Q 12-07-044JEE MainJEE Main 2025 (24 Jan, Shift 1)Easy
An alternating current is given by $I = I_A\sin\omega t + I_B\cos\omega t$. The r.m.s. current will be
Answer: (B) $\sqrt{\dfrac{I_A^2 + I_B^2}{2}}$
$I_A\sin\omega t + I_B\cos\omega t$ is a single sinusoid with amplitude $I_0 = \sqrt{I_A^2 + I_B^2}$ (the two parts are $90^\circ$ out of phase).
$$I_{rms} = \frac{I_0}{\sqrt2} = \sqrt{\frac{I_A^2 + I_B^2}{2}}$$
(Equivalently, the mean of $I^2$ over a cycle is $\tfrac{1}{2}I_A^2 + \tfrac{1}{2}I_B^2$, since the cross term averages to zero.)
Solution by Sreeraj P, M.Sc Physics