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Circuit Analysis: AC, Linear Circuits and Transients

In the parallel RLC circuit in the figure, R = 20 kΩ, L = 1 mH and C = 10 nF. Near the resonant frequency, which of the following is closest to the difference between the two frequencies at which the magnitude of the circuit impedance falls to 1/√2 of its value at resonance (the bandwidth)?

Parallel RLC circuit driven by a sinusoidal AC source V I IR IL IC R L C Y = 1/R + j(ωC − 1/(ωL))
Parallel RLC circuit driven by a sinusoidal AC source · View in diagrams
  1. AAbout 800 Hz
  2. BAbout 50 kHz
  3. CAbout 3.2 MHz
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Answer: A. About 800 Hz

Method 1 (from the Q factor): the resonant frequency is f0 = 1/(2π√(LC)) ≈ 50.3 kHz. The Q of a parallel resonant circuit is Q = R√(C/L) = 20×10³ × √(10×10⁻⁹/1×10⁻³) ≈ 63.2, so the bandwidth is BW = f0/Q ≈ 50.3 kHz/63.2 ≈ 796 Hz. Method 2 (directly from the admittance): the two angular frequencies at which the magnitude of the imaginary part of Y = 1/R + j(ωC − 1/(ωL)) equals the real part 1/R differ by 1/(RC), so BW = 1/(2πRC) = 1/(2π × 20×10³ × 10×10⁻⁹) ≈ 796 Hz. In parallel resonance, the larger R is, the higher the Q and the narrower the bandwidth.

  • A. Correct. BW = 1/(2πRC) ≈ 796 Hz.
  • B. About 50 kHz is the resonant frequency f0 itself.
  • C. About 3.2 MHz applies the series-resonance formula BW = R/(2πL). In parallel resonance the role of R is reversed, and Q = R√(C/L).
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