An alternating current is given by I = I A sin ω t + I B cos ω t . The r.m.s. current will be :-

Physics · Alternating Current

JEE Main 2025 — 24 January, Morning Shift — Question 62

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 :-

Option A:
$\sqrt{\mathrm{I}_{\mathrm{A}}^{2}+\mathrm{I}_{\mathrm{B}}^{2}}$
Option B:
$\frac{\sqrt{\mathrm{I}_{\mathrm{A}}^{2}+\mathrm{I}_{\mathrm{B}}^{2}}}{2}$
Option C:
$\sqrt{\frac{\mathrm{I}_{\mathrm{A}}^{2}+\mathrm{I}_{\mathrm{B}}^{2}}{2}}$
Correct
Option D:
$\frac{\left|I_{A}+I_{B}\right|}{\sqrt{2}}$

$\quad i_{\text {rms }}=\sqrt{\frac{\int I^{2} d t}{\int d t}}$

$\sqrt{\frac{I_{A}^{2}+I_{B}^{2}}{2}}=i_{\text {rms }}$

Answer key and solution verified before publishing.

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