{
"metadata": {
"name": "",
"signature": "sha256:9b0edaefe058f9c7f8efe48336cbbb08a13cafea026561f6301c02956a595ae7"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h1>Chapter 25: Application of complex numbers to parallel a.c. circuits</h1>"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 1, page no. 446</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"Z1 = 0 - 5j;# in ohms\n",
"Z2 = 25 + 40j;# in ohms\n",
"Z3 = 3 - 2j;# in ohms\n",
"r4 = 50;# in ohms\n",
"theta4 = 40;# in degrees\n",
"\n",
"#calculation:\n",
" #admittance Y\n",
"Y1 = 1/Z1\n",
" #conductance, G\n",
"G1 = Y1.real\n",
" #Suspectance, Bc\n",
"Bc1 = abs(Y1.imag)\n",
" #admittance Y\n",
"Y2 = 1/Z2\n",
" #conductance, G\n",
"G2 = Y2.real\n",
" #Suspectance, Bc\n",
"Bc2 = abs(Y2.imag)\n",
" #admittance Y\n",
"Y3 = 1/Z3\n",
" #conductance, G\n",
"G3 = Y3.real\n",
" #Suspectance, Bc\n",
"Bc3 = abs(Y3.imag)\n",
"Z4 = r4*math.cos(theta4*math.pi/180) + 1j*r4*math.sin(theta4*math.pi/180)\n",
" #admittance Y\n",
"Y4 = 1/Z4\n",
" #conductance, G\n",
"G4 = Y4.real\n",
" #Suspectance, Bc\n",
"Bc4 = abs(Y4.imag)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)admittance Y is (\",round(Y1.real,2),\" + (\",round(Y1.imag,2),\")i) S, \"\n",
"print \" conductance, G is \",round(G1,2),\" S, susceptance,Bc is \",round(Bc1,2),\" S\\n\"\n",
"print \"\\n (b)admittance Y is (\",round(Y2.real,2),\" + (\",round(Y2.imag,2),\")i) S, \"\n",
"print \" conductance, G is \",round(G2,2),\" S, susceptance,Bc is \",round(Bc2,2),\" S\\n\"\n",
"print \"\\n (c)admittance Y is (\",round(Y3.real,2),\" + (\",round(Y3.imag,2),\")i) S, \"\n",
"print \" conductance, G is \",round(G3,2),\" S, susceptance,Bc is \",round(Bc3,2),\" S\\n\"\n",
"print \"\\n (d)admittance Y is (\",round(Y4.real,2),\" + (\",round(Y4.imag,2),\")i) S, \"\n",
"print \" conductance, G is \",round(G4,2),\" S, susceptance,Bc is \",round(Bc4,2),\" S\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)admittance Y is ( -0.0 + ( 0.2 )i) S, \n",
" conductance, G is -0.0 S, susceptance,Bc is 0.2 S\n",
"\n",
"\n",
" (b)admittance Y is ( 0.01 + ( -0.02 )i) S, \n",
" conductance, G is 0.01 S, susceptance,Bc is 0.02 S\n",
"\n",
"\n",
" (c)admittance Y is ( 0.23 + ( 0.15 )i) S, \n",
" conductance, G is 0.23 S, susceptance,Bc is 0.15 S\n",
"\n",
"\n",
" (d)admittance Y is ( 0.02 + ( -0.01 )i) S, \n",
" conductance, G is 0.02 S, susceptance,Bc is 0.01 S\n",
"\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 2, page no. 447</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"Y2 = 0.001 - 0.002j;# in S\n",
"Y3 = 0.05 + 0.08j;# in S\n",
"r1 = 0.004;# in S\n",
"theta1 = 30;# in degrees\n",
"\n",
" #calculation:\n",
" #impedance, Z\n",
"Z2 = 1/Y2\n",
"Z3 = 1/Y3\n",
"Y1 = r1*math.cos(theta1*math.pi/180) + 1j*r1*math.sin(theta1*math.pi/180)\n",
"Z1 = 1/Y1\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)Impedance,Z is (\",round(Z1.real,2),\" + (\",round( Z1.imag,2),\")i) ohm\\n\"\n",
"print \"\\n (b)Impedance,Z is (\",round(Z2.real,2),\" + (\",round( Z2.imag,2),\")i) ohm\\n\"\n",
