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(B) The instantaneous current is given by the equation $i = i_0 \cos(\omega t + \phi)$,where $i_0$ is the peak current.Comparing this with the given e

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$2\sqrt{2} 	ext{ amperes}$

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(B) The instantaneous current is given by the equation $i = i_0 \cos(\omega t + \phi)$,where $i_0$ is the peak current.Comparing this with the given equation $i = 4 \cos(\omega t + \phi)$,we find the peak current $i_0 = 4 	ext{ A}$.The $r.m.s.$ value of an alternating current is related to the peak current by the formula $i_{r.m.s.} = \frac{i_0}{\sqrt{2}}$.Substituting the value of $i_0$,we get $i_{r.m.s.} = \frac{4}{\sqrt{2}} = \frac{4 	imes \sqrt{2}}{\sqrt{2} 	imes \sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} 	ext{ A}$.Therefore,the correct option is $B$.

If the instantaneous current is given by $i = 4 \cos(\omega t + \phi)$ amperes,then the $r.m.s.$ value of the current is:

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