{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter 36: Complex Waveforms
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 1, page no. 643
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#(a) Write down an expression to represent voltage v.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"V = 240; # in Volts\n",
"f = 50; # in Hz\n",
"x = 0.2;\n",
"phi3 = 3*math.pi/4; # in Rad\n",
"\n",
" #calculation:\n",
"Vamp = V*2**0.5\n",
"w = 2*math.pi*f\n",
"T = 1/f\n",
"V3 = Vamp*x\n",
"f3 = 3*f\n",
"w3 = 3*w\n",
"\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n voltage, V =\",round(Vamp,1),\"sin(\",round(w,1),\"t) + \",round(V3,1),\"sin(\", round(w3,1),\"t - \", round(phi3,1),\") volts\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" voltage, V = 339.4 sin( 314.2 t) + 67.9 sin( 942.5 t - 2.4 ) volts"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3, page no. 648
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the rms value of the current waveform\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"A1 = 0.100;# in amperes\n",
"A3 = 0.020;# in amperes\n",
"A5 = 0.010;# in amperes\n",
"\n",
" #calculation:\n",
" #the rms value of current is given by\n",
"Irms = ((A1**2 + A3**2 + A5**2)/2)**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the rms value of current is \",round(Irms*1000,2),\" mA\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the rms value of current is 72.46 mA"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 4, page no. 649
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine for the voltage, (a) the rms value, (b) the mean value and (c) the form factor.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"A1 = 10;# in volts\n",
"A3 = 3;# in volts\n",
"A5 = 2;# in volts\n",
"\n",
"#calculation:\n",
" #the rms value of voltage is given by\n",
"Vrms = ((A1**2 + A3**2 + A5**2)/2)**0.5\n",
" #the mean value of voltage is given by\n",
" #x = wt\n",
"Vav = (1/math.pi)*((10 + 1 + 2/5)-(-10 - 1 - 2/5))\n",
" #form factor is given by\n",
"ff = Vrms/Vav\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)the rms value of voltage is \",round(Vrms,2),\" V\"\n",
"print \"\\n (b)the mean value of voltage is \",round(Vav,2),\" V\"\n",
"print \"\\n (c)form factor is \",round(ff,3),\" \""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)the rms value of voltage is 7.52 V\n",
"\n",
" (b)the mean value of voltage is 7.26 V\n",
"\n",
" (c)form factor is 1.036 "
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5, page no. 649
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#(a) Determine the rms value of the fundamental and each harmonic. \n",
"#(b) Write down an expression to represent the complex voltage waveform if the frequency of the fundamental is 31.83 Hz.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"V = 240;# in volts\n",
"x = 0.3;# for third harmonic\n",
"y = 0.1;# for fifth harmonic\n",
"f = 31.83;# in Hz\n",
"\n",
" #calculation:\n",
" #V3 = x*V1\n",
" #V5 = y*V1\n",
" #the rms value of the fundamental,\n",
"V1 = ((V**2)/(1 + x**2 + y**2))**0.5\n",
" #Rms value of the third harmonic\n",
"V3 = x*V1\n",
" #the rms value of the fifth harmonic,\n",
"V5 = y*V1\n",
" #Maximum value of the fundamental,\n",
"V1m = V1*2**0.5\n",
" #Maximum value of the third harmonic,\n",
"V3m = V3*2**0.5\n",
" #Maximum value of the fifth harmonic,\n",
"V5m = V5*2**0.5\n",
"w = 2*math.pi*f\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"v = \",round(V1m,2),\"sin\",round(w,2),\"t + \",round(V3m,2),\"sin\",round((3*w),2),\"t + \",round(V5m,2),\"sin\",round((5*w),2),\"t Volts\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"v = 323.62 sin 199.99 t + 97.08 sin 599.98 t + 32.36 sin 999.97 t Volts\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 6, page no. 652
