A certain series resonant RLC circuit has bandwidth BW = 2 kHz. If the inductor is replaced by one with a higher quality factor Q (keeping resonant frequency the same), what happens to the bandwidth?

Introduction / Context:
Quality factor Q captures how underdamped a resonant circuit is. In series resonance, higher Q corresponds to sharper resonance and narrower bandwidth. This principle is used in narrowband filters and tuned circuits.

Given Data / Assumptions:

• Series resonant circuit with initial bandwidth 2 kHz.
• Resonant frequency f0 unchanged (component adjustments preserve f0).
• Inductor replaced by one with higher Q (lower loss).

Concept / Approach:

For series resonance, Q = f0 / BW. Increasing Q at fixed f0 necessarily reduces BW. Selectivity improves (not degrades) because the passband becomes narrower and more frequency-discriminating.

Step-by-Step Solution:

Verification / Alternative check:

Bode plots: higher Q shows a taller, narrower peak in |Z| or |V| transfer characteristic, confirming reduced bandwidth.

Why Other Options Are Wrong:

'Increase' and 'remain the same' contradict Q = f0/BW. 'Be less selective' is the opposite of what a higher-Q resonator provides.

Common Pitfalls:

Mixing series and parallel Q relations; assuming Q affects only peak amplitude but not bandwidth.

Final Answer:

decrease