RL low-pass cutoff: For a 5.6 mH inductor in series with a 3.3 kΩ resistor (output taken across the resistor), what is the critical frequency fc of this RL low-pass filter?

Introduction / Context:This question evaluates your ability to compute the cutoff frequency of an RL low-pass filter. In such a filter, the resistor is the load across which the output is taken; low frequencies pass with little attenuation, while high frequencies are impeded by the inductor's reactance.

Given Data / Assumptions:

• Inductance L = 5.6 mH = 0.0056 H.
• Resistance R = 3.3 kΩ = 3,300 Ω.
• Output is across R (standard RL low-pass configuration).
• Ideal components; sinusoidal steady-state.

Concept / Approach:For an RL low-pass, the cutoff (critical) frequency is fc = R / (2 * pi * L). At fc, the output power has dropped by 3 dB relative to the low-frequency passband, and the magnitudes of R and XL are equal.

Step-by-Step Solution:

Verification / Alternative check:Another quick check uses ωc = R / L. Then fc = ωc / (2 * pi) = (3300 / 0.0056) / (2 * pi) ≈ 589,286 / 6.283 ≈ 93,800 Hz, which matches the earlier result.

Why Other Options Are Wrong:

• 93.8 Hz: Off by 10^3; forgotten milli- vs kilo- scale.
• 861 Hz and 86.12 kHz: From arithmetic or 2 * pi slip-ups.

Common Pitfalls:

• Using fc = 1 / (2 * pi * L / R) instead of the correct R / (2 * pi * L).
• Incorrect unit conversions (mH to H, kΩ to Ω).

Final Answer:93.8 kHz