An alternating current is given by $\mathrm{i}=(3 \sin \omega t+4 \cos \omega t)$ A. The $r m s$ current will be

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Aicharya · Accepted Answer

$i=(3 \sin w t+4 \cos w k) A$ $	herefore$ Peak current, $\mathrm{I}_0=\sqrt{3^2+4^2}=5 \mathrm{~A}$ $	herefore \quad$ RMS current, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{\mathrm{o}}}{\sqrt{2}}=\frac{5}{\sqrt{2}} \mathrm{~A}$

An alternating current is given by $\mathrm{i}=(3 \sin \omega t+4 \cos \omega t)$ A. The $r m s$ current…

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