{
"metadata": {
"name": "",
"signature": "sha256:46dd88e506b63134e24582a78d5a791547c8824340e8bb4b0b9ee7239c438783"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter 43: Magnetically coupled circuits
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 1, page no. 842
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"Na = 1200; \n",
"Nb = 1000;\n",
"Ia = 0.8;# in amperes\n",
"Phia = 100E-6;# in Wb\n",
"xb = 0.75;\n",
"\n",
" #calculation:\n",
" #self inductance of coil A\n",
"La = Na*Phia/Ia\n",
" #mutual inductance, M\n",
"Phib = xb*Phia\n",
"M = Nb*Phib/Ia\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n self inductance of coil A is \",La,\" H\"\n",
"print \"\\n mutual inductance, M is \",M,\"H\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" self inductance of coil A is 0.15 H\n",
"\n",
" mutual inductance, M is 0.09375 H"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2, page no. 843
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"M = 600E-3;# in Henry\n",
"Ia = 5;# in amperes\n",
"dt = 0.2;# in secs\n",
"\n",
" #calculation:\n",
" #change of current\n",
"dIa = 2*Ia \n",
"dIadt = dIa/dt\n",
" #secondary induced e.m.f., E2\n",
"E2 = -1*M*dIadt\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n secondary induced e.m.f., E2 is \",E2,\" V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" secondary induced e.m.f., E2 is -30.0 V"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3, page no. 844
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"La = 250E-3;# in Henry\n",
"Lb = 400E-3;# in Henry\n",
"M = 80E-3;# in Henry\n",
"\n",
" #calculation:\n",
" #coupling coefficient,\n",
"k = M/(La*Lb)**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n coupling coefficient, is \",round(k,3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" coupling coefficient, is 0.253"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 4, page no. 845
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"Lx = 80E-3;# in Henry\n",
"Ly = 60E-3;# in Henry\n",
"Nx = 200;# turns\n",
"Ny = 100;# turns\n",
"Ix = 4;# in Amperes\n",
"Phiy = 0.005;# in Wb\n",
"\n",
"#calculation:\n",
" #mutual inductance, M\n",
"M = Ny*Phiy/(2*Ix)\n",
" #coupling coefficient,\n",
"k = M/(Lx*Ly)**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n mutual inductance, M is \",M*1E3,\"mH\"\n",
"print \"\\n coupling coefficient, is \",round(k,3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" mutual inductance, M is 62.5 mH\n",
"\n",
" coupling coefficient, is 0.902"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5, page no. 846
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"La = 40E-3;# in Henry\n",
"Lb = 10E-3;# in Henry\n",
"L = 60E-3;# in Henry\n",
"\n",
" #calculation:\n",
" #mutual inductance, M\n",
"M = (L - La - Lb)/2\n",
" #coupling coefficient,\n",
"k = M/(La*Lb)**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n mutual inductance, M is \",M*1E3,\"mH\"\n",
"print \"\\n coupling coefficient, is \",k"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" mutual inductance, M is 5.0 mH\n",
"\n",
" coupling coefficient, is 0.25"
]
}
],
"prompt_number": 5
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 6, page no. 847
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"V = 240;# in Volts\n",
"Ra = 5;# in Ohm\n",
"La = 1;# in Henry\n",
"Rb = 10;# in Ohm\n",
"Lb = 5;# in Henry\n",
"I = 8;# in amperes\n",
"dIdt = 15;# in A/sec\n",
