# Source free RL circuit confusion! When this switch closes, does the current end?

Hello everyone. I'm getting confused on what my professor did. Everything else makes sense to me but this one part. THe directions say:
Aftering having been closed for a long time, the switch in the circuit is opened at t= 0. (a) find IL(t) for t>0, (b) Evaluate IL(10ms).

I got both parts right. Here is my work and the circuit:
http://img119.imageshack.us/img119/7361/lastscan2at.jpg [Broken]

Okay if you look at t<0, you assume the inductor is short circuit, thats what those 2 little dots mean instead of the curly inductor symbol. THe current Io = IL. So when the switch is closed, if you notice the circuit is redrawn, with the 20 and 10 resistors in series not even in the circuit. When that switch is closed for along time, its DC current. But wouldn't the current branch off into 2 parts? 1 going into the 20 + 10 resistors on the left and another current going to the right through the 50 ohm resistor and then going into where the inductor was?

What i'm confused on is why U can just say screw those other resistors, Io isn't affected by them. Any explanation? Thanks!

Last edited by a moderator:

nrqed
Homework Helper
Gold Member
mr_coffee said:
Hello everyone. I'm getting confused on what my professor did. Everything else makes sense to me but this one part. THe directions say:
Aftering having been closed for a long time, the switch in the circuit is opened at t= 0. (a) find IL(t) for t>0, (b) Evaluate IL(10ms).

I got both parts right. Here is my work and the circuit:
http://img119.imageshack.us/img119/7361/lastscan2at.jpg [Broken]

Okay if you look at t<0, you assume the inductor is short circuit, thats what those 2 little dots mean instead of the curly inductor symbol. THe current Io = IL. So when the switch is closed, if you notice the circuit is redrawn, with the 20 and 10 resistors in series not even in the circuit. When that switch is closed for along time, its DC current. But wouldn't the current branch off into 2 parts? 1 going into the 20 + 10 resistors on the left and another current going to the right through the 50 ohm resistor and then going into where the inductor was?

What i'm confused on is why U can just say screw those other resistors, Io isn't affected by them. Any explanation? Thanks!

Its not that the 10 and 20 ohms disappear. It is simply that the potential difference across the 50 ohms is 100 V (and the potential difference across the 10 and 20 ohms combined is also 100 volts). Do you see this? (its clear since the terminals of the battery are directly connected across the 50 ohms). So if you only want the current across the 50 ohms, you simply divide 100V by 50 ohms. However, at the same time, the current through the other two resistors is 100V/30 ohms = 3.33 A. But you did not need that information to proceed so there was no need to calculate it.

Hope this helps

Pat

Last edited by a moderator:
ahhh i c now! thanks Pat, helped alot!