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$$
\begin{array}{l}{\text { a) For the circuit shown determine the impedance } \mathrm{Z_L}} \\ {\text { that result in maximum average power transferred to } \mathrm{Z_L} \text { . }} \\ {\text { b) What is the maximum average power transferred }} \\ {\text { to the load impedance determined in (a)? }}\end{array}
$$

$$
R_{e q}=\frac{R_{1} R_{2}}{R_{1}+ R_{2}}=\frac{20 *5}{20+5}=4 \Omega
$$

$$
Z_{e q}=\frac{Z_{1} * Z_{2}}{Z_{1}+Z_{2}}=\frac{(4*3J)(-6J)}{4+3J-6J}=5.76-1.68J
$$

$$
Z_{1}=Z_{t h}^{*}
$$

$$
Z_{L}=5.76+J{1.68}
$$

$$
R_{eq}=\frac{5*20}{20+5}=4 \Omega
$$

$$
V_{th}=\frac{Z_{1} *V}{Z_{1}+Z_{2}}
$$

$$
V_{th}=\frac{-6J * 16}{4+3J-6J}
$$

$$
V_{th}=11.52-J{15.36}) {v}
$$

$$
P_{max}=\frac{V_{th}|_{rms}^{2}}{4 R_L}=\frac{\left(\frac{19.2}{\sqrt{2}}\right)^{2}}{4*5.76}=8w
$$

$$
R_{L}=5.76 \Omega
$$

$$
V_{th}|_{rms}=\frac{V_{th}|_{max}}{\sqrt{2}}=\frac{19.2}{\sqrt{2}}
$$

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