My intuition is that the current upon an inductor (say, a solenoid) will always be time-continuous, without "sudden changes". But below is a case that seemingly contradicts this point of view. enter image description here
There are two wires $W_1$ and $W_2$, both intensely wrapped around a common core whose section area is $S$, length is $L$ ($L>>\sqrt S$), and permeability is $\mu_0$ (and it is not ferro-magnetic material !). At first the two switches $S_1$ and $S_2$ are both open. Now, close them both. After a time that is long enough, we know that the currents go through $W_1$ and $W_2$ will respectively stabilize at $$I_1=\frac{\mathscr E_1}{R_1}$$ and $$I_2=\frac{\mathscr E_2}{R_2}$$ So the ratio between $I_1$ and $I_2$ is $$\frac{I_1}{I_2}=\frac{\mathscr E_1 R_2}{\mathscr E_2 R_1}\tag{*}$$ Now, let's simultaneously switch $S_1$ and $S_2$ open. Then the current $I_1$ that $W_1$ is carrying will have no choice but to go through $R'_1$ and form a loop, which is just the same with $W_2$ side, the other half of the system. And the systems become time-varying. Let's just specify the moment when we switch $S_1$ and $S_2$ open as $t=0$, and let the currents going through $W_1$ and $W_2$ be denoted by $I_1(t)$ and $I_2(t)$, respectively, and the magnetic flux through the core denoted by $\Phi (t)$.
First, let's take a look at the $W_1$ loop. Because $\Phi (t)$ is time-varying, there will be an magnetically-induced emf (denoted by $\mathscr E^i_1(t)$) whose magnitude will be $$\mathscr E^i_1(t)=Ln_1\frac{d\Phi(t)}{dt}$$ where $n_1$ denotes the number of turns of $W_1$ wrapped around the core per unit length.
However, by applying KVL to the $W_1$ loop we will also get $$\mathscr E^i_1(t)=I_1(t)R'_1$$ Therefore $$I_1(t)=\frac{Ln_1}{R'_1}\frac{d\Phi(t)}{dt}$$ Then, just do the same analysis for $W_2$ loop, and likewise we get $$I_2(t)=\frac{Ln_2}{R'_2}\frac{d\Phi(t)}{dt}$$ So the ratio between them is $$\frac{I_1(t)}{I_2(t)}=\frac{n_1R'_2}{n_2R'_1}$$ which does not depend on $t$. Then, by the time-continuity of currents at $t=0$, we have $$\frac{I_1(0)}{I_2(0)}=\frac{n_1R'_2}{n_2R'_1}$$ But as I've specified, $I_1(0)$ is exactly $I_1$, and $I_2(0)$ exactly $I_2$, therefore $$\frac{I_1}{I_2}=\frac{n_1R'_2}{n_2R'_1}\tag{**}$$ Comparing (*) to (**), we'll find it's not possible. Because $\mathscr E, R, R', n$ are all arbitrary, they don't necessarily make the two ratios fit.
So could you please explain to me where I'm wrong? Best regards!

EDIT I think the reasons are probably of the following:
1).I've made too many ideal hypothesis, such as "intensely wrapped", "$L>>\sqrt S$" etc, which may make our assumptions too far away from the fact.
2).(*) is perhaps wrong (which I haven't proven rigorously but wrote by my intuition), i.e. $I_1$ and $I_2$ just won't stabilize at those expected values.
3).Or my "intuition" is just wrong. Perhaps there are sudden changes of currents in this case.

  • $\begingroup$ Suddenly disconnecting a coil usually results in a spark. A perfect switch, hence, is probably not possible. Assuming that you can switch perfectly, results in the contradiction. Reminds me of the simpler example where you discharge a capacitor on an equal but empty one and cut the wire when half the charge is transferred. The result is a contradiction in electrostatic energy. $\endgroup$ Commented Apr 14, 2015 at 12:12
  • $\begingroup$ @mikuszefski. Well the capacitor example also seems confusing to me. By $W=\frac{Q^2}{2C}$ it seems that half of the electrostatic energy would be lost during the discharge process.... But why? Is there any other form of energy into which the lost electrostatic energy is converted? $\endgroup$
    – Vim
    Commented Apr 14, 2015 at 12:27
  • $\begingroup$ @mikuszefski what's more, does sudden switch necessarily result in a spark? I think that's not 100% certain. $\endgroup$
    – Vim
    Commented Apr 14, 2015 at 12:30
  • $\begingroup$ The discharging of the capacitor is a dynamic process, so it is not correct to compare energies in a static case that is actually never reached. In a real experiment you would dissipate energy of course due to resistivity of the wires. $\endgroup$ Commented Apr 14, 2015 at 15:21
  • $\begingroup$ Concerning a spark "always", I guess there are situations where you do not get a spark, but those probably do not leave you troubled with an impossible equation. $\endgroup$ Commented Apr 14, 2015 at 15:22

1 Answer 1


The resolution is that currents can change instantaneously. Remember that the reason that the current in a single solenoid cannot change instantaneously is that a change in current causes a change in flux, which causes an electromotive force (emf) to oppose the original change in current.

