In [1]:

```
#Variable declaration
N = 100; # Number of turns
delta_phi = 10e-03; # Flux linked with coil, Wb
delta_t = 2e-03; # Time during which flux changes, s
#Calculations
e =((-N)*delta_phi)/delta_t; # Average induced emf, V
#Result
print "The average emf induced in the coil = %3d V"%e
```

In [2]:

```
#Variable declaration
N = 250; # Number of turns
delta_phi1 = 20e-03; # Flux linked with coil, Wb
delta_phi2 = -16e-03; # Flux linked with coil, Wb
delta_t1 = 0.05; # Time, s
delta_t2 = 0.01; # Time, s
#Calculations
e_1 =((-N)*delta_phi1)/delta_t1; # Average induced emf, V
e_2 =((-N)*delta_phi2)/delta_t2; # Average induced emf, V
#Results
print "Change in flux in first case = %4.2f weber"%delta_phi1
print "Emf induced in first case = %3d volts"%e_1
print "Change in flux in second case = %4.2f weber"%delta_phi2
print "Emf induced in second case = %3d volts"%e_2
```

In [3]:

```
#Variable declaration
e = 100; # Induced emf, V
#Calculations
# For simplification let (delta_phi)/(delta_t) = k
k = 0.1; # Rate of chage of flux linked with coil, Wb/s
# Since e =((-N)*delta_phi)/delta_t, soling for N
N = (e)/k; # Number of turns
#Result
print "The number of turns on the coil = %4d"%N
```

In [36]:

```
import math
#Variable declaration
v = 5; # Velocity, m^2
theta =(math.pi/3); # Angle, degrees
phi = 1.6e-03; # Flux, Wb
l = 0.1; # Length of pole face, m
d = 0.4; # Breadth of pole face, m
#Calculations
A = l*d; # Cross-sectional area of pole face, m^2
B = phi/ A; # Flux density, T
e =( B*l*v)*math.sin(theta); # Induced emf, V
#Result
print "The emf induced = %5.4f V"%e
```

In [37]:

```
import math
#Variable declaration
l = 0.15; # Effective length of conductor, m
v = 8; # Velocity, m^2
theta = 55; # Angle, degrees
e = 2.5; # Induced emf, V
#Calculations
# Since e = B*l*v*sin(theta), solving for B
B = e/(l*v*math.sin(theta*math.pi/180)); # Flux density, T
#Result
print "The density of the field = %5.3f Tesla"%B
```

In [38]:

```
import math
#Variable declaration
l = 2.2; # Effective length of conductor, m
B =38e-06; # Flux density, T
theta = (math.pi/2); # Angle, degrees
v = 800/36; # Velocity, m^2
#Calculations
e = B*l*v*math.sin(theta); # Induced emf, V
#Result
print "The emf induced in the axle = %4.2f mV"%(e/1e-03)
```

In [11]:

```
import math
#Variable declaration
l = 0.22; # Effective length of conductor, m
B = 0.35; # Flux density, T
I = 3; # Current, A
theta = (math.pi/2); # Angle, degrees
#Calculations
# Since the force exerted on the conductor placed in magnetic field is directly proportional to the flux density ,
#the value of current flowing through the conductor, and the length of conductor lying inside the field, therefore
F = B*I*l*math.sin(theta); # Force, N
#Result
print "The force exerted on the conductor = %5.3f N"%F
```

In [39]:

```
import math
#Variable declaration
phi = 2.5e-03; # Flux, Wb
l = 0.05; # Effective length of pole, m
d = 0.03; # Effective width of pole, m
F = 1.25; # Force exerted on conductor, N
A = l*d; # Cross-sectional area of pole face, m^2
B = phi/A; # Flux density, T
theta = (math.pi/2); # Angle, degrees
#Calculations
# Since F = B*I*l*sin(theta), solving for I
I = F/(B*l*math.sin(theta)); # Current in conductor, A
theta_2 = (math.pi/4); # New angle, degrees
F_2 = B*I*l*math.sin(theta_2); # Force exerted on conductor, N
#Results
print "The value of the current = %2d A"%I
print "The force exerted on conductor when placed at 45 degrees to the field = %5.3f newton"%F_2
```

