The r.m.s value of an a.c of `59 Hz` is `10 A`. The time taken by the alternating current in reaching from zero to maximum valuef and the peak value of current will be

Options

A. `2 xx 10^(-2) "sec"` and `14.14 A`
B. `1 xx 10^(-2) "sec"` and `7.07 A`
C. `5 xx 10^(-3) "sec"` and `7.07 A`
D. `5 xx 10^(-3) "sec"` and `14.14 A`

Answer

4

Explanation

`i_(0) = sqrt(2) i_("rms")` , `T = (1)/(f), t = (T)/(4)`