Current in circuit after joining two inductors with different currents through them [duplicate]

Question

Let's take two inductors: one has its peak current (say, I_o) flowing through itself and the other has no current through itself. Now if these two inductors are joined in series in a closed circuit, the first inductor will try to keep the current at I_o while the second one will try to keep it at zero.

To solve this, I created an analogy between inductor and mass. When two masses with different velocities collide, the inertia of each mass tries to keep its velocity unchanged. Thereafter we use principle of conservation of linear momentum and value of coefficient of restitution to find the final velocities of the two masses. But here, in case of inductors, I do not know which physical quantity is analogous to linear momentum and hence I can not continue any further.

My Physics teacher told me that we need to equate the total magnetic flux in the inductors before and after joining them but he did not give any reason behind doing so.

So my questions are:

1. Why is the total magnetic flux, before and after joining the inductors, equal?
2. Since I tried creating analogy between collision of two masses and joining two inductors, will there be energy loss in the latter case? And is there any quantity analogous to the coefficient of restitution on which the magnitude of loss of energy depends?

PS:

1. By loss of energy, I mean reduction in energy stored in the inductors.
2. The actual question that I was solving is this: I do not need a solution of this question. This question is just for reference.

Edit: There are two other questions similar to my question (I tried searching for related questions before asking my own question but could not find one):

1. Why is there a sudden change in current between $t=0^{-}$ and $t=0^{+}$ when an active inductor is connected in series with a relaxed inductor?
   2. Magnetic Flux conservation

As for the first one, I have not been yet taught about terms such as Inverse Laplace transformation and parasitic interwinding capacitance. Hence I can't grasp what is being discussed there.

The second question has a greater similarity to my question (as I have also asked about reason behind conservation of magnetic flux) but there is only one answer to that question which does not tell why flux is conserved. Hence my question doesn't have an answer there too.

Moreover, I have also asked about analogy between mechanical system and electrical system which neither of the above two questions have talked about. And I have already got an answer here which has solved the question using that analogy.

Answer 1

To solve this, I created an analogy between inductor and mass.

I like the way you think!

However, note that such an ideal circuit that you describe is somewhat pathological and certainly non-physical - the voltage across each ideal inductor has a delta 'function' at the moment the switch is thrown (since the current through is discontinuous there).

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Why is the total magnetic flux, before and after joining the inductors, equal?

In the Impedance analogy (where voltage $v$ and current $i$ are analogous to force $F$ and velocity $u$ respectively), inductance $L$ is analogous to mass $M$ and (magnetic) flux linkage $\lambda = Li$ is analogous to momentum $Mu$

Thus, conservation of flux linkage is analogous to conservation of momentum.

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Since I tried creating analogy between collision of two masses and joining two inductors, will there be energy loss in the latter case?

There will be an energy loss because, after the switch is thrown, the inductors (masses) are series connected and so, must have identical current through (velocity). That is, the mechanical analogy is of a perfectly inelastic collision with maximum loss of kinetic energy.

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Is there a method to prove that the flux will be conserved in this case without using the impedance analogy?

One approach is to 'regularize' the two inductor circuit by, e.g., placing a resistor with resistance $R'$ in parallel with the $L$ inductor. For simplicity, stipulate that the series resistor has resistance $R \ll R'$ so that we can ignore it (we'll justify this later). If you solve the resulting differential equation (for the circuit after the switch is thrown), you'll find that

$$i_L = \frac{2}{3}I_0\left(1 - e^{-\frac{2R}{3L}t}\right)$$ $$i_{2L} = \frac{1}{3}I_0\left(2 + e^{-\frac{2R}{3L}t}\right)$$ $$i_{R'} = I_0e^{-\frac{2R}{3L}t}$$

where $I_0$ is the current through the $2L$ inductor just before the switch is thrown.

The total flux linkage after the switch is thrown is then:

$$\lambda_{2L} + \lambda_{L} = 2Li_{2L} + Li_L = 2LI_0$$

But this equals the total flux linkage just before the switch is thrown and so the flux linkage is conserved across the switching event.

Now, we can make $R'$ arbitrarily large thus making the time constant (for the currents to settle to their final value) arbitrarily small which justifies our ignoring the series resistor $R$ in the analysis. We recover the original circuit in the limit as $R' \rightarrow\infty$.