{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter 16: Single-phase parallel a.c. circuits
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 1, page no. 239
"
]
},
{
"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": [
"Example 2, page no. 240
"
]
},
{
"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": [
"Example 3, page no. 241
"
]
},
{
"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": [
"Example 4, page no. 242
"
]
},
{
"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": [
"Example 5, page no. 242
"
]
},
{
"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": [
"Example 6, page no. 244
"
]
},
{
"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": [
"Example 7, page no. 246
"
]
},
{
"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": [
"Example 8, page no. 249
"
]
},
{
"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": [
"Example 9, page no. 250
"
]
},
{
"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": [
"Example 10, page no. 251
"
]
},
{
"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": [
"Example 11, page no. 252
"
]
},
{
"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": [
"Example 12, page no. 253
"
]
},
{
"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": [
"Example 13, page no. 254
"
]
},
{
"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": {}
}
]
}