\n",
- "I1mag = abs(I1)\n",
- " #Power P dissipated in the 10 ohm resistance of Figure 34.7 is given by\n",
- "Pr10 = ZE*I1mag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the equivalent circuit impedance across terminals AB is \",round(Zab.real,2),\" + (\",round(Zab.imag,2),\")i ohm\"\n",
- "print \"\\n (b)supply current I is \",round(I.real,2),\" + (\",round(I.imag,2),\")i A\"\n",
- "print \"\\n (c)power P dissipated in the 10 ohm resistor is \",round(Pr10,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the equivalent circuit impedance across terminals AB is 2.5 + ( 2.5 )i ohm\n",
- "\n",
- " (b)supply current I is 8.0 + ( -8.0 )i A\n",
- "\n",
- " (c)power P dissipated in the 10 ohm resistor is 320.0 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 607</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) the value of the single equivalent resistance that replaces the network between terminals A and B, \n",
- "#(b) the current supplied by the 52 V source, and (c) the current flowing in the 8 ohm resistance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 52;# in volts\n",
- "ZA = 8;# in ohm\n",
- "ZB = 16;# in ohm\n",
- "ZC = 40;# in ohm\n",
- "ZD = 1;# in ohm\n",
- "ZE = 4;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #In Figure 34.12, no resistances are directly in parallel or directly in series with each other. \n",
- " #However, ACD and BCD are both delta connections and either may be converted into an equivalent star connection. The delta network BCD is redrawn in Figure 34.13(a) and is transformed into an equivalent star connection as shown in Figure 34.13(b), where\n",
- "Z1 = ZA*ZB/(ZA + ZB + ZC)\n",
- "Z2 = ZC*ZB/(ZA + ZB + ZC)\n",
- "Z3 = ZA*ZC/(ZA + ZB + ZC)\n",
- " #The network of Figure 34.12 may thus be redrawn as shown in Figure 34.14. \n",
- " #The Z1 and ZE are in series with each other, as are the ZD and Z3 resistors. \n",
- " #Hence the equivalent network is as shown in Figure 34.15. \n",
- " #The total equivalent resistance across terminals A and B is given by\n",
- "Zab = (Z1 + ZE)*(ZD + Z3)/(Z1 + ZE + ZD + Z3) + Z2\n",
- " #Current supplied by the source, i.e., current I in Figure 34.15, is given by\n",
- "I = V/Zab\n",
- " #From Figure 34.15, current I1\n",
- "I1 = ((ZD + Z3)/(Z1 + ZE + ZD + Z3))*I\n",
- " #current I2\n",
- "I2 = I - I1\n",
- " #From Figure 34.14, p.d. across AC,\n",
- "Vac = I1*ZE\n",
- " #p.d. across AD\n",
- "Vad = I2*ZD\n",
- " #Hence p.d. between C and D is given\n",
- "Vcd = Vac - Vad\n",
- " #current in the 8 ohm resistance\n",
- "Ir8 = Vcd/ZA\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the equivalent circuit impedance across terminals AB is \",round(Zab,2),\"ohm\"\n",
- "print \"\\n (b)the current supplied by the 52 V source is \",round(I,2),\" A\"\n",
- "print \"\\n (c)the current flowing in the 8 ohm resistance is \",round(Ir8,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the equivalent circuit impedance across terminals AB is 13.0 ohm\n",
- "\n",
- " (b)the current supplied by the 52 V source is 4.0 A\n",
- "\n",
- " (c)the current flowing in the 8 ohm resistance is 0.75 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 608</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(b)determine the values of RX and LX at balance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R2 = 1000;# in ohm\n",
- "R3 = 1000;# in ohm\n",
- "R4 = 500;# in ohm\n",
- "R5 = 200;# in ohm\n",
- "C = 2E-6; # in Farad\n",
- "\n",
- "#calculation:\n",
- "Rx = R2*R4/R3\n",
- "Lx = R2*C*(R4 + R5 + R4*R5/R3)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (b)Rx is \",round(Rx,2),\"ohm, Lx is \",round(Lx,2),\"H\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (b)Rx is 500.0 ohm, Lx is 1.6 H"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 610</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current flowing in the j10 ohm impedance, and \n",
- "#(b) the power dissipated in the \u000720 ohm impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 120;# in volts\n",
- "thetav = 0;# in degrees\n",
- "ZA = 25 - 5j;# in ohm\n",
- "ZB = 15 + 10j;# in ohm\n",
- "ZC = 20 - 30j;# in ohm\n",
- "ZD = 20 + 0j;# in ohm\n",
- "ZE = 0 + 10j;# in ohm\n",
- "ZF = 2.5 - 5j;# in ohm\n",
- "\n",
- "#calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The network may initially be simplified by transforming the delta PQR to its equivalent star connection \n",
- " #as represented by impedances Z1, Z2 and Z3 in Figure 34.21. From equation (34.7),\n",
- "Z1 = ZA*ZB/(ZA + ZB + ZC)\n",
- "Z2 = ZC*ZB/(ZA + ZB + ZC)\n",
- "Z3 = ZA*ZC/(ZA + ZB + ZC)\n",
- " #The network is shown redrawn in Figure 34.22 and further simplified in Figure 34.23, from which,\n",
- "Zab = ((Z3 + ZE)*(ZD + Z2)/(Z2 + ZE + ZD + Z3)) + (Z1 + ZF)\n",
- " #Current I1\n",
- "I1 = V/Zab\n",
- " #current I2\n",
- "I2 = ((ZE + Z3)/(Z2 + ZE + ZD + Z3))*I1\n",
- " #current I3\n",
- "I3 = I1 - I2\n",
- " #The power P dissipated in the ZD impedance of Figure 34.20 is given by\n",
- "Pzd = ZD*I2**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the current flowing in the (0+i10) ohm impedance is \",round(abs(I3),2),\" A\"\n",
- "print \"\\n (b) the power dissipated in the (20 + i0) ohm impedance is \",round(abs(Pzd),2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the current flowing in the (0+i10) ohm impedance is 6.0 A\n",
- "\n",
- " (b) the power dissipated in the (20 + i0) ohm impedance is 80.0 W\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 613</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the delta-connected equivalent network for the star-connected impedances.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 10;# in ohm\n",
- "Z2 = 20;# in ohm\n",
- "Z3 = 5j;# in ohm\n",
- "\n",
- "#calculation:\n",
- "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
- "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
- "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,2),\" + (\",round(ZA.imag,2),\")i ohmn, ZB =\",round(ZB.real,2),\" + (\",round(ZB.imag,2),\")i Ohm and ZC =\",round(ZC.real,2),\" + (\",round(ZC.imag,2),\")i Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the Resistances of equivalent Delta connection are, ZA = 10.0 + ( 7.5 )i ohmn, ZB = 30.0 + ( -40.0 )i Ohm and ZC = 20.0 + ( 15.0 )i Ohm"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 613</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Convert the star to an equivalent delta connection\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1r = 100;# in ohm\n",
- "Z1a = 0; # in Deg\n",
- "Z2r = 63.25;# in ohm\n",
- "Z2a = 18.43; # in Deg\n",
- "Z3r = 100;# in ohm\n",
- "Z3a = -90; # in Deg\n",
- "\n",
- "#calculation\n",
- "Z1 = Z1r*math.cos(Z1a*math.pi/180) + 1j*Z1r*math.sin(Z1a*math.pi/180)\n",
- "Z2 = Z2r*math.cos(Z2a*math.pi/180) + 1j*Z2r*math.sin(Z2a*math.pi/180)\n",
- "Z3 = Z3r*math.cos(Z3a*math.pi/180) + 1j*Z3r*math.sin(Z3a*math.pi/180)\n",
- "\n",
- "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
- "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
- "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,0),\" + (\",round(ZA.imag,0),\")i ohmn, ZB =\",round(ZB.real,0),\" + (\",round(ZB.imag,0),\")i Ohm and ZC =\",round(ZC.real,0),\" - (\",round(ZC.imag,0),\")i Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the Resistances of equivalent Delta connection are, ZA = 50.0 + ( -250.0 )i ohmn, ZB = 140.0 + ( 80.0 )i Ohm and ZC = 80.0 - ( -140.0 )i Ohm"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_2.ipynb deleted file mode 100755 index ce641b05..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_2.ipynb +++ /dev/null @@ -1,449 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 34: Delta-star and star-delta transformations</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 605</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Replace the delta-connected network by an equivalent star connection.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ZA = 20;# in ohm\n",
- "ZB = 10 + 10j;# in ohm\n",
- "ZC = -20j;# in ohm\n",
- "\n",
- "#calculation:\n",
- "Z1 = ZA*ZB/(ZA+ZB+ZC)\n",
- "Z2 = ZB*ZC/(ZA+ZB+ZC)\n",
- "Z3 = ZA*ZC/(ZA+ZB+ZC)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the Resistances of equivalent star connection are, Z1 =\",round(Z1.real,2),\" + (\",round(Z1.imag,2),\")i ohmn, Z2 =\",round(Z2.real,2),\" + (\",round(Z2.imag,2),\")i Ohm and Z3 =\",round(Z3.real,2),\" + (\",round(Z3.imag,2),\")i Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the Resistances of equivalent star connection are, Z1 = 4.0 + ( 8.0 )i ohmn, Z2 = 8.0 + ( -4.0 )i Ohm and Z3 = 4.0 + ( -12.0 )i Ohm"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 606</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the equivalent circuit impedance across terminals AB,\n",
- "#(b) supply current I and (c) the power dissipated in the 10 ohm resistor\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 40;# in volts\n",
- "thetav = 0;# in degrees\n",
- "ZA = 10j;# in ohm\n",
- "ZB = 15j;# in ohm\n",
- "ZC = 25j;# in ohm\n",
- "ZD = -8j;# in ohm\n",
- "ZE = 10;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The network of Figure 34.7 is redrawn, as in Figure 34.8, \n",
- " #showing more clearly the part of the network 1, 2, 3 forming a delta connection \n",
- " #This may he transformed into a star connection as shown in Figure 34.9.\n",
- "Z1 = ZA*ZB/(ZA + ZB + ZC)\n",
- "Z2 = ZC*ZB/(ZA + ZB + ZC)\n",
- "Z3 = ZA*ZC/(ZA + ZB + ZC)\n",
- " #The equivalent network is shown in Figure 34.10 and is further simplified in Figure 34.11\n",
- " #(ZE + Z3) in parallel with (Z1 + ZD) gives an equivalent impedance of\n",
- "z = (ZE + Z3)*(Z1 + ZD)/(Z1 + ZD + ZE + Z3)\n",
- " #Hence the total circuit equivalent impedance across terminals AB is given by\n",
- "Zab = z + Z2\n",
- " #Supply current I\n",
- "I = V/Zab\n",
- "I1 = ((Z1 + ZD)/(Z1 + ZD + ZE + Z3))*I\n",
- "I1mag = abs(I1)\n",
- " #Power P dissipated in the 10 ohm resistance of Figure 34.7 is given by\n",
- "Pr10 = ZE*I1mag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the equivalent circuit impedance across terminals AB is \",round(Zab.real,2),\" + (\",round(Zab.imag,2),\")i ohm\"\n",
- "print \"\\n (b)supply current I is \",round(I.real,2),\" + (\",round(I.imag,2),\")i A\"\n",
- "print \"\\n (c)power P dissipated in the 10 ohm resistor is \",round(Pr10,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the equivalent circuit impedance across terminals AB is 2.5 + ( 2.5 )i ohm\n",
- "\n",
- " (b)supply current I is 8.0 + ( -8.0 )i A\n",
- "\n",
- " (c)power P dissipated in the 10 ohm resistor is 320.0 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 607</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) the value of the single equivalent resistance that replaces the network between terminals A and B, \n",
- "#(b) the current supplied by the 52 V source, and (c) the current flowing in the 8 ohm resistance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 52;# in volts\n",
- "ZA = 8;# in ohm\n",
- "ZB = 16;# in ohm\n",
- "ZC = 40;# in ohm\n",
- "ZD = 1;# in ohm\n",
- "ZE = 4;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #In Figure 34.12, no resistances are directly in parallel or directly in series with each other. \n",
- " #However, ACD and BCD are both delta connections and either may be converted into an equivalent star connection. The delta network BCD is redrawn in Figure 34.13(a) and is transformed into an equivalent star connection as shown in Figure 34.13(b), where\n",
- "Z1 = ZA*ZB/(ZA + ZB + ZC)\n",
- "Z2 = ZC*ZB/(ZA + ZB + ZC)\n",
- "Z3 = ZA*ZC/(ZA + ZB + ZC)\n",
- " #The network of Figure 34.12 may thus be redrawn as shown in Figure 34.14. \n",
- " #The Z1 and ZE are in series with each other, as are the ZD and Z3 resistors. \n",
- " #Hence the equivalent network is as shown in Figure 34.15. \n",
- " #The total equivalent resistance across terminals A and B is given by\n",
- "Zab = (Z1 + ZE)*(ZD + Z3)/(Z1 + ZE + ZD + Z3) + Z2\n",
- " #Current supplied by the source, i.e., current I in Figure 34.15, is given by\n",
- "I = V/Zab\n",
- " #From Figure 34.15, current I1\n",
- "I1 = ((ZD + Z3)/(Z1 + ZE + ZD + Z3))*I\n",
- " #current I2\n",
- "I2 = I - I1\n",
- " #From Figure 34.14, p.d. across AC,\n",
- "Vac = I1*ZE\n",
- " #p.d. across AD\n",
- "Vad = I2*ZD\n",
- " #Hence p.d. between C and D is given\n",
- "Vcd = Vac - Vad\n",
- " #current in the 8 ohm resistance\n",
- "Ir8 = Vcd/ZA\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the equivalent circuit impedance across terminals AB is \",round(Zab,2),\"ohm\"\n",
- "print \"\\n (b)the current supplied by the 52 V source is \",round(I,2),\" A\"\n",
- "print \"\\n (c)the current flowing in the 8 ohm resistance is \",round(Ir8,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the equivalent circuit impedance across terminals AB is 13.0 ohm\n",
- "\n",
- " (b)the current supplied by the 52 V source is 4.0 A\n",
- "\n",
- " (c)the current flowing in the 8 ohm resistance is 0.75 A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 608</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(b)determine the values of RX and LX at balance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R2 = 1000;# in ohm\n",
- "R3 = 1000;# in ohm\n",
- "R4 = 500;# in ohm\n",
- "R5 = 200;# in ohm\n",
- "C = 2E-6; # in Farad\n",
- "\n",
- "#calculation:\n",
- "Rx = R2*R4/R3\n",
- "Lx = R2*C*(R4 + R5 + R4*R5/R3)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (b)Rx is \",round(Rx,2),\"ohm, Lx is \",round(Lx,2),\"H\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (b)Rx is 500.0 ohm, Lx is 1.6 H"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 610</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current flowing in the j10 ohm impedance, and \n",
- "#(b) the power dissipated in the \u000720 ohm impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 120;# in volts\n",
- "thetav = 0;# in degrees\n",
- "ZA = 25 - 5j;# in ohm\n",
- "ZB = 15 + 10j;# in ohm\n",
- "ZC = 20 - 30j;# in ohm\n",
- "ZD = 20 + 0j;# in ohm\n",
- "ZE = 0 + 10j;# in ohm\n",
- "ZF = 2.5 - 5j;# in ohm\n",
- "\n",
- "#calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #The network may initially be simplified by transforming the delta PQR to its equivalent star connection \n",
- " #as represented by impedances Z1, Z2 and Z3 in Figure 34.21. From equation (34.7),\n",
- "Z1 = ZA*ZB/(ZA + ZB + ZC)\n",
- "Z2 = ZC*ZB/(ZA + ZB + ZC)\n",
- "Z3 = ZA*ZC/(ZA + ZB + ZC)\n",
- " #The network is shown redrawn in Figure 34.22 and further simplified in Figure 34.23, from which,\n",
- "Zab = ((Z3 + ZE)*(ZD + Z2)/(Z2 + ZE + ZD + Z3)) + (Z1 + ZF)\n",
- " #Current I1\n",
- "I1 = V/Zab\n",
- " #current I2\n",
- "I2 = ((ZE + Z3)/(Z2 + ZE + ZD + Z3))*I1\n",
- " #current I3\n",
- "I3 = I1 - I2\n",
- " #The power P dissipated in the ZD impedance of Figure 34.20 is given by\n",
- "Pzd = ZD*I2**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the current flowing in the (0+i10) ohm impedance is \",round(abs(I3),2),\" A\"\n",
- "print \"\\n (b) the power dissipated in the (20 + i0) ohm impedance is \",round(abs(Pzd),2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the current flowing in the (0+i10) ohm impedance is 6.0 A\n",
- "\n",
- " (b) the power dissipated in the (20 + i0) ohm impedance is 80.0 W\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 613</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the delta-connected equivalent network for the star-connected impedances.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1 = 10;# in ohm\n",
- "Z2 = 20;# in ohm\n",
- "Z3 = 5j;# in ohm\n",
- "\n",
- "#calculation:\n",
- "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
- "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
- "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,2),\" + (\",round(ZA.imag,2),\")i ohmn, ZB =\",round(ZB.real,2),\" + (\",round(ZB.imag,2),\")i Ohm and ZC =\",round(ZC.real,2),\" + (\",round(ZC.imag,2),\")i Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the Resistances of equivalent Delta connection are, ZA = 10.0 + ( 7.5 )i ohmn, ZB = 30.0 + ( -40.0 )i Ohm and ZC = 20.0 + ( 15.0 )i Ohm"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 613</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Convert the star to an equivalent delta connection\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Z1r = 100;# in ohm\n",
- "Z1a = 0; # in Deg\n",
- "Z2r = 63.25;# in ohm\n",
- "Z2a = 18.43; # in Deg\n",
- "Z3r = 100;# in ohm\n",
- "Z3a = -90; # in Deg\n",
- "\n",
- "#calculation\n",
- "Z1 = Z1r*math.cos(Z1a*math.pi/180) + 1j*Z1r*math.sin(Z1a*math.pi/180)\n",
- "Z2 = Z2r*math.cos(Z2a*math.pi/180) + 1j*Z2r*math.sin(Z2a*math.pi/180)\n",
- "Z3 = Z3r*math.cos(Z3a*math.pi/180) + 1j*Z3r*math.sin(Z3a*math.pi/180)\n",
- "\n",
- "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
- "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
- "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,0),\" + (\",round(ZA.imag,0),\")i ohmn, ZB =\",round(ZB.real,0),\" + (\",round(ZB.imag,0),\")i Ohm and ZC =\",round(ZC.real,0),\" - (\",round(ZC.imag,0),\")i Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the Resistances of equivalent Delta connection are, ZA = 50.0 + ( -250.0 )i ohmn, ZB = 140.0 + ( 80.0 )i Ohm and ZC = 80.0 - ( -140.0 )i Ohm"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_34.ipynb index 4c4d28d9..4c4d28d9 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_34.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint.ipynb deleted file mode 100755 index 7462b3a9..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint.ipynb +++ /dev/null @@ -1,637 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 35: Maximum power transfer theorems and impedance matching</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 620</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of R for maximum power to be transferred from the source to the load,\n",
- "#and (b) the value of the maximum power delivered to R\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 120;# in volts\n",
- "thetav = 0;# in degrees\n",
- "Z = 15 + 1j*20;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #maximum power transfer occurs when R = mod(Z)\n",
- "R = abs(Z)\n",
- " #the total circuit impedance\n",
- "ZT = Z + R\n",
- " #Current I flowing in the load is given by\n",
- "I = V/ZT\n",
- "Imag = abs(I)\n",
- " #maximum power delivered\n",
- "P = R*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",abs(P),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 25.0 ohm\n",
- "\n",
- " (b) maximum power delivered is 180.0 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 620</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine\n",
- "#(a) the value of Z that results in maximum power transfer, and\n",
- "#(b) the value of the maximum power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 120;# in volts\n",
- "thetav = 0;# in degrees\n",
- "Z = 15 + 1j*20;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #maximum power transfer occurs when X = -1*imag(Z) and R = real(Z)\n",
- "z = Z.real - 1j*Z.imag\n",
- " #Total circuit impedance at maximum power transfer condition,\n",
- "ZT = Z + z\n",
- " #Current I flowing in the load is given by\n",
- "I = V/ZT\n",
- "Imag = abs(I)\n",
- " #maximum power delivered\n",
- "P = Z.real*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when Z is \",z.real,\" + (\", z.imag,\")i ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",abs(P),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when Z is 15.0 + ( -20.0 )i ohm\n",
- "\n",
- " (b) maximum power delivered is 240.0 W"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 620</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine\n",
- "#(a) the value of the load resistance R required for maximum power transfer, and\n",
- "#(b) the value of the maximum power transferred.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 200;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 100;# in ohm\n",
- "C = 1E-6;# in farad\n",
- "f = 1000;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Capacitive reactance, Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #Hence source impedance,\n",
- "z = R1*(1j*Xc)/(R1 + 1j*Xc)\n",
- " #maximum power transfer is achieved when R = mod(z)\n",
- "R = abs(z)\n",
- " #Total circuit impedance at maximum power transfer condition,\n",
- "ZT = z + R\n",
- " #Current I flowing in the load is given by\n",
- "I = V/ZT\n",
- "Imag = abs(I)\n",
- " #maximum power transferred,\n",
- "P = R*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",round(R,2),\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 84.67 ohm\n",
- "\n",
- " (b) maximum power delivered is 127.9 W"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 621</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of R for which the power transferred to the load is a maximum, and \n",
- "#(b) the value of the maximum power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 60;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "XL = 10;# in ohm\n",
- "Xc = 7;# in ohm\n",
- "R2 = XL*1j;# in ohm\n",
- "R3 = -1j*Xc;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #maximum power transfer is achieved when\n",
- "R = (R1**2 + (XL - Xc)**2)**0.5\n",
- " #Hence source impedance,\n",
- "ZT = R1 + R2 + R3 + R\n",
- " #Current I flowing in the load is given by\n",
- "I = V/ZT\n",
- "Imag = abs(I)\n",
- " #maximum power transferred,\n",
- "P = R*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",P,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 5.0 ohm\n",
- "\n",
- " (b) maximum power delivered is 200.0 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 622</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the load resistance R\n",
- "#calculate the value of this power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 20;# in volts\n",
- "R1 = 5;# in ohm\n",
- "R2 = 15;# in ohm\n",
- "\n",
- "#calculation: \n",
- " #R is removed from the network as shown in Figure 35.6\n",
- " #P.d. across AB, E\n",
- "E = (R2/(R1 + R2))*V\n",
- " #Impedance \u2018looking-in\u2019 at terminals AB with the source removed is given by\n",
- "r = R1*R2/(R1 + R2)\n",
- " #The equivalent Th\u00b4evenin circuit supplying terminals AB is shown in Figure 35.7. \n",
- " #From condition (2), for maximum power transfer\n",
- "R = r\n",
- " #Current I flowing in the load is given by\n",
- "I = E/(R + r)\n",
- " #maximum power transferred,\n",
- "P = R*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",P,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 3.75 ohm\n",
- "\n",
- " (b) maximum power delivered is 15.0 W"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 622</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the values of R and X that will result in maximum power being transferred across terminals AB, and \n",
- "#(b) the value of the maximum power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 100;# in volts\n",
- "thetav = 30;# in degrees\n",
- "R1 = 5;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = 10j;# in ohm\n",
- "\n",
- "#calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Resistance R and reactance X are removed from the network as shown in Figure 35.9\n",
- " #P.d. across AB,\n",
- "E = ((R2 + R3)/(R1 + R2 + R3))*V\n",
- " #With the source removed the impedance, z, \u2018looking in\u2019 at terminals AB is given by:\n",
- "z = (R2 + R3)*R1/(R1 + R2 + R3)\n",
- " #The equivalent Th\u00b4evenin circuit is shown in Figure 35.10. From condition 3, \n",
- " #maximum power transfer is achieved when X = -1*imag(z) and R = real(z)\n",
- "X = -1*z.imag\n",
- "R = z.real\n",
- "Z = R + 1j*X\n",
- " #Current I flowing in the load is given by\n",
- "I = E/(z + Z)\n",
- "Imag = abs(I)\n",
- " #maximum power transferred,\n",
- "P = R*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm and X is \", X,\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 3.75 ohm and X is -1.25 ohm\n",
- "\n",
- " (b) maximum power delivered is 416.67 W"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 624</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the optimum value of load resistance for maximum power transfer\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ro = 448;# in ohm\n",
- "tr = 8;# turn ratio N1/N2\n",
- "\n",
- " #calculation: \n",
- " #The equivalent input resistance r of the transformer must be Ro for maximum power transfer.\n",
- "r = Ro\n",
- "RL = r*(1/tr)**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the optimum value of load resistance is \",RL,\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the optimum value of load resistance is 7.0 ohm"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 624</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the turns ratio of an ideal transformer necessary to match the generator to a load of (40 + j19) ohm \n",
- "#for maximum transfer of power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Zo = 450 + 1j*60;# in ohm\n",
- "ZL = 40 + 1j*19;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #transformer turns ratio tr = (N1/N2)\n",
- "Zomag = abs(Zo)\n",
- "ZLmag = abs(ZL)\n",
- "tr = (Zomag/ZLmag)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the transformer turns ratio is \",round(tr,2),\"\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the transformer turns ratio is 3.2 "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 625</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the primary current flowing, and \n",
- "#(b) the power dissipated in the load resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 240;# in volts\n",
- "V2 = 1920;# in volts\n",
- "R1 = 5;# in ohms\n",
- "R2 = 1600;# in ohms\n",
- "\n",
- "#calculation: \n",
- " #The network is shown in Figure 35.12.\n",
- " #turn ratio N1/N2 = V1/V2\n",
- "tr = V1/V2\n",
- " #Equivalent input resistance of the transformer,\n",
- "RL = R2\n",
- "r = RL*tr**2\n",
- " #Total input resistance,\n",
- "Rin = R1 + r\n",
- " #primary current, I1\n",
- "I1 = V1/Rin\n",
- " #For an ideal transformer V1/V2 = I2/I1\n",
- "I2 = I1*(V1/V2)\n",
- " #Power dissipated in the load resistance\n",
- "P = RL*I2**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) primary current flowing is \",I1,\" A\"\n",
- "print \"\\n (b) Power dissipated in the load resistance is \",P,\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) primary current flowing is 8.0 A\n",
- "\n",
- " (b) Power dissipated in the load resistance is 1600.0 W"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 625</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for maximum power transfer (a) the value of the load resistance,\n",
- "# and (b) the power dissipated in the load.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 30;# in volts\n",
- "thetav = 0;# in degrees\n",
- "r = 20000;# in ohms\n",
- "tr = 20;# turn ratio\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180) \n",
- " #The network diagram is shown in Figure 35.13.\n",
- " #For maximum power transfer, r1 must be equal to\n",
- "r1 = r\n",
- " #load resistance RL\n",
- "RL = r1/tr**2\n",
- " #The total input resistance when the source is connected to the matching transformer is\n",
- "RT = r + r1\n",
- " #Primary current\n",
- "I1 = V/RT\n",
- " #N1/N2 = I2/I1\n",
- "I2 = I1*tr\n",
- " #Power dissipated in load resistance RL is given by\n",
- "P = RL*I2**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the value of the load resistance is \",RL,\" ohm\"\n",
- "print \"\\n (b) Power dissipated in the load resistance is \",abs(P*1000),\"mW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the value of the load resistance is 50.0 ohm\n",
- "\n",
- " (b) Power dissipated in the load resistance is 11.25 mW"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_1.ipynb deleted file mode 100755 index 7462b3a9..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_1.ipynb +++ /dev/null @@ -1,637 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 35: Maximum power transfer theorems and impedance matching</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 620</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of R for maximum power to be transferred from the source to the load,\n",
- "#and (b) the value of the maximum power delivered to R\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 120;# in volts\n",
- "thetav = 0;# in degrees\n",
- "Z = 15 + 1j*20;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #maximum power transfer occurs when R = mod(Z)\n",
- "R = abs(Z)\n",
- " #the total circuit impedance\n",
- "ZT = Z + R\n",
- " #Current I flowing in the load is given by\n",
- "I = V/ZT\n",
- "Imag = abs(I)\n",
- " #maximum power delivered\n",
- "P = R*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",abs(P),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 25.0 ohm\n",
- "\n",
- " (b) maximum power delivered is 180.0 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 620</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine\n",
- "#(a) the value of Z that results in maximum power transfer, and\n",
- "#(b) the value of the maximum power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 120;# in volts\n",
- "thetav = 0;# in degrees\n",
- "Z = 15 + 1j*20;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #maximum power transfer occurs when X = -1*imag(Z) and R = real(Z)\n",
- "z = Z.real - 1j*Z.imag\n",
- " #Total circuit impedance at maximum power transfer condition,\n",
- "ZT = Z + z\n",
- " #Current I flowing in the load is given by\n",
- "I = V/ZT\n",
- "Imag = abs(I)\n",
- " #maximum power delivered\n",
- "P = Z.real*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when Z is \",z.real,\" + (\", z.imag,\")i ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",abs(P),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when Z is 15.0 + ( -20.0 )i ohm\n",
- "\n",
- " (b) maximum power delivered is 240.0 W"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 620</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine\n",
- "#(a) the value of the load resistance R required for maximum power transfer, and\n",
- "#(b) the value of the maximum power transferred.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 200;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 100;# in ohm\n",
- "C = 1E-6;# in farad\n",
- "f = 1000;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Capacitive reactance, Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #Hence source impedance,\n",
- "z = R1*(1j*Xc)/(R1 + 1j*Xc)\n",
- " #maximum power transfer is achieved when R = mod(z)\n",
- "R = abs(z)\n",
- " #Total circuit impedance at maximum power transfer condition,\n",
- "ZT = z + R\n",
- " #Current I flowing in the load is given by\n",
- "I = V/ZT\n",
- "Imag = abs(I)\n",
- " #maximum power transferred,\n",
- "P = R*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",round(R,2),\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 84.67 ohm\n",
- "\n",
- " (b) maximum power delivered is 127.9 W"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 621</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of R for which the power transferred to the load is a maximum, and \n",
- "#(b) the value of the maximum power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 60;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "XL = 10;# in ohm\n",
- "Xc = 7;# in ohm\n",
- "R2 = XL*1j;# in ohm\n",
- "R3 = -1j*Xc;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #maximum power transfer is achieved when\n",
- "R = (R1**2 + (XL - Xc)**2)**0.5\n",
- " #Hence source impedance,\n",
- "ZT = R1 + R2 + R3 + R\n",
- " #Current I flowing in the load is given by\n",
- "I = V/ZT\n",
- "Imag = abs(I)\n",
- " #maximum power transferred,\n",
- "P = R*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",P,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 5.0 ohm\n",
- "\n",
- " (b) maximum power delivered is 200.0 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 622</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the load resistance R\n",
- "#calculate the value of this power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 20;# in volts\n",
- "R1 = 5;# in ohm\n",
- "R2 = 15;# in ohm\n",
- "\n",
- "#calculation: \n",
- " #R is removed from the network as shown in Figure 35.6\n",
- " #P.d. across AB, E\n",
- "E = (R2/(R1 + R2))*V\n",
- " #Impedance \u2018looking-in\u2019 at terminals AB with the source removed is given by\n",
- "r = R1*R2/(R1 + R2)\n",
- " #The equivalent Th\u00b4evenin circuit supplying terminals AB is shown in Figure 35.7. \n",
- " #From condition (2), for maximum power transfer\n",
- "R = r\n",
- " #Current I flowing in the load is given by\n",
- "I = E/(R + r)\n",
- " #maximum power transferred,\n",
- "P = R*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",P,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 3.75 ohm\n",
- "\n",
- " (b) maximum power delivered is 15.0 W"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 622</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the values of R and X that will result in maximum power being transferred across terminals AB, and \n",
- "#(b) the value of the maximum power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 100;# in volts\n",
- "thetav = 30;# in degrees\n",
- "R1 = 5;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = 10j;# in ohm\n",
- "\n",
- "#calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Resistance R and reactance X are removed from the network as shown in Figure 35.9\n",
- " #P.d. across AB,\n",
- "E = ((R2 + R3)/(R1 + R2 + R3))*V\n",
- " #With the source removed the impedance, z, \u2018looking in\u2019 at terminals AB is given by:\n",
- "z = (R2 + R3)*R1/(R1 + R2 + R3)\n",
- " #The equivalent Th\u00b4evenin circuit is shown in Figure 35.10. From condition 3, \n",
- " #maximum power transfer is achieved when X = -1*imag(z) and R = real(z)\n",
- "X = -1*z.imag\n",
- "R = z.real\n",
- "Z = R + 1j*X\n",
- " #Current I flowing in the load is given by\n",
- "I = E/(z + Z)\n",
- "Imag = abs(I)\n",
- " #maximum power transferred,\n",
- "P = R*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm and X is \", X,\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 3.75 ohm and X is -1.25 ohm\n",
- "\n",
- " (b) maximum power delivered is 416.67 W"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 624</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the optimum value of load resistance for maximum power transfer\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ro = 448;# in ohm\n",
- "tr = 8;# turn ratio N1/N2\n",
- "\n",
- " #calculation: \n",
- " #The equivalent input resistance r of the transformer must be Ro for maximum power transfer.\n",
- "r = Ro\n",
- "RL = r*(1/tr)**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the optimum value of load resistance is \",RL,\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the optimum value of load resistance is 7.0 ohm"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 624</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the turns ratio of an ideal transformer necessary to match the generator to a load of (40 + j19) ohm \n",
- "#for maximum transfer of power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Zo = 450 + 1j*60;# in ohm\n",
- "ZL = 40 + 1j*19;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #transformer turns ratio tr = (N1/N2)\n",
- "Zomag = abs(Zo)\n",
- "ZLmag = abs(ZL)\n",
- "tr = (Zomag/ZLmag)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the transformer turns ratio is \",round(tr,2),\"\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the transformer turns ratio is 3.2 "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 625</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the primary current flowing, and \n",
- "#(b) the power dissipated in the load resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 240;# in volts\n",
- "V2 = 1920;# in volts\n",
- "R1 = 5;# in ohms\n",
- "R2 = 1600;# in ohms\n",
- "\n",
- "#calculation: \n",
- " #The network is shown in Figure 35.12.\n",
- " #turn ratio N1/N2 = V1/V2\n",
- "tr = V1/V2\n",
- " #Equivalent input resistance of the transformer,\n",
- "RL = R2\n",
- "r = RL*tr**2\n",
- " #Total input resistance,\n",
- "Rin = R1 + r\n",
- " #primary current, I1\n",
- "I1 = V1/Rin\n",
- " #For an ideal transformer V1/V2 = I2/I1\n",
- "I2 = I1*(V1/V2)\n",
- " #Power dissipated in the load resistance\n",
- "P = RL*I2**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) primary current flowing is \",I1,\" A\"\n",
- "print \"\\n (b) Power dissipated in the load resistance is \",P,\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) primary current flowing is 8.0 A\n",
- "\n",
- " (b) Power dissipated in the load resistance is 1600.0 W"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 625</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for maximum power transfer (a) the value of the load resistance,\n",
- "# and (b) the power dissipated in the load.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 30;# in volts\n",
- "thetav = 0;# in degrees\n",
- "r = 20000;# in ohms\n",
- "tr = 20;# turn ratio\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180) \n",
- " #The network diagram is shown in Figure 35.13.\n",
- " #For maximum power transfer, r1 must be equal to\n",
- "r1 = r\n",
- " #load resistance RL\n",
- "RL = r1/tr**2\n",
- " #The total input resistance when the source is connected to the matching transformer is\n",
- "RT = r + r1\n",
- " #Primary current\n",
- "I1 = V/RT\n",
- " #N1/N2 = I2/I1\n",
- "I2 = I1*tr\n",
- " #Power dissipated in load resistance RL is given by\n",
- "P = RL*I2**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the value of the load resistance is \",RL,\" ohm\"\n",
- "print \"\\n (b) Power dissipated in the load resistance is \",abs(P*1000),\"mW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the value of the load resistance is 50.0 ohm\n",
- "\n",
- " (b) Power dissipated in the load resistance is 11.25 mW"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_2.ipynb deleted file mode 100755 index 7462b3a9..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_2.ipynb +++ /dev/null @@ -1,637 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 35: Maximum power transfer theorems and impedance matching</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 620</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of R for maximum power to be transferred from the source to the load,\n",
- "#and (b) the value of the maximum power delivered to R\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 120;# in volts\n",
- "thetav = 0;# in degrees\n",
- "Z = 15 + 1j*20;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #maximum power transfer occurs when R = mod(Z)\n",
- "R = abs(Z)\n",
- " #the total circuit impedance\n",
- "ZT = Z + R\n",
- " #Current I flowing in the load is given by\n",
- "I = V/ZT\n",
- "Imag = abs(I)\n",
- " #maximum power delivered\n",
- "P = R*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",abs(P),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 25.0 ohm\n",
- "\n",
- " (b) maximum power delivered is 180.0 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 620</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine\n",
- "#(a) the value of Z that results in maximum power transfer, and\n",
- "#(b) the value of the maximum power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 120;# in volts\n",
- "thetav = 0;# in degrees\n",
- "Z = 15 + 1j*20;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #maximum power transfer occurs when X = -1*imag(Z) and R = real(Z)\n",
- "z = Z.real - 1j*Z.imag\n",
- " #Total circuit impedance at maximum power transfer condition,\n",
- "ZT = Z + z\n",
- " #Current I flowing in the load is given by\n",
- "I = V/ZT\n",
- "Imag = abs(I)\n",
- " #maximum power delivered\n",
- "P = Z.real*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when Z is \",z.real,\" + (\", z.imag,\")i ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",abs(P),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when Z is 15.0 + ( -20.0 )i ohm\n",
- "\n",
- " (b) maximum power delivered is 240.0 W"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 620</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine\n",
- "#(a) the value of the load resistance R required for maximum power transfer, and\n",
- "#(b) the value of the maximum power transferred.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 200;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 100;# in ohm\n",
- "C = 1E-6;# in farad\n",
- "f = 1000;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Capacitive reactance, Xc\n",
- "Xc = 1/(2*math.pi*f*C)\n",
- " #Hence source impedance,\n",
- "z = R1*(1j*Xc)/(R1 + 1j*Xc)\n",
- " #maximum power transfer is achieved when R = mod(z)\n",
- "R = abs(z)\n",
- " #Total circuit impedance at maximum power transfer condition,\n",
- "ZT = z + R\n",
- " #Current I flowing in the load is given by\n",
- "I = V/ZT\n",
- "Imag = abs(I)\n",
- " #maximum power transferred,\n",
- "P = R*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",round(R,2),\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 84.67 ohm\n",
- "\n",
- " (b) maximum power delivered is 127.9 W"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 621</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of R for which the power transferred to the load is a maximum, and \n",
- "#(b) the value of the maximum power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 60;# in volts\n",
- "thetav = 0;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "XL = 10;# in ohm\n",
- "Xc = 7;# in ohm\n",
- "R2 = XL*1j;# in ohm\n",
- "R3 = -1j*Xc;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #maximum power transfer is achieved when\n",
- "R = (R1**2 + (XL - Xc)**2)**0.5\n",
- " #Hence source impedance,\n",
- "ZT = R1 + R2 + R3 + R\n",
- " #Current I flowing in the load is given by\n",
- "I = V/ZT\n",
- "Imag = abs(I)\n",
- " #maximum power transferred,\n",
- "P = R*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",P,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 5.0 ohm\n",
- "\n",
- " (b) maximum power delivered is 200.0 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 622</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the load resistance R\n",
- "#calculate the value of this power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 20;# in volts\n",
- "R1 = 5;# in ohm\n",
- "R2 = 15;# in ohm\n",
- "\n",
- "#calculation: \n",
- " #R is removed from the network as shown in Figure 35.6\n",
- " #P.d. across AB, E\n",
- "E = (R2/(R1 + R2))*V\n",
- " #Impedance \u2018looking-in\u2019 at terminals AB with the source removed is given by\n",
- "r = R1*R2/(R1 + R2)\n",
- " #The equivalent Th\u00b4evenin circuit supplying terminals AB is shown in Figure 35.7. \n",
- " #From condition (2), for maximum power transfer\n",
- "R = r\n",
- " #Current I flowing in the load is given by\n",
- "I = E/(R + r)\n",
- " #maximum power transferred,\n",
- "P = R*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",P,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 3.75 ohm\n",
- "\n",
- " (b) maximum power delivered is 15.0 W"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 622</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the values of R and X that will result in maximum power being transferred across terminals AB, and \n",
- "#(b) the value of the maximum power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 100;# in volts\n",
- "thetav = 30;# in degrees\n",
- "R1 = 5;# in ohm\n",
- "R2 = 5;# in ohm\n",
- "R3 = 10j;# in ohm\n",
- "\n",
- "#calculation: \n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
- " #Resistance R and reactance X are removed from the network as shown in Figure 35.9\n",
- " #P.d. across AB,\n",
- "E = ((R2 + R3)/(R1 + R2 + R3))*V\n",
- " #With the source removed the impedance, z, \u2018looking in\u2019 at terminals AB is given by:\n",
- "z = (R2 + R3)*R1/(R1 + R2 + R3)\n",
- " #The equivalent Th\u00b4evenin circuit is shown in Figure 35.10. From condition 3, \n",
- " #maximum power transfer is achieved when X = -1*imag(z) and R = real(z)\n",
- "X = -1*z.imag\n",
- "R = z.real\n",
- "Z = R + 1j*X\n",
- " #Current I flowing in the load is given by\n",
- "I = E/(z + Z)\n",
- "Imag = abs(I)\n",
- " #maximum power transferred,\n",
- "P = R*Imag**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm and X is \", X,\" ohm\"\n",
- "print \"\\n (b) maximum power delivered is \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)maximum power transfer occurs when R is 3.75 ohm and X is -1.25 ohm\n",
- "\n",
- " (b) maximum power delivered is 416.67 W"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 624</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the optimum value of load resistance for maximum power transfer\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ro = 448;# in ohm\n",
- "tr = 8;# turn ratio N1/N2\n",
- "\n",
- " #calculation: \n",
- " #The equivalent input resistance r of the transformer must be Ro for maximum power transfer.\n",
- "r = Ro\n",
- "RL = r*(1/tr)**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the optimum value of load resistance is \",RL,\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the optimum value of load resistance is 7.0 ohm"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 624</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the turns ratio of an ideal transformer necessary to match the generator to a load of (40 + j19) ohm \n",
- "#for maximum transfer of power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Zo = 450 + 1j*60;# in ohm\n",
- "ZL = 40 + 1j*19;# in ohm\n",
- "\n",
- " #calculation: \n",
- " #transformer turns ratio tr = (N1/N2)\n",
- "Zomag = abs(Zo)\n",
- "ZLmag = abs(ZL)\n",
- "tr = (Zomag/ZLmag)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the transformer turns ratio is \",round(tr,2),\"\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the transformer turns ratio is 3.2 "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 625</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the primary current flowing, and \n",
- "#(b) the power dissipated in the load resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 240;# in volts\n",
- "V2 = 1920;# in volts\n",
- "R1 = 5;# in ohms\n",
- "R2 = 1600;# in ohms\n",
- "\n",
- "#calculation: \n",
- " #The network is shown in Figure 35.12.\n",
- " #turn ratio N1/N2 = V1/V2\n",
- "tr = V1/V2\n",
- " #Equivalent input resistance of the transformer,\n",
- "RL = R2\n",
- "r = RL*tr**2\n",
- " #Total input resistance,\n",
- "Rin = R1 + r\n",
- " #primary current, I1\n",
- "I1 = V1/Rin\n",
- " #For an ideal transformer V1/V2 = I2/I1\n",
- "I2 = I1*(V1/V2)\n",
- " #Power dissipated in the load resistance\n",
- "P = RL*I2**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) primary current flowing is \",I1,\" A\"\n",
- "print \"\\n (b) Power dissipated in the load resistance is \",P,\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) primary current flowing is 8.0 A\n",
- "\n",
- " (b) Power dissipated in the load resistance is 1600.0 W"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 625</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for maximum power transfer (a) the value of the load resistance,\n",
- "# and (b) the power dissipated in the load.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv = 30;# in volts\n",
- "thetav = 0;# in degrees\n",
- "r = 20000;# in ohms\n",
- "tr = 20;# turn ratio\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180) \n",
- " #The network diagram is shown in Figure 35.13.\n",
- " #For maximum power transfer, r1 must be equal to\n",
- "r1 = r\n",
- " #load resistance RL\n",
- "RL = r1/tr**2\n",
- " #The total input resistance when the source is connected to the matching transformer is\n",
- "RT = r + r1\n",
- " #Primary current\n",
- "I1 = V/RT\n",
- " #N1/N2 = I2/I1\n",
- "I2 = I1*tr\n",
- " #Power dissipated in load resistance RL is given by\n",
- "P = RL*I2**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the value of the load resistance is \",RL,\" ohm\"\n",
- "print \"\\n (b) Power dissipated in the load resistance is \",abs(P*1000),\"mW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the value of the load resistance is 50.0 ohm\n",
- "\n",
- " (b) Power dissipated in the load resistance is 11.25 mW"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_35.ipynb index d4ffa9bf..d4ffa9bf 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_35.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint.ipynb deleted file mode 100755 index 39a16efe..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint.ipynb +++ /dev/null @@ -1,1017 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 36: Complex Waveforms</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 643</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Write down an expression to represent voltage v.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 240; # in Volts\n",
- "f = 50; # in Hz\n",
- "x = 0.2;\n",
- "phi3 = 3*math.pi/4; # in Rad\n",
- "\n",
- " #calculation:\n",
- "Vamp = V*2**0.5\n",
- "w = 2*math.pi*f\n",
- "T = 1/f\n",
- "V3 = Vamp*x\n",
- "f3 = 3*f\n",
- "w3 = 3*w\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage, V =\",round(Vamp,1),\"sin(\",round(w,1),\"t) + \",round(V3,1),\"sin(\", round(w3,1),\"t - \", round(phi3,1),\") volts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage, V = 339.4 sin( 314.2 t) + 67.9 sin( 942.5 t - 2.4 ) volts"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 648</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the rms value of the current waveform\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "A1 = 0.100;# in amperes\n",
- "A3 = 0.020;# in amperes\n",
- "A5 = 0.010;# in amperes\n",
- "\n",
- " #calculation:\n",
- " #the rms value of current is given by\n",
- "Irms = ((A1**2 + A3**2 + A5**2)/2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the rms value of current is \",round(Irms*1000,2),\" mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the rms value of current is 72.46 mA"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 649</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the voltage, (a) the rms value, (b) the mean value and (c) the form factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "A1 = 10;# in volts\n",
- "A3 = 3;# in volts\n",
- "A5 = 2;# in volts\n",
- "\n",
- "#calculation:\n",
- " #the rms value of voltage is given by\n",
- "Vrms = ((A1**2 + A3**2 + A5**2)/2)**0.5\n",
- " #the mean value of voltage is given by\n",
- " #x = wt\n",
- "Vav = (1/math.pi)*((10 + 1 + 2/5)-(-10 - 1 - 2/5))\n",
- " #form factor is given by\n",
- "ff = Vrms/Vav\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the rms value of voltage is \",round(Vrms,2),\" V\"\n",
- "print \"\\n (b)the mean value of voltage is \",round(Vav,2),\" V\"\n",
- "print \"\\n (c)form factor is \",round(ff,3),\" \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the rms value of voltage is 7.52 V\n",
- "\n",
- " (b)the mean value of voltage is 7.26 V\n",
- "\n",
- " (c)form factor is 1.036 "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 649</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the rms value of the fundamental and each harmonic. \n",
- "#(b) Write down an expression to represent the complex voltage waveform if the frequency of the fundamental is 31.83 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 240;# in volts\n",
- "x = 0.3;# for third harmonic\n",
- "y = 0.1;# for fifth harmonic\n",
- "f = 31.83;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #V3 = x*V1\n",
- " #V5 = y*V1\n",
- " #the rms value of the fundamental,\n",
- "V1 = ((V**2)/(1 + x**2 + y**2))**0.5\n",
- " #Rms value of the third harmonic\n",
- "V3 = x*V1\n",
- " #the rms value of the fifth harmonic,\n",
- "V5 = y*V1\n",
- " #Maximum value of the fundamental,\n",
- "V1m = V1*2**0.5\n",
- " #Maximum value of the third harmonic,\n",
- "V3m = V3*2**0.5\n",
- " #Maximum value of the fifth harmonic,\n",
- "V5m = V5*2**0.5\n",
- "w = 2*math.pi*f\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"v = \",round(V1m,2),\"sin\",round(w,2),\"t + \",round(V3m,2),\"sin\",round((3*w),2),\"t + \",round(V5m,2),\"sin\",round((5*w),2),\"t Volts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "v = 323.62 sin 199.99 t + 97.08 sin 599.98 t + 32.36 sin 999.97 t Volts\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 652</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the average power in a 20 ohm resistance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "A1 = 12;# in amperes\n",
- "A3 = 5;# in amperes\n",
- "A5 = 2;# in amperes\n",
- "R = 20;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #rms current\n",
- "Irms = ((A1**2 + A3**2 + A5**2)/2)**0.5\n",
- " #average power\n",
- "P = R*Irms**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n average power \",P,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " average power 1730.0 W"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 652</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the total active power supplied to the circuit, and (b) the overall power factor\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ia1 = 2;# in amperes\n",
- "Ia3 = 0.3;# in amperes\n",
- "Ia5 = 0.1;# in amperes\n",
- "Va1 = 60;# in volts\n",
- "Va3 = 15;# in volts\n",
- "Va5 = 10;# in volts\n",
- "Phii1 = -1*math.pi/6;# in radians\n",
- "Phii3 = -1*math.pi/12;# in radians\n",
- "Phii5 = -8*math.pi/9;# in radians\n",
- "Phiv1 = 0;# in radians\n",
- "Phiv3 = math.pi/4;# in radians\n",
- "Phiv5 = -1*math.pi/2;# in radians\n",
- "\n",
- "\n",
- " #calculation:\n",
- " #rms values;\n",
- "I1 = Ia1/(2**0.5);# in amperes\n",
- "I3 = Ia3/(2**0.5);# in amperes\n",
- "I5 = Ia5/(2**0.5);# in amperes\n",
- "V1 = Va1/(2**0.5);# in volts\n",
- "V3 = Va3/(2**0.5);# in volts\n",
- "V5 = Va5/(2**0.5);# in volts\n",
- " #total power supplied,\n",
- "P = V1*I1*math.cos(Phiv1 - Phii1) + V3*I3*math.cos(Phiv3 - Phii3) + V5*I5*math.cos(Phiv5 - Phii5)\n",
- " #rms current\n",
- "Irms = ((I1**2 + I3**2 + I5**2))**0.5\n",
- " #rms voltage\n",
- "Vrms = ((V1**2 + V3**2 + V5**2))**0.5\n",
- " #overall power factor\n",
- "pf = P/(Vrms*Irms)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)the total active power supplied to the circuit \",round(P,2),\" W\"\n",
- "print \"\\n(b)overall power factor \",round(pf,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)the total active power supplied to the circuit 53.26 W\n",
- "\n",
- "(b)overall power factor 0.84"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 655</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for each case an expression for the current flowing if the fundamental frequency is 1 kHz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 40;# in ohm\n",
- "L = 7.96E-3;# in Henry\n",
- "C = 25E-6; # in Farad\n",
- "f = 1000; # in Hx\n",
- "\n",
- "#calculation:\n",
- "wL = 2*math.pi*1000*L\n",
- "wC = 2*math.pi*1000*C\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)i = \",round(100/R1,2),\"sin(wt) +\",round(30/R1,2),\"sin(3wt - pi/3) +\",round(10/R1,2),\"sin(5wt - pi/6) A\"\n",
- "print \"(b)i = \",round(100/wL,2),\"sin(wt - pi/2) +\",round(30/(3*wL),2),\"sin(3wt - pi/6) +\",round(10/(5*wL),2),\"sin(5wt - 2pi/3) A\"\n",
- "print \"(c)i = \",round(100*wC,2),\"sin(wt + pi/2) +\",round(30*3*wC,2),\"sin(3wt + 5pi/6) +\",round(10*5*wC,2),\"sin(5wt + pi/3) A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)i = 2.5 sin(wt) + 0.75 sin(3wt - pi/3) + 0.25 sin(5wt - pi/6) A\n",
- "(b)i = 2.0 sin(wt - pi/2) + 0.2 sin(3wt - pi/6) + 0.04 sin(5wt - 2pi/3) A\n",
- "(c)i = 15.71 sin(wt + pi/2) + 14.14 sin(3wt + 5pi/6) + 7.85 sin(5wt + pi/3) A\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 656</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) an expression to represent the instantaneous value of the current, \n",
- "#(b) the rms voltage, (c) the rms current, (d) the power dissipated, and (e) the overall power factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1m = 240;# in volts\n",
- "V3m = 40;# in volts\n",
- "V5m = 30;# in volts\n",
- "w1 = 314;# fundamental\n",
- "R = 12;# in ohm\n",
- "L = 0.00955;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #fundamental or first harmonic\n",
- " #inductive reactance,\n",
- "XL1 = w1*L\n",
- " #impedance at the fundamental frequency,\n",
- "Z1 = R + 1j*XL1\n",
- " #Maximum current at fundamental frequency\n",
- "I1m = V1m/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- " #Third harmonic\n",
- "XL3 = 3*XL1\n",
- " #impedance at the third harmonic frequency,\n",
- "Z3 = R + 1j*XL3\n",
- " #Maximum current at third harmonic frequency\n",
- "I3m = V3m/Z3\n",
- "I3mag = abs(I3m)\n",
- "phii3 = cmath.phase(complex(I3m.real,I3m.imag))\n",
- " #fifth harmonic\n",
- "XL5 = 5*XL1\n",
- " #impedance at the third harmonic frequency,\n",
- "Z5 = R + 1j*XL5\n",
- " #Maximum current at third harmonic frequency\n",
- "I5m = V5m/Z5\n",
- "I5mag = abs(I5m)\n",
- "phii5 = cmath.phase(complex(I5m.real,I5m.imag))\n",
- " #rms voltage\n",
- "Vrms = ((V1m**2 + V3m**2 + V5m**2)/2)**0.5\n",
- " #rms current\n",
- "Irms = ((I1mag**2 + I3mag**2 + I5mag**2)/2)**0.5\n",
- " #power dissipated\n",
- "P = R*Irms**2\n",
- " #overall power factor\n",
- "pf = P/(Vrms*Irms)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)i = \",round(I1mag,2),\"sin(\",round(w1,2),\"t + (\",round(phii1,2),\")) + \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t + (\",round(phii3,2),\")) + \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t + (\",round(phii5,2),\")) A\"\n",
- "print \"\\n(b)the rms value of current is \",round(Irms,2),\" A\"\n",
- "print \"\\n(c)the rms value of voltage is \",round(Vrms,2),\" V\"\n",
- "print \"\\n(d)the total power dissipated \",round(P,2),\" W\"\n",
- "print \"\\n(e)overall power factor \",round(pf,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)i = 19.4 sin( 314.0 t + ( -0.24 )) + 2.67 sin( 942.0 t + ( -0.64 )) + 1.56 sin( 1570.0 t + ( -0.9 )) A\n",
- "\n",
- "(b)the rms value of current is 13.89 A\n",
- "\n",
- "(c)the rms value of voltage is 173.35 V\n",
- "\n",
- "(d)the total power dissipated 2316.26 W\n",
- "\n",
- "(e)overall power factor 0.96"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 658</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Obtain an expression for the current flowing and hence determine the rms value of current.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vom = 50;# in volts\n",
- "V1m = 200;# in volts\n",
- "V2m = 40;# in volts\n",
- "V4m = 5;# in volts\n",
- "f = 50;# in Hz\n",
- "R = 50;# in ohm\n",
- "C = 100E-6;# in farad\n",
- "phiv1 = 0;# in rad\n",
- "phiv2 = -1*math.pi/2;# in rad\n",
- "phiv4 = math.pi/4;# in rad\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = V1m*math.cos(phiv1) + 1j*V1m*math.sin(phiv1)\n",
- "V2 = V2m*math.cos(phiv2) + 1j*V2m*math.sin(phiv2)\n",
- "V4 = V4m*math.cos(phiv4) + 1j*V4m*math.sin(phiv4)\n",
- " #Inductance has no effect on a steady current. Hence the d.c. component of the current, i0, is given by\n",
- "Iom = 0\n",
- " #fundamental or first harmonic\n",
- "w1 = 2*math.pi*f\n",
- " #inductive reactance,\n",
- "Xc1 = 1/(w1*C)\n",
- " #impedance at the fundamental frequency,\n",
- "Z1 = R + 1j*Xc1\n",
- " #Maximum current at fundamental frequency\n",
- "I1m = V1/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- " #second harmonic\n",
- "Xc2 = Xc1/2\n",
- " #impedance at the third harmonic frequency,\n",
- "Z2 = R + 1j*Xc2\n",
- " #Maximum current at third harmonic frequency\n",
- "I2m = V2/Z2\n",
- "I2mag = abs(I2m)\n",
- "phii2 = cmath.phase(complex(I2m.real,I2m.imag))\n",
- " #fourth harmonic\n",
- "Xc4 = Xc1/4\n",
- " #impedance at the third harmonic frequency,\n",
- "Z4 = R + 1j*Xc4\n",
- " #Maximum current at third harmonic frequency\n",
- "I4m = V4/Z4\n",
- "I4mag = abs(I4m)\n",
- "phii4 = cmath.phase(complex(I4m.real,I4m.imag))\n",
- " #rms current\n",
- "Irms = (Iom**2 + (I1mag**2 + I2mag**2 + I4mag**2)/2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t + (\",round(phii1,2),\")) + \",round(I2mag,2),\"sin(\",round((w1*2),2),\"t + (\",round(phii2,2),\")) + \",round(I4mag,2),\"sin(\",round((w1*4),2),\"t + (\",round(phii4,2),\")) A\"\n",
- "print \"(b)the rms value of current is \",round(Irms,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)i = 0.0 + 3.37 sin( 314.16 t + ( -0.57 )) + 0.76 sin( 628.32 t + ( -1.88 )) + 0.1 sin( 1256.64 t + ( 0.63 )) A\n",
- "(b)the rms value of current is 2.45 A\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 659</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) an expression to represent the current flowing in the circuit, \n",
- "#(b) the rms value of current, correct to two decimal places, and \n",
- "#(c) the power dissipated in the circuit, correct to three significant figures.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vom = 25;# in volts\n",
- "V1m = 100;# in volts\n",
- "V3m = 40;# in volts\n",
- "V5m = 20;# in volts\n",
- "w1 = 10000;# fundamental\n",
- "R = 5;# in ohm\n",
- "L = 500E-6;# in Henry\n",
- "phiv1 = 0;# in rad\n",
- "phiv3 = math.pi/6;# in rad\n",
- "phiv5 = math.pi/12;# in rad\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = V1m*math.cos(phiv1) + 1j*V1m*math.sin(phiv1)\n",
- "V3 = V3m*math.cos(phiv3) + 1j*V3m*math.sin(phiv3)\n",
- "V5 = V5m*math.cos(phiv5) + 1j*V5m*math.sin(phiv5)\n",
- " #Inductance has no effect on a steady current. Hence the d.c. component of the current, i0, is given by\n",
- "Iom = Vom/R\n",
- " #fundamental or first harmonic\n",
- " #inductive reactance,\n",
- "XL1 = w1*L\n",
- " #impedance at the fundamental frequency,\n",
- "Z1 = R + 1j*XL1\n",
- " #Maximum current at fundamental frequency\n",
- "I1m = V1/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- "#Third harmonic\n",
- "XL3 = 3*XL1\n",
- " #impedance at the third harmonic frequency,\n",
- "Z3 = R + 1j*XL3\n",
- " #Maximum current at third harmonic frequency\n",
- "I3m = V3/Z3\n",
- "I3mag = abs(I3m)\n",
- "phii3 = cmath.phase(complex(I3m.real,I3m.imag))\n",
- " #fifth harmonic\n",
- "XL5 = 5*XL1\n",
- " #impedance at the third harmonic frequency,\n",
- "Z5 = R + 1j*XL5\n",
- " #Maximum current at third harmonic frequency\n",
- "I5m = V5/Z5\n",
- "I5mag = abs(I5m)\n",
- "phii5 = cmath.phase(complex(I5m.real,I5m.imag))\n",
- " #rms current\n",
- "Irms = (Iom**2 + (I1mag**2 + I3mag**2 + I5mag**2)/2)**0.5\n",
- " #power dissipated\n",
- "P = R*Irms**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t + (\",round(phii1,2),\")) + \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t + (\",round(phii3,2),\")) + \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t + (\",round(phii5,2),\")) A\"\n",
- "print \"\\n(b)the rms value of current is \",round(Irms,2),\" A\"\n",
- "print \"\\n(c)the total power dissipated \",round(P,3),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)i = 5.0 + 14.14 sin( 10000.0 t + ( -0.79 )) + 2.53 sin( 30000.0 t + ( -0.73 )) + 0.78 sin( 50000.0 t + ( -1.11 )) A\n",
