Question

An ac source is rated at 220V, 50 Hz. The time taken for voltage to change from its peak value to zero is, 

(a) 50 sec 

(b) 0.02 sec 

(c) 5 sec 

(d) 5 × 10^–3sec

Answer 1

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.