"\n",
" #calculation:\n",
" #Kirchhoff\u2019s voltage law\n",
"L = (V - I*(Ra + Rb))/dIdt\n",
" #mutual inductance, M\n",
"M = (L - La - Lb)/2\n",
" #coupling coefficient,\n",
"k = M/(La*Lb)**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n mutual inductance, M is \",M,\"H\"\n",
"print \"\\n coupling coefficient, is \",round(k,3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" mutual inductance, M is 1.0 H\n",
"\n",
" coupling coefficient, is 0.447"
]
}
],
"prompt_number": 6
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 7, page no. 848
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"k = 0.7;# coefficient of coupling\n",
"L1 = 15E-3;# in Henry\n",
"L2 = 10E-3;# in Henry\n",
"\n",
"#calculation:\n",
" #L1 = La + Lb + 2*k*(La*Lb)**0.5\n",
" #L2 = La + Lb - 2*k*(La*Lb)**0.5\n",
" #self inductance of coils\n",
"a = ((L1 - (L1 + L2)/2)/(2*k))**2\n",
"La1 =((L1 + L2)/2 + (((L1 + L2)/2)**2 - 4*a)**0.5)/2\n",
"La2 =((L1 + L2)/2 - (((L1 + L2)/2)**2 - 4*a)**0.5)/2\n",
"Lb1 = (L1 + L2)/2 - La1\n",
"Lb2 = (L1 + L2)/2 - La2\n",
" #mutual inductance, M\n",
"M = (L1 - L2)/4\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\nself inductance of coils are \",round(La1*1E3,2),\"mH and \",round( Lb1*1E3,2),\"mH\"\n",
"print \"\\n mutual inductance, M is \",round(M*1E3,2),\"mH\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"self inductance of coils are 12.24 mH and 0.26 mH\n",
"\n",
" mutual inductance, M is 1.25 mH\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 8, page no. 850
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"E1 = 8;# in Volts\n",
"thetae1 = 0;# in degrees\n",
"w = 2500;# in rad/sec\n",
"R = 15;# in ohm\n",
"L = 5E-3;# in Henry\n",
"M = 0.1E-3;# in Henry\n",
"\n",
" #calculation:\n",
" #voltage\n",
"E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
" #Impedance of primary\n",
"Z1 = R + 1j*w*L\n",
" #Primary current I1\n",
"I1 = E1/Z1\n",
" #E2\n",
"E2 = 1j*w*M*I1\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\nE2 is \",round(abs(E2),3),\"/_\",round(cmath.phase(complex(E2.real,E2.imag))*180/math.pi,2),\"deg V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"E2 is 0.102 /_ 50.19 deg V"
]
}
],
"prompt_number": 8
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 9, page no. 850
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"Lx = 20E-3;# in Henry\n",
"Ly = 80E-3;# in Henry\n",
"k = 0.75;# coupling coeff.\n",
"Ex = 5;# in Volts\n",
"\n",
" #calculation:\n",
" #mutual inductance\n",
"M = k*(Lx*Ly)**0.5\n",
" #magnitude of the open circuit e.m.f. induced\n",
"Ey = M*Ex/Lx\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n mutual inductance is \",M,\" H\"\n",
"print \"\\n magnitude of the open circuit e.m.f. induced is \",Ey,\" V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" mutual inductance is 0.03 H\n",
"\n",
" magnitude of the open circuit e.m.f. induced is 7.5 V"
]
}
],
"prompt_number": 9
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 10, page no. 852
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"E1 = 2;# in Volts\n",
"thetae1 = 0;# in degrees\n",
"f = 1000/math.pi;# in Hz\n",
"R1 = 4;# in ohm\n",
"R2 = 16;# in ohm\n",
"R3 = 16;# in ohm\n",
"R4 = 50;# in ohm\n",
"L = 10E-3;# in Henry\n",
"M = 2E-3;# in Henry\n",
"\n",
"#calculation:\n",
"w = 2*math.pi*f\n",
" #R1e is the real part of Z1e\n",
"R1e = R1 + R2 + ((R3 + R4)*(M**2)*(w**2))/((R3 + R4)**2 + (w*L)**2)\n",
