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,
$i=(2\sin \omega t+6\cos \omega t)A$

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