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

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Question

$I_{r m s}=\sqrt{\frac{1}{T} \int_{0}^{T}\left(I_{0}+I_{1} \sin \omega t\right)^{2} d t}$

$\sqrt{\frac{1}{T} \int_{0}^{T} I_{0}^{2} d t+\int_{0}^{T

Vedclass · Accepted Answer

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

Vedclass · Answer

$I_{r m s}=\sqrt{\frac{1}{T} \int_{0}^{T}\left(I_{0}+I_{1} \sin \omega t\right)^{2} d t}$

$\sqrt{\frac{1}{T} \int_{0}^{T} I_{0}^{2} d t+\int_{0}^{T} I_{1}^{2} \sin ^{2} \omega t d t+\int_{0}^{T} \frac{2 I_{0} I_{1}}{T} \sin 2 \omega t d t}$

$=\sqrt{I_{0}^{2}+\left(I_{1}^{2} / 2\right)}=\sqrt{I_{0}^{2}+0.5 I_{1}^{2}}$

An $AC$ current is given by $I = I_0 + I_1$ $sin\, wt$ then its $rms$ value will be

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