"print \"\\n (c)Impedance,Z is (\",round(Z3.real,2),\" + (\",round( Z3.imag,2),\")i) ohm\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)Impedance,Z is ( 216.51 + ( -125.0 )i) ohm\n",
"\n",
"\n",
" (b)Impedance,Z is ( 200.0 + ( 400.0 )i) ohm\n",
"\n",
"\n",
" (c)Impedance,Z is ( 5.62 + ( -8.99 )i) ohm"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 3, page no. 448</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"Y = 0.040 - 1j*0.025;# in S\n",
"\n",
"#calculation:\n",
" #impedance, Z\n",
"Z = 1/Y\n",
" #conductance, G\n",
"G = Y.real\n",
" #Suspectance, Bc\n",
"Bc = abs(Y.imag)\n",
" #parallrl \n",
" #resistance, R\n",
"Rp = 1/G\n",
" #capacitive reactance\n",
"Xcp = 1/Bc\n",
" #series\n",
" #resistance, R\n",
"Rs = Z.real\n",
" #capacitive reactance\n",
"Xcs = abs(Z.imag)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)for parallel, resistance,R is \",round(Rp,2),\" ohm and capacitive reactance, Xc is \",round(Xcp,2),\" ohm\\n\"\n",
"print \"\\n (b)forseries, resistance,R is \",round(Rs,2),\" ohm and capacitive reactance, Xc is \",round(Xcs,2),\" ohm\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)for parallel, resistance,R is 25.0 ohm and capacitive reactance, Xc is 40.0 ohm\n",
"\n",
"\n",
" (b)forseries, resistance,R is 17.98 ohm and capacitive reactance, Xc is 11.24 ohm"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 4, page no. 449</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"R1 = 8;# in ohm\n",
"R = 5;# in ohm\n",
"R2 = 6;# ohm\n",
"rv = 50;# in volts\n",
"thetav = 0;# in degrees\n",
"\n",
"#calculation:\n",
" #voltage,V\n",
"V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
" #circuit impedance, ZT\n",
"ZT = R + (R1*1j*R2/(R1 + 1j*R2))\n",
" #Current I\n",
"I = V/ZT\n",
" #current,I1\n",
"I1 = I*(1j*R2/(R1 + 1j*R2))\n",
" #current, I2\n",
"I2 = I*(R1/(R1 + 1j*R2))\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n current, I = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg A,\"\n",
"print \"current,I1 = \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real, I1.imag))*180/math.pi,2),\"deg A, \"\n",
"print \"current, I2 = \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real, I2.imag))*180/math.pi,2),\"deg A\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" current, I = 5.7 /_ -25.98 deg A,\n",
"current,I1 = 3.42 /_ 27.15 deg A, \n",
"current, I2 = 4.56 /_ -62.85 deg A\n",
"\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 5, page no. 450</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"R1 = 5;# in ohm\n",
"R2 = 3;# in ohm \n",
"R3 = 8;# ohm\n",
"Xc = 4;# in ohms\n",
"XL = 12;# in Ohms\n",
"V = 40;# in volts\n",
"f = 50;# in Hz\n",
"\n",
"#calculation:\n",
"Z1 = R1 + 1j*XL\n",
"Z2 = R2 - 1j*Xc\n",
"Z3 = R3\n",
" #circuit admittance, YT = 1/ZT\n",
"YT = (1/Z1) + (1/Z2) + (1/Z3)\n",
" #Current I\n",
"I = V*YT\n",
"I1 = V/Z1\n",
"I2 = V/Z2\n",
"I3 = V/Z2\n",
"thetav = 0\n",
"thetai = cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
"phi = thetav - thetai \n",
"if (phi>0):\n",
" a = \"lagging\"\n",
"else:\n",
