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"`