A current in circuit is given by i = 3 + 4, s | vedclass

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Question

Current is given as $i=I_{o}+I_{1} \sin w t$

Rms current is given by $I_{r m s}=\sqrt{\frac{\int_{o}^{T} i^{2} d t}{T}}$

$	herefore I_{r m s}=\

Vedclass · Accepted Answer

$\sqrt {17} $

Vedclass · Answer

Current is given as $i=I_{o}+I_{1} \sin w t$

Rms current is given by $I_{r m s}=\sqrt{\frac{\int_{o}^{T} i^{2} d t}{T}}$

$	herefore I_{r m s}=\sqrt{\frac{\int_{o}^{T}\left(I_{o}^{2}+I_{1}^{2} \sin ^{2} w t+2 I_{o} I_{1} \sin w tight) d t}{T}}$

Or $I_{r m s}=\sqrt{I_{o}^{2} \frac{\int_{o}^{T} d t}{T}+I_{1}^{2} \frac{\int_{o}^{T} \sin ^{2} w t d t}{T}+2 I_{o} I_{1} \frac{\int_{o}^{T} \sin w t d t}{T}}$

Using $\frac{\int_{o}^{T} \sin ^{2} w t}{T}=0.5$ and $\frac{\int_{o}^{T} \sin w t}{T}=0$

We get $I_{r m s}=\sqrt{I_{o}^{2} \frac{T}{T}+I_{1}^{2} 	imes 0.5+2 I_{o} I_{1} 	imes 0}$

$\Longrightarrow I_{r m s}=\sqrt{I_{o}^{2}+0.5 I_{1}^{2}}$

Given $: I_{o}=3 \quad I_{1}=4$

$\Longrightarrow I_{r m s}=\sqrt{3^{2}+0.5 	imes 4^{2}}=\sqrt{17}$

A current in circuit is given by $i = 3 + 4\, sin\, \omega t$. Then the effective value of current is

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