Question 6.1: Obtain (6.2) and (6.3)....
Obtain (6.2) and (6.3).
Z_L=Z_G^*. \hspace{30 pt} \text{(6.2)}
P_{ \text{av} }=\frac{\left|V_0\right|^2}{4 R_G}. \hspace{30 pt} \text{(6.3)}
The maximum of P_L vs. R_L and X_L corresponds to a zero of the partial derivatives of P_L vs. the two variables, i.e., to conditions:
\left\{\begin{array}{l} \frac{\partial P_L}{\partial R_L}=\left|V_0\right|^2 \frac{\left|Z_G+Z_L\right|^2-2 R_L\left(R_L+R_G\right)}{\left|Z_G+Z_L\right|^4}=0\\ \\ \frac{\partial P_L}{\partial X_L}=-\left|V_0\right|^2 \frac{2 R_L\left(X_L+X_G\right)}{\left|Z_G+Z_L\right|^4}=0 ; \end{array}\right.
from the second equation we find X_L=-X_G; substituting in the first we obtain R_L=R_G and therefore the load corresponding to the maximum power transfer from a generator with internal impedance Z_G is Z_L=Z_G^*. (The value R_L=R_G clearly corresponds to a maximum since the P_L vanishes for R = 0 and for R_L \rightarrow \infty.) Substituting, we find the generator maximum or available power in (6.3). A similar procedure holds for a generator in its parallel (Norton) form, leading to (6.4).
P_{ \text{av} }=\frac{\left|I_0\right|^2}{4 G_G}, \hspace{30 pt} \text{(6.4)}