- "\n",
- "(b)the rms value of current is 11.34 A\n",
- "\n",
- "(c)the total power dissipated 642.538 W\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 661</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the average power supplied, (b) the type of components present, and (c) the values of the components.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vom = 30;# in volts\n",
- "V1m = 40;# in volts\n",
- "V2m = 25;# in volts\n",
- "V4m = 15;# in volts\n",
- "Iom = 0;# in amperes\n",
- "I1m = 0.743;# in Amperes\n",
- "I2m = 0.781;# in Amperes\n",
- "I4m = 0.636;# in Amperes\n",
- "phii1 = 1.190;# in rad\n",
- "phii2 = 0.896;# in rad\n",
- "phii4 = 0.559;# in rad\n",
- "w = 1000;# in rad\n",
- "\n",
- " #calculation:\n",
- " #the average power P is given by\n",
- "P = Vom*Iom+(0.707*V1m)*(0.707*I1m)*math.cos(phii1)+(0.707*V2m)*(0.707*I2m)*math.cos(phii2) + (0.707*V4m)*(0.707*I4m)*math.cos(phii4)\n",
- " #rms current\n",
- "Irms = (Iom**2 + (I1m**2 + I2m**2 + I4m**2)/2)**0.5\n",
- " #resistance R\n",
- "R = P/(Irms**2)\n",
- " #impedance\n",
- "Z1 = V1m/I1m\n",
- " #Xc1\n",
- "Xc1 = (Z1**2 - R**2)**0.5\n",
- " #capacitance\n",
- "C = 1/(w*Xc1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)the average power P is \",round(P,2),\" W\"\n",
- "print \"\\n(c)the resistance R \",round(R,2),\" ohm and capacitance \",round(C*1E6,2),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)the average power P is 15.66 W\n",
- "\n",
- "(c)the resistance R 19.99 ohm and capacitance 20.01 uF\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 662</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) an expression for the supply current, i, (b) the percentage harmonic content of the supply current, (c) the total power dissipated, (d) an expression for the p.d. shown as \t1, and (e) an expression for current ic.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1m = 300;# in volts\n",
- "V3m = 120;# in volts\n",
- "phiv1 = 0;# in rad\n",
- "phiv2 = 0.698;# in rad\n",
- "w1 = 314;# in rad\n",
- "C = 2.123E-6;# in farads\n",
- "R1 = 560;# in ohms\n",
- "R2 = 2000;# in Ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = V1m*math.cos(phiv1) + 1j*V1m*math.sin(phiv1)\n",
- "V3 = V3m*math.cos(phiv2) + 1j*V3m*math.sin(phiv2)\n",
- " #capacitive reactance,\n",
- "Xc1 = 1/(w1*C)\n",
- " #impedance at the fundamental frequency,\n",
- "Z1 = R1 - 1j*Xc1*R2/(R2 - 1j*Xc1)\n",
- " #Maximum current at fundamental frequency\n",
- "I1m = V1/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- " #Third harmonic\n",
- "Xc3 = Xc1/3\n",
- " #impedance at the third harmonic frequency,\n",
- "Z3 = R1 - 1j*Xc3*R2/(R2 - 1j*Xc3)\n",
- "I1m = V1m/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- " #Maximum current at third harmonic frequency\n",
- "I3m = V3/Z3\n",
- "I3mag = abs(I3m)\n",
- "phii3 = cmath.phase(complex(I3m.real,I3m.imag))\n",
- " #Percentage harmonic content of the supply current is given by\n",
- "percent = I3mag*100/I1mag\n",
- " #total active power\n",
- "P = (0.707*V1m)*(0.707*I1mag)*math.cos(phiv1 - phii1) + (0.707*V3m)*(0.707*I3m)*math.cos(phiv2 - phii3)\n",
- "\n",
- "I1 = I1m*R2/(R2 - 1j*Xc1)\n",
- "I3 = I3m*R2/(R2 - 1j*Xc3)\n",
- "\n",
- "I1nmag = abs(I1)\n",
- "phini1 = cmath.phase(complex(I1.real,I1.imag))\n",
- "I3nmag = abs(I3)\n",
- "phini3 = cmath.phase(complex(I3.real,I3.imag))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)supply current, i=\", round(I1mag,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
- "print \"\\n(b)Percentage harmonic content of the supply current is \",round(percent,2),\" percent\"\n",
- "print \"\\n(c)total active power is \",round(abs(P),2),\" W\"\n",
- "print \"\\n(d)Voltage, v1 =\", round(I1mag*R1,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag*R1,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
- "print \"\\n(e)current, ic =\", round(I1nmag,3),\"sin(\", w1,\"t +\",round(phini1,3),\") + \",round(I3nmag,3),\"sin(\", 3*w1,\"t +\",round(phini3,3),\") A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)supply current, i= 0.187 sin( 314 t + 0.643 ) + 0.145 sin( 942 t + 1.305 ) A\n",
- "\n",
- "(b)Percentage harmonic content of the supply current is 77.57 percent\n",
- "\n",
- "(c)total active power is 25.34 W\n",
- "\n",
- "(d)Voltage, v1 = 104.996 sin( 314 t + 0.643 ) + 81.45 sin( 942 t + 1.305 ) A\n",
- "\n",
- "(e)current, ic = 0.15 sin( 314 t + 1.287 ) + 0.141 sin( 942 t + 1.55 ) A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 664</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the fundamental frequency for resonance with the third harmonic, and \n",
- "#(b) the maximum value of the fundamental and third harmonic components of current\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1m = 400;# in volts\n",
- "V3m = 10;# in volts\n",
- "C = 0.2E-6;# in farads\n",
- "R = 2;# in ohms\n",
- "L = 0.5;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #Resonance with the third harmonic means that\n",
- "w = (1/(9*L*C))**0.5\n",
- " #fundamental frequency, f\n",
- "f = w/(2*math.pi)\n",
- " #At the fundamental frequency,\n",
- " #impedance Z1\n",
- "Z1 = R + 1j*(w*L - 1/(w*C))\n",
- "Z1mag = abs(Z1)\n",
- "phiZ1 = cmath.phase(complex(Z1.real,Z1.imag))\n",
- " #Maximum value of current at the fundamental frequency,\n",
- "I1m = V1m/Z1mag\n",
- " #At the third harmonic frequency,\n",
- "Z3 = R + 1j*(3*w*L - 1/(3*w*C))\n",
- "Z3mag = abs(Z3)\n",
- "phiZ3 = cmath.phase(complex(Z3.real,Z3.imag))\n",
- " #Maximum value of current at the third harmonic frequency,\n",
- "I3m = V3m/Z3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)fundamental frequency for resonance with the third harmonic is \",round(f,2),\" Hz\"\n",
- "print \"(b)Maximum value of current at fundamental freq. is\",round(abs(I1m),3),\"A \"\n",
- "print \"and at the third harmonic frequency \", abs(I3m),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)fundamental frequency for resonance with the third harmonic is 167.76 Hz\n",
- "(b)Maximum value of current at fundamental freq. is 0.095 A \n",
- "and at the third harmonic frequency 5.0 A\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 665</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of n, (b) the maximum value of current at the nth harmonic, \n",
- "#(c) the p.d. across the capacitor at the nth harmonic and\n",
- "#(d) the maximum value of the fundamental current.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1m = 800;# in volts\n",
- "f = 50;# in Hz\n",
- "x = 0.015;\n",
- "C = 0.122E-6;# in farads\n",
- "R = 5;# in ohms\n",
- "L = 0.369;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #voltage at nth harmonic\n",
- "Vnm = x*V1m\n",
- "w = 2*math.pi*f\n",
- " #For resonance at the nth harmonic nwL = 1/nwC\n",
- "n = 1/(w*(L*C)**0.5)\n",
- " #At resonance, impedance\n",
- "Zn = R\n",
- " #the maximum value of current at the nth harmonic\n",
- "Inm = Vnm/Zn\n",
- " #capacitive reactance, at nth harmonic\n",
- "Xcn = 1/(n*w*C)\n",
- " #the p.d. across the capacitor at the nth harmonic\n",
- "Vcn = Inm*Xcn\n",
- " #At the fundamental frequency, inductive reactance,\n",
- "XL1 = w*L\n",
- " #capacitive reactance\n",
- "Xc1 = 1/(w*C)\n",
- " #Impedance at the fundamental frequency,\n",
- "Z1 = R + 1j*(XL1 - Xc1)\n",
- "Z1mag = abs(Z1)\n",
- "phiZ1 = cmath.phase(complex(Z1.real,Z1.imag))\n",
- " #Maximum value of current at the fundamental frequency,\n",
- "I1m = V1m/Z1mag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)n = \",round(n,2),\"\"\n",
- "print \"\\n(b)the maximum value of current at the nth harmonic \",round(Inm,2),\" A\"\n",
- "print \"\\n(c)the p.d. across the capacitor at the nth harmonic is \",round(Vcn,2),\"\"\n",
- "print \"\\n(d)the maximum value of the fundamental current. \",round(I1m,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)n = 15.0 \n",
- "\n",
- "(b)the maximum value of current at the nth harmonic 2.4 A\n",
- "\n",
- "(c)the p.d. across the capacitor at the nth harmonic is 4173.92 \n",
- "\n",
- "(d)the maximum value of the fundamental current. 0.03 A"
- ]
- }
- ],
- "prompt_number": 14
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_1.ipynb deleted file mode 100755 index 39a16efe..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_1.ipynb +++ /dev/null @@ -1,1017 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 36: Complex Waveforms</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 643</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Write down an expression to represent voltage v.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 240; # in Volts\n",
- "f = 50; # in Hz\n",
- "x = 0.2;\n",
- "phi3 = 3*math.pi/4; # in Rad\n",
- "\n",
- " #calculation:\n",
- "Vamp = V*2**0.5\n",
- "w = 2*math.pi*f\n",
- "T = 1/f\n",
- "V3 = Vamp*x\n",
- "f3 = 3*f\n",
- "w3 = 3*w\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage, V =\",round(Vamp,1),\"sin(\",round(w,1),\"t) + \",round(V3,1),\"sin(\", round(w3,1),\"t - \", round(phi3,1),\") volts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage, V = 339.4 sin( 314.2 t) + 67.9 sin( 942.5 t - 2.4 ) volts"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 648</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the rms value of the current waveform\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "A1 = 0.100;# in amperes\n",
- "A3 = 0.020;# in amperes\n",
- "A5 = 0.010;# in amperes\n",
- "\n",
- " #calculation:\n",
- " #the rms value of current is given by\n",
- "Irms = ((A1**2 + A3**2 + A5**2)/2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the rms value of current is \",round(Irms*1000,2),\" mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the rms value of current is 72.46 mA"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 649</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the voltage, (a) the rms value, (b) the mean value and (c) the form factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "A1 = 10;# in volts\n",
- "A3 = 3;# in volts\n",
- "A5 = 2;# in volts\n",
- "\n",
- "#calculation:\n",
- " #the rms value of voltage is given by\n",
- "Vrms = ((A1**2 + A3**2 + A5**2)/2)**0.5\n",
- " #the mean value of voltage is given by\n",
- " #x = wt\n",
- "Vav = (1/math.pi)*((10 + 1 + 2/5)-(-10 - 1 - 2/5))\n",
- " #form factor is given by\n",
- "ff = Vrms/Vav\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the rms value of voltage is \",round(Vrms,2),\" V\"\n",
- "print \"\\n (b)the mean value of voltage is \",round(Vav,2),\" V\"\n",
- "print \"\\n (c)form factor is \",round(ff,3),\" \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the rms value of voltage is 7.52 V\n",
- "\n",
- " (b)the mean value of voltage is 7.26 V\n",
- "\n",
- " (c)form factor is 1.036 "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 649</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the rms value of the fundamental and each harmonic. \n",
- "#(b) Write down an expression to represent the complex voltage waveform if the frequency of the fundamental is 31.83 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 240;# in volts\n",
- "x = 0.3;# for third harmonic\n",
- "y = 0.1;# for fifth harmonic\n",
- "f = 31.83;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #V3 = x*V1\n",
- " #V5 = y*V1\n",
- " #the rms value of the fundamental,\n",
- "V1 = ((V**2)/(1 + x**2 + y**2))**0.5\n",
- " #Rms value of the third harmonic\n",
- "V3 = x*V1\n",
- " #the rms value of the fifth harmonic,\n",
- "V5 = y*V1\n",
- " #Maximum value of the fundamental,\n",
- "V1m = V1*2**0.5\n",
- " #Maximum value of the third harmonic,\n",
- "V3m = V3*2**0.5\n",
- " #Maximum value of the fifth harmonic,\n",
- "V5m = V5*2**0.5\n",
- "w = 2*math.pi*f\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"v = \",round(V1m,2),\"sin\",round(w,2),\"t + \",round(V3m,2),\"sin\",round((3*w),2),\"t + \",round(V5m,2),\"sin\",round((5*w),2),\"t Volts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "v = 323.62 sin 199.99 t + 97.08 sin 599.98 t + 32.36 sin 999.97 t Volts\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 652</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the average power in a 20 ohm resistance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "A1 = 12;# in amperes\n",
- "A3 = 5;# in amperes\n",
- "A5 = 2;# in amperes\n",
- "R = 20;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #rms current\n",
- "Irms = ((A1**2 + A3**2 + A5**2)/2)**0.5\n",
- " #average power\n",
- "P = R*Irms**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n average power \",P,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " average power 1730.0 W"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 652</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the total active power supplied to the circuit, and (b) the overall power factor\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ia1 = 2;# in amperes\n",
- "Ia3 = 0.3;# in amperes\n",
- "Ia5 = 0.1;# in amperes\n",
- "Va1 = 60;# in volts\n",
- "Va3 = 15;# in volts\n",
- "Va5 = 10;# in volts\n",
- "Phii1 = -1*math.pi/6;# in radians\n",
- "Phii3 = -1*math.pi/12;# in radians\n",
- "Phii5 = -8*math.pi/9;# in radians\n",
- "Phiv1 = 0;# in radians\n",
- "Phiv3 = math.pi/4;# in radians\n",
- "Phiv5 = -1*math.pi/2;# in radians\n",
- "\n",
- "\n",
- " #calculation:\n",
- " #rms values;\n",
- "I1 = Ia1/(2**0.5);# in amperes\n",
- "I3 = Ia3/(2**0.5);# in amperes\n",
- "I5 = Ia5/(2**0.5);# in amperes\n",
- "V1 = Va1/(2**0.5);# in volts\n",
- "V3 = Va3/(2**0.5);# in volts\n",
- "V5 = Va5/(2**0.5);# in volts\n",
- " #total power supplied,\n",
- "P = V1*I1*math.cos(Phiv1 - Phii1) + V3*I3*math.cos(Phiv3 - Phii3) + V5*I5*math.cos(Phiv5 - Phii5)\n",
- " #rms current\n",
- "Irms = ((I1**2 + I3**2 + I5**2))**0.5\n",
- " #rms voltage\n",
- "Vrms = ((V1**2 + V3**2 + V5**2))**0.5\n",
- " #overall power factor\n",
- "pf = P/(Vrms*Irms)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)the total active power supplied to the circuit \",round(P,2),\" W\"\n",
- "print \"\\n(b)overall power factor \",round(pf,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)the total active power supplied to the circuit 53.26 W\n",
- "\n",
- "(b)overall power factor 0.84"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 655</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for each case an expression for the current flowing if the fundamental frequency is 1 kHz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 40;# in ohm\n",
- "L = 7.96E-3;# in Henry\n",
- "C = 25E-6; # in Farad\n",
- "f = 1000; # in Hx\n",
- "\n",
- "#calculation:\n",
- "wL = 2*math.pi*1000*L\n",
- "wC = 2*math.pi*1000*C\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)i = \",round(100/R1,2),\"sin(wt) +\",round(30/R1,2),\"sin(3wt - pi/3) +\",round(10/R1,2),\"sin(5wt - pi/6) A\"\n",
- "print \"(b)i = \",round(100/wL,2),\"sin(wt - pi/2) +\",round(30/(3*wL),2),\"sin(3wt - pi/6) +\",round(10/(5*wL),2),\"sin(5wt - 2pi/3) A\"\n",
- "print \"(c)i = \",round(100*wC,2),\"sin(wt + pi/2) +\",round(30*3*wC,2),\"sin(3wt + 5pi/6) +\",round(10*5*wC,2),\"sin(5wt + pi/3) A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)i = 2.5 sin(wt) + 0.75 sin(3wt - pi/3) + 0.25 sin(5wt - pi/6) A\n",
- "(b)i = 2.0 sin(wt - pi/2) + 0.2 sin(3wt - pi/6) + 0.04 sin(5wt - 2pi/3) A\n",
- "(c)i = 15.71 sin(wt + pi/2) + 14.14 sin(3wt + 5pi/6) + 7.85 sin(5wt + pi/3) A\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 656</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) an expression to represent the instantaneous value of the current, \n",
- "#(b) the rms voltage, (c) the rms current, (d) the power dissipated, and (e) the overall power factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1m = 240;# in volts\n",
- "V3m = 40;# in volts\n",
- "V5m = 30;# in volts\n",
- "w1 = 314;# fundamental\n",
- "R = 12;# in ohm\n",
- "L = 0.00955;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #fundamental or first harmonic\n",
- " #inductive reactance,\n",
- "XL1 = w1*L\n",
- " #impedance at the fundamental frequency,\n",
- "Z1 = R + 1j*XL1\n",
- " #Maximum current at fundamental frequency\n",
- "I1m = V1m/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- " #Third harmonic\n",
- "XL3 = 3*XL1\n",
- " #impedance at the third harmonic frequency,\n",
- "Z3 = R + 1j*XL3\n",
- " #Maximum current at third harmonic frequency\n",
- "I3m = V3m/Z3\n",
- "I3mag = abs(I3m)\n",
- "phii3 = cmath.phase(complex(I3m.real,I3m.imag))\n",
- " #fifth harmonic\n",
- "XL5 = 5*XL1\n",
- " #impedance at the third harmonic frequency,\n",
- "Z5 = R + 1j*XL5\n",
- " #Maximum current at third harmonic frequency\n",
- "I5m = V5m/Z5\n",
- "I5mag = abs(I5m)\n",
- "phii5 = cmath.phase(complex(I5m.real,I5m.imag))\n",
- " #rms voltage\n",
- "Vrms = ((V1m**2 + V3m**2 + V5m**2)/2)**0.5\n",
- " #rms current\n",
- "Irms = ((I1mag**2 + I3mag**2 + I5mag**2)/2)**0.5\n",
- " #power dissipated\n",
- "P = R*Irms**2\n",
- " #overall power factor\n",
- "pf = P/(Vrms*Irms)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)i = \",round(I1mag,2),\"sin(\",round(w1,2),\"t + (\",round(phii1,2),\")) + \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t + (\",round(phii3,2),\")) + \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t + (\",round(phii5,2),\")) A\"\n",
- "print \"\\n(b)the rms value of current is \",round(Irms,2),\" A\"\n",
- "print \"\\n(c)the rms value of voltage is \",round(Vrms,2),\" V\"\n",
- "print \"\\n(d)the total power dissipated \",round(P,2),\" W\"\n",
- "print \"\\n(e)overall power factor \",round(pf,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)i = 19.4 sin( 314.0 t + ( -0.24 )) + 2.67 sin( 942.0 t + ( -0.64 )) + 1.56 sin( 1570.0 t + ( -0.9 )) A\n",
- "\n",
- "(b)the rms value of current is 13.89 A\n",
- "\n",
- "(c)the rms value of voltage is 173.35 V\n",
- "\n",
- "(d)the total power dissipated 2316.26 W\n",
- "\n",
- "(e)overall power factor 0.96"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 658</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Obtain an expression for the current flowing and hence determine the rms value of current.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vom = 50;# in volts\n",
- "V1m = 200;# in volts\n",
- "V2m = 40;# in volts\n",
- "V4m = 5;# in volts\n",
- "f = 50;# in Hz\n",
- "R = 50;# in ohm\n",
- "C = 100E-6;# in farad\n",
- "phiv1 = 0;# in rad\n",
- "phiv2 = -1*math.pi/2;# in rad\n",
- "phiv4 = math.pi/4;# in rad\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = V1m*math.cos(phiv1) + 1j*V1m*math.sin(phiv1)\n",
- "V2 = V2m*math.cos(phiv2) + 1j*V2m*math.sin(phiv2)\n",
- "V4 = V4m*math.cos(phiv4) + 1j*V4m*math.sin(phiv4)\n",
- " #Inductance has no effect on a steady current. Hence the d.c. component of the current, i0, is given by\n",
- "Iom = 0\n",
- " #fundamental or first harmonic\n",
- "w1 = 2*math.pi*f\n",
- " #inductive reactance,\n",
- "Xc1 = 1/(w1*C)\n",
- " #impedance at the fundamental frequency,\n",
- "Z1 = R + 1j*Xc1\n",
- " #Maximum current at fundamental frequency\n",
- "I1m = V1/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- " #second harmonic\n",
- "Xc2 = Xc1/2\n",
- " #impedance at the third harmonic frequency,\n",
- "Z2 = R + 1j*Xc2\n",
- " #Maximum current at third harmonic frequency\n",
- "I2m = V2/Z2\n",
- "I2mag = abs(I2m)\n",
- "phii2 = cmath.phase(complex(I2m.real,I2m.imag))\n",
- " #fourth harmonic\n",
- "Xc4 = Xc1/4\n",
- " #impedance at the third harmonic frequency,\n",
- "Z4 = R + 1j*Xc4\n",
- " #Maximum current at third harmonic frequency\n",
- "I4m = V4/Z4\n",
- "I4mag = abs(I4m)\n",
- "phii4 = cmath.phase(complex(I4m.real,I4m.imag))\n",
- " #rms current\n",
- "Irms = (Iom**2 + (I1mag**2 + I2mag**2 + I4mag**2)/2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t + (\",round(phii1,2),\")) + \",round(I2mag,2),\"sin(\",round((w1*2),2),\"t + (\",round(phii2,2),\")) + \",round(I4mag,2),\"sin(\",round((w1*4),2),\"t + (\",round(phii4,2),\")) A\"\n",
- "print \"(b)the rms value of current is \",round(Irms,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)i = 0.0 + 3.37 sin( 314.16 t + ( -0.57 )) + 0.76 sin( 628.32 t + ( -1.88 )) + 0.1 sin( 1256.64 t + ( 0.63 )) A\n",
- "(b)the rms value of current is 2.45 A\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 659</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) an expression to represent the current flowing in the circuit, \n",
- "#(b) the rms value of current, correct to two decimal places, and \n",
- "#(c) the power dissipated in the circuit, correct to three significant figures.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vom = 25;# in volts\n",
- "V1m = 100;# in volts\n",
- "V3m = 40;# in volts\n",
- "V5m = 20;# in volts\n",
- "w1 = 10000;# fundamental\n",
- "R = 5;# in ohm\n",
- "L = 500E-6;# in Henry\n",
- "phiv1 = 0;# in rad\n",
- "phiv3 = math.pi/6;# in rad\n",
- "phiv5 = math.pi/12;# in rad\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = V1m*math.cos(phiv1) + 1j*V1m*math.sin(phiv1)\n",
- "V3 = V3m*math.cos(phiv3) + 1j*V3m*math.sin(phiv3)\n",
- "V5 = V5m*math.cos(phiv5) + 1j*V5m*math.sin(phiv5)\n",
- " #Inductance has no effect on a steady current. Hence the d.c. component of the current, i0, is given by\n",
- "Iom = Vom/R\n",
- " #fundamental or first harmonic\n",
- " #inductive reactance,\n",
- "XL1 = w1*L\n",
- " #impedance at the fundamental frequency,\n",
- "Z1 = R + 1j*XL1\n",
- " #Maximum current at fundamental frequency\n",
- "I1m = V1/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- "#Third harmonic\n",
- "XL3 = 3*XL1\n",
- " #impedance at the third harmonic frequency,\n",
- "Z3 = R + 1j*XL3\n",
- " #Maximum current at third harmonic frequency\n",
- "I3m = V3/Z3\n",
- "I3mag = abs(I3m)\n",
- "phii3 = cmath.phase(complex(I3m.real,I3m.imag))\n",
- " #fifth harmonic\n",
- "XL5 = 5*XL1\n",
- " #impedance at the third harmonic frequency,\n",
- "Z5 = R + 1j*XL5\n",
- " #Maximum current at third harmonic frequency\n",
- "I5m = V5/Z5\n",
- "I5mag = abs(I5m)\n",
- "phii5 = cmath.phase(complex(I5m.real,I5m.imag))\n",
- " #rms current\n",
- "Irms = (Iom**2 + (I1mag**2 + I3mag**2 + I5mag**2)/2)**0.5\n",
- " #power dissipated\n",
- "P = R*Irms**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t + (\",round(phii1,2),\")) + \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t + (\",round(phii3,2),\")) + \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t + (\",round(phii5,2),\")) A\"\n",
- "print \"\\n(b)the rms value of current is \",round(Irms,2),\" A\"\n",
- "print \"\\n(c)the total power dissipated \",round(P,3),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)i = 5.0 + 14.14 sin( 10000.0 t + ( -0.79 )) + 2.53 sin( 30000.0 t + ( -0.73 )) + 0.78 sin( 50000.0 t + ( -1.11 )) A\n",
- "\n",
- "(b)the rms value of current is 11.34 A\n",
- "\n",
- "(c)the total power dissipated 642.538 W\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 661</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the average power supplied, (b) the type of components present, and (c) the values of the components.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vom = 30;# in volts\n",
- "V1m = 40;# in volts\n",
- "V2m = 25;# in volts\n",
- "V4m = 15;# in volts\n",
- "Iom = 0;# in amperes\n",
- "I1m = 0.743;# in Amperes\n",
- "I2m = 0.781;# in Amperes\n",
- "I4m = 0.636;# in Amperes\n",
- "phii1 = 1.190;# in rad\n",
- "phii2 = 0.896;# in rad\n",
- "phii4 = 0.559;# in rad\n",
- "w = 1000;# in rad\n",
- "\n",
- " #calculation:\n",
- " #the average power P is given by\n",
- "P = Vom*Iom+(0.707*V1m)*(0.707*I1m)*math.cos(phii1)+(0.707*V2m)*(0.707*I2m)*math.cos(phii2) + (0.707*V4m)*(0.707*I4m)*math.cos(phii4)\n",
- " #rms current\n",
- "Irms = (Iom**2 + (I1m**2 + I2m**2 + I4m**2)/2)**0.5\n",
- " #resistance R\n",
- "R = P/(Irms**2)\n",
- " #impedance\n",
- "Z1 = V1m/I1m\n",
- " #Xc1\n",
- "Xc1 = (Z1**2 - R**2)**0.5\n",
- " #capacitance\n",
- "C = 1/(w*Xc1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)the average power P is \",round(P,2),\" W\"\n",
- "print \"\\n(c)the resistance R \",round(R,2),\" ohm and capacitance \",round(C*1E6,2),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)the average power P is 15.66 W\n",
- "\n",
- "(c)the resistance R 19.99 ohm and capacitance 20.01 uF\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 662</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) an expression for the supply current, i, (b) the percentage harmonic content of the supply current, (c) the total power dissipated, (d) an expression for the p.d. shown as \t1, and (e) an expression for current ic.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1m = 300;# in volts\n",
- "V3m = 120;# in volts\n",
- "phiv1 = 0;# in rad\n",
- "phiv2 = 0.698;# in rad\n",
- "w1 = 314;# in rad\n",
- "C = 2.123E-6;# in farads\n",
- "R1 = 560;# in ohms\n",
- "R2 = 2000;# in Ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = V1m*math.cos(phiv1) + 1j*V1m*math.sin(phiv1)\n",
- "V3 = V3m*math.cos(phiv2) + 1j*V3m*math.sin(phiv2)\n",
- " #capacitive reactance,\n",
- "Xc1 = 1/(w1*C)\n",
- " #impedance at the fundamental frequency,\n",
- "Z1 = R1 - 1j*Xc1*R2/(R2 - 1j*Xc1)\n",
- " #Maximum current at fundamental frequency\n",
- "I1m = V1/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- " #Third harmonic\n",
- "Xc3 = Xc1/3\n",
- " #impedance at the third harmonic frequency,\n",
- "Z3 = R1 - 1j*Xc3*R2/(R2 - 1j*Xc3)\n",
- "I1m = V1m/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- " #Maximum current at third harmonic frequency\n",
- "I3m = V3/Z3\n",
- "I3mag = abs(I3m)\n",
- "phii3 = cmath.phase(complex(I3m.real,I3m.imag))\n",
- " #Percentage harmonic content of the supply current is given by\n",
- "percent = I3mag*100/I1mag\n",
- " #total active power\n",
- "P = (0.707*V1m)*(0.707*I1mag)*math.cos(phiv1 - phii1) + (0.707*V3m)*(0.707*I3m)*math.cos(phiv2 - phii3)\n",
- "\n",
- "I1 = I1m*R2/(R2 - 1j*Xc1)\n",
- "I3 = I3m*R2/(R2 - 1j*Xc3)\n",
- "\n",
- "I1nmag = abs(I1)\n",
- "phini1 = cmath.phase(complex(I1.real,I1.imag))\n",
- "I3nmag = abs(I3)\n",
- "phini3 = cmath.phase(complex(I3.real,I3.imag))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)supply current, i=\", round(I1mag,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
- "print \"\\n(b)Percentage harmonic content of the supply current is \",round(percent,2),\" percent\"\n",
- "print \"\\n(c)total active power is \",round(abs(P),2),\" W\"\n",
- "print \"\\n(d)Voltage, v1 =\", round(I1mag*R1,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag*R1,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
- "print \"\\n(e)current, ic =\", round(I1nmag,3),\"sin(\", w1,\"t +\",round(phini1,3),\") + \",round(I3nmag,3),\"sin(\", 3*w1,\"t +\",round(phini3,3),\") A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)supply current, i= 0.187 sin( 314 t + 0.643 ) + 0.145 sin( 942 t + 1.305 ) A\n",
- "\n",
- "(b)Percentage harmonic content of the supply current is 77.57 percent\n",
- "\n",
- "(c)total active power is 25.34 W\n",
- "\n",
- "(d)Voltage, v1 = 104.996 sin( 314 t + 0.643 ) + 81.45 sin( 942 t + 1.305 ) A\n",
- "\n",
- "(e)current, ic = 0.15 sin( 314 t + 1.287 ) + 0.141 sin( 942 t + 1.55 ) A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 664</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the fundamental frequency for resonance with the third harmonic, and \n",
- "#(b) the maximum value of the fundamental and third harmonic components of current\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1m = 400;# in volts\n",
- "V3m = 10;# in volts\n",
- "C = 0.2E-6;# in farads\n",
- "R = 2;# in ohms\n",
- "L = 0.5;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #Resonance with the third harmonic means that\n",
- "w = (1/(9*L*C))**0.5\n",
- " #fundamental frequency, f\n",
- "f = w/(2*math.pi)\n",
- " #At the fundamental frequency,\n",
- " #impedance Z1\n",
- "Z1 = R + 1j*(w*L - 1/(w*C))\n",
- "Z1mag = abs(Z1)\n",
- "phiZ1 = cmath.phase(complex(Z1.real,Z1.imag))\n",
- " #Maximum value of current at the fundamental frequency,\n",
- "I1m = V1m/Z1mag\n",
- " #At the third harmonic frequency,\n",
- "Z3 = R + 1j*(3*w*L - 1/(3*w*C))\n",
- "Z3mag = abs(Z3)\n",
- "phiZ3 = cmath.phase(complex(Z3.real,Z3.imag))\n",
- " #Maximum value of current at the third harmonic frequency,\n",
- "I3m = V3m/Z3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)fundamental frequency for resonance with the third harmonic is \",round(f,2),\" Hz\"\n",
- "print \"(b)Maximum value of current at fundamental freq. is\",round(abs(I1m),3),\"A \"\n",
- "print \"and at the third harmonic frequency \", abs(I3m),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)fundamental frequency for resonance with the third harmonic is 167.76 Hz\n",
- "(b)Maximum value of current at fundamental freq. is 0.095 A \n",
- "and at the third harmonic frequency 5.0 A\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 665</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of n, (b) the maximum value of current at the nth harmonic, \n",
- "#(c) the p.d. across the capacitor at the nth harmonic and\n",
- "#(d) the maximum value of the fundamental current.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1m = 800;# in volts\n",
- "f = 50;# in Hz\n",
- "x = 0.015;\n",
- "C = 0.122E-6;# in farads\n",
- "R = 5;# in ohms\n",
- "L = 0.369;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #voltage at nth harmonic\n",
- "Vnm = x*V1m\n",
- "w = 2*math.pi*f\n",
- " #For resonance at the nth harmonic nwL = 1/nwC\n",
- "n = 1/(w*(L*C)**0.5)\n",
- " #At resonance, impedance\n",
- "Zn = R\n",
- " #the maximum value of current at the nth harmonic\n",
- "Inm = Vnm/Zn\n",
- " #capacitive reactance, at nth harmonic\n",
- "Xcn = 1/(n*w*C)\n",
- " #the p.d. across the capacitor at the nth harmonic\n",
- "Vcn = Inm*Xcn\n",
- " #At the fundamental frequency, inductive reactance,\n",
- "XL1 = w*L\n",
- " #capacitive reactance\n",
- "Xc1 = 1/(w*C)\n",
- " #Impedance at the fundamental frequency,\n",
- "Z1 = R + 1j*(XL1 - Xc1)\n",
- "Z1mag = abs(Z1)\n",
- "phiZ1 = cmath.phase(complex(Z1.real,Z1.imag))\n",
- " #Maximum value of current at the fundamental frequency,\n",
- "I1m = V1m/Z1mag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)n = \",round(n,2),\"\"\n",
- "print \"\\n(b)the maximum value of current at the nth harmonic \",round(Inm,2),\" A\"\n",
- "print \"\\n(c)the p.d. across the capacitor at the nth harmonic is \",round(Vcn,2),\"\"\n",
- "print \"\\n(d)the maximum value of the fundamental current. \",round(I1m,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)n = 15.0 \n",
- "\n",
- "(b)the maximum value of current at the nth harmonic 2.4 A\n",
- "\n",
- "(c)the p.d. across the capacitor at the nth harmonic is 4173.92 \n",
- "\n",
- "(d)the maximum value of the fundamental current. 0.03 A"
- ]
- }
- ],
- "prompt_number": 14
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_2.ipynb deleted file mode 100755 index 39a16efe..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_2.ipynb +++ /dev/null @@ -1,1017 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 36: Complex Waveforms</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 643</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Write down an expression to represent voltage v.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 240; # in Volts\n",
- "f = 50; # in Hz\n",
- "x = 0.2;\n",
- "phi3 = 3*math.pi/4; # in Rad\n",
- "\n",
- " #calculation:\n",
- "Vamp = V*2**0.5\n",
- "w = 2*math.pi*f\n",
- "T = 1/f\n",
- "V3 = Vamp*x\n",
- "f3 = 3*f\n",
- "w3 = 3*w\n",
- "\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage, V =\",round(Vamp,1),\"sin(\",round(w,1),\"t) + \",round(V3,1),\"sin(\", round(w3,1),\"t - \", round(phi3,1),\") volts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage, V = 339.4 sin( 314.2 t) + 67.9 sin( 942.5 t - 2.4 ) volts"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 648</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the rms value of the current waveform\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "A1 = 0.100;# in amperes\n",
- "A3 = 0.020;# in amperes\n",
- "A5 = 0.010;# in amperes\n",
- "\n",
- " #calculation:\n",
- " #the rms value of current is given by\n",
- "Irms = ((A1**2 + A3**2 + A5**2)/2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the rms value of current is \",round(Irms*1000,2),\" mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the rms value of current is 72.46 mA"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 649</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the voltage, (a) the rms value, (b) the mean value and (c) the form factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "A1 = 10;# in volts\n",
- "A3 = 3;# in volts\n",
- "A5 = 2;# in volts\n",
- "\n",
- "#calculation:\n",
- " #the rms value of voltage is given by\n",
- "Vrms = ((A1**2 + A3**2 + A5**2)/2)**0.5\n",
- " #the mean value of voltage is given by\n",
- " #x = wt\n",
- "Vav = (1/math.pi)*((10 + 1 + 2/5)-(-10 - 1 - 2/5))\n",
- " #form factor is given by\n",
- "ff = Vrms/Vav\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the rms value of voltage is \",round(Vrms,2),\" V\"\n",
- "print \"\\n (b)the mean value of voltage is \",round(Vav,2),\" V\"\n",
- "print \"\\n (c)form factor is \",round(ff,3),\" \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the rms value of voltage is 7.52 V\n",
- "\n",
- " (b)the mean value of voltage is 7.26 V\n",
- "\n",
- " (c)form factor is 1.036 "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 649</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the rms value of the fundamental and each harmonic. \n",
- "#(b) Write down an expression to represent the complex voltage waveform if the frequency of the fundamental is 31.83 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 240;# in volts\n",
- "x = 0.3;# for third harmonic\n",
- "y = 0.1;# for fifth harmonic\n",
- "f = 31.83;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #V3 = x*V1\n",
- " #V5 = y*V1\n",
- " #the rms value of the fundamental,\n",
- "V1 = ((V**2)/(1 + x**2 + y**2))**0.5\n",
- " #Rms value of the third harmonic\n",
- "V3 = x*V1\n",
- " #the rms value of the fifth harmonic,\n",
- "V5 = y*V1\n",
- " #Maximum value of the fundamental,\n",
- "V1m = V1*2**0.5\n",
- " #Maximum value of the third harmonic,\n",
- "V3m = V3*2**0.5\n",
- " #Maximum value of the fifth harmonic,\n",
- "V5m = V5*2**0.5\n",
- "w = 2*math.pi*f\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"v = \",round(V1m,2),\"sin\",round(w,2),\"t + \",round(V3m,2),\"sin\",round((3*w),2),\"t + \",round(V5m,2),\"sin\",round((5*w),2),\"t Volts\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "v = 323.62 sin 199.99 t + 97.08 sin 599.98 t + 32.36 sin 999.97 t Volts\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 652</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the average power in a 20 ohm resistance\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "A1 = 12;# in amperes\n",
- "A3 = 5;# in amperes\n",
- "A5 = 2;# in amperes\n",
- "R = 20;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #rms current\n",
- "Irms = ((A1**2 + A3**2 + A5**2)/2)**0.5\n",
- " #average power\n",
- "P = R*Irms**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n average power \",P,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " average power 1730.0 W"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 652</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the total active power supplied to the circuit, and (b) the overall power factor\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ia1 = 2;# in amperes\n",
- "Ia3 = 0.3;# in amperes\n",
- "Ia5 = 0.1;# in amperes\n",
- "Va1 = 60;# in volts\n",
- "Va3 = 15;# in volts\n",
- "Va5 = 10;# in volts\n",
- "Phii1 = -1*math.pi/6;# in radians\n",
- "Phii3 = -1*math.pi/12;# in radians\n",
- "Phii5 = -8*math.pi/9;# in radians\n",
- "Phiv1 = 0;# in radians\n",
- "Phiv3 = math.pi/4;# in radians\n",
- "Phiv5 = -1*math.pi/2;# in radians\n",
- "\n",
- "\n",
- " #calculation:\n",
- " #rms values;\n",
- "I1 = Ia1/(2**0.5);# in amperes\n",
- "I3 = Ia3/(2**0.5);# in amperes\n",
- "I5 = Ia5/(2**0.5);# in amperes\n",
- "V1 = Va1/(2**0.5);# in volts\n",
- "V3 = Va3/(2**0.5);# in volts\n",
- "V5 = Va5/(2**0.5);# in volts\n",
- " #total power supplied,\n",
- "P = V1*I1*math.cos(Phiv1 - Phii1) + V3*I3*math.cos(Phiv3 - Phii3) + V5*I5*math.cos(Phiv5 - Phii5)\n",
- " #rms current\n",
- "Irms = ((I1**2 + I3**2 + I5**2))**0.5\n",
- " #rms voltage\n",
- "Vrms = ((V1**2 + V3**2 + V5**2))**0.5\n",
- " #overall power factor\n",
- "pf = P/(Vrms*Irms)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)the total active power supplied to the circuit \",round(P,2),\" W\"\n",
- "print \"\\n(b)overall power factor \",round(pf,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)the total active power supplied to the circuit 53.26 W\n",
- "\n",
- "(b)overall power factor 0.84"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 655</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for each case an expression for the current flowing if the fundamental frequency is 1 kHz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 40;# in ohm\n",
- "L = 7.96E-3;# in Henry\n",
- "C = 25E-6; # in Farad\n",
- "f = 1000; # in Hx\n",
- "\n",
- "#calculation:\n",
- "wL = 2*math.pi*1000*L\n",
- "wC = 2*math.pi*1000*C\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)i = \",round(100/R1,2),\"sin(wt) +\",round(30/R1,2),\"sin(3wt - pi/3) +\",round(10/R1,2),\"sin(5wt - pi/6) A\"\n",
- "print \"(b)i = \",round(100/wL,2),\"sin(wt - pi/2) +\",round(30/(3*wL),2),\"sin(3wt - pi/6) +\",round(10/(5*wL),2),\"sin(5wt - 2pi/3) A\"\n",
- "print \"(c)i = \",round(100*wC,2),\"sin(wt + pi/2) +\",round(30*3*wC,2),\"sin(3wt + 5pi/6) +\",round(10*5*wC,2),\"sin(5wt + pi/3) A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)i = 2.5 sin(wt) + 0.75 sin(3wt - pi/3) + 0.25 sin(5wt - pi/6) A\n",
- "(b)i = 2.0 sin(wt - pi/2) + 0.2 sin(3wt - pi/6) + 0.04 sin(5wt - 2pi/3) A\n",
- "(c)i = 15.71 sin(wt + pi/2) + 14.14 sin(3wt + 5pi/6) + 7.85 sin(5wt + pi/3) A\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 656</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) an expression to represent the instantaneous value of the current, \n",
- "#(b) the rms voltage, (c) the rms current, (d) the power dissipated, and (e) the overall power factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1m = 240;# in volts\n",
- "V3m = 40;# in volts\n",
- "V5m = 30;# in volts\n",
- "w1 = 314;# fundamental\n",
- "R = 12;# in ohm\n",
- "L = 0.00955;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #fundamental or first harmonic\n",
- " #inductive reactance,\n",
- "XL1 = w1*L\n",
- " #impedance at the fundamental frequency,\n",
- "Z1 = R + 1j*XL1\n",
- " #Maximum current at fundamental frequency\n",
- "I1m = V1m/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- " #Third harmonic\n",
- "XL3 = 3*XL1\n",
- " #impedance at the third harmonic frequency,\n",
- "Z3 = R + 1j*XL3\n",
- " #Maximum current at third harmonic frequency\n",
- "I3m = V3m/Z3\n",
- "I3mag = abs(I3m)\n",
- "phii3 = cmath.phase(complex(I3m.real,I3m.imag))\n",
- " #fifth harmonic\n",
- "XL5 = 5*XL1\n",
- " #impedance at the third harmonic frequency,\n",
- "Z5 = R + 1j*XL5\n",
- " #Maximum current at third harmonic frequency\n",
- "I5m = V5m/Z5\n",
- "I5mag = abs(I5m)\n",
- "phii5 = cmath.phase(complex(I5m.real,I5m.imag))\n",
- " #rms voltage\n",
- "Vrms = ((V1m**2 + V3m**2 + V5m**2)/2)**0.5\n",
- " #rms current\n",
- "Irms = ((I1mag**2 + I3mag**2 + I5mag**2)/2)**0.5\n",
- " #power dissipated\n",
- "P = R*Irms**2\n",
- " #overall power factor\n",
- "pf = P/(Vrms*Irms)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)i = \",round(I1mag,2),\"sin(\",round(w1,2),\"t + (\",round(phii1,2),\")) + \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t + (\",round(phii3,2),\")) + \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t + (\",round(phii5,2),\")) A\"\n",
- "print \"\\n(b)the rms value of current is \",round(Irms,2),\" A\"\n",
- "print \"\\n(c)the rms value of voltage is \",round(Vrms,2),\" V\"\n",
- "print \"\\n(d)the total power dissipated \",round(P,2),\" W\"\n",
- "print \"\\n(e)overall power factor \",round(pf,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)i = 19.4 sin( 314.0 t + ( -0.24 )) + 2.67 sin( 942.0 t + ( -0.64 )) + 1.56 sin( 1570.0 t + ( -0.9 )) A\n",
- "\n",
- "(b)the rms value of current is 13.89 A\n",
- "\n",
- "(c)the rms value of voltage is 173.35 V\n",
- "\n",
- "(d)the total power dissipated 2316.26 W\n",
- "\n",
- "(e)overall power factor 0.96"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 658</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Obtain an expression for the current flowing and hence determine the rms value of current.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vom = 50;# in volts\n",
- "V1m = 200;# in volts\n",
- "V2m = 40;# in volts\n",
- "V4m = 5;# in volts\n",
- "f = 50;# in Hz\n",
- "R = 50;# in ohm\n",
- "C = 100E-6;# in farad\n",
- "phiv1 = 0;# in rad\n",
- "phiv2 = -1*math.pi/2;# in rad\n",
- "phiv4 = math.pi/4;# in rad\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = V1m*math.cos(phiv1) + 1j*V1m*math.sin(phiv1)\n",
- "V2 = V2m*math.cos(phiv2) + 1j*V2m*math.sin(phiv2)\n",
- "V4 = V4m*math.cos(phiv4) + 1j*V4m*math.sin(phiv4)\n",
- " #Inductance has no effect on a steady current. Hence the d.c. component of the current, i0, is given by\n",
- "Iom = 0\n",
- " #fundamental or first harmonic\n",
- "w1 = 2*math.pi*f\n",
- " #inductive reactance,\n",
- "Xc1 = 1/(w1*C)\n",
- " #impedance at the fundamental frequency,\n",
- "Z1 = R + 1j*Xc1\n",
- " #Maximum current at fundamental frequency\n",
- "I1m = V1/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- " #second harmonic\n",
- "Xc2 = Xc1/2\n",
- " #impedance at the third harmonic frequency,\n",
- "Z2 = R + 1j*Xc2\n",
- " #Maximum current at third harmonic frequency\n",
- "I2m = V2/Z2\n",
- "I2mag = abs(I2m)\n",
- "phii2 = cmath.phase(complex(I2m.real,I2m.imag))\n",
- " #fourth harmonic\n",
- "Xc4 = Xc1/4\n",
- " #impedance at the third harmonic frequency,\n",
- "Z4 = R + 1j*Xc4\n",
- " #Maximum current at third harmonic frequency\n",
- "I4m = V4/Z4\n",
- "I4mag = abs(I4m)\n",
- "phii4 = cmath.phase(complex(I4m.real,I4m.imag))\n",
- " #rms current\n",
- "Irms = (Iom**2 + (I1mag**2 + I2mag**2 + I4mag**2)/2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t + (\",round(phii1,2),\")) + \",round(I2mag,2),\"sin(\",round((w1*2),2),\"t + (\",round(phii2,2),\")) + \",round(I4mag,2),\"sin(\",round((w1*4),2),\"t + (\",round(phii4,2),\")) A\"\n",
- "print \"(b)the rms value of current is \",round(Irms,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)i = 0.0 + 3.37 sin( 314.16 t + ( -0.57 )) + 0.76 sin( 628.32 t + ( -1.88 )) + 0.1 sin( 1256.64 t + ( 0.63 )) A\n",
- "(b)the rms value of current is 2.45 A\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 659</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) an expression to represent the current flowing in the circuit, \n",
- "#(b) the rms value of current, correct to two decimal places, and \n",
- "#(c) the power dissipated in the circuit, correct to three significant figures.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vom = 25;# in volts\n",
- "V1m = 100;# in volts\n",
- "V3m = 40;# in volts\n",
- "V5m = 20;# in volts\n",
- "w1 = 10000;# fundamental\n",
- "R = 5;# in ohm\n",
- "L = 500E-6;# in Henry\n",
- "phiv1 = 0;# in rad\n",
- "phiv3 = math.pi/6;# in rad\n",
- "phiv5 = math.pi/12;# in rad\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "V1 = V1m*math.cos(phiv1) + 1j*V1m*math.sin(phiv1)\n",
- "V3 = V3m*math.cos(phiv3) + 1j*V3m*math.sin(phiv3)\n",
- "V5 = V5m*math.cos(phiv5) + 1j*V5m*math.sin(phiv5)\n",
- " #Inductance has no effect on a steady current. Hence the d.c. component of the current, i0, is given by\n",
- "Iom = Vom/R\n",
- " #fundamental or first harmonic\n",
- " #inductive reactance,\n",
- "XL1 = w1*L\n",
- " #impedance at the fundamental frequency,\n",
- "Z1 = R + 1j*XL1\n",
- " #Maximum current at fundamental frequency\n",
- "I1m = V1/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- "#Third harmonic\n",
- "XL3 = 3*XL1\n",
- " #impedance at the third harmonic frequency,\n",
- "Z3 = R + 1j*XL3\n",
- " #Maximum current at third harmonic frequency\n",
- "I3m = V3/Z3\n",
- "I3mag = abs(I3m)\n",
- "phii3 = cmath.phase(complex(I3m.real,I3m.imag))\n",
- " #fifth harmonic\n",
- "XL5 = 5*XL1\n",
- " #impedance at the third harmonic frequency,\n",
- "Z5 = R + 1j*XL5\n",
- " #Maximum current at third harmonic frequency\n",
- "I5m = V5/Z5\n",
- "I5mag = abs(I5m)\n",
- "phii5 = cmath.phase(complex(I5m.real,I5m.imag))\n",
- " #rms current\n",
- "Irms = (Iom**2 + (I1mag**2 + I3mag**2 + I5mag**2)/2)**0.5\n",
- " #power dissipated\n",
- "P = R*Irms**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t + (\",round(phii1,2),\")) + \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t + (\",round(phii3,2),\")) + \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t + (\",round(phii5,2),\")) A\"\n",
- "print \"\\n(b)the rms value of current is \",round(Irms,2),\" A\"\n",
- "print \"\\n(c)the total power dissipated \",round(P,3),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)i = 5.0 + 14.14 sin( 10000.0 t + ( -0.79 )) + 2.53 sin( 30000.0 t + ( -0.73 )) + 0.78 sin( 50000.0 t + ( -1.11 )) A\n",
- "\n",
- "(b)the rms value of current is 11.34 A\n",
- "\n",
- "(c)the total power dissipated 642.538 W\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 661</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the average power supplied, (b) the type of components present, and (c) the values of the components.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vom = 30;# in volts\n",
- "V1m = 40;# in volts\n",
- "V2m = 25;# in volts\n",
- "V4m = 15;# in volts\n",
- "Iom = 0;# in amperes\n",
- "I1m = 0.743;# in Amperes\n",
- "I2m = 0.781;# in Amperes\n",
- "I4m = 0.636;# in Amperes\n",
- "phii1 = 1.190;# in rad\n",
- "phii2 = 0.896;# in rad\n",
- "phii4 = 0.559;# in rad\n",
- "w = 1000;# in rad\n",
- "\n",
- " #calculation:\n",
- " #the average power P is given by\n",
- "P = Vom*Iom+(0.707*V1m)*(0.707*I1m)*math.cos(phii1)+(0.707*V2m)*(0.707*I2m)*math.cos(phii2) + (0.707*V4m)*(0.707*I4m)*math.cos(phii4)\n",
- " #rms current\n",
- "Irms = (Iom**2 + (I1m**2 + I2m**2 + I4m**2)/2)**0.5\n",
- " #resistance R\n",
- "R = P/(Irms**2)\n",
- " #impedance\n",
- "Z1 = V1m/I1m\n",
- " #Xc1\n",
- "Xc1 = (Z1**2 - R**2)**0.5\n",
- " #capacitance\n",
- "C = 1/(w*Xc1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)the average power P is \",round(P,2),\" W\"\n",
- "print \"\\n(c)the resistance R \",round(R,2),\" ohm and capacitance \",round(C*1E6,2),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)the average power P is 15.66 W\n",
- "\n",
- "(c)the resistance R 19.99 ohm and capacitance 20.01 uF\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 662</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) an expression for the supply current, i, (b) the percentage harmonic content of the supply current, (c) the total power dissipated, (d) an expression for the p.d. shown as \t1, and (e) an expression for current ic.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1m = 300;# in volts\n",
- "V3m = 120;# in volts\n",
- "phiv1 = 0;# in rad\n",
- "phiv2 = 0.698;# in rad\n",
- "w1 = 314;# in rad\n",
- "C = 2.123E-6;# in farads\n",
- "R1 = 560;# in ohms\n",
- "R2 = 2000;# in Ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = V1m*math.cos(phiv1) + 1j*V1m*math.sin(phiv1)\n",
- "V3 = V3m*math.cos(phiv2) + 1j*V3m*math.sin(phiv2)\n",
- " #capacitive reactance,\n",
- "Xc1 = 1/(w1*C)\n",
- " #impedance at the fundamental frequency,\n",
- "Z1 = R1 - 1j*Xc1*R2/(R2 - 1j*Xc1)\n",
- " #Maximum current at fundamental frequency\n",
- "I1m = V1/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- " #Third harmonic\n",
- "Xc3 = Xc1/3\n",
- " #impedance at the third harmonic frequency,\n",
- "Z3 = R1 - 1j*Xc3*R2/(R2 - 1j*Xc3)\n",
- "I1m = V1m/Z1\n",
- "I1mag = abs(I1m)\n",
- "phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
- " #Maximum current at third harmonic frequency\n",
- "I3m = V3/Z3\n",
- "I3mag = abs(I3m)\n",
- "phii3 = cmath.phase(complex(I3m.real,I3m.imag))\n",
- " #Percentage harmonic content of the supply current is given by\n",
- "percent = I3mag*100/I1mag\n",
- " #total active power\n",
- "P = (0.707*V1m)*(0.707*I1mag)*math.cos(phiv1 - phii1) + (0.707*V3m)*(0.707*I3m)*math.cos(phiv2 - phii3)\n",
- "\n",
- "I1 = I1m*R2/(R2 - 1j*Xc1)\n",
- "I3 = I3m*R2/(R2 - 1j*Xc3)\n",
- "\n",
- "I1nmag = abs(I1)\n",
- "phini1 = cmath.phase(complex(I1.real,I1.imag))\n",
- "I3nmag = abs(I3)\n",
- "phini3 = cmath.phase(complex(I3.real,I3.imag))\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)supply current, i=\", round(I1mag,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
- "print \"\\n(b)Percentage harmonic content of the supply current is \",round(percent,2),\" percent\"\n",
- "print \"\\n(c)total active power is \",round(abs(P),2),\" W\"\n",
- "print \"\\n(d)Voltage, v1 =\", round(I1mag*R1,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag*R1,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
- "print \"\\n(e)current, ic =\", round(I1nmag,3),\"sin(\", w1,\"t +\",round(phini1,3),\") + \",round(I3nmag,3),\"sin(\", 3*w1,\"t +\",round(phini3,3),\") A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)supply current, i= 0.187 sin( 314 t + 0.643 ) + 0.145 sin( 942 t + 1.305 ) A\n",
- "\n",
- "(b)Percentage harmonic content of the supply current is 77.57 percent\n",
- "\n",
- "(c)total active power is 25.34 W\n",
- "\n",
- "(d)Voltage, v1 = 104.996 sin( 314 t + 0.643 ) + 81.45 sin( 942 t + 1.305 ) A\n",
- "\n",
- "(e)current, ic = 0.15 sin( 314 t + 1.287 ) + 0.141 sin( 942 t + 1.55 ) A"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 664</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the fundamental frequency for resonance with the third harmonic, and \n",
- "#(b) the maximum value of the fundamental and third harmonic components of current\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1m = 400;# in volts\n",
- "V3m = 10;# in volts\n",
- "C = 0.2E-6;# in farads\n",
- "R = 2;# in ohms\n",
- "L = 0.5;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #Resonance with the third harmonic means that\n",
- "w = (1/(9*L*C))**0.5\n",
- " #fundamental frequency, f\n",
- "f = w/(2*math.pi)\n",
- " #At the fundamental frequency,\n",
- " #impedance Z1\n",
- "Z1 = R + 1j*(w*L - 1/(w*C))\n",
- "Z1mag = abs(Z1)\n",
- "phiZ1 = cmath.phase(complex(Z1.real,Z1.imag))\n",
- " #Maximum value of current at the fundamental frequency,\n",
- "I1m = V1m/Z1mag\n",
- " #At the third harmonic frequency,\n",
- "Z3 = R + 1j*(3*w*L - 1/(3*w*C))\n",
- "Z3mag = abs(Z3)\n",
- "phiZ3 = cmath.phase(complex(Z3.real,Z3.imag))\n",
- " #Maximum value of current at the third harmonic frequency,\n",
- "I3m = V3m/Z3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"(a)fundamental frequency for resonance with the third harmonic is \",round(f,2),\" Hz\"\n",
- "print \"(b)Maximum value of current at fundamental freq. is\",round(abs(I1m),3),\"A \"\n",
- "print \"and at the third harmonic frequency \", abs(I3m),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "(a)fundamental frequency for resonance with the third harmonic is 167.76 Hz\n",
- "(b)Maximum value of current at fundamental freq. is 0.095 A \n",
- "and at the third harmonic frequency 5.0 A\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 665</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of n, (b) the maximum value of current at the nth harmonic, \n",
- "#(c) the p.d. across the capacitor at the nth harmonic and\n",
- "#(d) the maximum value of the fundamental current.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1m = 800;# in volts\n",
- "f = 50;# in Hz\n",
- "x = 0.015;\n",
- "C = 0.122E-6;# in farads\n",
- "R = 5;# in ohms\n",
- "L = 0.369;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #voltage at nth harmonic\n",
- "Vnm = x*V1m\n",
- "w = 2*math.pi*f\n",
- " #For resonance at the nth harmonic nwL = 1/nwC\n",
- "n = 1/(w*(L*C)**0.5)\n",
- " #At resonance, impedance\n",
- "Zn = R\n",
- " #the maximum value of current at the nth harmonic\n",
- "Inm = Vnm/Zn\n",
- " #capacitive reactance, at nth harmonic\n",
- "Xcn = 1/(n*w*C)\n",
- " #the p.d. across the capacitor at the nth harmonic\n",
- "Vcn = Inm*Xcn\n",
- " #At the fundamental frequency, inductive reactance,\n",
- "XL1 = w*L\n",
- " #capacitive reactance\n",
- "Xc1 = 1/(w*C)\n",
- " #Impedance at the fundamental frequency,\n",
- "Z1 = R + 1j*(XL1 - Xc1)\n",
- "Z1mag = abs(Z1)\n",
- "phiZ1 = cmath.phase(complex(Z1.real,Z1.imag))\n",
- " #Maximum value of current at the fundamental frequency,\n",
- "I1m = V1m/Z1mag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)n = \",round(n,2),\"\"\n",
- "print \"\\n(b)the maximum value of current at the nth harmonic \",round(Inm,2),\" A\"\n",
- "print \"\\n(c)the p.d. across the capacitor at the nth harmonic is \",round(Vcn,2),\"\"\n",
- "print \"\\n(d)the maximum value of the fundamental current. \",round(I1m,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)n = 15.0 \n",
- "\n",
- "(b)the maximum value of current at the nth harmonic 2.4 A\n",
- "\n",
- "(c)the p.d. across the capacitor at the nth harmonic is 4173.92 \n",
- "\n",
- "(d)the maximum value of the fundamental current. 0.03 A"
- ]
- }
- ],
- "prompt_number": 14
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_36.ipynb index 3cc6ab7f..3cc6ab7f 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_36.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_37.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_37.ipynb index de8c7108..de8c7108 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/Chapter_37.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_37.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint.ipynb deleted file mode 100755 index c4a1d2a6..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint.ipynb +++ /dev/null @@ -1,631 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 38: Magnetic materials</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 694</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the hysteresis loss per m3 per cycle, and (b) the hysteresis loss per m3 at a frequency of 50 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "A = 12.5;# in cm2\n",
- "x = 500;# horizontal axis 1 cm = 500 A/m\n",
- "y = 0.2;# vertical axis 1 cm = 0.2 T\n",
- "f = 50;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #hysteresis loss per cycle\n",
- "HL = A*x*y\n",
- " #At 50 Hz frequency, hysteresis loss\n",
- "HLf = HL*f\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)hysteresis loss per cycle is = \",HL,\" J/m3\"\n",
- "print \"\\n(b)At 50 Hz frequency, hysteresis loss \",HLf,\" W/m3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)hysteresis loss per cycle is = 1250.0 J/m3\n",
- "\n",
- "(b)At 50 Hz frequency, hysteresis loss 62500.0 W/m3"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 695</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the hysteresis loss per m3 for a maximum flux density of 1.1 T and frequency of 25 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "n = 1.6;# the Steinmetz index\n",
- "f1 = 50;# in Hz\n",
- "f2 = 25;# in Hz\n",
- "Bm1 = 1.5;# in Tesla\n",
- "Bm2 = 1.1;# in Tesla\n",
- "Ph1 = 62500;# in W/m3\n",
- "v = 1;\n",
- "\n",
- "#calculation: \n",
- " #hysteresis loss Ph = kh*v*f*(Bm)**n\n",
- "kh = Ph1/(v*f1*(Bm1)**n)\n",
- " #When f = 25 Hz and Bm = 1.1 T,\n",
- "Ph2 = kh*v*f2*(Bm2)**n\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n hysteresis loss When f = 25 Hz and Bm = 1.1 T, is = \",round(Ph2,2),\" W/m3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " hysteresis loss When f = 25 Hz and Bm = 1.1 T, is = 19025.33 W/m3"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 695</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the hysteresis loss at a frequency of 80 Hz\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "csa = 0.002;# in m2\n",
- "l = 1;# in m\n",
- "a = 400/0.01;# 10 mm = 400 A/m \n",
- "b = 0.1/0.01;# 10 mm = 0.1 T \n",
- "A = 0.01;# in m2\n",
- "f = 80;# in Hz\n",
- "\n",
- "#calculation: \n",
- " #hysteresis loss per cycle\n",
- "HL = A*a*b\n",
- " #At a frequency of 80 Hz,\n",
- " #hysteresis loss\n",
- "HLf = HL*f\n",
- " #Volume of ring\n",
- "v = csa*l\n",
- " #hysteresis loss\n",
- "Ph = HLf*v\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the hysteresis loss at a frequency of 80 Hz is \",Ph,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the hysteresis loss at a frequency of 80 Hz is 640.0 W"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 696</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the hysteresis loss when the maximum core flux is 8 mWb and the frequency is 50 Hz\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Phi1 = 0.01;# in Wb\n",
- "Phi2 = 0.008;# in Wb\n",
- "csa = 0.008;# in m2\n",
- "v = 0.005;# in m3\n",
- "f = 50;# in Hz\n",
- "n = 1.7;# the Steinmetz constant\n",
- "Ph1 = 100;# in Watt\n",
- "\n",
- " #calculation: \n",
- " #maximum flux density\n",
- "Bm1 = Phi1/csa\n",
- " #hysteresis loss Ph1 = kh*v*f*(Bm1)**n\n",
- "kh = Ph1/(v*f*(Bm1)**n)\n",
- " #When the maximum core flux is 8 mWb,\n",
- "Bm2 = Phi2/csa\n",
- " #hysteresis loss, Ph2\n",
- "Ph2 = kh*v*f*(Bm2)**n\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"value of hysteresis loss when maximum core flux is 8 mWb and the frequency is 50 Hz is \",round(Ph2,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "value of hysteresis loss when maximum core flux is 8 mWb and the frequency is 50 Hz is 68.43 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 699</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the new value of eddy current loss per cubic metre.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Pe1 = 10;# in W/m3\n",
- "f1 = 50;# in Hz\n",
- "f2 = 30;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #When the eddy current loss is 10 W/m3, frequency f is 50 Hz.\n",
- " #constant k\n",
- "k = Pe1/(f1**2)\n",
- " #When the frequency is 30 Hz, eddy current loss,\n",
- "Pe2 = k*(f2**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\neddy current loss per cubic metre is \",Pe2,\" W/m3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "eddy current loss per cubic metre is 3.6 W/m3"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 699</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the necessary new thickness of the laminations.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Pe = 100;# in W/m3\n",
- "f1 = 50;# in Hz\n",
- "t1 = 0.0005;# in m\n",
- "x = 1/3;\n",
- "f2 = 250;# in Hz\n",
- "Bm1 = 1;\n",
- " #calculation: \n",
- " #Pe = ke*(Bm1*f1*t1)**2\n",
- " #Hence, at 50 Hz frequency\n",
- "ke = Pe/(Bm1*f1*t1)**2\n",
- " #At 250 Hz frequency\n",
- "Bm2 = x*Bm1\n",
- "t2 = ((Pe/ke)**0.5)/(Bm2*f2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nlamination thickness is \",t2,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "lamination thickness is 0.0003 m"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 700</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the values of the losses if the frequency is increased to 60 Hz. \n",
- "#(b) What will be the total core loss if the frequency is 50 Hz and the lamination are made one-half of their original thickness?\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ph1 = 40;# in W\n",
- "Pe1 = 20;# in W\n",
- "f1 = 50;# in Hz\n",
- "x = 1/2;\n",
- "f2 = 60;# in Hz\n",
- "t1 = 1;\n",
- " #calculation: \n",
- " #hysteresis loss Ph = kh*v*f*(Bm)**n = k1*f\n",
- " #Thus when the hysteresis is 40 W and the frequency 50 Hz,\n",
- "k1 = Ph1/f1\n",
- " #If the frequency is increased to 60 Hz,\n",
- "Ph2 = k1*f2\n",
- " #eddy current loss, Pe = ke*(Bm1*f1*t1)**2 = k2*f**2\n",
- " #since the flux density and lamination thickness are constant.\n",
- " #When the eddy current loss is 20 W the frequency is 50 Hz. Thus\n",
- "k2 = Pe1/(f1**2)\n",
- " #If the frequency is increased to 60 Hz,\n",
- "Pe2 = k2*(f2**2)\n",
- " #hysteresis loss Ph = kh*v*f*(Bm)**n, is independent of the thickness of the laminations. \n",
- " #Thus, if the thickness of the laminations is halved, the hysteresis loss remains at \n",
- "Phb2 = Ph1\n",
- " #eddy current loss, Pe = ke*(Bm1*f1*t1)**2 = k2*t**3\n",
- "k3 = Pe1/(t1**3)\n",
- "t2 = 0.5*t1\n",
- "Peb2 = k3*t2**3\n",
- " #total core loss when the thickness of the laminations is halved is given by\n",
- "TL = Phb2 + Peb2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)If the frequency is increased to 60 Hz,hysteresis loss is \",Ph2,\" W and eddy current loss \", Pe2,\" W\"\n",
- "print \"\\n(b)the total core loss when the thickness of the laminations is halved \",TL,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)If the frequency is increased to 60 Hz,hysteresis loss is 48.0 W and eddy current loss 28.8 W\n",
- "\n",
- "(b)the total core loss when the thickness of the laminations is halved 42.5 W\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 701</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the new total core loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 500;# in Volts\n",
- "V2 = 1000;# in Volts\n",
- "Ph1 = 400;# in W\n",
- "Pe1 = 150;# in W\n",
- "f1 = 50;# in Hz\n",
- "n = 1.6;# Steinmetz index\n",
- "f2 = 100;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #hysteresis loss Ph = k1*f*(E/f)**n\n",
- " #At 500 V and 50 Hz\n",
- "k1 = Ph1/(f1*(V1/f1)**1.6)\n",
- " #At 1000 V and 100 Hz,\n",
- "Ph2 = k1*f2*(V2/f2)**1.6\n",
- " #eddy current loss, Pe = k2*E**2\n",
- " #At 500 V,\n",
- "k2 = Pe1/(V1**2)\n",
- " #At 1000 V,\n",
- "Pe2 = k2*(V2**2)\n",
- " #the new total core loss\n",
- "TL = Ph2 + Pe2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the new total core loss \",TL,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the new total core loss 1400.0 W"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 702</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the separate values of the hysteresis and eddy current losses at frequencies of (a) 50 Hz and (b) 60 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#from pylab import *\n",
- "%pylab inline\n",
- "#initializing the variables:\n",
- "f1 = 50;# in Hz\n",
- "f2 = 60;# in Hz\n",
- "\n",
- "#calculation:\n",
- "k1 = 0.5\n",
- "k2 = 0.032\n",
- "f = [30, 50, 70,90]\n",
- "Pcf = [1.5, 2.1, 2.7, 3.4]\n",
- "fig = plt.figure()\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(f,Pcf,'-')\n",
- "xlabel('f')\n",
- "ylabel('Pc/f')\n",
- "show()\n",
- "HL1 = k1*f1\n",
- "ECL1 = k2*f1**2\n",
- "\n",
- "HL2 = k1*f2\n",
- "ECL2 = k2*f2**2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) at Frequency = 50 Hz, hysteresis loss is\", HL1,\" W and eddy current loss is\", ECL1,\" W\"\n",
- "print \"\\n (b) at Frequency = 60 Hz, hysteresis loss is\", HL2,\" W and eddy current loss is\", ECL2,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
- "For more information, type 'help(pylab)'."
