A resistance of `20 Omega` is connected to a source of an alternating potential `V=220 sin (100 pi t)`. The time taken by the current to change from the peak value to rms value is

Options

A. 0.25 sec
B. `25 xx 10^(-3) sec`
C. `2.5 xx 10^(-3) sec`
D. 0.2 sec

Answer

C

Explanation

Peak current = `I_(0)` and r.m.s. current = `(I_0)/(sqrt(2))`
We want to make `i=(I_0)/(sqrt(2))`
`:. (I_0)/(sqrt(2)) = I_(0) sin (100 pi t)`
`:. sin (100 pi t) = 1/(sqrt(2)) = sin 45^(@) = sin (pi)/4`
`:. 100 pi t = (pi)/4`
`:. t = 1/(4 xx 100) = 1/(400) = 2.5 xx 10^(-3)s`.