- #1
mr_coffee
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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, that's 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!
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, that's 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!
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