In an ac circuit, the r.m.s. value of current, $I_{r m s}$ is related to the peak current, $I_{0}$ by the relation

I_{r m s} = \frac{1}{π} I_{0}

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I_{r m s} = \frac{1}{\sqrt{2}} I_{0}

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I_{r m s} = \sqrt{2} I_{0}

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I_{r m s} = π I_{0}

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Solution

The correct option is B $I_{r m s} = \frac{1}{\sqrt{2}} I_{0}$

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If $i_{0}, i_{r m s} a n d i_{m e a n}$ are the values of peak, rms and mean currents of a.c, then

(A)

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(R)

(I = I_{0} sin ω t) and I_{rms} = \frac{I_{0}}{\sqrt{2}}

i = 0.5 s i n (314 t + 60^{o})

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