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Circuit and electromagnetism basics

Which is closest to the bandwidth (the difference between the two frequencies at which the current falls to 1/√2 of its resonant value) of a series RLC resonant circuit with R = 10 Ω, L = 1 mH and C = 10 nF?

  1. AAbout 1.6 kHz
  2. BAbout 10 kHz
  3. CAbout 50 kHz
Show answer and explanation

Answer: A. About 1.6 kHz

The resonant frequency is f0 = 1/(2π√(LC)) ≈ 50.3 kHz and the Q factor is Q = (1/R)√(L/C) = (1/10) × √(1×10⁻³/(10×10⁻⁹)) ≈ 316/10 ≈ 31.6. The bandwidth is BW = f0/Q ≈ 50.3 kHz/31.6 ≈ 1.59 kHz. For a series RLC circuit it can also be found directly as BW = R/(2πL) = 10/(2π × 1×10⁻³) ≈ 1.59 kHz. A smaller resistance narrows the bandwidth and sharpens frequency selectivity.

  • A. Correct. f0/Q = R/(2πL) ≈ 1.59 kHz.
  • B. About 10 kHz takes R/L = 1×10⁴ rad/s as a frequency, without dividing by 2π.
  • C. About 50 kHz is the resonant frequency itself, not the bandwidth.
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