In an A.C. circuit,I_{text{rms}} and I_{0} ar | vedclass

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(B) The root mean square $(I_{	ext{rms}})$ value of an alternating current is defined as the square root of the mean of the squares of the instantane

Vedclass · Accepted Answer

$I_{	ext{rms}} = \frac{1}{\sqrt{2}} I_{0}$

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(B) The root mean square $(I_{	ext{rms}})$ value of an alternating current is defined as the square root of the mean of the squares of the instantaneous current over one complete cycle.For a sinusoidal alternating current given by $I = I_{0} \sin(\omega t)$,the $I_{	ext{rms}}$ is calculated as:$I_{	ext{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} I^{2} dt}$Substituting $I = I_{0} \sin(\omega t)$:$I_{	ext{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} I_{0}^{2} \sin^{2}(\omega t) dt}$$I_{	ext{rms}} = I_{0} \sqrt{\frac{1}{T} \int_{0}^{T} \frac{1 - \cos(2\omega t)}{2} dt}$$I_{	ext{rms}} = I_{0} \sqrt{\frac{1}{2T} [t - \frac{\sin(2\omega t)}{2\omega}]_{0}^{T}}$Since $\sin(2\omega T) = \sin(4\pi) = 0$,we get:$I_{	ext{rms}} = I_{0} \sqrt{\frac{T}{2T}} = \frac{I_{0}}{\sqrt{2}}$Therefore,the correct relation is $I_{	ext{rms}} = \frac{1}{\sqrt{2}} I_{0}$.

In an $A.C.$ circuit,$I_{\text{rms}}$ and $I_{0}$ are related as:

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