What is the current in these 3 circuits right after the switches are opened?

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SUMMARY

The current through the resistor marked 'R' in the three circuits just after the switch is reopened is determined by the behavior of the inductor. When the switch has been closed for a long time, the current is I = V/R, as the ideal inductor has zero resistance and effectively shorts the resistors. Upon opening the switch, the current through the inductor remains I = V/R, and in cases where resistors are in parallel, the current splits accordingly. The key takeaway is that the inductor maintains the initial current through the circuit immediately after the switch is opened.

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L_landau
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Homework Statement


What is the current through the resistor marked 'R' in all three circuits just after the switch is reopened after being closed for a long time?

The Attempt at a Solution


The explanation that we were given for (1) is that I = V/2R and (2) I = V/3R and (3) I = V/(R/2) but I don't understand how this makes sense because when the circuit is opened, current shouldn't flow in unclosed parts. As I understand it the inductor should act as an energy source and form a closed loop with the resistors so that in (1) I = V/R and (2) I = V/2R and (3) I = V/(R/2). What am I missing here??
 

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Well, I disagree with both of you. In all three cases, the current through the inductor when the switch has been closed for a long time is I = V/R. Remember that an ideal inductor has zero resistance, so the resistors in parallel with the inductor are effectively shorted out. When you open the switch, the current through the inductor cannot change instantaneously, so it is still I = V/R. What current goes through the resistor R in the three cases?
 
phyzguy said:
Well, I disagree with both of you.
Yeah, and the "R" we need to define the initial current is not labeled (it is the resistor in the upper left). The resistor labeled "R" is shorted by the inductor out while the switch is closed...
 
Okay so after the switch has been closed a long time, since ideal inductors have zero resistance, then I = V/R because all the current goes through the inductor.
Once the switch is opened, the current through the inductor is still V/R, so for (1) and (2) the current is V/R and for (3) the current just gets split between the two resistors? That seems right.
 
L_landau said:
Okay so after the switch has been closed a long time, since ideal inductors have zero resistance, then I = V/R because all the current goes through the inductor.
Once the switch is opened, the current through the inductor is still V/R, so for (1) and (2) the current is V/R and for (3) the current just gets split between the two resistors? That seems right.
That would be right if the upper left resistor were labeled "R", but it's blank. Are you told that all of the resistors are identical?
 
No we are just told that the labeled R is what we are trying to find the current through. How could current flow through the other resistor once the circuit is open? I thought a circuit had to be closed for current to run through it.
 
L_landau said:
No we are just told that the labeled R is what we are trying to find the current through. How could current flow through the other resistor once the circuit is open? I thought a circuit had to be closed for current to run through it.
It doesn't. When the switch is closed for a long time, the current flows from the voltage source through the upper left resistor (call it Rul) and through the inductor (which has V=0 across it since it's ideal). That defines the initial current through the inductor, Ii = V / Rul. That is the same current that flows in the right-hand circuits right after the switch is opened.
 
Yes, I just assumed they were all R. You can't do the problem otherwise, but you need to ask.
 
Okay. Thanks for the help everyone!
 

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