Q 12-07-125JEE MainJEE Main 2021 (26 Feb, Shift 1)Easy
An alternating current is given by the equation $i = i_1\sin\omega t + i_2\cos\omega t$. The rms current will be:
Answer: (D) $\frac{1}{\sqrt{2}}\left(i_1^2+i_2^2\right)^{\frac{1}{2}}$
The current can be written as a single sinusoid $i = i_0\sin(\omega t + \phi)$ with $i_0 = \sqrt{i_1^2 + i_2^2}$.
So $i_{rms} = \dfrac{i_0}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\left(i_1^2+i_2^2\right)^{1/2}$.
Solution by Sreeraj P, M.Sc Physics