" #X1e is the imaginary part of Z1e\n",
"X1e = w*L - (L*(M**2)*(w**3))/((R3 + R4)**2 + (w*L)**2)\n",
"Z1e = R1e + 1j*X1e\n",
"Z2e = R3 + R4 + 1j*w*L\n",
" #primary current, I1\n",
"I1 = E1/Z1e\n",
" #E2\n",
"E2 = 1j*w*M*I1\n",
" #secondary current I2\n",
"I2 = E2/Z2e\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"secondary current I2 is \",round(abs(I2)*1E3,3),\"/_\", round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg mA\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"secondary current I2 is 4.085 /_ 28.55 deg mA\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 11, page no. 853
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"E1 = 50;# in Volts\n",
"thetae1 = 0;# in degrees\n",
"w = 500;# in rad/sec\n",
"R1 = 300;# in ohm\n",
"L1 = 0.2;# in Henry\n",
"L2 = 0.5;# in Henry\n",
"L3 = 0.3;# in Henry\n",
"R2 = 500;# in ohm\n",
"C = 5E-6;# in farad\n",
"M = 0.2;# in Henry\n",
"\n",
"#calculation:\n",
" # Self impedance of primary circuit\n",
"Z1 = R1 + 1j*w*(L1 + L2)\n",
" #Self impedance of secondary circuit,\n",
"Z2 = R2 + 1j*(w*L3 - 1/(w*C))\n",
" #reflected impedance, Zr\n",
"Zr = (w*M)**2/Z2\n",
" #Effective primary impedance,\n",
"Z1e = Z1 + Zr\n",
" #Primary current I1 \n",
"I1 = E1/Z1e\n",
" #Secondary current I2\n",
"E2 = 1j*w*M*I1\n",
"I2 = E2/Z2\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n Self impedance of primary circuit, Z1 is \",Z1.real,\" + (\", Z1.imag,\")i ohm\"\n",
"print \"\\n Self impedance of secondary circuit, Z2 is \",Z2.real,\" + (\", Z2.imag,\")i ohm\"\n",
"print \"\\n reflected impedance, Zr is \",Zr.real,\" +(\", Zr.imag,\")i ohm\"\n",
"print \"\\n Effective primary impedance Z1(eff) is \",Z1e.real,\" +(\",Z1e.imag,\")i ohm\"\n",
"print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
"print \"\\n secondary current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" Self impedance of primary circuit, Z1 is 300.0 + ( 350.0 )i ohm\n",
"\n",
" Self impedance of secondary circuit, Z2 is 500.0 + ( -250.0 )i ohm\n",
"\n",
" reflected impedance, Zr is 16.0 +( 8.0 )i ohm\n",
"\n",
" Effective primary impedance Z1(eff) is 316.0 +( 358.0 )i ohm\n",
"\n",
" primary current I1 is 0.1 /_ -48.57 deg A\n",
"\n",
" secondary current I2 is 0.02 /_ 68.0 deg A"
]
}
],
"prompt_number": 11
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 12, page no. 855
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"E1 = 20;# in Volts\n",
"thetae1 = 0;# in degrees\n",
"R1 = 15;# in ohm\n",
"C1 = 400E-12;# in farad\n",
"R2 = 30;# in ohm\n",
"L1 = 0.001;# in Henry\n",
"L2 = 0.0002;# in Henry\n",
"R3 = 50;# in ohm\n",
"M = 10E-6;# in Henry\n",
"\n",
"#calculation:\n",
" #voltage\n",
"E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
" #the resonant frequency, fr \n",
"fr = 1/(2*math.pi*(L1*C1)**0.5)\n",
" #The secondary is also tuned to a resonant frequency\n",
" #capacitance,C2\n",
"C2 = 1/(L2*(2*math.pi*fr)**2)\n",
" #the effective primary impedance Z1eff\n",
"w = 2*math.pi*fr\n",
"Z1e = R1 + R2 + ((w*M)**2)/R3\n",
" #Primary current I1 \n",
"I1 = E1/Z1e\n",
" #Secondary current I2\n",
"E2 = 1j*w*M*I1\n",
"I2 = E2/Z1e\n",
" #voltage across capacitor C2\n",
"Vc2 = I2*(-1*1j/(w*C2))\n",
" #coefficient of coupling, k \n",
"k = M/(L1*L2)**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the resonant frequency,fr is \",round(fr/1000,2),\"KHz\"\n",
"print \"\\n capacitance,C2 is \",round(C2*1E9,2),\"nF\"\n",
