We started discussing DC RL-circuits in class and were analyzing the following circuit

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The scenario is the switch has been closed for a long time so the current through both 2-ohm resistors is the same flowing from top to bottom. The switch is then opened. Since the inductor resists the current change, the current in the 2-ohm resistor on the left now flows from bottom to top. This was fine. Then a student asked, what if there is another inductor in the leg which currently has just the 2-ohm resistor.

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I have to admit I'm stumped on this one. Both inductors will resist any current change, but as they are now in series one (or both of them) is going to have to change... My first guess would be, given the symmetry of the situation the current would somehow immediately go to 0 A... But if the situation we were not symmetric (differing resistor/inductor values)... Tried doing a bunch of Googling to no avail.

  • $\begingroup$ What if in the first diagram the other 2 ohm resistor were not there? $\endgroup$
    – Bob D
    Commented Feb 23, 2023 at 10:02
  • $\begingroup$ Check out this link: ciprian.com/wp-content/uploads/2018/01/… $\endgroup$
    – Bob D
    Commented Feb 23, 2023 at 10:22

1 Answer 1


This is an interesting question and let me start by discussing the first circuit which is relatively easy to analyse resulting in an exponential decay of current in the right hand loop after the switch is opened with an initial value $V_{\rm supply}/2$, a time constant of $0.2/4 = 0.05\,\rm s$.

The circuit can be simulated and the result $\bf 1$ is a graph of current against time with axes of current in amperes and time in seconds.

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However note that the current through the middle $2\Omega$ reverses direction "instantaneously" from downwards $(+1\rm A)$ before the switch is opened to upwards $(-1\rm A)$ after the switch is opened. The reason for this is that the circuit is limited in its modelling of the real world and "parasitic" inductance (there are loops in the circuit) and capacitance (eg the open switch) have not been included.

The simulations when there are two inductors in the circuit now follow.
Firstly when the inductors are of equal value, $\bf 2$.

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As predicted by the OP becoming zero - immediately the switch is opened???

And now with the inductors not of equal value $\bf 3$.

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Exponential decay but the initial currents no longer $1\rm A$ but $\pm \frac 23 \bf A$ and the instantaneous reversal of one of the currents which actually did start at $+1\bf A$ but "instantaneously" changed to $-\frac 23\bf A$.

To illustrate the limitations of the circuits (models) a capacitor of capacitance $1\rm fF$ has been added to the circuit.
The value was the first I chose and since it resulted in nice graphs I kept it that value.

That small value capacitor illustrates that there are no "instantaneous" changes only that some changes occur in a time scale much less than other which are being observed.

So first for the inductors having equal values $\bf 4$.

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The oscillations (ringing) are there because there is now an LCR circuit with a natural frequency of vibration which has been excited.
In this circuit the current does decay to zero over a time scale of $1.5\rm \mu s$ and that it was not seen in graph $\bf 2$ which was over a time scale of $100\rm ms$.

Now we come to the inductors of unequal value.

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Again there is ringing and then the decay to the initial values $\pm \frac 23 \rm A$ seen on graph $\bf 3$ or graph $\bf 6$ with the capacitor in place.

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Now I need to explain where the $\pm \frac 23 \rm A$ comes from.

The circuits which have been drawn with inductors in them have an associated magnetic flux when currents are passing through the inductors, $\Phi = LI$.

If changes are made to the circuits the magnetic flux cannot change instantaneously, and this is where the idea of conservation of magnetic flux has come akin to the idea of conservation of charge for capacitor.

In circuit $\bf 5$ before the switch is opened, and ignoring the small effect of any parasitic inductance, the magnetic flux is $0.5 \times 1 \color{red}- 0.1 \times 1 = 0.4\,\rm Wb$. The minus sign is there because as you look at the two inductors their windings are assumed to be the same way around but when they are considered to be in series the windings are reversed relative to one another and so produce magnetic fields in opposite directions.
After the switch is closed and the ringing has finished the magnetic flux is $(0.5+0.1) I\,\rm Wb$ where $I$ is the new current in the circuit.
Assuming that magnetic flux is conserved, $(0.5+0.1) I = 0.4 \Rightarrow I = \frac 23 \rm A$.

  • $\begingroup$ Thanks so much. This made it much clearer. My students were quite happy with the explanation. $\endgroup$
    – mkim153
    Commented Feb 25, 2023 at 14:02

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