In [13]:

```
#Variable declaration
l = 0.015; # Length of coil, m
d = 0.006; # Width of coil, m
B = 1.2; # Flux density, T
I = 1e-02; # Current, a
r = d/2; # Radius of rotation, m
#Calculations
# Since torque is given by the product of force and distance, therefore, we have
T = 2*B*I*l*r; # Torque, Nm
#Result
print "The torque exerted on the coil = %4.2f micro-Nm"%(T/1e-06)
```

In [14]:

```
#Variable declaration
N = 80; # Number of turns
l = 0.02; # Length of coil, m
r = 0.012; # Radius of coil, m
I = 45e-06; # Current in coil, A
T = 1.4e-06; # Torque exerted on coil, Nm
A = l*r; # Cross-sectional area of coil, m^2
#Calculations
# Since T = 2*B*I*l*r, solving for B
B = T/(2*A*N*I); # Flux density, T
#Result
print "The flux density produced by the pole pieces = %4.2f T"%B
```

In [15]:

```
#Variable declaration
d = 0.035; # Distance between two parallel conductors, m
I_1 = 50; # Electric current in first coil, A
I_2 = 40; # Electric current in second coil, A
#Calculations
F = ((2e-07)*I_1*I_2)/d; # Force exerted by conductors, N
#Result
print "The force exerted between the conductors = %4.1f mN"%(F/1e-03);
```

In [40]:

```
import math
#Variable declaration
d = 2; # Distance between two parallel conductors, m
I_1 = 1000; # Electric current in first coil, A
I_2 = 300; # Electric current in second coil, A
#Calculations
mew_o = 4*(math.pi)*1e-07; # Permeability for free space
B = (mew_o*I_1)/d; # Flux density due to first coil, T
F = ((2e-07)*I_1*I_2)/d; # Force exerted by conductors, N
#Result
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);
print "Force exerted by conductors = %2d mN"%(F/1e-03);
```

In [17]:

```
#Variable declaration
R_c = 40; # Resistance of coil, ohm
I_fsd = 5e-04; # Full-scale deflection current, A
I = 3; # Current reading, A
#Calculations
V_c = I_fsd*R_c; # Potential difference, V
# Since I = I_s + I_fsd, solving for I_s
I_s = I-I_fsd; # Shunt current, A
# From Ohm's law, V_c = I_s*R_s, solving for R_s
R_s = V_c/I_s; # Shunt resistance, ohm
#Result
print "The value of required shunt resistance = %4.2f milli-ohm"%(R_s/1e-03);
```

In [18]:

```
#Variable declaration
R_c = 40; # Resistance of coil, ohm
I_fsd = 5e-04; # Full-scale deflection current, A
I_fsd = 5e-04; # Full-scale deflection current, A
V = 10; # Voltage reading range, V
V_c = 0.02; # Potential difference across coil resistance, V
#Calculations
# From Ohm's law, V = I_fsd*R, solving for R
R = V/I_fsd; # Total resistance, ohm
# Since R = R_m + R_c, solving R_m
R_m = R - R_c; # Multiplier resistance, ohm
#Result
print "The required value of multiplier resistance = %5.2f kilo-ohms"%(R_m*1e-03);
```

In [19]:

```
#Variable declaration
R_c = 1500; # Coil resistance, ohm
I_fsd = 75e-06; # Full-scale deflection current, A
I = 5; # Current range, A
V = 10; # Voltage range, V
#Calculations
# Part (a)
# Using Ohm's law,
V_c = I_fsd*R_c; # Potential difference across coil resistance, V
# Since I = I_s + I_fsd, solving for I_s
I_s = I-I_fsd; # Shunt current, A
# From Ohm's law, V_c = I_s*R_s, solving for R_s
R_s = V_c/I_s; # Shunt resistance, ohm
# Part (b)
# Since = V = V_m + V_c, solving for V_m
V_m = V - V_c; # Potential difference across multiplier resistance, V
# From Ohm's law, V_m = I_fsd*R_m, solving for R_m
R_m = V_m/I_fsd # Multiplier resistance, ohm
#Results
print "The required value of shunt resistance = %4.1f mega-ohm"%(R_s/1e-03);
print "The required value of multiplier resistance = %4.1f mega-ohm"%(R_m*1e-03);
```

