For an AC current I = 50 cos(100t + 45^{circ} | vedclass

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(D) The given equation for the alternating current is $I = I_m \cos(\omega t + \phi)$,where $I_m$ is the peak current.Comparing this with the given eq

Vedclass · Accepted Answer

$25 \sqrt{2}$

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(D) The given equation for the alternating current is $I = I_m \cos(\omega t + \phi)$,where $I_m$ is the peak current.Comparing this with the given equation $I = 50 \cos(100t + 45^{\circ}) \ A$,we find the peak current $I_m = 50 \ A$.The root mean square current $I_{rms}$ is related to the peak current $I_m$ by the formula $I_{rms} = \frac{I_m}{\sqrt{2}}$.Substituting the value of $I_m$,we get $I_{rms} = \frac{50}{\sqrt{2}}$.To simplify,multiply the numerator and denominator by $\sqrt{2}$: $I_{rms} = \frac{50 \sqrt{2}}{2} = 25 \sqrt{2} \ A$.Therefore,the correct option is $D$.

Currents	$r.m.s.$ values
$(A) \ x_0 \sin \omega t$	$(i) \ x_0$
$(B) \ x_0 \sin \omega t \cos \omega t$	$(ii) \ \frac{x_0}{\sqrt{2}}$
$(C) \ x_0 \sin \omega t + x_0 \cos \omega t$	$(iii) \ \frac{x_0}{2\sqrt{2}}$

For an $AC$ current $I = 50 \cos(100t + 45^{\circ}) \ A$. The value of $I_{rms} =$ . . . . . . $A$.

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