Calculate iinf/i0, the ratio of the two currents.

After having been in the open position for a very long time, the switch S in the following circuit is closed (t=0). The current through the battery immediately after the switch is closed is i0(t=0) and its final value (after a very long time) is iinf (t → infinity.) Calculate iinf/i0, the ratio of the two currents. Use the following data: EMF = 3.0 V, R1 = 83.0 Ω, R2 = 35.0 Ω, L = 32.0 mH.

http://img831.imageshack.us/i/prob08lrc.gif/ figure

Since. the inductor current lags the resistive current; initially the
inductance appears as open circuit and io determined by resistance
only:

Io = E/(R1+R2) = 3/(83+35) = 25 mA

After a very long time the inductance appears as a short circuit and
so the voltage across it is zero.

I∞ = E/R1, v_L → 0, (s.c.)

I∞ = 3/83 = 36 mA

I∞/io = (E/R1)/(E/(R1+R2)) = (R1+R2)/(R1) = (83+35)/83 = 1.42