{
"metadata": {
"name": "",
"signature": "sha256:4381350f1f5861e39c88a88906555fe70746a56edfa9636dec1dfc702e44302a"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 5: Electromagnetism"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.1, Page 145"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"N = 100; # Number of turns\n",
"delta_phi = 10e-03; # Flux linked with coil, Wb\n",
"delta_t = 2e-03; # Time during which flux changes, s\n",
"\n",
"#Calculations\n",
"e =((-N)*delta_phi)/delta_t; # Average induced emf, V\n",
"\n",
"#Result\n",
"print \"The average emf induced in the coil = %3d V\"%e\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The average emf induced in the coil = -500 V\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.2, Page 146"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"N = 250; # Number of turns\n",
"delta_phi1 = 20e-03; # Flux linked with coil, Wb\n",
"delta_phi2 = -16e-03; # Flux linked with coil, Wb\n",
"delta_t1 = 0.05; # Time, s\n",
"delta_t2 = 0.01; # Time, s\n",
"\n",
"#Calculations\n",
"e_1 =((-N)*delta_phi1)/delta_t1; # Average induced emf, V\n",
"e_2 =((-N)*delta_phi2)/delta_t2; # Average induced emf, V\n",
"\n",
"#Results\n",
"print \"Change in flux in first case = %4.2f weber\"%delta_phi1\n",
"print \"Emf induced in first case = %3d volts\"%e_1\n",
"print \"Change in flux in second case = %4.2f weber\"%delta_phi2\n",
"print \"Emf induced in second case = %3d volts\"%e_2\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in flux in first case = 0.02 weber\n",
"Emf induced in first case = -100 volts\n",
"Change in flux in second case = -0.02 weber\n",
"Emf induced in second case = 400 volts\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.3, Page 147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"e = 100; # Induced emf, V\n",
"\n",
"#Calculations\n",
"# For simplification let (delta_phi)/(delta_t) = k\n",
"k = 0.1; # Rate of chage of flux linked with coil, Wb/s\n",
"# Since e =((-N)*delta_phi)/delta_t, soling for N\n",
"N = (e)/k; # Number of turns\n",
"\n",
"#Result\n",
"print \"The number of turns on the coil = %4d\"%N"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The number of turns on the coil = 1000\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.4, Page 149"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"v = 5; # Velocity, m^2\n",
"theta =(math.pi/3); # Angle, degrees\n",
"phi = 1.6e-03; # Flux, Wb\n",
"l = 0.1; # Length of pole face, m\n",
"d = 0.4; # Breadth of pole face, m\n",
"\n",
"#Calculations\n",
"A = l*d; # Cross-sectional area of pole face, m^2\n",
"B = phi/ A; # Flux density, T\n",
"e =( B*l*v)*math.sin(theta); # Induced emf, V\n",
"\n",
"#Result\n",
"print \"The emf induced = %5.4f V\"%e\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The emf induced = 0.0173 V\n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.5, Page 149"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"l = 0.15; # Effective length of conductor, m\n",
"v = 8; # Velocity, m^2\n",
"theta = 55; # Angle, degrees\n",
"e = 2.5; # Induced emf, V\n",
"\n",
"#Calculations\n",
"# Since e = B*l*v*sin(theta), solving for B\n",
"B = e/(l*v*math.sin(theta*math.pi/180)); # Flux density, T\n",
"\n",
"#Result\n",
"print \"The density of the field = %5.3f Tesla\"%B\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The density of the field = 2.543 Tesla\n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.6, Page 149"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"l = 2.2; # Effective length of conductor, m\n",
"B =38e-06; # Flux density, T\n",
"theta = (math.pi/2); # Angle, degrees\n",
"v = 800/36; # Velocity, m^2\n",
"\n",
"#Calculations\n",
"e = B*l*v*math.sin(theta); # Induced emf, V\n",
"\n",
"#Result\n",
"print \"The emf induced in the axle = %4.2f mV\"%(e/1e-03)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The emf induced in the axle = 1.84 mV\n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.7, Page 152"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"l = 0.22; # Effective length of conductor, m\n",
"B = 0.35; # Flux density, T\n",
"I = 3; # Current, A\n",
"theta = (math.pi/2); # Angle, degrees\n",
"\n",
"#Calculations\n",
