# Trivial RL Transient Circuit Question...

Discussion in 'Homework Help' started by blah2222, Dec 7, 2010.

1. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
552
33
Hi, so this is so trivial, but I just can't grasp it.

I have the circuit below:

The switch opens at t = 0, and I am trying to find $I_{L}(0^{-})$.

Assumed that the circuit has been closed for a long time, so the inductor is a short. Based on this, the circuit turns into this:

I thought that $I_{L}(0^{-})$ would be 6 A, (5 + 1), but it is not and I'm not really sure how this works. Isn't the short technically just ground?

Bah, oh well, anyone lend a hand?

Thanks,
JP

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I agree that the inductor current at t=0 would be 6A. It's quite clear this would be true - so I'm not sure what answer is stated elsewhere as being correct.

3. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
552
33
Ah just found the solution to this problem, turns out they wrote the question down wrong... The switch is initially open then it is closed.

Oh well... At least my logic is still correct, darn textbooks.

Thanks