- ]
- },
- {
- "output_type": "display_data",
- "png": 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- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) at Frequency = 50 Hz, hysteresis loss is 25.0 W and eddy current loss is 80.0 W\n",
- "\n",
- " (b) at Frequency = 60 Hz, hysteresis loss is 30.0 W and eddy current loss is 115.2 W"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 703</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the hysteresis and eddy current losses at 50 Hz (b) determine the hysteresis and eddy current losses at 50 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "TL1 = 400;# in Watt\n",
- "TL2 = 498;# in Watt\n",
- "x = 0.25;\n",
- "y = 0.4;\n",
- "f1 = 50;# in Hz\n",
- "n = 1.7;# Steinmetz index\n",
- "f2 = 60;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #if volume v and the maximum flux density are constant\n",
- " #hysteresis loss Ph = kh*v*f*(Bm)**n = k1*f\n",
- " #(if the maximum flux density and the lamination thickness are constant)\n",
- " #eddy current loss, Pe = ke*(Bm1*f1*t1)**2 = k2*f**2\n",
- " #At 50 Hz frequency, TL1 = k1*f1 + k2*f1**2\n",
- " #At 60 Hz frequency, TL2 = k1*f2 + k2*f2**2\n",
- " #Solving equations gives the values of k1 and k2.\n",
- "k2 = (5*TL2 - 6*TL1)/(5*(f2**2) - 6*(f1**2))\n",
- "k1 = (TL1 - k2*f1**2)/f1\n",
- " #hysteresis loss Ph = k1*f\n",
- "Ph1 = k1*f1\n",
- " #eddy current loss\n",
- "Pe1 = k2*f1**2\n",
- " #Since at 50 Hz the flux density is increased by 25%, the new hysteresis loss is\n",
- "Ph2 = Ph1*(1 + x)**1.7\n",
- " #Since at 50 Hz the flux density is increased by 25%, and the lamination thickness is increased by 40%, \n",
- " #the new eddy current loss is\n",
- "Pe2 = Pe1*((1 + x)**2)*(1 + y)**3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the hysteresis and eddy current losses at 50 Hz are \",round(Ph1,2),\" W and \",round( Pe1,2),\" W resp.\"\n",
- "print \"\\n (b)the hysteresis and eddy current losses at 50 Hz after increement are \",round(Ph2,2),\" W and \",round( Pe2,2),\" W resp.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the hysteresis and eddy current losses at 50 Hz are 325.0 W and 75.0 W resp.\n",
- "\n",
- " (b)the hysteresis and eddy current losses at 50 Hz after increement are 474.93 W and 321.56 W resp.\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_1.ipynb deleted file mode 100755 index c4a1d2a6..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_1.ipynb +++ /dev/null @@ -1,631 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 38: Magnetic materials</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 694</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the hysteresis loss per m3 per cycle, and (b) the hysteresis loss per m3 at a frequency of 50 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "A = 12.5;# in cm2\n",
- "x = 500;# horizontal axis 1 cm = 500 A/m\n",
- "y = 0.2;# vertical axis 1 cm = 0.2 T\n",
- "f = 50;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #hysteresis loss per cycle\n",
- "HL = A*x*y\n",
- " #At 50 Hz frequency, hysteresis loss\n",
- "HLf = HL*f\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)hysteresis loss per cycle is = \",HL,\" J/m3\"\n",
- "print \"\\n(b)At 50 Hz frequency, hysteresis loss \",HLf,\" W/m3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)hysteresis loss per cycle is = 1250.0 J/m3\n",
- "\n",
- "(b)At 50 Hz frequency, hysteresis loss 62500.0 W/m3"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 695</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the hysteresis loss per m3 for a maximum flux density of 1.1 T and frequency of 25 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "n = 1.6;# the Steinmetz index\n",
- "f1 = 50;# in Hz\n",
- "f2 = 25;# in Hz\n",
- "Bm1 = 1.5;# in Tesla\n",
- "Bm2 = 1.1;# in Tesla\n",
- "Ph1 = 62500;# in W/m3\n",
- "v = 1;\n",
- "\n",
- "#calculation: \n",
- " #hysteresis loss Ph = kh*v*f*(Bm)**n\n",
- "kh = Ph1/(v*f1*(Bm1)**n)\n",
- " #When f = 25 Hz and Bm = 1.1 T,\n",
- "Ph2 = kh*v*f2*(Bm2)**n\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n hysteresis loss When f = 25 Hz and Bm = 1.1 T, is = \",round(Ph2,2),\" W/m3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " hysteresis loss When f = 25 Hz and Bm = 1.1 T, is = 19025.33 W/m3"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 695</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the hysteresis loss at a frequency of 80 Hz\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "csa = 0.002;# in m2\n",
- "l = 1;# in m\n",
- "a = 400/0.01;# 10 mm = 400 A/m \n",
- "b = 0.1/0.01;# 10 mm = 0.1 T \n",
- "A = 0.01;# in m2\n",
- "f = 80;# in Hz\n",
- "\n",
- "#calculation: \n",
- " #hysteresis loss per cycle\n",
- "HL = A*a*b\n",
- " #At a frequency of 80 Hz,\n",
- " #hysteresis loss\n",
- "HLf = HL*f\n",
- " #Volume of ring\n",
- "v = csa*l\n",
- " #hysteresis loss\n",
- "Ph = HLf*v\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the hysteresis loss at a frequency of 80 Hz is \",Ph,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the hysteresis loss at a frequency of 80 Hz is 640.0 W"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 696</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the hysteresis loss when the maximum core flux is 8 mWb and the frequency is 50 Hz\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Phi1 = 0.01;# in Wb\n",
- "Phi2 = 0.008;# in Wb\n",
- "csa = 0.008;# in m2\n",
- "v = 0.005;# in m3\n",
- "f = 50;# in Hz\n",
- "n = 1.7;# the Steinmetz constant\n",
- "Ph1 = 100;# in Watt\n",
- "\n",
- " #calculation: \n",
- " #maximum flux density\n",
- "Bm1 = Phi1/csa\n",
- " #hysteresis loss Ph1 = kh*v*f*(Bm1)**n\n",
- "kh = Ph1/(v*f*(Bm1)**n)\n",
- " #When the maximum core flux is 8 mWb,\n",
- "Bm2 = Phi2/csa\n",
- " #hysteresis loss, Ph2\n",
- "Ph2 = kh*v*f*(Bm2)**n\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"value of hysteresis loss when maximum core flux is 8 mWb and the frequency is 50 Hz is \",round(Ph2,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "value of hysteresis loss when maximum core flux is 8 mWb and the frequency is 50 Hz is 68.43 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 699</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the new value of eddy current loss per cubic metre.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Pe1 = 10;# in W/m3\n",
- "f1 = 50;# in Hz\n",
- "f2 = 30;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #When the eddy current loss is 10 W/m3, frequency f is 50 Hz.\n",
- " #constant k\n",
- "k = Pe1/(f1**2)\n",
- " #When the frequency is 30 Hz, eddy current loss,\n",
- "Pe2 = k*(f2**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\neddy current loss per cubic metre is \",Pe2,\" W/m3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "eddy current loss per cubic metre is 3.6 W/m3"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 699</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the necessary new thickness of the laminations.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Pe = 100;# in W/m3\n",
- "f1 = 50;# in Hz\n",
- "t1 = 0.0005;# in m\n",
- "x = 1/3;\n",
- "f2 = 250;# in Hz\n",
- "Bm1 = 1;\n",
- " #calculation: \n",
- " #Pe = ke*(Bm1*f1*t1)**2\n",
- " #Hence, at 50 Hz frequency\n",
- "ke = Pe/(Bm1*f1*t1)**2\n",
- " #At 250 Hz frequency\n",
- "Bm2 = x*Bm1\n",
- "t2 = ((Pe/ke)**0.5)/(Bm2*f2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nlamination thickness is \",t2,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "lamination thickness is 0.0003 m"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 700</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the values of the losses if the frequency is increased to 60 Hz. \n",
- "#(b) What will be the total core loss if the frequency is 50 Hz and the lamination are made one-half of their original thickness?\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ph1 = 40;# in W\n",
- "Pe1 = 20;# in W\n",
- "f1 = 50;# in Hz\n",
- "x = 1/2;\n",
- "f2 = 60;# in Hz\n",
- "t1 = 1;\n",
- " #calculation: \n",
- " #hysteresis loss Ph = kh*v*f*(Bm)**n = k1*f\n",
- " #Thus when the hysteresis is 40 W and the frequency 50 Hz,\n",
- "k1 = Ph1/f1\n",
- " #If the frequency is increased to 60 Hz,\n",
- "Ph2 = k1*f2\n",
- " #eddy current loss, Pe = ke*(Bm1*f1*t1)**2 = k2*f**2\n",
- " #since the flux density and lamination thickness are constant.\n",
- " #When the eddy current loss is 20 W the frequency is 50 Hz. Thus\n",
- "k2 = Pe1/(f1**2)\n",
- " #If the frequency is increased to 60 Hz,\n",
- "Pe2 = k2*(f2**2)\n",
- " #hysteresis loss Ph = kh*v*f*(Bm)**n, is independent of the thickness of the laminations. \n",
- " #Thus, if the thickness of the laminations is halved, the hysteresis loss remains at \n",
- "Phb2 = Ph1\n",
- " #eddy current loss, Pe = ke*(Bm1*f1*t1)**2 = k2*t**3\n",
- "k3 = Pe1/(t1**3)\n",
- "t2 = 0.5*t1\n",
- "Peb2 = k3*t2**3\n",
- " #total core loss when the thickness of the laminations is halved is given by\n",
- "TL = Phb2 + Peb2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)If the frequency is increased to 60 Hz,hysteresis loss is \",Ph2,\" W and eddy current loss \", Pe2,\" W\"\n",
- "print \"\\n(b)the total core loss when the thickness of the laminations is halved \",TL,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)If the frequency is increased to 60 Hz,hysteresis loss is 48.0 W and eddy current loss 28.8 W\n",
- "\n",
- "(b)the total core loss when the thickness of the laminations is halved 42.5 W\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 701</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the new total core loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 500;# in Volts\n",
- "V2 = 1000;# in Volts\n",
- "Ph1 = 400;# in W\n",
- "Pe1 = 150;# in W\n",
- "f1 = 50;# in Hz\n",
- "n = 1.6;# Steinmetz index\n",
- "f2 = 100;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #hysteresis loss Ph = k1*f*(E/f)**n\n",
- " #At 500 V and 50 Hz\n",
- "k1 = Ph1/(f1*(V1/f1)**1.6)\n",
- " #At 1000 V and 100 Hz,\n",
- "Ph2 = k1*f2*(V2/f2)**1.6\n",
- " #eddy current loss, Pe = k2*E**2\n",
- " #At 500 V,\n",
- "k2 = Pe1/(V1**2)\n",
- " #At 1000 V,\n",
- "Pe2 = k2*(V2**2)\n",
- " #the new total core loss\n",
- "TL = Ph2 + Pe2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the new total core loss \",TL,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the new total core loss 1400.0 W"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 702</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the separate values of the hysteresis and eddy current losses at frequencies of (a) 50 Hz and (b) 60 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#from pylab import *\n",
- "%pylab inline\n",
- "#initializing the variables:\n",
- "f1 = 50;# in Hz\n",
- "f2 = 60;# in Hz\n",
- "\n",
- "#calculation:\n",
- "k1 = 0.5\n",
- "k2 = 0.032\n",
- "f = [30, 50, 70,90]\n",
- "Pcf = [1.5, 2.1, 2.7, 3.4]\n",
- "fig = plt.figure()\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(f,Pcf,'-')\n",
- "xlabel('f')\n",
- "ylabel('Pc/f')\n",
- "show()\n",
- "HL1 = k1*f1\n",
- "ECL1 = k2*f1**2\n",
- "\n",
- "HL2 = k1*f2\n",
- "ECL2 = k2*f2**2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) at Frequency = 50 Hz, hysteresis loss is\", HL1,\" W and eddy current loss is\", ECL1,\" W\"\n",
- "print \"\\n (b) at Frequency = 60 Hz, hysteresis loss is\", HL2,\" W and eddy current loss is\", ECL2,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
- "For more information, type 'help(pylab)'."
- ]
- },
- {
- "output_type": "display_data",
- "png": 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- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) at Frequency = 50 Hz, hysteresis loss is 25.0 W and eddy current loss is 80.0 W\n",
- "\n",
- " (b) at Frequency = 60 Hz, hysteresis loss is 30.0 W and eddy current loss is 115.2 W"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 703</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the hysteresis and eddy current losses at 50 Hz (b) determine the hysteresis and eddy current losses at 50 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "TL1 = 400;# in Watt\n",
- "TL2 = 498;# in Watt\n",
- "x = 0.25;\n",
- "y = 0.4;\n",
- "f1 = 50;# in Hz\n",
- "n = 1.7;# Steinmetz index\n",
- "f2 = 60;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #if volume v and the maximum flux density are constant\n",
- " #hysteresis loss Ph = kh*v*f*(Bm)**n = k1*f\n",
- " #(if the maximum flux density and the lamination thickness are constant)\n",
- " #eddy current loss, Pe = ke*(Bm1*f1*t1)**2 = k2*f**2\n",
- " #At 50 Hz frequency, TL1 = k1*f1 + k2*f1**2\n",
- " #At 60 Hz frequency, TL2 = k1*f2 + k2*f2**2\n",
- " #Solving equations gives the values of k1 and k2.\n",
- "k2 = (5*TL2 - 6*TL1)/(5*(f2**2) - 6*(f1**2))\n",
- "k1 = (TL1 - k2*f1**2)/f1\n",
- " #hysteresis loss Ph = k1*f\n",
- "Ph1 = k1*f1\n",
- " #eddy current loss\n",
- "Pe1 = k2*f1**2\n",
- " #Since at 50 Hz the flux density is increased by 25%, the new hysteresis loss is\n",
- "Ph2 = Ph1*(1 + x)**1.7\n",
- " #Since at 50 Hz the flux density is increased by 25%, and the lamination thickness is increased by 40%, \n",
- " #the new eddy current loss is\n",
- "Pe2 = Pe1*((1 + x)**2)*(1 + y)**3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the hysteresis and eddy current losses at 50 Hz are \",round(Ph1,2),\" W and \",round( Pe1,2),\" W resp.\"\n",
- "print \"\\n (b)the hysteresis and eddy current losses at 50 Hz after increement are \",round(Ph2,2),\" W and \",round( Pe2,2),\" W resp.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the hysteresis and eddy current losses at 50 Hz are 325.0 W and 75.0 W resp.\n",
- "\n",
- " (b)the hysteresis and eddy current losses at 50 Hz after increement are 474.93 W and 321.56 W resp.\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_2.ipynb deleted file mode 100755 index c4a1d2a6..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_2.ipynb +++ /dev/null @@ -1,631 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 38: Magnetic materials</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 694</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the hysteresis loss per m3 per cycle, and (b) the hysteresis loss per m3 at a frequency of 50 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "A = 12.5;# in cm2\n",
- "x = 500;# horizontal axis 1 cm = 500 A/m\n",
- "y = 0.2;# vertical axis 1 cm = 0.2 T\n",
- "f = 50;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #hysteresis loss per cycle\n",
- "HL = A*x*y\n",
- " #At 50 Hz frequency, hysteresis loss\n",
- "HLf = HL*f\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)hysteresis loss per cycle is = \",HL,\" J/m3\"\n",
- "print \"\\n(b)At 50 Hz frequency, hysteresis loss \",HLf,\" W/m3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)hysteresis loss per cycle is = 1250.0 J/m3\n",
- "\n",
- "(b)At 50 Hz frequency, hysteresis loss 62500.0 W/m3"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 695</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the hysteresis loss per m3 for a maximum flux density of 1.1 T and frequency of 25 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "n = 1.6;# the Steinmetz index\n",
- "f1 = 50;# in Hz\n",
- "f2 = 25;# in Hz\n",
- "Bm1 = 1.5;# in Tesla\n",
- "Bm2 = 1.1;# in Tesla\n",
- "Ph1 = 62500;# in W/m3\n",
- "v = 1;\n",
- "\n",
- "#calculation: \n",
- " #hysteresis loss Ph = kh*v*f*(Bm)**n\n",
- "kh = Ph1/(v*f1*(Bm1)**n)\n",
- " #When f = 25 Hz and Bm = 1.1 T,\n",
- "Ph2 = kh*v*f2*(Bm2)**n\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n hysteresis loss When f = 25 Hz and Bm = 1.1 T, is = \",round(Ph2,2),\" W/m3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " hysteresis loss When f = 25 Hz and Bm = 1.1 T, is = 19025.33 W/m3"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 695</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the hysteresis loss at a frequency of 80 Hz\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "csa = 0.002;# in m2\n",
- "l = 1;# in m\n",
- "a = 400/0.01;# 10 mm = 400 A/m \n",
- "b = 0.1/0.01;# 10 mm = 0.1 T \n",
- "A = 0.01;# in m2\n",
- "f = 80;# in Hz\n",
- "\n",
- "#calculation: \n",
- " #hysteresis loss per cycle\n",
- "HL = A*a*b\n",
- " #At a frequency of 80 Hz,\n",
- " #hysteresis loss\n",
- "HLf = HL*f\n",
- " #Volume of ring\n",
- "v = csa*l\n",
- " #hysteresis loss\n",
- "Ph = HLf*v\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the hysteresis loss at a frequency of 80 Hz is \",Ph,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the hysteresis loss at a frequency of 80 Hz is 640.0 W"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 696</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the value of the hysteresis loss when the maximum core flux is 8 mWb and the frequency is 50 Hz\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Phi1 = 0.01;# in Wb\n",
- "Phi2 = 0.008;# in Wb\n",
- "csa = 0.008;# in m2\n",
- "v = 0.005;# in m3\n",
- "f = 50;# in Hz\n",
- "n = 1.7;# the Steinmetz constant\n",
- "Ph1 = 100;# in Watt\n",
- "\n",
- " #calculation: \n",
- " #maximum flux density\n",
- "Bm1 = Phi1/csa\n",
- " #hysteresis loss Ph1 = kh*v*f*(Bm1)**n\n",
- "kh = Ph1/(v*f*(Bm1)**n)\n",
- " #When the maximum core flux is 8 mWb,\n",
- "Bm2 = Phi2/csa\n",
- " #hysteresis loss, Ph2\n",
- "Ph2 = kh*v*f*(Bm2)**n\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"value of hysteresis loss when maximum core flux is 8 mWb and the frequency is 50 Hz is \",round(Ph2,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "value of hysteresis loss when maximum core flux is 8 mWb and the frequency is 50 Hz is 68.43 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 699</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the new value of eddy current loss per cubic metre.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Pe1 = 10;# in W/m3\n",
- "f1 = 50;# in Hz\n",
- "f2 = 30;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #When the eddy current loss is 10 W/m3, frequency f is 50 Hz.\n",
- " #constant k\n",
- "k = Pe1/(f1**2)\n",
- " #When the frequency is 30 Hz, eddy current loss,\n",
- "Pe2 = k*(f2**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\neddy current loss per cubic metre is \",Pe2,\" W/m3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "eddy current loss per cubic metre is 3.6 W/m3"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 699</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the necessary new thickness of the laminations.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Pe = 100;# in W/m3\n",
- "f1 = 50;# in Hz\n",
- "t1 = 0.0005;# in m\n",
- "x = 1/3;\n",
- "f2 = 250;# in Hz\n",
- "Bm1 = 1;\n",
- " #calculation: \n",
- " #Pe = ke*(Bm1*f1*t1)**2\n",
- " #Hence, at 50 Hz frequency\n",
- "ke = Pe/(Bm1*f1*t1)**2\n",
- " #At 250 Hz frequency\n",
- "Bm2 = x*Bm1\n",
- "t2 = ((Pe/ke)**0.5)/(Bm2*f2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nlamination thickness is \",t2,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "lamination thickness is 0.0003 m"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 700</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the values of the losses if the frequency is increased to 60 Hz. \n",
- "#(b) What will be the total core loss if the frequency is 50 Hz and the lamination are made one-half of their original thickness?\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ph1 = 40;# in W\n",
- "Pe1 = 20;# in W\n",
- "f1 = 50;# in Hz\n",
- "x = 1/2;\n",
- "f2 = 60;# in Hz\n",
- "t1 = 1;\n",
- " #calculation: \n",
- " #hysteresis loss Ph = kh*v*f*(Bm)**n = k1*f\n",
- " #Thus when the hysteresis is 40 W and the frequency 50 Hz,\n",
- "k1 = Ph1/f1\n",
- " #If the frequency is increased to 60 Hz,\n",
- "Ph2 = k1*f2\n",
- " #eddy current loss, Pe = ke*(Bm1*f1*t1)**2 = k2*f**2\n",
- " #since the flux density and lamination thickness are constant.\n",
- " #When the eddy current loss is 20 W the frequency is 50 Hz. Thus\n",
- "k2 = Pe1/(f1**2)\n",
- " #If the frequency is increased to 60 Hz,\n",
- "Pe2 = k2*(f2**2)\n",
- " #hysteresis loss Ph = kh*v*f*(Bm)**n, is independent of the thickness of the laminations. \n",
- " #Thus, if the thickness of the laminations is halved, the hysteresis loss remains at \n",
- "Phb2 = Ph1\n",
- " #eddy current loss, Pe = ke*(Bm1*f1*t1)**2 = k2*t**3\n",
- "k3 = Pe1/(t1**3)\n",
- "t2 = 0.5*t1\n",
- "Peb2 = k3*t2**3\n",
- " #total core loss when the thickness of the laminations is halved is given by\n",
- "TL = Phb2 + Peb2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)If the frequency is increased to 60 Hz,hysteresis loss is \",Ph2,\" W and eddy current loss \", Pe2,\" W\"\n",
- "print \"\\n(b)the total core loss when the thickness of the laminations is halved \",TL,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)If the frequency is increased to 60 Hz,hysteresis loss is 48.0 W and eddy current loss 28.8 W\n",
- "\n",
- "(b)the total core loss when the thickness of the laminations is halved 42.5 W\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 701</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the new total core loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V1 = 500;# in Volts\n",
- "V2 = 1000;# in Volts\n",
- "Ph1 = 400;# in W\n",
- "Pe1 = 150;# in W\n",
- "f1 = 50;# in Hz\n",
- "n = 1.6;# Steinmetz index\n",
- "f2 = 100;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #hysteresis loss Ph = k1*f*(E/f)**n\n",
- " #At 500 V and 50 Hz\n",
- "k1 = Ph1/(f1*(V1/f1)**1.6)\n",
- " #At 1000 V and 100 Hz,\n",
- "Ph2 = k1*f2*(V2/f2)**1.6\n",
- " #eddy current loss, Pe = k2*E**2\n",
- " #At 500 V,\n",
- "k2 = Pe1/(V1**2)\n",
- " #At 1000 V,\n",
- "Pe2 = k2*(V2**2)\n",
- " #the new total core loss\n",
- "TL = Ph2 + Pe2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the new total core loss \",TL,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the new total core loss 1400.0 W"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 702</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the separate values of the hysteresis and eddy current losses at frequencies of (a) 50 Hz and (b) 60 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#from pylab import *\n",
- "%pylab inline\n",
- "#initializing the variables:\n",
- "f1 = 50;# in Hz\n",
- "f2 = 60;# in Hz\n",
- "\n",
- "#calculation:\n",
- "k1 = 0.5\n",
- "k2 = 0.032\n",
- "f = [30, 50, 70,90]\n",
- "Pcf = [1.5, 2.1, 2.7, 3.4]\n",
- "fig = plt.figure()\n",
- "ax = fig.add_subplot(1, 1, 1)\n",
- "ax.plot(f,Pcf,'-')\n",
- "xlabel('f')\n",
- "ylabel('Pc/f')\n",
- "show()\n",
- "HL1 = k1*f1\n",
- "ECL1 = k2*f1**2\n",
- "\n",
- "HL2 = k1*f2\n",
- "ECL2 = k2*f2**2\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) at Frequency = 50 Hz, hysteresis loss is\", HL1,\" W and eddy current loss is\", ECL1,\" W\"\n",
- "print \"\\n (b) at Frequency = 60 Hz, hysteresis loss is\", HL2,\" W and eddy current loss is\", ECL2,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
- "For more information, type 'help(pylab)'."
- ]
- },
- {
- "output_type": "display_data",
- "png": 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- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) at Frequency = 50 Hz, hysteresis loss is 25.0 W and eddy current loss is 80.0 W\n",
- "\n",
- " (b) at Frequency = 60 Hz, hysteresis loss is 30.0 W and eddy current loss is 115.2 W"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 703</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the hysteresis and eddy current losses at 50 Hz (b) determine the hysteresis and eddy current losses at 50 Hz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "TL1 = 400;# in Watt\n",
- "TL2 = 498;# in Watt\n",
- "x = 0.25;\n",
- "y = 0.4;\n",
- "f1 = 50;# in Hz\n",
- "n = 1.7;# Steinmetz index\n",
- "f2 = 60;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #if volume v and the maximum flux density are constant\n",
- " #hysteresis loss Ph = kh*v*f*(Bm)**n = k1*f\n",
- " #(if the maximum flux density and the lamination thickness are constant)\n",
- " #eddy current loss, Pe = ke*(Bm1*f1*t1)**2 = k2*f**2\n",
- " #At 50 Hz frequency, TL1 = k1*f1 + k2*f1**2\n",
- " #At 60 Hz frequency, TL2 = k1*f2 + k2*f2**2\n",
- " #Solving equations gives the values of k1 and k2.\n",
- "k2 = (5*TL2 - 6*TL1)/(5*(f2**2) - 6*(f1**2))\n",
- "k1 = (TL1 - k2*f1**2)/f1\n",
- " #hysteresis loss Ph = k1*f\n",
- "Ph1 = k1*f1\n",
- " #eddy current loss\n",
- "Pe1 = k2*f1**2\n",
- " #Since at 50 Hz the flux density is increased by 25%, the new hysteresis loss is\n",
- "Ph2 = Ph1*(1 + x)**1.7\n",
- " #Since at 50 Hz the flux density is increased by 25%, and the lamination thickness is increased by 40%, \n",
- " #the new eddy current loss is\n",
- "Pe2 = Pe1*((1 + x)**2)*(1 + y)**3\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the hysteresis and eddy current losses at 50 Hz are \",round(Ph1,2),\" W and \",round( Pe1,2),\" W resp.\"\n",
- "print \"\\n (b)the hysteresis and eddy current losses at 50 Hz after increement are \",round(Ph2,2),\" W and \",round( Pe2,2),\" W resp.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the hysteresis and eddy current losses at 50 Hz are 325.0 W and 75.0 W resp.\n",
- "\n",
- " (b)the hysteresis and eddy current losses at 50 Hz after increement are 474.93 W and 321.56 W resp.\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_38.ipynb index 68af6e5d..68af6e5d 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_38.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint.ipynb deleted file mode 100755 index 33240359..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint.ipynb +++ /dev/null @@ -1,200 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 39: Dielectrics and dielectric loss</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 717</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the capacitor, at a frequency of 8 MHz, \n",
- "#(a) the loss angle, (b) the power factor, (c) the Q-factor, and (d) the dissipation factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rs = 1.5;# in ohms\n",
- "Cs = 400E-12;# in Farads\n",
- "f = 8E6;# in Hz\n",
- "\n",
- "#calculation: \n",
- " #for a series equivalent circuit,\n",
- " #tan(del) = Rs*w*Cs\n",
- " #loss angle,\n",
- "de = math.atan(Rs*Cs*(2*math.pi*f))\n",
- " #power factor\n",
- "pf = math.cos(de)\n",
- " #the Q-factor\n",
- "Q = 1/math.tan(de)\n",
- " #dissipation factor,\n",
- "D = 1/Q\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)loss angle \",round(de,2),\" rad.\"\n",
- "print \"\\n (b)power factor \",round(de,2),\" rad.\"\n",
- "print \"\\n (c)Q-factor is \",round(Q,2)\n",
- "print \"\\n (d)dissipation factor \",round(D,2),\" rad.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)loss angle 0.03 rad.\n",
- "\n",
- " (b)power factor 0.03 rad.\n",
- "\n",
- " (c)Q-factor is 33.16\n",
- "\n",
- " (d)dissipation factor 0.03 rad."
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 718</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the component values of the equivalent parallel circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "de = 0.025;# in rad.\n",
- "V = 5000;# in Volts\n",
- "PL = 20;# power loss\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation: \n",
- " #power loss = w*C*V**2*tan(del)\n",
- "Cp = PL/(2*math.pi*f*V*V*math.tan(de))\n",
- " #for a parallel equivalent circuit,\n",
- " #tan(del) = 1/(Rp*w*Cp)\n",
- "Rp = 1/(2*math.pi*f*Cp*math.tan(de))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance C \",round(Cp*1E6,2),\"uF and parallel resistance \",round(Rp,2),\"ohm.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance C 0.1 uF and parallel resistance 1250000.0 ohm."
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 718</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the loss angle, (b) the equivalent series loss resistance, and (c) the equivalent parallel loss resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "P = 500E-6;# in Watt\n",
- "C = 2000E-12;# in Farads\n",
- "V = 20;# in Volts\n",
- "f = 10000;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #power loss = w*C*V**2*tan(del)\n",
- " #loss angle\n",
- "de = math.atan(P/(2*math.pi*f*V*V*C))\n",
- " #for an equivalent series circuit,\n",
- " #tan(del) = (Rs*w*Cs)\n",
- "Cs = C\n",
- "Rs = (math.tan(de))/(2*math.pi*f*Cs)\n",
- " #for an equivalent parallel circuit\n",
- " #tan(del) = 1/(Rp*w*Cp)\n",
- "Cp = C\n",
- "Rp = 1/(2*math.pi*f*Cp*math.tan(de))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)loss angle \",round(de*180/math.pi,2),\"deg\"\n",
- "print \"\\n (b)series resistance \",round(Rs,2),\" ohm.\"\n",
- "print \"\\n (c)parallel resistance \",round(Rp/1000,2),\"Kohm.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)loss angle 0.57 deg\n",
- "\n",
- " (b)series resistance 79.16 ohm.\n",
- "\n",
- " (c)parallel resistance 800.0 Kohm."
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_1.ipynb deleted file mode 100755 index 33240359..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_1.ipynb +++ /dev/null @@ -1,200 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 39: Dielectrics and dielectric loss</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 717</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the capacitor, at a frequency of 8 MHz, \n",
- "#(a) the loss angle, (b) the power factor, (c) the Q-factor, and (d) the dissipation factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rs = 1.5;# in ohms\n",
- "Cs = 400E-12;# in Farads\n",
- "f = 8E6;# in Hz\n",
- "\n",
- "#calculation: \n",
- " #for a series equivalent circuit,\n",
- " #tan(del) = Rs*w*Cs\n",
- " #loss angle,\n",
- "de = math.atan(Rs*Cs*(2*math.pi*f))\n",
- " #power factor\n",
- "pf = math.cos(de)\n",
- " #the Q-factor\n",
- "Q = 1/math.tan(de)\n",
- " #dissipation factor,\n",
- "D = 1/Q\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)loss angle \",round(de,2),\" rad.\"\n",
- "print \"\\n (b)power factor \",round(de,2),\" rad.\"\n",
- "print \"\\n (c)Q-factor is \",round(Q,2)\n",
- "print \"\\n (d)dissipation factor \",round(D,2),\" rad.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)loss angle 0.03 rad.\n",
- "\n",
- " (b)power factor 0.03 rad.\n",
- "\n",
- " (c)Q-factor is 33.16\n",
- "\n",
- " (d)dissipation factor 0.03 rad."
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 718</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the component values of the equivalent parallel circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "de = 0.025;# in rad.\n",
- "V = 5000;# in Volts\n",
- "PL = 20;# power loss\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation: \n",
- " #power loss = w*C*V**2*tan(del)\n",
- "Cp = PL/(2*math.pi*f*V*V*math.tan(de))\n",
- " #for a parallel equivalent circuit,\n",
- " #tan(del) = 1/(Rp*w*Cp)\n",
- "Rp = 1/(2*math.pi*f*Cp*math.tan(de))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance C \",round(Cp*1E6,2),\"uF and parallel resistance \",round(Rp,2),\"ohm.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance C 0.1 uF and parallel resistance 1250000.0 ohm."
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 718</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the loss angle, (b) the equivalent series loss resistance, and (c) the equivalent parallel loss resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "P = 500E-6;# in Watt\n",
- "C = 2000E-12;# in Farads\n",
- "V = 20;# in Volts\n",
- "f = 10000;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #power loss = w*C*V**2*tan(del)\n",
- " #loss angle\n",
- "de = math.atan(P/(2*math.pi*f*V*V*C))\n",
- " #for an equivalent series circuit,\n",
- " #tan(del) = (Rs*w*Cs)\n",
- "Cs = C\n",
- "Rs = (math.tan(de))/(2*math.pi*f*Cs)\n",
- " #for an equivalent parallel circuit\n",
- " #tan(del) = 1/(Rp*w*Cp)\n",
- "Cp = C\n",
- "Rp = 1/(2*math.pi*f*Cp*math.tan(de))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)loss angle \",round(de*180/math.pi,2),\"deg\"\n",
- "print \"\\n (b)series resistance \",round(Rs,2),\" ohm.\"\n",
- "print \"\\n (c)parallel resistance \",round(Rp/1000,2),\"Kohm.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)loss angle 0.57 deg\n",
- "\n",
- " (b)series resistance 79.16 ohm.\n",
- "\n",
- " (c)parallel resistance 800.0 Kohm."
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_2.ipynb deleted file mode 100755 index 33240359..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_2.ipynb +++ /dev/null @@ -1,200 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 39: Dielectrics and dielectric loss</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 717</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the capacitor, at a frequency of 8 MHz, \n",
- "#(a) the loss angle, (b) the power factor, (c) the Q-factor, and (d) the dissipation factor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rs = 1.5;# in ohms\n",
- "Cs = 400E-12;# in Farads\n",
- "f = 8E6;# in Hz\n",
- "\n",
- "#calculation: \n",
- " #for a series equivalent circuit,\n",
- " #tan(del) = Rs*w*Cs\n",
- " #loss angle,\n",
- "de = math.atan(Rs*Cs*(2*math.pi*f))\n",
- " #power factor\n",
- "pf = math.cos(de)\n",
- " #the Q-factor\n",
- "Q = 1/math.tan(de)\n",
- " #dissipation factor,\n",
- "D = 1/Q\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)loss angle \",round(de,2),\" rad.\"\n",
- "print \"\\n (b)power factor \",round(de,2),\" rad.\"\n",
- "print \"\\n (c)Q-factor is \",round(Q,2)\n",
- "print \"\\n (d)dissipation factor \",round(D,2),\" rad.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)loss angle 0.03 rad.\n",
- "\n",
- " (b)power factor 0.03 rad.\n",
- "\n",
- " (c)Q-factor is 33.16\n",
- "\n",
- " (d)dissipation factor 0.03 rad."
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 718</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the component values of the equivalent parallel circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "de = 0.025;# in rad.\n",
- "V = 5000;# in Volts\n",
- "PL = 20;# power loss\n",
- "f = 50;# in Hz\n",
- "\n",
- "#calculation: \n",
- " #power loss = w*C*V**2*tan(del)\n",
- "Cp = PL/(2*math.pi*f*V*V*math.tan(de))\n",
- " #for a parallel equivalent circuit,\n",
- " #tan(del) = 1/(Rp*w*Cp)\n",
- "Rp = 1/(2*math.pi*f*Cp*math.tan(de))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance C \",round(Cp*1E6,2),\"uF and parallel resistance \",round(Rp,2),\"ohm.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance C 0.1 uF and parallel resistance 1250000.0 ohm."
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 718</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the loss angle, (b) the equivalent series loss resistance, and (c) the equivalent parallel loss resistance.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "P = 500E-6;# in Watt\n",
- "C = 2000E-12;# in Farads\n",
- "V = 20;# in Volts\n",
- "f = 10000;# in Hz\n",
- "\n",
- " #calculation: \n",
- " #power loss = w*C*V**2*tan(del)\n",
- " #loss angle\n",
- "de = math.atan(P/(2*math.pi*f*V*V*C))\n",
- " #for an equivalent series circuit,\n",
- " #tan(del) = (Rs*w*Cs)\n",
- "Cs = C\n",
- "Rs = (math.tan(de))/(2*math.pi*f*Cs)\n",
- " #for an equivalent parallel circuit\n",
- " #tan(del) = 1/(Rp*w*Cp)\n",
- "Cp = C\n",
- "Rp = 1/(2*math.pi*f*Cp*math.tan(de))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)loss angle \",round(de*180/math.pi,2),\"deg\"\n",
- "print \"\\n (b)series resistance \",round(Rs,2),\" ohm.\"\n",
- "print \"\\n (c)parallel resistance \",round(Rp/1000,2),\"Kohm.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)loss angle 0.57 deg\n",
- "\n",
- " (b)series resistance 79.16 ohm.\n",
- "\n",
- " (c)parallel resistance 800.0 Kohm."
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_39.ipynb index 6edb5773..6edb5773 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_39.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint.ipynb deleted file mode 100755 index 3f3b710b..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint.ipynb +++ /dev/null @@ -1,1328 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 40: Field theory</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 725</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitance per metre length of the system\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 2.8;\n",
- "l = 1;# in m\n",
- "\n",
- "#calculation: \n",
- " #From Figure 40.9\n",
- "m = 16;# number of parallel squares measured along each equipotential\n",
- "n = 6;# the number of series squares measured along each line of force\n",
- "C = e0*er*l*m/n\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E12,2),\"pFarad.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 66.08 pFarad."
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 725</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitance of a 100 m length of the cable.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 3.4;\n",
- "l = 100;# in m\n",
- "\n",
- " #calculation: \n",
- " #From Figure 40.10\n",
- "m = 13;# number of parallel squares measured along each equipotential\n",
- "n = 4;# the number of series squares measured along each line of force\n",
- "C = e0*er*l*m/n\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E9,2),\"nFarad.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 9.78 nFarad."
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 726</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitance per metre length of the cable\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 2.7;\n",
- "ri = 0.0005;# in m\n",
- "ro = 0.006;# in m\n",
- "\n",
- " #calculation: \n",
- " #capacitance C\n",
- "C = 2*math.pi*e0*er/(math.log(ro/ri))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E12,2),\"pFarad.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 60.42 pFarad."
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 727</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the internal diameter of the sheath.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 80E-12;# in Farads\n",
- "e0 = 8.85E-12; \n",
- "er = 3.5;\n",
- "d0 = 0.008;# in m\n",
- "\n",
- " #calculation: \n",
- " #internal diameter\n",
- "di = d0*(math.e**(2*math.pi*e0*er/C))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n internal diameter is \",round(di,2),\" m.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " internal diameter is 0.09 m."
