## Solution to 1992 Problem 81

 We use phasor analysis.\begin{align*}\mathbf{Z}_R &= R \\\mathbf{Z}_C &= \frac{1}{j \omega C} \\\mathbf{Z}_L &= j \omega L \\\end{align*...The total impedance is then\begin{align*}\mathbf{Z} = \mathbf{Z}_R + \mathbf{Z}_C + \mathbf{Z}_L = R - \frac{j}{\omega C} + j \omega L\end{align*}The phasor voltage is \begin{align*}\mathbf{V} = \varepsilon_m\end{align*}Therefore, the steady state phasor current is\begin{align*}\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} = \frac{\varepsilon_m}{R - \frac{j}{\omega C} + j \omega L}\end{alig...The magnitude of the phasor current (which equals the amplitude of the sinusoidal current) is\begin{align}\left|\mathbf{I} \right| = \frac{\varepsilon_m}{\displaystyle \sqrt{R^2 + \left(\frac{-1}{\omega C} + \omega L \...From equation (\ref{eqn81:1}), we see that the amplitude of the current is maximized when \begin{align*}\frac{-1}{\omega C} + \omega L = 0 \Rightarrow \omega = \boxed{\frac{1}{\sqrt{C L}}}\end{align*}Therefore, answer (C) is correct.