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An inductance of 1mH a condenser of 0.5 ...

An inductance of `1mH` a condenser of `0.5` henry and capacitance of `10xx10^(-6)F` are connected in series through `50Hz` AC supply, then impedence is

A

`100 Omega`

B

`30 Omega`

C

`3.2 Omega`

D

`10 Omega`

Text Solution

Verified by Experts

The correct Answer is:
D
Given `omegaL=1/(omegaC)implies omega^(2)=1/(LC)`
or `omega=1/(sqrt(10^(-3)xx10xx10^(-6)))=1/(sqrt(10^(-8)))=10^(4)`
`X_(L)=omega L =10^(4)xx10^(-3)=10Omega`
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