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A coil has L=0.04H and R=12Omega. When i...

A coil has `L=0.04H` and `R=12Omega`. When it is connected to `220V, 50Hz` supply the current flowing through the coil, in ampere is

A

`10.7`

B

`11.7`

C

`14.7`

D

`12.7`

Text Solution

Verified by Experts

The correct Answer is:
D
Impendence `Z=sqrt(R^(2)+4pi^(2)v^(2)L^(2))`
`=sqrt((12)^(2)+4xx(3.14)^(2)xx(50)^(2)xx(0.04))=17.37A`
Now current `i=V/Z =220/17.37=12.7 Omega`
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