Quality Factor And Bandwidth
A series resonant circuit used in a signal filter has resistance 5 (\Omega), inductance 0.1 H and resonates at an angular frequency of 1000 rad/s. What is the quality factor of this circuit at resonance?
Select the correct option:
Solution
20
The quality factor of a series resonant circuit measures the sharpness of its resonance, comparing the reactance at resonance to the resistance, and is written (Q = \dfrac{\omega_0 L}{R}). A higher Q indicates a narrower bandwidth and a more selective circuit. Substituting (\omega_0 = 1000) rad/s, (L = 0.1) H and (R = 5;\Omega) gives (Q = (1000 \times 0.1)/5 = 100/5 = 20). The option 10 halves the inductive reactance incorrectly. The option 100 reports the resonant reactance (\omega_0 L) itself without dividing by resistance. The option 5 simply echoes the resistance value. This is the NCERT measure relating selectivity to how lightly damped the oscillation is, and an equivalent expression (Q = (1/R)\sqrt{L/C}) shows that lower resistance or a larger inductance-to-capacitance ratio sharpens the resonance. Physically, Q also equals (2\pi) times the ratio of energy stored to energy dissipated per cycle, so a high-Q circuit loses only a small fraction of its stored energy in each oscillation. A plausibility check confirms Q is dimensionless since ohms cancel, and a value of twenty indicates a fairly sharp resonance, meaning the bandwidth (\Delta\omega = \omega_0/Q = 50) rad/s is small compared with the centre frequency, exactly the behaviour wanted in a frequency-selective filter.
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About This Question
- Subject
- physics
- Chapter
- electromagnetic induction and alternating currents
- Topic
- quality factor and bandwidth
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
20
The quality factor of a series resonant circuit measures the sharpness of its resonance, comparing the reactance at resonance to the resistance, and is written (Q = \dfrac{\omega_0 L}{R}). A higher Q indicates a narrower bandwidth and a more selective circuit. Substituting (\omega_0 = 1000) rad/s, (L = 0.1) H and (R = 5;\Omega) gives (Q = (1000 \times 0.1)/5 = 100/5 = 20). The option 10 halves the inductive reactance incorrectly. The option 100 reports the resonant reactance (\omega_0 L) itself without dividing by resistance. The option 5 simply echoes the resistance value. This is the NCERT measure relating selectivity to how lightly damped the oscillation is, and an equivalent expression (Q = (1/R)\sqrt{L/C}) shows that lower resistance or a larger inductance-to-capacitance ratio sharpens the resonance. Physically, Q also equals (2\pi) times the ratio of energy stored to energy dissipated per cycle, so a high-Q circuit loses only a small fraction of its stored energy in each oscillation. A plausibility check confirms Q is dimensionless since ohms cancel, and a value of twenty indicates a fairly sharp resonance, meaning the bandwidth (\Delta\omega = \omega_0/Q = 50) rad/s is small compared with the centre frequency, exactly the behaviour wanted in a frequency-selective filter.
This hard difficulty physics question is from the chapter electromagnetic induction and alternating currents, covering the topic of quality factor and bandwidth. It appeared in the 2025 exam.
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