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

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(D) The time taken by an alternating current to reach its maximum value from zero is $\frac{T}{4}$,where $T$ is the time period.Given frequency $f = 5

Vedclass · Accepted Answer

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

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(D) The time taken by an alternating current to reach its maximum value from zero is $\frac{T}{4}$,where $T$ is the time period.Given frequency $f = 50\,Hz$.The time period $T = \frac{1}{f} = \frac{1}{50} = 0.02\,s$.Time taken to reach maximum value $= \frac{T}{4} = \frac{0.02}{4} = 0.005\,s = 5 	imes 10^{-3}\,s$.The peak value of current $I_0$ is related to the $r.m.s.$ value by $I_{rms} = \frac{I_0}{\sqrt{2}}$.Therefore,$I_0 = I_{rms} 	imes \sqrt{2} = 10 	imes 1.414 = 14.14\,A$.

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

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