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MHT CET 2025 (22 Apr, Shift 1) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from MHT CET 2025 (22 Apr, Shift 1)

Q33MHT CET 2025 · 22 APR · SHIFT 1Medium

Resonance, Bandwidth & Q Factor

An a.c. e.m.f. of peak value 230 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 ' CC ' for the current in the circuit to be maximum. The value of ' CC ' and maximum current are respectively ( π2=10\pi^2 = 10)

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Answer: (a)

  1. Current is maximum at resonance: C=1ω2LC = \frac{1}{\omega^2 L}, ω=100π\omega = 100\pi.
  2. C=1104×10×2.5=4×10−6C = \frac{1}{10^4\times10\times2.5} = 4\times10^{-6} F
  3. Imax=V0R=23011.5=20I_{max} = \frac{V_0}{R} = \frac{230}{11.5} = 20 A
  4. So the answer is (a) 4 μ4\ \muF, 20 A.
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