We can use phasor analysis to determine which of the circuits is a high-pass filter. A high-pass filter is defined as a circuit where the magnitude of the output signal and the magnitude of the input signal are identical when the input frequency goes to $\infty$ , and the magnitude of the output signal is $0$ when input frequency goes to $0$ .

The voltage gain (which means the phasor input voltage divided by the phasor output voltage) for the four circuits are

$\begin{align*}G_I &= \frac{R}{j \omega L + R} \\G_{II} &= \frac{j \omega L}{j \omega L + R} \\G_{III} &= \frac{R}...$
For a high-pass filter, the magnitude of $\omega$ should go to $1$ as $\omega \to \infty$ and to $0$ as $\omega \to 0$ . This is only true for $G_{II}$ and $G_{III}$ . The other two circuits are low-pass filters. Therefore, answer (D) is correct.

Solution to 2001 Problem 39