Transformer Exercise (Part 3)

Very nice problem.

The basic $8$ equations are: $E_{1}=(0+j4)I_{1}+(0+j2)I_{2}$ $E_{2}= (0+j2)I_{1}+(0+j4)I_{2}$ $E_{3}= (0+j6)I_{3}+(0+j2)I_{4}$ $E_{4}=(0+j2)I_{3}+(0+j6)I_{4}$ $E_{S1}-I_{1}Z_{1}=E_{1}$ $I_{2}=-I_{3}$ $E_{2}=E_{3}$ $E_{S2}-I_{4}Z_{2}=E_{4}$
Solving this 8 simple equations gives $I_{1}=\frac{5083}{5654}-j\frac{5795}{5654}$ $I_{2}=\frac{-617}{5654}-j\frac{33}{514}$ $I_{3}=\frac{617}{5654}+j\frac{33}{514}$ $I_{4}=\frac{999}{2827}-j\frac{3805}{2827}$ $E_{1}=\frac{11953}{2827}+j\frac{9549}{2827}$ $E_{2}=\frac{6521}{2827}+j\frac{3849}{2827}$ $E_{3}=\frac{6521}{2827}+j\frac{3849}{2827}$ $E_{4}=\frac{22467}{2827}+j\frac{601}{257}$
Now the read the $7^{th}$ and $8^{th}$ line of the problem carefully.
To calculate active power , we only calculate the real part of $P_{active}=E_{1}\bar{I_{1}}$
Where the bar line over the $\bar{I_{1}}$ tells the complex conjugate of $I_{1}$
Suppose we take take a complex number $N=x+jy$
Then it's complex conjugate is $\bar{N}=x-jy$
Now come to the problem $P_{active}=(\frac{11953}{2827}+j\frac{9549}{2827})(\frac{5083}{5654}+j\frac{5795}{5654})$ $P_{active} \approx 0.3391324+j7.3702608$
Our job is to take real part only $\Large \boxed{\textcolor{#3D99F6}{P_{active} \approx 0.3391324}}$

Great problem as always. Thanks for posting. The circuit equations as per Kirchoff's voltage law are:

$-E_{S1} + I_1Z_1 + E_1 = 0$ $-E_2 + E_3 = 0$ $-E_4 -I_4Z_2 + E_{S2}=0$

According to Kirchoff's current law for the second loop:

$I_2 + I_3 = 0$

Rearranging these four equations in a matrix form:

$\left[\begin{matrix}Z_1+Z_{S1}&Z_{M1}&0&0\\-Z_{M1}&-Z_{S1}&Z_{S2}&Z_{M2}\\0&0&Z_{M2}&Z_{S2}+Z_2\\0&1&1&0\end{matrix}\right]\left[\begin{matrix}I_1\\I_2\\I_3\\I_4\end{matrix}\right] = \left[\begin{matrix}E_{S1}\\0\\E_{S2}\\0\end{matrix}\right]$ $\implies \left[\begin{matrix}I_1\\I_2\\I_3\\I_4\end{matrix}\right] = \left[\begin{matrix}Z_1+Z_{S1}&Z_{M1}&0&0\\-Z_{M1}&-Z_{S1}&Z_{S2}&Z_{M2}\\0&0&Z_{M2}&Z_{S2}+Z_2\\0&1&1&0\end{matrix}\right]^{-1}\left[\begin{matrix}E_{S1}\\0\\E_{S2}\\0\end{matrix}\right]$

The active power is the real part of the product of the current through the terminal and the terminal voltage:

$P_{a} = \mathrm{real}\left(E_1I_1^*\right)$

Here, $I_1^*$ is the complex conjugate of current. The required answer evaluates to: $\boxed{P_a \approx 0.3391}$