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-{
- "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