Question

An alternating current is given by $=I_1\cos \omega t+I_2\sin \omega t.$ The r.m.s. current is given by-

• A. $\frac{I_1+I_2}{2}$
• B. $\frac{I_1+I_2}{\sqrt{2}}$
• C. $\sqrt{\frac{I_1^2+I_2^2}{2}}$
• D. $\frac{I_1^2+I_2^2}{2}$

Answer 1

The correct option is C $\sqrt{\frac{I_1^2+I_2^2}{2}}$

Given,

$I=I_1\cos \omega t+I_2\sin \omega t$
this can be re framed as
$I=\sqrt{I_1^2+I_2^2}(\frac{I_1}{\sqrt{I_1^2+I_2^2}}\cos \omega t+\frac{I_2}{\sqrt{I_1^2+I_2^2}}\sin \omega t)$
Let $\sin \varphi =\frac{I_1}{\sqrt{I_1^2+I_2^2}},\cos \varphi =\frac{I_2}{\sqrt{I_1^2+I_2^2}}$
$I=\sqrt{I_1^2+I_2^2}\sin (\omega t+\varphi )$


Hence
$I_{rms}=\frac{I_{peak}}{\sqrt{2}}$

$\Rightarrow I_{\text{rms}}=\sqrt{\frac{I_1^2+I_2^2}{2}}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.