{
"metadata": {
"name": "",
"signature": "sha256:2c3da80fc1d074df861310f0cc84dab7f94d8474cb6362025e07b853bd81e258"
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"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter 31: Mesh-current and nodal analysis
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 1, page no. 546
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import numpy\n",
"import cmath\n",
"#initializing the variables:\n",
"V1 = 4;# in volts\n",
"V2 = 5;# in volts\n",
"R1 = 3;# in ohm\n",
"R2 = 5;# in ohm\n",
"R3 = 4;# in ohm\n",
"R4 = 1;# in ohm\n",
"R5 = 6;# in ohm\n",
"R6 = 8;# in ohm\n",
"\n",
"#calculation:\n",
" #The mesh currents I1, I2 and I3 are shown in Figure 31.2. Using Kirchhoff\u2019s voltage law in 3 loops\n",
" #three eqns obtained\n",
" #(R1 + R2)*I1 - R2*I2 = V1\n",
" #-1*R2*I1 + (R2 + R3 + R4 + R5)*I2 - R4*I3 = 0\n",
" # -1*R4*I2 + (R4 + R6)*I3 = -1*V2\n",
" #using determinants\n",
"d1 = [[V1, -1*R2, 0],[0, (R2 + R3 + R4 + R5), -1*R4],[-1*V2, -1*R4, (R4 + R6)]]\n",
"D1 = numpy.linalg.det(d1)\n",
"d2 = [[(R1 + R2), V1, 0],[-1*R2, 0, -1*R4],[0, -1*V2, (R4 + R6)]]\n",
"D2 = numpy.linalg.det(d2)\n",
"d3 = [[(R1 + R2), -1*R2, V1],[-1*R2, (R2 + R3 + R4 + R5), 0],[0, -1*R4, -1*V2]]\n",
"D3 = numpy.linalg.det(d3)\n",
"d = [[(R1 + R2), -1*R2, 0],[-1*R2, (R2 + R3 + R4 + R5), -1*R4],[0, -1*R4, (R4 + R6)]]\n",
"D = numpy.linalg.det(d)\n",
"I1 = D1/D\n",
"I2 = D2/D\n",
"I3 = D3/D \n",
"IR2 = I1 - I2\n",
"IR4 = I2 - I3\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)current in the 5 ohm resistance is \",round(IR2,2),\" A\"\n",
"print \"\\n (b)current in the 1 ohm resistance is \",round(IR4,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)current in the 5 ohm resistance is 0.44 A\n",
"\n",
" (b)current in the 1 ohm resistance is 0.69 A"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2, page no. 547
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import numpy\n",
"import cmath\n",
"#initializing the variables:\n",
"rv = 100;# in volts\n",
"thetav = 0;# in degrees\n",
"R1 = 5;# in ohm\n",
"R2 = -1j*4;# in ohm\n",
"R3 = 4;# in ohm\n",
"R4 = 3j;# in ohm\n",
"\n",
" #calculation:\n",
" #voltages\n",
"V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
" #Currents I1, I2 with their directions are shown in Figure 31.03.\n",
" #Two loops are chosen. The choice of loop directions is arbitrary.\n",
" #using kirchoff rule in 2 loops\n",
" #two eqns obtained\n",
" #(R1 + R2)*I1 - R2*I2 = V\n",
" #-1*R2*I1 + (R3 + R2 + R4)*I2 = 0\n",
" #using determinants\n",
"d1 = [[V, -1*R2],[0, (R3 + R2 + R4)]]\n",
"D1 = numpy.linalg.det(d1)\n",
"d2 = [[(R1 + R2), V],[-1*R2, 0]]\n",
"D2 = numpy.linalg.det(d2)\n",
"d = [[(R1 + R2), -1*R2],[-1*R2, (R3 + R2 + R4)]]\n",
"D = numpy.linalg.det(d)\n",
"I1 = D1/D\n",
"I2 = D2/D\n",
"I1mag = abs(I1)\n",
" #Current flowing in capacitor\n",
"Ic = I1 - I2\n",
" #Source power P\n",
"phi = cmath.phase(complex(I1.real,I1.imag))\n",
"P = V*I1mag*math.cos(phi)\n",
"Icmag = abs(Ic)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"(a)current,I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A, current, I2 is\",round(I2.real,2),\" + (\",round(I2.imag,2),\")i A\"\n",
"print \"(b)current in the capacitor is \",round(Icmag,2),\" A\"\n",
"print \"(c)Source power P is \",round(abs(P),2),\" W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"(a)current,I1 is 10.17 + ( 3.55 )i A, current, I2 is 5.73 + ( -8.74 )i A\n",
"(b)current in the capacitor is 13.06 A\n",
