Q 12-07-132JEE MainJEE Main 2021 (27 Aug, Shift 1)Medium
The alternating current is given by, $i = \left\{\sqrt{42}\sin\left(\frac{2\pi}{T}t\right) + 10\right\}$ A. The R.M.S. value of this current is ______ A.
Numerical value type. Enter your answer.
Answer: 11
Mean of $i^2$ over a cycle: the cross term averages to zero, $\langle\sin^2\rangle = \frac{1}{2}$.
$$i_{rms}^2 = \frac{42}{2} + 10^2 = 121 \Rightarrow i_{rms} = 11\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics