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

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

5×10−35 \times 10^{- 3} sec

Answer

5×10−35 \times 10^{- 3} sec

Explanation

Solution

Required time

t=T/4=14×50=5×10−3sec⁡t = T/4 = \frac{1}{4 \times 50} = 5 \times 10^{- 3}\sec