An alternating voltage `E=200sqrt(2)sin(100t)` is connected to a `1` microfarad capacitor through an AC ammeter. The reading of the ammeter shall be

Options

A. 5 mA
B. 10 mA
C. 15 mA
D. 20 mA

Answer

D

Explanation

Given, `e=200sqrt(2)sin(100t)` . . .(i)
Capacitance of capacitor `(C)=1muF=1xx10^(-6)F`
The standard equation of voltage of AC is given by
`V=V_(0)sinomegat` . . .(ii)
On comparing Eqs. (i) (ii), we get
`V_(0)=200sqrt(2)`
`omega=100`
We know that,
`l_(rms)=(V_(mrs))/(X_(C))`
But `V_(rms)=(V_(0))/(sqrt(2))" "L_(rms)=(V_(0))/(sqrt(2)X_(c))`
`L_(mrs)=(V_(0)omegaC)/(sqrt(2))" "(becauseX_(C)=(1)/(omegaC))`
`l_(rms)=(200xxsqrt(2)xx100xx1xx10^(-6))/(sqrt(2))`
`l_(rms)=20xx10^(-3)A=20mA`