In the circuit shown below, the frequency of the voltage source $$v_s$$ can be varied over a broad range of frequencies while the amplitude of its voltage is kept constant. The frequency (kHz) at which $$v_o$$, has maximum amplitude is most nearly:

A.   1.59
B.   3.18
C.   15.9
D.   31.8

The frequency at which the output voltage will have maximum amplitude is at the resonant frequency. At the resonant frequency, the imaginary parts of our total impedance will effectively cancel out. Since the capacitor and inductor are the only circuit components that have imaginary parts to their total impedance, these are the only items we will be concerned about:

$At resonant frequency: X_L=X_C$

The reactance of an inductor and capacitor are:

$$\begin{gathered} X_L=2\mathrm{\pi fL} \\ {\mathrm{X}}_{\mathrm{C}}=\frac{1}{2\mathrm{\pi fC}} \end{gathered}$$

After plugging them into the equation, we can solve for the frequency where this equations holds true. This will be our resonant frequency:

$$\begin{gathered} X_L=X_C \\ 2\mathrm{\pi fL} = \frac{1}{2\mathrm{\pi fC}} \\ \left(2\mathrm{\pi fL}\right)\left(2\mathrm{\pi fC}\right) = 1 \\ 4{\mathrm{\pi }}^2{\mathrm{f}}^2\mathrm{LC}=1 \\ {\mathrm{f}}^2=\frac{1}{4{\mathrm{\pi }}^2\mathrm{LC}} \\ \mathrm{f}=\sqrt{\frac{1}{4{\mathrm{\pi }}^2\mathrm{LC}}} \end{gathered}$$

Now plugging in the capacitance and inductance values we are given within our circuit:

$$\begin{gathered} \mathrm{f}=\sqrt{\frac{1}{4{\mathrm{\pi }}^2\left(5\mathrm{mH}\right(0.5\mathrm{uF})}} \\ f = 3183.10Hz \end{gathered}$$

Therefore, the frequency at which our amplitude will be maximum is most nearly 3.18kHz, answer B.