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 728</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the capacitance per kilometre length of the cable, \n",
- "#(b) the dielectric stress at a radius of 30 mm, and\n",
- "#(c) the maximum and minimum values of dielectric stress.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 3.5;\n",
- "di = 0.08;# in m\n",
- "d0 = 0.032;# in m\n",
- "r = 0.03;# in m\n",
- "V = 40000;# in Volts\n",
- "\n",
- "#calculation: \n",
- " #capacitance C\n",
- "C = 2*math.pi*e0*er/(math.log(di/d0))\n",
- " #dielectric stress at radius r,\n",
- "E = V/(r*math.log(di/d0))\n",
- " #maximum dielectric stress,\n",
- "Emax = V/((d0/2)*(math.log((di/d0))))\n",
- " #minimum dielectric stress,\n",
- "Emin = V/((di/2)*(math.log((di/d0))))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E12,2),\"pF/km\"\n",
- "print \"\\n dielectric stress at radius r is \",round(E,2),\"V/m\"\n",
- "print \"\\n maximum dielectric stress, is \",round(Emax,2),\"V/m minimum dielectric stress \",round( Emin,2),\"V/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 212.4 pF/km\n",
- "\n",
- " dielectric stress at radius r is 1455142.22 V/m\n",
- "\n",
- " maximum dielectric stress, is 2728391.67 V/m minimum dielectric stress 1091356.67 V/m"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 729</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the core and inner sheath radii for the most economical cable,\n",
- "#(b) the capacitance per metre length, and (c) the charging current per kilometre run.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 3.5;\n",
- "V = 60000;# in Volts\n",
- "f = 50;# in Hz\n",
- "Em = 10E6;# in V/m\n",
- "\n",
- "\n",
- "#calculation: \n",
- " #core radius, a\n",
- "a = V/Em\n",
- " #internal sheath radius,\n",
- "b = a*math.e**1\n",
- " #capacitance\n",
- "C = 2*math.pi*e0*er/(math.log(b/a))\n",
- " #Charging current\n",
- "I = V*2*math.pi*f*C\n",
- " #charging current per kilometre\n",
- "Ipkm = I*1000\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n core radius is \",round(a*1000,2),\"mm and internal sheath radius \",round(b*1000,1),\"mm\"\n",
- "print \"\\n capacitance is \",round(C*1E12,0),\"pF/m\"\n",
- "print \"\\n the charging current per kilometre \",round(Ipkm,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " core radius is 6.0 mm and internal sheath radius 16.3 mm\n",
- "\n",
- " capacitance is 195.0 pF/m\n",
- "\n",
- " the charging current per kilometre 3.67 A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 730</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for a 1 km length of the cable (a) the capacitance, (b) the charging current and (c) the power loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 2.5;\n",
- "di = 0.08;# in m\n",
- "d0 = 0.025;# in m\n",
- "r = 1000;# in m\n",
- "V = 132000;# in Volts\n",
- "f = 50;# in Hz\n",
- "de = 3.5E-3;# rad.\n",
- "\n",
- " #calculation:\n",
- " #core radius, a\n",
- "a = d0/2\n",
- " #internal sheath radius,\n",
- "b = di/2\n",
- " #capacitance\n",
- "C = 2*math.pi*e0*er*1E3/(math.log(b/a))\n",
- " #Charging current\n",
- "I = V*2*math.pi*f*C\n",
- " #power loss\n",
- "P = (2*math.pi*f*C*math.tan(de))*V**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)capacitance for a 1 km length is \",round(C*1E6,2),\"uF\"\n",
- "print \"\\n (b)the charging current \",round(I,2),\"A/km\"\n",
- "print \"\\n (c)power loss \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)capacitance for a 1 km length is 0.12 uF\n",
- "\n",
- " (b)the charging current 4.96 A/km\n",
- "\n",
- " (c)power loss 2289.78 W"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 732</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitance of the cable per metre length by the method of curvilinear squares\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 3.2;\n",
- "di = 0.06;# in m\n",
- "d0 = 0.020;# in m\n",
- "\n",
- " #calculation:\n",
- " #core radius, a\n",
- "a = d0/2\n",
- " #internal sheath radius,\n",
- "b = di/2\n",
- " #capacitance\n",
- "C = 2*math.pi*e0*er/(math.log(b/a))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance per m of length is \",round(C*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance per m of length is 0.16 nF"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 736</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitance of the line if the total length is 200 m.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "D = 0.05;# in m\n",
- "d = 0.005;# in m\n",
- "l = 200;# in m\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = math.pi*e0*er/(math.log(D/(d/2)))\n",
- " #capacitance of a 200 m length\n",
- "C200 = C*l\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance of a 200 m length is \",round(C200*1E6,5),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance of a 200 m length is 0.00186 uF"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 736</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for a 1 km length of line, (a) the capacitance of the conductors, \n",
- "#(b) the value of charge carried by each conductor, and (c) the charging current\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "D = 1.2;# in m\n",
- "r = 0.004;# in m\n",
- "f = 50;# in Hz\n",
- "V = 15000;# in Volts\n",
- "l = 1000;# in m\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = math.pi*e0*er/(math.log(D/r))\n",
- " #capacitance of a 1 km length\n",
- "Cpkm = C*l\n",
- " #Charge Q\n",
- "Q = Cpkm*V\n",
- " #Charging current\n",
- "I = V*2*math.pi*f*Cpkm\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance per 1km length is \",round(Cpkm*1E9,2),\"nF\"\n",
- "print \"\\n Charge Q is \",round(Q*1E6,2),\"uC\"\n",
- "print \"\\n Charging current is \",round(I,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance per 1km length is 4.87 nF\n",
- "\n",
- " Charge Q is 73.12 uC\n",
- "\n",
- " Charging current is 0.02 A"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 737</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the maximum value required for the capacitance per metre length,\n",
- "#and (b) the maximum diameter of each conductor if their distance between centres is 1.25 m.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "I = 0.015;# in Amperes\n",
- "d = 1.25;# in m\n",
- "r = 800;# in m\n",
- "f = 50;# in Hz\n",
- "V = 10000;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = I/(2*math.pi*f*V)\n",
- " #required maximum value of capacitance\n",
- "Cmax = C/r\n",
- " #maximum diameter of each conductor\n",
- "D = 2*d/(math.e**(math.pi*e0*er/Cmax))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n required maximum value of capacitance is \",round(Cmax*1E12,2),\"pF/m\"\n",
- "print \"\\nthe maximum diameter of each conductor is \",round(D,2),\" m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " required maximum value of capacitance is 5.97 pF/m\n",
- "\n",
- "the maximum diameter of each conductor is 0.02 m"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 739</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the energy stored in a 10 nF capacitor when charged to 1 kV, and \n",
- "#the average power developed if this energy is dissipated in 10 \u03bcs\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "C = 10E-9;# in Farad\n",
- "V = 1000;# in Volts\n",
- "t = 10E-6;# in sec\n",
- "\n",
- " #calculation:\n",
- " #energy stored,Wf\n",
- "Wf = C*V*V/2\n",
- " #average power developed\n",
- "Pav = Wf/t\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the energy stored is \",Wf,\"J\"\n",
- "print \"\\nthe average power developed is \",Pav,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the energy stored is 0.005 J\n",
- "\n",
- "the average power developed is 500.0 W"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 739</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the voltage across the plates and (b) the capacitance of the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "Q = 5E-3;# in Coulomb\n",
- "W = 0.625;# in Joules\n",
- "\n",
- " #calculation:\n",
- " #voltage across the plates\n",
- "V = 2*W/Q\n",
- " #Capacitance C\n",
- "C = Q/V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage across the plates is \",V,\" V\"\n",
- "print \"\\n Capacitance C is \",C*1E6,\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage across the plates is 250.0 V\n",
- "\n",
- " Capacitance C is 20.0 uF"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 740</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the required thickness of the ceramic dielectric, \n",
- "#(b) the area of plate required if the relative permittivity of the ceramic is 10, and \n",
- "#(c) the maximum energy stored by the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 10;\n",
- "C = 0.01E-6;# in Farad\n",
- "E = 10E6;# in V/m\n",
- "V = 2500;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #thickness of ceramic dielectric,\n",
- "d = V/E\n",
- " #cross-sectional area of plate\n",
- "A = C*d/(e0*er)\n",
- " #Maximum energy stored,\n",
- "W = C*V*V/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n thickness of ceramic dielectric is \",d*1000,\"mm\"\n",
- "print \"\\n cross-sectional area of plate, is \",round(A,2),\"m2\"\n",
- "print \"\\n Maximum energy stored is \",round(W,3),\" J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " thickness of ceramic dielectric is 0.25 mm\n",
- "\n",
- " cross-sectional area of plate, is 0.03 m2\n",
- "\n",
- " Maximum energy stored is 0.031 J"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 740</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the energy stored per cubic metre of the dielectric.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 2.3;\n",
- "A = 0.02;# in m2\n",
- "C = 400E-12;# in Farad\n",
- "V = 100;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #energy stored per unit volume of dielectric,\n",
- "W = ((C*V)**2)/(2*e0*er*A**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n energy stored per unit volume of dielectric is \",round(W,2),\" J/m3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " energy stored per unit volume of dielectric is 0.1 J/m3"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 744</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the inductance of the cable per metre length.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "a = 0.001;# in m\n",
- "b = 0.004;# in m\n",
- "\n",
- " #calculation:\n",
- " #inductance L\n",
- "L = (u0*ur/(2*math.pi))*(0.25 + math.log(b/a))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n inductance L is \",round(L*1E6,2),\"uH/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " inductance L is 0.33 uH/m"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 744</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the diameter of the sheath.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "da = 0.010;# in m\n",
- "L = 4E-7;# in H/m\n",
- "\n",
- " #calculation:\n",
- " #diameter of the sheath\n",
- "db = da*(math.e**(L/(u0*ur/(2*math.pi))))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n diameter of the sheath is \",round(db,2),\" m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " diameter of the sheath is 0.07 m"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 745</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the cable (a) the inductance, assuming nonmagnetic materials, and \n",
- "#(b) the capacitance, assuming a dielectric of relative permittivity 3.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "e0 = 8.85E-12;\n",
- "er = 3;\n",
- "da = 0.010;# in m\n",
- "db = 0.025;# in m\n",
- "l = 7500;# in m\n",
- "\n",
- "#calculation:\n",
- " #inductance per metre length\n",
- "L = (u0*ur/(2*math.pi))*(0.25 + math.log(db/da))\n",
- " #Since the cable is 7500 m long,\n",
- "L7500 = L*7500\n",
- " #capacitance C\n",
- "C = 2*math.pi*e0*er/(math.log(db/da))\n",
- " #//Since the cable is 7500 m long,\n",
- "C7500 = C*7500\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ninductance is \",round(L7500*1000,2),\" mH\"\n",
- "print \"\\ncapCItance is \",round(C7500*1E6,2),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "inductance is 1.75 mH\n",
- "\n",
- "capCItance is 1.37 uF"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 748</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the inductance of the line per metre length ignoring internal linkages\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "e0 = 8.85E-12;\n",
- "er = 3;\n",
- "D = 1.2;# in m\n",
- "a = 0.008;# in m\n",
- "\n",
- " #calculation:\n",
- " #inductance per metre length\n",
- "L = (u0*ur/(math.pi))*(math.log(D/a))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ninductance is \",round(L*1E6,2),\"uH/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "inductance is 2.0 uH/m"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 748</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the loop inductance, and \n",
- "#(b) the capacitance of a 1 km length of single-phase twin line having conductors of diameter 10 mm and spaced 800 mm apart in air.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "e0 = 8.85E-12;\n",
- "er = 1;\n",
- "l = 1000;# in m\n",
- "D = 0.8;# in m\n",
- "a = 0.01/2;# in m\n",
- "\n",
- " #calculation:\n",
- " #inductance per metre length\n",
- "L = (u0*ur/(math.pi))*(0.25 + math.log(D/a))\n",
- " #Since the cable is 1000 m long,\n",
- "L1k = L*l\n",
- " #capacitance C\n",
- "C = math.pi*e0*er/(math.log(D/a))\n",
- " #//Since the cable is 1000 m long,\n",
- "C1k = C*l\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ninductance is \",round(L1k*1000,2),\" mH\"\n",
- "print \"\\ncapcitance is \",round(C1k*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "inductance is 2.13 mH\n",
- "\n",
- "capcitance is 5.48 nF\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 749</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the distance between their centres.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 2.185E-6;# in H/m\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "a = 0.012/2;# in m\n",
- "\n",
- " #calculation:\n",
- " #distance D\n",
- "D = a*math.e**((L*math.pi)/(u0*ur) - 0.25)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ndistance D is \",round(D,2),\" m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "distance D is 1.1 m"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 752</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What value of current would double the energy stored?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.2;# in H\n",
- "I = 0.05;# in Amperes\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "\n",
- "#calculation:\n",
- " #energy stored in inductor\n",
- "W = L*I*I/2\n",
- " #current I\n",
- "I = (2*2*W/L)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nenergy stored in inductor is \",round(W*1000,2),\"mJ\"\n",
- "print \"\\ncurrent I is \",round(I,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "energy stored in inductor is 0.25 mJ\n",
- "\n",
- "current I is 0.07 A"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 752</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the total energy stored in the magnetic field of the airgap.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "B = 0.05;# in Tesla\n",
- "A = 500E-6;# in m2\n",
- "l = 0.002;# in m\n",
- "u0 = 4*math.pi*1E-7; \n",
- "\n",
- "#calculation:\n",
- " #energy stored\n",
- "W = (B**2)/(2*u0)\n",
- " #Volume of airgap\n",
- "v = A*l\n",
- " #energy stored in airgap\n",
- "W = W*v\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nenergy stored in the airgap is \",round(W*1E6,2),\"uJ\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "energy stored in the airgap is 994.72 uJ"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 752</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the strength of a uniform electric fi\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "B = 0.8;# in Tesla\n",
- "A = 500E-6;# in m2\n",
- "l = 0.002;# in m\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "e0 = 8.85E-12;\n",
- "er = 1;\n",
- "\n",
- "#calculation:\n",
- " #energy stored in mag. field\n",
- "W = (B**2)/(2*u0)\n",
- " #electric field\n",
- "E = (2*W/(e0*er))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nelectric field strength is \",round(E/1E6,2),\"MV/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "electric field strength is 239.89 MV/m"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_1.ipynb deleted file mode 100755 index 3f3b710b..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_1.ipynb +++ /dev/null @@ -1,1328 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 40: Field theory</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 725</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitance per metre length of the system\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 2.8;\n",
- "l = 1;# in m\n",
- "\n",
- "#calculation: \n",
- " #From Figure 40.9\n",
- "m = 16;# number of parallel squares measured along each equipotential\n",
- "n = 6;# the number of series squares measured along each line of force\n",
- "C = e0*er*l*m/n\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E12,2),\"pFarad.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 66.08 pFarad."
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 725</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitance of a 100 m length of the cable.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 3.4;\n",
- "l = 100;# in m\n",
- "\n",
- " #calculation: \n",
- " #From Figure 40.10\n",
- "m = 13;# number of parallel squares measured along each equipotential\n",
- "n = 4;# the number of series squares measured along each line of force\n",
- "C = e0*er*l*m/n\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E9,2),\"nFarad.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 9.78 nFarad."
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 726</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitance per metre length of the cable\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 2.7;\n",
- "ri = 0.0005;# in m\n",
- "ro = 0.006;# in m\n",
- "\n",
- " #calculation: \n",
- " #capacitance C\n",
- "C = 2*math.pi*e0*er/(math.log(ro/ri))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E12,2),\"pFarad.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 60.42 pFarad."
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 727</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the internal diameter of the sheath.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 80E-12;# in Farads\n",
- "e0 = 8.85E-12; \n",
- "er = 3.5;\n",
- "d0 = 0.008;# in m\n",
- "\n",
- " #calculation: \n",
- " #internal diameter\n",
- "di = d0*(math.e**(2*math.pi*e0*er/C))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n internal diameter is \",round(di,2),\" m.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " internal diameter is 0.09 m."
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 728</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the capacitance per kilometre length of the cable, \n",
- "#(b) the dielectric stress at a radius of 30 mm, and\n",
- "#(c) the maximum and minimum values of dielectric stress.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 3.5;\n",
- "di = 0.08;# in m\n",
- "d0 = 0.032;# in m\n",
- "r = 0.03;# in m\n",
- "V = 40000;# in Volts\n",
- "\n",
- "#calculation: \n",
- " #capacitance C\n",
- "C = 2*math.pi*e0*er/(math.log(di/d0))\n",
- " #dielectric stress at radius r,\n",
- "E = V/(r*math.log(di/d0))\n",
- " #maximum dielectric stress,\n",
- "Emax = V/((d0/2)*(math.log((di/d0))))\n",
- " #minimum dielectric stress,\n",
- "Emin = V/((di/2)*(math.log((di/d0))))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E12,2),\"pF/km\"\n",
- "print \"\\n dielectric stress at radius r is \",round(E,2),\"V/m\"\n",
- "print \"\\n maximum dielectric stress, is \",round(Emax,2),\"V/m minimum dielectric stress \",round( Emin,2),\"V/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 212.4 pF/km\n",
- "\n",
- " dielectric stress at radius r is 1455142.22 V/m\n",
- "\n",
- " maximum dielectric stress, is 2728391.67 V/m minimum dielectric stress 1091356.67 V/m"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 729</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the core and inner sheath radii for the most economical cable,\n",
- "#(b) the capacitance per metre length, and (c) the charging current per kilometre run.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 3.5;\n",
- "V = 60000;# in Volts\n",
- "f = 50;# in Hz\n",
- "Em = 10E6;# in V/m\n",
- "\n",
- "\n",
- "#calculation: \n",
- " #core radius, a\n",
- "a = V/Em\n",
- " #internal sheath radius,\n",
- "b = a*math.e**1\n",
- " #capacitance\n",
- "C = 2*math.pi*e0*er/(math.log(b/a))\n",
- " #Charging current\n",
- "I = V*2*math.pi*f*C\n",
- " #charging current per kilometre\n",
- "Ipkm = I*1000\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n core radius is \",round(a*1000,2),\"mm and internal sheath radius \",round(b*1000,1),\"mm\"\n",
- "print \"\\n capacitance is \",round(C*1E12,0),\"pF/m\"\n",
- "print \"\\n the charging current per kilometre \",round(Ipkm,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " core radius is 6.0 mm and internal sheath radius 16.3 mm\n",
- "\n",
- " capacitance is 195.0 pF/m\n",
- "\n",
- " the charging current per kilometre 3.67 A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 730</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for a 1 km length of the cable (a) the capacitance, (b) the charging current and (c) the power loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 2.5;\n",
- "di = 0.08;# in m\n",
- "d0 = 0.025;# in m\n",
- "r = 1000;# in m\n",
- "V = 132000;# in Volts\n",
- "f = 50;# in Hz\n",
- "de = 3.5E-3;# rad.\n",
- "\n",
- " #calculation:\n",
- " #core radius, a\n",
- "a = d0/2\n",
- " #internal sheath radius,\n",
- "b = di/2\n",
- " #capacitance\n",
- "C = 2*math.pi*e0*er*1E3/(math.log(b/a))\n",
- " #Charging current\n",
- "I = V*2*math.pi*f*C\n",
- " #power loss\n",
- "P = (2*math.pi*f*C*math.tan(de))*V**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)capacitance for a 1 km length is \",round(C*1E6,2),\"uF\"\n",
- "print \"\\n (b)the charging current \",round(I,2),\"A/km\"\n",
- "print \"\\n (c)power loss \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)capacitance for a 1 km length is 0.12 uF\n",
- "\n",
- " (b)the charging current 4.96 A/km\n",
- "\n",
- " (c)power loss 2289.78 W"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 732</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitance of the cable per metre length by the method of curvilinear squares\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 3.2;\n",
- "di = 0.06;# in m\n",
- "d0 = 0.020;# in m\n",
- "\n",
- " #calculation:\n",
- " #core radius, a\n",
- "a = d0/2\n",
- " #internal sheath radius,\n",
- "b = di/2\n",
- " #capacitance\n",
- "C = 2*math.pi*e0*er/(math.log(b/a))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance per m of length is \",round(C*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance per m of length is 0.16 nF"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 736</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitance of the line if the total length is 200 m.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "D = 0.05;# in m\n",
- "d = 0.005;# in m\n",
- "l = 200;# in m\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = math.pi*e0*er/(math.log(D/(d/2)))\n",
- " #capacitance of a 200 m length\n",
- "C200 = C*l\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance of a 200 m length is \",round(C200*1E6,5),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance of a 200 m length is 0.00186 uF"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 736</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for a 1 km length of line, (a) the capacitance of the conductors, \n",
- "#(b) the value of charge carried by each conductor, and (c) the charging current\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "D = 1.2;# in m\n",
- "r = 0.004;# in m\n",
- "f = 50;# in Hz\n",
- "V = 15000;# in Volts\n",
- "l = 1000;# in m\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = math.pi*e0*er/(math.log(D/r))\n",
- " #capacitance of a 1 km length\n",
- "Cpkm = C*l\n",
- " #Charge Q\n",
- "Q = Cpkm*V\n",
- " #Charging current\n",
- "I = V*2*math.pi*f*Cpkm\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance per 1km length is \",round(Cpkm*1E9,2),\"nF\"\n",
- "print \"\\n Charge Q is \",round(Q*1E6,2),\"uC\"\n",
- "print \"\\n Charging current is \",round(I,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance per 1km length is 4.87 nF\n",
- "\n",
- " Charge Q is 73.12 uC\n",
- "\n",
- " Charging current is 0.02 A"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 737</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the maximum value required for the capacitance per metre length,\n",
- "#and (b) the maximum diameter of each conductor if their distance between centres is 1.25 m.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "I = 0.015;# in Amperes\n",
- "d = 1.25;# in m\n",
- "r = 800;# in m\n",
- "f = 50;# in Hz\n",
- "V = 10000;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = I/(2*math.pi*f*V)\n",
- " #required maximum value of capacitance\n",
- "Cmax = C/r\n",
- " #maximum diameter of each conductor\n",
- "D = 2*d/(math.e**(math.pi*e0*er/Cmax))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n required maximum value of capacitance is \",round(Cmax*1E12,2),\"pF/m\"\n",
- "print \"\\nthe maximum diameter of each conductor is \",round(D,2),\" m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " required maximum value of capacitance is 5.97 pF/m\n",
- "\n",
- "the maximum diameter of each conductor is 0.02 m"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 739</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the energy stored in a 10 nF capacitor when charged to 1 kV, and \n",
- "#the average power developed if this energy is dissipated in 10 \u03bcs\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "C = 10E-9;# in Farad\n",
- "V = 1000;# in Volts\n",
- "t = 10E-6;# in sec\n",
- "\n",
- " #calculation:\n",
- " #energy stored,Wf\n",
- "Wf = C*V*V/2\n",
- " #average power developed\n",
- "Pav = Wf/t\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the energy stored is \",Wf,\"J\"\n",
- "print \"\\nthe average power developed is \",Pav,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the energy stored is 0.005 J\n",
- "\n",
- "the average power developed is 500.0 W"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 739</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the voltage across the plates and (b) the capacitance of the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "Q = 5E-3;# in Coulomb\n",
- "W = 0.625;# in Joules\n",
- "\n",
- " #calculation:\n",
- " #voltage across the plates\n",
- "V = 2*W/Q\n",
- " #Capacitance C\n",
- "C = Q/V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage across the plates is \",V,\" V\"\n",
- "print \"\\n Capacitance C is \",C*1E6,\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage across the plates is 250.0 V\n",
- "\n",
- " Capacitance C is 20.0 uF"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 740</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the required thickness of the ceramic dielectric, \n",
- "#(b) the area of plate required if the relative permittivity of the ceramic is 10, and \n",
- "#(c) the maximum energy stored by the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 10;\n",
- "C = 0.01E-6;# in Farad\n",
- "E = 10E6;# in V/m\n",
- "V = 2500;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #thickness of ceramic dielectric,\n",
- "d = V/E\n",
- " #cross-sectional area of plate\n",
- "A = C*d/(e0*er)\n",
- " #Maximum energy stored,\n",
- "W = C*V*V/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n thickness of ceramic dielectric is \",d*1000,\"mm\"\n",
- "print \"\\n cross-sectional area of plate, is \",round(A,2),\"m2\"\n",
- "print \"\\n Maximum energy stored is \",round(W,3),\" J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " thickness of ceramic dielectric is 0.25 mm\n",
- "\n",
- " cross-sectional area of plate, is 0.03 m2\n",
- "\n",
- " Maximum energy stored is 0.031 J"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 740</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the energy stored per cubic metre of the dielectric.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 2.3;\n",
- "A = 0.02;# in m2\n",
- "C = 400E-12;# in Farad\n",
- "V = 100;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #energy stored per unit volume of dielectric,\n",
- "W = ((C*V)**2)/(2*e0*er*A**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n energy stored per unit volume of dielectric is \",round(W,2),\" J/m3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " energy stored per unit volume of dielectric is 0.1 J/m3"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 744</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the inductance of the cable per metre length.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "a = 0.001;# in m\n",
- "b = 0.004;# in m\n",
- "\n",
- " #calculation:\n",
- " #inductance L\n",
- "L = (u0*ur/(2*math.pi))*(0.25 + math.log(b/a))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n inductance L is \",round(L*1E6,2),\"uH/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " inductance L is 0.33 uH/m"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 744</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the diameter of the sheath.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "da = 0.010;# in m\n",
- "L = 4E-7;# in H/m\n",
- "\n",
- " #calculation:\n",
- " #diameter of the sheath\n",
- "db = da*(math.e**(L/(u0*ur/(2*math.pi))))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n diameter of the sheath is \",round(db,2),\" m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " diameter of the sheath is 0.07 m"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 745</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the cable (a) the inductance, assuming nonmagnetic materials, and \n",
- "#(b) the capacitance, assuming a dielectric of relative permittivity 3.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "e0 = 8.85E-12;\n",
- "er = 3;\n",
- "da = 0.010;# in m\n",
- "db = 0.025;# in m\n",
- "l = 7500;# in m\n",
- "\n",
- "#calculation:\n",
- " #inductance per metre length\n",
- "L = (u0*ur/(2*math.pi))*(0.25 + math.log(db/da))\n",
- " #Since the cable is 7500 m long,\n",
- "L7500 = L*7500\n",
- " #capacitance C\n",
- "C = 2*math.pi*e0*er/(math.log(db/da))\n",
- " #//Since the cable is 7500 m long,\n",
- "C7500 = C*7500\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ninductance is \",round(L7500*1000,2),\" mH\"\n",
- "print \"\\ncapCItance is \",round(C7500*1E6,2),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "inductance is 1.75 mH\n",
- "\n",
- "capCItance is 1.37 uF"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 748</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the inductance of the line per metre length ignoring internal linkages\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "e0 = 8.85E-12;\n",
- "er = 3;\n",
- "D = 1.2;# in m\n",
- "a = 0.008;# in m\n",
- "\n",
- " #calculation:\n",
- " #inductance per metre length\n",
- "L = (u0*ur/(math.pi))*(math.log(D/a))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ninductance is \",round(L*1E6,2),\"uH/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "inductance is 2.0 uH/m"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 748</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the loop inductance, and \n",
- "#(b) the capacitance of a 1 km length of single-phase twin line having conductors of diameter 10 mm and spaced 800 mm apart in air.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "e0 = 8.85E-12;\n",
- "er = 1;\n",
- "l = 1000;# in m\n",
- "D = 0.8;# in m\n",
- "a = 0.01/2;# in m\n",
- "\n",
- " #calculation:\n",
- " #inductance per metre length\n",
- "L = (u0*ur/(math.pi))*(0.25 + math.log(D/a))\n",
- " #Since the cable is 1000 m long,\n",
- "L1k = L*l\n",
- " #capacitance C\n",
- "C = math.pi*e0*er/(math.log(D/a))\n",
- " #//Since the cable is 1000 m long,\n",
- "C1k = C*l\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ninductance is \",round(L1k*1000,2),\" mH\"\n",
- "print \"\\ncapcitance is \",round(C1k*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "inductance is 2.13 mH\n",
- "\n",
- "capcitance is 5.48 nF\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 749</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the distance between their centres.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 2.185E-6;# in H/m\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "a = 0.012/2;# in m\n",
- "\n",
- " #calculation:\n",
- " #distance D\n",
- "D = a*math.e**((L*math.pi)/(u0*ur) - 0.25)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ndistance D is \",round(D,2),\" m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "distance D is 1.1 m"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 752</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What value of current would double the energy stored?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.2;# in H\n",
- "I = 0.05;# in Amperes\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "\n",
- "#calculation:\n",
- " #energy stored in inductor\n",
- "W = L*I*I/2\n",
- " #current I\n",
- "I = (2*2*W/L)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nenergy stored in inductor is \",round(W*1000,2),\"mJ\"\n",
- "print \"\\ncurrent I is \",round(I,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "energy stored in inductor is 0.25 mJ\n",
- "\n",
- "current I is 0.07 A"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 752</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the total energy stored in the magnetic field of the airgap.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "B = 0.05;# in Tesla\n",
- "A = 500E-6;# in m2\n",
- "l = 0.002;# in m\n",
- "u0 = 4*math.pi*1E-7; \n",
- "\n",
- "#calculation:\n",
- " #energy stored\n",
- "W = (B**2)/(2*u0)\n",
- " #Volume of airgap\n",
- "v = A*l\n",
- " #energy stored in airgap\n",
- "W = W*v\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nenergy stored in the airgap is \",round(W*1E6,2),\"uJ\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "energy stored in the airgap is 994.72 uJ"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 752</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the strength of a uniform electric fi\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "B = 0.8;# in Tesla\n",
- "A = 500E-6;# in m2\n",
- "l = 0.002;# in m\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "e0 = 8.85E-12;\n",
- "er = 1;\n",
- "\n",
- "#calculation:\n",
- " #energy stored in mag. field\n",
- "W = (B**2)/(2*u0)\n",
- " #electric field\n",
- "E = (2*W/(e0*er))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nelectric field strength is \",round(E/1E6,2),\"MV/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "electric field strength is 239.89 MV/m"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_2.ipynb deleted file mode 100755 index 3f3b710b..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_2.ipynb +++ /dev/null @@ -1,1328 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 40: Field theory</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 725</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitance per metre length of the system\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 2.8;\n",
- "l = 1;# in m\n",
- "\n",
- "#calculation: \n",
- " #From Figure 40.9\n",
- "m = 16;# number of parallel squares measured along each equipotential\n",
- "n = 6;# the number of series squares measured along each line of force\n",
- "C = e0*er*l*m/n\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E12,2),\"pFarad.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 66.08 pFarad."
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 725</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitance of a 100 m length of the cable.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 3.4;\n",
- "l = 100;# in m\n",
- "\n",
- " #calculation: \n",
- " #From Figure 40.10\n",
- "m = 13;# number of parallel squares measured along each equipotential\n",
- "n = 4;# the number of series squares measured along each line of force\n",
- "C = e0*er*l*m/n\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E9,2),\"nFarad.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 9.78 nFarad."
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 726</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitance per metre length of the cable\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 2.7;\n",
- "ri = 0.0005;# in m\n",
- "ro = 0.006;# in m\n",
- "\n",
- " #calculation: \n",
- " #capacitance C\n",
- "C = 2*math.pi*e0*er/(math.log(ro/ri))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E12,2),\"pFarad.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 60.42 pFarad."
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 727</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the internal diameter of the sheath.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "C = 80E-12;# in Farads\n",
- "e0 = 8.85E-12; \n",
- "er = 3.5;\n",
- "d0 = 0.008;# in m\n",
- "\n",
- " #calculation: \n",
- " #internal diameter\n",
- "di = d0*(math.e**(2*math.pi*e0*er/C))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n internal diameter is \",round(di,2),\" m.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " internal diameter is 0.09 m."
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 728</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the capacitance per kilometre length of the cable, \n",
- "#(b) the dielectric stress at a radius of 30 mm, and\n",
- "#(c) the maximum and minimum values of dielectric stress.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 3.5;\n",
- "di = 0.08;# in m\n",
- "d0 = 0.032;# in m\n",
- "r = 0.03;# in m\n",
- "V = 40000;# in Volts\n",
- "\n",
- "#calculation: \n",
- " #capacitance C\n",
- "C = 2*math.pi*e0*er/(math.log(di/d0))\n",
- " #dielectric stress at radius r,\n",
- "E = V/(r*math.log(di/d0))\n",
- " #maximum dielectric stress,\n",
- "Emax = V/((d0/2)*(math.log((di/d0))))\n",
- " #minimum dielectric stress,\n",
- "Emin = V/((di/2)*(math.log((di/d0))))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance is \",round(C*1E12,2),\"pF/km\"\n",
- "print \"\\n dielectric stress at radius r is \",round(E,2),\"V/m\"\n",
- "print \"\\n maximum dielectric stress, is \",round(Emax,2),\"V/m minimum dielectric stress \",round( Emin,2),\"V/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance is 212.4 pF/km\n",
- "\n",
- " dielectric stress at radius r is 1455142.22 V/m\n",
- "\n",
- " maximum dielectric stress, is 2728391.67 V/m minimum dielectric stress 1091356.67 V/m"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 729</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the core and inner sheath radii for the most economical cable,\n",
- "#(b) the capacitance per metre length, and (c) the charging current per kilometre run.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 3.5;\n",
- "V = 60000;# in Volts\n",
- "f = 50;# in Hz\n",
- "Em = 10E6;# in V/m\n",
- "\n",
- "\n",
- "#calculation: \n",
- " #core radius, a\n",
- "a = V/Em\n",
- " #internal sheath radius,\n",
- "b = a*math.e**1\n",
- " #capacitance\n",
- "C = 2*math.pi*e0*er/(math.log(b/a))\n",
- " #Charging current\n",
- "I = V*2*math.pi*f*C\n",
- " #charging current per kilometre\n",
- "Ipkm = I*1000\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n core radius is \",round(a*1000,2),\"mm and internal sheath radius \",round(b*1000,1),\"mm\"\n",
- "print \"\\n capacitance is \",round(C*1E12,0),\"pF/m\"\n",
- "print \"\\n the charging current per kilometre \",round(Ipkm,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " core radius is 6.0 mm and internal sheath radius 16.3 mm\n",
- "\n",
- " capacitance is 195.0 pF/m\n",
- "\n",
- " the charging current per kilometre 3.67 A"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 730</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for a 1 km length of the cable (a) the capacitance, (b) the charging current and (c) the power loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 2.5;\n",
- "di = 0.08;# in m\n",
- "d0 = 0.025;# in m\n",
- "r = 1000;# in m\n",
- "V = 132000;# in Volts\n",
- "f = 50;# in Hz\n",
- "de = 3.5E-3;# rad.\n",
- "\n",
- " #calculation:\n",
- " #core radius, a\n",
- "a = d0/2\n",
- " #internal sheath radius,\n",
- "b = di/2\n",
- " #capacitance\n",
- "C = 2*math.pi*e0*er*1E3/(math.log(b/a))\n",
- " #Charging current\n",
- "I = V*2*math.pi*f*C\n",
- " #power loss\n",
- "P = (2*math.pi*f*C*math.tan(de))*V**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)capacitance for a 1 km length is \",round(C*1E6,2),\"uF\"\n",
- "print \"\\n (b)the charging current \",round(I,2),\"A/km\"\n",
- "print \"\\n (c)power loss \",round(P,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)capacitance for a 1 km length is 0.12 uF\n",
- "\n",
- " (b)the charging current 4.96 A/km\n",
- "\n",
- " (c)power loss 2289.78 W"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 732</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitance of the cable per metre length by the method of curvilinear squares\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 3.2;\n",
- "di = 0.06;# in m\n",
- "d0 = 0.020;# in m\n",
- "\n",
- " #calculation:\n",
- " #core radius, a\n",
- "a = d0/2\n",
- " #internal sheath radius,\n",
- "b = di/2\n",
- " #capacitance\n",
- "C = 2*math.pi*e0*er/(math.log(b/a))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance per m of length is \",round(C*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance per m of length is 0.16 nF"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 736</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitance of the line if the total length is 200 m.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "D = 0.05;# in m\n",
- "d = 0.005;# in m\n",
- "l = 200;# in m\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = math.pi*e0*er/(math.log(D/(d/2)))\n",
- " #capacitance of a 200 m length\n",
- "C200 = C*l\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance of a 200 m length is \",round(C200*1E6,5),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance of a 200 m length is 0.00186 uF"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 736</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for a 1 km length of line, (a) the capacitance of the conductors, \n",
- "#(b) the value of charge carried by each conductor, and (c) the charging current\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "D = 1.2;# in m\n",
- "r = 0.004;# in m\n",
- "f = 50;# in Hz\n",
- "V = 15000;# in Volts\n",
- "l = 1000;# in m\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = math.pi*e0*er/(math.log(D/r))\n",
- " #capacitance of a 1 km length\n",
- "Cpkm = C*l\n",
- " #Charge Q\n",
- "Q = Cpkm*V\n",
- " #Charging current\n",
- "I = V*2*math.pi*f*Cpkm\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance per 1km length is \",round(Cpkm*1E9,2),\"nF\"\n",
- "print \"\\n Charge Q is \",round(Q*1E6,2),\"uC\"\n",
- "print \"\\n Charging current is \",round(I,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance per 1km length is 4.87 nF\n",
- "\n",
- " Charge Q is 73.12 uC\n",
- "\n",
- " Charging current is 0.02 A"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 737</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the maximum value required for the capacitance per metre length,\n",
- "#and (b) the maximum diameter of each conductor if their distance between centres is 1.25 m.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "I = 0.015;# in Amperes\n",
- "d = 1.25;# in m\n",
- "r = 800;# in m\n",
- "f = 50;# in Hz\n",
- "V = 10000;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = I/(2*math.pi*f*V)\n",
- " #required maximum value of capacitance\n",
- "Cmax = C/r\n",
- " #maximum diameter of each conductor\n",
- "D = 2*d/(math.e**(math.pi*e0*er/Cmax))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n required maximum value of capacitance is \",round(Cmax*1E12,2),\"pF/m\"\n",
- "print \"\\nthe maximum diameter of each conductor is \",round(D,2),\" m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " required maximum value of capacitance is 5.97 pF/m\n",
- "\n",
- "the maximum diameter of each conductor is 0.02 m"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 739</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the energy stored in a 10 nF capacitor when charged to 1 kV, and \n",
- "#the average power developed if this energy is dissipated in 10 \u03bcs\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "C = 10E-9;# in Farad\n",
- "V = 1000;# in Volts\n",
- "t = 10E-6;# in sec\n",
- "\n",
- " #calculation:\n",
- " #energy stored,Wf\n",
- "Wf = C*V*V/2\n",
- " #average power developed\n",
- "Pav = Wf/t\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the energy stored is \",Wf,\"J\"\n",
- "print \"\\nthe average power developed is \",Pav,\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the energy stored is 0.005 J\n",
- "\n",
- "the average power developed is 500.0 W"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 739</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the voltage across the plates and (b) the capacitance of the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 1;\n",
- "Q = 5E-3;# in Coulomb\n",
- "W = 0.625;# in Joules\n",
- "\n",
- " #calculation:\n",
- " #voltage across the plates\n",
- "V = 2*W/Q\n",
- " #Capacitance C\n",
- "C = Q/V\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n voltage across the plates is \",V,\" V\"\n",
- "print \"\\n Capacitance C is \",C*1E6,\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " voltage across the plates is 250.0 V\n",
- "\n",
- " Capacitance C is 20.0 uF"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 740</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the required thickness of the ceramic dielectric, \n",
- "#(b) the area of plate required if the relative permittivity of the ceramic is 10, and \n",
- "#(c) the maximum energy stored by the capacitor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 10;\n",
- "C = 0.01E-6;# in Farad\n",
- "E = 10E6;# in V/m\n",
- "V = 2500;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #thickness of ceramic dielectric,\n",
- "d = V/E\n",
- " #cross-sectional area of plate\n",
- "A = C*d/(e0*er)\n",
- " #Maximum energy stored,\n",
- "W = C*V*V/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n thickness of ceramic dielectric is \",d*1000,\"mm\"\n",
- "print \"\\n cross-sectional area of plate, is \",round(A,2),\"m2\"\n",
- "print \"\\n Maximum energy stored is \",round(W,3),\" J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " thickness of ceramic dielectric is 0.25 mm\n",
- "\n",
- " cross-sectional area of plate, is 0.03 m2\n",
- "\n",
- " Maximum energy stored is 0.031 J"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 740</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the energy stored per cubic metre of the dielectric.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "e0 = 8.85E-12; \n",
- "er = 2.3;\n",
- "A = 0.02;# in m2\n",
- "C = 400E-12;# in Farad\n",
- "V = 100;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #energy stored per unit volume of dielectric,\n",
- "W = ((C*V)**2)/(2*e0*er*A**2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n energy stored per unit volume of dielectric is \",round(W,2),\" J/m3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " energy stored per unit volume of dielectric is 0.1 J/m3"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 744</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the inductance of the cable per metre length.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "a = 0.001;# in m\n",
- "b = 0.004;# in m\n",
- "\n",
- " #calculation:\n",
- " #inductance L\n",
- "L = (u0*ur/(2*math.pi))*(0.25 + math.log(b/a))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n inductance L is \",round(L*1E6,2),\"uH/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " inductance L is 0.33 uH/m"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 744</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the diameter of the sheath.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "da = 0.010;# in m\n",
- "L = 4E-7;# in H/m\n",
- "\n",
- " #calculation:\n",
- " #diameter of the sheath\n",
- "db = da*(math.e**(L/(u0*ur/(2*math.pi))))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n diameter of the sheath is \",round(db,2),\" m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " diameter of the sheath is 0.07 m"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 745</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the cable (a) the inductance, assuming nonmagnetic materials, and \n",
- "#(b) the capacitance, assuming a dielectric of relative permittivity 3.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "e0 = 8.85E-12;\n",
- "er = 3;\n",
- "da = 0.010;# in m\n",
- "db = 0.025;# in m\n",
- "l = 7500;# in m\n",
- "\n",
- "#calculation:\n",
- " #inductance per metre length\n",
- "L = (u0*ur/(2*math.pi))*(0.25 + math.log(db/da))\n",
- " #Since the cable is 7500 m long,\n",
- "L7500 = L*7500\n",
- " #capacitance C\n",
- "C = 2*math.pi*e0*er/(math.log(db/da))\n",
- " #//Since the cable is 7500 m long,\n",
- "C7500 = C*7500\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ninductance is \",round(L7500*1000,2),\" mH\"\n",
- "print \"\\ncapCItance is \",round(C7500*1E6,2),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "inductance is 1.75 mH\n",
- "\n",
- "capCItance is 1.37 uF"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 748</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the inductance of the line per metre length ignoring internal linkages\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "e0 = 8.85E-12;\n",
- "er = 3;\n",
- "D = 1.2;# in m\n",
- "a = 0.008;# in m\n",
- "\n",
- " #calculation:\n",
- " #inductance per metre length\n",
- "L = (u0*ur/(math.pi))*(math.log(D/a))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ninductance is \",round(L*1E6,2),\"uH/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "inductance is 2.0 uH/m"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 748</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the loop inductance, and \n",
- "#(b) the capacitance of a 1 km length of single-phase twin line having conductors of diameter 10 mm and spaced 800 mm apart in air.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "e0 = 8.85E-12;\n",
- "er = 1;\n",
- "l = 1000;# in m\n",
- "D = 0.8;# in m\n",
- "a = 0.01/2;# in m\n",
- "\n",
- " #calculation:\n",
- " #inductance per metre length\n",
- "L = (u0*ur/(math.pi))*(0.25 + math.log(D/a))\n",
- " #Since the cable is 1000 m long,\n",
- "L1k = L*l\n",
- " #capacitance C\n",
- "C = math.pi*e0*er/(math.log(D/a))\n",
- " #//Since the cable is 1000 m long,\n",
- "C1k = C*l\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ninductance is \",round(L1k*1000,2),\" mH\"\n",
- "print \"\\ncapcitance is \",round(C1k*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "inductance is 2.13 mH\n",
- "\n",
- "capcitance is 5.48 nF\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 749</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the distance between their centres.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 2.185E-6;# in H/m\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "a = 0.012/2;# in m\n",
- "\n",
- " #calculation:\n",
- " #distance D\n",
- "D = a*math.e**((L*math.pi)/(u0*ur) - 0.25)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ndistance D is \",round(D,2),\" m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "distance D is 1.1 m"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 752</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What value of current would double the energy stored?\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 0.2;# in H\n",
- "I = 0.05;# in Amperes\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "\n",
- "#calculation:\n",
- " #energy stored in inductor\n",
- "W = L*I*I/2\n",
- " #current I\n",
- "I = (2*2*W/L)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nenergy stored in inductor is \",round(W*1000,2),\"mJ\"\n",
- "print \"\\ncurrent I is \",round(I,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "energy stored in inductor is 0.25 mJ\n",
- "\n",
- "current I is 0.07 A"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 752</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the total energy stored in the magnetic field of the airgap.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "B = 0.05;# in Tesla\n",
- "A = 500E-6;# in m2\n",
- "l = 0.002;# in m\n",
- "u0 = 4*math.pi*1E-7; \n",
- "\n",
- "#calculation:\n",
- " #energy stored\n",
- "W = (B**2)/(2*u0)\n",
- " #Volume of airgap\n",
- "v = A*l\n",
- " #energy stored in airgap\n",
- "W = W*v\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nenergy stored in the airgap is \",round(W*1E6,2),\"uJ\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "energy stored in the airgap is 994.72 uJ"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 752</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the strength of a uniform electric fi\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "B = 0.8;# in Tesla\n",
- "A = 500E-6;# in m2\n",
- "l = 0.002;# in m\n",
- "u0 = 4*math.pi*1E-7; \n",
- "ur = 1;\n",
- "e0 = 8.85E-12;\n",
- "er = 1;\n",
- "\n",
- "#calculation:\n",
- " #energy stored in mag. field\n",
- "W = (B**2)/(2*u0)\n",
- " #electric field\n",
- "E = (2*W/(e0*er))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nelectric field strength is \",round(E/1E6,2),\"MV/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "electric field strength is 239.89 MV/m"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_40.ipynb index 181b437b..181b437b 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_40.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint.ipynb deleted file mode 100755 index 788f6f5e..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint.ipynb +++ /dev/null @@ -1,1005 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 41: Attenuators</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 763</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the power ratio in each case (i) in decibels and (ii) in nepers.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- " #ratio of output power to input power\n",
- "rp1 = 2;\n",
- "rp2 = 25; \n",
- "rp3 = 1000;\n",
- "rp4 = 0.01;\n",
- "\n",
- "#calculation:\n",
- " #power ratio in decibels\n",
- "rpd1 = 10*(1/2.303)*math.log(rp1)\n",
- "rpd2 = 10*(1/2.303)*math.log(rp2)\n",
- "rpd3 = 10*(1/2.303)*math.log(rp3)\n",
- "rpd4 = 10*(1/2.303)*math.log(rp4)\n",
- " #power ratio in nepers\n",
- "rpn1 = (math.log(rp1))/2\n",
- "rpn2 = (math.log(rp2))/2\n",
- "rpn3 = (math.log(rp3))/2\n",
- "rpn4 = (math.log(rp4))/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n power ratio in decibels are (a)\",round(rpd1,2),\" dB (b)\",round(rpd2,2),\" dB \"\n",
- "print \"(c) \",round(rpd3,2),\" dB and (d) \",round(rpd4,2),\" dB\"\n",
- "print \"\\n power ratio in nepers are (a)\",round(rpn1,2),\" Np (b)\",round(rpn2,2),\" Np\"\n",
- "print \"(c) \",round(rpn3,2),\" Np and (d) \",round(rpn4,2),\" Np\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " power ratio in decibels are (a) 3.01 dB (b) 13.98 dB \n",
- "(c) 29.99 dB and (d) -20.0 dB\n",
- "\n",
- " power ratio in nepers are (a) 0.35 Np (b) 1.61 Np\n",
- "(c) 3.45 Np and (d) -2.3 Np\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 763</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the attenuation in decibels\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rp = 0.05;# power ratio P2/P1\n",
- "\n",
- " #calculation:\n",
- " #power ratio in decibels\n",
- "rpd = 10*(1/2.303)*math.log(rp)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nthe attenuation is \",round(abs(rpd),2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "the attenuation is 13.01 dB"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 764</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the output power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "gain = 1.5;# in dB\n",
- "Pi = 0.012;# in Watt\n",
- "\n",
- "#calculation:\n",
- " #output power\n",
- "Po = Pi*10**gain\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\noutput power is \",round(Po,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "output power is 0.38 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 764</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current input and (b) the current ratio expressed in decibels.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "I2 = 0.05;# in Amperes\n",
- "rin = 1.32;# in Np\n",
- "\n",
- " #calculation:\n",
- " #current input, I1\n",
- "I1 = I2*math.e**(rin)\n",
- " #current ratio in decibels\n",
- "rid = 20*(1/2.303)*math.log(I2/I1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ncurrent input, I1 is \",round(I1,2),\" A\"\n",
- "print \"\\ncurrent ratio in decibels is \",round(rid,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "current input, I1 is 0.19 A\n",
- "\n",
- "current ratio in decibels is -11.46 dB"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 769</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the characteristic impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ra1 = 8; # in ohms\n",
- "Ra2 = 8; # in ohms\n",
- "Ra3 = 21; # in ohms\n",
- "Rb1 = 10; # in ohms\n",
- "Rb2 = 10; # in ohms\n",
- "Rb3 = 15; # in ohms\n",
- "Rc1 = 200; # in ohms\n",
- "Rc2 = 200; # in ohms\n",
- "Rc3 = 56.25; # in ohms\n",
- "\n",
- "#calculation:\n",
- "R01 = (Ra1**2 + 2*Ra2*Ra3)**0.5\n",
- "R02 = (Rb1**2 + 2*Rb2*Rb3)**0.5\n",
- "R03 = (Rc1**2 + 2*Rc2*Rc3)**0.5\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) the characteristic impedance, R0 is \",R01,\" ohm\"\n",
- "print \"\\n(b) the characteristic impedance, R0 is \",R02,\" ohm\"\n",
- "print \"\\n(c) the characteristic impedance, R0 is \",R03,\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) the characteristic impedance, R0 is 20.0 ohm\n",
- "\n",
- "(b) the characteristic impedance, R0 is 20.0 ohm\n",
- "\n",
- "(c) the characteristic impedance, R0 is 250.0 ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 769</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the characteristic impedance, and (b) the attenuation (in dB) produced by the pad\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 500;# in ohm\n",
- "R2 = 1000;# in ohm\n",
- "I1 = 1;# in ampere (lets say)\n",
- "\n",
- "#calculation:\n",
- " # for symmetrical pi-attenuator section\n",
- " #characteristic impedance, R0\n",
- "R0 = (R1*(R2**2)/(R1 + 2*R2))**0.5\n",
- " #current Ix\n",
- "Ix = (R2/(R2 + R1 + (R2*R0/(R2 + R0))))*I1\n",
- " #current I2\n",
- "I2 = (R2/(R2 + R0))*Ix\n",
- "ri = I1/I2;# retio of currents\n",
- " #attenuation\n",
- "attn = 20*(1/2.303)*math.log(ri)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the characteristic impedance is \",round(R0,2),\" ohm\"\n",
- "print \"\\n attenuation is \",round(attn,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the characteristic impedance is 447.21 ohm\n",
- "\n",
- " attenuation is 8.36 dB"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 770</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the input resistance when the output port is open-circuited, \n",
- "#(b) the input resistance when the output port is short-circuited, and (c) the characteristic impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ra1 = 15;# in ohm\n",
- "Ra2 = 15;# in ohm\n",
- "Ra3 = 10;# in ohm\n",
- "Rb1 = 15;# in ohm\n",
- "Rb2 = 5;# in ohm\n",
- "Rb3 = 5;# in ohm\n",
- "\n",
- "#calculation:\n",
- "Roc1 = Ra1 + Ra3\n",
- "Rsc1 = Ra1 + Ra2*Ra3/(Ra2+Ra3)\n",
- "R01 = (Roc1*Rsc1)**0.5\n",
- "\n",
- "Roc2 = Rb2* (Rb1 + Rb3)/(Rb1 + Rb2 + Rb3)\n",
- "Rsc2 = Rb2*Rb1/(Rb2+Rb1)\n",
- "R02 = (Roc2*Rsc2)**0.5\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the input resistance when the output port is open-circuited is\", Roc1,\" ohm for T-Network\"\n",
- "print \"and \",Roc2,\" ohm for pi-Network \"\n",
