Two inductors L i (inductance 1   mH , internal resistance 3   Ω ) and L 2 (inductance 2…

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

$I_{max}=\frac{\epsilon }{R}=\frac{5}{12}A$ (Initially at t = 0)

Upon reaching steady state, the inductances will act as pure resistors and hence the current will be minimum.
$I_{min}=\frac{\epsilon }{R_{eq}}= \epsilon \left(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{R}\right)$ (Finally in steady state)
$=5\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{12}\right)$
$=\frac{10}{3}A$
$\frac{I_{max}}{I_{min}}=8$