RC high-pass filter design: For a 0.2 µF capacitor in series with a 220 Ω resistor (output taken across the resistor), what is the critical (cutoff) frequency fc of the circuit?

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:

  • Capacitance C = 0.2 µF.
  • Resistance R = 220 Ω.
  • Output is taken across R (standard RC high-pass).
  • Ideal components and sinusoidal steady-state conditions.


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:

  • 723 Hz and 362 Hz: These come from using incorrect R or C values, or mixing milli/µ prefixes.
  • 7,234 Hz: Roughly twice the correct value (e.g., missing the 2 * pi factor).


Common Pitfalls:

  • Forgetting to convert µF to F before calculation.
  • Using fc = 1 / (pi * RC) instead of fc = 1 / (2 * pi * RC).


Final Answer:
3,617 Hz

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