- "print \"\\n (b) the input resistance when the output port is short-circuited is,\", Rsc1,\" ohm for T-Network\"\n",
- "print \"and \",Rsc2,\" ohm for pi-Network \"\n",
- "print \"\\n (c) the characteristic impedance. is,\",round(R01,1),\" ohm for T-Network and ,\",round(R02,2),\" ohm for pi-Network \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the input resistance when the output port is open-circuited is 25 ohm for T-Network\n",
- "and 4.0 ohm for pi-Network \n",
- "\n",
- " (b) the input resistance when the output port is short-circuited is, 21.0 ohm for T-Network\n",
- "and 3.75 ohm for pi-Network \n",
- "\n",
- " (c) the characteristic impedance. is, 22.9 ohm for T-Network and , 3.87 ohm for pi-Network \n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 770</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design a T-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n",
- "#having a characteristic impedance of 600 ohm.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vat = 20; # in db\n",
- "R0 = 600;# in ohm\n",
- "\n",
- "#calculation:\n",
- "N = math.e**(Vat*2.3/20)\n",
- "R1 = R0*(N-1)/(N+1)\n",
- "R2 = R0*2*N/(N**2 - 1)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,0),\" ohm and Resistance R2 is\",round(R2,1),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For a T-section symmetrical attenuator pad, Resistance R1 is 491.0 ohm and Resistance R2 is 121.5 ohm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 771</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design a pi-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n",
- "#having a characteristic impedance of 600 ohm.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vat = 20; # in db\n",
- "R0 = 600;# in ohm\n",
- "\n",
- "#calculation:\n",
- "N = math.e**(Vat*2.303/20)\n",
- "R1 = R0*(N**2 - 1)/(2*N)\n",
- "R2 = R0*(N+1)/(N-1)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For a pi-section symmetrical attenuator pad, Resistance R1 is\",round(R1/1000,2),\" Kohm\"\n",
- "print \"and Resistance R2 is\",round(R2,0),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For a pi-section symmetrical attenuator pad, Resistance R1 is 2.97 Kohm\n",
- "and Resistance R2 is 733.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 772</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the characteristic impedance R0, and (b) the insertion loss in decibels\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 300;# in ohm\n",
- "R2 = 450;# in ohm\n",
- "I1 = 1;# in ampere (lets say)\n",
- "\n",
- "#calculation:\n",
- " #the characteristic impedance of a symmetric T-pad attenuator is given by\n",
- "R0 = (R1**2 + 2*R1*R2)**0.5\n",
- " #By current division\n",
- " #current I2\n",
- "I2 = (R2/(R2 + R1+ R0))*I1\n",
- "ri = I1/I2;# ratio of currents\n",
- " #insertion loss\n",
- "il = 20*(1/2.303)*math.log(ri)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the characteristic impedance is \",R0,\" ohm\"\n",
- "print \"\\n insertion loss is \",round(il,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the characteristic impedance is 600.0 ohm\n",
- "\n",
- " insertion loss is 9.54 dB"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 773</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the insertion loss at a tapping of (a) 2 kohm, (b) 1 kohm.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r = 500;# in ohm\n",
- "Rhm = 3000;# in ohm\n",
- "RL = 2000;# in ohm\n",
- "r1 = 2000;# in ohm\n",
- "r2 = 1000;# in ohm\n",
- "E = 1;# in volts (lets say)\n",
- "\n",
- " #calculation:\n",
- " #Without the rheostat in the circuit the voltage across the 2 kohm\u0018 load, VL\n",
- "VL = (RL/(RL + r))*E\n",
- " #voltage V2 with 2kohm tapping\n",
- "V2 = ((RL*r1/(r1 + RL))/((RL*r1/(r1 + RL)) + Rhm - r1 + r))*E\n",
- "rv1 = VL/V2;# ratio of currents\n",
- " #insertion loss \n",
- "il1 = 20*(1/2.303)*math.log(rv1)\n",
- " #voltage V1 with 1kohm tapping\n",
- "V1 = ((RL*r2/(r2 + RL))/((RL*r2/(r2 + RL)) + Rhm - r2 + r))*E\n",
- "rv2 = VL/V1;# ratio of currents\n",
- " #insertion loss \n",
- "il2 = 20*(1/2.303)*math.log(rv2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n insertion loss for 2kohm tap is \",round(il1,2),\" dB\"\n",
- "print \"\\n insertion loss for 1kohm tap is \",round(il2,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " insertion loss for 2kohm tap is 6.02 dB\n",
- "\n",
- " insertion loss for 1kohm tap is 11.59 dB"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 774</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) its characteristic impedance, and (b) the insertion loss (in decibels) when feeding a matched load.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 1000;# in ohm\n",
- "R2 = 500;# in ohm\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- "#calculation:\n",
- " #characteristic impedance of a symmetrical attenuator\n",
- "R0 = (R1*(R2**2)/(R1 + 2*R2))**0.5\n",
- " #current Ix\n",
- "Ix = (R2/(R2 + R1 + (R2*R0/(R2 + R0))))*I1\n",
- " #current I2\n",
- "I2 = (R2/(R2 + R0))*Ix\n",
- "ri = I1/I2;# retio of currents\n",
- " #insertion loss \n",
- "il = 20*(1/2.303)*math.log(ri)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n characteristic impedance is \",round(R0,2),\" ohm\"\n",
- "print \"\\n insertion loss is \",round(il,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " characteristic impedance is 353.55 ohm\n",
- "\n",
- " insertion loss is 15.31 dB"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 776</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 100;# in ohm\n",
- "R2 = 200;# in ohm\n",
- "R3 = 300;# in ohm\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- "#calculation:\n",
- " #image impedance Roa\n",
- "Roa = ((R1 + R2)*(R2 + (R1*R3/(R1 + R3))))**0.5\n",
- " #image impedance Rob\n",
- "Rob = ((R1 + R3)*(R3 + (R1*R2/(R1 + R2))))**0.5\n",
- " #The iterative impedance at port 1\n",
- "Ri1 = (-1*R1 + (R1**2 - (-1*4*((R2*(R1 + R3)) + (R3*R1))))**0.5)/2\n",
- " #The iterative impedance at port 2\n",
- "Ri2 = (R1 + (R1**2 - (-1*4*((R3*(R1 + R2)) + (R2*R1))))**0.5)/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n image impedance are \",round(Roa,2),\" ohm and \",round(Rob,2),\" ohm \"\n",
- "print \"\\n iterative impedances are \",round(Ri1,2),\" ohm and \",round(Ri2,2),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " image impedance are 287.23 ohm and 382.97 ohm \n",
- "\n",
- " iterative impedances are 285.41 ohm and 385.41 ohm "
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 777</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 1000;# in ohm\n",
- "R2 = 2000;# in ohm\n",
- "R3 = 3000;# in ohm\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- "#calculation:\n",
- " #image impedance Roa\n",
- "Roa = (((R3 + R2)*R1/(R1 + R2 + R3))*(R1*R3/(R1 + R3)))**0.5\n",
- " #image impedance Rob\n",
- "Rob = (((R3 + R1)*R2/(R1 + R2 + R3))*(R2*R3/(R2 + R3)))**0.5\n",
- " #The iterative impedance at port 1\n",
- "Ri1 = (-1*R1 + ((R1**2) - (-1*4*2*R2*R1))**0.5)/(2*2)\n",
- " #The iterative impedance at port 2\n",
- "Ri2 = (R1 + ((-1*R1)**2 - (-1*4*2*R2*R1))**0.5)/(2*2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n image impedance are \",round(Roa,2),\" ohm and \",round(Rob,2),\" ohm \"\n",
- "print \"\\n iterative impedances are \",round(Ri1,2),\" ohm and \",round(Ri2,2),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " image impedance are 790.57 ohm and 1264.91 ohm \n",
- "\n",
- " iterative impedances are 780.78 ohm and 1280.78 ohm "
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 780</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the values of resistance R1 and R2, (b) the attenuation of the pad in decibels, and (c) its insertion loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r = 500;# in ohm\n",
- "RL = 100;# in ohm\n",
- "E = 1;# in volts (lets say)\n",
- "\n",
- "#calculation:\n",
- " #res.\n",
- "R1 = (r*(r - RL))**0.5\n",
- "R2 = (r*RL**2/(r - RL))**0.5\n",
- " #current I1\n",
- "I1 = E/(r + R1 + R2*RL/(RL + R2))\n",
- " #current I2\n",
- "I2 = (R2/(R2 + RL))*I1\n",
- " #input power\n",
- "P1 = r*I1**2\n",
- " #output power\n",
- "P2 = RL*I2**2\n",
- " #attenuation\n",
- "attn = 10*(1/2.303)*math.log(P1/P2)\n",
- " #Load current, IL\n",
- "IL = E/(r + RL)\n",
- " #voltage, VL\n",
- "VL = IL*RL\n",
- " #voltage, V1\n",
- "V1 = E - I1*r\n",
- " #voltage, V2\n",
- "V2 = V1 - I1*R1\n",
- " #insertion loss\n",
- "il = 20*(1/2.303)*math.log(VL/V2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n R1 = \",round(R1,2),\" ohm and R2 = \",round(R2,2),\" ohm \"\n",
- "print \"\\n attenuation is \",round(attn,2),\" dB \"\n",
- "print \"\\n In decibels, the insertion loss is \",round(il,2),\" dB \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " R1 = 447.21 ohm and R2 = 111.8 ohm \n",
- "\n",
- " attenuation is 12.54 dB \n",
- "\n",
- " In decibels, the insertion loss is 9.98 dB "
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 783</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the attenuation of each individual attenuation section, \n",
- "#(b) the voltage output of the final stage, and (c) the voltage output of the third stage.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "attnO = 70;# in dB\n",
- "n = 5;# numbers of identical atteneurs\n",
- "V1 = 0.02;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #attenuation of each section\n",
- "attn = attnO/n\n",
- " #output of the final stage\n",
- "Vo = V1/(10**(attnO/20))\n",
- " #voltage output of the third stage\n",
- "V3 = V1/(10**(3*attn/20))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n attenuation of each section = \",round(attn,2),\" dB \"\n",
- "print \"\\n output of the final stage is \",round(Vo*1E6,2),\"uV \"\n",
- "print \"\\n voltage output of the third stage is \",round(V3*1E3,3),\"mV \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " attenuation of each section = 14.0 dB \n",
- "\n",
- " output of the final stage is 6.32 uV \n",
- "\n",
- " voltage output of the third stage is 0.159 mV "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 784</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)Design a T-network attenuator pad\n",
- "#(b)determine the fraction of the initial current \n",
- "#(c) Determine the attenuation in decibels given by four such sections\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r = 450; # in ohm\n",
- "R0 = 450; # in ohms\n",
- "x = 1/8\n",
- "\n",
- "#calculation:\n",
- "N = 1/x\n",
- "R1 = R0*(N-1)/(N+1)\n",
- "R2 = R0*2*N/(N**2 - 1)\n",
- "\n",
- "Io = x*x\n",
- "\n",
- "attn = 20*math.log10(N)\n",
- "attnO = 4*attn\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,2),\" ohm \"\n",
- "print \"and Resistance R2 is\",round(R2,0),\" ohm\"\n",
- "print \"\\n (b)current flows in the load = \",round(Io,2),\"of the original current.\"\n",
- "print \"\\n (c)overall attenuation is \",round(attnO,2),\"dB \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)For a T-section symmetrical attenuator pad, Resistance R1 is 350.0 ohm \n",
- "and Resistance R2 is 114.0 ohm\n",
- "\n",
- " (b)current flows in the load = 0.02 of the original current.\n",
- "\n",
- " (c)overall attenuation is 72.25 dB \n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_1.ipynb deleted file mode 100755 index 788f6f5e..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_1.ipynb +++ /dev/null @@ -1,1005 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 41: Attenuators</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 763</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the power ratio in each case (i) in decibels and (ii) in nepers.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- " #ratio of output power to input power\n",
- "rp1 = 2;\n",
- "rp2 = 25; \n",
- "rp3 = 1000;\n",
- "rp4 = 0.01;\n",
- "\n",
- "#calculation:\n",
- " #power ratio in decibels\n",
- "rpd1 = 10*(1/2.303)*math.log(rp1)\n",
- "rpd2 = 10*(1/2.303)*math.log(rp2)\n",
- "rpd3 = 10*(1/2.303)*math.log(rp3)\n",
- "rpd4 = 10*(1/2.303)*math.log(rp4)\n",
- " #power ratio in nepers\n",
- "rpn1 = (math.log(rp1))/2\n",
- "rpn2 = (math.log(rp2))/2\n",
- "rpn3 = (math.log(rp3))/2\n",
- "rpn4 = (math.log(rp4))/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n power ratio in decibels are (a)\",round(rpd1,2),\" dB (b)\",round(rpd2,2),\" dB \"\n",
- "print \"(c) \",round(rpd3,2),\" dB and (d) \",round(rpd4,2),\" dB\"\n",
- "print \"\\n power ratio in nepers are (a)\",round(rpn1,2),\" Np (b)\",round(rpn2,2),\" Np\"\n",
- "print \"(c) \",round(rpn3,2),\" Np and (d) \",round(rpn4,2),\" Np\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " power ratio in decibels are (a) 3.01 dB (b) 13.98 dB \n",
- "(c) 29.99 dB and (d) -20.0 dB\n",
- "\n",
- " power ratio in nepers are (a) 0.35 Np (b) 1.61 Np\n",
- "(c) 3.45 Np and (d) -2.3 Np\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 763</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the attenuation in decibels\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rp = 0.05;# power ratio P2/P1\n",
- "\n",
- " #calculation:\n",
- " #power ratio in decibels\n",
- "rpd = 10*(1/2.303)*math.log(rp)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nthe attenuation is \",round(abs(rpd),2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "the attenuation is 13.01 dB"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 764</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the output power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "gain = 1.5;# in dB\n",
- "Pi = 0.012;# in Watt\n",
- "\n",
- "#calculation:\n",
- " #output power\n",
- "Po = Pi*10**gain\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\noutput power is \",round(Po,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "output power is 0.38 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 764</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current input and (b) the current ratio expressed in decibels.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "I2 = 0.05;# in Amperes\n",
- "rin = 1.32;# in Np\n",
- "\n",
- " #calculation:\n",
- " #current input, I1\n",
- "I1 = I2*math.e**(rin)\n",
- " #current ratio in decibels\n",
- "rid = 20*(1/2.303)*math.log(I2/I1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ncurrent input, I1 is \",round(I1,2),\" A\"\n",
- "print \"\\ncurrent ratio in decibels is \",round(rid,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "current input, I1 is 0.19 A\n",
- "\n",
- "current ratio in decibels is -11.46 dB"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 769</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the characteristic impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ra1 = 8; # in ohms\n",
- "Ra2 = 8; # in ohms\n",
- "Ra3 = 21; # in ohms\n",
- "Rb1 = 10; # in ohms\n",
- "Rb2 = 10; # in ohms\n",
- "Rb3 = 15; # in ohms\n",
- "Rc1 = 200; # in ohms\n",
- "Rc2 = 200; # in ohms\n",
- "Rc3 = 56.25; # in ohms\n",
- "\n",
- "#calculation:\n",
- "R01 = (Ra1**2 + 2*Ra2*Ra3)**0.5\n",
- "R02 = (Rb1**2 + 2*Rb2*Rb3)**0.5\n",
- "R03 = (Rc1**2 + 2*Rc2*Rc3)**0.5\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) the characteristic impedance, R0 is \",R01,\" ohm\"\n",
- "print \"\\n(b) the characteristic impedance, R0 is \",R02,\" ohm\"\n",
- "print \"\\n(c) the characteristic impedance, R0 is \",R03,\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) the characteristic impedance, R0 is 20.0 ohm\n",
- "\n",
- "(b) the characteristic impedance, R0 is 20.0 ohm\n",
- "\n",
- "(c) the characteristic impedance, R0 is 250.0 ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 769</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the characteristic impedance, and (b) the attenuation (in dB) produced by the pad\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 500;# in ohm\n",
- "R2 = 1000;# in ohm\n",
- "I1 = 1;# in ampere (lets say)\n",
- "\n",
- "#calculation:\n",
- " # for symmetrical pi-attenuator section\n",
- " #characteristic impedance, R0\n",
- "R0 = (R1*(R2**2)/(R1 + 2*R2))**0.5\n",
- " #current Ix\n",
- "Ix = (R2/(R2 + R1 + (R2*R0/(R2 + R0))))*I1\n",
- " #current I2\n",
- "I2 = (R2/(R2 + R0))*Ix\n",
- "ri = I1/I2;# retio of currents\n",
- " #attenuation\n",
- "attn = 20*(1/2.303)*math.log(ri)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the characteristic impedance is \",round(R0,2),\" ohm\"\n",
- "print \"\\n attenuation is \",round(attn,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the characteristic impedance is 447.21 ohm\n",
- "\n",
- " attenuation is 8.36 dB"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 770</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the input resistance when the output port is open-circuited, \n",
- "#(b) the input resistance when the output port is short-circuited, and (c) the characteristic impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ra1 = 15;# in ohm\n",
- "Ra2 = 15;# in ohm\n",
- "Ra3 = 10;# in ohm\n",
- "Rb1 = 15;# in ohm\n",
- "Rb2 = 5;# in ohm\n",
- "Rb3 = 5;# in ohm\n",
- "\n",
- "#calculation:\n",
- "Roc1 = Ra1 + Ra3\n",
- "Rsc1 = Ra1 + Ra2*Ra3/(Ra2+Ra3)\n",
- "R01 = (Roc1*Rsc1)**0.5\n",
- "\n",
- "Roc2 = Rb2* (Rb1 + Rb3)/(Rb1 + Rb2 + Rb3)\n",
- "Rsc2 = Rb2*Rb1/(Rb2+Rb1)\n",
- "R02 = (Roc2*Rsc2)**0.5\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the input resistance when the output port is open-circuited is\", Roc1,\" ohm for T-Network\"\n",
- "print \"and \",Roc2,\" ohm for pi-Network \"\n",
- "print \"\\n (b) the input resistance when the output port is short-circuited is,\", Rsc1,\" ohm for T-Network\"\n",
- "print \"and \",Rsc2,\" ohm for pi-Network \"\n",
- "print \"\\n (c) the characteristic impedance. is,\",round(R01,1),\" ohm for T-Network and ,\",round(R02,2),\" ohm for pi-Network \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the input resistance when the output port is open-circuited is 25 ohm for T-Network\n",
- "and 4.0 ohm for pi-Network \n",
- "\n",
- " (b) the input resistance when the output port is short-circuited is, 21.0 ohm for T-Network\n",
- "and 3.75 ohm for pi-Network \n",
- "\n",
- " (c) the characteristic impedance. is, 22.9 ohm for T-Network and , 3.87 ohm for pi-Network \n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 770</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design a T-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n",
- "#having a characteristic impedance of 600 ohm.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vat = 20; # in db\n",
- "R0 = 600;# in ohm\n",
- "\n",
- "#calculation:\n",
- "N = math.e**(Vat*2.3/20)\n",
- "R1 = R0*(N-1)/(N+1)\n",
- "R2 = R0*2*N/(N**2 - 1)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,0),\" ohm and Resistance R2 is\",round(R2,1),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For a T-section symmetrical attenuator pad, Resistance R1 is 491.0 ohm and Resistance R2 is 121.5 ohm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 771</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design a pi-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n",
- "#having a characteristic impedance of 600 ohm.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vat = 20; # in db\n",
- "R0 = 600;# in ohm\n",
- "\n",
- "#calculation:\n",
- "N = math.e**(Vat*2.303/20)\n",
- "R1 = R0*(N**2 - 1)/(2*N)\n",
- "R2 = R0*(N+1)/(N-1)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For a pi-section symmetrical attenuator pad, Resistance R1 is\",round(R1/1000,2),\" Kohm\"\n",
- "print \"and Resistance R2 is\",round(R2,0),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For a pi-section symmetrical attenuator pad, Resistance R1 is 2.97 Kohm\n",
- "and Resistance R2 is 733.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 772</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the characteristic impedance R0, and (b) the insertion loss in decibels\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 300;# in ohm\n",
- "R2 = 450;# in ohm\n",
- "I1 = 1;# in ampere (lets say)\n",
- "\n",
- "#calculation:\n",
- " #the characteristic impedance of a symmetric T-pad attenuator is given by\n",
- "R0 = (R1**2 + 2*R1*R2)**0.5\n",
- " #By current division\n",
- " #current I2\n",
- "I2 = (R2/(R2 + R1+ R0))*I1\n",
- "ri = I1/I2;# ratio of currents\n",
- " #insertion loss\n",
- "il = 20*(1/2.303)*math.log(ri)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the characteristic impedance is \",R0,\" ohm\"\n",
- "print \"\\n insertion loss is \",round(il,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the characteristic impedance is 600.0 ohm\n",
- "\n",
- " insertion loss is 9.54 dB"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 773</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the insertion loss at a tapping of (a) 2 kohm, (b) 1 kohm.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r = 500;# in ohm\n",
- "Rhm = 3000;# in ohm\n",
- "RL = 2000;# in ohm\n",
- "r1 = 2000;# in ohm\n",
- "r2 = 1000;# in ohm\n",
- "E = 1;# in volts (lets say)\n",
- "\n",
- " #calculation:\n",
- " #Without the rheostat in the circuit the voltage across the 2 kohm\u0018 load, VL\n",
- "VL = (RL/(RL + r))*E\n",
- " #voltage V2 with 2kohm tapping\n",
- "V2 = ((RL*r1/(r1 + RL))/((RL*r1/(r1 + RL)) + Rhm - r1 + r))*E\n",
- "rv1 = VL/V2;# ratio of currents\n",
- " #insertion loss \n",
- "il1 = 20*(1/2.303)*math.log(rv1)\n",
- " #voltage V1 with 1kohm tapping\n",
- "V1 = ((RL*r2/(r2 + RL))/((RL*r2/(r2 + RL)) + Rhm - r2 + r))*E\n",
- "rv2 = VL/V1;# ratio of currents\n",
- " #insertion loss \n",
- "il2 = 20*(1/2.303)*math.log(rv2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n insertion loss for 2kohm tap is \",round(il1,2),\" dB\"\n",
- "print \"\\n insertion loss for 1kohm tap is \",round(il2,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " insertion loss for 2kohm tap is 6.02 dB\n",
- "\n",
- " insertion loss for 1kohm tap is 11.59 dB"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 774</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) its characteristic impedance, and (b) the insertion loss (in decibels) when feeding a matched load.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 1000;# in ohm\n",
- "R2 = 500;# in ohm\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- "#calculation:\n",
- " #characteristic impedance of a symmetrical attenuator\n",
- "R0 = (R1*(R2**2)/(R1 + 2*R2))**0.5\n",
- " #current Ix\n",
- "Ix = (R2/(R2 + R1 + (R2*R0/(R2 + R0))))*I1\n",
- " #current I2\n",
- "I2 = (R2/(R2 + R0))*Ix\n",
- "ri = I1/I2;# retio of currents\n",
- " #insertion loss \n",
- "il = 20*(1/2.303)*math.log(ri)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n characteristic impedance is \",round(R0,2),\" ohm\"\n",
- "print \"\\n insertion loss is \",round(il,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " characteristic impedance is 353.55 ohm\n",
- "\n",
- " insertion loss is 15.31 dB"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 776</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 100;# in ohm\n",
- "R2 = 200;# in ohm\n",
- "R3 = 300;# in ohm\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- "#calculation:\n",
- " #image impedance Roa\n",
- "Roa = ((R1 + R2)*(R2 + (R1*R3/(R1 + R3))))**0.5\n",
- " #image impedance Rob\n",
- "Rob = ((R1 + R3)*(R3 + (R1*R2/(R1 + R2))))**0.5\n",
- " #The iterative impedance at port 1\n",
- "Ri1 = (-1*R1 + (R1**2 - (-1*4*((R2*(R1 + R3)) + (R3*R1))))**0.5)/2\n",
- " #The iterative impedance at port 2\n",
- "Ri2 = (R1 + (R1**2 - (-1*4*((R3*(R1 + R2)) + (R2*R1))))**0.5)/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n image impedance are \",round(Roa,2),\" ohm and \",round(Rob,2),\" ohm \"\n",
- "print \"\\n iterative impedances are \",round(Ri1,2),\" ohm and \",round(Ri2,2),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " image impedance are 287.23 ohm and 382.97 ohm \n",
- "\n",
- " iterative impedances are 285.41 ohm and 385.41 ohm "
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 777</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 1000;# in ohm\n",
- "R2 = 2000;# in ohm\n",
- "R3 = 3000;# in ohm\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- "#calculation:\n",
- " #image impedance Roa\n",
- "Roa = (((R3 + R2)*R1/(R1 + R2 + R3))*(R1*R3/(R1 + R3)))**0.5\n",
- " #image impedance Rob\n",
- "Rob = (((R3 + R1)*R2/(R1 + R2 + R3))*(R2*R3/(R2 + R3)))**0.5\n",
- " #The iterative impedance at port 1\n",
- "Ri1 = (-1*R1 + ((R1**2) - (-1*4*2*R2*R1))**0.5)/(2*2)\n",
- " #The iterative impedance at port 2\n",
- "Ri2 = (R1 + ((-1*R1)**2 - (-1*4*2*R2*R1))**0.5)/(2*2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n image impedance are \",round(Roa,2),\" ohm and \",round(Rob,2),\" ohm \"\n",
- "print \"\\n iterative impedances are \",round(Ri1,2),\" ohm and \",round(Ri2,2),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " image impedance are 790.57 ohm and 1264.91 ohm \n",
- "\n",
- " iterative impedances are 780.78 ohm and 1280.78 ohm "
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 780</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the values of resistance R1 and R2, (b) the attenuation of the pad in decibels, and (c) its insertion loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r = 500;# in ohm\n",
- "RL = 100;# in ohm\n",
- "E = 1;# in volts (lets say)\n",
- "\n",
- "#calculation:\n",
- " #res.\n",
- "R1 = (r*(r - RL))**0.5\n",
- "R2 = (r*RL**2/(r - RL))**0.5\n",
- " #current I1\n",
- "I1 = E/(r + R1 + R2*RL/(RL + R2))\n",
- " #current I2\n",
- "I2 = (R2/(R2 + RL))*I1\n",
- " #input power\n",
- "P1 = r*I1**2\n",
- " #output power\n",
- "P2 = RL*I2**2\n",
- " #attenuation\n",
- "attn = 10*(1/2.303)*math.log(P1/P2)\n",
- " #Load current, IL\n",
- "IL = E/(r + RL)\n",
- " #voltage, VL\n",
- "VL = IL*RL\n",
- " #voltage, V1\n",
- "V1 = E - I1*r\n",
- " #voltage, V2\n",
- "V2 = V1 - I1*R1\n",
- " #insertion loss\n",
- "il = 20*(1/2.303)*math.log(VL/V2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n R1 = \",round(R1,2),\" ohm and R2 = \",round(R2,2),\" ohm \"\n",
- "print \"\\n attenuation is \",round(attn,2),\" dB \"\n",
- "print \"\\n In decibels, the insertion loss is \",round(il,2),\" dB \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " R1 = 447.21 ohm and R2 = 111.8 ohm \n",
- "\n",
- " attenuation is 12.54 dB \n",
- "\n",
- " In decibels, the insertion loss is 9.98 dB "
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 783</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the attenuation of each individual attenuation section, \n",
- "#(b) the voltage output of the final stage, and (c) the voltage output of the third stage.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "attnO = 70;# in dB\n",
- "n = 5;# numbers of identical atteneurs\n",
- "V1 = 0.02;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #attenuation of each section\n",
- "attn = attnO/n\n",
- " #output of the final stage\n",
- "Vo = V1/(10**(attnO/20))\n",
- " #voltage output of the third stage\n",
- "V3 = V1/(10**(3*attn/20))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n attenuation of each section = \",round(attn,2),\" dB \"\n",
- "print \"\\n output of the final stage is \",round(Vo*1E6,2),\"uV \"\n",
- "print \"\\n voltage output of the third stage is \",round(V3*1E3,3),\"mV \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " attenuation of each section = 14.0 dB \n",
- "\n",
- " output of the final stage is 6.32 uV \n",
- "\n",
- " voltage output of the third stage is 0.159 mV "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 784</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)Design a T-network attenuator pad\n",
- "#(b)determine the fraction of the initial current \n",
- "#(c) Determine the attenuation in decibels given by four such sections\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r = 450; # in ohm\n",
- "R0 = 450; # in ohms\n",
- "x = 1/8\n",
- "\n",
- "#calculation:\n",
- "N = 1/x\n",
- "R1 = R0*(N-1)/(N+1)\n",
- "R2 = R0*2*N/(N**2 - 1)\n",
- "\n",
- "Io = x*x\n",
- "\n",
- "attn = 20*math.log10(N)\n",
- "attnO = 4*attn\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,2),\" ohm \"\n",
- "print \"and Resistance R2 is\",round(R2,0),\" ohm\"\n",
- "print \"\\n (b)current flows in the load = \",round(Io,2),\"of the original current.\"\n",
- "print \"\\n (c)overall attenuation is \",round(attnO,2),\"dB \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)For a T-section symmetrical attenuator pad, Resistance R1 is 350.0 ohm \n",
- "and Resistance R2 is 114.0 ohm\n",
- "\n",
- " (b)current flows in the load = 0.02 of the original current.\n",
- "\n",
- " (c)overall attenuation is 72.25 dB \n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_2.ipynb deleted file mode 100755 index 788f6f5e..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_2.ipynb +++ /dev/null @@ -1,1005 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 41: Attenuators</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 763</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the power ratio in each case (i) in decibels and (ii) in nepers.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- " #ratio of output power to input power\n",
- "rp1 = 2;\n",
- "rp2 = 25; \n",
- "rp3 = 1000;\n",
- "rp4 = 0.01;\n",
- "\n",
- "#calculation:\n",
- " #power ratio in decibels\n",
- "rpd1 = 10*(1/2.303)*math.log(rp1)\n",
- "rpd2 = 10*(1/2.303)*math.log(rp2)\n",
- "rpd3 = 10*(1/2.303)*math.log(rp3)\n",
- "rpd4 = 10*(1/2.303)*math.log(rp4)\n",
- " #power ratio in nepers\n",
- "rpn1 = (math.log(rp1))/2\n",
- "rpn2 = (math.log(rp2))/2\n",
- "rpn3 = (math.log(rp3))/2\n",
- "rpn4 = (math.log(rp4))/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n power ratio in decibels are (a)\",round(rpd1,2),\" dB (b)\",round(rpd2,2),\" dB \"\n",
- "print \"(c) \",round(rpd3,2),\" dB and (d) \",round(rpd4,2),\" dB\"\n",
- "print \"\\n power ratio in nepers are (a)\",round(rpn1,2),\" Np (b)\",round(rpn2,2),\" Np\"\n",
- "print \"(c) \",round(rpn3,2),\" Np and (d) \",round(rpn4,2),\" Np\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " power ratio in decibels are (a) 3.01 dB (b) 13.98 dB \n",
- "(c) 29.99 dB and (d) -20.0 dB\n",
- "\n",
- " power ratio in nepers are (a) 0.35 Np (b) 1.61 Np\n",
- "(c) 3.45 Np and (d) -2.3 Np\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 763</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the attenuation in decibels\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rp = 0.05;# power ratio P2/P1\n",
- "\n",
- " #calculation:\n",
- " #power ratio in decibels\n",
- "rpd = 10*(1/2.303)*math.log(rp)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nthe attenuation is \",round(abs(rpd),2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "the attenuation is 13.01 dB"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 764</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the output power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "gain = 1.5;# in dB\n",
- "Pi = 0.012;# in Watt\n",
- "\n",
- "#calculation:\n",
- " #output power\n",
- "Po = Pi*10**gain\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\noutput power is \",round(Po,2),\" W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "output power is 0.38 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 764</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current input and (b) the current ratio expressed in decibels.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "I2 = 0.05;# in Amperes\n",
- "rin = 1.32;# in Np\n",
- "\n",
- " #calculation:\n",
- " #current input, I1\n",
- "I1 = I2*math.e**(rin)\n",
- " #current ratio in decibels\n",
- "rid = 20*(1/2.303)*math.log(I2/I1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ncurrent input, I1 is \",round(I1,2),\" A\"\n",
- "print \"\\ncurrent ratio in decibels is \",round(rid,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "current input, I1 is 0.19 A\n",
- "\n",
- "current ratio in decibels is -11.46 dB"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 769</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the characteristic impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ra1 = 8; # in ohms\n",
- "Ra2 = 8; # in ohms\n",
- "Ra3 = 21; # in ohms\n",
- "Rb1 = 10; # in ohms\n",
- "Rb2 = 10; # in ohms\n",
- "Rb3 = 15; # in ohms\n",
- "Rc1 = 200; # in ohms\n",
- "Rc2 = 200; # in ohms\n",
- "Rc3 = 56.25; # in ohms\n",
- "\n",
- "#calculation:\n",
- "R01 = (Ra1**2 + 2*Ra2*Ra3)**0.5\n",
- "R02 = (Rb1**2 + 2*Rb2*Rb3)**0.5\n",
- "R03 = (Rc1**2 + 2*Rc2*Rc3)**0.5\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) the characteristic impedance, R0 is \",R01,\" ohm\"\n",
- "print \"\\n(b) the characteristic impedance, R0 is \",R02,\" ohm\"\n",
- "print \"\\n(c) the characteristic impedance, R0 is \",R03,\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) the characteristic impedance, R0 is 20.0 ohm\n",
- "\n",
- "(b) the characteristic impedance, R0 is 20.0 ohm\n",
- "\n",
- "(c) the characteristic impedance, R0 is 250.0 ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 769</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the characteristic impedance, and (b) the attenuation (in dB) produced by the pad\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 500;# in ohm\n",
- "R2 = 1000;# in ohm\n",
- "I1 = 1;# in ampere (lets say)\n",
- "\n",
- "#calculation:\n",
- " # for symmetrical pi-attenuator section\n",
- " #characteristic impedance, R0\n",
- "R0 = (R1*(R2**2)/(R1 + 2*R2))**0.5\n",
- " #current Ix\n",
- "Ix = (R2/(R2 + R1 + (R2*R0/(R2 + R0))))*I1\n",
- " #current I2\n",
- "I2 = (R2/(R2 + R0))*Ix\n",
- "ri = I1/I2;# retio of currents\n",
- " #attenuation\n",
- "attn = 20*(1/2.303)*math.log(ri)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the characteristic impedance is \",round(R0,2),\" ohm\"\n",
- "print \"\\n attenuation is \",round(attn,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the characteristic impedance is 447.21 ohm\n",
- "\n",
- " attenuation is 8.36 dB"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 770</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the input resistance when the output port is open-circuited, \n",
- "#(b) the input resistance when the output port is short-circuited, and (c) the characteristic impedance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Ra1 = 15;# in ohm\n",
- "Ra2 = 15;# in ohm\n",
- "Ra3 = 10;# in ohm\n",
- "Rb1 = 15;# in ohm\n",
- "Rb2 = 5;# in ohm\n",
- "Rb3 = 5;# in ohm\n",
- "\n",
- "#calculation:\n",
- "Roc1 = Ra1 + Ra3\n",
- "Rsc1 = Ra1 + Ra2*Ra3/(Ra2+Ra3)\n",
- "R01 = (Roc1*Rsc1)**0.5\n",
- "\n",
- "Roc2 = Rb2* (Rb1 + Rb3)/(Rb1 + Rb2 + Rb3)\n",
- "Rsc2 = Rb2*Rb1/(Rb2+Rb1)\n",
- "R02 = (Roc2*Rsc2)**0.5\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) the input resistance when the output port is open-circuited is\", Roc1,\" ohm for T-Network\"\n",
- "print \"and \",Roc2,\" ohm for pi-Network \"\n",
- "print \"\\n (b) the input resistance when the output port is short-circuited is,\", Rsc1,\" ohm for T-Network\"\n",
- "print \"and \",Rsc2,\" ohm for pi-Network \"\n",
- "print \"\\n (c) the characteristic impedance. is,\",round(R01,1),\" ohm for T-Network and ,\",round(R02,2),\" ohm for pi-Network \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) the input resistance when the output port is open-circuited is 25 ohm for T-Network\n",
- "and 4.0 ohm for pi-Network \n",
- "\n",
- " (b) the input resistance when the output port is short-circuited is, 21.0 ohm for T-Network\n",
- "and 3.75 ohm for pi-Network \n",
- "\n",
- " (c) the characteristic impedance. is, 22.9 ohm for T-Network and , 3.87 ohm for pi-Network \n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 770</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design a T-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n",
- "#having a characteristic impedance of 600 ohm.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vat = 20; # in db\n",
- "R0 = 600;# in ohm\n",
- "\n",
- "#calculation:\n",
- "N = math.e**(Vat*2.3/20)\n",
- "R1 = R0*(N-1)/(N+1)\n",
- "R2 = R0*2*N/(N**2 - 1)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,0),\" ohm and Resistance R2 is\",round(R2,1),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For a T-section symmetrical attenuator pad, Resistance R1 is 491.0 ohm and Resistance R2 is 121.5 ohm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 771</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design a pi-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n",
- "#having a characteristic impedance of 600 ohm.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vat = 20; # in db\n",
- "R0 = 600;# in ohm\n",
- "\n",
- "#calculation:\n",
- "N = math.e**(Vat*2.303/20)\n",
- "R1 = R0*(N**2 - 1)/(2*N)\n",
- "R2 = R0*(N+1)/(N-1)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For a pi-section symmetrical attenuator pad, Resistance R1 is\",round(R1/1000,2),\" Kohm\"\n",
- "print \"and Resistance R2 is\",round(R2,0),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For a pi-section symmetrical attenuator pad, Resistance R1 is 2.97 Kohm\n",
- "and Resistance R2 is 733.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 772</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the characteristic impedance R0, and (b) the insertion loss in decibels\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R1 = 300;# in ohm\n",
- "R2 = 450;# in ohm\n",
- "I1 = 1;# in ampere (lets say)\n",
- "\n",
- "#calculation:\n",
- " #the characteristic impedance of a symmetric T-pad attenuator is given by\n",
- "R0 = (R1**2 + 2*R1*R2)**0.5\n",
- " #By current division\n",
- " #current I2\n",
- "I2 = (R2/(R2 + R1+ R0))*I1\n",
- "ri = I1/I2;# ratio of currents\n",
- " #insertion loss\n",
- "il = 20*(1/2.303)*math.log(ri)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the characteristic impedance is \",R0,\" ohm\"\n",
- "print \"\\n insertion loss is \",round(il,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the characteristic impedance is 600.0 ohm\n",
- "\n",
- " insertion loss is 9.54 dB"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 773</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the insertion loss at a tapping of (a) 2 kohm, (b) 1 kohm.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r = 500;# in ohm\n",
- "Rhm = 3000;# in ohm\n",
- "RL = 2000;# in ohm\n",
- "r1 = 2000;# in ohm\n",
- "r2 = 1000;# in ohm\n",
- "E = 1;# in volts (lets say)\n",
- "\n",
- " #calculation:\n",
- " #Without the rheostat in the circuit the voltage across the 2 kohm\u0018 load, VL\n",
- "VL = (RL/(RL + r))*E\n",
- " #voltage V2 with 2kohm tapping\n",
- "V2 = ((RL*r1/(r1 + RL))/((RL*r1/(r1 + RL)) + Rhm - r1 + r))*E\n",
- "rv1 = VL/V2;# ratio of currents\n",
- " #insertion loss \n",
- "il1 = 20*(1/2.303)*math.log(rv1)\n",
- " #voltage V1 with 1kohm tapping\n",
- "V1 = ((RL*r2/(r2 + RL))/((RL*r2/(r2 + RL)) + Rhm - r2 + r))*E\n",
- "rv2 = VL/V1;# ratio of currents\n",
- " #insertion loss \n",
- "il2 = 20*(1/2.303)*math.log(rv2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n insertion loss for 2kohm tap is \",round(il1,2),\" dB\"\n",
- "print \"\\n insertion loss for 1kohm tap is \",round(il2,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " insertion loss for 2kohm tap is 6.02 dB\n",
- "\n",
- " insertion loss for 1kohm tap is 11.59 dB"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 774</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) its characteristic impedance, and (b) the insertion loss (in decibels) when feeding a matched load.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 1000;# in ohm\n",
- "R2 = 500;# in ohm\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- "#calculation:\n",
- " #characteristic impedance of a symmetrical attenuator\n",
- "R0 = (R1*(R2**2)/(R1 + 2*R2))**0.5\n",
- " #current Ix\n",
- "Ix = (R2/(R2 + R1 + (R2*R0/(R2 + R0))))*I1\n",
- " #current I2\n",
- "I2 = (R2/(R2 + R0))*Ix\n",
- "ri = I1/I2;# retio of currents\n",
- " #insertion loss \n",
- "il = 20*(1/2.303)*math.log(ri)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n characteristic impedance is \",round(R0,2),\" ohm\"\n",
- "print \"\\n insertion loss is \",round(il,2),\" dB\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " characteristic impedance is 353.55 ohm\n",
- "\n",
- " insertion loss is 15.31 dB"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 776</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 100;# in ohm\n",
- "R2 = 200;# in ohm\n",
- "R3 = 300;# in ohm\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- "#calculation:\n",
- " #image impedance Roa\n",
- "Roa = ((R1 + R2)*(R2 + (R1*R3/(R1 + R3))))**0.5\n",
- " #image impedance Rob\n",
- "Rob = ((R1 + R3)*(R3 + (R1*R2/(R1 + R2))))**0.5\n",
- " #The iterative impedance at port 1\n",
- "Ri1 = (-1*R1 + (R1**2 - (-1*4*((R2*(R1 + R3)) + (R3*R1))))**0.5)/2\n",
- " #The iterative impedance at port 2\n",
- "Ri2 = (R1 + (R1**2 - (-1*4*((R3*(R1 + R2)) + (R2*R1))))**0.5)/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n image impedance are \",round(Roa,2),\" ohm and \",round(Rob,2),\" ohm \"\n",
- "print \"\\n iterative impedances are \",round(Ri1,2),\" ohm and \",round(Ri2,2),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " image impedance are 287.23 ohm and 382.97 ohm \n",
- "\n",
- " iterative impedances are 285.41 ohm and 385.41 ohm "
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 777</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R1 = 1000;# in ohm\n",
- "R2 = 2000;# in ohm\n",
- "R3 = 3000;# in ohm\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- "#calculation:\n",
- " #image impedance Roa\n",
- "Roa = (((R3 + R2)*R1/(R1 + R2 + R3))*(R1*R3/(R1 + R3)))**0.5\n",
- " #image impedance Rob\n",
- "Rob = (((R3 + R1)*R2/(R1 + R2 + R3))*(R2*R3/(R2 + R3)))**0.5\n",
- " #The iterative impedance at port 1\n",
- "Ri1 = (-1*R1 + ((R1**2) - (-1*4*2*R2*R1))**0.5)/(2*2)\n",
- " #The iterative impedance at port 2\n",
- "Ri2 = (R1 + ((-1*R1)**2 - (-1*4*2*R2*R1))**0.5)/(2*2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n image impedance are \",round(Roa,2),\" ohm and \",round(Rob,2),\" ohm \"\n",
- "print \"\\n iterative impedances are \",round(Ri1,2),\" ohm and \",round(Ri2,2),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " image impedance are 790.57 ohm and 1264.91 ohm \n",
- "\n",
- " iterative impedances are 780.78 ohm and 1280.78 ohm "
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 780</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the values of resistance R1 and R2, (b) the attenuation of the pad in decibels, and (c) its insertion loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r = 500;# in ohm\n",
- "RL = 100;# in ohm\n",
- "E = 1;# in volts (lets say)\n",
- "\n",
- "#calculation:\n",
- " #res.\n",
- "R1 = (r*(r - RL))**0.5\n",
- "R2 = (r*RL**2/(r - RL))**0.5\n",
- " #current I1\n",
- "I1 = E/(r + R1 + R2*RL/(RL + R2))\n",
- " #current I2\n",
- "I2 = (R2/(R2 + RL))*I1\n",
- " #input power\n",
- "P1 = r*I1**2\n",
- " #output power\n",
- "P2 = RL*I2**2\n",
- " #attenuation\n",
- "attn = 10*(1/2.303)*math.log(P1/P2)\n",
- " #Load current, IL\n",
- "IL = E/(r + RL)\n",
- " #voltage, VL\n",
- "VL = IL*RL\n",
- " #voltage, V1\n",
- "V1 = E - I1*r\n",
- " #voltage, V2\n",
- "V2 = V1 - I1*R1\n",
- " #insertion loss\n",
- "il = 20*(1/2.303)*math.log(VL/V2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n R1 = \",round(R1,2),\" ohm and R2 = \",round(R2,2),\" ohm \"\n",
- "print \"\\n attenuation is \",round(attn,2),\" dB \"\n",
- "print \"\\n In decibels, the insertion loss is \",round(il,2),\" dB \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " R1 = 447.21 ohm and R2 = 111.8 ohm \n",
- "\n",
- " attenuation is 12.54 dB \n",
- "\n",
- " In decibels, the insertion loss is 9.98 dB "
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 783</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the attenuation of each individual attenuation section, \n",
- "#(b) the voltage output of the final stage, and (c) the voltage output of the third stage.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "attnO = 70;# in dB\n",
- "n = 5;# numbers of identical atteneurs\n",
- "V1 = 0.02;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #attenuation of each section\n",
- "attn = attnO/n\n",
- " #output of the final stage\n",
- "Vo = V1/(10**(attnO/20))\n",
- " #voltage output of the third stage\n",
- "V3 = V1/(10**(3*attn/20))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n attenuation of each section = \",round(attn,2),\" dB \"\n",
- "print \"\\n output of the final stage is \",round(Vo*1E6,2),\"uV \"\n",
- "print \"\\n voltage output of the third stage is \",round(V3*1E3,3),\"mV \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " attenuation of each section = 14.0 dB \n",
- "\n",
- " output of the final stage is 6.32 uV \n",
- "\n",
- " voltage output of the third stage is 0.159 mV "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 784</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a)Design a T-network attenuator pad\n",
- "#(b)determine the fraction of the initial current \n",
- "#(c) Determine the attenuation in decibels given by four such sections\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r = 450; # in ohm\n",
- "R0 = 450; # in ohms\n",
- "x = 1/8\n",
- "\n",
- "#calculation:\n",
- "N = 1/x\n",
- "R1 = R0*(N-1)/(N+1)\n",
- "R2 = R0*2*N/(N**2 - 1)\n",
- "\n",
- "Io = x*x\n",
- "\n",
- "attn = 20*math.log10(N)\n",
- "attnO = 4*attn\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,2),\" ohm \"\n",
- "print \"and Resistance R2 is\",round(R2,0),\" ohm\"\n",
- "print \"\\n (b)current flows in the load = \",round(Io,2),\"of the original current.\"\n",
- "print \"\\n (c)overall attenuation is \",round(attnO,2),\"dB \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)For a T-section symmetrical attenuator pad, Resistance R1 is 350.0 ohm \n",
- "and Resistance R2 is 114.0 ohm\n",
- "\n",
- " (b)current flows in the load = 0.02 of the original current.\n",
- "\n",
- " (c)overall attenuation is 72.25 dB \n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_41.ipynb index 4bdaf71c..4bdaf71c 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_41.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint.ipynb deleted file mode 100755 index fcac8b6a..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint.ipynb +++ /dev/null @@ -1,959 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 42: Filter networks</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 799</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the cut-off frequency and the nominal impedance of each of the low-pass filter sections\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L1 = 2*100E-3;# in Henry\n",
- "C1 = 0.2E-6;# in Fareads\n",
- "L2 = 0.4;# in Henry\n",
- "C2 = 2*200E-12;# in Fareads\n",
- "\n",
- "#calculation:\n",
- " #cut-off frequency\n",
- "fc1 = 1/(math.pi*(L1*C1)**0.5)\n",
- " #nominal impedance\n",
- "R01 = (L1/C1)**0.5\n",
- " #cut-off frequency\n",
- "fc2 = 1/(math.pi*(L2*C2)**0.5)\n",
- " #nominal impedance\n",
- "R02 = (L2/C2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n cut-off frequency \",round(fc1,2),\" Hz and the nominal impedance is \",round( R01,2),\" ohm \"\n",
- "print \"\\n cut-off frequency \",round(fc2,2),\" Hz and the nominal impedance is \",round( R02,2),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " cut-off frequency 1591.55 Hz and the nominal impedance is 1000.0 ohm \n",
- "\n",
- " cut-off frequency 25164.61 Hz and the nominal impedance is 31622.78 ohm "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 801</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) a low-pass T section filter, and (b) a low-pass \u0003 section filter to meet these requirements.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 5E6;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = 1/(math.pi*R0*fc)\n",
- " #inductance\n",
- "L = R0/(math.pi*fc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"A low-pass T section filter capcitance is \",round(C*1E12,2),\"pfarad and inductance is\",round( L/2*1E6,2),\"uHenry\"\n",
- "print \"A low-pass pi section filter capcitance is \",round(C/2*1E12,2),\"pfarad and inductance is\",round( L*1E6,2),\"uHenry\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "A low-pass T section filter capcitance is 106.1 pfarad and inductance is 19.1 uHenry\n",
- "A low-pass pi section filter capcitance is 53.05 pfarad and inductance is 38.2 uHenry\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 805</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of the characteristic impedance of the section at a frequency of 90 kHz, and \n",
- "#(b) the value of the characteristic impedance of the equivalent low-pass T section filter.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 500;# in ohm\n",
- "fc = 100000;# in Hz\n",
- "f = 90000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #characteristic impedance of the pi section\n",
- "Zpi = R0/(1 - (f/fc)**2)**0.5\n",
- " #characteristic impedance of the T section\n",
- "Zt = R0*(1 - (f/fc)**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ncharacteristic impedance of the pi section is \",round(Zpi,2),\" ohm\"\n",
- "print \"\\ncharacteristic impedance of the T section is \",round(Zt,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "characteristic impedance of the pi section is 1147.08 ohm\n",
- "\n",
- "characteristic impedance of the T section is 217.94 ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 806</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the frequency at which the characteristic impedance of the section is (a) 600ohm (b) 1 kohm(c) 10 kohm\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 2E6;# in Hz\n",
- "Z1 = 600;# in ohm\n",
- "Z2 = 1000;# in ohm\n",
- "Z3 = 10000;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #frequency\n",
- "f1 = fc*(1 - (R0/Z1)**2)**0.5\n",
- "f2 = fc*(1 - (R0/Z2)**2)**0.5\n",
- "f3 = fc*(1 - (R0/Z3)**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"frequency at which the characteristic impedance of the section is 600 ohm is \",f1,\" Hz \"\n",
- "print \"and 1000 Ohm is \",f2*1E-3,\"kHz and 10000 ohm is \",round(f3*1E-3,2),\"kHz \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "frequency at which the characteristic impedance of the section is 600 ohm is 0.0 Hz \n",
- "and 1000 Ohm is 1600.0 kHz and 10000 ohm is 1996.4 kHz "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 809</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for each of the high-pass filter sections (i) the cut-off frequency, and (ii) the nominal impedance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L1 = 100*1E-3;# in Henry\n",
- "C1 = 0.2*1E-6;# in Fareads\n",
- "L2 = 200*1E-6;# in Henry\n",
- "C2 = 4000*1E-12;# in Fareads\n",
- "\n",
- "#calculation:\n",
- " #cut-off frequency\n",
- "fc1 = 1/(4*math.pi*(L1*C1/2)**0.5)\n",
- " #nominal impedance\n",
- "R01 = (L1*2/C1)**0.5\n",
- " #cut-off frequency\n",
- "fc2 = 1/(4*math.pi*(L2*C2/2)**0.5)\n",
- " #nominal impedance\n",
- "R02 = (L2/(C2*2))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n cut-off frequency \",round(fc1,0),\" Hz and the nominal impedance is \",round( R01,0),\" ohm\"\n",
- "print \"\\n cut-off frequency \",round(fc2/1000,0),\"KHz and the nominal impedance is \",round( R02,0),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " cut-off frequency 796.0 Hz and the nominal impedance is 1000.0 ohm\n",
- "\n",
- " cut-off frequency 126.0 KHz and the nominal impedance is 158.0 ohm "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 811</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) a high-pass T section filter and (b) a high-pass pi-section filter to meet these requirements\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 25000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #capacitance\n",
- "C1 = 2/(4*math.pi*R0*fc)\n",
- " #inductance\n",
- "L1 = R0/(4*math.pi*fc)\n",
- " #capacitance\n",
- "C2 = 1/(4*math.pi*R0*fc)\n",
- " #inductance\n",
- "L2 = 2*R0/(4*math.pi*fc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n A low-pass T section filter capcitance is \",round(C1*1E9,2),\"nfarad and inductance is\",round(L1*1E3,2),\"mHenry\"\n",
- "print \"\\n A high-pass pi section filter capcitance is \",round(C2*1E9,3),\"nfarad and inductance is\",round(L2*1E3,2),\"mHenry\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " A low-pass T section filter capcitance is 10.61 nfarad and inductance is 1.91 mHenry\n",
- "\n",
- " A high-pass pi section filter capcitance is 5.305 nfarad and inductance is 3.82 mHenry"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 814</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the frequency at which the characteristic impedance of the section is (a) zero, (b) 300 ohm, (c) 590 ohm\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 500;# in Hz\n",
- "Z1 = 0;# in ohm\n",
- "Z2 = 300;# in ohm\n",
- "Z3 = 590;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #frequency\n",
- "f1 = fc\n",
- "f2 = fc/(1 - (Z2/R0)**2)**0.5\n",
- "f3 = fc/(1 - (Z3/R0)**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"requency at which the characteristic impedance of the section is 0 ohm is \",f1,\" Hz \"\n",
- "print \"and 300 Ohm is \",round(f2,2),\" Hz and 590 ohm is \",round(f3,2),\" Hz \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "requency at which the characteristic impedance of the section is 0 ohm is 500 Hz \n",
- "and 300 Ohm is 577.35 Hz and 590 ohm is 2750.1 Hz "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 817</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for each (i) the attenuation coefficient, and (ii) the phase shift coefficient.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r1 = 1.25 + 0.52j;# propagation coefficients \n",
- "rr = 1.794;# propagation coefficients \n",
- "thetar = -39.4;# in ddegrees\n",
- "\n",
- "#calculation:\n",
- " #r\n",
- "r2 = rr*math.cos(thetar*math.pi/180) + 1j*rr*math.sin(thetar*math.pi/180)\n",
- " #attenuation coefficient\n",
- "a1 = r1.real\n",
- "a2 = r2.real\n",
- " #phase shift coefficient\n",
- "b1 = r1.imag\n",
- "b2 = r2.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nattenuation coefficient are for (a) is \",a1,\" N and for (b) is \",round(a2,2),\" N \"\n",
- "print \"\\nphase shift coefficient are for (a) is \",b1,\" rad and for (b) is \",round(b2,2),\" rad \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "attenuation coefficient are for (a) is 1.25 N and for (b) is 1.39 N \n",
- "\n",
- "phase shift coefficient are for (a) is 0.52 rad and for (b) is -1.14 rad "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 818</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the section (a) the attenuation coefficient, \n",
- "#(b) the phase shift coefficient, and (c) the propagation coefficient. \n",
- "#(d) If five such sections are cascaded determine the output current of the fifth stage and \n",
- "#the overall propagation constant of the network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ri1 = 0.024;# in amperes\n",
- "ri2 = 0.008;# in amperes\n",
- "thetai1 = 10;# in ddegrees\n",
- "thetai2 = -45;# in ddegrees\n",
- "\n",
- "#calculation:\n",
- " #currents\n",
- "I1 = ri1*math.cos(thetai1*math.pi/180) + 1j*ri1*math.sin(thetai1*math.pi/180)\n",
- "I2 = ri2*math.cos(thetai2*math.pi/180) + 1j*ri2*math.sin(thetai2*math.pi/180)\n",
- " #ir\n",
- "ir = I1/I2\n",
- "irmag = ri1/ri2\n",
- "thetai = thetai1-thetai2\n",
- " #attenuation coefficient\n",
- "a = math.log(irmag)\n",
- " #phase shift coefficient\n",
- "b = thetai*math.pi/180\n",
- " #propagation coefficient \n",
- "r = a + 1j*b\n",
- " #output current of the fifth stage\n",
- "I6 = I1/(ir**5)\n",
- "x = ir**5\n",
- "xmg = abs(x)\n",
- " #overall attenuation coefficient\n",
- "ad = math.log(xmg)\n",
- " #overall phase shift coefficient\n",
- "bd = cmath.phase(complex(x.real,x.imag))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nattenuation coefficient is \",round(a,3),\" N \"\n",
- "print \"\\nphase shift coefficient is \",round(b,3),\" rad \"\n",
- "print \"\\npropagation coefficient is \",round(a,3),\" + (\",round(b,3),\")i \"\n",
- "print \"\\nthe output current of the fifth stage is \",round(abs(I6*1E6),1),\"/_\",round(cmath.phase(complex(I6.real,I6.imag))*180/math.pi,2),\"deg mA \"\n",
- "print \"and the overall propagation coefficient is \",round(ad,2),\" + (\",round(bd+(2*math.pi),2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "attenuation coefficient is 1.099 N \n",
- "\n",
- "phase shift coefficient is 0.96 rad \n",
- "\n",
- "propagation coefficient is 1.099 + ( 0.96 )i \n",
- "\n",
- "the output current of the fifth stage is 98.8 /_ 95.0 deg mA \n",
- "and the overall propagation coefficient is 5.49 + ( 4.8 )i\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 819</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the attenuation coefficient, (b) the phase shift coefficient and (c) the propagation coefficient gamma\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "XL = 5j;# in ohms\n",
- "Xc = -1j*10;# in ohms\n",
- "RL = 12;# in ohms\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- " #calculation:\n",
- " #current I2\n",
- "I2 = (Xc/(Xc + XL + RL))*I1\n",
- " #current ratio\n",
- "Ir = I1/I2\n",
- "Irmg = abs(Ir)\n",
- " #attenuation coefficient\n",
- "a = math.log(Irmg)\n",
- " #phase shift coefficient\n",
- "b = cmath.phase(complex(Ir.real, Ir.imag))\n",
- " #propagation coefficient \n",
- "r = a + 1j*b\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nattenuation coefficient is \",round(a,2),\" N \"\n",
- "print \"\\nphase shift coefficient is \",round(b,2),\" rad \"\n",
- "print \"\\npropagation coefficient is \",round(a,2),\" + (\",round(b,2),\")i \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "attenuation coefficient is 0.26 N \n",
- "\n",
- "phase shift coefficient is 1.18 rad \n",
- "\n",
- "propagation coefficient is 0.26 + ( 1.18 )i "
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 823</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) the time delay for the signal to pass through the filter, assuming the phase shift is small, and \n",
- "#(b) the time delay for a signal to pass through the section at the cut-off frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 2*0.5;# in Henry\n",
- "C = 2E-9;# in Farad\n",
- "\n",
- "#calculation:\n",
- " #time delay\n",
- "t = (L*C)**0.5\n",
- " #time delay at the cut-off frequency\n",
- "tfc = t*math.pi/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n time delay is \",round(t*1E6,2),\"usec \"\n",
- "print \"\\ntime delay at the cut-off frequency is \",round(tfc*1E6,2),\"usec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " time delay is 44.72 usec \n",
- "\n",
- "time delay at the cut-off frequency is 70.25 usec"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 824</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the values of the elements in each section, and (b) the value of n.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fc = 500000;# in Hz\n",
- "t1 = 9.55E-6;# in secs\n",
- "R0 = 1000;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #for a low-pass filter section, capacitance\n",
- "C = 1/(math.pi*R0*fc)\n",
- " #inductance\n",
- "L = R0/(math.pi*fc)\n",
- " #time delay\n",
- "t2 = (L*C)**0.5\n",
- " #number of cascaded sections required\n",
- "n = t1/t2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n for low-pass T section inductance is \",round(L/2*1E6,2),\"uH and capacitance is \",round(C*1E12,2),\"pF\"\n",
- "print \"\\n for low-pass pi section inductance is \",round(L*1E6,2),\"uH and capacitance is \",round(C/2*1E12,2),\"pF\"\n",
- "print \"\\nnumber of cascaded sections required is \",round(n,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " for low-pass T section inductance is 318.31 uH and capacitance is 636.62 pF\n",
- "\n",
- " for low-pass pi section inductance is 636.62 uH and capacitance is 318.31 pF\n",
- "\n",
- "number of cascaded sections required is 15.0"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 824</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the component values for each section if the filter is (a) a low-pass T network, and (b) a high-pass \u0003 network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "n = 8;# sections in cascade\n",
- "R0 = 1000;# in ohm\n",
- "t1 = 4E-6;# in secs\n",
- "\n",
- "\n",
- "#calculation:\n",
- " #time delay\n",
- "t2 = t1/n\n",
- " #capacitance\n",
- "C = t2/R0\n",
- " #inductance\n",
- "L = t2*R0\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n for low-pass T section inductance is \",L/2*1E6,\"uH and capacitance is \",C*1E12,\"pF\"\n",
