Two inductors L1 (inductance 1 mH, internal resistance 3Ω) and L2 (inductance 2 mH, internal resistance 4Ω), and a resistor R (resistance 12Ω) are all connected in parallel across a 5 V battery. The circuit is switched on at time t=0. The ratio of the maximum to the minimum current (Imax/Imin) drawn from the battery is :

At t=0, the inductors act as open circuits, so the current flows only through the resistor R.
Therefore, the minimum current Imin is given by:
Imin = V/R = 5V / 12Ω = 5/12 A

As t approaches infinity, the inductors act as short circuits with only their internal resistances.
The equivalent resistance of the parallel combination of L1, L2 and R is:
1/Req = 1/3 + 1/4 + 1/12 = (4 + 3 + 1)/12 = 8/12 = 2/3
Req = 3/2 Ω

The maximum current Imax is given by:
Imax = V/Req = 5V / (3/2 Ω) = 10/3 A

The ratio of the maximum to the minimum current is:
Imax/Imin = (10/3) / (5/12) = (10/3) * (12/5) = 8

Therefore, the ratio of the maximum to the minimum current drawn from the battery is 8.