"# Since the force exerted on the conductor placed in magnetic field is directly proportional to the flux density , \n",
"#the value of current flowing through the conductor, and the length of conductor lying inside the field, therefore\n",
"F = B*I*l*math.sin(theta); # Force, N\n",
"\n",
"#Result\n",
"print \"The force exerted on the conductor = %5.3f N\"%F\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The force exerted on the conductor = 0.231 N\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.8, Page 153"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"phi = 2.5e-03; # Flux, Wb\n",
"l = 0.05; # Effective length of pole, m\n",
"d = 0.03; # Effective width of pole, m\n",
"F = 1.25; # Force exerted on conductor, N\n",
"A = l*d; # Cross-sectional area of pole face, m^2\n",
"B = phi/A; # Flux density, T\n",
"theta = (math.pi/2); # Angle, degrees\n",
"\n",
"#Calculations\n",
"# Since F = B*I*l*sin(theta), solving for I\n",
"I = F/(B*l*math.sin(theta)); # Current in conductor, A\n",
"theta_2 = (math.pi/4); # New angle, degrees\n",
"F_2 = B*I*l*math.sin(theta_2); # Force exerted on conductor, N\n",
"\n",
"#Results\n",
"print \"The value of the current = %2d A\"%I\n",
"print \"The force exerted on conductor when placed at 45 degrees to the field = %5.3f newton\"%F_2\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of the current = 14 A\n",
"The force exerted on conductor when placed at 45 degrees to the field = 0.884 newton\n"
]
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.9, Page 154"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"l = 0.015; # Length of coil, m\n",
"d = 0.006; # Width of coil, m\n",
"B = 1.2; # Flux density, T\n",
"I = 1e-02; # Current, a\n",
"r = d/2; # Radius of rotation, m\n",
"\n",
"#Calculations\n",
"# Since torque is given by the product of force and distance, therefore, we have\n",
"T = 2*B*I*l*r; # Torque, Nm\n",
"\n",
"#Result\n",
"print \"The torque exerted on the coil = %4.2f micro-Nm\"%(T/1e-06)\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The torque exerted on the coil = 1.08 micro-Nm\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.10, Page 155"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"N = 80; # Number of turns\n",
"l = 0.02; # Length of coil, m\n",
"r = 0.012; # Radius of coil, m\n",
"I = 45e-06; # Current in coil, A\n",
"T = 1.4e-06; # Torque exerted on coil, Nm\n",
"A = l*r; # Cross-sectional area of coil, m^2\n",
"\n",
"#Calculations\n",
"# Since T = 2*B*I*l*r, solving for B\n",
"B = T/(2*A*N*I); # Flux density, T\n",
"\n",
"#Result\n",
"print \"The flux density produced by the pole pieces = %4.2f T\"%B\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The flux density produced by the pole pieces = 0.81 T\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.11, Page 158"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"d = 0.035; # Distance between two parallel conductors, m\n",
"I_1 = 50; # Electric current in first coil, A\n",
"I_2 = 40; # Electric current in second coil, A\n",
"\n",
"#Calculations\n",
"F = ((2e-07)*I_1*I_2)/d; # Force exerted by conductors, N\n",
"\n",
"#Result\n",
"print \"The force exerted between the conductors = %4.1f mN\"%(F/1e-03);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The force exerted between the conductors = 11.4 mN\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.12, Page 158"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"d = 2; # Distance between two parallel conductors, m\n",
"I_1 = 1000; # Electric current in first coil, A\n",
"I_2 = 300; # Electric current in second coil, A\n",
"\n",
"#Calculations\n",
"mew_o = 4*(math.pi)*1e-07; # Permeability for free space\n",
"B = (mew_o*I_1)/d; # Flux density due to first coil, T\n",
"F = ((2e-07)*I_1*I_2)/d; # Force exerted by conductors, N\n",
"\n",
"#Result\n",
"print \"The flux density at a distance of %1d m from the centre of a conductor carrying a current of %4d A = %5.3f mT\"%(d, I_1, B/1e-03);\n",
"print \"Force exerted by conductors = %2d mN\"%(F/1e-03);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The flux density at a distance of 2 m from the centre of a conductor carrying a current of 1000 A = 0.628 mT\n",
"Force exerted by conductors = 30 mN\n"