"(c)Source power P is 1017.06 W\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3, page no. 548
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import numpy\n",
"import cmath\n",
"#initializing the variables:\n",
"rv1 = 415;# in volts\n",
"rv2 = 415;# in volts\n",
"thetav1 = 120;# in degrees\n",
"thetav2 = 0;# in degrees\n",
"R = 3 + 4j;# in ohm\n",
"\n",
" #calculation:\n",
" #voltages\n",
"V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
"V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
" #Two mesh currents I1 and I2 are chosen as shown in Figure 31.4.\n",
" #Two loops are chosen. The choice of loop directions is arbitrary.\n",
" #using kirchoff rule in 2 loops\n",
" #two eqns obtained\n",
" #2*R*I1 - R*I2 = V1\n",
" #-1*R*I1 + 2*R*I2 = V2\n",
" #using determinants\n",
"d1 = [[V1, -1*R],[V2, 2*R]]\n",
"D1 = numpy.linalg.det(d1)\n",
"d2 = [[2*R, V1],[-1*R, V2]]\n",
"D2 = numpy.linalg.det(d2)\n",
"d = [[2*R, -1*R],[-1*R, 2*R]]\n",
"D = numpy.linalg.det(d)\n",
"I1 = D1/D\n",
"I2 = D2/D\n",
"I1mag = abs(I1)\n",
" #line current IR\n",
"IR = I1\n",
" #line current IB\n",
"IB = -1*I2\n",
" #line current IY\n",
"IY = I2 - I1\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"current, IR is\",round(IR.real,2),\" + (\",round( IR.imag,2),\")i A, current, IB is\",round(IB.real,2),\" + (\",round( IB.imag,2),\")i A and current, IY is \",round(IY.real,2),\" + (\",round(IY.imag,2),\")i A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"current, IR is 38.34 + ( 28.75 )i A, current, IB is -44.07 + ( 18.82 )i A and current, IY is 5.73 + ( -47.58 )i A\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 4, page no. 551
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"ri = 20;# in amperes\n",
"thetai = 0;# in degrees\n",
"R1 = 10;# in ohm\n",
"R2 = 3j;# in ohm\n",
"R3 = 4;# in ohm\n",
"R4 = 16;# in ohm\n",
"\n",
"#calculation:\n",
" #current\n",
"I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
" #Figure 31.8 contains two principal nodes (at 1 and B) and thus only one nodal equation is required. \n",
" #B is taken as the reference node and the equation for node 1 is obtained as follows. \n",
" #Applying Kirchhoff\u2019s current law to node 1 gives:\n",
" #IX + IY = I\n",
"V1 = I/((1/R4) +(1/(R2 +R3)))\n",
"IY = V1/(R2 + R3)\n",
"VAB = IY*R3\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n voltage VAB is \",round(VAB.real,2),\" + (\",round( VAB.imag,2),\")i V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" voltage VAB is 62.59 + ( -9.39 )i V"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5, page no. 552
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"rv1 = 8;# in volts\n",
"rv2 = 8;# in volts\n",
"thetav1 = 0;# in degrees\n",
"thetav2 = 90;# in degrees\n",
"R1 = 5;# in ohm\n",
"R2 = 6j;# in ohm\n",
"R3 = 4;# in ohm\n",
"R4 = 3;# in ohm\n",
"\n",
"#calculation:\n",
" #voltages\n",
"V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
"V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
" #The circuit contains no principal nodes. \n",
" #However, if point Y is chosen as the reference node then an equation \n",
" #may be written for node X assuming that current leaves point X by both branches\n",
"VX = ((V1/(R1 + R3) + V2/(R2 + R4))/(1/(R1 + R3) + 1/(R2 + R4)))\n",
"VXY = VX\n",
"\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n voltage VXY = \",round(abs(VXY),2),\"/_\",round(cmath.phase(complex(VXY.real, VXY.imag))*180/math.pi,2),\"deg V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" voltage VXY = 9.12 /_ 52.13 deg V"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 6, page no. 553
