Question:
At time t=0, an initially unenergized 10-H inductor is connected to a 12-V battery for two seconds, then short-circuited. What is the current in the inductor \(-\infty<t<\infty \)

Below is the work I have done so far and I'm not sure if it is correct.
I assume the method to use is for inductors
For the two seconds when it is connected to the battery, i have determined the current follows
\(V=L {di\over dt}
V=12
i=\int {V \over L} dt
i(t)={12 \over 10} t
\)
But this doesn't seem right because it's unreasonable that the current is unbounded and can just increase indefinitely.
I haven't a clue what the current in the inductor is like when it is short-circuited.

 

The inductor is only connected for 2s, so the current is not unbounded. In an inductor connected to a voltage source, the current does increase forever... (neglecting any resistance).

di/dt = V/L = 12/10 = 1.2A/s. That is the rate of change of the current; not the current itself.

 

Oh okay, I got it, thanks so much. So could you explain to me what occurs when the inductor is disconnected and then short-circuited? I assume the current goes to 0 immediately because there is no resistance.

 

.. So could you explain to me what occurs when the inductor is disconnected and then short-circuited? I assume the current goes to 0 immediately because there is no resistance.

Your assumption is incorrect.

A little memory crutch for you to remember.

1. Current in an inductor wants to keep flowing...
2. It takes an infinite voltage to make a step change in inductor current...

So with respect to establishing a current in an inductor, then suddenly removing the voltage source and replacing it with a short circuit, apply #1, above...

 

Understood. Thanks so much for your help!

The correct answer would be that since there is no resistance, the current through the inductor stays at the current that it was at the instant the battery was removed and the short circuit was established.
I will mark this problem as solved as it has been.
SOLVED