An ac source is rated at 220V, 50 Hz. The time taken for voltage to change from its peak value to zero is, ( … .02 sec (c) 5 sec …
Answers:
The correct option (d) 5 × 10–3sec
Explanation:
Time to change from peak value to zero is nothing but
(T/2) – (T/4) = (2T/8) = (T/4)
Given f = 50 Hz hence T = (1/50) sec
∴ t = (T/4)
= [1/{50 × 4}]
= (1/200)
= 5 × 10–3sec.