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An alternating voltage v(t)=220sin100πt volt is applied to a purely resistive load of 50Ω .The time taken for the current to rise from half of the peak value to the peak value is (in milli-second)

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answer is 3.33.

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Detailed Solution

As V(t)= 220sin100πtSo, I(t)= 22050sin 100πti.e; I= Imsin(100πt)For I= Im t1=π2×1100π=1200secAnd for I= Im2⇒Im2=Imsin(100πt2)⇒t2=1600s∴treq=1200−1600=2600=1300s=3.33ms

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