Solution to 2008 Problem 69

We use phasor analysis. The impedance of the resistor is \mathbf{Z}_R= R. The impedance of the capacitor is \mathbf{Z}_C = 1/(i \omega C). The total impedance is
\begin{align*}\mathbf{Z} = \mathbf{Z}_R + \mathbf{Z}_C = R + \frac{1}{i \omega C}\end{align*}
The phasor input voltage is
\begin{align*}\mathbf{V}_i = V_i\end{align*}
Therefore, the steady state phasor current is
\begin{align*}\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} = \frac{V_i}{R + \frac{1}{i\omega C}}\end{align*}
Therefore, the phasor output voltage is
\begin{align*}\mathbf{V}_o = \mathbf{I} \mathbf{Z}_C = \frac{V_i}{R + \frac{1}{i\omega C}} \cdot \frac{1}{i \omega C} = \frac...
The magnitude of the phasor output voltage is therefore,
\begin{align*}V_o = \left|\mathbf{V}_o \right| = \frac{V_i}{\sqrt{1 + R^2 C^2 \omega^2}}\end{align*}
From this we can see that answer (D) is correct.

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