In [20]:

```
#Variable declaration
R_1 = 30.; # Resistance, ohm
R_2 = 70.; # Resistance, ohm
R_in = 200.; # Internal resistance of meter, ohm
V = 12.; # Supply voltage, V
#Calculations
# Using voltage divider rule, we have
V_2t = (R_2 /(R_1 + R_2))*V # True value of p.d across resistance R_2, V
# Since the resistances R_2 and R-in are parallel, so their equivalent resistance is given their parallel combination
R_BC = (R_2 * R_in)/(R_2 + R_in); # Resistance, ohms
# Using the potential divider technique,
V_2i = (R_BC / ( R_BC + R_1 ))*V # Indicated value of p.d across by voltmetre, volts
err = (( V_2i-V_2t ) / V_2t)*100 # Percentage error in the reading
#Results
print "The p.d. indicated by the meter = %3.1f V"%V_2i
print "The percentage error in the reading = %4.2f percent"%err
```

In [21]:

```
#Variable declaration
R_in = 200.; # Internal resistance of meter, kilo-ohms
V = 10.; # Supply voltage, volts
R_1 = 10.; # Resistance, kilo-ohms
R_2 = 47.; # Resistance, kilo-ohms
#Calculations
V_1 = R_1/(R_1+R_2)*V # P.d across resistance R_1, V
V_2 = R_2/(R_1+R_2)*V # P.d across resistance R_2, V
# Part (a)
R_AB = (R_1 * R_in)/(R_1 + R_in); # Resistance, kilo-ohms
V_AB = (R_AB / ( R_AB + R_2 ))*V # True value of p.d across by voltmetre, V
R_BC = (R_2 * R_in)/(R_2 + R_in); # Resistance, kilo-ohms
V_BC = (R_BC / ( R_BC + R_1 ))*V # Indicated value of p.d across by voltmetre, V
# Part (b)
# Error for V_1 measurement
error_AB = (V_AB - V_1)/V_1*100 # Percentage error in the reading
#Error for V_2 measurement
error_BC = (V_BC-V_2)/V_2*100 # Percentage error in the reading
#Results
print "The p.d. indicated by the meter across first resistor = %4.2f V"%V_AB
print "The p.d. indicated by the meter across second resistor = %4.2f V"%V_BC
print "Percentage error for V_1 measurement = %4.2f percent"%error_AB
print "Percentage error for V_2 measurement = %4.2f percent"%error_BC
```

In [23]:

```
#Variable declaration
L = 0.25; # Self-inductance, H
delta_I = 250e-03; # Change in current, A
delta_t = 25e-03; # Time, s
#Calculations
e = ((L)*delta_I)/(delta_t); # Induced emf, V
#Result
print "The value of emf induced = %3.1f V"%e
```

In [24]:

```
#Variable declaration
e = 30.; # Induced emf, V
#Calculations
# For simplicity, let rate of change of current i.e delta_I/delta_t = k
k = 200; # Rate of change of current, ampere-second
# Since e = ((-L)*delta_I)/(delta_t), solving for L
L = e/k; # Self-inductance, H
#Result
print "The inductance of the circuit = %4.2f H"%L
```

In [25]:

```
#Variable declaration
L = 50e-03; # Self-inductance, H
e = 8; # Induced emf, V
#Calculations
# Since e = ((-L)*delta_I)/(delta_t), solving for delta_I/delta_t,and for simplicity letting the rate of change of
#current i.e delta_I/delta_t = k
k = e/L; # Rate of change of current, As
#Result
print "The rate of change of current = %3d A/s"%k
```

