Say the 3H branch of the circuit was connected to the 1V source for a long time such that 500mA current is flowing through the 3H inductor. circuit was in this stage for a long time

Then after long time the switch was connected to 1H inductor branch

Then the circuit is connected to another branch

Now, the currents can't change abruptly in an inductors. Therefore 500 mA can't flow through 1H branch just after switch transition, neither the current can drop to zero because 3H inductor will prevent it from happening.

My question is what will be the current in the circuit just after switch transitioning happen?

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    \$\begingroup\$ The ideal circuit is not physical. In reality you will get some sparks. \$\endgroup\$ – Eugene Sh. Jan 19 '18 at 17:53
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    \$\begingroup\$ It's just the same as connecting one charged capacitor to another one... try to think about this... \$\endgroup\$ – carloc Jan 19 '18 at 17:54
  • \$\begingroup\$ @carloc I think you won't get sparks with caps :) \$\endgroup\$ – Eugene Sh. Jan 19 '18 at 17:56
  • \$\begingroup\$ The current cannot change abruptly in the inductor, correct. But if your switching a charged inductor, it creates a large voltage across the inductor. \$\endgroup\$ – Voltage Spike Jan 19 '18 at 17:57
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    \$\begingroup\$ @EugeneSh. Well it may spark or not but that's not the point. The analogy is that when you connect the two capacitors together a sharp current spike will equalize voltages.Here, dually, a sharp voltage spike will equalize currents. Then a lot could be said about what "spike" actually means with different degrees of ideality considered, but this applies equally to both cases :) \$\endgroup\$ – carloc Jan 19 '18 at 19:09

In real world there is always some resistance. At worst it is resistance through air, or through the PCB substrate, across either inductor or across the switch back to the source. (Most likely inside the switch.)

Since that resistance is very large, the voltage induced across it is very large also and in this instance is negative. If it is through the air, the voltage is large enough to ionize the air and an arc, spark, will form.

Since R is large the current decays very very quickly, meanwhile the current in the second inductor will also grows quickly till the point where the arc extinguishes.

  • \$\begingroup\$ What you say it's true, but even the ideal case can be easily solved just considering Dirac's delta... This problem is exactly dual to two capacitors charged with different voltages connected together. \$\endgroup\$ – carloc Jan 19 '18 at 19:15
  • \$\begingroup\$ @carloc what capacitors? \$\endgroup\$ – Trevor_G Jan 19 '18 at 19:17
  • \$\begingroup\$ The op's problem is about two inductances initially charged with different currents and suddenly connected together and I found it is often seen as "strange" and misty when idealized. Thinking of its dual , i.e two capacitor initially charged with different voltages suddenly connected together, clarifies the topic (this was not in the op's post but I has been introduced by me) \$\endgroup\$ – carloc Jan 19 '18 at 20:57
  • \$\begingroup\$ @Carloc... ah.. sorry I c, I thought you meant dual capacitors... Indeed you are right. \$\endgroup\$ – Trevor_G Jan 20 '18 at 1:28
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    \$\begingroup\$ I used Laplace transform approach to solve this. From that initial current comes out to be 3/8 A. Now this result is also replicated if we assume the quantity LI to be constant. i.e. if L(initial) x I(initial)=L(final) x I(final) That will give 3 x 0.5 = 4 x I(0+) so I(0+) = 3/8. This can be thought analogous to law of conservation of momentum in mechanics where one ball collide with another and stick to it. The quantity mv remain conserved. Is there a law of conservation of electrical momentum(LI)? \$\endgroup\$ – user175306 Jan 21 '18 at 4:38

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