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Consider a circuit as shown in this figure below-

enter image description here

Assuming switch S1 is closed such that current flows in loop ABCDE for a long time and steady state is reached. Now, my question is simple, suppose (after when the steady state is reached) the switch S2 is flipped, thereby disconnecting from point D and reconnecting to point F then, what will be the current in the loop BFGH?

Here's my problem, After the steady state is reached a nearly constant current of magnitude V/R1 will flow in loop ABCDE.

Suppose there is a current (say i) in the loop BFGH just after the switch is flipped(say at t=5sec), The inductor L2 will oppose this change in current. But if there was NO current in the loop BFGH at t =5- sec (minus sign denotes 'just before' 5seconds) and a finite current (i) at t = 5+ sec this would mean that derivative of current at t = 5 is infinity in wire GH! As a result the magnitude of 'potential drop' across inductor will become infinite! This can't happen so I concluded that the current in the loop BFGH must be zero.

But wait! If there's zero current in loop BFGH at t = 5+ and a finite current of magnitude V/R1 in wire BC implies that current derivative is again infinite in wire BC! which would mean that the magnitude of 'potential drop' across inductor L1 must be infinity! So the current in loop BFGH must not be zero.

enter image description here

So you guys can see that I'm in a fix. Can someone tell what's the problem here? What will be the current in wire GH and where am I going wrong?

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  • $\begingroup$ If you want to know what happens in a physical realization of this circuit, you should include the inter-winding capacitance of each inductor in your model. $\endgroup$ – The Photon Dec 17 '18 at 18:59
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    $\begingroup$ This is the dual of the two capacitor paradox. Like the two-capacitor paradox, the answer depends which parts of the lumped-circuit approximation you choose to relax to solve the problem. $\endgroup$ – The Photon Dec 17 '18 at 19:07
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    $\begingroup$ Can you explain in the simplest words possible, please? I'm saying this because I'm not able to understand terms like 'lumped-circuit approximation'. Looks like pretty advanced thing to me. $\endgroup$ – Shivansh J Dec 17 '18 at 19:14
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    $\begingroup$ Sorry, if there were no resistor in there, I could explain why flux is conserved in this circuit. With the resistor present, I can't convince myself that it must be. Although I expect the "conservation of flux" argument is in fact what the instructor was looking for when they put this question in an exam. $\endgroup$ – The Photon Dec 18 '18 at 5:21
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    $\begingroup$ @The Photon "'conservation of flux' argument is in fact what the instructor was looking for them when they put this question in an exam". You're ABSOLUTELY right! When I asked about this question to my instructor, he told me to conserve magnetic flux. But what I don't understand is WHY flux is being conserved here. :( $\endgroup$ – Shivansh J Dec 18 '18 at 9:28
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You got paralysed by demanding impossible in an overidealized model. Of course, if current is cut out instantaneously, the voltage on inductor will be infinite. That is a correct result. If you do not want that, you must not cut out current instantaneously.

In practice, infinite voltage does not happen because current everywhere actually cannot be cut out instantaneously. Even if the switch disconnects two wires very quickly so current ceases in the joint point between them to zero in that short time, current may continue to flow in the wires for a longer while.

The hanging ends of disconnected wires that both carried current just an instant ago become plates of a capacitor and will get charged up by the continuing current.

Charge will accumulate on the surface of the wires where the capacitance is the highest; it is actually this additional charge that is the source of high voltage that will arise between the terminals of the inductor. The current just after disconnection has an unstable value (too high), so the system will quickly but continuously change it to a stable value (corresponding to the new circuit). This is a so-called "transition phenomenon".

You may object that this shouldn't be possible with ideal wires, since they have no capacitance, but in reality, wires do have capacitance and it is what is happening. So you should redo the analysis on an enhanced model, where disconnection and connection of another wire does not cut out or create current instantaneously, but allows for variation of current to be continuous. For example, you can put additional ideal capacitor in parallel with the switch, and another (much bigger) in parallel with the inductor (since the loops of the inductor do have some capacitance between each other). Then, when the original path of the current is removed, the current will continue to flow, to charge up the added capacitors, and the system will transition to a stable stationary state for the new circuit, while all voltages remaining finite during the whole process.

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Based on comments, your instructor was looking for a solution based on conservation of magnetic flux. I'll try to explain how you'd get such a solution, but also why I think it doesn't actually apply to this problem.

Let's say we have a simpler circuit, without the series resistor shown in your example:

enter image description here

Now, in this circuit, when the switch flips from the B to A, we don't really know how high a voltage will be generated on node A, but we can imagine there's a very short impulse voltage, that equalizes the currents of the two inductors (so that KCL is satisfied after the impulse settles down).

By impulse I mean a short duration high-voltage pulse, where the actual shape of the pulse is not known, but we do know (or will work out) the integral

$$v' = \int_{-\epsilon}^{\epsilon} V\ {\rm d}t$$

for $\epsilon > 0$

Now, because of Kirchhoff's voltage law, we know the impulse applied to L2 is exactly equal to the impulse applied to L1. And since magnetic flux in the inductor is equal to the integral of voltage applied over time, we find the total flux in the two inductors after the switch flips is equal to the flux before the switch flipped.

This problem is something like the well-known two-capacitor paradox, where a switch instantaneously connects two capacitors initially charged to different voltages. In the capacitor case, we solve the problem using conservation of charge. And it's somewhat obvious that charge must be conserved, since charge is associated with actual physical particles (electrons) that can't be created or destroyed. I'm not aware of any reason that magnetic flux must be conserved the same way as charge must. Like in the two-capacitor paradox, we'll find in this inductor circuit that energy isn't conserved when the switch is flipped.

Now, what about the circuit your instructor presented?

enter image description here

The problem with analyzing this circuit is that the way we develop our understanding of an impulse signal is to consider a signal with a finite width and finite peak voltage. Then we take the limit as the pulse width is shortened but the peak voltage is increased to keep the integral (above) constant.

So thinking about the case with a finite width pulse, we'll find in the new circuit, that the voltage across L2 is not identical to the voltage across L1, because as soon as the current through L2 is non-zero, there will be a non-zero voltage across R2, and the voltage across L2 will be reduced. As far as I can see, this applies no matter how narrow a pulse we consider, since R2 is an ideal resistor whose voltage is instantaneously proportional to its current. Since we have a problem no matter how narrow the pulse gets, we also have the same problem in the limit as the pulse width goes to 0.

So, I don't think conservation of flux applies to the example problem.

And I'd be happy to learn from anybody who has a solution for this version of the problem.

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