"print \"\\n Effective primary impedance Z1(eff) is \",round(abs(Z1e),2),\" ohm\"\n",
"print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag)),0),\"deg A\"\n",
"print \"\\n voltage across capacitor C2 is \",round(abs(Vc2),2),\"/_\",round(abs(cmath.phase(complex(Vc2.real,Vc2.imag))),0),\"deg V\"\n",
"print \"\\n coefficient of coupling, k is \",round(k,4),\"\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the resonant frequency,fr is 251.65 KHz\n",
"\n",
" capacitance,C2 is 2.0 nF\n",
"\n",
" Effective primary impedance Z1(eff) is 50.0 ohm\n",
"\n",
" primary current I1 is 0.4 /_ 0.0 deg A\n",
"\n",
" voltage across capacitor C2 is 40.0 /_ 0.0 deg V\n",
"\n",
" coefficient of coupling, k is 0.0224 \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 13, page no. 858
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"E1 = 250;# in Volts\n",
"thetae1 = 0;# in degrees\n",
"R1 = 50j;# in ohm\n",
"R2 = 10;# in ohm\n",
"R3 = 10;# in ohm\n",
"R4 = 50j;# in ohm\n",
"R5 = 50;# in ohm\n",
"M = 10j;# in ohm\n",
"\n",
" #calculation:\n",
" #voltage\n",
"E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
" #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
" #(R1 + R2)*I1 - M*I2 = E1\n",
" #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
" #-1*M*I1 + ( R3 + R4 + R5)*I2 = 0\n",
" #solving these two\n",
"I2 = E1/((R1 + R2)*(R3 + R4 + R5)/(-1*M) + (-1*M))\n",
"I1 = I2*(R3 + R4 + R5)/(-1*M)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n primary current I1 is \",round(I1.real,2),\" +(\",round( I1.imag,2),\")i A\"\n",
"print \"\\n secondary current I2 is \",round(I2.real,2),\" +(\",round( I2.imag,2),\")i A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" primary current I1 is 0.85 +( -4.77 )i A\n",
"\n",
" secondary current I2 is -0.54 +( 0.31 )i A"
]
}
],
"prompt_number": 13
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14, page no. 859
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"E1 = 40;# in Volts\n",
"thetae1 = 0;# in degrees\n",
"R1 = 5;# in ohm\n",
"L1 = 0.001;# in Henry\n",
"L2 = 0.006;# in Henry\n",
"R2 = 40;# in ohm\n",
"rzl = 200;# in ohm\n",
"thetazl = -60;# in degrees\n",
"k = 0.70\n",
"f = 20000;# in Hz\n",
"\n",
" #calculation:\n",
"w = 2*math.pi*f\n",
" #voltage\n",
"#E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n",
" #impedance\n",
"ZL = rzl*math.cos(thetazl*math.pi/180) + 1j*rzl*math.sin(thetazl*math.pi/180)\n",
" #mutual inductance, M\n",
"M = k*(L1*L2)**0.5\n",
" #Applying Kirchhoff\u00e2\u20ac\u2122s voltage law to the primary circuit gives\n",
" #(R1 + 1j*w*L1)*I1 - 1j*w*M*I2 = E1\n",
" #Applying Kirchhoff\u00e2\u20ac\u2122s voltage law to the secondary circuit gives\n",
" #-1j*w*M*I1 + ( R2 + ZL + 1j*w*L2)*I2 = 0\n",
" #solving these two\n",
"\n",
"a = R1 + 1j*w*L1\n",
"b = 1j*w*M\n",
"c = R2 + ZL + 1j*w*L2\n",
"I1 = E1/(1*a - (b**2)/c)\n",
"d = -1*cmath.phase(complex(I1.real,I1.imag))\n",
"e = abs(I1)\n",
"I2 = (b/c)*(e*math.cos(d) + 1j*e*math.sin(d))\n",
"pd2 = I2*ZL\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n mutual induction M is \",round(M*1E3,3),\"mH\"\n",
"print \"\\n primary current I1 is \",round(abs(I1),3),\"/_\",round(-1*cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
"print \"\\n secondary current I2 is \",round(abs(pd2),1),\"/_\",round(cmath.phase(complex(pd2.real,pd2.imag))*180/math.pi,2),\"deg V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" mutual induction M is 1.715 mH\n",
