An alternating e.m.f. `e=50sqrt(2) sin(100 t)`, its connected to a capacitor `C=1 mu F`. Then the reading shown by the a.c. ammeter connected in the circuit is

Options

A. `2.5 mA`
B. `5 sqrt(2) mA`
C. `5 mA`
D. `5/(sqrt2) mA`

Answer

C

Explanation

`e_(rms)= (e_0)/(sqrt2) = (50sqrt2)/2 = 50V`
`sin(omega t) = sin (2pift) = sin (100t)`
`:. F = 100/(2pi)`
and `X_(C) = 1/(omegaC) = 1/(2 pifC)`
`=(1)/(2pixx100/(2pi)xx10^(-6))= 10^(4)`
`:. I_(rms) = (e_rms)/(X_C) = 50/(10^4) = 5 xx 10^(-3) = 5mA`.