Question 15.11: Consider the circuit shown in Figure 15.26(a). Find the phas...

First, we reflect the load impedance Z_L to the primary side of the transformer as shown in Figure 15.26(b). The impedance seen from the primary side is
Z^′_L = \left(\frac{N_1} {N_2} \right) ^2 Z_L = (10)^2(10 + j20) = 1000 + j2000
The total impedance seen by the source is
Z_s = R_1 + Z^′_L = 1000 + 1000 + j2000 = 2000 + j2000
Converting to polar form, we have
Z_s = 2828∠45°

Now, we can compute the primary current and voltage:
\textbf{I}_1 = \frac{\textbf{V}_s} {Z_s} = \frac{1000∠0°}{ 2828∠45°} = 0.3536∠−45° A peak
\textbf{V}_1 = \textbf{I}_1Z^′_L = 0.3536∠−45° × (1000 + j2000)
= 0.3536∠−45° × (2236∠63.43°) = 790.6∠18.43° V peak
Next, we can use the turns ratio to compute the secondary current and voltage:
\textbf{I}_2 = \frac{N_2} {N_1}\textbf{I}_1 =\frac{10}{1}0.3536∠−45° = 3.536∠−45° A peak
\textbf{V}_2 = \frac{N_2} {N_1}\textbf{V}_1 = \frac{1}{10} 790.6∠18.43° = 79.06∠18.43° V peak
Finally, we compute the power delivered to the load:
P_L = I^2_{2rms}R_L = \left(\frac{3.536}{ \sqrt{2}}\right) ^2 (10) = 62.51 W