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MHT CET 2026 (19 Apr, Shift 1) · Pure R, L and C Circuits

Pure R, L and C Circuits question from MHT CET 2026 (19 Apr, Shift 1)

Q12MHT CET 2026 · 19 APR · SHIFT 1Easy

Pure R, L and C Circuits

An alternating voltage e=1502sin⁡100te = 150\sqrt{2} \sin 100t volt is applied to a capacitor of capacity 2 μ2\ \muF. The root mean square value of current in the circuit is

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Answer: (c)

  1. ω=100\omega = 100 rad/s, XC=1ωC=1100×2×10−6=5000 ΩX_C = \frac{1}{\omega C} = \frac{1}{100 \times 2\times10^{-6}} = 5000\ \Omega
  2. Vrms=15022=150V_{rms} = \frac{150\sqrt{2}}{\sqrt{2}} = 150 V
  3. Irms=1505000=0.03I_{rms} = \frac{150}{5000} = 0.03 A =30= 30 mA
  4. So the answer is (c) 30 mA.
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