Bandwidth of a series RLC circuit: A 15 Ω resistor, a 220 µH inductor, and a 60 pF capacitor are connected in series across an AC source. What is the circuit bandwidth (in Hz)?

Introduction / Context:For a series RLC circuit, the -3 dB bandwidth around resonance is directly related to the series resistance and the inductance. This relationship is used widely in selecting component Q, shaping selectivity, and predicting filter performance.

Given Data / Assumptions:

• Series elements: R = 15 Ω; L = 220 µH = 220 × 10^-6 H; C = 60 pF.
• We seek the bandwidth (BW) of the series resonant circuit.
• Ideal components aside from the given R.

Concept / Approach:For a series RLC, the (linear) bandwidth in Hz is BW = R / (2π L). Note that BW does not depend on C explicitly; it is implicitly tied through the resonant frequency and Q = ω0 L / R, but the closed-form for BW uses R and L only.

Step-by-Step Solution:

Verification / Alternative check:Using Q = ω0 L / R and f0 = 1/(2π√(LC)) yields BW = f0 / Q = R/(2πL) again, confirming the formula regardless of C’s exact value.

Why Other Options Are Wrong:

• 138 kHz / 138 MHz: Orders of magnitude too high for the given R and L.
• 1,907 Hz: Too small; inconsistent with R/(2πL).

Common Pitfalls:

• Confusing series with parallel bandwidth expressions (which involve R_p and C).

Final Answer:10,866 Hz