Two inductors L _ i (inductance 1mH , internal resistance 3Ω ) and...

Initally, the inductances will offer infinite resistance and the current will be maximum.

{I}_\max = \dfrac{ε}{R} = \dfrac{5}{12}A (Initially at t = 0)

Upon reaching steady state, the inductances will act as pure resistors and hence the current will be minimum.
{I}_\min = \dfrac{ε}{{R}_{eq}} = ε\left(\dfrac{1}{{r}_{1}} + \dfrac{1}{{r}_{2}} + \dfrac{1}{R}\right)$(Finally in steady state)<br >$= 5\left(\dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{12}\right)$<br >$= \dfrac{10}{3}A$<br >$\dfrac{{I}_\max}{{I}_\min} = 8