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I'm not sure whether this has already been asked, but I post here as I couldn't find a satisfactory answer anywhere.

This classic example was used by my teacher to illustrate the effect of an inductor in a circuit.

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Consider the circuit given above.

What happens to the bulbs when you turn the switch off?

Here's my teacher's argument.

Here, when you switch the circuit off, inductor opposes a change in current and hence, bulb A is supposed to glow for a longer time.

But, I had a different answer.

The inductor resists the current decay and makes it still flow through the circuit. Since the external circuit is open, the circuit containing both the bulbs would act as a separate unit, with flowing current. Hence, both the bulbs would glow for a while.

When my teacher insisted on his answer, I had to convince myself that the current due to the inductor is just enough to feed bulb A.

I am very sure that that this is not the case, then what is?

Am I wrong in considering the inductor as a emf source?

By the way, please explain, from where the inductor pulls charges to maintain the current in the circuit?

Is it from the conducting wire ? Or from the battery ?!

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The inductor resists the current decay and makes it still flow through the circuit. Since the external circuit is open, the circuit containing both the bulbs would act as a separate unit, with flowing current. Hence, both the bulbs would glow for a while.

This is correct. After the switch is open, there is only one closed loop, and equal current must flow through the two bulbs.

By the way, please explain, from where the inductor pulls charges to maintain the current in the circuit?

The wire contains many free electrons at all times, whether a current is flowing through it or not.

As the magnetic field in the inductor core collapses (as explained in another answer), it is these electrons that are driven around the circuit.

Am I wrong in considering the inductor as a emf source?

From a physics point of view, it's correct to consider the inductor as an EMF source. We also consider it an EMF source when the battery is pushing current through it and it generates an opposing EMF proportional to its $\frac{{\rm d}i}{{\rm d}t}$.

From a circuit analysis point of view, we normally consider the inductor as current source rather than an EMF source.

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  • $\begingroup$ Can you please explain similarly for the circuit described in the comment below? $\endgroup$ – Krishnanand J Oct 8 '17 at 15:30
  • $\begingroup$ Without the resistor-bulb arm, the only way for current to continue to flow would be for an arc to form across the switch as the switch opens. This is not good for your switch and its why switches are rated for fewer cycles when switching inductive loads. Where the charge comes from is the same: it's in the wires the whole time. $\endgroup$ – The Photon Oct 8 '17 at 15:37
  • $\begingroup$ For that matter it's the same when the battery drives the circuit: The battery pushes on charge that's already in the wire, though it does so by injecting "new" charge in one end of the wire and taking it away from the other. This produces a current in the wire using the "pre-existing" charge before the "new" charge has time to move along the length of the wire. We've had many questions about this (in the case of battery as source) $\endgroup$ – The Photon Oct 8 '17 at 15:37
  • $\begingroup$ That meets my logic. Thank you for the nice explanation. $\endgroup$ – Krishnanand J Oct 8 '17 at 15:46
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When a current $I$ is flowing though the inductor $L$ there is energy stored in the inductor $\frac 12 LI^2$.

When the switch is opened that energy stored in the inductor decreases as now a decreasing current will be passing through a series circuit consisting of the inductor, resistor and two light bulbs producing light and heat in that circuit.
The rate of decay of the current will be controlled by the time constant of that circuit $\frac LR$ where $R$ is the sum of the resistances of the two bulbs, the inductor and the resistor.

There is no store of charge, rather the collapsing magnetic field in the inductor induces an emf which drives charges (a current) around the circuit consisting of the inductor, two bulbs and the resistor).

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  • $\begingroup$ Clear and precise answer. That makes everything clear, for THIS example. But , what about the same circuit without the resistor-bulb arm? From where does the inductor pull charges in this case? $\endgroup$ – Krishnanand J Oct 8 '17 at 15:25
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    $\begingroup$ @KRISHNANANDJ That comes down to a situation of the form "this circuit diagram made by ideal components cannot be analyzed completely in an ideal context," like connecting a battery's terminals with a plain wire. Instead a small amount of charge will build up on the right hand side compared to the left hand side, driving a very large voltage difference. In other words the break in the circuit acts as a capacitor with a very low capacitance and what you'd get is an RLC circuit whose natural frequency is very high. $\endgroup$ – CR Drost Oct 8 '17 at 15:32
  • $\begingroup$ Yes, I understood the idea from 'The Photon' . The idea of circuit break behaving like a capacitor is fantastic!! $\endgroup$ – Krishnanand J Oct 8 '17 at 15:51
  • $\begingroup$ @KRISHNANANDJ Carrying on from what CR Drost has written because the “resistance” in the circuit is so high when the switch is opened and hence the time constant so low, the emf induced in the circuit on opening the switch could be so high that a spark is produced across the switch contacts. $\endgroup$ – Farcher Oct 8 '17 at 16:13
  • $\begingroup$ So, RLC oscillations won't result here, likely. Now, not seriously, would it occur if the switch is replaced by a circuit break, with enough separation? $\endgroup$ – Krishnanand J Oct 8 '17 at 16:24

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