A series capacitor–resistor network with the output taken across the resistor is driven by an AC signal. What filter action results?

Introduction:
Simple RC networks can implement first-order filters. Whether the network is low-pass or high-pass depends on which component's voltage is used as the output. This question checks understanding of a series C–R with the output across the resistor.

Given Data / Assumptions:

• Input: AC signal applied in series to capacitor and resistor.
• Output: Voltage across the resistor (not across the capacitor).
• Linear, time-invariant components and small-signal conditions.

Concept / Approach:
In a series C–R network, the capacitor's impedance is Xc = 1 / (2 * pi * f * C). At low frequencies, Xc is large, most input drops across the capacitor, and the resistor sees little voltage. At high frequencies, Xc becomes small, more voltage appears across R, so the output increases with frequency. Therefore, the configuration acts as a high-pass filter with cutoff fc = 1 / (2 * pi * R * C).

Step-by-Step Solution:

Verification / Alternative check:
Bode magnitude plot of |V_R/V_in| shows a +20 dB/decade slope below cutoff and 0 dB above cutoff, confirming a first-order high-pass behavior.

Why Other Options Are Wrong:

• Bandpass: Requires at least second-order behavior or RLC; not present here.
• Low-pass: Would occur if output were taken across the capacitor.
• Band-stop: Needs a topology that attenuates a band; not applicable.

Common Pitfalls:
Confusing which node is the output or mixing up capacitor and resistor positions; also assuming cutoff uses 0.707 amplitude without linking it to the correct fc expression.

Final Answer:
High-pass.