- "print \"\\n for high-pass pi section inductance is \",2*L*1E6,\"uH and capacitance is \",C*1E12,\"pF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " for low-pass T section inductance is 250.0 uH and capacitance is 500.0 pF\n",
- "\n",
- " for high-pass pi section inductance is 1000.0 uH and capacitance is 500.0 pF"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 829</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) an appropriate \u2018mderived\u2019 T section, and (b) an appropriate \u2018m-derived\u2019 pi section.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 5000; # in Hz\n",
- "finf = 5500; #in Hz\n",
- "\n",
- "#calculation:\n",
- "m = (1 - (fc/finf)**2)**0.5\n",
- "C = 1/(math.pi*R0*fc)\n",
- "L = R0/(math.pi*fc)\n",
- "\n",
- "LT = m*L/2\n",
- "CT = m*C\n",
- "Ls = (1- m**2)*L/(4*m)\n",
- "\n",
- "Cpi = m*C/2\n",
- "Lpi = m*L\n",
- "Cp = (1- (m**2))*C/(4*m)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n for mderived T section inductance is \",round(Ls*1000,2),\"mH and capacitance is \",round(CT*1E9,2),\"nF\"\n",
- "print \"\\n for mderived pi section inductance is \",round(Lpi*1000,2),\"mH and capacitance is \",round(Cp*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " for mderived T section inductance is 18.94 mH and capacitance is 44.2 nF\n",
- "\n",
- " for mderived pi section inductance is 15.91 mH and capacitance is 52.62 nF"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 832</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) a suitable \u2018m-derived\u2019 T section, and (b) a suitable \u2018m-derived\u2019 pi section having a cut-off frequency of 20 kHz,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 500;# in ohm\n",
- "fc = 20000; # in Hz\n",
- "finf = 16000; #in Hz\n",
- "\n",
- "#calculation:\n",
- "m = (1 - (finf/fc)**2)**0.5\n",
- "C = 1/(4*math.pi*R0*fc)\n",
- "L = R0/(4*math.pi*fc)\n",
- "\n",
- "LT = L/m\n",
- "CT = 4*m*C/(1- m**2)\n",
- "Csa = 2*C/m\n",
- "\n",
- "Cpi = C/m\n",
- "Lpi = 4*m*L/(1- m**2)\n",
- "Lsa = 2*L/m\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For an 'm-derived' high-pass T section: series arm contains a capacitance of \",round(Csa*1E9,2),\"nF\"\n",
- "print \"the shunt arm contains an inductance of\",round(LT*1000,3),\" mH in series with a capacitance of\",round(CT*1E9,2),\"nF\"\n",
- "print \"\\n For an 'm-derived' high pass pi section: shunt arms each contain inductance of \",round(Lsa*1000,2),\"mH\"\n",
- "print \"series arm contains a capacitance of \",round(Cpi*1E9,2),\"nF in parallel with an inductance of\",round(Lpi*1E3,3),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For an 'm-derived' high-pass T section: series arm contains a capacitance of 26.53 nF\n",
- "the shunt arm contains an inductance of 3.316 mH in series with a capacitance of 29.84 nF\n",
- "\n",
- " For an 'm-derived' high pass pi section: shunt arms each contain inductance of 6.63 mH\n",
- "series arm contains a capacitance of 13.26 nF in parallel with an inductance of 7.46 mH\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 835</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the component values needed if the filter is to comprise a prototype T section, \n",
- "#an \u2018m-derived\u2019 T section and two terminating \u2018m-derived\u2019 halfsections.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 10000; # in Hz\n",
- "finf = 11800; #in Hz\n",
- "\n",
- "#calculation:\n",
- "m = (1 - (fc/finf)**2)**0.5\n",
- "C = 1/(math.pi*R0*fc)\n",
- "L = R0/(math.pi*fc)\n",
- "\n",
- "LmT = (1- m**2)*L/(4*m)\n",
- "\n",
- "mH = 0.6\n",
- "LmH = (1- mH**2)*L/(2*mH)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For an Prototype T section: series arm contains a Inductance of \",round(L*1000/2,1),\"mH\"\n",
- "print \"the shunt arm contains an Capacitance of\",round(C*1E6,4),\" uF\"\n",
- "print \"\\n For an 'm-derived' T section: Series arms each contain inductance of \",round(m*L*1000/2,2),\"mH \"\n",
- "print \"Shunt arm contains a capacitance of \",round(m*C*1E6,4),\"uF in Series with an inductance of\",round(LmT*1E3,2),\"mH\"\n",
- "print \"\\n For an 'm-derived' Half section: Series arms each contain inductance of \",round(mH*L*1000/2,1),\"mH\"\n",
- "print \"Shunt arm contains a capacitance of \",round(mH*C*1E6/2,4),\"uF in Series with an inductance of\",round(LmH*1E3,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For an Prototype T section: series arm contains a Inductance of 9.5 mH\n",
- "the shunt arm contains an Capacitance of 0.0531 uF\n",
- "\n",
- " For an 'm-derived' T section: Series arms each contain inductance of 5.07 mH \n",
- "Shunt arm contains a capacitance of 0.0282 uF in Series with an inductance of 6.46 mH\n",
- "\n",
- " For an 'm-derived' Half section: Series arms each contain inductance of 5.7 mH\n",
- "Shunt arm contains a capacitance of 0.0159 uF in Series with an inductance of 10.19 mH\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_1.ipynb deleted file mode 100755 index fcac8b6a..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_1.ipynb +++ /dev/null @@ -1,959 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 42: Filter networks</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 799</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the cut-off frequency and the nominal impedance of each of the low-pass filter sections\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L1 = 2*100E-3;# in Henry\n",
- "C1 = 0.2E-6;# in Fareads\n",
- "L2 = 0.4;# in Henry\n",
- "C2 = 2*200E-12;# in Fareads\n",
- "\n",
- "#calculation:\n",
- " #cut-off frequency\n",
- "fc1 = 1/(math.pi*(L1*C1)**0.5)\n",
- " #nominal impedance\n",
- "R01 = (L1/C1)**0.5\n",
- " #cut-off frequency\n",
- "fc2 = 1/(math.pi*(L2*C2)**0.5)\n",
- " #nominal impedance\n",
- "R02 = (L2/C2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n cut-off frequency \",round(fc1,2),\" Hz and the nominal impedance is \",round( R01,2),\" ohm \"\n",
- "print \"\\n cut-off frequency \",round(fc2,2),\" Hz and the nominal impedance is \",round( R02,2),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " cut-off frequency 1591.55 Hz and the nominal impedance is 1000.0 ohm \n",
- "\n",
- " cut-off frequency 25164.61 Hz and the nominal impedance is 31622.78 ohm "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 801</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) a low-pass T section filter, and (b) a low-pass \u0003 section filter to meet these requirements.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 5E6;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = 1/(math.pi*R0*fc)\n",
- " #inductance\n",
- "L = R0/(math.pi*fc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"A low-pass T section filter capcitance is \",round(C*1E12,2),\"pfarad and inductance is\",round( L/2*1E6,2),\"uHenry\"\n",
- "print \"A low-pass pi section filter capcitance is \",round(C/2*1E12,2),\"pfarad and inductance is\",round( L*1E6,2),\"uHenry\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "A low-pass T section filter capcitance is 106.1 pfarad and inductance is 19.1 uHenry\n",
- "A low-pass pi section filter capcitance is 53.05 pfarad and inductance is 38.2 uHenry\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 805</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of the characteristic impedance of the section at a frequency of 90 kHz, and \n",
- "#(b) the value of the characteristic impedance of the equivalent low-pass T section filter.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 500;# in ohm\n",
- "fc = 100000;# in Hz\n",
- "f = 90000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #characteristic impedance of the pi section\n",
- "Zpi = R0/(1 - (f/fc)**2)**0.5\n",
- " #characteristic impedance of the T section\n",
- "Zt = R0*(1 - (f/fc)**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ncharacteristic impedance of the pi section is \",round(Zpi,2),\" ohm\"\n",
- "print \"\\ncharacteristic impedance of the T section is \",round(Zt,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "characteristic impedance of the pi section is 1147.08 ohm\n",
- "\n",
- "characteristic impedance of the T section is 217.94 ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 806</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the frequency at which the characteristic impedance of the section is (a) 600ohm (b) 1 kohm(c) 10 kohm\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 2E6;# in Hz\n",
- "Z1 = 600;# in ohm\n",
- "Z2 = 1000;# in ohm\n",
- "Z3 = 10000;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #frequency\n",
- "f1 = fc*(1 - (R0/Z1)**2)**0.5\n",
- "f2 = fc*(1 - (R0/Z2)**2)**0.5\n",
- "f3 = fc*(1 - (R0/Z3)**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"frequency at which the characteristic impedance of the section is 600 ohm is \",f1,\" Hz \"\n",
- "print \"and 1000 Ohm is \",f2*1E-3,\"kHz and 10000 ohm is \",round(f3*1E-3,2),\"kHz \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "frequency at which the characteristic impedance of the section is 600 ohm is 0.0 Hz \n",
- "and 1000 Ohm is 1600.0 kHz and 10000 ohm is 1996.4 kHz "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 809</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for each of the high-pass filter sections (i) the cut-off frequency, and (ii) the nominal impedance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L1 = 100*1E-3;# in Henry\n",
- "C1 = 0.2*1E-6;# in Fareads\n",
- "L2 = 200*1E-6;# in Henry\n",
- "C2 = 4000*1E-12;# in Fareads\n",
- "\n",
- "#calculation:\n",
- " #cut-off frequency\n",
- "fc1 = 1/(4*math.pi*(L1*C1/2)**0.5)\n",
- " #nominal impedance\n",
- "R01 = (L1*2/C1)**0.5\n",
- " #cut-off frequency\n",
- "fc2 = 1/(4*math.pi*(L2*C2/2)**0.5)\n",
- " #nominal impedance\n",
- "R02 = (L2/(C2*2))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n cut-off frequency \",round(fc1,0),\" Hz and the nominal impedance is \",round( R01,0),\" ohm\"\n",
- "print \"\\n cut-off frequency \",round(fc2/1000,0),\"KHz and the nominal impedance is \",round( R02,0),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " cut-off frequency 796.0 Hz and the nominal impedance is 1000.0 ohm\n",
- "\n",
- " cut-off frequency 126.0 KHz and the nominal impedance is 158.0 ohm "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 811</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) a high-pass T section filter and (b) a high-pass pi-section filter to meet these requirements\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 25000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #capacitance\n",
- "C1 = 2/(4*math.pi*R0*fc)\n",
- " #inductance\n",
- "L1 = R0/(4*math.pi*fc)\n",
- " #capacitance\n",
- "C2 = 1/(4*math.pi*R0*fc)\n",
- " #inductance\n",
- "L2 = 2*R0/(4*math.pi*fc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n A low-pass T section filter capcitance is \",round(C1*1E9,2),\"nfarad and inductance is\",round(L1*1E3,2),\"mHenry\"\n",
- "print \"\\n A high-pass pi section filter capcitance is \",round(C2*1E9,3),\"nfarad and inductance is\",round(L2*1E3,2),\"mHenry\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " A low-pass T section filter capcitance is 10.61 nfarad and inductance is 1.91 mHenry\n",
- "\n",
- " A high-pass pi section filter capcitance is 5.305 nfarad and inductance is 3.82 mHenry"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 814</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the frequency at which the characteristic impedance of the section is (a) zero, (b) 300 ohm, (c) 590 ohm\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 500;# in Hz\n",
- "Z1 = 0;# in ohm\n",
- "Z2 = 300;# in ohm\n",
- "Z3 = 590;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #frequency\n",
- "f1 = fc\n",
- "f2 = fc/(1 - (Z2/R0)**2)**0.5\n",
- "f3 = fc/(1 - (Z3/R0)**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"requency at which the characteristic impedance of the section is 0 ohm is \",f1,\" Hz \"\n",
- "print \"and 300 Ohm is \",round(f2,2),\" Hz and 590 ohm is \",round(f3,2),\" Hz \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "requency at which the characteristic impedance of the section is 0 ohm is 500 Hz \n",
- "and 300 Ohm is 577.35 Hz and 590 ohm is 2750.1 Hz "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 817</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for each (i) the attenuation coefficient, and (ii) the phase shift coefficient.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r1 = 1.25 + 0.52j;# propagation coefficients \n",
- "rr = 1.794;# propagation coefficients \n",
- "thetar = -39.4;# in ddegrees\n",
- "\n",
- "#calculation:\n",
- " #r\n",
- "r2 = rr*math.cos(thetar*math.pi/180) + 1j*rr*math.sin(thetar*math.pi/180)\n",
- " #attenuation coefficient\n",
- "a1 = r1.real\n",
- "a2 = r2.real\n",
- " #phase shift coefficient\n",
- "b1 = r1.imag\n",
- "b2 = r2.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nattenuation coefficient are for (a) is \",a1,\" N and for (b) is \",round(a2,2),\" N \"\n",
- "print \"\\nphase shift coefficient are for (a) is \",b1,\" rad and for (b) is \",round(b2,2),\" rad \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "attenuation coefficient are for (a) is 1.25 N and for (b) is 1.39 N \n",
- "\n",
- "phase shift coefficient are for (a) is 0.52 rad and for (b) is -1.14 rad "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 818</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the section (a) the attenuation coefficient, \n",
- "#(b) the phase shift coefficient, and (c) the propagation coefficient. \n",
- "#(d) If five such sections are cascaded determine the output current of the fifth stage and \n",
- "#the overall propagation constant of the network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ri1 = 0.024;# in amperes\n",
- "ri2 = 0.008;# in amperes\n",
- "thetai1 = 10;# in ddegrees\n",
- "thetai2 = -45;# in ddegrees\n",
- "\n",
- "#calculation:\n",
- " #currents\n",
- "I1 = ri1*math.cos(thetai1*math.pi/180) + 1j*ri1*math.sin(thetai1*math.pi/180)\n",
- "I2 = ri2*math.cos(thetai2*math.pi/180) + 1j*ri2*math.sin(thetai2*math.pi/180)\n",
- " #ir\n",
- "ir = I1/I2\n",
- "irmag = ri1/ri2\n",
- "thetai = thetai1-thetai2\n",
- " #attenuation coefficient\n",
- "a = math.log(irmag)\n",
- " #phase shift coefficient\n",
- "b = thetai*math.pi/180\n",
- " #propagation coefficient \n",
- "r = a + 1j*b\n",
- " #output current of the fifth stage\n",
- "I6 = I1/(ir**5)\n",
- "x = ir**5\n",
- "xmg = abs(x)\n",
- " #overall attenuation coefficient\n",
- "ad = math.log(xmg)\n",
- " #overall phase shift coefficient\n",
- "bd = cmath.phase(complex(x.real,x.imag))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nattenuation coefficient is \",round(a,3),\" N \"\n",
- "print \"\\nphase shift coefficient is \",round(b,3),\" rad \"\n",
- "print \"\\npropagation coefficient is \",round(a,3),\" + (\",round(b,3),\")i \"\n",
- "print \"\\nthe output current of the fifth stage is \",round(abs(I6*1E6),1),\"/_\",round(cmath.phase(complex(I6.real,I6.imag))*180/math.pi,2),\"deg mA \"\n",
- "print \"and the overall propagation coefficient is \",round(ad,2),\" + (\",round(bd+(2*math.pi),2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "attenuation coefficient is 1.099 N \n",
- "\n",
- "phase shift coefficient is 0.96 rad \n",
- "\n",
- "propagation coefficient is 1.099 + ( 0.96 )i \n",
- "\n",
- "the output current of the fifth stage is 98.8 /_ 95.0 deg mA \n",
- "and the overall propagation coefficient is 5.49 + ( 4.8 )i\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 819</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the attenuation coefficient, (b) the phase shift coefficient and (c) the propagation coefficient gamma\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "XL = 5j;# in ohms\n",
- "Xc = -1j*10;# in ohms\n",
- "RL = 12;# in ohms\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- " #calculation:\n",
- " #current I2\n",
- "I2 = (Xc/(Xc + XL + RL))*I1\n",
- " #current ratio\n",
- "Ir = I1/I2\n",
- "Irmg = abs(Ir)\n",
- " #attenuation coefficient\n",
- "a = math.log(Irmg)\n",
- " #phase shift coefficient\n",
- "b = cmath.phase(complex(Ir.real, Ir.imag))\n",
- " #propagation coefficient \n",
- "r = a + 1j*b\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nattenuation coefficient is \",round(a,2),\" N \"\n",
- "print \"\\nphase shift coefficient is \",round(b,2),\" rad \"\n",
- "print \"\\npropagation coefficient is \",round(a,2),\" + (\",round(b,2),\")i \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "attenuation coefficient is 0.26 N \n",
- "\n",
- "phase shift coefficient is 1.18 rad \n",
- "\n",
- "propagation coefficient is 0.26 + ( 1.18 )i "
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 823</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) the time delay for the signal to pass through the filter, assuming the phase shift is small, and \n",
- "#(b) the time delay for a signal to pass through the section at the cut-off frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 2*0.5;# in Henry\n",
- "C = 2E-9;# in Farad\n",
- "\n",
- "#calculation:\n",
- " #time delay\n",
- "t = (L*C)**0.5\n",
- " #time delay at the cut-off frequency\n",
- "tfc = t*math.pi/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n time delay is \",round(t*1E6,2),\"usec \"\n",
- "print \"\\ntime delay at the cut-off frequency is \",round(tfc*1E6,2),\"usec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " time delay is 44.72 usec \n",
- "\n",
- "time delay at the cut-off frequency is 70.25 usec"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 824</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the values of the elements in each section, and (b) the value of n.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fc = 500000;# in Hz\n",
- "t1 = 9.55E-6;# in secs\n",
- "R0 = 1000;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #for a low-pass filter section, capacitance\n",
- "C = 1/(math.pi*R0*fc)\n",
- " #inductance\n",
- "L = R0/(math.pi*fc)\n",
- " #time delay\n",
- "t2 = (L*C)**0.5\n",
- " #number of cascaded sections required\n",
- "n = t1/t2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n for low-pass T section inductance is \",round(L/2*1E6,2),\"uH and capacitance is \",round(C*1E12,2),\"pF\"\n",
- "print \"\\n for low-pass pi section inductance is \",round(L*1E6,2),\"uH and capacitance is \",round(C/2*1E12,2),\"pF\"\n",
- "print \"\\nnumber of cascaded sections required is \",round(n,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " for low-pass T section inductance is 318.31 uH and capacitance is 636.62 pF\n",
- "\n",
- " for low-pass pi section inductance is 636.62 uH and capacitance is 318.31 pF\n",
- "\n",
- "number of cascaded sections required is 15.0"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 824</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the component values for each section if the filter is (a) a low-pass T network, and (b) a high-pass \u0003 network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "n = 8;# sections in cascade\n",
- "R0 = 1000;# in ohm\n",
- "t1 = 4E-6;# in secs\n",
- "\n",
- "\n",
- "#calculation:\n",
- " #time delay\n",
- "t2 = t1/n\n",
- " #capacitance\n",
- "C = t2/R0\n",
- " #inductance\n",
- "L = t2*R0\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n for low-pass T section inductance is \",L/2*1E6,\"uH and capacitance is \",C*1E12,\"pF\"\n",
- "print \"\\n for high-pass pi section inductance is \",2*L*1E6,\"uH and capacitance is \",C*1E12,\"pF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " for low-pass T section inductance is 250.0 uH and capacitance is 500.0 pF\n",
- "\n",
- " for high-pass pi section inductance is 1000.0 uH and capacitance is 500.0 pF"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 829</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) an appropriate \u2018mderived\u2019 T section, and (b) an appropriate \u2018m-derived\u2019 pi section.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 5000; # in Hz\n",
- "finf = 5500; #in Hz\n",
- "\n",
- "#calculation:\n",
- "m = (1 - (fc/finf)**2)**0.5\n",
- "C = 1/(math.pi*R0*fc)\n",
- "L = R0/(math.pi*fc)\n",
- "\n",
- "LT = m*L/2\n",
- "CT = m*C\n",
- "Ls = (1- m**2)*L/(4*m)\n",
- "\n",
- "Cpi = m*C/2\n",
- "Lpi = m*L\n",
- "Cp = (1- (m**2))*C/(4*m)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n for mderived T section inductance is \",round(Ls*1000,2),\"mH and capacitance is \",round(CT*1E9,2),\"nF\"\n",
- "print \"\\n for mderived pi section inductance is \",round(Lpi*1000,2),\"mH and capacitance is \",round(Cp*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " for mderived T section inductance is 18.94 mH and capacitance is 44.2 nF\n",
- "\n",
- " for mderived pi section inductance is 15.91 mH and capacitance is 52.62 nF"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 832</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) a suitable \u2018m-derived\u2019 T section, and (b) a suitable \u2018m-derived\u2019 pi section having a cut-off frequency of 20 kHz,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 500;# in ohm\n",
- "fc = 20000; # in Hz\n",
- "finf = 16000; #in Hz\n",
- "\n",
- "#calculation:\n",
- "m = (1 - (finf/fc)**2)**0.5\n",
- "C = 1/(4*math.pi*R0*fc)\n",
- "L = R0/(4*math.pi*fc)\n",
- "\n",
- "LT = L/m\n",
- "CT = 4*m*C/(1- m**2)\n",
- "Csa = 2*C/m\n",
- "\n",
- "Cpi = C/m\n",
- "Lpi = 4*m*L/(1- m**2)\n",
- "Lsa = 2*L/m\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For an 'm-derived' high-pass T section: series arm contains a capacitance of \",round(Csa*1E9,2),\"nF\"\n",
- "print \"the shunt arm contains an inductance of\",round(LT*1000,3),\" mH in series with a capacitance of\",round(CT*1E9,2),\"nF\"\n",
- "print \"\\n For an 'm-derived' high pass pi section: shunt arms each contain inductance of \",round(Lsa*1000,2),\"mH\"\n",
- "print \"series arm contains a capacitance of \",round(Cpi*1E9,2),\"nF in parallel with an inductance of\",round(Lpi*1E3,3),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For an 'm-derived' high-pass T section: series arm contains a capacitance of 26.53 nF\n",
- "the shunt arm contains an inductance of 3.316 mH in series with a capacitance of 29.84 nF\n",
- "\n",
- " For an 'm-derived' high pass pi section: shunt arms each contain inductance of 6.63 mH\n",
- "series arm contains a capacitance of 13.26 nF in parallel with an inductance of 7.46 mH\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 835</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the component values needed if the filter is to comprise a prototype T section, \n",
- "#an \u2018m-derived\u2019 T section and two terminating \u2018m-derived\u2019 halfsections.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 10000; # in Hz\n",
- "finf = 11800; #in Hz\n",
- "\n",
- "#calculation:\n",
- "m = (1 - (fc/finf)**2)**0.5\n",
- "C = 1/(math.pi*R0*fc)\n",
- "L = R0/(math.pi*fc)\n",
- "\n",
- "LmT = (1- m**2)*L/(4*m)\n",
- "\n",
- "mH = 0.6\n",
- "LmH = (1- mH**2)*L/(2*mH)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For an Prototype T section: series arm contains a Inductance of \",round(L*1000/2,1),\"mH\"\n",
- "print \"the shunt arm contains an Capacitance of\",round(C*1E6,4),\" uF\"\n",
- "print \"\\n For an 'm-derived' T section: Series arms each contain inductance of \",round(m*L*1000/2,2),\"mH \"\n",
- "print \"Shunt arm contains a capacitance of \",round(m*C*1E6,4),\"uF in Series with an inductance of\",round(LmT*1E3,2),\"mH\"\n",
- "print \"\\n For an 'm-derived' Half section: Series arms each contain inductance of \",round(mH*L*1000/2,1),\"mH\"\n",
- "print \"Shunt arm contains a capacitance of \",round(mH*C*1E6/2,4),\"uF in Series with an inductance of\",round(LmH*1E3,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For an Prototype T section: series arm contains a Inductance of 9.5 mH\n",
- "the shunt arm contains an Capacitance of 0.0531 uF\n",
- "\n",
- " For an 'm-derived' T section: Series arms each contain inductance of 5.07 mH \n",
- "Shunt arm contains a capacitance of 0.0282 uF in Series with an inductance of 6.46 mH\n",
- "\n",
- " For an 'm-derived' Half section: Series arms each contain inductance of 5.7 mH\n",
- "Shunt arm contains a capacitance of 0.0159 uF in Series with an inductance of 10.19 mH\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_2.ipynb deleted file mode 100755 index fcac8b6a..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_2.ipynb +++ /dev/null @@ -1,959 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 42: Filter networks</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 799</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the cut-off frequency and the nominal impedance of each of the low-pass filter sections\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L1 = 2*100E-3;# in Henry\n",
- "C1 = 0.2E-6;# in Fareads\n",
- "L2 = 0.4;# in Henry\n",
- "C2 = 2*200E-12;# in Fareads\n",
- "\n",
- "#calculation:\n",
- " #cut-off frequency\n",
- "fc1 = 1/(math.pi*(L1*C1)**0.5)\n",
- " #nominal impedance\n",
- "R01 = (L1/C1)**0.5\n",
- " #cut-off frequency\n",
- "fc2 = 1/(math.pi*(L2*C2)**0.5)\n",
- " #nominal impedance\n",
- "R02 = (L2/C2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n cut-off frequency \",round(fc1,2),\" Hz and the nominal impedance is \",round( R01,2),\" ohm \"\n",
- "print \"\\n cut-off frequency \",round(fc2,2),\" Hz and the nominal impedance is \",round( R02,2),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " cut-off frequency 1591.55 Hz and the nominal impedance is 1000.0 ohm \n",
- "\n",
- " cut-off frequency 25164.61 Hz and the nominal impedance is 31622.78 ohm "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 801</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) a low-pass T section filter, and (b) a low-pass \u0003 section filter to meet these requirements.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 5E6;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = 1/(math.pi*R0*fc)\n",
- " #inductance\n",
- "L = R0/(math.pi*fc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"A low-pass T section filter capcitance is \",round(C*1E12,2),\"pfarad and inductance is\",round( L/2*1E6,2),\"uHenry\"\n",
- "print \"A low-pass pi section filter capcitance is \",round(C/2*1E12,2),\"pfarad and inductance is\",round( L*1E6,2),\"uHenry\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "A low-pass T section filter capcitance is 106.1 pfarad and inductance is 19.1 uHenry\n",
- "A low-pass pi section filter capcitance is 53.05 pfarad and inductance is 38.2 uHenry\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 805</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of the characteristic impedance of the section at a frequency of 90 kHz, and \n",
- "#(b) the value of the characteristic impedance of the equivalent low-pass T section filter.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 500;# in ohm\n",
- "fc = 100000;# in Hz\n",
- "f = 90000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #characteristic impedance of the pi section\n",
- "Zpi = R0/(1 - (f/fc)**2)**0.5\n",
- " #characteristic impedance of the T section\n",
- "Zt = R0*(1 - (f/fc)**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ncharacteristic impedance of the pi section is \",round(Zpi,2),\" ohm\"\n",
- "print \"\\ncharacteristic impedance of the T section is \",round(Zt,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "characteristic impedance of the pi section is 1147.08 ohm\n",
- "\n",
- "characteristic impedance of the T section is 217.94 ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 806</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the frequency at which the characteristic impedance of the section is (a) 600ohm (b) 1 kohm(c) 10 kohm\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 2E6;# in Hz\n",
- "Z1 = 600;# in ohm\n",
- "Z2 = 1000;# in ohm\n",
- "Z3 = 10000;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #frequency\n",
- "f1 = fc*(1 - (R0/Z1)**2)**0.5\n",
- "f2 = fc*(1 - (R0/Z2)**2)**0.5\n",
- "f3 = fc*(1 - (R0/Z3)**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"frequency at which the characteristic impedance of the section is 600 ohm is \",f1,\" Hz \"\n",
- "print \"and 1000 Ohm is \",f2*1E-3,\"kHz and 10000 ohm is \",round(f3*1E-3,2),\"kHz \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "frequency at which the characteristic impedance of the section is 600 ohm is 0.0 Hz \n",
- "and 1000 Ohm is 1600.0 kHz and 10000 ohm is 1996.4 kHz "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 809</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for each of the high-pass filter sections (i) the cut-off frequency, and (ii) the nominal impedance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L1 = 100*1E-3;# in Henry\n",
- "C1 = 0.2*1E-6;# in Fareads\n",
- "L2 = 200*1E-6;# in Henry\n",
- "C2 = 4000*1E-12;# in Fareads\n",
- "\n",
- "#calculation:\n",
- " #cut-off frequency\n",
- "fc1 = 1/(4*math.pi*(L1*C1/2)**0.5)\n",
- " #nominal impedance\n",
- "R01 = (L1*2/C1)**0.5\n",
- " #cut-off frequency\n",
- "fc2 = 1/(4*math.pi*(L2*C2/2)**0.5)\n",
- " #nominal impedance\n",
- "R02 = (L2/(C2*2))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n cut-off frequency \",round(fc1,0),\" Hz and the nominal impedance is \",round( R01,0),\" ohm\"\n",
- "print \"\\n cut-off frequency \",round(fc2/1000,0),\"KHz and the nominal impedance is \",round( R02,0),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " cut-off frequency 796.0 Hz and the nominal impedance is 1000.0 ohm\n",
- "\n",
- " cut-off frequency 126.0 KHz and the nominal impedance is 158.0 ohm "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 811</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) a high-pass T section filter and (b) a high-pass pi-section filter to meet these requirements\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 25000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #capacitance\n",
- "C1 = 2/(4*math.pi*R0*fc)\n",
- " #inductance\n",
- "L1 = R0/(4*math.pi*fc)\n",
- " #capacitance\n",
- "C2 = 1/(4*math.pi*R0*fc)\n",
- " #inductance\n",
- "L2 = 2*R0/(4*math.pi*fc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n A low-pass T section filter capcitance is \",round(C1*1E9,2),\"nfarad and inductance is\",round(L1*1E3,2),\"mHenry\"\n",
- "print \"\\n A high-pass pi section filter capcitance is \",round(C2*1E9,3),\"nfarad and inductance is\",round(L2*1E3,2),\"mHenry\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " A low-pass T section filter capcitance is 10.61 nfarad and inductance is 1.91 mHenry\n",
- "\n",
- " A high-pass pi section filter capcitance is 5.305 nfarad and inductance is 3.82 mHenry"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 814</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the frequency at which the characteristic impedance of the section is (a) zero, (b) 300 ohm, (c) 590 ohm\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 500;# in Hz\n",
- "Z1 = 0;# in ohm\n",
- "Z2 = 300;# in ohm\n",
- "Z3 = 590;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #frequency\n",
- "f1 = fc\n",
- "f2 = fc/(1 - (Z2/R0)**2)**0.5\n",
- "f3 = fc/(1 - (Z3/R0)**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"requency at which the characteristic impedance of the section is 0 ohm is \",f1,\" Hz \"\n",
- "print \"and 300 Ohm is \",round(f2,2),\" Hz and 590 ohm is \",round(f3,2),\" Hz \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "requency at which the characteristic impedance of the section is 0 ohm is 500 Hz \n",
- "and 300 Ohm is 577.35 Hz and 590 ohm is 2750.1 Hz "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 817</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for each (i) the attenuation coefficient, and (ii) the phase shift coefficient.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r1 = 1.25 + 0.52j;# propagation coefficients \n",
- "rr = 1.794;# propagation coefficients \n",
- "thetar = -39.4;# in ddegrees\n",
- "\n",
- "#calculation:\n",
- " #r\n",
- "r2 = rr*math.cos(thetar*math.pi/180) + 1j*rr*math.sin(thetar*math.pi/180)\n",
- " #attenuation coefficient\n",
- "a1 = r1.real\n",
- "a2 = r2.real\n",
- " #phase shift coefficient\n",
- "b1 = r1.imag\n",
- "b2 = r2.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nattenuation coefficient are for (a) is \",a1,\" N and for (b) is \",round(a2,2),\" N \"\n",
- "print \"\\nphase shift coefficient are for (a) is \",b1,\" rad and for (b) is \",round(b2,2),\" rad \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "attenuation coefficient are for (a) is 1.25 N and for (b) is 1.39 N \n",
- "\n",
- "phase shift coefficient are for (a) is 0.52 rad and for (b) is -1.14 rad "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 818</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the section (a) the attenuation coefficient, \n",
- "#(b) the phase shift coefficient, and (c) the propagation coefficient. \n",
- "#(d) If five such sections are cascaded determine the output current of the fifth stage and \n",
- "#the overall propagation constant of the network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ri1 = 0.024;# in amperes\n",
- "ri2 = 0.008;# in amperes\n",
- "thetai1 = 10;# in ddegrees\n",
- "thetai2 = -45;# in ddegrees\n",
- "\n",
- "#calculation:\n",
- " #currents\n",
- "I1 = ri1*math.cos(thetai1*math.pi/180) + 1j*ri1*math.sin(thetai1*math.pi/180)\n",
- "I2 = ri2*math.cos(thetai2*math.pi/180) + 1j*ri2*math.sin(thetai2*math.pi/180)\n",
- " #ir\n",
- "ir = I1/I2\n",
- "irmag = ri1/ri2\n",
- "thetai = thetai1-thetai2\n",
- " #attenuation coefficient\n",
- "a = math.log(irmag)\n",
- " #phase shift coefficient\n",
- "b = thetai*math.pi/180\n",
- " #propagation coefficient \n",
- "r = a + 1j*b\n",
- " #output current of the fifth stage\n",
- "I6 = I1/(ir**5)\n",
- "x = ir**5\n",
- "xmg = abs(x)\n",
- " #overall attenuation coefficient\n",
- "ad = math.log(xmg)\n",
- " #overall phase shift coefficient\n",
- "bd = cmath.phase(complex(x.real,x.imag))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nattenuation coefficient is \",round(a,3),\" N \"\n",
- "print \"\\nphase shift coefficient is \",round(b,3),\" rad \"\n",
- "print \"\\npropagation coefficient is \",round(a,3),\" + (\",round(b,3),\")i \"\n",
- "print \"\\nthe output current of the fifth stage is \",round(abs(I6*1E6),1),\"/_\",round(cmath.phase(complex(I6.real,I6.imag))*180/math.pi,2),\"deg mA \"\n",
- "print \"and the overall propagation coefficient is \",round(ad,2),\" + (\",round(bd+(2*math.pi),2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "attenuation coefficient is 1.099 N \n",
- "\n",
- "phase shift coefficient is 0.96 rad \n",
- "\n",
- "propagation coefficient is 1.099 + ( 0.96 )i \n",
- "\n",
- "the output current of the fifth stage is 98.8 /_ 95.0 deg mA \n",
- "and the overall propagation coefficient is 5.49 + ( 4.8 )i\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 819</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the attenuation coefficient, (b) the phase shift coefficient and (c) the propagation coefficient gamma\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "XL = 5j;# in ohms\n",
- "Xc = -1j*10;# in ohms\n",
- "RL = 12;# in ohms\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- " #calculation:\n",
- " #current I2\n",
- "I2 = (Xc/(Xc + XL + RL))*I1\n",
- " #current ratio\n",
- "Ir = I1/I2\n",
- "Irmg = abs(Ir)\n",
- " #attenuation coefficient\n",
- "a = math.log(Irmg)\n",
- " #phase shift coefficient\n",
- "b = cmath.phase(complex(Ir.real, Ir.imag))\n",
- " #propagation coefficient \n",
- "r = a + 1j*b\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nattenuation coefficient is \",round(a,2),\" N \"\n",
- "print \"\\nphase shift coefficient is \",round(b,2),\" rad \"\n",
- "print \"\\npropagation coefficient is \",round(a,2),\" + (\",round(b,2),\")i \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "attenuation coefficient is 0.26 N \n",
- "\n",
- "phase shift coefficient is 1.18 rad \n",
- "\n",
- "propagation coefficient is 0.26 + ( 1.18 )i "
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 823</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) the time delay for the signal to pass through the filter, assuming the phase shift is small, and \n",
- "#(b) the time delay for a signal to pass through the section at the cut-off frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 2*0.5;# in Henry\n",
- "C = 2E-9;# in Farad\n",
- "\n",
- "#calculation:\n",
- " #time delay\n",
- "t = (L*C)**0.5\n",
- " #time delay at the cut-off frequency\n",
- "tfc = t*math.pi/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n time delay is \",round(t*1E6,2),\"usec \"\n",
- "print \"\\ntime delay at the cut-off frequency is \",round(tfc*1E6,2),\"usec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " time delay is 44.72 usec \n",
- "\n",
- "time delay at the cut-off frequency is 70.25 usec"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 824</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the values of the elements in each section, and (b) the value of n.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fc = 500000;# in Hz\n",
- "t1 = 9.55E-6;# in secs\n",
- "R0 = 1000;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #for a low-pass filter section, capacitance\n",
- "C = 1/(math.pi*R0*fc)\n",
- " #inductance\n",
- "L = R0/(math.pi*fc)\n",
- " #time delay\n",
- "t2 = (L*C)**0.5\n",
- " #number of cascaded sections required\n",
- "n = t1/t2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n for low-pass T section inductance is \",round(L/2*1E6,2),\"uH and capacitance is \",round(C*1E12,2),\"pF\"\n",
- "print \"\\n for low-pass pi section inductance is \",round(L*1E6,2),\"uH and capacitance is \",round(C/2*1E12,2),\"pF\"\n",
- "print \"\\nnumber of cascaded sections required is \",round(n,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " for low-pass T section inductance is 318.31 uH and capacitance is 636.62 pF\n",
- "\n",
- " for low-pass pi section inductance is 636.62 uH and capacitance is 318.31 pF\n",
- "\n",
- "number of cascaded sections required is 15.0"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 824</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the component values for each section if the filter is (a) a low-pass T network, and (b) a high-pass \u0003 network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "n = 8;# sections in cascade\n",
- "R0 = 1000;# in ohm\n",
- "t1 = 4E-6;# in secs\n",
- "\n",
- "\n",
- "#calculation:\n",
- " #time delay\n",
- "t2 = t1/n\n",
- " #capacitance\n",
- "C = t2/R0\n",
- " #inductance\n",
- "L = t2*R0\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n for low-pass T section inductance is \",L/2*1E6,\"uH and capacitance is \",C*1E12,\"pF\"\n",
- "print \"\\n for high-pass pi section inductance is \",2*L*1E6,\"uH and capacitance is \",C*1E12,\"pF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " for low-pass T section inductance is 250.0 uH and capacitance is 500.0 pF\n",
- "\n",
- " for high-pass pi section inductance is 1000.0 uH and capacitance is 500.0 pF"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 829</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) an appropriate \u2018mderived\u2019 T section, and (b) an appropriate \u2018m-derived\u2019 pi section.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 5000; # in Hz\n",
- "finf = 5500; #in Hz\n",
- "\n",
- "#calculation:\n",
- "m = (1 - (fc/finf)**2)**0.5\n",
- "C = 1/(math.pi*R0*fc)\n",
- "L = R0/(math.pi*fc)\n",
- "\n",
- "LT = m*L/2\n",
- "CT = m*C\n",
- "Ls = (1- m**2)*L/(4*m)\n",
- "\n",
- "Cpi = m*C/2\n",
- "Lpi = m*L\n",
- "Cp = (1- (m**2))*C/(4*m)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n for mderived T section inductance is \",round(Ls*1000,2),\"mH and capacitance is \",round(CT*1E9,2),\"nF\"\n",
- "print \"\\n for mderived pi section inductance is \",round(Lpi*1000,2),\"mH and capacitance is \",round(Cp*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " for mderived T section inductance is 18.94 mH and capacitance is 44.2 nF\n",
- "\n",
- " for mderived pi section inductance is 15.91 mH and capacitance is 52.62 nF"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 832</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) a suitable \u2018m-derived\u2019 T section, and (b) a suitable \u2018m-derived\u2019 pi section having a cut-off frequency of 20 kHz,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 500;# in ohm\n",
- "fc = 20000; # in Hz\n",
- "finf = 16000; #in Hz\n",
- "\n",
- "#calculation:\n",
- "m = (1 - (finf/fc)**2)**0.5\n",
- "C = 1/(4*math.pi*R0*fc)\n",
- "L = R0/(4*math.pi*fc)\n",
- "\n",
- "LT = L/m\n",
- "CT = 4*m*C/(1- m**2)\n",
- "Csa = 2*C/m\n",
- "\n",
- "Cpi = C/m\n",
- "Lpi = 4*m*L/(1- m**2)\n",
- "Lsa = 2*L/m\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For an 'm-derived' high-pass T section: series arm contains a capacitance of \",round(Csa*1E9,2),\"nF\"\n",
- "print \"the shunt arm contains an inductance of\",round(LT*1000,3),\" mH in series with a capacitance of\",round(CT*1E9,2),\"nF\"\n",
- "print \"\\n For an 'm-derived' high pass pi section: shunt arms each contain inductance of \",round(Lsa*1000,2),\"mH\"\n",
- "print \"series arm contains a capacitance of \",round(Cpi*1E9,2),\"nF in parallel with an inductance of\",round(Lpi*1E3,3),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For an 'm-derived' high-pass T section: series arm contains a capacitance of 26.53 nF\n",
- "the shunt arm contains an inductance of 3.316 mH in series with a capacitance of 29.84 nF\n",
- "\n",
- " For an 'm-derived' high pass pi section: shunt arms each contain inductance of 6.63 mH\n",
- "series arm contains a capacitance of 13.26 nF in parallel with an inductance of 7.46 mH\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 835</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the component values needed if the filter is to comprise a prototype T section, \n",
- "#an \u2018m-derived\u2019 T section and two terminating \u2018m-derived\u2019 halfsections.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 10000; # in Hz\n",
- "finf = 11800; #in Hz\n",
- "\n",
- "#calculation:\n",
- "m = (1 - (fc/finf)**2)**0.5\n",
- "C = 1/(math.pi*R0*fc)\n",
- "L = R0/(math.pi*fc)\n",
- "\n",
- "LmT = (1- m**2)*L/(4*m)\n",
- "\n",
- "mH = 0.6\n",
- "LmH = (1- mH**2)*L/(2*mH)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For an Prototype T section: series arm contains a Inductance of \",round(L*1000/2,1),\"mH\"\n",
- "print \"the shunt arm contains an Capacitance of\",round(C*1E6,4),\" uF\"\n",
- "print \"\\n For an 'm-derived' T section: Series arms each contain inductance of \",round(m*L*1000/2,2),\"mH \"\n",
- "print \"Shunt arm contains a capacitance of \",round(m*C*1E6,4),\"uF in Series with an inductance of\",round(LmT*1E3,2),\"mH\"\n",
- "print \"\\n For an 'm-derived' Half section: Series arms each contain inductance of \",round(mH*L*1000/2,1),\"mH\"\n",
- "print \"Shunt arm contains a capacitance of \",round(mH*C*1E6/2,4),\"uF in Series with an inductance of\",round(LmH*1E3,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For an Prototype T section: series arm contains a Inductance of 9.5 mH\n",
- "the shunt arm contains an Capacitance of 0.0531 uF\n",
- "\n",
- " For an 'm-derived' T section: Series arms each contain inductance of 5.07 mH \n",
- "Shunt arm contains a capacitance of 0.0282 uF in Series with an inductance of 6.46 mH\n",
- "\n",
- " For an 'm-derived' Half section: Series arms each contain inductance of 5.7 mH\n",
- "Shunt arm contains a capacitance of 0.0159 uF in Series with an inductance of 10.19 mH\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_42.ipynb index d475997f..d475997f 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_42.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint.ipynb deleted file mode 100755 index 618fd99a..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint.ipynb +++ /dev/null @@ -1,1037 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 43: Magnetically coupled circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 842</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the self inductance of coil A, and (b) the mutual inductance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Na = 1200; \n",
- "Nb = 1000;\n",
- "Ia = 0.8;# in amperes\n",
- "Phia = 100E-6;# in Wb\n",
- "xb = 0.75;\n",
- "\n",
- " #calculation:\n",
- " #self inductance of coil A\n",
- "La = Na*Phia/Ia\n",
- " #mutual inductance, M\n",
- "Phib = xb*Phia\n",
- "M = Nb*Phib/Ia\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n self inductance of coil A is \",La,\" H\"\n",
- "print \"\\n mutual inductance, M is \",M,\"H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " self inductance of coil A is 0.15 H\n",
- "\n",
- " mutual inductance, M is 0.09375 H"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 843</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. induced in the secondary,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "M = 600E-3;# in Henry\n",
- "Ia = 5;# in amperes\n",
- "dt = 0.2;# in secs\n",
- "\n",
- " #calculation:\n",
- " #change of current\n",
- "dIa = 2*Ia \n",
- "dIadt = dIa/dt\n",
- " #secondary induced e.m.f., E2\n",
- "E2 = -1*M*dIadt\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n secondary induced e.m.f., E2 is \",E2,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " secondary induced e.m.f., E2 is -30.0 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 844</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the magnetic coupling coefficient of the pair of coils\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "La = 250E-3;# in Henry\n",
- "Lb = 400E-3;# in Henry\n",
- "M = 80E-3;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #coupling coefficient,\n",
- "k = M/(La*Lb)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n coupling coefficient, is \",round(k,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " coupling coefficient, is 0.253"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 845</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the mutual inductance between the coils, and (b) the coefficient of coupling.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Lx = 80E-3;# in Henry\n",
- "Ly = 60E-3;# in Henry\n",
- "Nx = 200;# turns\n",
- "Ny = 100;# turns\n",
- "Ix = 4;# in Amperes\n",
- "Phiy = 0.005;# in Wb\n",
- "\n",
- "#calculation:\n",
- " #mutual inductance, M\n",
- "M = Ny*Phiy/(2*Ix)\n",
- " #coupling coefficient,\n",
- "k = M/(Lx*Ly)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance, M is \",M*1E3,\"mH\"\n",
- "print \"\\n coupling coefficient, is \",round(k,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance, M is 62.5 mH\n",
- "\n",
- " coupling coefficient, is 0.902"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 846</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the mutual inductance between the two coils, and (b) the coefficient of coupling.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "La = 40E-3;# in Henry\n",
- "Lb = 10E-3;# in Henry\n",
- "L = 60E-3;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #mutual inductance, M\n",
- "M = (L - La - Lb)/2\n",
- " #coupling coefficient,\n",
- "k = M/(La*Lb)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance, M is \",M*1E3,\"mH\"\n",
- "print \"\\n coupling coefficient, is \",k"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance, M is 5.0 mH\n",
- "\n",
- " coupling coefficient, is 0.25"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 847</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the mutual inductance between the coils and (b) the coefficient of coupling.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 240;# in Volts\n",
- "Ra = 5;# in Ohm\n",
- "La = 1;# in Henry\n",
- "Rb = 10;# in Ohm\n",
- "Lb = 5;# in Henry\n",
- "I = 8;# in amperes\n",
- "dIdt = 15;# in A/sec\n",
- "\n",
- " #calculation:\n",
- " #Kirchhoff\u2019s voltage law\n",
- "L = (V - I*(Ra + Rb))/dIdt\n",
- " #mutual inductance, M\n",
- "M = (L - La - Lb)/2\n",
- " #coupling coefficient,\n",
- "k = M/(La*Lb)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance, M is \",M,\"H\"\n",
- "print \"\\n coupling coefficient, is \",round(k,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance, M is 1.0 H\n",
- "\n",
- " coupling coefficient, is 0.447"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 848</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the self inductance of each coil, and (b) the mutual inductance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "k = 0.7;# coefficient of coupling\n",
- "L1 = 15E-3;# in Henry\n",
- "L2 = 10E-3;# in Henry\n",
- "\n",
- "#calculation:\n",
- " #L1 = La + Lb + 2*k*(La*Lb)**0.5\n",
- " #L2 = La + Lb - 2*k*(La*Lb)**0.5\n",
- " #self inductance of coils\n",
- "a = ((L1 - (L1 + L2)/2)/(2*k))**2\n",
- "La1 =((L1 + L2)/2 + (((L1 + L2)/2)**2 - 4*a)**0.5)/2\n",
- "La2 =((L1 + L2)/2 - (((L1 + L2)/2)**2 - 4*a)**0.5)/2\n",
- "Lb1 = (L1 + L2)/2 - La1\n",
- "Lb2 = (L1 + L2)/2 - La2\n",
- " #mutual inductance, M\n",
- "M = (L1 - L2)/4\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nself inductance of coils are \",round(La1*1E3,2),\"mH and \",round( Lb1*1E3,2),\"mH\"\n",
- "print \"\\n mutual inductance, M is \",round(M*1E3,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "self inductance of coils are 12.24 mH and 0.26 mH\n",
- "\n",
- " mutual inductance, M is 1.25 mH"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 850</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the p.d. E2 which appears across the open-circuited secondary winding,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 8;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "w = 2500;# in rad/sec\n",
- "R = 15;# in ohm\n",
- "L = 5E-3;# in Henry\n",
- "M = 0.1E-3;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
- " #Impedance of primary\n",
- "Z1 = R + 1j*w*L\n",
- " #Primary current I1\n",
- "I1 = E1/Z1\n",
- " #E2\n",
- "E2 = 1j*w*M*I1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nE2 is \",round(abs(E2),3),\"/_\",round(cmath.phase(complex(E2.real,E2.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "E2 is 0.102 /_ 50.19 deg V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 850</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the magnitude of the open circuit e.m.f. induced in y.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Lx = 20E-3;# in Henry\n",
- "Ly = 80E-3;# in Henry\n",
- "k = 0.75;# coupling coeff.\n",
- "Ex = 5;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #mutual inductance\n",
- "M = k*(Lx*Ly)**0.5\n",
- " #magnitude of the open circuit e.m.f. induced\n",
- "Ey = M*Ex/Lx\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance is \",M,\" H\"\n",
- "print \"\\n magnitude of the open circuit e.m.f. induced is \",Ey,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance is 0.03 H\n",
- "\n",
- " magnitude of the open circuit e.m.f. induced is 7.5 V"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 852</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of the secondary current I2\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 2;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "f = 1000/math.pi;# in Hz\n",
- "R1 = 4;# in ohm\n",
- "R2 = 16;# in ohm\n",
- "R3 = 16;# in ohm\n",
- "R4 = 50;# in ohm\n",
- "L = 10E-3;# in Henry\n",
- "M = 2E-3;# in Henry\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #R1e is the real part of Z1e\n",
- "R1e = R1 + R2 + ((R3 + R4)*(M**2)*(w**2))/((R3 + R4)**2 + (w*L)**2)\n",
- " #X1e is the imaginary part of Z1e\n",
- "X1e = w*L - (L*(M**2)*(w**3))/((R3 + R4)**2 + (w*L)**2)\n",
- "Z1e = R1e + 1j*X1e\n",
- "Z2e = R3 + R4 + 1j*w*L\n",
- " #primary current, I1\n",
- "I1 = E1/Z1e\n",
- " #E2\n",
- "E2 = 1j*w*M*I1\n",
- " #secondary current I2\n",
- "I2 = E2/Z2e\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"secondary current I2 is \",round(abs(I2)*1E3,3),\"/_\", round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "secondary current I2 is 4.085 /_ 28.55 deg mA\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 853</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate (a) the self impedance of the primary circuit, \n",
- "#(b) the self impedance of the secondary circuit, \n",
- "#(c) the impedance reflected the primary circuit, \n",
- "#(d) the effective primary impedance, \n",
- "#(e) the primary current, and \n",
- "#(f) the secondary current\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 50;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "w = 500;# in rad/sec\n",
- "R1 = 300;# in ohm\n",
- "L1 = 0.2;# in Henry\n",
- "L2 = 0.5;# in Henry\n",
- "L3 = 0.3;# in Henry\n",
- "R2 = 500;# in ohm\n",
- "C = 5E-6;# in farad\n",
- "M = 0.2;# in Henry\n",
- "\n",
- "#calculation:\n",
- " # Self impedance of primary circuit\n",
- "Z1 = R1 + 1j*w*(L1 + L2)\n",
- " #Self impedance of secondary circuit,\n",
- "Z2 = R2 + 1j*(w*L3 - 1/(w*C))\n",
- " #reflected impedance, Zr\n",
- "Zr = (w*M)**2/Z2\n",
- " #Effective primary impedance,\n",
- "Z1e = Z1 + Zr\n",
- " #Primary current I1 \n",
- "I1 = E1/Z1e\n",
- " #Secondary current I2\n",
- "E2 = 1j*w*M*I1\n",
- "I2 = E2/Z2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Self impedance of primary circuit, Z1 is \",Z1.real,\" + (\", Z1.imag,\")i ohm\"\n",
- "print \"\\n Self impedance of secondary circuit, Z2 is \",Z2.real,\" + (\", Z2.imag,\")i ohm\"\n",
- "print \"\\n reflected impedance, Zr is \",Zr.real,\" +(\", Zr.imag,\")i ohm\"\n",
- "print \"\\n Effective primary impedance Z1(eff) is \",Z1e.real,\" +(\",Z1e.imag,\")i ohm\"\n",
- "print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n secondary current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Self impedance of primary circuit, Z1 is 300.0 + ( 350.0 )i ohm\n",
- "\n",
- " Self impedance of secondary circuit, Z2 is 500.0 + ( -250.0 )i ohm\n",
- "\n",
- " reflected impedance, Zr is 16.0 +( 8.0 )i ohm\n",
- "\n",
- " Effective primary impedance Z1(eff) is 316.0 +( 358.0 )i ohm\n",
- "\n",
- " primary current I1 is 0.1 /_ -48.57 deg A\n",
- "\n",
- " secondary current I2 is 0.02 /_ 68.0 deg A"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 855</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, \n",
- "#(b) the value of capacitor C2, \n",
- "#(c) the effective primary impedance, \n",
- "#(d) the primary current, (e) the voltage across capacitor C2 and \n",
- "#(f) the coefficient of coupling.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 20;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "R1 = 15;# in ohm\n",
- "C1 = 400E-12;# in farad\n",
- "R2 = 30;# in ohm\n",
- "L1 = 0.001;# in Henry\n",
- "L2 = 0.0002;# in Henry\n",
- "R3 = 50;# in ohm\n",
- "M = 10E-6;# in Henry\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
- " #the resonant frequency, fr \n",
- "fr = 1/(2*math.pi*(L1*C1)**0.5)\n",
- " #The secondary is also tuned to a resonant frequency\n",
- " #capacitance,C2\n",
- "C2 = 1/(L2*(2*math.pi*fr)**2)\n",
- " #the effective primary impedance Z1eff\n",
- "w = 2*math.pi*fr\n",
- "Z1e = R1 + R2 + ((w*M)**2)/R3\n",
- " #Primary current I1 \n",
- "I1 = E1/Z1e\n",
- " #Secondary current I2\n",
- "E2 = 1j*w*M*I1\n",
- "I2 = E2/Z1e\n",
- " #voltage across capacitor C2\n",
- "Vc2 = I2*(-1*1j/(w*C2))\n",
- " #coefficient of coupling, k \n",
- "k = M/(L1*L2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the resonant frequency,fr is \",round(fr/1000,2),\"KHz\"\n",
- "print \"\\n capacitance,C2 is \",round(C2*1E9,2),\"nF\"\n",
- "print \"\\n Effective primary impedance Z1(eff) is \",round(abs(Z1e),2),\" ohm\"\n",
- "print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag)),0),\"deg A\"\n",
- "print \"\\n voltage across capacitor C2 is \",round(abs(Vc2),2),\"/_\",round(abs(cmath.phase(complex(Vc2.real,Vc2.imag))),0),\"deg V\"\n",
- "print \"\\n coefficient of coupling, k is \",round(k,4),\"\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the resonant frequency,fr is 251.65 KHz\n",
- "\n",
- " capacitance,C2 is 2.0 nF\n",
- "\n",
- " Effective primary impedance Z1(eff) is 50.0 ohm\n",
- "\n",
- " primary current I1 is 0.4 /_ 0.0 deg A\n",
- "\n",
- " voltage across capacitor C2 is 40.0 /_ 0.0 deg V\n",
- "\n",
- " coefficient of coupling, k is 0.0224 \n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 858</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the values of currents I1 and I2\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 250;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "R1 = 50j;# in ohm\n",
- "R2 = 10;# in ohm\n",
- "R3 = 10;# in ohm\n",
- "R4 = 50j;# in ohm\n",
- "R5 = 50;# in ohm\n",
- "M = 10j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
- " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
- " #(R1 + R2)*I1 - M*I2 = E1\n",
- " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
- " #-1*M*I1 + ( R3 + R4 + R5)*I2 = 0\n",
- " #solving these two\n",
- "I2 = E1/((R1 + R2)*(R3 + R4 + R5)/(-1*M) + (-1*M))\n",
- "I1 = I2*(R3 + R4 + R5)/(-1*M)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n primary current I1 is \",round(I1.real,2),\" +(\",round( I1.imag,2),\")i A\"\n",
- "print \"\\n secondary current I2 is \",round(I2.real,2),\" +(\",round( I2.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " primary current I1 is 0.85 +( -4.77 )i A\n",
- "\n",
- " secondary current I2 is -0.54 +( 0.31 )i A"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 859</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the circuit (a) the mutual inductance M, (b) the primary current I1 and (c) the secondary terminal p.d.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 40;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "R1 = 5;# in ohm\n",
- "L1 = 0.001;# in Henry\n",
- "L2 = 0.006;# in Henry\n",
- "R2 = 40;# in ohm\n",
- "rzl = 200;# in ohm\n",
- "thetazl = -60;# in degrees\n",
- "k = 0.70\n",
- "f = 20000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #voltage\n",
- "#E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
- " #impedance\n",
- "ZL = rzl*math.cos(thetazl*math.pi/180) + 1j*rzl*math.sin(thetazl*math.pi/180)\n",
- " #mutual inductance, M\n",
- "M = k*(L1*L2)**0.5\n",
- " #Applying Kirchhoff\u00e2\u20ac\u2122s voltage law to the primary circuit gives\n",
- " #(R1 + 1j*w*L1)*I1 - 1j*w*M*I2 = E1\n",
- " #Applying Kirchhoff\u00e2\u20ac\u2122s voltage law to the secondary circuit gives\n",
- " #-1j*w*M*I1 + ( R2 + ZL + 1j*w*L2)*I2 = 0\n",
- " #solving these two\n",
- "\n",
- "a = R1 + 1j*w*L1\n",
- "b = 1j*w*M\n",
- "c = R2 + ZL + 1j*w*L2\n",
- "I1 = E1/(1*a - (b**2)/c)\n",
- "d = -1*cmath.phase(complex(I1.real,I1.imag))\n",
- "e = abs(I1)\n",
- "I2 = (b/c)*(e*math.cos(d) + 1j*e*math.sin(d))\n",
- "pd2 = I2*ZL\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual induction M is \",round(M*1E3,3),\"mH\"\n",
- "print \"\\n primary current I1 is \",round(abs(I1),3),\"/_\",round(-1*cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n secondary current I2 is \",round(abs(pd2),1),\"/_\",round(cmath.phase(complex(pd2.real,pd2.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual induction M is 1.715 mH\n",
- "\n",
- " primary current I1 is 0.724 /_ 65.15 deg A\n",
- "\n",
- " secondary current I2 is 52.2 /_ 18.7 deg V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 860</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the generator current I1 and (b) the load current I2.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 50;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "r = 5;# in ohm\n",
- "R1 = 20;# in ohm\n",
- "L1 = 0.2;# in Henry\n",
- "L2 = 0.4;# in Henry\n",
- "R2 = 25;# in ohm\n",
- "RL = 20;# in ohm\n",
- "M = 0.1;# in Henry\n",
- "f = 75/math.pi;# in Hz\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
- " #(r + R1 + 1j*w*L1)*I1 - 1j*w*M*I2 = E1\n",
- " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
- " #-1*1j*w*M*I1 + ( R2 + RL + 1j*w*L2)*I2 = 0\n",
- " #solving these two\n",
- "I2 = E1/((r + R1 + 1j*w*L1)*(R2 + RL + 1j*w*L2)/(1j*w*M) + (-1*1j*w*M))\n",
- "I1 = I2*(R2 + RL + 1j*w*L2)/(1j*w*M)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n load current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " primary current I1 is 1.3 /_ -45.84 deg A\n",
- "\n",
- " load current I2 is 0.26 /_ -8.97 deg A"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 862</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the source and load currents for (a) the windings as shown , and (b) with one winding reversed\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 50;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "r = 5;# in ohm\n",
- "R1 = 20;# in ohm\n",
- "L1 = 0.2;# in Henry\n",
