Question

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.

Answer 1

When the coil is connected to `dc` source the final current is decided by the resistance of the coil. `therefore r=10/2.5=4 Omega`
When the coil is connected to `ac` source, 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)=3 Omega`
`therefore omegaL=2pifL=3`
`therefore 2pi50L=3`
`therefore L=3//100pi` Henry