For a first-order RC low-pass filter with resistance R and capacitance C, what is the cutoff frequency?

• A. f_c = 1/(RC)
• B. f_c = 1/(2πRC)
• C. f_c = π/(RC)
• D. f_c = 2πRC

Answer: (B)

For a first-order RC low-pass, the cutoff is the -3 dB point, where the output is reduced to 1/√2 of its low-frequency value. The transfer function is H(jω) = 1/(1 + jωRC), so the magnitude is |H| = 1/√(1 + (ωRC)²). Setting |H| = 1/√2 gives (ωRC)² = 1, so ω_c = 1/RC. Converting to frequency in hertz, f_c = ω_c/(2π) = 1/(2πRC). Therefore, the correct expression is 1/(2πRC). The angular cutoff would be 1/RC, and 2πRC or π/(RC) do not match the correct frequency relationship or units.