Question

Current in circuit,
$\left(i\right)$ Just after closing key
$\left(ii\right)$ some time after closing the key

Answer 1

Immediately after the key is closed, the cell wants to drive a current of E/R through the circuit. But the inductor opposes the sudden change in magnetic flux through it. So, it slowly allows the current through it to build up until, after a longgg time, the current in circuit becomes E/R.

Applying Faraday’s Law through the loop containing the inductors,

let though L1, current is i1 and through L2, current is i2.

-L1 (di1/dt) + L2(di2/dt) = 0

so, L1 x i1 = L2 x i2

so current though L1 at steaddy state ,

i1 = (L2 x E ) / R(L1 + L2) and i2 = (L1 x E)/ R(L1 + L2)

Options A and B are correct.