Difficulty: Easy
Correct Answer: 3,617 Hz
Explanation:
Introduction / Context:This problem checks fundamental filter theory for an RC high-pass network. In a simple series RC high-pass filter with output across the resistor, the critical (cutoff) frequency fc is the frequency at which the output voltage equals the input divided by √2, corresponding to a −3 dB point and a 45° phase shift.
Given Data / Assumptions:
Concept / Approach:For an RC high-pass filter, the cutoff frequency is defined by fc = 1 / (2 * pi * R * C). At fc, the magnitudes of the capacitive reactance and the resistance are equal, so the output is at −3 dB relative to the high-frequency passband.
Step-by-Step Solution:
Convert C: 0.2 µF = 0.2 * 10^-6 F = 2.0 * 10^-7 F.Compute RC: RC = 220 * 2.0 * 10^-7 = 4.4 * 10^-5 s.Apply formula: fc = 1 / (2 * pi * RC) = 1 / (2 * pi * 4.4 * 10^-5).Numerical value: fc ≈ 1 / (2.764 * 10^-4) ≈ 3.617 * 10^3 Hz.Therefore, fc ≈ 3,617 Hz.Verification / Alternative check:As a quick check, note that for R ≈ 220 Ω and C ≈ 0.2 µF, the time constant τ = RC ≈ 44 µs. Using the shortcut fc ≈ 1 / (6.283 * 44 µs) gives ≈ 3.6 kHz, confirming the precise calculation.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:3,617 Hz
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