Question

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`

Answer 1

Correct Answer - 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`