In [26]:

```
#Variable declaration
N = 150; # Number of turns in a coil
I = 10; # Electric current flowing through coil, A
phi = 0.10; # Flux, Wb
delta_t = 0.1; # Time, s
#Calculations
# Part (a)
L = (N * phi)/I # Self-inductance, H
delta_I = 20; # Change in current, A
# Part (b)
e = abs((-L*delta_I)/(delta_t)); # Induced emf, V
#Results
print "The inductance of the coi = %3.1f H"%L
print "The emf induced in the coil = %2d V"%e
```

In [27]:

```
#Variable declaration
I_1 = 8; # Electric current, A
I_2 = 2; # Electric current, A
N = 3000; # Number of turns in a coil
phi_1 = 4e-03; # Flux, Wb
delta_t = 0.1; # Reversal time of current, s
#Calculations
L = (N * phi_1)/I_1; # Self-inductance, H
delta_I = I_1 - I_2; # Change in current, A
e = ((L)*delta_I)/(delta_t); # Induced emf, V
#Result
print "The emf induced in the coil = %2d volts"%e
```

In [41]:

```
import math
#Variable declaration
N_1 = 600; # Number of turns in a coil in first case
N_2 = 900; # Number of turns in a coil in second case
N_3 = 900; # Number of turns in a coil in third case
l = 45e-03; # Effective length of coil, m
A = 4e-04; # Cross-sectional area of coil, m^2
mew_o = 4*(math.pi)*1e-07; # Permeability for free space
mew_r1 = 1; # Relative permeability in first case
mew_r2 = 1; # Relative permeability in second case
#Calculations
# Part (a)
mew_r3 = 75; # Relative permeability in third case
L_1 = (mew_o*mew_r1*(N_1**2)*A)/l; # Self-inductance of coil in first case, H
# Part (b)
# Since self-inductance of a coil is directly proportional to the number of turns in a coil, therefore,
#we have L_2/L_1 = (N_2^2)/(N_1^2), solving for L_2
L_2 = (L_1*(N_2**2))/(N_1**2); # Self-inductance of coil in second case, H
# Part (c)
# Since mew_r3 = 75*mew_r2, keeping all other quantities same we have
L_3 = mew_r3*L_2; # Self-inductance of coil in third case, H
#Results
print "Self-inductance of coil in first case = %4.2f mH"%(L_1/1e-03);
print "Self-inductance of coil in second case = %5.3f mH"%(L_2/1e-03);
print "Self-inductance of coil in third case = %5.3f H"%L_3
```

In [29]:

```
#Variable declaration
N_A = 2000; # Number of turns in a coil A
N_B = 1500; # Number of turns in a coil B
I_A = 0.5; # Electric current in coil A, A
phi_A = 60e-06; # Flux linked with coil A, Wb
#Calculations
# Part (a)
L_A = (N_A*phi_A)/I_A; # Self-inductance of coil A
phi_B = 0.83*(60e-06); # Flux linked with coil B, Wb
# Part (b)
M = (N_B*phi_B)/I_A; # Mutual inductance of the two coils, H
#Results
print "Self-inductance of coil A = %4.2f H"%L_A
print "Mutual inductance of the two coils = %5.3f H"%M
```

In [42]:

```
import math
#Variable declaration
N = 400; # Number of turns in a coil
l = 0.25; # Effective length of coil, m
A = 4.5e-04; # Cross-sectional area, m^2
mew_r = 180; # Relative permeability
#Calculations
mew_o = 4*(math.pi)*1e-07; # Pemeability for free space
L = (mew_o*mew_r*(N**2)*A)/l # Self-inductance of coil, H
#Result
print "The self inductance of the coil = %2d milli-henry"%(L/1e-03);
```

