Question

An alternating current (in $A$) varies with time as $I=3\sin \omega t+4\cos \omega t$. The rms value of the current is $x\sqrt{2}A$. Then $x$ is
[Answer upto two decimal points]

Answer 1

Given value of alternating current can be simplified to$I=5\sin (\omega t+\varphi )\text{where}\varphi ={53}^{\circ }$
So $I_{rms}=\frac{5}{\sqrt{2}}$.
Since rms value of $\sin \theta$ is $\frac{1}{\sqrt{2}}$.
$\therefore x\sqrt{2}=\frac{5}{\sqrt{2}}$
$\Rightarrow x=2.5$