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

<p>$I_{\text{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}$</p> <p>To determine the root mean square (r.m.s) value of the alternating current given by</p> <p>$I(t) = I_A \sin (\omega t) + I_B \cos (\omega t),$</p> <p>we start by squaring the current:</p> <p>$$ I(t)^2 = I_A^2 \sin^2 (\omega t) + I_B^2 \cos^2 (\omega t) + 2 I_A I_B \sin (\omega t) \cos (\omega t). $$</p> <p>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:</p> <p>$$ \langle \sin^2 (\omega t) \rangle = \frac{1}{2}, \quad \langle \cos^2 (\omega t) \rangle = \frac{1}{2}, \quad \langle \sin (\omega t) \cos (\omega t) \rangle = 0. $$</p> <p>Thus, the time-averaged square of the current becomes:</p> <p>$$ \langle I(t)^2 \rangle = I_A^2 \frac{1}{2} + I_B^2 \frac{1}{2} = \frac{I_A^2 + I_B^2}{2}. $$</p> <p>Taking the square root gives the r.m.s current:</p> <p>$I_{\text{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}.$</p> <p>This corresponds to Option B.</p>