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JEE Main 2022 (24 Jun, Shift 1) · Pure R, L and C Circuits

Pure R, L and C Circuits question from JEE Main 2022 (24 Jun, Shift 1)

Q49JEE Main 2022 · 24 JUN · SHIFT 1Medium

Pure R, L and C Circuits

As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is aπ\frac{\sqrt{a}}{\pi} A. The value of aa is ___________.

Figure

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

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Answer: 242

  1. In a pure inductor current lags voltage by 90∘90^\circ, so when the source voltage is 0 the current is at its peak.
  2. XL=2πfL=2π×50×0.2=20π ΩX_L = 2\pi f L = 2\pi \times 50 \times 0.2 = 20\pi\ \Omega
  3. I0=2×22020π=112π=242πI_0 = \frac{\sqrt{2} \times 220}{20\pi} = \frac{11\sqrt{2}}{\pi} = \frac{\sqrt{242}}{\pi} A
  4. So a=242a = 242.
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