RL high-pass network with output across the inductor: For R = 470 Ω and L = 600 mH driven by an AC source, what is the critical (cutoff) frequency of the circuit?

Introduction / Context:
This problem tests the cutoff (critical) frequency of a simple RL filter. Whether the output is taken across the resistor (high-pass behavior) or across the inductor (low-pass behavior), the pole frequency set by R and L is the same and is a key specification for design and analysis.

Given Data / Assumptions:

• Resistor R = 470 Ω.
• Inductor L = 600 mH = 0.6 H.
• Linear, time-invariant components; sinusoidal steady state.
• Critical frequency is defined by the RL pole.

Concept / Approach:
For an RL network, the critical (cutoff) frequency is f_c = R / (2 * pi * L). This is the frequency where the magnitude response is 1/sqrt(2) of the passband level (the -3 dB point). It depends only on the component values R and L.

Step-by-Step Solution:
R = 470 Ω, L = 0.6 Hf_c = R / (2 * pi * L)f_c = 470 / (2 * 3.1416 * 0.6)f_c ≈ 470 / 3.7699 ≈ 124.6 Hz ≈ 125 Hz

Verification / Alternative check:
Units: Ω / H gives s^-1; dividing by 2 * pi yields Hz. The numeric result sits well below 1 kHz, consistent with a relatively large inductance of 0.6 H and moderate resistance.

Why Other Options Are Wrong:

• 1,250 Hz and 5,644 Hz: Both are an order of magnitude higher than the computed result.
• 564 Hz: Not supported by the R/L ratio; it would require a different R or L.

Common Pitfalls:
Confusing whether the output node changes f_c (it does not), or mixing up L in mH versus H. Always convert 600 mH to 0.6 H before calculation.

Final Answer:
125 Hz