How is the cutoff frequency of an RC low-pass filter determined, and how does loading affect it?

• A. fc = 1/(2πRC); loading (source/driven impedance) in parallel with R changes effective resistance and thus fc.
• B. fc = 2πRC.
• C. fc = R/(2πC).
• D. fc = 1/(2πL/R).

Answer: (A)

The key idea is that the cutoff frequency is set by the time constant that the capacitor experiences. For an RC low-pass with an ideal (zero-impedance) source, the capacitor sees the series resistor R, so the time constant is τ = RC and the -3 dB point is at ωc = 1/RC, giving fc = 1/(2πRC).

If a load is connected across the output (in parallel with the capacitor), the resistor seen by the capacitor is no longer just R. The effective resistance is the parallel combination R_eff = R || R_load. The time constant becomes τ = R_eff C, so the cutoff frequency is fc = 1/(2π R_eff C) = 1/(2π (R || R_load) C).

This means loading changes fc by altering the resistance the capacitor “sees.” If the load is very large (nearly open), R_eff ≈ R and fc ≈ 1/(2πRC). If the load is smaller (lower impedance), R_eff decreases and fc moves higher accordingly.