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JEE Main 2021 (26 Feb, Shift 2) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2021 (26 Feb, Shift 2)

Q97JEE Main 2021 · 26 FEB · SHIFT 2Medium

Resonance, Bandwidth & Q Factor

Find the peak current and resonant frequency of the following circuit (as shown in figure).

Figure

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Answer: (d)

  1. Source: V0=30 VV_0 = 30\text{ V}, ω=100 rad/s\omega = 100\text{ rad/s}.
  2. XL=ωL=100×0.1=10 ΩX_L = \omega L = 100 \times 0.1 = 10\ \Omega, XC=1ωC=1100×10−4=100 ΩX_C = \frac{1}{\omega C} = \frac{1}{100\times10^{-4}} = 100\ \Omega
  3. Z=1202+902=150 ΩZ = \sqrt{120^2 + 90^2} = 150\ \Omega, so I0=30150=0.2 AI_0 = \frac{30}{150} = 0.2\text{ A}.
  4. f0=12πLC=12π0.1×10−4=3162π≈50 Hzf_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.1\times10^{-4}}} = \frac{316}{2\pi} \approx 50\text{ Hz}
  5. So the answer is (d) 0.2 A and 50 Hz.
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