When an inductor coil is connected to an ideal battery of emf `10V`, a constant current `2.5 A` flows. When the same inductor coil is connected to an ac source of `10V` and `50 HZ` then the current is `2A`. Find out the inductance of the coil.

Explanation

When the coil is connected to dc source, the final current is decided by the resitance of the coil.
`therefore r=(10)/(2.5)=4Omega`
When the coil is connected to ac, the final current is decided by the impedance of the coil.
`therefore Z=(10)/(2)=5Omega`
`"But" Z=sqrt((r)^(2)+(X_(L))^(2))=X_(L)^(2)=5^(2)-4^(2)=9`
`X_(L)=3Omega`
`therefore omegaL=2pifL=3`
`therefore 2pi 50 L =3 " " therefore L=3//100pi"Henry"`