- "R = 8;# in ohm\n",
- "L = 0.1;# in Henry\n",
- "L2 = 0.4;# in Henry\n",
- "R2 = 25;# in ohm\n",
- "RL = 20;# in ohm\n",
- "M = 0.1;# in Henry\n",
- "f = 75/math.pi;# in Hz\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
- " #(r + R1 + 1j*w*L1 + R + 1j*w*L)*I1 - (1j*w*M + R + 1j*w*L)*I2 = E1\n",
- " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
- " #-1*(1j*w*M + R + 1j*w*L)*I1 + (R2 + RL + 1j*w*L2 + R + 1j*w*L)*I2 = 0\n",
- " #solving these two\n",
- "I2 = E1/((r + R1 + 1j*w*L1 + R + 1j*w*L)*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(1j*w*M + R + 1j*w*L) + (-1*(1j*w*M + R + 1j*w*L)))\n",
- "I1 = I2*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(1j*w*M + R + 1j*w*L)\n",
- " #reversing\n",
- " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
- " #(r + R1 + 1j*w*L1 + R + 1j*w*L)*I1r - (-1*1j*w*M + R + 1j*w*L)*I2r = E1\n",
- " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
- " #-1*(-1*1j*w*M + R + 1j*w*L)*I1r + (R2 + RL + 1j*w*L2 + R + 1j*w*L)*I2r = 0\n",
- " #solving these two\n",
- "I2r = E1/((r + R1 + 1j*w*L1 + R + 1j*w*L)*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(-1*1j*w*M + R + 1j*w*L) + (-1*(-1*1j*w*M + R + 1j*w*L)))\n",
- "I1r = I2r*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(-1*1j*w*M + R + 1j*w*L)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"load current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"reversed primary current I1r is \",round(abs(I1r),2),\"/_\",round(cmath.phase(complex(I1r.real,I1r.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"reversed load current I2r is \",round(abs(I2r),2),\"/_\",round(cmath.phase(complex(I2r.real,I2r.imag))*180/math.pi,2),\"deg A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "primary current I1 is 1.03 /_ -45.47 deg A\n",
- "load current I2 is 0.35 /_ -25.16 deg A\n",
- "reversed primary current I1r is 0.89 /_ -54.42 deg A\n",
- "reversed load current I2r is 0.08 /_ -109.17 deg A\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_1.ipynb deleted file mode 100755 index 618fd99a..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_1.ipynb +++ /dev/null @@ -1,1037 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 43: Magnetically coupled circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 842</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the self inductance of coil A, and (b) the mutual inductance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Na = 1200; \n",
- "Nb = 1000;\n",
- "Ia = 0.8;# in amperes\n",
- "Phia = 100E-6;# in Wb\n",
- "xb = 0.75;\n",
- "\n",
- " #calculation:\n",
- " #self inductance of coil A\n",
- "La = Na*Phia/Ia\n",
- " #mutual inductance, M\n",
- "Phib = xb*Phia\n",
- "M = Nb*Phib/Ia\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n self inductance of coil A is \",La,\" H\"\n",
- "print \"\\n mutual inductance, M is \",M,\"H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " self inductance of coil A is 0.15 H\n",
- "\n",
- " mutual inductance, M is 0.09375 H"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 843</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. induced in the secondary,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "M = 600E-3;# in Henry\n",
- "Ia = 5;# in amperes\n",
- "dt = 0.2;# in secs\n",
- "\n",
- " #calculation:\n",
- " #change of current\n",
- "dIa = 2*Ia \n",
- "dIadt = dIa/dt\n",
- " #secondary induced e.m.f., E2\n",
- "E2 = -1*M*dIadt\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n secondary induced e.m.f., E2 is \",E2,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " secondary induced e.m.f., E2 is -30.0 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 844</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the magnetic coupling coefficient of the pair of coils\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "La = 250E-3;# in Henry\n",
- "Lb = 400E-3;# in Henry\n",
- "M = 80E-3;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #coupling coefficient,\n",
- "k = M/(La*Lb)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n coupling coefficient, is \",round(k,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " coupling coefficient, is 0.253"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 845</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the mutual inductance between the coils, and (b) the coefficient of coupling.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Lx = 80E-3;# in Henry\n",
- "Ly = 60E-3;# in Henry\n",
- "Nx = 200;# turns\n",
- "Ny = 100;# turns\n",
- "Ix = 4;# in Amperes\n",
- "Phiy = 0.005;# in Wb\n",
- "\n",
- "#calculation:\n",
- " #mutual inductance, M\n",
- "M = Ny*Phiy/(2*Ix)\n",
- " #coupling coefficient,\n",
- "k = M/(Lx*Ly)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance, M is \",M*1E3,\"mH\"\n",
- "print \"\\n coupling coefficient, is \",round(k,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance, M is 62.5 mH\n",
- "\n",
- " coupling coefficient, is 0.902"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 846</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the mutual inductance between the two coils, and (b) the coefficient of coupling.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "La = 40E-3;# in Henry\n",
- "Lb = 10E-3;# in Henry\n",
- "L = 60E-3;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #mutual inductance, M\n",
- "M = (L - La - Lb)/2\n",
- " #coupling coefficient,\n",
- "k = M/(La*Lb)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance, M is \",M*1E3,\"mH\"\n",
- "print \"\\n coupling coefficient, is \",k"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance, M is 5.0 mH\n",
- "\n",
- " coupling coefficient, is 0.25"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 847</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the mutual inductance between the coils and (b) the coefficient of coupling.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 240;# in Volts\n",
- "Ra = 5;# in Ohm\n",
- "La = 1;# in Henry\n",
- "Rb = 10;# in Ohm\n",
- "Lb = 5;# in Henry\n",
- "I = 8;# in amperes\n",
- "dIdt = 15;# in A/sec\n",
- "\n",
- " #calculation:\n",
- " #Kirchhoff\u2019s voltage law\n",
- "L = (V - I*(Ra + Rb))/dIdt\n",
- " #mutual inductance, M\n",
- "M = (L - La - Lb)/2\n",
- " #coupling coefficient,\n",
- "k = M/(La*Lb)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance, M is \",M,\"H\"\n",
- "print \"\\n coupling coefficient, is \",round(k,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance, M is 1.0 H\n",
- "\n",
- " coupling coefficient, is 0.447"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 848</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the self inductance of each coil, and (b) the mutual inductance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "k = 0.7;# coefficient of coupling\n",
- "L1 = 15E-3;# in Henry\n",
- "L2 = 10E-3;# in Henry\n",
- "\n",
- "#calculation:\n",
- " #L1 = La + Lb + 2*k*(La*Lb)**0.5\n",
- " #L2 = La + Lb - 2*k*(La*Lb)**0.5\n",
- " #self inductance of coils\n",
- "a = ((L1 - (L1 + L2)/2)/(2*k))**2\n",
- "La1 =((L1 + L2)/2 + (((L1 + L2)/2)**2 - 4*a)**0.5)/2\n",
- "La2 =((L1 + L2)/2 - (((L1 + L2)/2)**2 - 4*a)**0.5)/2\n",
- "Lb1 = (L1 + L2)/2 - La1\n",
- "Lb2 = (L1 + L2)/2 - La2\n",
- " #mutual inductance, M\n",
- "M = (L1 - L2)/4\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nself inductance of coils are \",round(La1*1E3,2),\"mH and \",round( Lb1*1E3,2),\"mH\"\n",
- "print \"\\n mutual inductance, M is \",round(M*1E3,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "self inductance of coils are 12.24 mH and 0.26 mH\n",
- "\n",
- " mutual inductance, M is 1.25 mH"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 850</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the p.d. E2 which appears across the open-circuited secondary winding,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 8;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "w = 2500;# in rad/sec\n",
- "R = 15;# in ohm\n",
- "L = 5E-3;# in Henry\n",
- "M = 0.1E-3;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
- " #Impedance of primary\n",
- "Z1 = R + 1j*w*L\n",
- " #Primary current I1\n",
- "I1 = E1/Z1\n",
- " #E2\n",
- "E2 = 1j*w*M*I1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nE2 is \",round(abs(E2),3),\"/_\",round(cmath.phase(complex(E2.real,E2.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "E2 is 0.102 /_ 50.19 deg V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 850</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the magnitude of the open circuit e.m.f. induced in y.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Lx = 20E-3;# in Henry\n",
- "Ly = 80E-3;# in Henry\n",
- "k = 0.75;# coupling coeff.\n",
- "Ex = 5;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #mutual inductance\n",
- "M = k*(Lx*Ly)**0.5\n",
- " #magnitude of the open circuit e.m.f. induced\n",
- "Ey = M*Ex/Lx\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance is \",M,\" H\"\n",
- "print \"\\n magnitude of the open circuit e.m.f. induced is \",Ey,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance is 0.03 H\n",
- "\n",
- " magnitude of the open circuit e.m.f. induced is 7.5 V"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 852</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of the secondary current I2\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 2;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "f = 1000/math.pi;# in Hz\n",
- "R1 = 4;# in ohm\n",
- "R2 = 16;# in ohm\n",
- "R3 = 16;# in ohm\n",
- "R4 = 50;# in ohm\n",
- "L = 10E-3;# in Henry\n",
- "M = 2E-3;# in Henry\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #R1e is the real part of Z1e\n",
- "R1e = R1 + R2 + ((R3 + R4)*(M**2)*(w**2))/((R3 + R4)**2 + (w*L)**2)\n",
- " #X1e is the imaginary part of Z1e\n",
- "X1e = w*L - (L*(M**2)*(w**3))/((R3 + R4)**2 + (w*L)**2)\n",
- "Z1e = R1e + 1j*X1e\n",
- "Z2e = R3 + R4 + 1j*w*L\n",
- " #primary current, I1\n",
- "I1 = E1/Z1e\n",
- " #E2\n",
- "E2 = 1j*w*M*I1\n",
- " #secondary current I2\n",
- "I2 = E2/Z2e\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"secondary current I2 is \",round(abs(I2)*1E3,3),\"/_\", round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "secondary current I2 is 4.085 /_ 28.55 deg mA\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 853</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate (a) the self impedance of the primary circuit, \n",
- "#(b) the self impedance of the secondary circuit, \n",
- "#(c) the impedance reflected the primary circuit, \n",
- "#(d) the effective primary impedance, \n",
- "#(e) the primary current, and \n",
- "#(f) the secondary current\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 50;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "w = 500;# in rad/sec\n",
- "R1 = 300;# in ohm\n",
- "L1 = 0.2;# in Henry\n",
- "L2 = 0.5;# in Henry\n",
- "L3 = 0.3;# in Henry\n",
- "R2 = 500;# in ohm\n",
- "C = 5E-6;# in farad\n",
- "M = 0.2;# in Henry\n",
- "\n",
- "#calculation:\n",
- " # Self impedance of primary circuit\n",
- "Z1 = R1 + 1j*w*(L1 + L2)\n",
- " #Self impedance of secondary circuit,\n",
- "Z2 = R2 + 1j*(w*L3 - 1/(w*C))\n",
- " #reflected impedance, Zr\n",
- "Zr = (w*M)**2/Z2\n",
- " #Effective primary impedance,\n",
- "Z1e = Z1 + Zr\n",
- " #Primary current I1 \n",
- "I1 = E1/Z1e\n",
- " #Secondary current I2\n",
- "E2 = 1j*w*M*I1\n",
- "I2 = E2/Z2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Self impedance of primary circuit, Z1 is \",Z1.real,\" + (\", Z1.imag,\")i ohm\"\n",
- "print \"\\n Self impedance of secondary circuit, Z2 is \",Z2.real,\" + (\", Z2.imag,\")i ohm\"\n",
- "print \"\\n reflected impedance, Zr is \",Zr.real,\" +(\", Zr.imag,\")i ohm\"\n",
- "print \"\\n Effective primary impedance Z1(eff) is \",Z1e.real,\" +(\",Z1e.imag,\")i ohm\"\n",
- "print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n secondary current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Self impedance of primary circuit, Z1 is 300.0 + ( 350.0 )i ohm\n",
- "\n",
- " Self impedance of secondary circuit, Z2 is 500.0 + ( -250.0 )i ohm\n",
- "\n",
- " reflected impedance, Zr is 16.0 +( 8.0 )i ohm\n",
- "\n",
- " Effective primary impedance Z1(eff) is 316.0 +( 358.0 )i ohm\n",
- "\n",
- " primary current I1 is 0.1 /_ -48.57 deg A\n",
- "\n",
- " secondary current I2 is 0.02 /_ 68.0 deg A"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 855</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, \n",
- "#(b) the value of capacitor C2, \n",
- "#(c) the effective primary impedance, \n",
- "#(d) the primary current, (e) the voltage across capacitor C2 and \n",
- "#(f) the coefficient of coupling.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 20;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "R1 = 15;# in ohm\n",
- "C1 = 400E-12;# in farad\n",
- "R2 = 30;# in ohm\n",
- "L1 = 0.001;# in Henry\n",
- "L2 = 0.0002;# in Henry\n",
- "R3 = 50;# in ohm\n",
- "M = 10E-6;# in Henry\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
- " #the resonant frequency, fr \n",
- "fr = 1/(2*math.pi*(L1*C1)**0.5)\n",
- " #The secondary is also tuned to a resonant frequency\n",
- " #capacitance,C2\n",
- "C2 = 1/(L2*(2*math.pi*fr)**2)\n",
- " #the effective primary impedance Z1eff\n",
- "w = 2*math.pi*fr\n",
- "Z1e = R1 + R2 + ((w*M)**2)/R3\n",
- " #Primary current I1 \n",
- "I1 = E1/Z1e\n",
- " #Secondary current I2\n",
- "E2 = 1j*w*M*I1\n",
- "I2 = E2/Z1e\n",
- " #voltage across capacitor C2\n",
- "Vc2 = I2*(-1*1j/(w*C2))\n",
- " #coefficient of coupling, k \n",
- "k = M/(L1*L2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the resonant frequency,fr is \",round(fr/1000,2),\"KHz\"\n",
- "print \"\\n capacitance,C2 is \",round(C2*1E9,2),\"nF\"\n",
- "print \"\\n Effective primary impedance Z1(eff) is \",round(abs(Z1e),2),\" ohm\"\n",
- "print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag)),0),\"deg A\"\n",
- "print \"\\n voltage across capacitor C2 is \",round(abs(Vc2),2),\"/_\",round(abs(cmath.phase(complex(Vc2.real,Vc2.imag))),0),\"deg V\"\n",
- "print \"\\n coefficient of coupling, k is \",round(k,4),\"\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the resonant frequency,fr is 251.65 KHz\n",
- "\n",
- " capacitance,C2 is 2.0 nF\n",
- "\n",
- " Effective primary impedance Z1(eff) is 50.0 ohm\n",
- "\n",
- " primary current I1 is 0.4 /_ 0.0 deg A\n",
- "\n",
- " voltage across capacitor C2 is 40.0 /_ 0.0 deg V\n",
- "\n",
- " coefficient of coupling, k is 0.0224 \n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 858</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the values of currents I1 and I2\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 250;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "R1 = 50j;# in ohm\n",
- "R2 = 10;# in ohm\n",
- "R3 = 10;# in ohm\n",
- "R4 = 50j;# in ohm\n",
- "R5 = 50;# in ohm\n",
- "M = 10j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
- " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
- " #(R1 + R2)*I1 - M*I2 = E1\n",
- " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
- " #-1*M*I1 + ( R3 + R4 + R5)*I2 = 0\n",
- " #solving these two\n",
- "I2 = E1/((R1 + R2)*(R3 + R4 + R5)/(-1*M) + (-1*M))\n",
- "I1 = I2*(R3 + R4 + R5)/(-1*M)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n primary current I1 is \",round(I1.real,2),\" +(\",round( I1.imag,2),\")i A\"\n",
- "print \"\\n secondary current I2 is \",round(I2.real,2),\" +(\",round( I2.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " primary current I1 is 0.85 +( -4.77 )i A\n",
- "\n",
- " secondary current I2 is -0.54 +( 0.31 )i A"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 859</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the circuit (a) the mutual inductance M, (b) the primary current I1 and (c) the secondary terminal p.d.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 40;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "R1 = 5;# in ohm\n",
- "L1 = 0.001;# in Henry\n",
- "L2 = 0.006;# in Henry\n",
- "R2 = 40;# in ohm\n",
- "rzl = 200;# in ohm\n",
- "thetazl = -60;# in degrees\n",
- "k = 0.70\n",
- "f = 20000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #voltage\n",
- "#E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
- " #impedance\n",
- "ZL = rzl*math.cos(thetazl*math.pi/180) + 1j*rzl*math.sin(thetazl*math.pi/180)\n",
- " #mutual inductance, M\n",
- "M = k*(L1*L2)**0.5\n",
- " #Applying Kirchhoff\u00e2\u20ac\u2122s voltage law to the primary circuit gives\n",
- " #(R1 + 1j*w*L1)*I1 - 1j*w*M*I2 = E1\n",
- " #Applying Kirchhoff\u00e2\u20ac\u2122s voltage law to the secondary circuit gives\n",
- " #-1j*w*M*I1 + ( R2 + ZL + 1j*w*L2)*I2 = 0\n",
- " #solving these two\n",
- "\n",
- "a = R1 + 1j*w*L1\n",
- "b = 1j*w*M\n",
- "c = R2 + ZL + 1j*w*L2\n",
- "I1 = E1/(1*a - (b**2)/c)\n",
- "d = -1*cmath.phase(complex(I1.real,I1.imag))\n",
- "e = abs(I1)\n",
- "I2 = (b/c)*(e*math.cos(d) + 1j*e*math.sin(d))\n",
- "pd2 = I2*ZL\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual induction M is \",round(M*1E3,3),\"mH\"\n",
- "print \"\\n primary current I1 is \",round(abs(I1),3),\"/_\",round(-1*cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n secondary current I2 is \",round(abs(pd2),1),\"/_\",round(cmath.phase(complex(pd2.real,pd2.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual induction M is 1.715 mH\n",
- "\n",
- " primary current I1 is 0.724 /_ 65.15 deg A\n",
- "\n",
- " secondary current I2 is 52.2 /_ 18.7 deg V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 860</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the generator current I1 and (b) the load current I2.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 50;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "r = 5;# in ohm\n",
- "R1 = 20;# in ohm\n",
- "L1 = 0.2;# in Henry\n",
- "L2 = 0.4;# in Henry\n",
- "R2 = 25;# in ohm\n",
- "RL = 20;# in ohm\n",
- "M = 0.1;# in Henry\n",
- "f = 75/math.pi;# in Hz\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
- " #(r + R1 + 1j*w*L1)*I1 - 1j*w*M*I2 = E1\n",
- " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
- " #-1*1j*w*M*I1 + ( R2 + RL + 1j*w*L2)*I2 = 0\n",
- " #solving these two\n",
- "I2 = E1/((r + R1 + 1j*w*L1)*(R2 + RL + 1j*w*L2)/(1j*w*M) + (-1*1j*w*M))\n",
- "I1 = I2*(R2 + RL + 1j*w*L2)/(1j*w*M)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n load current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " primary current I1 is 1.3 /_ -45.84 deg A\n",
- "\n",
- " load current I2 is 0.26 /_ -8.97 deg A"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 862</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the source and load currents for (a) the windings as shown , and (b) with one winding reversed\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 50;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "r = 5;# in ohm\n",
- "R1 = 20;# in ohm\n",
- "L1 = 0.2;# in Henry\n",
- "R = 8;# in ohm\n",
- "L = 0.1;# in Henry\n",
- "L2 = 0.4;# in Henry\n",
- "R2 = 25;# in ohm\n",
- "RL = 20;# in ohm\n",
- "M = 0.1;# in Henry\n",
- "f = 75/math.pi;# in Hz\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
- " #(r + R1 + 1j*w*L1 + R + 1j*w*L)*I1 - (1j*w*M + R + 1j*w*L)*I2 = E1\n",
- " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
- " #-1*(1j*w*M + R + 1j*w*L)*I1 + (R2 + RL + 1j*w*L2 + R + 1j*w*L)*I2 = 0\n",
- " #solving these two\n",
- "I2 = E1/((r + R1 + 1j*w*L1 + R + 1j*w*L)*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(1j*w*M + R + 1j*w*L) + (-1*(1j*w*M + R + 1j*w*L)))\n",
- "I1 = I2*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(1j*w*M + R + 1j*w*L)\n",
- " #reversing\n",
- " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
- " #(r + R1 + 1j*w*L1 + R + 1j*w*L)*I1r - (-1*1j*w*M + R + 1j*w*L)*I2r = E1\n",
- " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
- " #-1*(-1*1j*w*M + R + 1j*w*L)*I1r + (R2 + RL + 1j*w*L2 + R + 1j*w*L)*I2r = 0\n",
- " #solving these two\n",
- "I2r = E1/((r + R1 + 1j*w*L1 + R + 1j*w*L)*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(-1*1j*w*M + R + 1j*w*L) + (-1*(-1*1j*w*M + R + 1j*w*L)))\n",
- "I1r = I2r*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(-1*1j*w*M + R + 1j*w*L)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"load current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"reversed primary current I1r is \",round(abs(I1r),2),\"/_\",round(cmath.phase(complex(I1r.real,I1r.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"reversed load current I2r is \",round(abs(I2r),2),\"/_\",round(cmath.phase(complex(I2r.real,I2r.imag))*180/math.pi,2),\"deg A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "primary current I1 is 1.03 /_ -45.47 deg A\n",
- "load current I2 is 0.35 /_ -25.16 deg A\n",
- "reversed primary current I1r is 0.89 /_ -54.42 deg A\n",
- "reversed load current I2r is 0.08 /_ -109.17 deg A\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_2.ipynb deleted file mode 100755 index 618fd99a..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_2.ipynb +++ /dev/null @@ -1,1037 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 43: Magnetically coupled circuits</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 842</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the self inductance of coil A, and (b) the mutual inductance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Na = 1200; \n",
- "Nb = 1000;\n",
- "Ia = 0.8;# in amperes\n",
- "Phia = 100E-6;# in Wb\n",
- "xb = 0.75;\n",
- "\n",
- " #calculation:\n",
- " #self inductance of coil A\n",
- "La = Na*Phia/Ia\n",
- " #mutual inductance, M\n",
- "Phib = xb*Phia\n",
- "M = Nb*Phib/Ia\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n self inductance of coil A is \",La,\" H\"\n",
- "print \"\\n mutual inductance, M is \",M,\"H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " self inductance of coil A is 0.15 H\n",
- "\n",
- " mutual inductance, M is 0.09375 H"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 843</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. induced in the secondary,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "M = 600E-3;# in Henry\n",
- "Ia = 5;# in amperes\n",
- "dt = 0.2;# in secs\n",
- "\n",
- " #calculation:\n",
- " #change of current\n",
- "dIa = 2*Ia \n",
- "dIadt = dIa/dt\n",
- " #secondary induced e.m.f., E2\n",
- "E2 = -1*M*dIadt\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n secondary induced e.m.f., E2 is \",E2,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " secondary induced e.m.f., E2 is -30.0 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 844</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the magnetic coupling coefficient of the pair of coils\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "La = 250E-3;# in Henry\n",
- "Lb = 400E-3;# in Henry\n",
- "M = 80E-3;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #coupling coefficient,\n",
- "k = M/(La*Lb)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n coupling coefficient, is \",round(k,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " coupling coefficient, is 0.253"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 845</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the mutual inductance between the coils, and (b) the coefficient of coupling.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Lx = 80E-3;# in Henry\n",
- "Ly = 60E-3;# in Henry\n",
- "Nx = 200;# turns\n",
- "Ny = 100;# turns\n",
- "Ix = 4;# in Amperes\n",
- "Phiy = 0.005;# in Wb\n",
- "\n",
- "#calculation:\n",
- " #mutual inductance, M\n",
- "M = Ny*Phiy/(2*Ix)\n",
- " #coupling coefficient,\n",
- "k = M/(Lx*Ly)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance, M is \",M*1E3,\"mH\"\n",
- "print \"\\n coupling coefficient, is \",round(k,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance, M is 62.5 mH\n",
- "\n",
- " coupling coefficient, is 0.902"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 846</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the mutual inductance between the two coils, and (b) the coefficient of coupling.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "La = 40E-3;# in Henry\n",
- "Lb = 10E-3;# in Henry\n",
- "L = 60E-3;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #mutual inductance, M\n",
- "M = (L - La - Lb)/2\n",
- " #coupling coefficient,\n",
- "k = M/(La*Lb)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance, M is \",M*1E3,\"mH\"\n",
- "print \"\\n coupling coefficient, is \",k"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance, M is 5.0 mH\n",
- "\n",
- " coupling coefficient, is 0.25"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 847</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the mutual inductance between the coils and (b) the coefficient of coupling.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "V = 240;# in Volts\n",
- "Ra = 5;# in Ohm\n",
- "La = 1;# in Henry\n",
- "Rb = 10;# in Ohm\n",
- "Lb = 5;# in Henry\n",
- "I = 8;# in amperes\n",
- "dIdt = 15;# in A/sec\n",
- "\n",
- " #calculation:\n",
- " #Kirchhoff\u2019s voltage law\n",
- "L = (V - I*(Ra + Rb))/dIdt\n",
- " #mutual inductance, M\n",
- "M = (L - La - Lb)/2\n",
- " #coupling coefficient,\n",
- "k = M/(La*Lb)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance, M is \",M,\"H\"\n",
- "print \"\\n coupling coefficient, is \",round(k,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance, M is 1.0 H\n",
- "\n",
- " coupling coefficient, is 0.447"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 848</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the self inductance of each coil, and (b) the mutual inductance.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "k = 0.7;# coefficient of coupling\n",
- "L1 = 15E-3;# in Henry\n",
- "L2 = 10E-3;# in Henry\n",
- "\n",
- "#calculation:\n",
- " #L1 = La + Lb + 2*k*(La*Lb)**0.5\n",
- " #L2 = La + Lb - 2*k*(La*Lb)**0.5\n",
- " #self inductance of coils\n",
- "a = ((L1 - (L1 + L2)/2)/(2*k))**2\n",
- "La1 =((L1 + L2)/2 + (((L1 + L2)/2)**2 - 4*a)**0.5)/2\n",
- "La2 =((L1 + L2)/2 - (((L1 + L2)/2)**2 - 4*a)**0.5)/2\n",
- "Lb1 = (L1 + L2)/2 - La1\n",
- "Lb2 = (L1 + L2)/2 - La2\n",
- " #mutual inductance, M\n",
- "M = (L1 - L2)/4\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nself inductance of coils are \",round(La1*1E3,2),\"mH and \",round( Lb1*1E3,2),\"mH\"\n",
- "print \"\\n mutual inductance, M is \",round(M*1E3,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "self inductance of coils are 12.24 mH and 0.26 mH\n",
- "\n",
- " mutual inductance, M is 1.25 mH"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 850</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the p.d. E2 which appears across the open-circuited secondary winding,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 8;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "w = 2500;# in rad/sec\n",
- "R = 15;# in ohm\n",
- "L = 5E-3;# in Henry\n",
- "M = 0.1E-3;# in Henry\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
- " #Impedance of primary\n",
- "Z1 = R + 1j*w*L\n",
- " #Primary current I1\n",
- "I1 = E1/Z1\n",
- " #E2\n",
- "E2 = 1j*w*M*I1\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nE2 is \",round(abs(E2),3),\"/_\",round(cmath.phase(complex(E2.real,E2.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "E2 is 0.102 /_ 50.19 deg V"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 850</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the magnitude of the open circuit e.m.f. induced in y.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Lx = 20E-3;# in Henry\n",
- "Ly = 80E-3;# in Henry\n",
- "k = 0.75;# coupling coeff.\n",
- "Ex = 5;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #mutual inductance\n",
- "M = k*(Lx*Ly)**0.5\n",
- " #magnitude of the open circuit e.m.f. induced\n",
- "Ey = M*Ex/Lx\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual inductance is \",M,\" H\"\n",
- "print \"\\n magnitude of the open circuit e.m.f. induced is \",Ey,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual inductance is 0.03 H\n",
- "\n",
- " magnitude of the open circuit e.m.f. induced is 7.5 V"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 852</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of the secondary current I2\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 2;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "f = 1000/math.pi;# in Hz\n",
- "R1 = 4;# in ohm\n",
- "R2 = 16;# in ohm\n",
- "R3 = 16;# in ohm\n",
- "R4 = 50;# in ohm\n",
- "L = 10E-3;# in Henry\n",
- "M = 2E-3;# in Henry\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #R1e is the real part of Z1e\n",
- "R1e = R1 + R2 + ((R3 + R4)*(M**2)*(w**2))/((R3 + R4)**2 + (w*L)**2)\n",
- " #X1e is the imaginary part of Z1e\n",
- "X1e = w*L - (L*(M**2)*(w**3))/((R3 + R4)**2 + (w*L)**2)\n",
- "Z1e = R1e + 1j*X1e\n",
- "Z2e = R3 + R4 + 1j*w*L\n",
- " #primary current, I1\n",
- "I1 = E1/Z1e\n",
- " #E2\n",
- "E2 = 1j*w*M*I1\n",
- " #secondary current I2\n",
- "I2 = E2/Z2e\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"secondary current I2 is \",round(abs(I2)*1E3,3),\"/_\", round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "secondary current I2 is 4.085 /_ 28.55 deg mA\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 853</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate (a) the self impedance of the primary circuit, \n",
- "#(b) the self impedance of the secondary circuit, \n",
- "#(c) the impedance reflected the primary circuit, \n",
- "#(d) the effective primary impedance, \n",
- "#(e) the primary current, and \n",
- "#(f) the secondary current\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 50;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "w = 500;# in rad/sec\n",
- "R1 = 300;# in ohm\n",
- "L1 = 0.2;# in Henry\n",
- "L2 = 0.5;# in Henry\n",
- "L3 = 0.3;# in Henry\n",
- "R2 = 500;# in ohm\n",
- "C = 5E-6;# in farad\n",
- "M = 0.2;# in Henry\n",
- "\n",
- "#calculation:\n",
- " # Self impedance of primary circuit\n",
- "Z1 = R1 + 1j*w*(L1 + L2)\n",
- " #Self impedance of secondary circuit,\n",
- "Z2 = R2 + 1j*(w*L3 - 1/(w*C))\n",
- " #reflected impedance, Zr\n",
- "Zr = (w*M)**2/Z2\n",
- " #Effective primary impedance,\n",
- "Z1e = Z1 + Zr\n",
- " #Primary current I1 \n",
- "I1 = E1/Z1e\n",
- " #Secondary current I2\n",
- "E2 = 1j*w*M*I1\n",
- "I2 = E2/Z2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Self impedance of primary circuit, Z1 is \",Z1.real,\" + (\", Z1.imag,\")i ohm\"\n",
- "print \"\\n Self impedance of secondary circuit, Z2 is \",Z2.real,\" + (\", Z2.imag,\")i ohm\"\n",
- "print \"\\n reflected impedance, Zr is \",Zr.real,\" +(\", Zr.imag,\")i ohm\"\n",
- "print \"\\n Effective primary impedance Z1(eff) is \",Z1e.real,\" +(\",Z1e.imag,\")i ohm\"\n",
- "print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n secondary current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Self impedance of primary circuit, Z1 is 300.0 + ( 350.0 )i ohm\n",
- "\n",
- " Self impedance of secondary circuit, Z2 is 500.0 + ( -250.0 )i ohm\n",
- "\n",
- " reflected impedance, Zr is 16.0 +( 8.0 )i ohm\n",
- "\n",
- " Effective primary impedance Z1(eff) is 316.0 +( 358.0 )i ohm\n",
- "\n",
- " primary current I1 is 0.1 /_ -48.57 deg A\n",
- "\n",
- " secondary current I2 is 0.02 /_ 68.0 deg A"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 855</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the resonant frequency, \n",
- "#(b) the value of capacitor C2, \n",
- "#(c) the effective primary impedance, \n",
- "#(d) the primary current, (e) the voltage across capacitor C2 and \n",
- "#(f) the coefficient of coupling.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 20;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "R1 = 15;# in ohm\n",
- "C1 = 400E-12;# in farad\n",
- "R2 = 30;# in ohm\n",
- "L1 = 0.001;# in Henry\n",
- "L2 = 0.0002;# in Henry\n",
- "R3 = 50;# in ohm\n",
- "M = 10E-6;# in Henry\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
- " #the resonant frequency, fr \n",
- "fr = 1/(2*math.pi*(L1*C1)**0.5)\n",
- " #The secondary is also tuned to a resonant frequency\n",
- " #capacitance,C2\n",
- "C2 = 1/(L2*(2*math.pi*fr)**2)\n",
- " #the effective primary impedance Z1eff\n",
- "w = 2*math.pi*fr\n",
- "Z1e = R1 + R2 + ((w*M)**2)/R3\n",
- " #Primary current I1 \n",
- "I1 = E1/Z1e\n",
- " #Secondary current I2\n",
- "E2 = 1j*w*M*I1\n",
- "I2 = E2/Z1e\n",
- " #voltage across capacitor C2\n",
- "Vc2 = I2*(-1*1j/(w*C2))\n",
- " #coefficient of coupling, k \n",
- "k = M/(L1*L2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the resonant frequency,fr is \",round(fr/1000,2),\"KHz\"\n",
- "print \"\\n capacitance,C2 is \",round(C2*1E9,2),\"nF\"\n",
- "print \"\\n Effective primary impedance Z1(eff) is \",round(abs(Z1e),2),\" ohm\"\n",
- "print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag)),0),\"deg A\"\n",
- "print \"\\n voltage across capacitor C2 is \",round(abs(Vc2),2),\"/_\",round(abs(cmath.phase(complex(Vc2.real,Vc2.imag))),0),\"deg V\"\n",
- "print \"\\n coefficient of coupling, k is \",round(k,4),\"\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the resonant frequency,fr is 251.65 KHz\n",
- "\n",
- " capacitance,C2 is 2.0 nF\n",
- "\n",
- " Effective primary impedance Z1(eff) is 50.0 ohm\n",
- "\n",
- " primary current I1 is 0.4 /_ 0.0 deg A\n",
- "\n",
- " voltage across capacitor C2 is 40.0 /_ 0.0 deg V\n",
- "\n",
- " coefficient of coupling, k is 0.0224 \n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 858</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the values of currents I1 and I2\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 250;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "R1 = 50j;# in ohm\n",
- "R2 = 10;# in ohm\n",
- "R3 = 10;# in ohm\n",
- "R4 = 50j;# in ohm\n",
- "R5 = 50;# in ohm\n",
- "M = 10j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
- " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
- " #(R1 + R2)*I1 - M*I2 = E1\n",
- " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
- " #-1*M*I1 + ( R3 + R4 + R5)*I2 = 0\n",
- " #solving these two\n",
- "I2 = E1/((R1 + R2)*(R3 + R4 + R5)/(-1*M) + (-1*M))\n",
- "I1 = I2*(R3 + R4 + R5)/(-1*M)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n primary current I1 is \",round(I1.real,2),\" +(\",round( I1.imag,2),\")i A\"\n",
- "print \"\\n secondary current I2 is \",round(I2.real,2),\" +(\",round( I2.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " primary current I1 is 0.85 +( -4.77 )i A\n",
- "\n",
- " secondary current I2 is -0.54 +( 0.31 )i A"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 859</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the circuit (a) the mutual inductance M, (b) the primary current I1 and (c) the secondary terminal p.d.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 40;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "R1 = 5;# in ohm\n",
- "L1 = 0.001;# in Henry\n",
- "L2 = 0.006;# in Henry\n",
- "R2 = 40;# in ohm\n",
- "rzl = 200;# in ohm\n",
- "thetazl = -60;# in degrees\n",
- "k = 0.70\n",
- "f = 20000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #voltage\n",
- "#E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
- " #impedance\n",
- "ZL = rzl*math.cos(thetazl*math.pi/180) + 1j*rzl*math.sin(thetazl*math.pi/180)\n",
- " #mutual inductance, M\n",
- "M = k*(L1*L2)**0.5\n",
- " #Applying Kirchhoff\u00e2\u20ac\u2122s voltage law to the primary circuit gives\n",
- " #(R1 + 1j*w*L1)*I1 - 1j*w*M*I2 = E1\n",
- " #Applying Kirchhoff\u00e2\u20ac\u2122s voltage law to the secondary circuit gives\n",
- " #-1j*w*M*I1 + ( R2 + ZL + 1j*w*L2)*I2 = 0\n",
- " #solving these two\n",
- "\n",
- "a = R1 + 1j*w*L1\n",
- "b = 1j*w*M\n",
- "c = R2 + ZL + 1j*w*L2\n",
- "I1 = E1/(1*a - (b**2)/c)\n",
- "d = -1*cmath.phase(complex(I1.real,I1.imag))\n",
- "e = abs(I1)\n",
- "I2 = (b/c)*(e*math.cos(d) + 1j*e*math.sin(d))\n",
- "pd2 = I2*ZL\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n mutual induction M is \",round(M*1E3,3),\"mH\"\n",
- "print \"\\n primary current I1 is \",round(abs(I1),3),\"/_\",round(-1*cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n secondary current I2 is \",round(abs(pd2),1),\"/_\",round(cmath.phase(complex(pd2.real,pd2.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " mutual induction M is 1.715 mH\n",
- "\n",
- " primary current I1 is 0.724 /_ 65.15 deg A\n",
- "\n",
- " secondary current I2 is 52.2 /_ 18.7 deg V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 860</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the generator current I1 and (b) the load current I2.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 50;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "r = 5;# in ohm\n",
- "R1 = 20;# in ohm\n",
- "L1 = 0.2;# in Henry\n",
- "L2 = 0.4;# in Henry\n",
- "R2 = 25;# in ohm\n",
- "RL = 20;# in ohm\n",
- "M = 0.1;# in Henry\n",
- "f = 75/math.pi;# in Hz\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
- " #(r + R1 + 1j*w*L1)*I1 - 1j*w*M*I2 = E1\n",
- " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
- " #-1*1j*w*M*I1 + ( R2 + RL + 1j*w*L2)*I2 = 0\n",
- " #solving these two\n",
- "I2 = E1/((r + R1 + 1j*w*L1)*(R2 + RL + 1j*w*L2)/(1j*w*M) + (-1*1j*w*M))\n",
- "I1 = I2*(R2 + RL + 1j*w*L2)/(1j*w*M)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"\\n load current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " primary current I1 is 1.3 /_ -45.84 deg A\n",
- "\n",
- " load current I2 is 0.26 /_ -8.97 deg A"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 862</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the source and load currents for (a) the windings as shown , and (b) with one winding reversed\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 50;# in Volts\n",
- "thetae1 = 0;# in degrees\n",
- "r = 5;# in ohm\n",
- "R1 = 20;# in ohm\n",
- "L1 = 0.2;# in Henry\n",
- "R = 8;# in ohm\n",
- "L = 0.1;# in Henry\n",
- "L2 = 0.4;# in Henry\n",
- "R2 = 25;# in ohm\n",
- "RL = 20;# in ohm\n",
- "M = 0.1;# in Henry\n",
- "f = 75/math.pi;# in Hz\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
- " #(r + R1 + 1j*w*L1 + R + 1j*w*L)*I1 - (1j*w*M + R + 1j*w*L)*I2 = E1\n",
- " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
- " #-1*(1j*w*M + R + 1j*w*L)*I1 + (R2 + RL + 1j*w*L2 + R + 1j*w*L)*I2 = 0\n",
- " #solving these two\n",
- "I2 = E1/((r + R1 + 1j*w*L1 + R + 1j*w*L)*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(1j*w*M + R + 1j*w*L) + (-1*(1j*w*M + R + 1j*w*L)))\n",
- "I1 = I2*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(1j*w*M + R + 1j*w*L)\n",
- " #reversing\n",
- " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
- " #(r + R1 + 1j*w*L1 + R + 1j*w*L)*I1r - (-1*1j*w*M + R + 1j*w*L)*I2r = E1\n",
- " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
- " #-1*(-1*1j*w*M + R + 1j*w*L)*I1r + (R2 + RL + 1j*w*L2 + R + 1j*w*L)*I2r = 0\n",
- " #solving these two\n",
- "I2r = E1/((r + R1 + 1j*w*L1 + R + 1j*w*L)*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(-1*1j*w*M + R + 1j*w*L) + (-1*(-1*1j*w*M + R + 1j*w*L)))\n",
- "I1r = I2r*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(-1*1j*w*M + R + 1j*w*L)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"load current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"reversed primary current I1r is \",round(abs(I1r),2),\"/_\",round(cmath.phase(complex(I1r.real,I1r.imag))*180/math.pi,2),\"deg A\"\n",
- "print \"reversed load current I2r is \",round(abs(I2r),2),\"/_\",round(cmath.phase(complex(I2r.real,I2r.imag))*180/math.pi,2),\"deg A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "primary current I1 is 1.03 /_ -45.47 deg A\n",
- "load current I2 is 0.35 /_ -25.16 deg A\n",
- "reversed primary current I1r is 0.89 /_ -54.42 deg A\n",
- "reversed load current I2r is 0.08 /_ -109.17 deg A\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_43.ipynb index 628db606..628db606 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_43.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint.ipynb deleted file mode 100755 index 03d830a4..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint.ipynb +++ /dev/null @@ -1,1093 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 44: Transmission lines</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 873</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the wavelength on the line, and (b) the speed of transmission of a signal.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "f = 1910;# in Hz\n",
- "b = 0.05;# in rad/km\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #wavelength \n",
- "Y = 2*math.pi/b\n",
- " #speed of transmission\n",
- "u = f*Y\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n wavelength Y is \",round(Y,1),\" km\"\n",
- "print \"\\n speed of transmission \",round(u,1),\"km/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " wavelength Y is 125.7 km\n",
- "\n",
- " speed of transmission 240017.7 km/sec"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 873</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for a frequency of operation of 1 kHz, \n",
- "#(a) the phase delay, \n",
- "#(b) the wavelength on the line, and \n",
- "#(c) the velocity of propagation (in metres per second) of the signal.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.004;# in Henry/loop\n",
- "C = 0.004E-6;# in F/loop\n",
- "f = 1000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #phase delay\n",
- "b = w*(L*C)**0.5\n",
- " #wavelength \n",
- "Y = 2*math.pi/b\n",
- " #speed of transmission\n",
- "u = f*Y\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n phase delay is \",round(b,3),\" rad/km\"\n",
- "print \"\\n wavelength Y is \",Y,\" km\"\n",
- "print \"\\n speed of transmission \",u,\"km/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " phase delay is 0.025 rad/km\n",
- "\n",
- " wavelength Y is 250.0 km\n",
- "\n",
- " speed of transmission 250000.0 km/sec"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 874</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the voltage at a point 10 km down the line,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "a = 0.25;# in Np/km\n",
- "b = 0.20;# in rad/km\n",
- "Vs = 5;# in Volts\n",
- "n = 10;# in km\n",
- "f = 2000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #the voltage 10 km down the line\n",
- "r = a + 1j*b\n",
- "VR = Vs*cmath.e**(-1*n*r)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n Result \\n\\n\"\n",
- "print \"voltage 10 km down the line is \",round(abs(VR),2),\"/_\",round(cmath.phase(complex(VR.real,VR.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " Result \n",
- "\n",
- "\n",
- "voltage 10 km down the line is 0.41 /_ -114.59 deg V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 875</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the magnitude and phase of the current at the receiving end,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "a = 0.5;# in Np/km\n",
- "b = 0.25;# in rad/km\n",
- "rvs = 2;# in Volts\n",
- "thetavs = 0;# in degrees\n",
- "rzo = 800;# in ohm\n",
- "thetazo = -25;# in degrees\n",
- "n = 5;# in km\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "Vs = rvs*math.cos(thetavs*math.pi/180) + 1j*rvs*math.sin(thetavs*math.pi/180)\n",
- " #characteristic impedance\n",
- "Zo = rzo*math.cos(thetazo*math.pi/180) + 1j*rzo*math.sin(thetazo*math.pi/180)\n",
- " # receiving end voltage\n",
- "r = a + 1j*b\n",
- "VR = Vs*cmath.e**(-1*n*r)\n",
- " #Receiving end current,\n",
- "IR = VR/Zo\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"Receiving end current, IR is \",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "Receiving end current, IR is 0.205 /_ -46.62 deg mA\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 875</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the output voltage if the length of the line is doubled.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vs = 8;# in Volts\n",
- "VR = 2;# in Volts\n",
- "x = 2; \n",
- "\n",
- "#calculation:\n",
- " # receiving end voltage VR = Vs*e**(-nr)\n",
- " #e**-nr = p\n",
- "p = VR/Vs\n",
- " #If the line is doubled in length, then\n",
- "VR = Vs*(p)**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Receiving end voltage If the line is doubled in length, VR is \",abs(VR),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Receiving end voltage If the line is doubled in length, VR is 0.5 V"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 876</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the characteristic impedance of the line at this frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rzoc = 800;# in ohm\n",
- "thetazoc = -50;# in degrees\n",
- "rzsc = 413;# in ohm\n",
- "thetazsc = -20;# in degrees\n",
- "f = 1500;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #open circuit impedance\n",
- "Zoc = rzoc*math.cos(thetazoc*math.pi/180) + 1j*rzoc*math.sin(thetazoc*math.pi/180)\n",
- " #short circuit impedance\n",
- "Zsc = rzsc*math.cos(thetazsc*math.pi/180) + 1j*rzsc*math.sin(thetazsc*math.pi/180)\n",
- " #characteristic impedance Zo\n",
- "Zo = (Zoc*Zsc)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"characteristic impedance Zo is\",round(abs(Zo)),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "characteristic impedance Zo is 575.0 /_ -35.0 deg ohm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 877</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the characteristic impedance of the line when the frequency is 2 kHz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 15;# in ohm/loop km\n",
- "L = 0.0034;# in H/loop km\n",
- "C = 10E-9;# in F/km\n",
- "G = 3E-6;# in S/km\n",
- "f = 2000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #characteristic impedance Zo\n",
- "Zo = ((R + 1j*w*L)/(G + 1j*w*C))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"characteristic impedance Zo is \",round(abs(Zo),0),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "characteristic impedance Zo is 600.0 /_ -8.99 deg ohm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 879</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, at an operating frequency of 400 kHz, (a) the characteristic impedance,\n",
- "#(b) the propagation coefficient, (c) the wavelength on the line, and \n",
- "#(d) the velocity of propagation, in metres per second, of a signal.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.0005;# in H/loop km\n",
- "C = 0.12E-6;# in F/km\n",
- "f = 400000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #characteristic impedance Zo\n",
- "Zo = (L/C)**0.5\n",
- " #the propagation coefficient\n",
- "r = 1j*w*(L*C)**0.5\n",
- " #the attenuation coefficient \n",
- "a = r.real\n",
- " #the phaseshift coefficient\n",
- "b = r.imag\n",
- " #wavelength\n",
- "Y = 2*math.pi/b\n",
- " #velocity of propagation \n",
- "u = f*Y\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n characteristic impedance Zo is \",abs(Zo),\"ohm\"\n",
- "print \"\\n propagation coefficient is \",a,\" +(\",round(b,2),\")i\"\n",
- "print \"\\n wavelength Y is \",round(Y*1E3,0),\"m\"\n",
- "print \"\\n speed of transmission \",round(u,2),\"km/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " characteristic impedance Zo is 64.5497224368 ohm\n",
- "\n",
- " propagation coefficient is 0.0 +( 19.47 )i\n",
- "\n",
- " wavelength Y is 323.0 m\n",
- "\n",
- " speed of transmission 129099.44 km/sec"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 880</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the line (a) the characteristic impedance,\n",
- "#(b) the propagation coefficient, (c) the attenuation coefficient and\n",
- "#(d) the phase-shift coefficient\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 25;# in ohm/loop km\n",
- "L = 0.005;# in H/loop km\n",
- "C = 0.04E-6;# in F/km\n",
- "G = 80E-6;# in S/km\n",
- "f = 1000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #characteristic impedance Zo\n",
- "Zo = ((R + 1j*w*L)/(G + 1j*w*C))**0.5\n",
- " #the propagation coefficient\n",
- "r = ((R + 1j*w*L)*(G + 1j*w*C))**0.5\n",
- " #the attenuation coefficient \n",
- "a = r.real\n",
- " #the phaseshift coefficient\n",
- "b = r.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"characteristic impedance Zo is\",round(abs(Zo),2),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg ohm\"\n",
- "print \"\\n propagation coefficient is \",round(abs(r),4),\"/_\",round(cmath.phase(complex(a,b))*180/math.pi,2),\"deg\"\n",
- "print \"\\n attenuation coefficient is \",round(a,4),\" Np/km\"\n",
- "print \"\\n the phase-shift coefficient \",round(b,4),\" rad/km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "characteristic impedance Zo is 390.16 /_ -10.43 deg ohm\n",
- "\n",
- " propagation coefficient is 0.1029 /_ 61.92 deg\n",
- "\n",
- " attenuation coefficient is 0.0484 Np/km\n",
- "\n",
- " the phase-shift coefficient 0.0908 rad/km\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 881</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the characteristic impedance, \n",
- "#(b) the propagation coefficient, \n",
- "#(c) the attenuation and phase-shift coefficients, \n",
- "#(d) the sending-end current, \n",
- "#(e) the receiving-end current, \n",
- "#(f) the wavelength on the line, and \n",
- "#(g) the speed of transmission of signal.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 8;# in ohm/loop km\n",
- "L = 0.003;# in H/loop km\n",
- "C = 7500E-12;# in F/km\n",
- "G = 0.25E-6;# in S/km\n",
- "f = 1000;# in Hz\n",
- "n = 300;# in km\n",
- "Zg = 400 + 1j*0;# in ohm\n",
- "Vg = 10;# in Volts\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #characteristic impedance Zo\n",
- "Zo = ((R + 1j*w*L)/(G + 1j*w*C))**0.5\n",
- " #the propagation coefficient\n",
- "r = ((R + 1j*w*L)*(G + 1j*w*C))**0.5\n",
- " #the attenuation coefficient \n",
- "a = r.real\n",
- " #the phaseshift coefficient\n",
- "b = r.imag\n",
- " #the sending-end current,\n",
- "Is = Vg/(Zg + Zo)\n",
- " #the receiving-end current,\n",
- "IR = Is*cmath.e**(-1*n*r)\n",
- " #wavelength\n",
- "Y = 2*math.pi/b\n",
- " #velocity of propagation \n",
- "u = f*Y\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"characteristic impedance Zo is\",round(abs(Zo),1),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg ohm\"\n",
- "print \"propagation coefficient is \",round(abs(r),5),\"/_\",round(cmath.phase(complex(r.real,r.imag))*180/math.pi,2),\"deg\"\n",
- "print \"attenuation coefficient is \",round(a,5),\" Np/km and the phaseshift coefficient \",round(b,5),\" rad/km\"\n",
- "print \"sending-end current Is is \",round(abs(Is)*1E3,3),\"/_\",round(cmath.phase(complex(Is.real,Is.imag))*180/math.pi,2),\"deg mA\"\n",
- "print \"receiving-end current IR is\",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg mA\"\n",
- "print \"wavelength Y is \",round(Y,1),\" km\"\n",
- "print \"speed of transmission \",round(u,1),\"km/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "characteristic impedance Zo is 659.2 /_ -11.35 deg ohm\n",
- "propagation coefficient is 0.03106 /_ 78.35 deg\n",
- "attenuation coefficient is 0.00627 Np/km and the phaseshift coefficient 0.03042 rad/km\n",
- "sending-end current Is is 9.485 /_ 7.07 deg mA\n",
- "receiving-end current IR is 1.445 /_ -155.88 deg mA\n",
- "wavelength Y is 206.5 km\n",
- "speed of transmission 206521.1 km/sec\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 884</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine by how much the inductance should be increased to satisfy the condition for minimum distortion.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 10;# in ohm/loop km\n",
- "L = 0.0015;# in H/loop km\n",
- "C = 0.06E-6;# in F/km\n",
- "G = 1.2E-6;# in S/km\n",
- "\n",
- " #calculation:\n",
- " #the condition for minimum distortion is given by LG = CR, from which,\n",
- "Lm = C*R/G\n",
- "dL = Lm - L\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n inductance should be increased by \",round(dL*1E3,1),\"mH/loop km for minimum distortion\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " inductance should be increased by 498.5 mH/loop km for minimum distortion"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 884</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for minimum distortion at a frequency of 1.5 kHz \n",
- "#(a) the value of inductance per loop kilometre required, \n",
- "#(b) the propagation coefficient, \n",
- "#(c) the velocity of propagation of signal, and \n",
- "#(d) the wavelength on the line\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 80;# in ohm/loop km\n",
- "C = 5E-9;# in F/km\n",
- "G = 2E-6;# in S/km\n",
- "f = 1500;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #the condition for minimum distortion is given by LG = CR, from which, inductance\n",
- "L = C*R/G\n",
- " #attenuation coefficient,\n",
- "a = (R*G)**0.5\n",
- " #phase shift coefficient,\n",
- "b = w*(L*C)**0.5\n",
- " #propagation coefficient,\n",
- "r = a + 1j*b\n",
- " #velocity of propagation,\n",
- "u = 1/(L*C)**0.5\n",
- " #wavelength\n",
- "Y = u/f\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n inductance is \",round(L,2),\" H\"\n",
- "print \"\\n propagation coefficient is \",round(a,2),\" +(\",round(b,2),\")i\"\n",
- "print \"\\n speed of transmission \",round(u,2),\"km/sec\"\n",
- "print \"\\n wavelength Y is \",round(Y,2),\" km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " inductance is 0.2 H\n",
- "\n",
- " propagation coefficient is 0.01 +( 0.3 )i\n",
- "\n",
- " speed of transmission 31622.78 km/sec\n",
- "\n",
- " wavelength Y is 21.08 km\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 888</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of (a) the reflection coefficient for the line, (b) the incident current, \n",
- "#(c) the incident voltage, (d) the reflected current, and (e) the reflected voltage \n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Zo = 75;# in ohm\n",
- "ZR = 250;# in ohm\n",
- "VR = 10;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #reflection coefficient\n",
- "p = (Zo - ZR)/(Zo + ZR)\n",
- " #Current flowing in the terminating load\n",
- "IR = VR/ZR\n",
- " #incident current, Ii\n",
- "Ii = IR/(1 + p)\n",
- " #incident voltage, Vi \n",
- "Vi = Ii*Zo\n",
- " #reflected current, Ir\n",
- "Ir = IR - Ii\n",
- " #reflected voltage, Vr\n",
- "Vr = -1*Ir*Zo\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reflection coefficient is \",round(p,3),\"\"\n",
- "print \"\\n incident current, Ii is \",round(Ii,4),\" A\"\n",
- "print \"\\n incident voltage, Vi is \",round(Vi,2),\" V\"\n",
- "print \"\\n reflected current, Ir is \",round(Ir,4),\" A\"\n",
- "print \"\\n reflected voltage, Vr is \",round(Vr,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reflection coefficient is -0.538 \n",
- "\n",
- " incident current, Ii is 0.0867 A\n",
- "\n",
- " incident voltage, Vi is 6.5 V\n",
- "\n",
- " reflected current, Ir is -0.0467 A\n",
- "\n",
- " reflected voltage, Vr is 3.5 V"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 889</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the magnitude of the reflection coefficient in each case.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Zo = 500 - 1j*40;# in ohm\n",
- "ZR1 = 500 + 1j*40;# in ohm\n",
- "ZR2 = 600 + 1j*0;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #reflection coefficient\n",
- "p1 = (Zo - ZR1)/(Zo + ZR1)\n",
- "p2 = (Zo - ZR2)/(Zo + ZR2)\n",
- "p1mag = abs(p1)\n",
- "p2mag = abs(p2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reflection coefficient (a)\",p1mag,\" and (b)\", round(p2mag,2),\"\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reflection coefficient (a) 0.08 and (b) 0.1 "
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 890</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the magnitude of the ratio of the reflected to the incident voltage wave, and \n",
- "#(b) the incident voltage\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rzo = 500;# in ohm\n",
- "thetazo = 0;# in degrees\n",
- "ZR = 320 + 1j*240;# in ohm\n",
- "rvr = 20;# in volts\n",
- "thetavr = 35;# in degrees\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "VR = rvr*math.cos(thetavr*math.pi/180) + 1j*rvr*math.sin(thetavr*math.pi/180)\n",
- " #characteristic impedance\n",
- "Zo = rzo*math.cos(thetazo*math.pi/180) + 1j*rzo*math.sin(thetazo*math.pi/180)\n",
- " #the ratio of the reflected to the incident voltage \n",
- " #vr = VR/Vi\n",
- "vr = (ZR - Zo)/(Zo + ZR)\n",
- "vrmag = abs(vr)\n",
- " #incident voltage, Vi\n",
- "Vi = VR/vr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the magnitude of the ratio Vr : Vi is \",round(vrmag,3),\"\"\n",
- "print \"\\n incident voltage, Vi is \",round(abs(Vi),1),\"/_\",round(cmath.phase(complex(Vi.real,Vi.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the magnitude of the ratio Vr : Vi is 0.351 \n",
- "\n",
- " incident voltage, Vi is 57.0 /_ -75.56 deg V\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 895</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the reflection coefficient and (b) the standing-wave ratio.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rzo = 600;# in ohm\n",
- "thetazo = 0;# in degrees\n",
- "ZR = 400 + 250j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #characteristic impedance\n",
- "Zo = rzo*math.cos(thetazo*math.pi/180) + 1j*rzo*math.sin(thetazo*math.pi/180)\n",
- " #reflection coefficient\n",
- "p = (Zo - ZR)/(Zo + ZR)\n",
- "pmag = abs(p)\n",
- " #standing-wave ratio,\n",
- "s = (1 + pmag)/(1 - pmag)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reflection coefficient, is \",round(abs(p),4),\"/_\",round(cmath.phase(complex(p.real,p.imag))*180/math.pi,2),\"deg\"\n",
- "print \"\\n standing-wave ratio, s is \",round(s,3),\"\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reflection coefficient, is 0.3106 /_ -65.38 deg\n",
- "\n",
- " standing-wave ratio, s is 1.901 "
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 896</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the standing-wave ratio, \n",
- "#(b) the load impedance, and\n",
- "#(c) the incident current flowing if the reflected current is 10 mA.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rp = 0.2; \n",
- "thetap = -120;# in degrees\n",
- "Zo = 80;# in ohm\n",
- "Ir = 0.01;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #reflection coefficient\n",
- "p = rp*math.cos(thetap*math.pi/180) + 1j*rp*math.sin(thetap*math.pi/180)\n",
- " #standing-wave ratio,\n",
- "s = (1 + rp)/(1 - rp)\n",
- " #load impedance ZR \n",
- "ZR = Zo*(1 - p)/(1 + p)\n",
- " #incident current\n",
- "Ii = Ir*(s + 1)/(s - 1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n standing-wave ratio, s is \",s,\"\"\n",
- "print \"\\n load impedance ZR is \",round(ZR.real,2),\" +(\",round(ZR.imag,1),\")i ohm\"\n",
- "print \"\\n incident current is \",Ii,\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " standing-wave ratio, s is 1.5 \n",
- "\n",
- " load impedance ZR is 91.43 +( 33.0 )i ohm\n",
- "\n",
- " incident current is 0.05 A\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 897</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of the reflected power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "s = 1.6;\n",
- "Pi = 0.2;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #reflected power, Pr\n",
- "Pr = Pi*((s - 1)/(s + 1))**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reflected power, Pr is \",round(Pr*1E3,2),\" mW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reflected power, Pr is 10.65 mW"
- ]
- }
- ],
- "prompt_number": 18
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_1.ipynb deleted file mode 100755 index 03d830a4..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_1.ipynb +++ /dev/null @@ -1,1093 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 44: Transmission lines</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 873</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the wavelength on the line, and (b) the speed of transmission of a signal.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "f = 1910;# in Hz\n",
- "b = 0.05;# in rad/km\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #wavelength \n",
- "Y = 2*math.pi/b\n",
- " #speed of transmission\n",
- "u = f*Y\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n wavelength Y is \",round(Y,1),\" km\"\n",
- "print \"\\n speed of transmission \",round(u,1),\"km/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " wavelength Y is 125.7 km\n",
- "\n",
- " speed of transmission 240017.7 km/sec"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 873</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for a frequency of operation of 1 kHz, \n",
- "#(a) the phase delay, \n",
- "#(b) the wavelength on the line, and \n",
