vmc workshop doubt 2

After some time ..the inductor will be short-circuited and then the flow of current is totally dependent on resistance..since current in both the circuits is same after a long time that means resistance is same in both of them...
Now draw a line at some time t u will find that current is greater in 1 than 2..
I_{1 } = I_{0}( 1 - e^{\frac{-Rt}{L_{1}}})....I_{2 } = I_{0}( 1 - e^{\frac{-Rt}{L_{2}}}) ..
since I₁ > I₂...
( 1 - e^{\frac{-Rt}{L_{1}}}) > ( 1 - e^{\frac{-Rt}{L_{2}}}) \Rightarrow e^{\frac{-Rt}{L_{1}}} < e^{\frac{-Rt}{L_{2}}}...

Now taking log on both sides..
\frac{-Rt}{L_{1}} < \frac{-Rt}{L_{2}} \Rightarrow \frac{Rt}{L_{1}} > \frac{Rt}{L_{2}} ..so ....L_{2} > L_{1}

Hence the answer is B and D...