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MHT CET 2025 (20 Apr, Shift 2) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from MHT CET 2025 (20 Apr, Shift 2)

Q31MHT CET 2025 · 20 APR · SHIFT 2Medium

Resonance, Bandwidth & Q Factor

An a.c. e.m.f. of peak value =230 V= 230\text{ V} and frequency 50 Hz is connected to a circuit with R=11.5 Ω, L=2.5 HR = 11.5\ \Omega,\ L = 2.5\text{ H} and a capacitor all in series. The value of capacitance is ' C ' for the current in the circuit to be maximum. The value of C and maximum current are respectively (π2=10)(\pi^2 = 10).

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Answer: (b)

  1. Current is maximum at resonance: C=1ω2LC = \frac{1}{\omega^2 L}, ω=100π\omega = 100\pi.
  2. C=1104×10×2.5=12.5×105=4 μC = \frac{1}{10^4\times10\times2.5} = \frac{1}{2.5\times10^{5}} = 4\ \muF
  3. At resonance Z=RZ = R: Imax=23011.5=20I_{max} = \frac{230}{11.5} = 20 A (peak)
  4. So the answer is (b) 4 μ4\ \muF, 20 A.
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