Solution to 2001 Problem 38

The phasor voltage is
\begin{align*}\mathbf{V} = \varepsilon_m e^{j \theta}\end{align*}
where \varepsilon_m = 40 \mbox{ V}. Therefore, the steady state phasor current is
\begin{align*}\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} = \frac{\varepsilon_m  e^{j \theta}}{R - \frac{j}{\omega C} + j \ome...
The magnitude of the phasor current (which equals the maximum amplitude of 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 (1), we see that the amplitude of the current is maximized when
\begin{align*}\frac{-1}{\omega C} + \omega L = 0 \Rightarrow C = \frac{1}{\omega^2 L} = \frac{1}{\left(1000 \mbox{ rad/s}\rig...
Therefore, answer (D) is correct.

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