RC low-pass with shorted reactance: A sinusoid of 18 V peak-to-peak feeds an RC low-pass whose reactance is effectively zero at the input frequency. With output across the capacitor, what is Vout?

Introduction / Context:
Understanding extreme-frequency behavior of RC filters is vital. In a low-pass with output across the capacitor, very high frequency (Xc → 0) causes the capacitor to behave like a short, severely attenuating the output.

Given Data / Assumptions:

• Low-pass RC, output across C.
• Input sine: 18 V peak-to-peak.
• At the input frequency, Xc ≈ 0 (capacitor acts like a short).

Concept / Approach:
If the capacitor is effectively a short at the operating frequency, the output node (across C) is nearly at ground potential relative to the source signal. Therefore, the measured output across the capacitor is essentially 0 V for AC at that frequency.

Step-by-Step Reasoning:
In low-pass with output across C, Vout ≈ Vin at low f and Vout → 0 as f → ∞Given Xc ≈ 0 at the chosen frequencyThus Vout ≈ 0 V at that frequency

Verification / Alternative check:
The transfer function magnitude |H(jω)| = 1 / sqrt(1 + (ωRC)^2). For very large ω, |H| → 0, confirming the zero output limit.

Why Other Options Are Wrong:

• 18 Vpp or 9 Vpp or 12.74 Vpp: These would imply finite or large output, contradicting the condition Xc ≈ 0.

Common Pitfalls:
Confusing high-pass versus low-pass behavior or mixing up where the output is taken (across R vs across C). Output across C in a low-pass collapses at high frequency.

Final Answer:
zero