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the average power in a 20 ohm resistance\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"A1 = 12;# in amperes\n",
"A3 = 5;# in amperes\n",
"A5 = 2;# in amperes\n",
"R = 20;# in ohms\n",
"\n",
"#calculation:\n",
" #rms current\n",
"Irms = ((A1**2 + A3**2 + A5**2)/2)**0.5\n",
" #average power\n",
"P = R*Irms**2\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n average power \",P,\" W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" average power 1730.0 W"
]
}
],
"prompt_number": 5
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 7, page no. 652
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the total active power supplied to the circuit, and (b) the overall power factor\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"Ia1 = 2;# in amperes\n",
"Ia3 = 0.3;# in amperes\n",
"Ia5 = 0.1;# in amperes\n",
"Va1 = 60;# in volts\n",
"Va3 = 15;# in volts\n",
"Va5 = 10;# in volts\n",
"Phii1 = -1*math.pi/6;# in radians\n",
"Phii3 = -1*math.pi/12;# in radians\n",
"Phii5 = -8*math.pi/9;# in radians\n",
"Phiv1 = 0;# in radians\n",
"Phiv3 = math.pi/4;# in radians\n",
"Phiv5 = -1*math.pi/2;# in radians\n",
"\n",
"\n",
" #calculation:\n",
" #rms values;\n",
"I1 = Ia1/(2**0.5);# in amperes\n",
"I3 = Ia3/(2**0.5);# in amperes\n",
"I5 = Ia5/(2**0.5);# in amperes\n",
"V1 = Va1/(2**0.5);# in volts\n",
"V3 = Va3/(2**0.5);# in volts\n",
"V5 = Va5/(2**0.5);# in volts\n",
" #total power supplied,\n",
"P = V1*I1*math.cos(Phiv1 - Phii1) + V3*I3*math.cos(Phiv3 - Phii3) + V5*I5*math.cos(Phiv5 - Phii5)\n",
" #rms current\n",
"Irms = ((I1**2 + I3**2 + I5**2))**0.5\n",
" #rms voltage\n",
"Vrms = ((V1**2 + V3**2 + V5**2))**0.5\n",
" #overall power factor\n",
"pf = P/(Vrms*Irms)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a)the total active power supplied to the circuit \",round(P,2),\" W\"\n",
"print \"\\n(b)overall power factor \",round(pf,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a)the total active power supplied to the circuit 53.26 W\n",
"\n",
"(b)overall power factor 0.84"
]
}
],
"prompt_number": 6
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 8, page no. 655
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine for each case an expression for the current flowing if the fundamental frequency is 1 kHz.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"R1 = 40;# in ohm\n",
"L = 7.96E-3;# in Henry\n",
"C = 25E-6; # in Farad\n",
"f = 1000; # in Hx\n",
"\n",
"#calculation:\n",
"wL = 2*math.pi*1000*L\n",
"wC = 2*math.pi*1000*C\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"(a)i = \",round(100/R1,2),\"sin(wt) +\",round(30/R1,2),\"sin(3wt - pi/3) +\",round(10/R1,2),\"sin(5wt - pi/6) A\"\n",
"print \"(b)i = \",round(100/wL,2),\"sin(wt - pi/2) +\",round(30/(3*wL),2),\"sin(3wt - pi/6) +\",round(10/(5*wL),2),\"sin(5wt - 2pi/3) A\"\n",
"print \"(c)i = \",round(100*wC,2),\"sin(wt + pi/2) +\",round(30*3*wC,2),\"sin(3wt + 5pi/6) +\",round(10*5*wC,2),\"sin(5wt + pi/3) A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"(a)i = 2.5 sin(wt) + 0.75 sin(3wt - pi/3) + 0.25 sin(5wt - pi/6) A\n",
"(b)i = 2.0 sin(wt - pi/2) + 0.2 sin(3wt - pi/6) + 0.04 sin(5wt - 2pi/3) A\n",
"(c)i = 15.71 sin(wt + pi/2) + 14.14 sin(3wt + 5pi/6) + 7.85 sin(5wt + pi/3) A\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 9, page no. 656
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) an expression to represent the instantaneous value of the current, \n",
"#(b) the rms voltage, (c) the rms current, (d) the power dissipated, and (e) the overall power factor.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"V1m = 240;# in volts\n",
