The peak value of an alternating current is 6 | vedclass

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(C) The relationship between the peak current $(I_0)$ and the root mean square current $(I_{rms})$ for a sinusoidal alternating current is given by th

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$3\sqrt{2} \ A$

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(C) The relationship between the peak current $(I_0)$ and the root mean square current $(I_{rms})$ for a sinusoidal alternating current is given by the formula:$I_{rms} = \frac{I_0}{\sqrt{2}}$Given that the peak value $I_0 = 6 \ A$,we substitute this value into the formula:$I_{rms} = \frac{6}{\sqrt{2}}$To simplify,multiply the numerator and denominator by $\sqrt{2}$:$I_{rms} = \frac{6 	imes \sqrt{2}}{\sqrt{2} 	imes \sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \ A$Therefore,the r.m.s. value of the current is $3\sqrt{2} \ A$.

$(A) \ x_0 \sin \omega t$	$(i) \ x_0$
$(B) \ x_0 \sin \omega t \cos \omega t$	$(ii) \ \frac{x_0}{\sqrt{2}}$
$(C) \ x_0 \sin \omega t + x_0 \cos \omega t$	$(iii) \ \frac{x_0}{2 \sqrt{2}}$

The peak value of an alternating current is $6 \ A$. What is the root mean square (r.m.s.) value of the current?

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