Let us take the following question as an example:

enter image description here

For the above question I drew the corresponding Laplace transform diagram, as follows (didn't draw the switch since it basically open circuit after $t=0$):


For the inductor on the upper right, note that I plugged in the value of $i(0^{-})$ that is, $3A$, as that was the current that was flowing through it when the switch was closed for a long time (as $\frac{12 V}{4\Omega}=3A$).

The loop equation thus turns out to be:

$$\frac{12}{s}-4I(s)-2sI(s)+6-sI(s)-4I(s)=0$$ $$\implies I(s)=\frac{12+6s}{8s+3s^2}$$

Which on Inverse Laplace transform gives me the actual loop current in time domain as $i(t)=\frac{3}{2}+\frac{1}{2}e^{-8t/3}$.

Clearly, $i(0^{+}) = \lim_{t\to 0^{+}}i(t)=\frac{3}{2}+\frac{1}{2}=2$. Thus, $i(0^{+})$ is quite different from $i(0^{-})$, which is $3$ (in amperes).

Can we logically explain the sudden jump in current when an active inductor is connected in series with an inactive inductor? Or, is my conclusion wrong?

  • 1
    $\begingroup$ It's not a great idea to accept the first answer you get right away. Wait 24 hours for people in all time zones to get a chance to answer. Having an accepted answer tells possible other answerers you're already satisfied with the answer you have, so there's no need for them to contribute. $\endgroup$ – The Photon Apr 22 '18 at 17:00
  • $\begingroup$ Okay, I'll keep that in mind next time. But Farcher's answer looks reasonable to me. Do you have any objection? I noticed that someone downvoted it (probably you) @ThePhoton $\endgroup$ – user193454 Apr 22 '18 at 17:02
  • $\begingroup$ Yes. His talk about the switch capacitance has nothing to do with the model you presented. $\endgroup$ – The Photon Apr 22 '18 at 17:05
  • $\begingroup$ @ThePhoton Okay, I unaccepted that answer for the time being. If you have the time, please elaborate on your viewpoint $\endgroup$ – user193454 Apr 22 '18 at 17:07
  • $\begingroup$ If you want to introduce a parasitic capacitance, consider the interwinding capacitances of the inductors. $\endgroup$ – The Photon Apr 22 '18 at 17:08

You have a current of $3$ A passing though the switch and no current through the lower resistor and inductor combination.

The switch is opened and that current of $3$ A cannot change instantaneously through the top resistor and inductor combination and the current in the bottom resistor and inductor combination cannot change from zero instantaneously.

The act of opening the switch means that the parasitic interwinding capacitance of the inductors becomes significant and that capacitor starts to charge up with the initial charging current being $3$ A.

The capacitance of the windings of the inductor allows for “smooth” changes in the currents which flow in the circuit.

It might also be the case that the voltage across the switch contacts becomes so large that the air between the contacts becomes a conductor and adding an extra resistance in the circuit.

  • $\begingroup$ I have two confusions. 1) You say that before the switch is opened, there is a current of $\frac{3}{2}$ A passing through the circuit. That is false. Initially, the bottom branch is short-circuited and the only resistance in the circuit is $3\Omega$. Hence, before the switch is opened the current in the $2$ H inductor is $3$ A, and not $3/2$ A (since $(12V)/(4\Omega) = 3 A$). Meanwhile, the current in the lower inductor is $0$ A (which you rightly said). 2) Do you agree with me or not that there is a sudden jump in current from 3 A to 2 A, after the switch is opened? Or are my equations wrong? $\endgroup$ – user193454 Apr 22 '18 at 16:13
  • $\begingroup$ I feel that at $t=0^{-}$ the current in the series circuit is $3$ A, but immediately after the switch is opened, at $t=0^{+}$, the current becomes $2$ A. Then it decays from $2$ A to $1.5$ A. $\endgroup$ – user193454 Apr 22 '18 at 16:13
  • $\begingroup$ @user554252 Sorry, my mistake about the currents which I have now corrected. Your equations are correct but the assumption as to what happens around the time the switch is opened leads to a situation which cannot happen. My point is that a current cannot change instantaneously from $3$ A to $2$ A and so there must be a way for the current to change smoothly and that is done by introducing a capacitor in the circuit as well as any other parasitic capacitance in the circuit. $\endgroup$ – Farcher Apr 22 '18 at 16:30
  • $\begingroup$ Thanks. Yes, I do agree that the current cannot instantaneously change from $3$ A to $2$ A. But, firstly do my equations look correct? I want to make sure that the fact I'm getting at $t\to 0^{+}$, $i(t)$ is $2$A, isn't due to incorrect math on my part. There is surely some small time $2\epsilon$ for the circuit to change from $t=0^{-}$ to $t=0^{+}$. $\endgroup$ – user193454 Apr 22 '18 at 16:33
  • $\begingroup$ Initially the lower branch was short-circuited. So the current there is 0. But the current through the conducting wire is indeed 3 A. $\endgroup$ – user193454 Apr 22 '18 at 16:51

Can we logically explain the sudden jump in current when an active inductor is connected in series with an inactive inductor?

In the context of ideal circuit theory, the solution to this transient problem includes a voltage impulse at $t=0$ such that the current can change instantaneously. To see this, 'regularize' the circuit by introducing a resistor of resistance $R$ in parallel with the $2\,\mathrm{H}$ inductor.

At $t=0+$, the current through the $1\,\mathrm{H}$ inductor must be zero and so the entire $3\,\mathrm{A}$ current through the $2\,\mathrm{H}$ inductor must be through the resistance $R$. The voltage across the $1\,\mathrm{H}$ inductor is then

$$v_{1\mathrm{H}}(0+) = 12\,\mathrm{V} + 3\,\mathrm{A}\cdot R$$

while the voltage across the $2\,\mathrm{H}$ inductor is

$$v_{2\mathrm{H}}(0+) = - 3\,\mathrm{A}\cdot R$$

Now, see that as $R \rightarrow \infty$, these instantaneous voltages go to infinity and the time constant involving $R$ goes to zero, i.e., there is a voltage impulse across each inductor when the switch is opened. Since the inductor voltage is proportional to the time derivative of the inductor current, a voltage impulse implies a current step.

Note that the voltage impulse is positive for the $1\,\mathrm{H}$ inductor implying that the current instantaneously increased. Similarly, since the voltage impulse is negative for the $2\,\mathrm{H}$ inductor, the current instantaneously decreased.

But a voltage impulse is unphysical so we know that this ideal circuit doesn't adequately model a physical version of this circuit. Physical inductors have parasitic capacitance that hasn't been included in this ideal circuit. Further, a physical switch cannot instantaneously open a circuit and more, the contacts can arc over if the voltage across becomes too large.

So, while this is an interesting exercise, it's important to keep in mind that the solution is unphysical.


Consider that to instantaneously change an inductor's current, you need to apply a voltage impulse ($v(t)=V_0 \delta(t)$). Opening the switch would cause the first inductor to generate an impulsive back-emf as its current changes rapidly. This impulse would be taken up by the other inductor, changing its current also. Net effect, like you said in other comments, will be conserving flux.

In the real world, the inductors' interwinding capacitances and/or the switch arcing would limit the maximum voltage that develops. The capacitive parasitics would also likely lead to ringing behavior.

Depending on your needs, the impulse model may be close enough for you.


protected by Qmechanic Apr 22 '18 at 17:51

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?