Answer 8.3

_psets/6.md

@@ -49,11 +49,31 @@ E_\text{max} &= \left( 2 \sqrt{\frac{\mu_0}{\epsilon_0}} \langle P \rangle \righ
 \end{align*}
 $$
 
+
 ## (8.3)
 
 {:.question}
 For what value of $$R_\text{load}$$ is the maximum power delivered to the load in Figure 8.3?
 
+By Ohm's Law $$V = I (R_r + R_l)$$ (where $$R_r$$ is the radiation resistance, and $$R_l$$ the load
+resistance). Thus $$I = V / (R_r + R_l)$$ and the power dissipated in the load resistance is
+
+$$
+P = I^2 R = \frac{V^2 R_l}{(R_r + R_l)^2}
+$$
+
+This is maximized where its derivative is zero.
+
+$$
+\frac{d P}{d R_l}  V^2 ( (R_r + R_l)^{-2} - 2 R_l (R_r + R_l)^{-3} )
+$$
+
+Setting this to zero implies $$R_r + R_l = 2 R_l$$, or
+
+$$
+R_l = R_r
+$$
+
 
 ## (8.4)