"V3m = 40;# in volts\n",
"V5m = 30;# in volts\n",
"w1 = 314;# fundamental\n",
"R = 12;# in ohm\n",
"L = 0.00955;# in Henry\n",
"\n",
" #calculation:\n",
" #fundamental or first harmonic\n",
" #inductive reactance,\n",
"XL1 = w1*L\n",
" #impedance at the fundamental frequency,\n",
"Z1 = R + 1j*XL1\n",
" #Maximum current at fundamental frequency\n",
"I1m = V1m/Z1\n",
"I1mag = abs(I1m)\n",
"phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
" #Third harmonic\n",
"XL3 = 3*XL1\n",
" #impedance at the third harmonic frequency,\n",
"Z3 = R + 1j*XL3\n",
" #Maximum current at third harmonic frequency\n",
"I3m = V3m/Z3\n",
"I3mag = abs(I3m)\n",
"phii3 = cmath.phase(complex(I3m.real,I3m.imag))\n",
" #fifth harmonic\n",
"XL5 = 5*XL1\n",
" #impedance at the third harmonic frequency,\n",
"Z5 = R + 1j*XL5\n",
" #Maximum current at third harmonic frequency\n",
"I5m = V5m/Z5\n",
"I5mag = abs(I5m)\n",
"phii5 = cmath.phase(complex(I5m.real,I5m.imag))\n",
" #rms voltage\n",
"Vrms = ((V1m**2 + V3m**2 + V5m**2)/2)**0.5\n",
" #rms current\n",
"Irms = ((I1mag**2 + I3mag**2 + I5mag**2)/2)**0.5\n",
" #power dissipated\n",
"P = R*Irms**2\n",
" #overall power factor\n",
"pf = P/(Vrms*Irms)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a)i = \",round(I1mag,2),\"sin(\",round(w1,2),\"t + (\",round(phii1,2),\")) + \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t + (\",round(phii3,2),\")) + \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t + (\",round(phii5,2),\")) A\"\n",
"print \"\\n(b)the rms value of current is \",round(Irms,2),\" A\"\n",
"print \"\\n(c)the rms value of voltage is \",round(Vrms,2),\" V\"\n",
"print \"\\n(d)the total power dissipated \",round(P,2),\" W\"\n",
"print \"\\n(e)overall power factor \",round(pf,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a)i = 19.4 sin( 314.0 t + ( -0.24 )) + 2.67 sin( 942.0 t + ( -0.64 )) + 1.56 sin( 1570.0 t + ( -0.9 )) A\n",
"\n",
"(b)the rms value of current is 13.89 A\n",
"\n",
"(c)the rms value of voltage is 173.35 V\n",
"\n",
"(d)the total power dissipated 2316.26 W\n",
"\n",
"(e)overall power factor 0.96"
]
}
],
"prompt_number": 7
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 10, page no. 658
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Obtain an expression for the current flowing and hence determine the rms value of current.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"Vom = 50;# in volts\n",
"V1m = 200;# in volts\n",
"V2m = 40;# in volts\n",
"V4m = 5;# in volts\n",
"f = 50;# in Hz\n",
"R = 50;# in ohm\n",
"C = 100E-6;# in farad\n",
"phiv1 = 0;# in rad\n",
"phiv2 = -1*math.pi/2;# in rad\n",
"phiv4 = math.pi/4;# in rad\n",
"\n",
" #calculation:\n",
" #voltage\n",
"V1 = V1m*math.cos(phiv1) + 1j*V1m*math.sin(phiv1)\n",
"V2 = V2m*math.cos(phiv2) + 1j*V2m*math.sin(phiv2)\n",
"V4 = V4m*math.cos(phiv4) + 1j*V4m*math.sin(phiv4)\n",
" #Inductance has no effect on a steady current. Hence the d.c. component of the current, i0, is given by\n",
"Iom = 0\n",
" #fundamental or first harmonic\n",
"w1 = 2*math.pi*f\n",
" #inductive reactance,\n",
"Xc1 = 1/(w1*C)\n",
" #impedance at the fundamental frequency,\n",
"Z1 = R + 1j*Xc1\n",
" #Maximum current at fundamental frequency\n",
"I1m = V1/Z1\n",
"I1mag = abs(I1m)\n",
"phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
" #second harmonic\n",
"Xc2 = Xc1/2\n",
" #impedance at the third harmonic frequency,\n",
"Z2 = R + 1j*Xc2\n",
" #Maximum current at third harmonic frequency\n",
"I2m = V2/Z2\n",
"I2mag = abs(I2m)\n",
"phii2 = cmath.phase(complex(I2m.real,I2m.imag))\n",
" #fourth harmonic\n",
"Xc4 = Xc1/4\n",
" #impedance at the third harmonic frequency,\n",
"Z4 = R + 1j*Xc4\n",
