Resonance In Series Lcr
A radio tuning circuit contains a fixed inductance of 2 mH in series with a variable capacitor. To resonate the circuit at an angular frequency of 5 (\times) 10^4 rad/s, what capacitance must be selected?
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About This Question
- Subject
- physics
- Chapter
- electromagnetic induction and alternating currents
- Topic
- resonance in series lcr
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
0.2 \(\mu\)F
Resonance in a series LCR circuit occurs when the inductive and capacitive reactances become equal, so the net reactance vanishes and the circuit behaves purely resistively, drawing maximum current. The resonant condition (\omega_0 = 1/\sqrt{LC}) rearranges to (C = 1/(\omega_0^2 L)). With (\omega_0 = 5\times10^4) rad/s, (\omega_0^2 = 2.5\times10^9), and (L = 2\times10^{-3}) H, the product (\omega_0^2 L = 5\times10^6). Hence (C = 1/(5\times10^6) = 2\times10^{-7}) F (= 0.2;\mu)F. The option 0.5 (\mu)F mishandles the squaring of frequency. The option 2 (\mu)F drops a factor of ten in the exponent. The option 0.05 (\mu)F overcorrects the power of ten the other way. This is the tuning principle NCERT uses to explain how a receiver selects one station: by adjusting the variable capacitor the resonant frequency is shifted until it coincides with the incoming carrier, at which point the circuit responds most strongly and rejects neighbouring frequencies. Note that the resonant frequency depends only on the product (LC) and is completely independent of the circuit resistance, which instead controls only the sharpness of the response. A plausibility check confirms that the units (1/[(rad/s)^2 H]) reduce to farads and that a sub-microfarad value is realistic for radio-frequency tuning; selecting this capacitance maximises the current response at the chosen frequency while minimising impedance.
This hard difficulty physics question is from the chapter electromagnetic induction and alternating currents, covering the topic of resonance in series lcr. It appeared in the 2025 exam.
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