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In the circuit below, the switch has been in position 1 indefinitely long and suddenly switches to position 2 at t=0. I'm asked to find the initial voltage of the inductor at t=0.

My professor lists the answer as 15-5(5)=-10 V (not sure where he's getting -5(5)), but I don't see why.

Here's how I'm thinking through the problem:

Since the initial current in the inductor is 5 A at t=0- (from when the switch is at position 1) and then 0 A at t=0+ (when the switch is at position 2, the capacitor prevents any current), di/dt is "infinite" (technically the derivative does not exist) meaning that the voltage should also be "infinitely high." This also makes sense intuitively as the sudden drop in current will produce an arbitrarily large induced EMF (at least in the ideal case).

Is there an error in my reasoning? I'd appreciate any feedback!

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  • $\begingroup$ The capacitor only has zero current under dc conditions. At the instant you move the switch you have transient conditions, so current will flow through the capacitor. $\endgroup$ Apr 29, 2022 at 16:12
  • $\begingroup$ Can you clarify on that? I know that realistically a current through the capacitor would appear a short time later, but theoretically wouldn't that happen after t=0 since the capacitor acts as an open? Supposing there was such a current, how would you compute it? $\endgroup$
    – James
    Apr 29, 2022 at 17:01
  • $\begingroup$ The current through the capacitor is determined by the rest of the circuit in this case. It appears at t=0+, as soon as the switch is moved to position 2. Not a "short time later" but instantly. What makes you think that it would not? $\endgroup$ Apr 29, 2022 at 17:28

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The whole idea of using inductors is the fact that their currents do not change instantly. You assumed infinite di/dt, but that doesn't happen. The inductor generates just as high voltage as needed to allow the current to continue so that it decays gradually with finite di/dt. The voltage is L(di/dt) if it really is an inductor.

Let's assume a moment the switch is a practical one. Even during the flight of the switch there's an arc inside. That's because, as I said, the inductor will not allow current changes to happen instantly. Otherwise it's not an inductor, but something else.

It's just like a permanent item which has non-zero mass doesn't allow instant changes of its velocity, its velocity changes always gradually. A guy named Newton (never heard?) became famous when he published a book (=Principia) which told it among other as popular and practically useful things.

Arc needs energy and if the flight of the switch lasts long enough the magnetic energy is used before the switch reaches position 2. In that case the inductor current is zero when the switch reaches position 2. But this obviously is an ideal switch. Its's flight time is zero, so at t=0 the inductor current is that already said 5A. The situation changes obeying a 2nd degree differential equation, because there's also a capacitor taken along.

The problem is not properly defined before one knows the initial charge of the capacitor at t=0. Of course, one can guess that it might be zero, but nothing already shown tells it.

BTW to avoid an analog problem with capacitors try to learn to start to think that the voltage between the poles of a capacitor never changes instantly. The voltage changes as current charges or discharges the capacitor. In this case the current goes through the inductor and you have already got to know that the initial value of that current is 5A, not zero. It's true that in practical circuits often the currents of the capacitors decay towards zero if there's no sources of permanently changing voltages, but here the initial value is 5A as the inductor declares if it really is an inductor.

In this old case https://electronics.stackexchange.com/questions/282053/how-does-the-inductor-really-induce-voltage you can find a practical induction self-teaching machine. Do not try it to others.

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