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MHT CET 2026 (17 Apr, Shift 2) · Pure R, L and C Circuits

Pure R, L and C Circuits question from MHT CET 2026 (17 Apr, Shift 2)

Q3MHT CET 2026 · 17 APR · SHIFT 2Easy

Pure R, L and C Circuits

In an a.c. circuit with pure capacitance 'C' and a.c. source E=E0sin⁡ωtE = E_0 \sin \omega t, the equation of instantaneous current is given by

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Answer: (b)

  1. Charge q=CE0sin⁡ωtq = CE_0\sin\omega t, so I=dqdt=E0ωCcos⁡ωtI = \frac{dq}{dt} = E_0\omega C\cos\omega t.
  2. cos⁡ωt=sin⁡(ωt+π2)\cos\omega t = \sin\left(\omega t + \frac{\pi}{2}\right)
  3. I=E0ωCsin⁡(ωt+π2)I = E_0\omega C\sin\left(\omega t + \frac{\pi}{2}\right)
  4. So the answer is (b).
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