Question

A 20 H inductor is placed in series with 10 W resistor and an emf of 100 V is suddenly applied to the combination. At t = 1 s from t = 1 s from the start, find the rate at which energy is being stored in the magnetic field around the inductor (givne `e^(-0.5)=0.61`).

Answer 1

`i_(0)=(epsilon)/(R)=10A`. For growth of current, `i=i_(c)(1-e^((-Rt)/(L)))`
`therefore" "e=10(1-e^(-0.5))=3.9A`
Energy `U=(1)/(2)Li^(2)rArr (dU)/(dt)=Li(di)/(dt)=(R)/(L)i_(0)(1-e^((-Rt)/(L)))(i_(0)e^((-Rt)/(L)))=237.9W.`