A capacitor of capacitance `10muF` is connected to an AC source and an AC Ammeter. If the source voltage varies as `V=50sqrt(2)sin100t`, the reading of the ammeter is

Options

A. 50 mA.
B. 70.7 mA
C. 5.0 mA
D. 7.07 mA.

Answer

A

Explanation

(a) `V= V_(0) sin omegat`
Compared with ` V= 50 sqrt(2) sin 100t `
`omega=100`
`X_(C) = (1)/(omegaC) = (1)/(100xx10xx10^(-6))= 10^(3) Omega`
`V_(0)= 50sqrt(2)`
`V_("rms") = (V_(0))/(sqrt(2))=50`
`I_("rms") = (V_(.rms"))/(X_(C))`
`implies I_("rms") = 50 mA`
AC ammeter reads rms value.