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"rv1 = 100;# in volts\n",
"rv2 = 50;# in volts\n",
"thetav1 = 0;# in degrees\n",
"thetav2 = 90;# in degrees\n",
"R1 = 25;# in ohm\n",
"R2 = 20;# in ohm\n",
"R3 = 10;# in ohm\n",
"\n",
"#calculation:\n",
" #voltages\n",
"V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
"V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
" #There are only two principal nodes in Figure 31.10 so only one nodal equation is required. \n",
" #Node 2 is taken as the reference node.\n",
" #The equation at node 1 is I1 + I2 + I3 = 0\n",
"Vn1 = ((V1/R1 + V2/R3)/(1/R1 + 1/R2 + 1/R3))\n",
"I1 = (Vn1 - V1)/R1\n",
"I2 = Vn1/R2\n",
"I3 = (Vn1 - V2)/R3\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n current, I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A, \\n current, I2 is \",round(I2.real,2),\" + (\",round( I2.imag,2),\")i A \\n and current, I3 is \",round(I3.real,2),\" + (\",round(I3.imag,2),\")i A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" current, I1 is -3.16 + ( 1.05 )i A, \n",
" current, I2 is 1.05 + ( 1.32 )i A \n",
" and current, I3 is 2.11 + ( -2.37 )i A"
]
}
],
"prompt_number": 6
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 7, page no. 554
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import numpy\n",
"import cmath\n",
"#initializing the variables:\n",
"rv1 = 25;# in volts\n",
"rv2 = 25;# in volts\n",
"thetav1 = 0;# in degrees\n",
"thetav2 = 90;# in degrees\n",
"R1 = 2;# in ohm\n",
"R2 = -1j*4;# in ohm\n",
"R3 = 5;# in ohm\n",
"R4 = 4j;# in ohm\n",
"R5 = 2.5;# in ohm\n",
"\n",
" #calculation:\n",
" #voltages\n",
"V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
"V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
" #The equation at node 1\n",
" #Vn1*(1/R1 + 1/R2 + 1/R3) - Vn2/R3 = V1/R1\n",
" #The equation at node 2\n",
" #Vn1*(-1/R3) + Vn2*(1/R4 + 1/R5 + 1/R3) = V2/R5\n",
" #using determinants\n",
"d1 = [[V1/R1, -1/R3],[V2/R5, (1/R4 + 1/R5 + 1/R3)]]\n",
"D1 = numpy.linalg.det(d1)\n",
"d2 = [[(1/R1 + 1/R2 + 1/R3), V1/R1],[-1/R3, V2/R5]]\n",
"D2 = numpy.linalg.det(d2)\n",
"d = [[(1/R1 + 1/R2 + 1/R3), -1/R3],[-1/R3, (1/R4 + 1/R5 + 1/R3)]]\n",
"D = numpy.linalg.det(d)\n",
"Vn1 = D1/D\n",
"Vn2 = D2/D\n",
" #current in the j4 ohm inductance is given by:\n",
"I4 = Vn2/R4\n",
" #current in the 5 ohm resistance is given by:\n",
"I3 = (Vn1 - Vn2)/R3\n",
" #active power dissipated in the 2.5 ohm resistor is given by\n",
"P5 = R5*((Vn2 - V2)/R5)**2\n",
" #magnitude of the active power dissipated\n",
"P5mag = abs(P5)\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a) the voltage at nodes 1 and 2 are \",round(abs(Vn1),1),\"/_\",round(cmath.phase(complex(Vn1.real, Vn1.imag))*180/math.pi,2),\"deg V and \",round(abs(Vn2),1),\"/_\",round(cmath.phase(complex(Vn2.real, Vn2.imag))*180/math.pi,1),\"deg V\"\n",
"print \"\\n (b)the current in the j4 ohm inductance = \",round(abs(I4),2),\"/_\",round(cmath.phase(complex(I4.real, I4.imag))*180/math.pi,2),\"deg A\"\n",
"print \"\\n (c)the current in the 5 ohm resistance = \",round(abs(I3),2),\"/_\",round(cmath.phase(complex(I3.real, I3.imag))*180/math.pi,2),\"deg A\"\n",
"print \"\\n (d) magnitude of the active power dissipated in the 2.5 ohm resistance is \",round(P5mag,2),\" W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a) the voltage at nodes 1 and 2 are 17.1 /_ -5.3 deg V and 15.8 /_ 93.2 deg V\n",