However, suppose now there are two currents, as in your example. Suppose that these currents can simultaneously change so that flux does not change. In your example, we would just need to change $I_1$ and $I_2$ in such a way that $I_1+I_2$ remains fixed. Then no emf will will be produced. This means an arbitrarily fast change is in fact allowed.

To see how this comes about mathematically, look at the differential equation describing your problem. We can write a current vector $\mathbf{I} = (I_1,I_2)$, an inductance matrix $\mathbf{L}=\begin{pmatrix} L_1 & M \\ M & L_2 \end{pmatrix}$, and a resistance matrix $\mathbf{R}=\begin{pmatrix} R'_1 & 0 \\ 0 & R'_2 \end{pmatrix}$. Our differential equation becomes

$$\mathbf{L \dot{I}=-RI}.$$ The solution is $\mathbf{I}=e^{\mathbf{L}^{-1}\, \, \mathbf{R}t}\, \,\mathbf{I_0}$, where $\mathbf{I_0}=(I_1(0),I_2(0)).$ However we have a problem in our case, which is that $\mathbf{L}$ is not invertible. To see what the solution should really be, lets look at the limit as the determinant of the matrix goes to zero.

Consider a specific example where the two systems are symmetric except one has a larger resistance, so $L_1=L_2$ and $I_1(0)=I_2(0)$, but $R_2 = 2 R_1$. Then the only control parameter to play with is $M$, the mutual inductance, which describes the coupling between the two systems. First consider the case where $M=0$, corresponding to infinite physical separation of the two systems. In this case both currents decay from the common initial value to their different equilibrium values. The time constant describing how long the decay takes is $\tau = L /R$. Since one system has a smaller $R$, it will have a larger $\tau$. I will refer to this system as the slow system and the other system as the fast system.

Now let's imagine we increase $M$ by bringing the two systems together. Then as the fast system is nearing equilibrium, it will still see the changing field from the slow system, which has not equilibrated yet. This changing field felt by the fast system will cause it to keep a current even when it would have otherwise died out.

Now lets increase $M$ even more. At some point, the fast system will be so sensitive to flux from the slow system, that its current will actually increase initially. To see this, notice that the derivative of flux from the slow system goes like $I/\tau \sim I R_\textrm{slow} $, so the induced current will be like $I R_\textrm{slow} / R_\textrm{fast}$, which will be greater than $I$ since $ R_\textrm{slow} >R_\textrm{fast}$. Now this increase in current will then create its own change in flux opposing the increase in current. The the slow system sees this flux from the fast system, it will feel an emf to oppose the increase it current. I.e., it will have its current decrease faster. The whole net effect is that the two system will equilibrate at a more similar rate.

Now consider the case where $M$ is just slightly smaller than $L$, so that $\mathbf{L}$ is just barely invertible. Then the two system equilibrate almost instantly to a state where the currents are given by the ratio of the resistances (your eqation **). Once they have reached this state, then current decays smoothly to zero. A plot is given below.

enter image description here

This log-linear plot has three regimes. In the first regime on the left, the currents stay equal to the initial value, because the time elapsed is smaller than the time constant for the currents to adjust so they can be consistent with (**). In the next regime, the currents have shift to be consistent with (**). Notice though that the total current has remained the same. This is because if this mechanism is to be as fast as it is, it must not cause an overall change in flux. This implies that one of the currents must actually increase. Once this state has been established, it transition to the third regime, where the currents have decayed to zero.

The final case then is the singular case where $M=L$ and $\mathbf{L}$ is not invertible. Here we can just say that the current changes instantaneously to the value given by your (**), and then decays smoothly.

Appendix: Code to generate mathematica plot

L := {{L1, M}, {M, L2}};
i[t_] := {i1[t], i2[t]};
R := {{R1, 0}, {0, R2}};

L1 = 1;
L2 = 1;
M = .99999;
i10 = 1;
i20 = 1;
R1 = 2;
R2 = 1;

differentialEquations := Thread[L.i'[t] == -R.i[t]];
initialConditions := Thread[i[0] == {i10, i20}];

solution[t_] = 
  i[t] /. DSolve[{differentialEquations, initialConditions}, 
     i[t], {t, 0, 100}][[1]];

LogLinearPlot[{solution[t][[1]], solution[t][[2]]}, {t, .000000001, 
  1000}, PlotRange -> All, AxesLabel -> {"time", "current"}, 
 PlotLegends -> {"i1", "i2"}]
  • $\begingroup$ Thank you so much! I couldn't expect a more detailed answer here! $\endgroup$
    – Vim
    Commented Apr 14, 2015 at 17:36
  • $\begingroup$ I got slow and fast backwards in my explanation. The slow system with smaller resistance is more susceptible to changes in flux, and it is the one whose current increases as a result of the fast system's fast change in flux. I will fix my answer later. $\endgroup$ Commented Apr 14, 2015 at 23:12

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