{
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"name": ""
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"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter 26: Power in a.c. circuits
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 1, page no. 466
"
]
},
{
"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": [
"Example 2, page no. 467
"
]
},
{
"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": [
"Example 3, page no. 468
"
]
},
{
"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": [
"Example 4, page no. 468
"
]
},
{
"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": [
"Example 5, page no. 469
"
]
},
{
"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": [
"Example 6, page no. 471
"
]
},
{
"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": [
"Example 7, page no. 471
"
]
},
{
"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": {}
}
]
}