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JEE Main 2026 (23 Jan, Shift 1) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2026 (23 Jan, Shift 1)

Q18JEE Main 2026 · 23 JAN · SHIFT 1Medium

Resonance, Bandwidth & Q Factor

Using a variable frequency a.c. voltage source the maximum current measured in the given LCR circuit is 50 mA for V=5sin⁡(100t)V = 5\sin(100t) The values of LL and RR are shown in the figure. The capacitance of the capacitor ( C) used is ‾ μF\underline{\qquad}\ \mu\text{F}.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 50

  1. At ω=100\omega = 100 rad/s: I0=50I_0 = 50 mA =5100= \frac{5}{100}, i.e. Z=RZ = R. So ω=100\omega = 100 rad/s is the resonance.
  2. C=1ω2L=11002×2C = \frac{1}{\omega^2 L} = \frac{1}{100^2\times2}
  3. C=5×10−5C = 5\times10^{-5} F =50 μ= 50\ \muF
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