RC low-pass cutoff calculation: A low-pass filter has R = 120 Ω and C = 0.002 µF, with output taken across the capacitor. What is the circuit’s critical (cutoff) frequency?

Introduction / Context:
Determining the cutoff frequency of an RC filter is a foundational task in signal conditioning and analog design. The first-order RC cutoff frequency depends only on the product RC.

Given Data / Assumptions:

• R = 120 Ω.
• C = 0.002 µF = 0.002 × 10^-6 F = 2 × 10^-9 F.
• First-order RC low-pass; output across C.

Concept / Approach:
The -3 dB (critical) frequency is fc = 1 / (2 * pi * R * C). Careful unit conversion is crucial: microfarads to farads must be accurate to avoid order-of-magnitude errors.

Step-by-Step Solution:
C = 0.002 µF = 2 × 10^-9 FRC = 120 * 2 × 10^-9 = 240 × 10^-9 s = 2.40 × 10^-7 sfc = 1 / (2 * pi * RC)fc ≈ 1 / (2 * 3.1416 * 2.40 × 10^-7) ≈ 1 / (1.508 × 10^-6) ≈ 6.63 × 10^5 Hzfc ≈ 663 kHz (closest option: 633 kHz)

Verification / Alternative check:
Using a calculator: 1 / (2π * 120 * 2e-9) ≈ 663 kHz; due to rounding and option granularity, 633 kHz is the nearest listed value.

Why Other Options Are Wrong:

• 333 kHz, 331 kHz, 60 kHz: Significantly different from the computed ≈663 kHz result.

Common Pitfalls:
Mistaking 0.002 µF for 2 × 10^-6 F (it is 2 × 10^-9 F), which would yield a thousandfold error. Always normalize units before calculation.

Final Answer:
633 kHz