{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "

Chapter 2: An Introduction to Electric Circuits

" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 1, page no. 12

" ] }, { "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": [ "

Example 2, page no. 12

" ] }, { "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": [ "

Example 3, page no. 14

" ] }, { "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": [ "

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": [ "

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": [ "

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": [ "

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": [ "

Example 8, page no. 16

" ] }, { "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": [ "

Example 9, page no. 17

" ] }, { "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": [ "

Example 10, page no. 17

" ] }, { "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": [ "

Example 11, page no. 17

" ] }, { "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": [ "

Example 12, page no. 17

" ] }, { "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": [ "

Example 13, page no. 18

" ] }, { "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": [ "

Example 14, page no. 18

" ] }, { "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": [ "

Example 15, page no. 18

" ] }, { "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": [ "

Example 16, page no. 18

" ] }, { "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": [ "

Example 17, page no. 19

" ] }, { "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": [ "

Example 18, page no. 19

" ] }, { "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": [ "

Example 19, page no. 19

" ] }, { "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": [ "

Example 20, page no. 20

" ] }, { "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": {} } ] }