The r.m.s. value of an AC of 50 Hz is 10 A. | vedclass

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(D) The time period $T$ of the alternating current is given by $T = \frac{1}{
u} = \frac{1}{50} \ s = 0.02 \ s$.The time taken to reach the maximum v

Vedclass · Accepted Answer

$5 	imes 10^{-3} \ s$ and $14.14 \ A$

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(D) The time period $T$ of the alternating current is given by $T = \frac{1}{
u} = \frac{1}{50} \ s = 0.02 \ s$.The time taken to reach the maximum value from zero is one-fourth of the time period:$t = \frac{T}{4} = \frac{0.02}{4} = 0.005 \ s = 5 	imes 10^{-3} \ s$.The peak value of the current $(i_0)$ is related to the $r.m.s.$ value $(i_{rms})$ by the formula:$i_0 = i_{rms} 	imes \sqrt{2} = 10 	imes 1.414 = 14.14 \ A$.Therefore,the time taken is $5 	imes 10^{-3} \ s$ and the peak value is $14.14 \ A$.

The $r.m.s.$ value of an $AC$ of $50 \ Hz$ is $10 \ A$. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be:

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