Simple RL Circuit: Verifying Current Flow After Switch Opened

[kdinser]

I posted this is the homework section, but it looks like EE homework problems are getting more attention here.

I'm pretty sure I've got this right, but I would like to make sure before I hand this is.

V=10V DC source
I1 = 1H inductor
I2 = 2H inductor
R = Resistor
S= a switch that has been closed for a long time and is opened at t=0


|----I1---R---|----|
| | |
V I2 S
| | |
|-------------|----|


Am I correct in assuming that opening the switch will have no effect on the current running through I1 and R? In other words, would I be right in assuming that after the switch has been closed for a long time, I2 will become fully charged and can be treated as a short circuit in this simple problem?

Thanks for any help.

 

[Corneo]

Yes your assumption of I2 is a short right before the switch is open. But once S is open something will happen to I1 and R.

PS: On a practical side point of view, do you see a problem with S being closed and parallel with a DC power supply?

 

[kdinser]

Thanks for the help corneo, but I forgot what HTML does to extra spaces.

Here is a link to the correct circuit.

https://www.physicsforums.com/attachment.php?attachmentid=11049&d=1191075148

If someone could get a chance to look at this before 3:00 pm EST and help me out, I would really appreciate it.

 

[alvaros]

If this is an ideal circuit, which means that Rs are Rs and Ls are Ls and nothing else, there is no answer for the question because: at time<0 the I2 inductor and the switch have 0 Ohms, so how big is the current throug the switch and trough I2 ?
I there is a current throug the switch the answer if the same that for the real case.

I this is a real circuit all the current is passing through the switch at time<0.
Once you open the switch there will be a spark across the switch, because the current can't change in inductors in 0 time.

V = L di/dt

I you want to change the current very fast in an inductor the voltage will rise until there is a spark anywhere.

The problem is not so simple beacuse you have two 0 ohms in parallel !

 

[kdinser]

Thanks alvaros, my professor agreed that it was not a well thought out question. According to him, the solution manual indicates that the 2H inductor should be considered uncharged at t=0. I'm wondering if there was supposed to be a resistor in series with the 2H inductor that didn't make it into the final problem.

Either way, when I assume the inductor to be uncharged, I'm able to get the math to come out right. At t=0 the inductor begins charging and after a few seconds, the current rises back up to 5 amps.

 

[CEL]

Thanks alvaros, my professor agreed that it was not a well thought out question. According to him, the solution manual indicates that the 2H inductor should be considered uncharged at t=0. I'm wondering if there was supposed to be a resistor in series with the 2H inductor that didn't make it into the final problem.

Either way, when I assume the inductor to be uncharged, I'm able to get the math to come out right. At t=0 the inductor begins charging and after a few seconds, the current rises back up to 5 amps.

An inductor is considered a short circuit only if a constant current flows through it. If the switch is closed at time -infinity, no current flows through I2, so it is uncharged at t = 0.
The problem with the circuit is that at t = 0+, you have two inductors in series, one with a 5A current through it and the other with 0A.
The solution to the problem is that inductor I2 is submitted to an impulse of voltage, so that its current rises instantaneously to 5A.