JEE Main 2025 — Alternating Current Question with Solution

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

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:

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

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:

$$⟨{\sin }^2(\omega t)⟩=\frac{1}{2},⟨{\cos }^2(\omega t)⟩=\frac{1}{2},⟨\sin (\omega t)\cos (\omega t)⟩=0.$$

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

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

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.