" a = \"leading\"\n",
"\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n current, I is (\",round(I.real,2),\" + (\",round(I.imag,2),\")i) A,\"\n",
"print \"and its phase relative to the 40 V supply is \",a,\"s by \",round(abs(phi),2),\"deg\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" current, I is ( 10.98 + ( 3.56 )i) A,\n",
"and its phase relative to the 40 V supply is leading s by 17.96 deg\n",
"\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 6, page no. 451</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"L = 0.07958;# in Henry\n",
"R = 18;# in ohm\n",
"C = 64.96E-6;# in Farad\n",
"rv = 250;# in volts\n",
"thetav = 0;# in degrees\n",
"f = 50;# in Hz\n",
"\n",
"#calculation:\n",
" #Inductive reactance\n",
"XL = 2*math.pi*f*L\n",
" #capacitive reactance\n",
"Xc = 1/(2*math.pi*f*C)\n",
" #impedance of the coil,\n",
"Zcoil = R + 1j*XL\n",
" #impedance presented by the capacitor,\n",
"Zc = -1j*Xc\n",
" #Total equivalent circuit impedance,\n",
"ZT = Zcoil*Zc/(Zcoil + Zc)\n",
" #voltage\n",
"V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
" #current, I\n",
"I = V/ZT\n",
"thetai = cmath.phase(complex(I.real,I.imag))*180/math.pi\n",
"phi = thetav - thetai\n",
"if (phi>0):\n",
" a = \"lagging\"\n",
"else:\n",
" a = \"leading\"\n",
"\n",
" #Current in the coil, ICOIL\n",
"Icoil = V/Zcoil\n",
" #Current in the capacitor, IC\n",
"Ic = V/Zc\n",
"\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)the circuit impedance is \",round(ZT.real,2),\" + (\",round( ZT.imag,2),\")i ohm\\n\"\n",
"print \"\\n (b)supply current, I = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg A\\n\"\n",
"print \"\\n (c)circuit phase relative is \",a,\"s by \",round(abs(phi),2),\"deg\\n\"\n",
"print \"\\n (d)current in coil, Icoil = \",round(abs(Icoil),2),\"/_\",round(cmath.phase(complex(Icoil.real, Icoil.imag))*180/math.pi,2),\"deg A\\n\"\n",
"print \"\\n (e)current in capacitor, Ic = \",round(abs(Ic),2),\"/_\",round(cmath.phase(complex(Ic.real, Ic.imag))*180/math.pi,2),\"deg A\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)the circuit impedance is 48.02 + ( 15.03 )i ohm\n",
"\n",
"\n",
" (b)supply current, I = 4.97 /_ -17.38 deg A\n",
"\n",
"\n",
" (c)circuit phase relative is lagging s by 17.38 deg\n",
"\n",
"\n",
" (d)current in coil, Icoil = 8.12 /_ -54.25 deg A\n",
"\n",
"\n",
" (e)current in capacitor, Ic = 5.1 /_ 90.0 deg A"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 7, page no. 452</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"RL = 6j;# in ohm\n",
"R2 = 8;# in ohm\n",
"Z3 = 10;# in ohm\n",
"rv = 50;# in volts\n",
"thetav = 30;# in degrees\n",
"ri = 31.4;# in amperes\n",
"thetai = 52.48;# in degrees\n",
"f = 5000;# in Hz\n",
"\n",
"#calculation:\n",
" #impedance, Z2\n",
"Z2 = R2 + RL\n",
" #voltage\n",
"V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
" #current, I\n",
"I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
" #Total circuit admittance,\n",
"YT = I/V\n",
" #admittance, Y3\n",
"Y3 = 1/Z3\n",
" #admittance, Y2\n",
"Y2 = 1/Z2\n",
" #admittance, Y1\n",
