Question

An alternating current is given by $i=2 \sin \omega t+6 \cos \omega t$. The rms current in amperes is

• A. $2 \sqrt{5}$
• B. $2 \sqrt{10}$
• C. $\sqrt{5}$
• D. $10 \sqrt{2}$

Answer 1

Answer: (A)

Given, equation of an alternating current,
$i=(2 \sin \omega t+6 \cos \omega t) A$

$ \therefore i=\sqrt{2^2+6^2}\left[\frac{2}{\sqrt{2^2+6^2}} \sin \omega t+\frac{6}{\sqrt{2^2+6^2}} \cos \omega t\right] $
$ =\sqrt{40}[\cos \theta \sin \omega t+\sin \theta \cos \omega t] $
$i=\sqrt{40} \sin (\omega t+\theta) $
$ \therefore i_{r m s}=\frac{i_0}{\sqrt{2}}=\frac{\sqrt{40}}{\sqrt{2}}=\sqrt{20}=2 \sqrt{5} A$