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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 |