refer to the circuit below for items 39 and 40. the transformer is ideal. $dot{v}_s = 24angle0^{circ}v$ $z_a…

Answer:

39. B. 3.46 + j0.69 VA
40. D. 0.90 A

Explanation:

Step1: Find the reflected impedance of $Z_B$

The turns - ratio $n = 2$. The reflected impedance $Z_{ref}$ of $Z_B$ to the primary side is given by $Z_{ref}=n^{2}Z_{B}$. $Z_{ref}=4(10 + j2)=40 + j8\Omega$.

Step2: Calculate the total impedance in the primary circuit

The total impedance $Z_{total}=Z_A+Z_{ref}=(16 + j4)+(40 + j8)=56 + j12\Omega$.

Step3: Find the primary current $\bar{I}_1$

The source voltage $\bar{V}_s = 24\angle0^{\circ}V$. Using Ohm's law $\bar{I}_1=\frac{\bar{V}s}{Z{total}}=\frac{24}{56 + j12}$. Rationalize the denominator: $\bar{I}_1=\frac{24(56 - j12)}{(56 + j12)(56 - j12)}=\frac{1344 - j288}{56^{2}+12^{2}}=\frac{1344 - j288}{3136 + 144}=\frac{1344 - j288}{3280}=0.41 + j(- 0.088)\approx0.90A$ (magnitude).

Step4: Find the secondary current $\bar{I}_2$

Since $\frac{\bar{I}_1}{\bar{I}_2}=n$, $\bar{I}_2=\frac{\bar{I}_1}{n}$. And the voltage across $Z_B$ is $\bar{V}_B=\bar{I}_2Z_B$. First, from $\bar{I}_1 = 0.90A$, $\bar{I}_2=\frac{0.90}{2}=0.45A$. $\bar{V}_B=\bar{I}_2Z_B=0.45(10 + j2)=4.5 + j0.9V$.

Step5: Calculate the complex power absorbed by $Z_B$

The complex power $S=\bar{V}_B\bar{I}_2^{}$. $\bar{I}_2 = 0.45 + j0A$, $\bar{I}_2^{}=0.45 - j0A$. $S=(4.5 + j0.9)(0.45 - j0)=3.46 + j0.69VA$.