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(B) The given current is $i = i_1 \sin \omega t + i_2 \cos \omega t$.We can rewrite this as $i = I_0 \sin(\omega t + \phi)$,where $I_0$ is the peak cu

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$\sqrt{\frac{i_1^2 + i_2^2}{2}}$

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(B) The given current is $i = i_1 \sin \omega t + i_2 \cos \omega t$.We can rewrite this as $i = I_0 \sin(\omega t + \phi)$,where $I_0$ is the peak current.The square of the peak current is $I_0^2 = i_1^2 + i_2^2$.Thus,$I_0 = \sqrt{i_1^2 + i_2^2}$.The $r.m.s.$ value of a sinusoidal current is given by $I_{rms} = \frac{I_0}{\sqrt{2}}$.Substituting the value of $I_0$,we get $I_{rms} = \frac{\sqrt{i_1^2 + i_2^2}}{\sqrt{2}} = \sqrt{\frac{i_1^2 + i_2^2}{2}}$.

The current in an $AC$ circuit is given by $i = i_1 \sin \omega t + i_2 \cos \omega t$. Its $r.m.s.$ value is:

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