]
}
],
"prompt_number": 40
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.13, Page 163"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"R_c = 40; # Resistance of coil, ohm\n",
"I_fsd = 5e-04; # Full-scale deflection current, A\n",
"I = 3; # Current reading, A\n",
"\n",
"#Calculations\n",
"V_c = I_fsd*R_c; # Potential difference, V\n",
"# Since I = I_s + I_fsd, solving for I_s\n",
"I_s = I-I_fsd; # Shunt current, A\n",
"# From Ohm's law, V_c = I_s*R_s, solving for R_s\n",
"R_s = V_c/I_s; # Shunt resistance, ohm\n",
"\n",
"#Result\n",
"print \"The value of required shunt resistance = %4.2f milli-ohm\"%(R_s/1e-03); \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of required shunt resistance = 6.67 milli-ohm\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.14, Page 163"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"R_c = 40; # Resistance of coil, ohm\n",
"I_fsd = 5e-04; # Full-scale deflection current, A\n",
"I_fsd = 5e-04; # Full-scale deflection current, A\n",
"V = 10; # Voltage reading range, V\n",
"V_c = 0.02; # Potential difference across coil resistance, V\n",
"\n",
"#Calculations\n",
"# From Ohm's law, V = I_fsd*R, solving for R\n",
"R = V/I_fsd; # Total resistance, ohm\n",
"# Since R = R_m + R_c, solving R_m\n",
"R_m = R - R_c; # Multiplier resistance, ohm\n",
"\n",
"#Result\n",
"print \"The required value of multiplier resistance = %5.2f kilo-ohms\"%(R_m*1e-03);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required value of multiplier resistance = 19.96 kilo-ohms\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.15, Page 164"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"R_c = 1500; # Coil resistance, ohm\n",
"I_fsd = 75e-06; # Full-scale deflection current, A\n",
"I = 5; # Current range, A\n",
"V = 10; # Voltage range, V\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"# Using Ohm's law,\n",
"V_c = I_fsd*R_c; # Potential difference across coil resistance, V\n",
"# Since I = I_s + I_fsd, solving for I_s\n",
"I_s = I-I_fsd; # Shunt current, A\n",
"# From Ohm's law, V_c = I_s*R_s, solving for R_s\n",
"R_s = V_c/I_s; # Shunt resistance, ohm\n",
"\n",
"# Part (b)\n",
"# Since = V = V_m + V_c, solving for V_m\n",
"V_m = V - V_c; # Potential difference across multiplier resistance, V\n",
"# From Ohm's law, V_m = I_fsd*R_m, solving for R_m\n",
"R_m = V_m/I_fsd # Multiplier resistance, ohm\n",
"\n",
"#Results\n",
"print \"The required value of shunt resistance = %4.1f mega-ohm\"%(R_s/1e-03);\n",
"print \"The required value of multiplier resistance = %4.1f mega-ohm\"%(R_m*1e-03);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required value of shunt resistance = 22.5 mega-ohm\n",
"The required value of multiplier resistance = 131.8 mega-ohm\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.16, Page 166"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"R_1 = 30.; # Resistance, ohm\n",
"R_2 = 70.; # Resistance, ohm\n",
"R_in = 200.; # Internal resistance of meter, ohm\n",
"V = 12.; # Supply voltage, V\n",
"\n",
"#Calculations\n",
"# Using voltage divider rule, we have\n",
"V_2t = (R_2 /(R_1 + R_2))*V # True value of p.d across resistance R_2, V\n",
"# Since the resistances R_2 and R-in are parallel, so their equivalent resistance is given their parallel combination\n",
"R_BC = (R_2 * R_in)/(R_2 + R_in); # Resistance, ohms\n",
"# Using the potential divider technique, \n",
"V_2i = (R_BC / ( R_BC + R_1 ))*V # Indicated value of p.d across by voltmetre, volts\n",
"err = (( V_2i-V_2t ) / V_2t)*100 # Percentage error in the reading\n",
"\n",
"#Results\n",
"print \"The p.d. indicated by the meter = %3.1f V\"%V_2i\n",
"print \"The percentage error in the reading = %4.2f percent\"%err\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The p.d. indicated by the meter = 7.6 V\n",
"The percentage error in the reading = -9.50 percent\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.17, Page 168"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"R_in = 200.; # Internal resistance of meter, kilo-ohms\n",
"V = 10.; # Supply voltage, volts\n",
"R_1 = 10.; # Resistance, kilo-ohms\n",
"R_2 = 47.; # Resistance, kilo-ohms\n",
"\n",
"#Calculations\n",
"V_1 = R_1/(R_1+R_2)*V # P.d across resistance R_1, V\n",