" #Maximum current at third harmonic frequency\n",
"I4m = V4/Z4\n",
"I4mag = abs(I4m)\n",
"phii4 = cmath.phase(complex(I4m.real,I4m.imag))\n",
" #rms current\n",
"Irms = (Iom**2 + (I1mag**2 + I2mag**2 + I4mag**2)/2)**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t + (\",round(phii1,2),\")) + \",round(I2mag,2),\"sin(\",round((w1*2),2),\"t + (\",round(phii2,2),\")) + \",round(I4mag,2),\"sin(\",round((w1*4),2),\"t + (\",round(phii4,2),\")) A\"\n",
"print \"(b)the rms value of current is \",round(Irms,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"(a)i = 0.0 + 3.37 sin( 314.16 t + ( -0.57 )) + 0.76 sin( 628.32 t + ( -1.88 )) + 0.1 sin( 1256.64 t + ( 0.63 )) A\n",
"(b)the rms value of current is 2.45 A\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 11, page no. 659
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) an expression to represent the current flowing in the circuit, \n",
"#(b) the rms value of current, correct to two decimal places, and \n",
"#(c) the power dissipated in the circuit, correct to three significant figures.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"Vom = 25;# in volts\n",
"V1m = 100;# in volts\n",
"V3m = 40;# in volts\n",
"V5m = 20;# in volts\n",
"w1 = 10000;# fundamental\n",
"R = 5;# in ohm\n",
"L = 500E-6;# in Henry\n",
"phiv1 = 0;# in rad\n",
"phiv3 = math.pi/6;# in rad\n",
"phiv5 = math.pi/12;# in rad\n",
"\n",
" #calculation:\n",
" #voltage\n",
"V1 = V1m*math.cos(phiv1) + 1j*V1m*math.sin(phiv1)\n",
"V3 = V3m*math.cos(phiv3) + 1j*V3m*math.sin(phiv3)\n",
"V5 = V5m*math.cos(phiv5) + 1j*V5m*math.sin(phiv5)\n",
" #Inductance has no effect on a steady current. Hence the d.c. component of the current, i0, is given by\n",
"Iom = Vom/R\n",
" #fundamental or first harmonic\n",
" #inductive reactance,\n",
"XL1 = w1*L\n",
" #impedance at the fundamental frequency,\n",
"Z1 = R + 1j*XL1\n",
" #Maximum current at fundamental frequency\n",
"I1m = V1/Z1\n",
"I1mag = abs(I1m)\n",
"phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
"#Third harmonic\n",
"XL3 = 3*XL1\n",
" #impedance at the third harmonic frequency,\n",
"Z3 = R + 1j*XL3\n",
" #Maximum current at third harmonic frequency\n",
"I3m = V3/Z3\n",
"I3mag = abs(I3m)\n",
"phii3 = cmath.phase(complex(I3m.real,I3m.imag))\n",
" #fifth harmonic\n",
"XL5 = 5*XL1\n",
" #impedance at the third harmonic frequency,\n",
"Z5 = R + 1j*XL5\n",
" #Maximum current at third harmonic frequency\n",
"I5m = V5/Z5\n",
"I5mag = abs(I5m)\n",
"phii5 = cmath.phase(complex(I5m.real,I5m.imag))\n",
" #rms current\n",
"Irms = (Iom**2 + (I1mag**2 + I3mag**2 + I5mag**2)/2)**0.5\n",
" #power dissipated\n",
"P = R*Irms**2\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t + (\",round(phii1,2),\")) + \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t + (\",round(phii3,2),\")) + \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t + (\",round(phii5,2),\")) A\"\n",
"print \"\\n(b)the rms value of current is \",round(Irms,2),\" A\"\n",
"print \"\\n(c)the total power dissipated \",round(P,3),\" W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a)i = 5.0 + 14.14 sin( 10000.0 t + ( -0.79 )) + 2.53 sin( 30000.0 t + ( -0.73 )) + 0.78 sin( 50000.0 t + ( -1.11 )) A\n",
"\n",
"(b)the rms value of current is 11.34 A\n",
"\n",
"(c)the total power dissipated 642.538 W\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 12, page no. 661
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the average power supplied, (b) the type of components present, and (c) the values of the components.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"Vom = 30;# in volts\n",
"V1m = 40;# in volts\n",
"V2m = 25;# in volts\n",
"V4m = 15;# in volts\n",
"Iom = 0;# in amperes\n",
"I1m = 0.743;# in Amperes\n",
