An AC current is given by I = I_0 + I_1 sin o | vedclass

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(A) The $rms$ value of a current $I(t)$ over a time period $T$ is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I(t)^2 dt}$.Given $I = I_0 + I_

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$\sqrt{I_0^2 + 0.5 I_1^2}$

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(A) The $rms$ value of a current $I(t)$ over a time period $T$ is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I(t)^2 dt}$.Given $I = I_0 + I_1 \sin \omega t$, we square the expression: $I^2 = I_0^2 + I_1^2 \sin^2 \omega t + 2 I_0 I_1 \sin \omega t$.Now, integrate over one full cycle $T = \frac{2\pi}{\omega}$:$I_{rms}^2 = \frac{1}{T} \int_{0}^{T} (I_0^2 + I_1^2 \sin^2 \omega t + 2 I_0 I_1 \sin \omega t) dt$.Since the average value of $\sin \omega t$ over a full cycle is $0$, the term $\int_{0}^{T} 2 I_0 I_1 \sin \omega t dt = 0$.The average value of $\sin^2 \omega t$ over a full cycle is $\frac{1}{2}$.Thus, $I_{rms}^2 = I_0^2 + I_1^2 \left( \frac{1}{2} \right) = I_0^2 + 0.5 I_1^2$.Therefore, $I_{rms} = \sqrt{I_0^2 + 0.5 I_1^2}$.

An $AC$ current is given by $I = I_0 + I_1 \sin \omega t$. Then its $rms$ value will be:

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