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

An alternating voltage v(t) = 220 sin 100πt volt is applied to a pure

1.	An alternating voltage v(t) = 220 sin 100πt volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is:
A.	5 ms
B.	2.2 ms
C.	7.2 ms
D.	3.3 ms
Answer» E.

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

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