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An alternating voltage E=200sqrt(2) sin ...

An alternating voltage `E=200sqrt(2) sin (100 t)V` is connected to a `1 (mu)F` capaacitor through an ac ammeter (it reads rms value). What will be the reading of he ammeter?

A

`10 mA`

B

`5 mA`

C

`5 sqrt(2)mA`

D

`10 sqrt(2)mA`

Text Solution

Verified by Experts

The correct Answer is:
B
`X_(C) = 1/(omega C) =(1)/(100 xx 10 ^(-6)) = 10^(4) Omega`
so, ` I_(rms) = (E_(rms))/(Z) = (50 sqrt(2))/(sqrt(2) xx 10^(4)) = 5 mA`.
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