# A contradiction regarding flux conservation

Consider a simple circuit Initially switch S was opened till steady state is reached and then it was closed. My understanding regarding inductors is that you can't have a SUDDEN change in current flowing through an inductor otherwise it would result in VERY LARGE (infinite) potential difference across it. (as the magnitude of time derivative of current will shoot to infinity!).

After switch S is closed, we're asked to find the current across inductor 1. My teacher told me to apply flux conservation to find the current just after the switch S is closed. Flux conservation would say that current is suddenly changed as soon as switch is closed.

But how is it possible to have a SUDDEN change in current flowing across inductor 1, moreover I still don't understand exactly WHY flux is conserved.

Is my understanding regarding inductors wrong?

1. How to resolve this contradiction?
2. Why Flux is conserved?

EDIT: The flux conservation says that current will suddenly change because- Clearly initial current (when switch S is opened and steady state is reached) isifferent from final (after switch S is closed).

• Why do you say that flux conservation means that the current would change suddenly? – Hearth Jan 18 '19 at 15:25
• Sorry, but your added equations don't make any sense. In a dc circuit an inductor acts like a short circuit, a resistance of zero ohms, and the inductance value is irrelevant. In a dc situation the current is not affected the slightest bit by the value of the inductors. – Elliot Alderson Jan 18 '19 at 15:33
• Whoever made this question was probably assuming that when you close the switch, it's the same as removing L2 from the circuit also (e.g, using a SPDT switch instead of using a SPST). Definitely ask your instructor. It may be worthwhile to solve the alternate problem, so you have a solution ready depending on what they say. – Justin Jan 18 '19 at 16:04

In steady state, the voltage across L1 and L2 is zero, each, simply because a inductor has zero effective resistance to DC.

So, closing the switch ironically won't do anything in this idealized model: with zero voltage across the switch, there's no reason for current to flow through the switch.

So, closing the switch changes nothing in this idealized model.

One could, however, just as well say: well, if I put two perfect conductors (the closed switch and L2 for DC) in parallel, then the current would be split half-and-half; but that's really just a fallacy, because you simply can't find a reason that this should be happening. The current-divider rules you know only apply to cases where there's some impedance in both branches; that's not the case here, and the math becomes ill-defined.
What will happen with this circuit in reality is that of course, an inductor has ohmic resistance, and so the voltage across L2 isn't actually zero. A current will hence flow through the switch once closed, in the opposite direction as the current through L2. That current will need to come from "somewhere". So, yes, the current through L2 will change in reality, but certainly not SUDDENly, to put it in your words. See Scott's answer for a discussion of the differential equation you'd need to solve.

• @ShivanshJ No. At dc the voltage across the inductor must be zero. Therefore, closing the switch does not change the voltage across the inductor. Therefore, closing the switch does not change the current through the inductor. – Elliot Alderson Jan 18 '19 at 15:35
• @ShivanshJ Sure, but it's still zero assuming ideal components. – Hearth Jan 18 '19 at 15:37
• @ShivanshJ it doesn't change, as I wrote, and as Elliot and Hearth both confirmed for you. The potential before, while and after changing is simply 0. – Marcus Müller Jan 18 '19 at 15:37
• I don't understand where your first equation even comes from, so I guess that yes, it's wrong. – Marcus Müller Jan 18 '19 at 15:39
• Then you didn't understand anything: the flux is conserved. There isn't a sudden change in current. There can't be. end of story. – Marcus Müller Jan 18 '19 at 15:46

$$\ V= L \frac{di}{dt}\$$, no two ways about it.

If $$\\frac{di}{dt}\$$ is infinite, so is the voltage across the inductor.

This is not a contradiction. It is very hard to produce a truly infinite rate of change of current -- especially when the inductor varies from ideal with an effective series resistance.