Chapter 15 : Mutual Inductance and Transformers

Example 15.4 Page No : 306

In [1]:
import math 
#Example 15.4")

# Given
#L1 = 0.1H L2 = 0.2H")
#i1 = 4A i2 = 10A")
L1 = 0.1;L2 = 0.2
i1 = 4;i2 = 10;
#The energy stored in coupled coils is
#W = (L1*i1**2)/2+(L2*i2**2)/2+M*i1*i2")

#a)")
M = 0.1;
W = (L1*i1**2)/2+(L2*i2**2)/2+M*i1*i2;
print "Total Energy in the coils = %3.2fJ"%(W);

#b)")
M = math.sqrt(2)/10;
W = (L1*i1**2)/2+(L2*i2**2)/2+M*i1*i2;
print "Total Energy in the coils = %3.2fJ"%(W);

#c)")
M = -0.1;
W = (L1*i1**2)/2+(L2*i2**2)/2+M*i1*i2;
print "Total Energy in the coils = %3.2fJ"%(W);

#a)")
M = -math.sqrt(2)/10;
W = (L1*i1**2)/2+(L2*i2**2)/2+M*i1*i2;
print "Total Energy in the coils = %3.2fJ"%(W);
Total Energy in the coils = 14.80J
Total Energy in the coils = 16.46J
Total Energy in the coils = 6.80J
Total Energy in the coils = 5.14J

Example 15.7 Page No : 311

In [4]:
import math 
from scipy.linalg import polar

#Example 15.7")

# Given
#N1 = 20 N2 = N3 = 10")
#I2 = 10(-53.13 deg) I3 = 10(-45 deg)")
N1 = 20;
N2 = 10;
N3 = 10;
I2mag = 10;
I2ph = -53.13;
I3mag = 10;
I3ph = -45;
#From figure 15.14
#N1*I1-N2*I2-N3*I3 = 0")
#Solving for I1
Xmag = N2*I2mag 
Xph = I2ph
x = Xmag*math.cos((Xph*math.pi)/180);
y = Xmag*math.sin((Xph*math.pi)/180);
z = complex(x,y)

Ymag = N3*I3mag 
Yph = I3ph
x1 = Ymag*math.cos((Yph*math.pi)/180);
y1 = Ymag*math.sin((Yph*math.pi)/180);
z1 = complex(x1,y1)

I1 = (z+z1)/N1
R,Theta = polar([[I1]]);
R = R[0][0].real
Theta = Theta[0][0].real

print "I1 = %3.2f%3.2f deg) A"%(R,(Theta*180)/math.pi);
I1 = 0.66571.52 deg) A

Example 15.8 Page No : 316

In [6]:
import math 
from scipy.linalg import polar
#Example 15.8")

# Given
#L1 = 0.2H L2 = 0.1H")
#M = 0.1H R = 10ohm")
#v1 = 142.3*math.sin(100*t)")
L1 = 0.2;L2 = 0.1
M = 0.1;R = 10;
v1mag = 142.3;
w = 100;
#Let Input impedance be Z1 and can be calculated as
#From the equations in 15.10
#Z1 = 1j*w*L1+((M*w)**2)/(Z2+1j*w*L2)")
Z1 = 1j*w*L1+((M*w)**2)/(R+1j*w*L2)
R,Theta = polar([[Z1]])
R = R[0][0].real
Theta = Theta[0][0].real

#If I1 is the input current
I1mag = v1mag/R
I1ph = -(Theta*180)/math.pi
#In time domain form
print "i1 = %3.1f*math.sin%d*t%3.1f deg) A)"%(I1mag,w,I1ph);
i1 = 450.0*math.sin100*t-905.9 deg) A)

Example 15.9 Page No : 318

In [8]:
import math 
from sympy import Symbol

s = Symbol('s')
# Given
#L1 = 0.2H L2 = 0.1H")
#M = 0.1H R = 10ohm")
#v1 = u(t) a unit step function")
L1 = 0.2;
L2 = 0.1
M = 0.1;
R = 10;
v1 = 1;
w = 100;
#Let Input impedance be Z1 and can be calculated as
#From the equations in 15.10
#Z1(s) = L1*s-((M*s)**2)/(R+L2*s)")
Z1 = L1*s-(((M*s)**2)/(R+L2*s))
#Proper rearranging of co-efficients
Num = Z1/0.01
Den = Z1*100

print "Z1(s)",Num/Den
Y1 = 1./Z1
print "Y1(s)",Den/Num

#As the input is unit step function the value is 1V for t>0
#In exponential form the value is represented as exp(s*t) with s = 0 as the pole of Y1(s)

#Therefore forced response
k = 1/L1;
print "Forced response i1,f = %d*t) A)"%(k);
Z1(s) 1
Y1(s) 1
Forced response i1,f = 5*t) A)