An alternating current is given by $\mathrm{I}=\mathrm{I}_{\mathrm{A}} \sin \omega \mathrm{t}+\mathrm{I}_{\mathrm{B}} \cos \omega \mathrm{t}$. The r.m.s. current will be :-

(C) $\sqrt{\frac{I_{A}^{2}+I_{B}^{2}}{2}}$

(A) $\sqrt{I_{\mathrm{A}}^{2}+I_{\mathrm{B}}^{2}}$

(B) $\frac{\sqrt{I_{A}^{2}+I_{B}^{2}}}{2}$

(D) $\frac{\left|I_{A}+I_{B}\right|}{\sqrt{2}}$

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Alternating Current

An alternating current is given by

I = I_{A} sin ω t + I_{B} cos ω t

.
The r.m.s. current will be :-

Answer: (C)

The rms value of a periodic current is

I_{r m s} = \frac{\int _{0}^{T} I ^{2} d t}{T}

With

I = I_{A} sin ω t + I_{B} cos ω t

I^{2} = I_{A}^{2} sin^{2} ω t + I_{B}^{2} cos^{2} ω t + 2 I_{A} I_{B} sin ω t cos ω t

Averaging over a full cycle:

⟨ sin^{2} ω t ⟩ = ⟨ cos^{2} ω t ⟩ = \frac{1}{2}

, and

⟨ sin ω t cos ω t ⟩ = 0

⟨ I^{2} ⟩ = \frac{I _{A}^{2}}{2} + \frac{I _{B}^{2}}{2} = \frac{I _{A}^{2} + I _{B}^{2}}{2}

I_{r m s} = \frac{I _{A}^{2} + I _{B}^{2}}{2}

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option