"\n",
" primary current I1 is 0.724 /_ 65.15 deg A\n",
"\n",
" secondary current I2 is 52.2 /_ 18.7 deg V"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 15, page no. 860
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"E1 = 50;# in Volts\n",
"thetae1 = 0;# in degrees\n",
"r = 5;# in ohm\n",
"R1 = 20;# in ohm\n",
"L1 = 0.2;# in Henry\n",
"L2 = 0.4;# in Henry\n",
"R2 = 25;# in ohm\n",
"RL = 20;# in ohm\n",
"M = 0.1;# in Henry\n",
"f = 75/math.pi;# in Hz\n",
"\n",
"#calculation:\n",
"w = 2*math.pi*f\n",
" #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
" #(r + R1 + 1j*w*L1)*I1 - 1j*w*M*I2 = E1\n",
" #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
" #-1*1j*w*M*I1 + ( R2 + RL + 1j*w*L2)*I2 = 0\n",
" #solving these two\n",
"I2 = E1/((r + R1 + 1j*w*L1)*(R2 + RL + 1j*w*L2)/(1j*w*M) + (-1*1j*w*M))\n",
"I1 = I2*(R2 + RL + 1j*w*L2)/(1j*w*M)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
"print \"\\n load current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" primary current I1 is 1.3 /_ -45.84 deg A\n",
"\n",
" load current I2 is 0.26 /_ -8.97 deg A"
]
}
],
"prompt_number": 15
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 16, page no. 862
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"E1 = 50;# in Volts\n",
"thetae1 = 0;# in degrees\n",
"r = 5;# in ohm\n",
"R1 = 20;# in ohm\n",
"L1 = 0.2;# in Henry\n",
"R = 8;# in ohm\n",
"L = 0.1;# in Henry\n",
"L2 = 0.4;# in Henry\n",
"R2 = 25;# in ohm\n",
"RL = 20;# in ohm\n",
"M = 0.1;# in Henry\n",
"f = 75/math.pi;# in Hz\n",
"\n",
"#calculation:\n",
"w = 2*math.pi*f\n",
" #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
" #(r + R1 + 1j*w*L1 + R + 1j*w*L)*I1 - (1j*w*M + R + 1j*w*L)*I2 = E1\n",
" #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
" #-1*(1j*w*M + R + 1j*w*L)*I1 + (R2 + RL + 1j*w*L2 + R + 1j*w*L)*I2 = 0\n",
" #solving these two\n",
"I2 = E1/((r + R1 + 1j*w*L1 + R + 1j*w*L)*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(1j*w*M + R + 1j*w*L) + (-1*(1j*w*M + R + 1j*w*L)))\n",
"I1 = I2*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(1j*w*M + R + 1j*w*L)\n",
" #reversing\n",
" #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n",
" #(r + R1 + 1j*w*L1 + R + 1j*w*L)*I1r - (-1*1j*w*M + R + 1j*w*L)*I2r = E1\n",
" #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n",
" #-1*(-1*1j*w*M + R + 1j*w*L)*I1r + (R2 + RL + 1j*w*L2 + R + 1j*w*L)*I2r = 0\n",
" #solving these two\n",
"I2r = E1/((r + R1 + 1j*w*L1 + R + 1j*w*L)*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(-1*1j*w*M + R + 1j*w*L) + (-1*(-1*1j*w*M + R + 1j*w*L)))\n",
"I1r = I2r*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(-1*1j*w*M + R + 1j*w*L)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n",
"print \"load current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\"\n",
"print \"reversed primary current I1r is \",round(abs(I1r),2),\"/_\",round(cmath.phase(complex(I1r.real,I1r.imag))*180/math.pi,2),\"deg A\"\n",
"print \"reversed load current I2r is \",round(abs(I2r),2),\"/_\",round(cmath.phase(complex(I2r.real,I2r.imag))*180/math.pi,2),\"deg A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"primary current I1 is 1.03 /_ -45.47 deg A\n",
"load current I2 is 0.35 /_ -25.16 deg A\n",
"reversed primary current I1r is 0.89 /_ -54.42 deg A\n",
"reversed load current I2r is 0.08 /_ -109.17 deg A\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}