"Y1 = YT - Y2 - Y3\n",
" #impedance, Z1\n",
"Z1 = 1/Y1\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)the impedance Z1 is \",round(Z1.real,2),\" + (\",round( Z1.imag,2),\")i ohm\\n\"\n",
"\n",
" #resistance, R1\n",
"R1 = Z1.real\n",
"X1 = Z1.imag \n",
"if ((R1>0)&(X1<0)):\n",
" C1 = -1/(2*math.pi*f*X1)\n",
" print \"\\n (b)The series circuit thus consists of a resistor of resistance \",round(R1,2),\" ohm\"\n",
" print \" and a capacitor of capacitance \",round(C1*1E6,2),\"uFarad\\n\"\n",
"elif ((R1>0)&(X1>0)):\n",
" L1 = 2*math.pi*f*X1\n",
" print \"\\n (b)The series circuit thus consists of a resistor of resistance \",round(R1,2),\" ohm \"\n",
" print \" and a inductor of insuctance \",round(L1*1000,2),\"mHenry\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)the impedance Z1 is 1.6 + ( -1.2 )i ohm\n",
"\n",
"\n",
" (b)The series circuit thus consists of a resistor of resistance 1.6 ohm\n",
" and a capacitor of capacitance 26.55 uFarad\n",
"\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 8, page no. 453</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"RL1 = 1.02j;# in ohm\n",
"R1 = 1.65;# in ohm\n",
"RLa = 7j;# in ohm\n",
"Ra = 5;# in ohm\n",
"Rcb = -1j*15;# in ohm\n",
"Rb = 4;# in ohm\n",
"rv = 91;# in volts\n",
"thetav = 0;# 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, Z1\n",
"Z1 = R1 + RL1\n",
" #impedance, Za\n",
"Za = Ra + RLa\n",
" #impedance, Zb\n",
"Zb = Rb + Rcb\n",
" #impedance, Z, of the two branches connected in parallel\n",
"Z = Za*Zb/(Za + Zb)\n",
" #Total circuit impedance\n",
"ZT = Z1 + Z\n",
" #Supply current, I\n",
"I = V/ZT\n",
"thetai = cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
"phi = thetav - thetai \n",
"if (phi>0):\n",
" a = \"lagging\"\n",
"else:\n",
" a = \"leading\"\n",
"\n",
" #Voltage V1\n",
"V1 = I*Z1\n",
" #Voltage V2\n",
"V2 = I*Z\n",
" #current Ia\n",
"Ia = V2/Za\n",
" #Current Ib\n",
"Ib = V2/Zb\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)equivalent series circuit impedance is \",round(ZT.real,2),\" + (\",round( ZT.imag,2),\")i ohm\\n\"\n",
"print \"\\n (b)supply current, I is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\\n\"\n",
"print \"\\n (c)circuit phase relative is \",a,\" by \",round(abs(phi),2),\"deg\\n\"\n",
"print \"\\n (d)voltage, V1 is (\",round(V1.real,2),\" + (\",round(V1.imag,2),\")i) V and V2 is(\",round(V2.real,2),\" + (\",round( V2.imag,2),\")i) V\\n\"\n",
"print \"\\n (e)current, Ia is (\",round(Ia.real,2),\" + (\",round( Ia.imag,2),\")i) A and Ib is(\",round(Ib.real,2),\" + (\",round( Ib.imag,2),\")i) A\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)equivalent series circuit impedance is 12.0 + ( 5.0 )i ohm\n",
"\n",
"\n",
" (b)supply current, I is 6.46 + ( -2.69 )i A\n",
"\n",
"\n",
" (c)circuit phase relative is lagging by 22.61 deg\n",
"\n",
"\n",
" (d)voltage, V1 is ( 13.41 + ( 2.15 )i) V and V2 is( 77.59 + ( -2.15 )i) V\n",
"\n",
"\n",
" (e)current, Ia is ( 5.04 + ( -7.49 )i) A and Ib is( 1.42 + ( 4.79 )i) A"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}