"V_2 = R_2/(R_1+R_2)*V # P.d across resistance R_2, V\n",
"\n",
"# Part (a)\n",
"R_AB = (R_1 * R_in)/(R_1 + R_in); # Resistance, kilo-ohms\n",
"V_AB = (R_AB / ( R_AB + R_2 ))*V # True value of p.d across by voltmetre, V\n",
"R_BC = (R_2 * R_in)/(R_2 + R_in); # Resistance, kilo-ohms\n",
"V_BC = (R_BC / ( R_BC + R_1 ))*V # Indicated value of p.d across by voltmetre, V\n",
"\n",
"# Part (b)\n",
"# Error for V_1 measurement\n",
"error_AB = (V_AB - V_1)/V_1*100 # Percentage error in the reading\n",
"#Error for V_2 measurement\n",
"error_BC = (V_BC-V_2)/V_2*100 # Percentage error in the reading \n",
"\n",
"#Results\n",
"print \"The p.d. indicated by the meter across first resistor = %4.2f V\"%V_AB\n",
"print \"The p.d. indicated by the meter across second resistor = %4.2f V\"%V_BC\n",
"print \"Percentage error for V_1 measurement = %4.2f percent\"%error_AB\n",
"print \"Percentage error for V_2 measurement = %4.2f percent\"%error_BC\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The p.d. indicated by the meter across first resistor = 1.68 V\n",
"The p.d. indicated by the meter across second resistor = 7.92 V\n",
"Percentage error for V_1 measurement = -3.96 percent\n",
"Percentage error for V_2 measurement = -3.96 percent\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.18, Page 176"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"L = 0.25; # Self-inductance, H\n",
"delta_I = 250e-03; # Change in current, A\n",
"delta_t = 25e-03; # Time, s\n",
"\n",
"#Calculations\n",
"e = ((L)*delta_I)/(delta_t); # Induced emf, V\n",
"\n",
"#Result\n",
"print \"The value of emf induced = %3.1f V\"%e\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of emf induced = 2.5 V\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.19, Page 176"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"e = 30.; # Induced emf, V\n",
"\n",
"#Calculations\n",
"# For simplicity, let rate of change of current i.e delta_I/delta_t = k\n",
"k = 200; # Rate of change of current, ampere-second\n",
"# Since e = ((-L)*delta_I)/(delta_t), solving for L\n",
"L = e/k; # Self-inductance, H\n",
"\n",
"#Result\n",
"print \"The inductance of the circuit = %4.2f H\"%L\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The inductance of the circuit = 0.15 H\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.20, Page 176"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"L = 50e-03; # Self-inductance, H\n",
"e = 8; # Induced emf, V\n",
"\n",
"#Calculations\n",
"# Since e = ((-L)*delta_I)/(delta_t), solving for delta_I/delta_t,and for simplicity letting the rate of change of \n",
"#current i.e delta_I/delta_t = k\n",
"k = e/L; # Rate of change of current, As\n",
"\n",
"#Result\n",
"print \"The rate of change of current = %3d A/s\"%k\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The rate of change of current = 160 A/s\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.21, Page 178"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"N = 150; # Number of turns in a coil\n",
"I = 10; # Electric current flowing through coil, A\n",
"phi = 0.10; # Flux, Wb\n",
"delta_t = 0.1; # Time, s\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"L = (N * phi)/I # Self-inductance, H\n",
"delta_I = 20; # Change in current, A\n",
"# Part (b)\n",
"e = abs((-L*delta_I)/(delta_t)); # Induced emf, V\n",
"\n",
"#Results\n",
"print \"The inductance of the coi = %3.1f H\"%L\n",
"print \"The emf induced in the coil = %2d V\"%e\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The inductance of the coi = 1.5 H\n",
"The emf induced in the coil = 300 V\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.22, Page 178"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"I_1 = 8; # Electric current, A\n",
"I_2 = 2; # Electric current, A\n",
"N = 3000; # Number of turns in a coil\n",
"phi_1 = 4e-03; # Flux, Wb\n",
"delta_t = 0.1; # Reversal time of current, s\n",
"\n",
"#Calculations\n",
"L = (N * phi_1)/I_1; # Self-inductance, H\n",
"delta_I = I_1 - I_2; # Change in current, A\n",
"e = ((L)*delta_I)/(delta_t); # Induced emf, V\n",
"\n",
"#Result\n",
