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If resistance of 100Omega , inductance o...

If resistance of `100Omega` , inductance of `0.5` henry and capacitor of `10xx10^(-6)F` are connected in series through `50Hz` AC supply, then impedence is

A

`1.876`

B

`18.76`

C

`189.72`

D

`101.3`

Text Solution

Verified by Experts

The correct Answer is:
C
`Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))`
`=sqrt(100^(2)+(0.5xx100pi-1/(10xx10^(-6)xx100pi))^(2))`
`=189.72 Omega`
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