Transformer Exercise (Part 2)

@Neeraj Anand Badgujar has shown how the solution works. I will post some Python code which uses the linear system solver to calculate the currents. I will re-iterate the equations for the sake of completeness.

Basic equations

$E_1 = Z_S I_1 + Z_M I_2 \\ E_2 = Z_M I_1 + Z_S I_2$

Substituting in for $E_1$ and $E_2$ :

$E_S - Z_1 I_1= Z_S I_1 + Z_M I_2 \\ -Z_2 I_2 = Z_M I_1 + Z_S I_2$

Consolidating terms:

$E_S = (Z_S + Z_1) I_1 + Z_M I_2 \\ 0 = Z_M I_1 + (Z_S + Z_2) I_2$

Solve the linear system for the currents. Then calculate the terminal voltages.

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import math
import numpy as np

#############################################################

# System parameters

ES = complex(0.0,10.0)

Z1 = complex(2.0,5.0)
Z2 = complex(5.0,5.0)

ZS = complex(0.0,10.0)
ZM = complex(0.0,5.0)

#############################################################

# Set up linear system
# Solve for currents

M11 = ZS + Z1
M12 = ZM

M21 = ZM
M22 = ZS + Z2

M =  np.array([[M11,M12],[M21,M22]])
vec = np.array([ES,0.0])

Sol = np.linalg.solve(M, vec)

I1 = Sol[0]
I2 = Sol[1]

#############################################################

# Calculate terminal voltages 
# Check basic equations

E1 = ES - Z1*I1
E2 = -I2*Z2

c1 = E1 - ZS*I1 - ZM*I2
c2 = E2 - ZM*I1 - ZS*I2

print abs(c1)
print abs(c2)
print ""

#9.155133597044475e-16
#4.440892098500626e-16

#############################################################

# Print E2 magnitude

print abs(E2)

#1.6286558549611407

Referring to the system of equations derived by @Steven Chase , the final equation to compute the absolute value of $E_2$ can be re-arranged in a matrix form to be:

$\lvert E_2 \rvert = \mathrm{mod}\left(\left[\begin{matrix}Z_M&Z_S\end{matrix}\right]\left[\begin{matrix}Z_1+Z_S&Z_M\\Z_M&Z_2+Z_S\end{matrix}\right]^{-1}\left[\begin{matrix}E_S\\0\end{matrix}\right] \right)$

This is effectively only one line of MATLAB code, excluding initialisations.

1

E2_abs = abs([Zm ZS]*inv([Z1+ZS Zm;Zm ZS+Z2])*[ES;0])

Nice problem.
The basic equations are $E_{1}=Z_{s}I_{1}+Z_{M}I_{2}$ $E_{1}=j(10I_{1}+5I_{2})$ $E_{2}=Z_{M}I_{1}+Z_{s}I_{2}$ $E_{2}=j(5I_{1}+10I_{2})$
Using Kirchoff's voltage rule $E_{s}-I_{1}Z_{1}=E_{1}$ $E_{2}=-I_{2}Z_{2}$
Solving the above $2$ equation for $I_{1}$ and $I_{2}$ gives $I_{2}=\frac{76}{377}+\frac{42}{377}j$
$E_{2}=-I_{2}Z_{2}$
Substituting the values of $Z_{2}$ and $I_{2}$ in the above equation, and taking it's modulus gives $\boxed{|E_{2}| \approx\textcolor{#20A900}{ 1.628}}$