"I2m = 0.781;# in Amperes\n",
"I4m = 0.636;# in Amperes\n",
"phii1 = 1.190;# in rad\n",
"phii2 = 0.896;# in rad\n",
"phii4 = 0.559;# in rad\n",
"w = 1000;# in rad\n",
"\n",
" #calculation:\n",
" #the average power P is given by\n",
"P = Vom*Iom+(0.707*V1m)*(0.707*I1m)*math.cos(phii1)+(0.707*V2m)*(0.707*I2m)*math.cos(phii2) + (0.707*V4m)*(0.707*I4m)*math.cos(phii4)\n",
" #rms current\n",
"Irms = (Iom**2 + (I1m**2 + I2m**2 + I4m**2)/2)**0.5\n",
" #resistance R\n",
"R = P/(Irms**2)\n",
" #impedance\n",
"Z1 = V1m/I1m\n",
" #Xc1\n",
"Xc1 = (Z1**2 - R**2)**0.5\n",
" #capacitance\n",
"C = 1/(w*Xc1)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a)the average power P is \",round(P,2),\" W\"\n",
"print \"\\n(c)the resistance R \",round(R,2),\" ohm and capacitance \",round(C*1E6,2),\"uF\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a)the average power P is 15.66 W\n",
"\n",
"(c)the resistance R 19.99 ohm and capacitance 20.01 uF\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 13, page no. 662
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) an expression for the supply current, i, (b) the percentage harmonic content of the supply current, (c) the total power dissipated, (d) an expression for the p.d. shown as \t1, and (e) an expression for current ic.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"V1m = 300;# in volts\n",
"V3m = 120;# in volts\n",
"phiv1 = 0;# in rad\n",
"phiv2 = 0.698;# in rad\n",
"w1 = 314;# in rad\n",
"C = 2.123E-6;# in farads\n",
"R1 = 560;# in ohms\n",
"R2 = 2000;# in Ohm\n",
"\n",
"#calculation:\n",
" #voltage\n",
"V1 = V1m*math.cos(phiv1) + 1j*V1m*math.sin(phiv1)\n",
"V3 = V3m*math.cos(phiv2) + 1j*V3m*math.sin(phiv2)\n",
" #capacitive reactance,\n",
"Xc1 = 1/(w1*C)\n",
" #impedance at the fundamental frequency,\n",
"Z1 = R1 - 1j*Xc1*R2/(R2 - 1j*Xc1)\n",
" #Maximum current at fundamental frequency\n",
"I1m = V1/Z1\n",
"I1mag = abs(I1m)\n",
"phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
" #Third harmonic\n",
"Xc3 = Xc1/3\n",
" #impedance at the third harmonic frequency,\n",
"Z3 = R1 - 1j*Xc3*R2/(R2 - 1j*Xc3)\n",
"I1m = V1m/Z1\n",
"I1mag = abs(I1m)\n",
"phii1 = cmath.phase(complex(I1m.real,I1m.imag))\n",
" #Maximum current at third harmonic frequency\n",
"I3m = V3/Z3\n",
"I3mag = abs(I3m)\n",
"phii3 = cmath.phase(complex(I3m.real,I3m.imag))\n",
" #Percentage harmonic content of the supply current is given by\n",
"percent = I3mag*100/I1mag\n",
" #total active power\n",
"P = (0.707*V1m)*(0.707*I1mag)*math.cos(phiv1 - phii1) + (0.707*V3m)*(0.707*I3m)*math.cos(phiv2 - phii3)\n",
"\n",
"I1 = I1m*R2/(R2 - 1j*Xc1)\n",
"I3 = I3m*R2/(R2 - 1j*Xc3)\n",
"\n",
"I1nmag = abs(I1)\n",
"phini1 = cmath.phase(complex(I1.real,I1.imag))\n",
"I3nmag = abs(I3)\n",
"phini3 = cmath.phase(complex(I3.real,I3.imag))\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a)supply current, i=\", round(I1mag,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
"print \"\\n(b)Percentage harmonic content of the supply current is \",round(percent,2),\" percent\"\n",
"print \"\\n(c)total active power is \",round(abs(P),2),\" W\"\n",
"print \"\\n(d)Voltage, v1 =\", round(I1mag*R1,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag*R1,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
"print \"\\n(e)current, ic =\", round(I1nmag,3),\"sin(\", w1,\"t +\",round(phini1,3),\") + \",round(I3nmag,3),\"sin(\", 3*w1,\"t +\",round(phini3,3),\") A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a)supply current, i= 0.187 sin( 314 t + 0.643 ) + 0.145 sin( 942 t + 1.305 ) A\n",
"\n",
"(b)Percentage harmonic content of the supply current is 77.57 percent\n",
"\n",
"(c)total active power is 25.34 W\n",
"\n",
