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An alternating voltage V(t)=220 100 t volt is applied to a purely resistive load of 50 . The time taken for the current to rise from half of the peak value to the peak value is:
(2) 3.3 ms
Pure R i=i0 100 t (in phase with V)
i= i02 at 100 t1= 6; i=i0 at 100 t2= 2
t2-t1= /2- /6100 = 1300 s
=3.3 ms
Answer: (2)
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