An alternating current of frequency 50 ,Hz ha | vedclass

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(C) Given frequency $f = 50 \,Hz$. The time period $T$ is given by $T = \frac{1}{f} = \frac{1}{50} = 0.02 \,s$.The time taken to reach from zero to th

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$0.005 \,s, 10 \,A$

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(C) Given frequency $f = 50 \,Hz$. The time period $T$ is given by $T = \frac{1}{f} = \frac{1}{50} = 0.02 \,s$.The time taken to reach from zero to the maximum value (peak) is one-quarter of the time period:$t = \frac{T}{4} = \frac{0.02}{4} = 0.005 \,s$.The peak current $I_0 = 14.14 \,A$. The r.m.s. value of the current $I_{rms}$ is given by:$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{14.14}{1.414} = 10 \,A$.Thus,the time taken is $0.005 \,s$ and the r.m.s. value is $10 \,A$.

An alternating current of frequency $50 \,Hz$ has a peak value of $14.14 \,A$. The time taken by the alternating current to reach from zero to its maximum value and the root mean square (r.m.s.) value of the current are respectively:

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