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An alternating voltage v(t) = 220 sin 10...

An alternating voltage v(t) = 220 sin 100 pt volt is applied to a purely resistive load of `50Omega` . The time taken for the current to rise from half of the peak value to the peak value is :

Text Solution

Verified by Experts

As `V(t)=22sin 10 pi t`
So `I(t)=220/50sin 100 pi t`
i.e. `I=I_(m)=sin(100pi t)`
For `I=I_(m)`
L `t_(1)=(pi)/2xx1/(100pi)=1/200` sec
and for `I=(I_(m))/2`
`implies(I_(m))/2=I_(m) sin(100pi t_(2))`
`implies (pi)/6=100pit_(2)impliest_(2)=1/600s`
`:.t_("req")=1/200-1/600=2/600=1/300s=3.3ms`
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