Question 1.SGPYQ.6: Two AC sources feed a common variable resistive load as show......

For maximum power transfer,

$R_{ L }=\left|Z_{ Th }\right| \quad \text { and } \quad P=I^2 R_{ L }$

To find the Thevenin impedance, short circuit the voltage source

$\begin{aligned} Z_{ Th } & =(6+8 j \Omega)||(6+8 j \Omega) \\ & =\frac{(6+8 j \Omega) \times(6+8 j \Omega)}{(6+8 j \Omega)+(6+8 j  \Omega)} \\ & =3+4 j  \Omega \end{aligned}$

Therefore, $R_{ L }=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5  \Omega$

To find the Thevenin’s voltage, open the load:

By nodal analysis method,

$\begin{gathered} \frac{V_{ Th }-110 \angle 0^{\circ}}{6+8 j}+\frac{V_{ Th }-90 \angle 0^{\circ}}{6+8 j}=0 \\ V_{ Th }-100 \angle 0^{\circ}  V \end{gathered}$

Therefore,

$\begin{aligned} P & =\frac{V_{ Th }}{R_{ Th }+R_{ L }} \cdot R_{ L }=\left|\frac{100}{(3+4 j)+5}\right|^2 \times 5 \\ & =\frac{(100)^2 \times 5}{80}=625  W \end{aligned}$