In [43]:

```
import math
#Variable declaration
L_1 = 65e-03; # Self-inductance of first coil, H
delta_I = 1.5; # Change in current, A
delta_t = 3e-03; # Time, s
k = 0.95; # 95 percent of flux produced
N_1 = 400; # Number of turns in a coil A
N_2 = 650; # Number of turns in a coil B
#Calculations
# Part (a)
# Since self-inductance of a coil is directly proportional to the number of turns in a coil, therefore,
#we have L_2/L_1 = (N_2^2)/(N_1^2), solving for L_2
L_2 = (L_1*(N_2**2))/(N_1**2) # Self-inductance of second coil , H
# Part (b)
M = k*math.sqrt(L_1*L_2); # Mutual inductance of two coils, H
# Part (c)
e_1 = ((L_1)*delta_I)/(delta_t); # Induced emf in first coil, V
# Part (d)
e_2 = (M*delta_I)/delta_t; # Induced emf in second coil, V
#Results
print "The self-inductance of coil 2 = %3d mH"%(L_2/1e-03)
print "The value of mutual inductance = %3d mH"%(M/1e-03)
print "The self-induced emf in coil 1 = %4.1f V"%e_1
print "The mutually induced emf in coil 2 = %2d V"%e_2
```

In [32]:

```
#Variable declaration
L = 50e-03; # Self-inductance of coil, H
I = 0.75; # Electric current in coil, A
#Calculations
W = (L*(I**2))/2 # Energy stored, J
#Result
print "Energy stored in the inductor = %4.1f mJ"%(W/1e-03)
```

In [44]:

```
import math
#Variable declaration
L_1 = 25e-03; # Self-inductance of first coil, H
L_2 = 40e-03; # Self-inductance of second coil, H
I = 0.25; # Electric current in coils, A
k =0.8; # Coupling coefficient
#Calculations
# Part (a)
W_1 = (L_1*(I**2))/2; # Energy stored in first coil, J
W_2 = (L_2*(I**2))/2; # Energy stored in second coil, J
M = k*math.sqrt(L_1*L_2); # Mutual inductance of coils
# Part (b)
W_M = M*(I)*(I); # Energy stored due to mutual inductance of coils, J
W_sa = W_1 + W_2 + W_M; # Energy stored by two inductors when connected in series aiding, J
W_so = W_1 + W_2 - W_M; # Energy stored by two inductors when connected in series opposition, J
#Results
print "Energy stored in first coil = %4.2f mJ"%(W_1/1e-03)
print "Energy stored in second coil = %4.2f mJ"%(W_2/1e-03)
print "Energy stored by two inductors when connected in series aiding = %3.1f mJ"%(W_sa/1e-03)
print "Energy stored by two inductors when connected in series opposition = %4.2f mJ"%(W_so/1e-03)
```

In [34]:

```
#Variable declaration
V_2 = 60; # Output voltage, V
V_1 = 240; # Input voltage, V
N_2 = 500; # Secondary turns
#Calculations
# Part (a)
# For simplicity let V_1/V_2 = N_1/N_2 = k
k = V_1/V_2 # Turns ratio
# Part (b)
# Since V_1/V_2 = N_1/N_2, solving for N_1
N_1 = k*N_2; # Primary turns
#Results
print "The required turns ratio = %1d:1"%k
print "The number of primary turns = %4d"%N_1
```

In [35]:

```
#Variable declaration
R_L = 15; # Load resistor, ohms
V_2 = 240.; # Terminal p.d at secondary, V
V_1 = 600; # Supply voltage, V
#Calculations
# Part (a)
# Since V_1/V_2 = N_1/N_2 = k
k = V_1/V_2; # Turns ratio
# Part (b)
I_2 = V_2/R_L; # Current drawn by the load, A
P_2 = V_2*I_2; # Power drawn by the load, W
# Part (c)
I_1 = P_2/V_1 # Current drawn from the supply, A
#Results
print "The transformer turns ratio = %3.1f:1"%k
print "The current drawn by the load = %2d A"%I_2
print "The power drawn by the load = %4.2f W"%(P_2*1e-03);
print "The current drawn from the supply = %3.1f A"%I_1
```