QJEEM25Ph1a3b: JEE Main 2025 Physics - Class 12 JEEM25Ph1a3b Solution: An alternating current is given by . The r.m.s current will be

$I_{\text{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}$

To determine the root mean square (r.m.s) value of the alternating current given by

$I(t) = I_A \sin (\omega t) + I_B \cos (\omega t),$

we start by squaring the current:

$I(t)^2 = I_A^2 \sin^2 (\omega t) + I_B^2 \cos^2 (\omega t) + 2 I_A I_B \sin (\omega t) \cos (\omega t).$

The r.m.s value is defined as the square root of the average of this squared current over one complete period. When averaging over a full cycle, we utilize the known averages:

$\langle \sin^2 (\omega t) \rangle = \frac{1}{2}, \quad \langle \cos^2 (\omega t) \rangle = \frac{1}{2}, \quad \langle \sin (\omega t) \cos (\omega t) \rangle = 0.$

Thus, the time-averaged square of the current becomes:

$\langle I(t)^2 \rangle = I_A^2 \frac{1}{2} + I_B^2 \frac{1}{2} = \frac{I_A^2 + I_B^2}{2}.$

Taking the square root gives the r.m.s current:

$I_{\text{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}.$

This corresponds to Option B.