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An alternating current is I=I_A sin ω t+I_B cos ω t. The r.m.s. current is

Asked in JEE Main 24th Jan 1st Shift 2025 · RMS and mean values

(1) √(I_A²+I_B²)/2
(2) (√I_A²+I_B²)/2
(3) (|I_A²+I_B²|)/(√2)
(4) √I_A²+I_B²

Answer: (1) √(I_A²+I_B²)/2

Step-by-step solution

I_A sin ω t+I_B cos ω t combines to a single sinusoid of amplitude I₀=√I_A²+I_B².

iᵣₘₛ=(I₀)/(√2)=(√I_A²+I_B²)/(√2)=√(I_A²+I_B²)/2.

Why the other options are wrong

(2) Divides the resultant amplitude by 2 instead of √2.
(3) Drops the outer square root over I_A²+I_B².
(4) Uses the resultant amplitude as the r.m.s. value (forgets the 1/√2).

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