An alternating is given by the equation $\mathrm{i}=\mathrm{i}_{1} \sin \omega \mathrm{t}+\mathrm{i}_{2} \cos \omega \mathrm{t}$. Then rms current will be:

(A) $\frac{1}{\sqrt{2}}\left(\dot{i}_{1}^{2}+\dot{i}_{2}^{2}\right)^{\frac{1}{2}}$

(B) $\frac{1}{\sqrt{2}}\left(\mathrm{i}_{1}+\mathrm{i}_{2}\right)^{2}$

(C) $\frac{1}{\sqrt{2}}\left(\mathrm{i}_{2}+\mathrm{i}_{2}\right)$

(D) $\frac{1}{2}\left(\mathrm{i}_{1}^{2}+\mathrm{i}_{2}^{2}\right)^{\frac{1}{2}}$

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Alternating Current

An alternating is given by the equation

i = i_{1} sin ω t + i_{2} cos ω t

. Then rms current will be:

Answer: (A)

The current can be written as

i = i_{1} sin ω t + i_{2} cos ω t = i_{1}^{2} + i_{2}^{2} sin (ω t + ϕ)

where

ϕ

is a phase constant. The peak current is

i_{0} = i_{1}^{2} + i_{2}^{2}

. The rms current is

i_{r m s} = \frac{i _{0}}{2} = \frac{1}{2} i_{1}^{2} + i_{2}^{2}

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option