An alternating current is given by I = IA sin ω t + IB cos ω t. The r.m.s current will be (a) (√(IA² + IB²))/(2) (b) √((IA² + IB²)/(2)) (c) (|IA + IB|)/(√(2)) (d) √(IA² + IB²)

Answer: (b). Combine: I = √(IA² + IB²) sin (ω t + φ), a single sinusoid. Peak I₀ = √(IA² + IB²) Irms = (I₀)/(√(2)) = √((IA² + IB²)/(2)) So the answer is (b).

An alternating current is given by I = IA sin ω t + IB cos ω t. The… | JEE Main 2025 (24 Jan, Shift 1)

Q13JEE Main 2025 · 24 JAN · SHIFT 1Easy

AC Basics: Peak, RMS & Average Values

An alternating current is given by $I = I_A \sin \omega t + I_B \cos \omega t$ . The r.m.s current will be

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Answer: (b)

Combine: $I = \sqrt{I_A^2 + I_B^2}\,\sin(\omega t + \phi)$ , a single sinusoid.
Peak $I_0 = \sqrt{I_A^2 + I_B^2}$
$I_{rms} = \frac{I_0}{\sqrt{2}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}$
So the answer is (b).

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