# Simple RL circuit

1. Sep 29, 2007

### kdinser

1. The problem statement, all variables and given/known data
The switch is closed for a long time and opened instantaneously at t=0. Find and sketch the current y(t).

I haven’t had a class where I needed to do DC transient circuit analysis for over a year, so this simple thing is giving me some problems.

The way I see it, either:

1. The inductor that is parallel with the switch (2nd inductor) is fully charged at t=0 and opening the switch has no effect on the current y(t) and y(t) = 5 amps at all time t. Which is what my solution using laplace transforms is showing and what pspice seems to confirm.

OR

2. The 2nd inductor has not been charged at all at t=0 because all the current is running through the closed switch. This would mean that I'm making some kind of serious error when I set up my equation, because when I try to make this assumption, I get a final current below 5 amps at t=infinity so I know that can't be right.

Thanks for any help, if I'm wrong on both accounts, I'll start posting equations so I can get this figured out.

Just noticed that my jpg says pending approval. To draw this circuit, put a 10V voltage source in series with a 1H inductor, a 2 ohm resistor, and a 2H inductor in a closed loop. Then add a switch that opens at t=0 and is closed at t < 0 in parallel with the 2H inductor.

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Last edited: Sep 29, 2007
2. Oct 1, 2007

### doodle

I think it is more of the latter, that the 2H inductor is not charged before t = 0. My reason for saying so is as follows: The voltage across the 2H is always 0V before the switch opens. Then Ldi/dt = 0, or in other words, there is no change in the current across this inductor, hence it remains at 0A.

This, however, presents a huge problem when the switch opens since you would then have both 5A (over the 1H) and 0A (over the 2H) flowing in the same loop at t = 0 (which is impossible, of course). That being so, I would rate this problem as an ill-posed problem.