"print \"The emf induced in the coil = %2d volts\"%e\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The emf induced in the coil = 90 volts\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.23, Page 179"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"N_1 = 600; # Number of turns in a coil in first case\n",
"N_2 = 900; # Number of turns in a coil in second case \n",
"N_3 = 900; # Number of turns in a coil in third case\n",
"l = 45e-03; # Effective length of coil, m\n",
"A = 4e-04; # Cross-sectional area of coil, m^2\n",
"mew_o = 4*(math.pi)*1e-07; # Permeability for free space\n",
"mew_r1 = 1; # Relative permeability in first case\n",
"mew_r2 = 1; # Relative permeability in second case\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"mew_r3 = 75; # Relative permeability in third case\n",
"L_1 = (mew_o*mew_r1*(N_1**2)*A)/l; # Self-inductance of coil in first case, H\n",
"\n",
"# Part (b)\n",
"# Since self-inductance of a coil is directly proportional to the number of turns in a coil, therefore, \n",
"#we have L_2/L_1 = (N_2^2)/(N_1^2), solving for L_2\n",
"L_2 = (L_1*(N_2**2))/(N_1**2); # Self-inductance of coil in second case, H\n",
"\n",
"# Part (c)\n",
"# Since mew_r3 = 75*mew_r2, keeping all other quantities same we have\n",
"L_3 = mew_r3*L_2; # Self-inductance of coil in third case, H\n",
"\n",
"#Results\n",
"print \"Self-inductance of coil in first case = %4.2f mH\"%(L_1/1e-03);\n",
"print \"Self-inductance of coil in second case = %5.3f mH\"%(L_2/1e-03);\n",
"print \"Self-inductance of coil in third case = %5.3f H\"%L_3\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Self-inductance of coil in first case = 4.02 mH\n",
"Self-inductance of coil in second case = 9.048 mH\n",
"Self-inductance of coil in third case = 0.679 H\n"
]
}
],
"prompt_number": 41
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.24, Page 182"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"N_A = 2000; # Number of turns in a coil A\n",
"N_B = 1500; # Number of turns in a coil B\n",
"I_A = 0.5; # Electric current in coil A, A\n",
"phi_A = 60e-06; # Flux linked with coil A, Wb\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"L_A = (N_A*phi_A)/I_A; # Self-inductance of coil A\n",
"phi_B = 0.83*(60e-06); # Flux linked with coil B, Wb\n",
"\n",
"# Part (b)\n",
"M = (N_B*phi_B)/I_A; # Mutual inductance of the two coils, H\n",
"\n",
"#Results\n",
"print \"Self-inductance of coil A = %4.2f H\"%L_A\n",
"print \"Mutual inductance of the two coils = %5.3f H\"%M\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Self-inductance of coil A = 0.24 H\n",
"Mutual inductance of the two coils = 0.149 H\n"
]
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.25, Page 183"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"N = 400; # Number of turns in a coil\n",
"l = 0.25; # Effective length of coil, m\n",
"A = 4.5e-04; # Cross-sectional area, m^2\n",
"mew_r = 180; # Relative permeability \n",
"\n",
"#Calculations\n",
"mew_o = 4*(math.pi)*1e-07; # Pemeability for free space\n",
"L = (mew_o*mew_r*(N**2)*A)/l # Self-inductance of coil, H\n",
"\n",
"#Result\n",
"print \"The self inductance of the coil = %2d milli-henry\"%(L/1e-03);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The self inductance of the coil = 65 milli-henry\n"
]
}
],
"prompt_number": 42
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.26, Page 183"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"L_1 = 65e-03; # Self-inductance of first coil, H\n",
"delta_I = 1.5; # Change in current, A\n",
"delta_t = 3e-03; # Time, s\n",
"k = 0.95; # 95 percent of flux produced\n",
"N_1 = 400; # Number of turns in a coil A\n",
"N_2 = 650; # Number of turns in a coil B \n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"# Since self-inductance of a coil is directly proportional to the number of turns in a coil, therefore, \n",
"#we have L_2/L_1 = (N_2^2)/(N_1^2), solving for L_2\n",
"L_2 = (L_1*(N_2**2))/(N_1**2) # Self-inductance of second coil , H\n",
"\n",
"# Part (b)\n",
"M = k*math.sqrt(L_1*L_2); # Mutual inductance of two coils, H\n",
"\n",
"# Part (c)\n",
"e_1 = ((L_1)*delta_I)/(delta_t); # Induced emf in first coil, V \n",
"\n",
"# Part (d)\n",
"e_2 = (M*delta_I)/delta_t; # Induced emf in second coil, V \n",
"\n",
"#Results\n",
"print \"The self-inductance of coil 2 = %3d mH\"%(L_2/1e-03)\n",
