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?
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Adecrease
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Bremain the same
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Cincrease
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Dbe less selective
Answer
Correct Answer: decrease
Explanation
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:
Initial: BW = f0 / Q.Increase Q ⇒ BW′ = f0 / Q′ with Q′ > Q ⇒ BW′ < BW.Conclusion: Bandwidth decreases and selectivity increases.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