"(d)Voltage, v1 = 104.996 sin( 314 t + 0.643 ) + 81.45 sin( 942 t + 1.305 ) A\n",
"\n",
"(e)current, ic = 0.15 sin( 314 t + 1.287 ) + 0.141 sin( 942 t + 1.55 ) A"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14, page no. 664
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the fundamental frequency for resonance with the third harmonic, and \n",
"#(b) the maximum value of the fundamental and third harmonic components of current\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"V1m = 400;# in volts\n",
"V3m = 10;# in volts\n",
"C = 0.2E-6;# in farads\n",
"R = 2;# in ohms\n",
"L = 0.5;# in Henry\n",
"\n",
" #calculation:\n",
" #Resonance with the third harmonic means that\n",
"w = (1/(9*L*C))**0.5\n",
" #fundamental frequency, f\n",
"f = w/(2*math.pi)\n",
" #At the fundamental frequency,\n",
" #impedance Z1\n",
"Z1 = R + 1j*(w*L - 1/(w*C))\n",
"Z1mag = abs(Z1)\n",
"phiZ1 = cmath.phase(complex(Z1.real,Z1.imag))\n",
" #Maximum value of current at the fundamental frequency,\n",
"I1m = V1m/Z1mag\n",
" #At the third harmonic frequency,\n",
"Z3 = R + 1j*(3*w*L - 1/(3*w*C))\n",
"Z3mag = abs(Z3)\n",
"phiZ3 = cmath.phase(complex(Z3.real,Z3.imag))\n",
" #Maximum value of current at the third harmonic frequency,\n",
"I3m = V3m/Z3\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"(a)fundamental frequency for resonance with the third harmonic is \",round(f,2),\" Hz\"\n",
"print \"(b)Maximum value of current at fundamental freq. is\",round(abs(I1m),3),\"A \"\n",
"print \"and at the third harmonic frequency \", abs(I3m),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"(a)fundamental frequency for resonance with the third harmonic is 167.76 Hz\n",
"(b)Maximum value of current at fundamental freq. is 0.095 A \n",
"and at the third harmonic frequency 5.0 A\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 15, page no. 665
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the value of n, (b) the maximum value of current at the nth harmonic, \n",
"#(c) the p.d. across the capacitor at the nth harmonic and\n",
"#(d) the maximum value of the fundamental current.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"V1m = 800;# in volts\n",
"f = 50;# in Hz\n",
"x = 0.015;\n",
"C = 0.122E-6;# in farads\n",
"R = 5;# in ohms\n",
"L = 0.369;# in Henry\n",
"\n",
" #calculation:\n",
" #voltage at nth harmonic\n",
"Vnm = x*V1m\n",
"w = 2*math.pi*f\n",
" #For resonance at the nth harmonic nwL = 1/nwC\n",
"n = 1/(w*(L*C)**0.5)\n",
" #At resonance, impedance\n",
"Zn = R\n",
" #the maximum value of current at the nth harmonic\n",
"Inm = Vnm/Zn\n",
" #capacitive reactance, at nth harmonic\n",
"Xcn = 1/(n*w*C)\n",
" #the p.d. across the capacitor at the nth harmonic\n",
"Vcn = Inm*Xcn\n",
" #At the fundamental frequency, inductive reactance,\n",
"XL1 = w*L\n",
" #capacitive reactance\n",
"Xc1 = 1/(w*C)\n",
" #Impedance at the fundamental frequency,\n",
"Z1 = R + 1j*(XL1 - Xc1)\n",
"Z1mag = abs(Z1)\n",
"phiZ1 = cmath.phase(complex(Z1.real,Z1.imag))\n",
" #Maximum value of current at the fundamental frequency,\n",
"I1m = V1m/Z1mag\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a)n = \",round(n,2),\"\"\n",
"print \"\\n(b)the maximum value of current at the nth harmonic \",round(Inm,2),\" A\"\n",
"print \"\\n(c)the p.d. across the capacitor at the nth harmonic is \",round(Vcn,2),\"\"\n",
"print \"\\n(d)the maximum value of the fundamental current. \",round(I1m,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a)n = 15.0 \n",
"\n",
"(b)the maximum value of current at the nth harmonic 2.4 A\n",
"\n",
"(c)the p.d. across the capacitor at the nth harmonic is 4173.92 \n",
"\n",
"(d)the maximum value of the fundamental current. 0.03 A"
]
}
],
"prompt_number": 14
}
],
"metadata": {}
}
]
}