"print \"The value of mutual inductance = %3d mH\"%(M/1e-03)\n",
"print \"The self-induced emf in coil 1 = %4.1f V\"%e_1\n",
"print \"The mutually induced emf in coil 2 = %2d V\"%e_2\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The self-inductance of coil 2 = 171 mH\n",
"The value of mutual inductance = 100 mH\n",
"The self-induced emf in coil 1 = 32.5 V\n",
"The mutually induced emf in coil 2 = 50 V\n"
]
}
],
"prompt_number": 43
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.27, Page 185"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"L = 50e-03; # Self-inductance of coil, H\n",
"I = 0.75; # Electric current in coil, A\n",
"\n",
"#Calculations\n",
"W = (L*(I**2))/2 # Energy stored, J\n",
"\n",
"#Result\n",
"print \"Energy stored in the inductor = %4.1f mJ\"%(W/1e-03)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Energy stored in the inductor = 14.1 mJ\n"
]
}
],
"prompt_number": 32
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.28, Page 185"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"L_1 = 25e-03; # Self-inductance of first coil, H\n",
"L_2 = 40e-03; # Self-inductance of second coil, H\n",
"I = 0.25; # Electric current in coils, A\n",
"k =0.8; # Coupling coefficient\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"W_1 = (L_1*(I**2))/2; # Energy stored in first coil, J\n",
"W_2 = (L_2*(I**2))/2; # Energy stored in second coil, J\n",
"M = k*math.sqrt(L_1*L_2); # Mutual inductance of coils\n",
"\n",
"# Part (b)\n",
"W_M = M*(I)*(I); # Energy stored due to mutual inductance of coils, J\n",
"W_sa = W_1 + W_2 + W_M; # Energy stored by two inductors when connected in series aiding, J\n",
"W_so = W_1 + W_2 - W_M; # Energy stored by two inductors when connected in series opposition, J\n",
"\n",
"#Results\n",
"print \"Energy stored in first coil = %4.2f mJ\"%(W_1/1e-03)\n",
"print \"Energy stored in second coil = %4.2f mJ\"%(W_2/1e-03)\n",
"print \"Energy stored by two inductors when connected in series aiding = %3.1f mJ\"%(W_sa/1e-03)\n",
"print \"Energy stored by two inductors when connected in series opposition = %4.2f mJ\"%(W_so/1e-03)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Energy stored in first coil = 0.78 mJ\n",
"Energy stored in second coil = 1.25 mJ\n",
"Energy stored by two inductors when connected in series aiding = 3.6 mJ\n",
"Energy stored by two inductors when connected in series opposition = 0.45 mJ\n"
]
}
],
"prompt_number": 44
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.29, Page 189"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"V_2 = 60; # Output voltage, V\n",
"V_1 = 240; # Input voltage, V\n",
"N_2 = 500; # Secondary turns\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"# For simplicity let V_1/V_2 = N_1/N_2 = k\n",
"k = V_1/V_2 # Turns ratio\n",
"\n",
"# Part (b)\n",
"# Since V_1/V_2 = N_1/N_2, solving for N_1\n",
"N_1 = k*N_2; # Primary turns\n",
"\n",
"#Results\n",
"print \"The required turns ratio = %1d:1\"%k\n",
"print \"The number of primary turns = %4d\"%N_1\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required turns ratio = 4:1\n",
"The number of primary turns = 2000\n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.30, Page 190"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"R_L = 15; # Load resistor, ohms\n",
"V_2 = 240.; # Terminal p.d at secondary, V\n",
"V_1 = 600; # Supply voltage, V\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"# Since V_1/V_2 = N_1/N_2 = k\n",
"k = V_1/V_2; # Turns ratio\n",
"\n",
"# Part (b)\n",
"I_2 = V_2/R_L; # Current drawn by the load, A\n",
"P_2 = V_2*I_2; # Power drawn by the load, W\n",
"\n",
"# Part (c)\n",
"I_1 = P_2/V_1 # Current drawn from the supply, A\n",
"\n",
"#Results\n",
"print \"The transformer turns ratio = %3.1f:1\"%k\n",
"print \"The current drawn by the load = %2d A\"%I_2\n",
"print \"The power drawn by the load = %4.2f W\"%(P_2*1e-03);\n",
"print \"The current drawn from the supply = %3.1f A\"%I_1\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The transformer turns ratio = 2.5:1\n",
"The current drawn by the load = 16 A\n",
"The power drawn by the load = 3.84 W\n",
"The current drawn from the supply = 6.4 A\n"
]
}
],
"prompt_number": 35
}
],
"metadata": {}
}
]
}