Calculate the back `e.m.f` of a `10H`, `200 Omega` coil `100 ms` after a `100V` d.c supply is connected to it.

Explanation

The value of current at `100 ms` after the switch is closed is
`I-I_(0)[1-e^((-t)/(T_(0)))]`. Here, `I_(0) = (100)/(200) = 0.5 amp`,
`tau_(0) = (L)/(R ) = (10)/(200) = 0.05 sec , t = 0.1 sec`
`I = 0.5(1-e^(-0.1//0.05)) = 0.5(1-e^(-2)) =0.4325A`
Now, `E = IR + L(dI)/(dt)`, or
`100 = 0.4325 xx 200 + L(dI)/(dt)`
Back `e.m.f = L(dI)/(dt) = 100-0.4325 xx 200 = 13.5V`