Substitution into $\epsilon=A_0U^2/(N\omega)$ gives
$$
\boxed{\epsilon=\frac{A_0B^2D_0}
{[(A+A_0)(D+D_0)-B^2][(A+A_0)D-B^2]}.}
$$
When $B^2\ll AD$ and $D\ll D_0$,
$$
\boxed{\epsilon\simeq\frac{A_0B^2}{D(A+A_0)^2}.}
$$
After scaling body size so that $A_0\mapsto\lambda A_0$, differentiation with respect to $\lambda A_0$ shows that the optimum impedance match is $\lambda A_0=A$. Therefore
$$
\boxed{\lambda_{\max}=\frac A{A_0},
\qquad \epsilon_{\max}=\frac{B^2}{4AD}.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 355 / 2 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-355/2/e/solution