Find the value of load impedance $Z_L$,

Solution:

$ \text { Part-1 - TO find " } V_{T H} \text { " } $

$ \begin{aligned} &V_{T H}=50 \mid 45-3 I \quad \text { OR } V_{T H}=(2+j 10) I\\\\ &\text { KVL to loop }\\\\ &50 \lfloor 45-3 I-2 I-j 10 I=0\\\\ &-(5+j 10) I=-50 \lfloor 45\\\\ &I=\frac{50 \lfloor 45}{(5+j 10)}=4.472 \frac{-18.434}{\text { Amp }}\\\\ &\text { from ea- (1) }\\\\ &V_{T H}=50\lfloor 45-3(4.472\lfloor-18.434)\\\\ &V_{T H}=45.60860 .25 \text { volts } \\ \end{aligned} \\ $

$ \text { Part-1 - TO find " } Z_{T H} \text { " } $

$ \begin{aligned} Z_{A B}=Z_{T H} &=(2+j 10) \|(3) \\\\ &=\frac{(2+j 10)(3)}{(2+j 10)+3}=\frac{6+j 30}{5+j 10} \\\\ Z_{T H} &=2.64+j 0.72=2.736 \lfloor 15.25 \Omega \end{aligned} \\ $

According to the maximum power transfer theorem

$ \begin{aligned} \\ Z_L=Z_{T H}^* &=2.64-j 0.72 \\\\ &=2.736 \lfloor{-15.25} \Omega \end{aligned} \\ $

And Maximum power is given by

$ \begin{aligned} \\ &P_{\text {max }}=\frac{\left|V_{T H}^2\right|}{4 R_{T H}}=\frac{(45.608)^2}{4(2.64)} \\\\ &P_{\text {max }}=196.97 \text { Walts } \\ \end{aligned} \\ $