An alternating current is given by i = (3 sin | vedclass

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(C) The given alternating current is $i = 3 \sin \omega t + 4 \cos \omega t$.We can express this in the form $i = I_0 \sin(\omega t + \phi)$,where $I_

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$\frac{5}{\sqrt{2}} \ A$

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(C) The given alternating current is $i = 3 \sin \omega t + 4 \cos \omega t$.We can express this in the form $i = I_0 \sin(\omega t + \phi)$,where $I_0$ is the peak current.Using the identity $a \sin 	heta + b \cos 	heta = \sqrt{a^2 + b^2} \sin(	heta + \phi)$,we get:$I_0 = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \ A$.The $rms$ current is given by the formula $I_{rms} = \frac{I_0}{\sqrt{2}}$.Substituting the value of $I_0$,we get $I_{rms} = \frac{5}{\sqrt{2}} \ A$.

An alternating current is given by $i = (3 \sin \omega t + 4 \cos \omega t) \ A$. The $rms$ current will be:

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