- "#(c) the velocity of propagation (in metres per second) of the signal.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.004;# in Henry/loop\n",
- "C = 0.004E-6;# in F/loop\n",
- "f = 1000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #phase delay\n",
- "b = w*(L*C)**0.5\n",
- " #wavelength \n",
- "Y = 2*math.pi/b\n",
- " #speed of transmission\n",
- "u = f*Y\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n phase delay is \",round(b,3),\" rad/km\"\n",
- "print \"\\n wavelength Y is \",Y,\" km\"\n",
- "print \"\\n speed of transmission \",u,\"km/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " phase delay is 0.025 rad/km\n",
- "\n",
- " wavelength Y is 250.0 km\n",
- "\n",
- " speed of transmission 250000.0 km/sec"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 874</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the voltage at a point 10 km down the line,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "a = 0.25;# in Np/km\n",
- "b = 0.20;# in rad/km\n",
- "Vs = 5;# in Volts\n",
- "n = 10;# in km\n",
- "f = 2000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #the voltage 10 km down the line\n",
- "r = a + 1j*b\n",
- "VR = Vs*cmath.e**(-1*n*r)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n Result \\n\\n\"\n",
- "print \"voltage 10 km down the line is \",round(abs(VR),2),\"/_\",round(cmath.phase(complex(VR.real,VR.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " Result \n",
- "\n",
- "\n",
- "voltage 10 km down the line is 0.41 /_ -114.59 deg V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 875</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the magnitude and phase of the current at the receiving end,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "a = 0.5;# in Np/km\n",
- "b = 0.25;# in rad/km\n",
- "rvs = 2;# in Volts\n",
- "thetavs = 0;# in degrees\n",
- "rzo = 800;# in ohm\n",
- "thetazo = -25;# in degrees\n",
- "n = 5;# in km\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "Vs = rvs*math.cos(thetavs*math.pi/180) + 1j*rvs*math.sin(thetavs*math.pi/180)\n",
- " #characteristic impedance\n",
- "Zo = rzo*math.cos(thetazo*math.pi/180) + 1j*rzo*math.sin(thetazo*math.pi/180)\n",
- " # receiving end voltage\n",
- "r = a + 1j*b\n",
- "VR = Vs*cmath.e**(-1*n*r)\n",
- " #Receiving end current,\n",
- "IR = VR/Zo\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"Receiving end current, IR is \",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "Receiving end current, IR is 0.205 /_ -46.62 deg mA\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 875</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the output voltage if the length of the line is doubled.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vs = 8;# in Volts\n",
- "VR = 2;# in Volts\n",
- "x = 2; \n",
- "\n",
- "#calculation:\n",
- " # receiving end voltage VR = Vs*e**(-nr)\n",
- " #e**-nr = p\n",
- "p = VR/Vs\n",
- " #If the line is doubled in length, then\n",
- "VR = Vs*(p)**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Receiving end voltage If the line is doubled in length, VR is \",abs(VR),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Receiving end voltage If the line is doubled in length, VR is 0.5 V"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 876</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the characteristic impedance of the line at this frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rzoc = 800;# in ohm\n",
- "thetazoc = -50;# in degrees\n",
- "rzsc = 413;# in ohm\n",
- "thetazsc = -20;# in degrees\n",
- "f = 1500;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #open circuit impedance\n",
- "Zoc = rzoc*math.cos(thetazoc*math.pi/180) + 1j*rzoc*math.sin(thetazoc*math.pi/180)\n",
- " #short circuit impedance\n",
- "Zsc = rzsc*math.cos(thetazsc*math.pi/180) + 1j*rzsc*math.sin(thetazsc*math.pi/180)\n",
- " #characteristic impedance Zo\n",
- "Zo = (Zoc*Zsc)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"characteristic impedance Zo is\",round(abs(Zo)),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "characteristic impedance Zo is 575.0 /_ -35.0 deg ohm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 877</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the characteristic impedance of the line when the frequency is 2 kHz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 15;# in ohm/loop km\n",
- "L = 0.0034;# in H/loop km\n",
- "C = 10E-9;# in F/km\n",
- "G = 3E-6;# in S/km\n",
- "f = 2000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #characteristic impedance Zo\n",
- "Zo = ((R + 1j*w*L)/(G + 1j*w*C))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"characteristic impedance Zo is \",round(abs(Zo),0),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "characteristic impedance Zo is 600.0 /_ -8.99 deg ohm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 879</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, at an operating frequency of 400 kHz, (a) the characteristic impedance,\n",
- "#(b) the propagation coefficient, (c) the wavelength on the line, and \n",
- "#(d) the velocity of propagation, in metres per second, of a signal.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.0005;# in H/loop km\n",
- "C = 0.12E-6;# in F/km\n",
- "f = 400000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #characteristic impedance Zo\n",
- "Zo = (L/C)**0.5\n",
- " #the propagation coefficient\n",
- "r = 1j*w*(L*C)**0.5\n",
- " #the attenuation coefficient \n",
- "a = r.real\n",
- " #the phaseshift coefficient\n",
- "b = r.imag\n",
- " #wavelength\n",
- "Y = 2*math.pi/b\n",
- " #velocity of propagation \n",
- "u = f*Y\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n characteristic impedance Zo is \",abs(Zo),\"ohm\"\n",
- "print \"\\n propagation coefficient is \",a,\" +(\",round(b,2),\")i\"\n",
- "print \"\\n wavelength Y is \",round(Y*1E3,0),\"m\"\n",
- "print \"\\n speed of transmission \",round(u,2),\"km/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " characteristic impedance Zo is 64.5497224368 ohm\n",
- "\n",
- " propagation coefficient is 0.0 +( 19.47 )i\n",
- "\n",
- " wavelength Y is 323.0 m\n",
- "\n",
- " speed of transmission 129099.44 km/sec"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 880</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the line (a) the characteristic impedance,\n",
- "#(b) the propagation coefficient, (c) the attenuation coefficient and\n",
- "#(d) the phase-shift coefficient\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 25;# in ohm/loop km\n",
- "L = 0.005;# in H/loop km\n",
- "C = 0.04E-6;# in F/km\n",
- "G = 80E-6;# in S/km\n",
- "f = 1000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #characteristic impedance Zo\n",
- "Zo = ((R + 1j*w*L)/(G + 1j*w*C))**0.5\n",
- " #the propagation coefficient\n",
- "r = ((R + 1j*w*L)*(G + 1j*w*C))**0.5\n",
- " #the attenuation coefficient \n",
- "a = r.real\n",
- " #the phaseshift coefficient\n",
- "b = r.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"characteristic impedance Zo is\",round(abs(Zo),2),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg ohm\"\n",
- "print \"\\n propagation coefficient is \",round(abs(r),4),\"/_\",round(cmath.phase(complex(a,b))*180/math.pi,2),\"deg\"\n",
- "print \"\\n attenuation coefficient is \",round(a,4),\" Np/km\"\n",
- "print \"\\n the phase-shift coefficient \",round(b,4),\" rad/km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "characteristic impedance Zo is 390.16 /_ -10.43 deg ohm\n",
- "\n",
- " propagation coefficient is 0.1029 /_ 61.92 deg\n",
- "\n",
- " attenuation coefficient is 0.0484 Np/km\n",
- "\n",
- " the phase-shift coefficient 0.0908 rad/km\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 881</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the characteristic impedance, \n",
- "#(b) the propagation coefficient, \n",
- "#(c) the attenuation and phase-shift coefficients, \n",
- "#(d) the sending-end current, \n",
- "#(e) the receiving-end current, \n",
- "#(f) the wavelength on the line, and \n",
- "#(g) the speed of transmission of signal.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 8;# in ohm/loop km\n",
- "L = 0.003;# in H/loop km\n",
- "C = 7500E-12;# in F/km\n",
- "G = 0.25E-6;# in S/km\n",
- "f = 1000;# in Hz\n",
- "n = 300;# in km\n",
- "Zg = 400 + 1j*0;# in ohm\n",
- "Vg = 10;# in Volts\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #characteristic impedance Zo\n",
- "Zo = ((R + 1j*w*L)/(G + 1j*w*C))**0.5\n",
- " #the propagation coefficient\n",
- "r = ((R + 1j*w*L)*(G + 1j*w*C))**0.5\n",
- " #the attenuation coefficient \n",
- "a = r.real\n",
- " #the phaseshift coefficient\n",
- "b = r.imag\n",
- " #the sending-end current,\n",
- "Is = Vg/(Zg + Zo)\n",
- " #the receiving-end current,\n",
- "IR = Is*cmath.e**(-1*n*r)\n",
- " #wavelength\n",
- "Y = 2*math.pi/b\n",
- " #velocity of propagation \n",
- "u = f*Y\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"characteristic impedance Zo is\",round(abs(Zo),1),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg ohm\"\n",
- "print \"propagation coefficient is \",round(abs(r),5),\"/_\",round(cmath.phase(complex(r.real,r.imag))*180/math.pi,2),\"deg\"\n",
- "print \"attenuation coefficient is \",round(a,5),\" Np/km and the phaseshift coefficient \",round(b,5),\" rad/km\"\n",
- "print \"sending-end current Is is \",round(abs(Is)*1E3,3),\"/_\",round(cmath.phase(complex(Is.real,Is.imag))*180/math.pi,2),\"deg mA\"\n",
- "print \"receiving-end current IR is\",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg mA\"\n",
- "print \"wavelength Y is \",round(Y,1),\" km\"\n",
- "print \"speed of transmission \",round(u,1),\"km/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "characteristic impedance Zo is 659.2 /_ -11.35 deg ohm\n",
- "propagation coefficient is 0.03106 /_ 78.35 deg\n",
- "attenuation coefficient is 0.00627 Np/km and the phaseshift coefficient 0.03042 rad/km\n",
- "sending-end current Is is 9.485 /_ 7.07 deg mA\n",
- "receiving-end current IR is 1.445 /_ -155.88 deg mA\n",
- "wavelength Y is 206.5 km\n",
- "speed of transmission 206521.1 km/sec\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 884</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine by how much the inductance should be increased to satisfy the condition for minimum distortion.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 10;# in ohm/loop km\n",
- "L = 0.0015;# in H/loop km\n",
- "C = 0.06E-6;# in F/km\n",
- "G = 1.2E-6;# in S/km\n",
- "\n",
- " #calculation:\n",
- " #the condition for minimum distortion is given by LG = CR, from which,\n",
- "Lm = C*R/G\n",
- "dL = Lm - L\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n inductance should be increased by \",round(dL*1E3,1),\"mH/loop km for minimum distortion\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " inductance should be increased by 498.5 mH/loop km for minimum distortion"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 884</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for minimum distortion at a frequency of 1.5 kHz \n",
- "#(a) the value of inductance per loop kilometre required, \n",
- "#(b) the propagation coefficient, \n",
- "#(c) the velocity of propagation of signal, and \n",
- "#(d) the wavelength on the line\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 80;# in ohm/loop km\n",
- "C = 5E-9;# in F/km\n",
- "G = 2E-6;# in S/km\n",
- "f = 1500;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #the condition for minimum distortion is given by LG = CR, from which, inductance\n",
- "L = C*R/G\n",
- " #attenuation coefficient,\n",
- "a = (R*G)**0.5\n",
- " #phase shift coefficient,\n",
- "b = w*(L*C)**0.5\n",
- " #propagation coefficient,\n",
- "r = a + 1j*b\n",
- " #velocity of propagation,\n",
- "u = 1/(L*C)**0.5\n",
- " #wavelength\n",
- "Y = u/f\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n inductance is \",round(L,2),\" H\"\n",
- "print \"\\n propagation coefficient is \",round(a,2),\" +(\",round(b,2),\")i\"\n",
- "print \"\\n speed of transmission \",round(u,2),\"km/sec\"\n",
- "print \"\\n wavelength Y is \",round(Y,2),\" km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " inductance is 0.2 H\n",
- "\n",
- " propagation coefficient is 0.01 +( 0.3 )i\n",
- "\n",
- " speed of transmission 31622.78 km/sec\n",
- "\n",
- " wavelength Y is 21.08 km\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 888</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of (a) the reflection coefficient for the line, (b) the incident current, \n",
- "#(c) the incident voltage, (d) the reflected current, and (e) the reflected voltage \n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Zo = 75;# in ohm\n",
- "ZR = 250;# in ohm\n",
- "VR = 10;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #reflection coefficient\n",
- "p = (Zo - ZR)/(Zo + ZR)\n",
- " #Current flowing in the terminating load\n",
- "IR = VR/ZR\n",
- " #incident current, Ii\n",
- "Ii = IR/(1 + p)\n",
- " #incident voltage, Vi \n",
- "Vi = Ii*Zo\n",
- " #reflected current, Ir\n",
- "Ir = IR - Ii\n",
- " #reflected voltage, Vr\n",
- "Vr = -1*Ir*Zo\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reflection coefficient is \",round(p,3),\"\"\n",
- "print \"\\n incident current, Ii is \",round(Ii,4),\" A\"\n",
- "print \"\\n incident voltage, Vi is \",round(Vi,2),\" V\"\n",
- "print \"\\n reflected current, Ir is \",round(Ir,4),\" A\"\n",
- "print \"\\n reflected voltage, Vr is \",round(Vr,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reflection coefficient is -0.538 \n",
- "\n",
- " incident current, Ii is 0.0867 A\n",
- "\n",
- " incident voltage, Vi is 6.5 V\n",
- "\n",
- " reflected current, Ir is -0.0467 A\n",
- "\n",
- " reflected voltage, Vr is 3.5 V"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 889</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the magnitude of the reflection coefficient in each case.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Zo = 500 - 1j*40;# in ohm\n",
- "ZR1 = 500 + 1j*40;# in ohm\n",
- "ZR2 = 600 + 1j*0;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #reflection coefficient\n",
- "p1 = (Zo - ZR1)/(Zo + ZR1)\n",
- "p2 = (Zo - ZR2)/(Zo + ZR2)\n",
- "p1mag = abs(p1)\n",
- "p2mag = abs(p2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reflection coefficient (a)\",p1mag,\" and (b)\", round(p2mag,2),\"\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reflection coefficient (a) 0.08 and (b) 0.1 "
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 890</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the magnitude of the ratio of the reflected to the incident voltage wave, and \n",
- "#(b) the incident voltage\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rzo = 500;# in ohm\n",
- "thetazo = 0;# in degrees\n",
- "ZR = 320 + 1j*240;# in ohm\n",
- "rvr = 20;# in volts\n",
- "thetavr = 35;# in degrees\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "VR = rvr*math.cos(thetavr*math.pi/180) + 1j*rvr*math.sin(thetavr*math.pi/180)\n",
- " #characteristic impedance\n",
- "Zo = rzo*math.cos(thetazo*math.pi/180) + 1j*rzo*math.sin(thetazo*math.pi/180)\n",
- " #the ratio of the reflected to the incident voltage \n",
- " #vr = VR/Vi\n",
- "vr = (ZR - Zo)/(Zo + ZR)\n",
- "vrmag = abs(vr)\n",
- " #incident voltage, Vi\n",
- "Vi = VR/vr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the magnitude of the ratio Vr : Vi is \",round(vrmag,3),\"\"\n",
- "print \"\\n incident voltage, Vi is \",round(abs(Vi),1),\"/_\",round(cmath.phase(complex(Vi.real,Vi.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the magnitude of the ratio Vr : Vi is 0.351 \n",
- "\n",
- " incident voltage, Vi is 57.0 /_ -75.56 deg V\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 895</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the reflection coefficient and (b) the standing-wave ratio.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rzo = 600;# in ohm\n",
- "thetazo = 0;# in degrees\n",
- "ZR = 400 + 250j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #characteristic impedance\n",
- "Zo = rzo*math.cos(thetazo*math.pi/180) + 1j*rzo*math.sin(thetazo*math.pi/180)\n",
- " #reflection coefficient\n",
- "p = (Zo - ZR)/(Zo + ZR)\n",
- "pmag = abs(p)\n",
- " #standing-wave ratio,\n",
- "s = (1 + pmag)/(1 - pmag)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reflection coefficient, is \",round(abs(p),4),\"/_\",round(cmath.phase(complex(p.real,p.imag))*180/math.pi,2),\"deg\"\n",
- "print \"\\n standing-wave ratio, s is \",round(s,3),\"\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reflection coefficient, is 0.3106 /_ -65.38 deg\n",
- "\n",
- " standing-wave ratio, s is 1.901 "
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 896</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the standing-wave ratio, \n",
- "#(b) the load impedance, and\n",
- "#(c) the incident current flowing if the reflected current is 10 mA.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rp = 0.2; \n",
- "thetap = -120;# in degrees\n",
- "Zo = 80;# in ohm\n",
- "Ir = 0.01;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #reflection coefficient\n",
- "p = rp*math.cos(thetap*math.pi/180) + 1j*rp*math.sin(thetap*math.pi/180)\n",
- " #standing-wave ratio,\n",
- "s = (1 + rp)/(1 - rp)\n",
- " #load impedance ZR \n",
- "ZR = Zo*(1 - p)/(1 + p)\n",
- " #incident current\n",
- "Ii = Ir*(s + 1)/(s - 1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n standing-wave ratio, s is \",s,\"\"\n",
- "print \"\\n load impedance ZR is \",round(ZR.real,2),\" +(\",round(ZR.imag,1),\")i ohm\"\n",
- "print \"\\n incident current is \",Ii,\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " standing-wave ratio, s is 1.5 \n",
- "\n",
- " load impedance ZR is 91.43 +( 33.0 )i ohm\n",
- "\n",
- " incident current is 0.05 A\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 897</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of the reflected power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "s = 1.6;\n",
- "Pi = 0.2;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #reflected power, Pr\n",
- "Pr = Pi*((s - 1)/(s + 1))**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reflected power, Pr is \",round(Pr*1E3,2),\" mW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reflected power, Pr is 10.65 mW"
- ]
- }
- ],
- "prompt_number": 18
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_2.ipynb deleted file mode 100755 index 03d830a4..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_2.ipynb +++ /dev/null @@ -1,1093 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 44: Transmission lines</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 873</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the wavelength on the line, and (b) the speed of transmission of a signal.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "f = 1910;# in Hz\n",
- "b = 0.05;# in rad/km\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #wavelength \n",
- "Y = 2*math.pi/b\n",
- " #speed of transmission\n",
- "u = f*Y\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n wavelength Y is \",round(Y,1),\" km\"\n",
- "print \"\\n speed of transmission \",round(u,1),\"km/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " wavelength Y is 125.7 km\n",
- "\n",
- " speed of transmission 240017.7 km/sec"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 873</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for a frequency of operation of 1 kHz, \n",
- "#(a) the phase delay, \n",
- "#(b) the wavelength on the line, and \n",
- "#(c) the velocity of propagation (in metres per second) of the signal.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.004;# in Henry/loop\n",
- "C = 0.004E-6;# in F/loop\n",
- "f = 1000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #phase delay\n",
- "b = w*(L*C)**0.5\n",
- " #wavelength \n",
- "Y = 2*math.pi/b\n",
- " #speed of transmission\n",
- "u = f*Y\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n phase delay is \",round(b,3),\" rad/km\"\n",
- "print \"\\n wavelength Y is \",Y,\" km\"\n",
- "print \"\\n speed of transmission \",u,\"km/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " phase delay is 0.025 rad/km\n",
- "\n",
- " wavelength Y is 250.0 km\n",
- "\n",
- " speed of transmission 250000.0 km/sec"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 874</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the voltage at a point 10 km down the line,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "a = 0.25;# in Np/km\n",
- "b = 0.20;# in rad/km\n",
- "Vs = 5;# in Volts\n",
- "n = 10;# in km\n",
- "f = 2000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #the voltage 10 km down the line\n",
- "r = a + 1j*b\n",
- "VR = Vs*cmath.e**(-1*n*r)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n Result \\n\\n\"\n",
- "print \"voltage 10 km down the line is \",round(abs(VR),2),\"/_\",round(cmath.phase(complex(VR.real,VR.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " Result \n",
- "\n",
- "\n",
- "voltage 10 km down the line is 0.41 /_ -114.59 deg V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 875</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the magnitude and phase of the current at the receiving end,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "a = 0.5;# in Np/km\n",
- "b = 0.25;# in rad/km\n",
- "rvs = 2;# in Volts\n",
- "thetavs = 0;# in degrees\n",
- "rzo = 800;# in ohm\n",
- "thetazo = -25;# in degrees\n",
- "n = 5;# in km\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "Vs = rvs*math.cos(thetavs*math.pi/180) + 1j*rvs*math.sin(thetavs*math.pi/180)\n",
- " #characteristic impedance\n",
- "Zo = rzo*math.cos(thetazo*math.pi/180) + 1j*rzo*math.sin(thetazo*math.pi/180)\n",
- " # receiving end voltage\n",
- "r = a + 1j*b\n",
- "VR = Vs*cmath.e**(-1*n*r)\n",
- " #Receiving end current,\n",
- "IR = VR/Zo\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"Receiving end current, IR is \",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "Receiving end current, IR is 0.205 /_ -46.62 deg mA\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 875</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the output voltage if the length of the line is doubled.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Vs = 8;# in Volts\n",
- "VR = 2;# in Volts\n",
- "x = 2; \n",
- "\n",
- "#calculation:\n",
- " # receiving end voltage VR = Vs*e**(-nr)\n",
- " #e**-nr = p\n",
- "p = VR/Vs\n",
- " #If the line is doubled in length, then\n",
- "VR = Vs*(p)**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Receiving end voltage If the line is doubled in length, VR is \",abs(VR),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Receiving end voltage If the line is doubled in length, VR is 0.5 V"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 876</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the characteristic impedance of the line at this frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rzoc = 800;# in ohm\n",
- "thetazoc = -50;# in degrees\n",
- "rzsc = 413;# in ohm\n",
- "thetazsc = -20;# in degrees\n",
- "f = 1500;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #open circuit impedance\n",
- "Zoc = rzoc*math.cos(thetazoc*math.pi/180) + 1j*rzoc*math.sin(thetazoc*math.pi/180)\n",
- " #short circuit impedance\n",
- "Zsc = rzsc*math.cos(thetazsc*math.pi/180) + 1j*rzsc*math.sin(thetazsc*math.pi/180)\n",
- " #characteristic impedance Zo\n",
- "Zo = (Zoc*Zsc)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"characteristic impedance Zo is\",round(abs(Zo)),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "characteristic impedance Zo is 575.0 /_ -35.0 deg ohm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 877</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the characteristic impedance of the line when the frequency is 2 kHz.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 15;# in ohm/loop km\n",
- "L = 0.0034;# in H/loop km\n",
- "C = 10E-9;# in F/km\n",
- "G = 3E-6;# in S/km\n",
- "f = 2000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #characteristic impedance Zo\n",
- "Zo = ((R + 1j*w*L)/(G + 1j*w*C))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"characteristic impedance Zo is \",round(abs(Zo),0),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "characteristic impedance Zo is 600.0 /_ -8.99 deg ohm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 879</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, at an operating frequency of 400 kHz, (a) the characteristic impedance,\n",
- "#(b) the propagation coefficient, (c) the wavelength on the line, and \n",
- "#(d) the velocity of propagation, in metres per second, of a signal.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.0005;# in H/loop km\n",
- "C = 0.12E-6;# in F/km\n",
- "f = 400000;# in Hz\n",
- "\n",
- "#calculation:\n",
- "w = 2*math.pi*f\n",
- " #characteristic impedance Zo\n",
- "Zo = (L/C)**0.5\n",
- " #the propagation coefficient\n",
- "r = 1j*w*(L*C)**0.5\n",
- " #the attenuation coefficient \n",
- "a = r.real\n",
- " #the phaseshift coefficient\n",
- "b = r.imag\n",
- " #wavelength\n",
- "Y = 2*math.pi/b\n",
- " #velocity of propagation \n",
- "u = f*Y\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n characteristic impedance Zo is \",abs(Zo),\"ohm\"\n",
- "print \"\\n propagation coefficient is \",a,\" +(\",round(b,2),\")i\"\n",
- "print \"\\n wavelength Y is \",round(Y*1E3,0),\"m\"\n",
- "print \"\\n speed of transmission \",round(u,2),\"km/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " characteristic impedance Zo is 64.5497224368 ohm\n",
- "\n",
- " propagation coefficient is 0.0 +( 19.47 )i\n",
- "\n",
- " wavelength Y is 323.0 m\n",
- "\n",
- " speed of transmission 129099.44 km/sec"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 880</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the line (a) the characteristic impedance,\n",
- "#(b) the propagation coefficient, (c) the attenuation coefficient and\n",
- "#(d) the phase-shift coefficient\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 25;# in ohm/loop km\n",
- "L = 0.005;# in H/loop km\n",
- "C = 0.04E-6;# in F/km\n",
- "G = 80E-6;# in S/km\n",
- "f = 1000;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #characteristic impedance Zo\n",
- "Zo = ((R + 1j*w*L)/(G + 1j*w*C))**0.5\n",
- " #the propagation coefficient\n",
- "r = ((R + 1j*w*L)*(G + 1j*w*C))**0.5\n",
- " #the attenuation coefficient \n",
- "a = r.real\n",
- " #the phaseshift coefficient\n",
- "b = r.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"characteristic impedance Zo is\",round(abs(Zo),2),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg ohm\"\n",
- "print \"\\n propagation coefficient is \",round(abs(r),4),\"/_\",round(cmath.phase(complex(a,b))*180/math.pi,2),\"deg\"\n",
- "print \"\\n attenuation coefficient is \",round(a,4),\" Np/km\"\n",
- "print \"\\n the phase-shift coefficient \",round(b,4),\" rad/km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "characteristic impedance Zo is 390.16 /_ -10.43 deg ohm\n",
- "\n",
- " propagation coefficient is 0.1029 /_ 61.92 deg\n",
- "\n",
- " attenuation coefficient is 0.0484 Np/km\n",
- "\n",
- " the phase-shift coefficient 0.0908 rad/km\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 881</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the characteristic impedance, \n",
- "#(b) the propagation coefficient, \n",
- "#(c) the attenuation and phase-shift coefficients, \n",
- "#(d) the sending-end current, \n",
- "#(e) the receiving-end current, \n",
- "#(f) the wavelength on the line, and \n",
- "#(g) the speed of transmission of signal.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 8;# in ohm/loop km\n",
- "L = 0.003;# in H/loop km\n",
- "C = 7500E-12;# in F/km\n",
- "G = 0.25E-6;# in S/km\n",
- "f = 1000;# in Hz\n",
- "n = 300;# in km\n",
- "Zg = 400 + 1j*0;# in ohm\n",
- "Vg = 10;# in Volts\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #characteristic impedance Zo\n",
- "Zo = ((R + 1j*w*L)/(G + 1j*w*C))**0.5\n",
- " #the propagation coefficient\n",
- "r = ((R + 1j*w*L)*(G + 1j*w*C))**0.5\n",
- " #the attenuation coefficient \n",
- "a = r.real\n",
- " #the phaseshift coefficient\n",
- "b = r.imag\n",
- " #the sending-end current,\n",
- "Is = Vg/(Zg + Zo)\n",
- " #the receiving-end current,\n",
- "IR = Is*cmath.e**(-1*n*r)\n",
- " #wavelength\n",
- "Y = 2*math.pi/b\n",
- " #velocity of propagation \n",
- "u = f*Y\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"characteristic impedance Zo is\",round(abs(Zo),1),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg ohm\"\n",
- "print \"propagation coefficient is \",round(abs(r),5),\"/_\",round(cmath.phase(complex(r.real,r.imag))*180/math.pi,2),\"deg\"\n",
- "print \"attenuation coefficient is \",round(a,5),\" Np/km and the phaseshift coefficient \",round(b,5),\" rad/km\"\n",
- "print \"sending-end current Is is \",round(abs(Is)*1E3,3),\"/_\",round(cmath.phase(complex(Is.real,Is.imag))*180/math.pi,2),\"deg mA\"\n",
- "print \"receiving-end current IR is\",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg mA\"\n",
- "print \"wavelength Y is \",round(Y,1),\" km\"\n",
- "print \"speed of transmission \",round(u,1),\"km/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "characteristic impedance Zo is 659.2 /_ -11.35 deg ohm\n",
- "propagation coefficient is 0.03106 /_ 78.35 deg\n",
- "attenuation coefficient is 0.00627 Np/km and the phaseshift coefficient 0.03042 rad/km\n",
- "sending-end current Is is 9.485 /_ 7.07 deg mA\n",
- "receiving-end current IR is 1.445 /_ -155.88 deg mA\n",
- "wavelength Y is 206.5 km\n",
- "speed of transmission 206521.1 km/sec\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 884</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine by how much the inductance should be increased to satisfy the condition for minimum distortion.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 10;# in ohm/loop km\n",
- "L = 0.0015;# in H/loop km\n",
- "C = 0.06E-6;# in F/km\n",
- "G = 1.2E-6;# in S/km\n",
- "\n",
- " #calculation:\n",
- " #the condition for minimum distortion is given by LG = CR, from which,\n",
- "Lm = C*R/G\n",
- "dL = Lm - L\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n inductance should be increased by \",round(dL*1E3,1),\"mH/loop km for minimum distortion\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " inductance should be increased by 498.5 mH/loop km for minimum distortion"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 884</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for minimum distortion at a frequency of 1.5 kHz \n",
- "#(a) the value of inductance per loop kilometre required, \n",
- "#(b) the propagation coefficient, \n",
- "#(c) the velocity of propagation of signal, and \n",
- "#(d) the wavelength on the line\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R = 80;# in ohm/loop km\n",
- "C = 5E-9;# in F/km\n",
- "G = 2E-6;# in S/km\n",
- "f = 1500;# in Hz\n",
- "\n",
- " #calculation:\n",
- "w = 2*math.pi*f\n",
- " #the condition for minimum distortion is given by LG = CR, from which, inductance\n",
- "L = C*R/G\n",
- " #attenuation coefficient,\n",
- "a = (R*G)**0.5\n",
- " #phase shift coefficient,\n",
- "b = w*(L*C)**0.5\n",
- " #propagation coefficient,\n",
- "r = a + 1j*b\n",
- " #velocity of propagation,\n",
- "u = 1/(L*C)**0.5\n",
- " #wavelength\n",
- "Y = u/f\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n inductance is \",round(L,2),\" H\"\n",
- "print \"\\n propagation coefficient is \",round(a,2),\" +(\",round(b,2),\")i\"\n",
- "print \"\\n speed of transmission \",round(u,2),\"km/sec\"\n",
- "print \"\\n wavelength Y is \",round(Y,2),\" km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " inductance is 0.2 H\n",
- "\n",
- " propagation coefficient is 0.01 +( 0.3 )i\n",
- "\n",
- " speed of transmission 31622.78 km/sec\n",
- "\n",
- " wavelength Y is 21.08 km\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 888</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of (a) the reflection coefficient for the line, (b) the incident current, \n",
- "#(c) the incident voltage, (d) the reflected current, and (e) the reflected voltage \n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Zo = 75;# in ohm\n",
- "ZR = 250;# in ohm\n",
- "VR = 10;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #reflection coefficient\n",
- "p = (Zo - ZR)/(Zo + ZR)\n",
- " #Current flowing in the terminating load\n",
- "IR = VR/ZR\n",
- " #incident current, Ii\n",
- "Ii = IR/(1 + p)\n",
- " #incident voltage, Vi \n",
- "Vi = Ii*Zo\n",
- " #reflected current, Ir\n",
- "Ir = IR - Ii\n",
- " #reflected voltage, Vr\n",
- "Vr = -1*Ir*Zo\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reflection coefficient is \",round(p,3),\"\"\n",
- "print \"\\n incident current, Ii is \",round(Ii,4),\" A\"\n",
- "print \"\\n incident voltage, Vi is \",round(Vi,2),\" V\"\n",
- "print \"\\n reflected current, Ir is \",round(Ir,4),\" A\"\n",
- "print \"\\n reflected voltage, Vr is \",round(Vr,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reflection coefficient is -0.538 \n",
- "\n",
- " incident current, Ii is 0.0867 A\n",
- "\n",
- " incident voltage, Vi is 6.5 V\n",
- "\n",
- " reflected current, Ir is -0.0467 A\n",
- "\n",
- " reflected voltage, Vr is 3.5 V"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 889</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the magnitude of the reflection coefficient in each case.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "Zo = 500 - 1j*40;# in ohm\n",
- "ZR1 = 500 + 1j*40;# in ohm\n",
- "ZR2 = 600 + 1j*0;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #reflection coefficient\n",
- "p1 = (Zo - ZR1)/(Zo + ZR1)\n",
- "p2 = (Zo - ZR2)/(Zo + ZR2)\n",
- "p1mag = abs(p1)\n",
- "p2mag = abs(p2)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reflection coefficient (a)\",p1mag,\" and (b)\", round(p2mag,2),\"\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reflection coefficient (a) 0.08 and (b) 0.1 "
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 890</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the magnitude of the ratio of the reflected to the incident voltage wave, and \n",
- "#(b) the incident voltage\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rzo = 500;# in ohm\n",
- "thetazo = 0;# in degrees\n",
- "ZR = 320 + 1j*240;# in ohm\n",
- "rvr = 20;# in volts\n",
- "thetavr = 35;# in degrees\n",
- "\n",
- " #calculation:\n",
- " #voltage\n",
- "VR = rvr*math.cos(thetavr*math.pi/180) + 1j*rvr*math.sin(thetavr*math.pi/180)\n",
- " #characteristic impedance\n",
- "Zo = rzo*math.cos(thetazo*math.pi/180) + 1j*rzo*math.sin(thetazo*math.pi/180)\n",
- " #the ratio of the reflected to the incident voltage \n",
- " #vr = VR/Vi\n",
- "vr = (ZR - Zo)/(Zo + ZR)\n",
- "vrmag = abs(vr)\n",
- " #incident voltage, Vi\n",
- "Vi = VR/vr\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the magnitude of the ratio Vr : Vi is \",round(vrmag,3),\"\"\n",
- "print \"\\n incident voltage, Vi is \",round(abs(Vi),1),\"/_\",round(cmath.phase(complex(Vi.real,Vi.imag))*180/math.pi,2),\"deg V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the magnitude of the ratio Vr : Vi is 0.351 \n",
- "\n",
- " incident voltage, Vi is 57.0 /_ -75.56 deg V\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 895</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the reflection coefficient and (b) the standing-wave ratio.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rzo = 600;# in ohm\n",
- "thetazo = 0;# in degrees\n",
- "ZR = 400 + 250j;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #characteristic impedance\n",
- "Zo = rzo*math.cos(thetazo*math.pi/180) + 1j*rzo*math.sin(thetazo*math.pi/180)\n",
- " #reflection coefficient\n",
- "p = (Zo - ZR)/(Zo + ZR)\n",
- "pmag = abs(p)\n",
- " #standing-wave ratio,\n",
- "s = (1 + pmag)/(1 - pmag)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reflection coefficient, is \",round(abs(p),4),\"/_\",round(cmath.phase(complex(p.real,p.imag))*180/math.pi,2),\"deg\"\n",
- "print \"\\n standing-wave ratio, s is \",round(s,3),\"\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reflection coefficient, is 0.3106 /_ -65.38 deg\n",
- "\n",
- " standing-wave ratio, s is 1.901 "
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 896</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the standing-wave ratio, \n",
- "#(b) the load impedance, and\n",
- "#(c) the incident current flowing if the reflected current is 10 mA.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rp = 0.2; \n",
- "thetap = -120;# in degrees\n",
- "Zo = 80;# in ohm\n",
- "Ir = 0.01;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #reflection coefficient\n",
- "p = rp*math.cos(thetap*math.pi/180) + 1j*rp*math.sin(thetap*math.pi/180)\n",
- " #standing-wave ratio,\n",
- "s = (1 + rp)/(1 - rp)\n",
- " #load impedance ZR \n",
- "ZR = Zo*(1 - p)/(1 + p)\n",
- " #incident current\n",
- "Ii = Ir*(s + 1)/(s - 1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n standing-wave ratio, s is \",s,\"\"\n",
- "print \"\\n load impedance ZR is \",round(ZR.real,2),\" +(\",round(ZR.imag,1),\")i ohm\"\n",
- "print \"\\n incident current is \",Ii,\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " standing-wave ratio, s is 1.5 \n",
- "\n",
- " load impedance ZR is 91.43 +( 33.0 )i ohm\n",
- "\n",
- " incident current is 0.05 A\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 897</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the value of the reflected power.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "s = 1.6;\n",
- "Pi = 0.2;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #reflected power, Pr\n",
- "Pr = Pi*((s - 1)/(s + 1))**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n reflected power, Pr is \",round(Pr*1E3,2),\" mW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " reflected power, Pr is 10.65 mW"
- ]
- }
- ],
- "prompt_number": 18
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_44.ipynb index 61794674..61794674 100755..100644 --- a/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_3.ipynb +++ b/Electrical_Circuit_Theory_And_Technology/chapter_44.ipynb diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint.ipynb deleted file mode 100755 index f30323f6..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint.ipynb +++ /dev/null @@ -1,435 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 45: Transients and Laplace\n",
- "transforms</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 903</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the initial value of current flowing, \n",
- "#(b) the value of current 150 ms after connection, \n",
- "#(c) the value of capacitor voltage 80 ms after connection, and \n",
- "#(d) the time after connection when the resistor voltage is 35 V.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 500E-9;# in Farad\n",
- "R = 100000;# in Ohm\n",
- "V = 50;# in VOlts\n",
- "ti = 0.15;# in sec\n",
- "tc = 0.08;# in sec\n",
- "Vrt = 35;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #Initial current, \n",
- "i0 = (V/R)\n",
- " #when time t = 150ms current is\n",
- "i150 = (V/R)*math.e**(-1*ti/(R*C))\n",
- " #capacitor voltage, Vc\n",
- "Vc = V*(1 - math.e**(-1*tc/(R*C)))\n",
- " #time, t\n",
- "tvr = -1*R*C*math.log(Vrt/V)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n initial value of current flowing is \",round(i0*1E3,2),\"mA\"\n",
- "print \"\\n current flowing at t = 150ms is \",round(i150*1E6,2),\"uA\"\n",
- "print \"\\n value of capacitor voltage at t = 80ms is \",round(Vc,2),\" V\"\n",
- "print \"\\n the time after connection when the resistor voltage is 35 V is \",round(tvr*1E3,2),\"msec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " initial value of current flowing is 0.5 mA\n",
- "\n",
- " current flowing at t = 150ms is 24.89 uA\n",
- "\n",
- " value of capacitor voltage at t = 80ms is 39.91 V\n",
- "\n",
- " the time after connection when the resistor voltage is 35 V is 17.83 msec\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 905</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the p.d. across the capacitor after 20 s\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 5E-6;# in Farad\n",
- "R = 2000000;# in Ohm\n",
- "V = 200;# in VOlts\n",
- "tc = 20;# in sec\n",
- "\n",
- " #calculation:\n",
- " #capacitor voltage, Vc\n",
- "Vc = V*(math.e**(-1*tc/(R*C)))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n value of capacitor voltage at t = 20s is \",round(Vc,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " value of capacitor voltage at t = 20s is 27.07 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 907</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the final value of current, (b) the value of current after 4 ms, \n",
- "#(c) the value of the voltage across the resistor after 6 ms,\n",
- "#(d) the value of the voltage across the inductance after 6 ms, and \n",
- "#(e) the time when the current reaches 15 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.05;# in Henry\n",
- "R = 5;# in Ohm\n",
- "V = 110;# in VOlts\n",
- "ti = 0.004;# in sec\n",
- "tvr = 0.006;# in sec\n",
- "tvl = 0.006;# in sec\n",
- "it = 15;# in amperes\n",
- "\n",
- " #calculation:\n",
- " #steady state current i\n",
- "i = V/R\n",
- " #when time t = 4ms current is\n",
- "i4 = (V/R)*(1 - math.e**(-1*ti*R/L))\n",
- " #resistor voltage, VR\n",
- "VR6 = V*(1 - math.e**(-1*tvr*R/L))\n",
- " #inductor voltage, VL\n",
- "VL6 = V*(math.e**(-1*tvl*R/L))\n",
- " #time, t\n",
- "ti = (-1*L/R)*math.log(1 - it*R/V)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n steady state current i is \",round(i,2),\" A\"\n",
- "print \"\\n when time t = 4ms current is is \",round(i4,2),\" A\"\n",
- "print \"\\n value of resistor voltage at t = 6ms is \",round(VR6,2),\" V\"\n",
- "print \"\\n value of inductor voltage at t = 6ms is \",round(VL6,2),\" V\"\n",
- "print \"\\n the time after connection when the current is 15 V is \",round(ti,5),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " steady state current i is 22.0 A\n",
- "\n",
- " when time t = 4ms current is is 7.25 A\n",
- "\n",
- " value of resistor voltage at t = 6ms is 49.63 V\n",
- "\n",
- " value of inductor voltage at t = 6ms is 60.37 V\n",
- "\n",
- " the time after connection when the current is 15 V is 0.01145 sec"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 909</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine\n",
- "#(a) the time for the current in the 2 H inductor to fall to 200 mA,\n",
- "#and (b) the maximum voltage appearing across the resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "i = 5;# in Amperes\n",
- "L = 2# in Henry\n",
- "i1 = 0.2;# in Amperes\n",
- "R = 10;# in Ohm\n",
- "\n",
- " #calculation:\n",
- " #time t\n",
- "ti = (-1*L/R)*math.log(i1/i)\n",
- " #voltage across the resistor is a maximum \n",
- "VRm = i*R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n time t for the current in the 2 H inductor to fall to 200 mA is \",round(ti,3),\" sec\"\n",
- "print \"\\n max voltage across the resistor is \",VRm,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " time t for the current in the 2 H inductor to fall to 200 mA is 0.644 sec\n",
- "\n",
- " max voltage across the resistor is 50 V"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 911</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine whether the circuit is over, critical or underdamped. (b) If C D 5 nF, determine the state of damping.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.002# in Henry\n",
- "R = 1000;# in Ohm\n",
- "C1 = 5E-6;# in farad\n",
- "C2 = 5E-9;# in farad\n",
- "\n",
- " #calculation:\n",
- "a = (R/(2*L))**2\n",
- "b = 1/(L*C1)\n",
- "if (a>b):\n",
- "\ts1 = \"overdamped\";\n",
- "elif (a<b):\n",
- "\ts1 = \"underdamped\";\n",
- "else:\n",
- "\ts1 = \"critically damped\";\n",
- "c = 1/(L*C2)\n",
- "if (a>c):\n",
- "\ts2 = \"overdamped\";\n",
- "elif (a<c):\n",
- "\ts2 = \"underdamped\";\n",
- "else:\n",
- "\ts2 = \"critically damped\";\n",
- "\t\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n circuit is \",s1\n",
- "print \"\\n if C = 5 nF, circuit is \",s2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " circuit is overdamped\n",
- "\n",
- " if C = 5 nF, circuit is underdamped"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 912</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#what value of capacitance will give critical damping ?\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.002# in Henry\n",
- "R = 1000;# in Ohm\n",
- "\n",
- " #calculation:\n",
- "a = (R/(2*L))**2\n",
- " #for critically damped\n",
- "C = 4*L/R**2\n",
- "\t\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance C is \",C*1E9,\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance C is 8.0 nF"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 913</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the nature of the response and obtain an expression for the current in the coil.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 1.5# in Henry\n",
- "R = 90;# in Ohm\n",
- "C = 5*1E-6; # in Farad\n",
- "V = 10; # in Volts\n",
- "\n",
- "#calculation:\n",
- "a = -1*R/(2*L)\n",
- "b = (1/(L*C) - (R/(2*L))**2)**0.5\n",
- "V0 = V\n",
- "I0 = 0\n",
- "A = V0\n",
- "B = (I0 - C*a*V0)/(C*b)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"Current, i = e^(\",a,\"t) (\",round((a*C*B - A*C*b),4),\"sin(\",round(b,1),\"t) + (\",round((-1*a*C*A + B*C*C*b),0),\"cos(\",round(b,1),\"t) Amps.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "Current, i = e^( -30.0 t) ( -0.0183 sin( 363.9 t) + ( 0.0 cos( 363.9 t) Amps.\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_1.ipynb deleted file mode 100755 index f30323f6..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_1.ipynb +++ /dev/null @@ -1,435 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 45: Transients and Laplace\n",
- "transforms</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 903</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the initial value of current flowing, \n",
- "#(b) the value of current 150 ms after connection, \n",
- "#(c) the value of capacitor voltage 80 ms after connection, and \n",
- "#(d) the time after connection when the resistor voltage is 35 V.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 500E-9;# in Farad\n",
- "R = 100000;# in Ohm\n",
- "V = 50;# in VOlts\n",
- "ti = 0.15;# in sec\n",
- "tc = 0.08;# in sec\n",
- "Vrt = 35;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #Initial current, \n",
- "i0 = (V/R)\n",
- " #when time t = 150ms current is\n",
- "i150 = (V/R)*math.e**(-1*ti/(R*C))\n",
- " #capacitor voltage, Vc\n",
- "Vc = V*(1 - math.e**(-1*tc/(R*C)))\n",
- " #time, t\n",
- "tvr = -1*R*C*math.log(Vrt/V)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n initial value of current flowing is \",round(i0*1E3,2),\"mA\"\n",
- "print \"\\n current flowing at t = 150ms is \",round(i150*1E6,2),\"uA\"\n",
- "print \"\\n value of capacitor voltage at t = 80ms is \",round(Vc,2),\" V\"\n",
- "print \"\\n the time after connection when the resistor voltage is 35 V is \",round(tvr*1E3,2),\"msec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " initial value of current flowing is 0.5 mA\n",
- "\n",
- " current flowing at t = 150ms is 24.89 uA\n",
- "\n",
- " value of capacitor voltage at t = 80ms is 39.91 V\n",
- "\n",
- " the time after connection when the resistor voltage is 35 V is 17.83 msec\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 905</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the p.d. across the capacitor after 20 s\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 5E-6;# in Farad\n",
- "R = 2000000;# in Ohm\n",
- "V = 200;# in VOlts\n",
- "tc = 20;# in sec\n",
- "\n",
- " #calculation:\n",
- " #capacitor voltage, Vc\n",
- "Vc = V*(math.e**(-1*tc/(R*C)))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n value of capacitor voltage at t = 20s is \",round(Vc,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " value of capacitor voltage at t = 20s is 27.07 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 907</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the final value of current, (b) the value of current after 4 ms, \n",
- "#(c) the value of the voltage across the resistor after 6 ms,\n",
- "#(d) the value of the voltage across the inductance after 6 ms, and \n",
- "#(e) the time when the current reaches 15 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.05;# in Henry\n",
- "R = 5;# in Ohm\n",
- "V = 110;# in VOlts\n",
- "ti = 0.004;# in sec\n",
- "tvr = 0.006;# in sec\n",
- "tvl = 0.006;# in sec\n",
- "it = 15;# in amperes\n",
- "\n",
- " #calculation:\n",
- " #steady state current i\n",
- "i = V/R\n",
- " #when time t = 4ms current is\n",
- "i4 = (V/R)*(1 - math.e**(-1*ti*R/L))\n",
- " #resistor voltage, VR\n",
- "VR6 = V*(1 - math.e**(-1*tvr*R/L))\n",
- " #inductor voltage, VL\n",
- "VL6 = V*(math.e**(-1*tvl*R/L))\n",
- " #time, t\n",
- "ti = (-1*L/R)*math.log(1 - it*R/V)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n steady state current i is \",round(i,2),\" A\"\n",
- "print \"\\n when time t = 4ms current is is \",round(i4,2),\" A\"\n",
- "print \"\\n value of resistor voltage at t = 6ms is \",round(VR6,2),\" V\"\n",
- "print \"\\n value of inductor voltage at t = 6ms is \",round(VL6,2),\" V\"\n",
- "print \"\\n the time after connection when the current is 15 V is \",round(ti,5),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " steady state current i is 22.0 A\n",
- "\n",
- " when time t = 4ms current is is 7.25 A\n",
- "\n",
- " value of resistor voltage at t = 6ms is 49.63 V\n",
- "\n",
- " value of inductor voltage at t = 6ms is 60.37 V\n",
- "\n",
- " the time after connection when the current is 15 V is 0.01145 sec"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 909</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine\n",
- "#(a) the time for the current in the 2 H inductor to fall to 200 mA,\n",
- "#and (b) the maximum voltage appearing across the resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "i = 5;# in Amperes\n",
- "L = 2# in Henry\n",
- "i1 = 0.2;# in Amperes\n",
- "R = 10;# in Ohm\n",
- "\n",
- " #calculation:\n",
- " #time t\n",
- "ti = (-1*L/R)*math.log(i1/i)\n",
- " #voltage across the resistor is a maximum \n",
- "VRm = i*R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n time t for the current in the 2 H inductor to fall to 200 mA is \",round(ti,3),\" sec\"\n",
- "print \"\\n max voltage across the resistor is \",VRm,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " time t for the current in the 2 H inductor to fall to 200 mA is 0.644 sec\n",
- "\n",
- " max voltage across the resistor is 50 V"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 911</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine whether the circuit is over, critical or underdamped. (b) If C D 5 nF, determine the state of damping.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.002# in Henry\n",
- "R = 1000;# in Ohm\n",
- "C1 = 5E-6;# in farad\n",
- "C2 = 5E-9;# in farad\n",
- "\n",
- " #calculation:\n",
- "a = (R/(2*L))**2\n",
- "b = 1/(L*C1)\n",
- "if (a>b):\n",
- "\ts1 = \"overdamped\";\n",
- "elif (a<b):\n",
- "\ts1 = \"underdamped\";\n",
- "else:\n",
- "\ts1 = \"critically damped\";\n",
- "c = 1/(L*C2)\n",
- "if (a>c):\n",
- "\ts2 = \"overdamped\";\n",
- "elif (a<c):\n",
- "\ts2 = \"underdamped\";\n",
- "else:\n",
- "\ts2 = \"critically damped\";\n",
- "\t\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n circuit is \",s1\n",
- "print \"\\n if C = 5 nF, circuit is \",s2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " circuit is overdamped\n",
- "\n",
- " if C = 5 nF, circuit is underdamped"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 912</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#what value of capacitance will give critical damping ?\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.002# in Henry\n",
- "R = 1000;# in Ohm\n",
- "\n",
- " #calculation:\n",
- "a = (R/(2*L))**2\n",
- " #for critically damped\n",
- "C = 4*L/R**2\n",
- "\t\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance C is \",C*1E9,\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance C is 8.0 nF"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 913</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the nature of the response and obtain an expression for the current in the coil.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 1.5# in Henry\n",
- "R = 90;# in Ohm\n",
- "C = 5*1E-6; # in Farad\n",
- "V = 10; # in Volts\n",
- "\n",
- "#calculation:\n",
- "a = -1*R/(2*L)\n",
- "b = (1/(L*C) - (R/(2*L))**2)**0.5\n",
- "V0 = V\n",
- "I0 = 0\n",
- "A = V0\n",
- "B = (I0 - C*a*V0)/(C*b)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"Current, i = e^(\",a,\"t) (\",round((a*C*B - A*C*b),4),\"sin(\",round(b,1),\"t) + (\",round((-1*a*C*A + B*C*C*b),0),\"cos(\",round(b,1),\"t) Amps.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "Current, i = e^( -30.0 t) ( -0.0183 sin( 363.9 t) + ( 0.0 cos( 363.9 t) Amps.\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_2.ipynb deleted file mode 100755 index f30323f6..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_2.ipynb +++ /dev/null @@ -1,435 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 45: Transients and Laplace\n",
- "transforms</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 903</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the initial value of current flowing, \n",
- "#(b) the value of current 150 ms after connection, \n",
- "#(c) the value of capacitor voltage 80 ms after connection, and \n",
- "#(d) the time after connection when the resistor voltage is 35 V.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 500E-9;# in Farad\n",
- "R = 100000;# in Ohm\n",
- "V = 50;# in VOlts\n",
- "ti = 0.15;# in sec\n",
- "tc = 0.08;# in sec\n",
- "Vrt = 35;# in Volts\n",
- "\n",
- " #calculation:\n",
- " #Initial current, \n",
- "i0 = (V/R)\n",
- " #when time t = 150ms current is\n",
- "i150 = (V/R)*math.e**(-1*ti/(R*C))\n",
- " #capacitor voltage, Vc\n",
- "Vc = V*(1 - math.e**(-1*tc/(R*C)))\n",
- " #time, t\n",
- "tvr = -1*R*C*math.log(Vrt/V)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n initial value of current flowing is \",round(i0*1E3,2),\"mA\"\n",
- "print \"\\n current flowing at t = 150ms is \",round(i150*1E6,2),\"uA\"\n",
- "print \"\\n value of capacitor voltage at t = 80ms is \",round(Vc,2),\" V\"\n",
- "print \"\\n the time after connection when the resistor voltage is 35 V is \",round(tvr*1E3,2),\"msec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " initial value of current flowing is 0.5 mA\n",
- "\n",
- " current flowing at t = 150ms is 24.89 uA\n",
- "\n",
- " value of capacitor voltage at t = 80ms is 39.91 V\n",
- "\n",
- " the time after connection when the resistor voltage is 35 V is 17.83 msec\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 905</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the p.d. across the capacitor after 20 s\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "C = 5E-6;# in Farad\n",
- "R = 2000000;# in Ohm\n",
- "V = 200;# in VOlts\n",
- "tc = 20;# in sec\n",
- "\n",
- " #calculation:\n",
- " #capacitor voltage, Vc\n",
- "Vc = V*(math.e**(-1*tc/(R*C)))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n value of capacitor voltage at t = 20s is \",round(Vc,2),\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " value of capacitor voltage at t = 20s is 27.07 V"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 907</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the final value of current, (b) the value of current after 4 ms, \n",
- "#(c) the value of the voltage across the resistor after 6 ms,\n",
- "#(d) the value of the voltage across the inductance after 6 ms, and \n",
- "#(e) the time when the current reaches 15 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.05;# in Henry\n",
- "R = 5;# in Ohm\n",
- "V = 110;# in VOlts\n",
- "ti = 0.004;# in sec\n",
- "tvr = 0.006;# in sec\n",
- "tvl = 0.006;# in sec\n",
- "it = 15;# in amperes\n",
- "\n",
- " #calculation:\n",
- " #steady state current i\n",
- "i = V/R\n",
- " #when time t = 4ms current is\n",
- "i4 = (V/R)*(1 - math.e**(-1*ti*R/L))\n",
- " #resistor voltage, VR\n",
- "VR6 = V*(1 - math.e**(-1*tvr*R/L))\n",
- " #inductor voltage, VL\n",
- "VL6 = V*(math.e**(-1*tvl*R/L))\n",
- " #time, t\n",
- "ti = (-1*L/R)*math.log(1 - it*R/V)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n steady state current i is \",round(i,2),\" A\"\n",
- "print \"\\n when time t = 4ms current is is \",round(i4,2),\" A\"\n",
- "print \"\\n value of resistor voltage at t = 6ms is \",round(VR6,2),\" V\"\n",
- "print \"\\n value of inductor voltage at t = 6ms is \",round(VL6,2),\" V\"\n",
- "print \"\\n the time after connection when the current is 15 V is \",round(ti,5),\" sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " steady state current i is 22.0 A\n",
- "\n",
- " when time t = 4ms current is is 7.25 A\n",
- "\n",
- " value of resistor voltage at t = 6ms is 49.63 V\n",
- "\n",
- " value of inductor voltage at t = 6ms is 60.37 V\n",
- "\n",
- " the time after connection when the current is 15 V is 0.01145 sec"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 909</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine\n",
- "#(a) the time for the current in the 2 H inductor to fall to 200 mA,\n",
- "#and (b) the maximum voltage appearing across the resistor.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "i = 5;# in Amperes\n",
- "L = 2# in Henry\n",
- "i1 = 0.2;# in Amperes\n",
- "R = 10;# in Ohm\n",
- "\n",
- " #calculation:\n",
- " #time t\n",
- "ti = (-1*L/R)*math.log(i1/i)\n",
- " #voltage across the resistor is a maximum \n",
- "VRm = i*R\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n time t for the current in the 2 H inductor to fall to 200 mA is \",round(ti,3),\" sec\"\n",
- "print \"\\n max voltage across the resistor is \",VRm,\" V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " time t for the current in the 2 H inductor to fall to 200 mA is 0.644 sec\n",
- "\n",
- " max voltage across the resistor is 50 V"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 911</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine whether the circuit is over, critical or underdamped. (b) If C D 5 nF, determine the state of damping.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.002# in Henry\n",
- "R = 1000;# in Ohm\n",
- "C1 = 5E-6;# in farad\n",
- "C2 = 5E-9;# in farad\n",
- "\n",
- " #calculation:\n",
- "a = (R/(2*L))**2\n",
- "b = 1/(L*C1)\n",
- "if (a>b):\n",
- "\ts1 = \"overdamped\";\n",
- "elif (a<b):\n",
- "\ts1 = \"underdamped\";\n",
- "else:\n",
- "\ts1 = \"critically damped\";\n",
- "c = 1/(L*C2)\n",
- "if (a>c):\n",
- "\ts2 = \"overdamped\";\n",
- "elif (a<c):\n",
- "\ts2 = \"underdamped\";\n",
- "else:\n",
- "\ts2 = \"critically damped\";\n",
- "\t\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n circuit is \",s1\n",
- "print \"\\n if C = 5 nF, circuit is \",s2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " circuit is overdamped\n",
- "\n",
- " if C = 5 nF, circuit is underdamped"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 912</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#what value of capacitance will give critical damping ?\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 0.002# in Henry\n",
- "R = 1000;# in Ohm\n",
- "\n",
- " #calculation:\n",
- "a = (R/(2*L))**2\n",
- " #for critically damped\n",
- "C = 4*L/R**2\n",
- "\t\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n capacitance C is \",C*1E9,\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " capacitance C is 8.0 nF"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 913</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the nature of the response and obtain an expression for the current in the coil.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "L = 1.5# in Henry\n",
- "R = 90;# in Ohm\n",
- "C = 5*1E-6; # in Farad\n",
- "V = 10; # in Volts\n",
- "\n",
- "#calculation:\n",
- "a = -1*R/(2*L)\n",
- "b = (1/(L*C) - (R/(2*L))**2)**0.5\n",
- "V0 = V\n",
- "I0 = 0\n",
- "A = V0\n",
- "B = (I0 - C*a*V0)/(C*b)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"Current, i = e^(\",a,\"t) (\",round((a*C*B - A*C*b),4),\"sin(\",round(b,1),\"t) + (\",round((-1*a*C*A + B*C*C*b),0),\"cos(\",round(b,1),\"t) Amps.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "Current, i = e^( -30.0 t) ( -0.0183 sin( 363.9 t) + ( 0.0 cos( 363.9 t) Amps.\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
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