1.

The frequency of an alternating current is 50 Hz. What is the minimum time taken by current to reach its peak value from rms value ?

Answer»

`0.02s`
`5 xx 10^(-3) S `
`10 xx 10^(-3) S `
`2.5 xx 10^(-3) S`

SOLUTION :`i=i_0 sinomega t`
` at= (PI)/(4) ,t_1 =(1)/( 8 f)`
Att `=t_2, I =i_0 ,impliest_2(1)/(4f)`
` Deltat=t_2 -t_1 = 2.5ms`


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