"\n",
" (b)the current in the j4 ohm inductance = 3.95 /_ 3.23 deg A\n",
"\n",
" (c)the current in the 5 ohm resistance = 4.99 /_ -44.06 deg A\n",
"\n",
" (d) magnitude of the active power dissipated in the 2.5 ohm resistance is 34.4 W"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 8, page no. 556
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import numpy\n",
"import cmath\n",
"#initializing the variables:\n",
"ri = 25;# in amperes\n",
"thetai = 0;# in degrees\n",
"R1 = 4;# in ohm\n",
"R2 = 3j;# in ohm\n",
"R3 = 5;# in ohm\n",
"R4 = 10j;# in ohm\n",
"R5 = 20j;# in ohm\n",
"\n",
" #calculation:\n",
" #current\n",
"I = ri*math.cos(thetai*math.pi/180) + 1j*ri*math.sin(thetai*math.pi/180)\n",
" #Node 3 is taken as the reference node.\n",
" #At node 1,\n",
" #V1*(1/(R1 + R2) + 1/R3) - V2/R3 = I\n",
" #The equation at node 2\n",
" #V1*(-1/R3) + V2*(1/R4 + 1/R5 + 1/R3) = 0\n",
" #using determinants\n",
"d1 = [[I, -1/R3],[0 , (1/R4 + 1/R5 + 1/R3)]]\n",
"D1 = numpy.linalg.det(d1)\n",
"d2 = [[(1/(R1 + R2) + 1/R3), I],[-1/R3, 0]]\n",
"D2 = numpy.linalg.det(d2)\n",
"d = [[(1/(R1 + R2) + 1/R3), -1/R3],[-1/R3, (1/R4 + 1/R5 + 1/R3)]]\n",
"D = numpy.linalg.det(d)\n",
"V1 = D1/D\n",
"V2 = D2/D\n",
" #the voltage between point X and node 3 is\n",
"VX = V1*R2/(R1 + R2)\n",
" #Thus the voltage\n",
"VY = V2\n",
"VXY = VX - VY\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n voltage VXY is \",round(VXY.real,2),\" + (\",round( VXY.imag,2),\")i V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" voltage VXY is -16.16 + ( -15.05 )i V"
]
}
],
"prompt_number": 8
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 9, page no. 557
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import numpy\n",
"import cmath\n",
"#initializing the variables:\n",
"V = 8;# in volts\n",
"R1 = 1;# in ohm\n",
"R2 = 2;# in ohm\n",
"R3 = 3;# in ohm\n",
"R4 = 4;# in ohm\n",
"R5 = 5;# in ohm\n",
"R6 = 6;# in ohm\n",
"\n",
"#calculation:\n",
" #In Figure 31.13, the reference node is shown at point A.\n",
" #At node 1,\n",
" #V1*(1/R1 + 1/R6 + 1/R5) - V2/R1 - V3/R5 = V/R5\n",
" #The equation at node 2\n",
" #V1*(-1/R1) + V2*(1/R2 + 1/R1 + 1/R3) - V3/R3 = 0\n",
" #At node 3\n",
" # - V1/R5 - V2/R3 + V3*(1/R4 + 1/R3 + 1/R5) = -1*V/R5\n",
"#using determinants\n",
"d1 = [[V/R5, -1/R1, -1/R5],[0, (1/R2 + 1/R1 + 1/R3), -1/R3],[-1*V/R5, -1/R3, (1/R4 + 1/R3 + 1/R5)]]\n",
"D1 = numpy.linalg.det(d1)\n",
"d2 = [[(1/R1 + 1/R6 + 1/R5), V/R5, -1/R5],[-1/R1, 0, -1/R3],[-1/R5, -1*V/R5, (1/R4 + 1/R3 + 1/R5)]]\n",
"D2 = numpy.linalg.det(d2)\n",
"d3 = [[(1/R1 + 1/R6 + 1/R5), -1/R1, V/R5],[-1/R1, (1/R2 + 1/R1 + 1/R3), 0],[-1/R5, -1/R3, -1*V/R5]]\n",
"D3 = numpy.linalg.det(d3)\n",
"d =[[(1/R1 + 1/R6 + 1/R5), -1/R1, -1/R5],[-1/R1, (1/R2 + 1/R1 + 1/R3), -1/R3],[-1/R5, -1/R3, (1/R4 + 1/R3 + 1/R5)]]\n",
"D = numpy.linalg.det(d)\n",
"Vn1 = D1/D\n",
"Vn2 = D2/D\n",
"Vn3 = D3/D \n",
" #the current in the 2 ohm resistor\n",
"I2 = Vn2/R2\n",
" #power dissipated in the 3 ohm resistance\n",
"P3 = R3*((Vn2 - Vn3)/R3)**2\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)current through 2 ohm resistor is \",round(I2,2),\" A\"\n",
"print \"\\n (b)power dissipated in the 3 ohm resistor is \",round(P3,2),\" W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)current through 2 ohm resistor is 0.19 A\n",
"\n",
" (b)power dissipated in the 3 ohm resistor is 1.27 W\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}