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MHT CET 2026 (19 May, Shift 2) · AC Generator & Transformer

AC Generator & Transformer question from MHT CET 2026 (19 May, Shift 2)

Q5MHT CET 2026 · 19 MAY · SHIFT 2Medium

AC Generator & Transformer

The output of step-down transformer is measured to be 3636 V when connected to 66 W light bulb. The value of peak current is

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Answer: (c)

  1. rms current: Irms=PV=636=16 AI_{rms} = \dfrac{P}{V} = \dfrac{6}{36} = \dfrac{1}{6}\text{ A}
  2. I0=2 Irms=26=132 AI_0 = \sqrt{2}\,I_{rms} = \dfrac{\sqrt{2}}{6} = \dfrac{1}{3\sqrt{2}}\text{ A}
  3. So the answer is (c) 132\dfrac{1}{3\sqrt{2}} A.
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