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JEE Main 2025 (4 Apr, Shift 1) · AC Basics: Peak, RMS & Average Values

AC Basics: Peak, RMS & Average Values question from JEE Main 2025 (4 Apr, Shift 1)

Q11JEE Main 2025 · 4 APR · SHIFT 1Easy

AC Basics: Peak, RMS & Average Values

An alternating current is represented by the equation, i=1002sin⁡(100πt)i = 100\sqrt{2} \sin(100\pi t) ampere. The RMS value of current and the frequency of the given alternating current are

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Answer: (d)

  1. Irms=I02=10022=100I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100 A
  2. ω=100π=2πf⇒f=50\omega = 100\pi = 2\pi f \Rightarrow f = 50 Hz
  3. So the answer is (d) 100100 A, 5050 Hz.
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