An alternating voltage v(t) = 220 sin 100 pt volt is applied to a purely resistive load of `50Omega` . The time taken for the current to rise from half of the peak value to the peak value is :

Options

A. 2.2 ms
B. 3.3 ms
C. 5 ms
D. 7.2 ms

Answer

B

Explanation

(b) : `V=V_(0) sin omegat `
Time taken toi reach half the peak value from zero can be calculated as follows :
`(V_(0))/(2) = V_(0) sin omegat_(1) implies omegat_(1)= (pi)/(6) implies t_(1) = (pi)/(6omega)`
Time taken to reach peak from zero can be written as follows :
`t_(2) = (T)/(4) = (2pi)/(4xxomega)= (pi)/(2omega)`
Time taken to reach peak from half the peak .
`Deltat = t_(2) -t_(1)=(pi)/(2omega) - (pi)/(6omega) = (pi)/(3omega)= (pi)/(3xx100pi)= 0.0033 s `
=3.3 ms
Hence option (b) is correct .