If an alternating current is given by i = a s | vedclass

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(C) The given equation is $i = a \sin(\omega t) + b \cos(\omega t)$.We can rewrite this as $i = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \si

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$\sqrt{\frac{a^2 + b^2}{2}}$

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(C) The given equation is $i = a \sin(\omega t) + b \cos(\omega t)$.We can rewrite this as $i = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \sin(\omega t) + \frac{b}{\sqrt{a^2 + b^2}} \cos(\omega t) \right)$.Let $\frac{a}{\sqrt{a^2 + b^2}} = \cos \phi$ and $\frac{b}{\sqrt{a^2 + b^2}} = \sin \phi$.Then $i = \sqrt{a^2 + b^2} \sin(\omega t + \phi)$.The peak value (amplitude) of the current is $i_0 = \sqrt{a^2 + b^2}$.The $rms$ value of an alternating current is given by $i_{rms} = \frac{i_0}{\sqrt{2}}$.Substituting the value of $i_0$,we get $i_{rms} = \frac{\sqrt{a^2 + b^2}}{\sqrt{2}} = \sqrt{\frac{a^2 + b^2}{2}}$.

If an alternating current is given by $i = a \sin(\omega t) + b \cos(\omega t)$,then the $rms$ value of the current is

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