I'm struggling to wrap my head around a 'should-be-simple' problem involving an RL circuit. Referring to the image I've pasted below, what happens if we were to remove switch 1, and instead created a scenario where switch 2 simply opens? Where would the current go in the inductor?

I know that the current in an inductor can't instantaneously change, however, if switch 2 opens, it is now located on an abandoned branch thus can't allow current to flow...

How is the energy in the inductor is released?

enter image description here

  • 2
    \$\begingroup\$ Nothing like a nice big inductor, an open blade-type switch, and and arc shield for illustrating this. Of course there's one on youtube (there may be dozens, but this will do) - youtu.be/aSmMFog10D0 \$\endgroup\$
    – Ecnerwal
    Apr 10, 2016 at 3:30

1 Answer 1


The answer is found in Ohm's Law, with due notice taken of the amount of energy stored in the magnetic field. An open circuit is effectively an infinite resistance. For the same current to flow through such a resistance, the coil must develop an infinite voltage across itself.

In practice, this doesn't happen, of course. However, the coil does develop a high enough voltage to arc across the switch contacts as they open. This immediately discharges most of the energy in the magnetic field. The remaining energy dissipates rapidly as the magnetic field collapses.

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    \$\begingroup\$ To elaborate just a bit as to what to do with this knowledge, this is why 'flyback diodes' are used in DC motors, DC/DC converters, and other circuits that use an inductor as an integral part of the architecture. It gives the energy a path that doesn't require an arc event. Based on your answer, I'm sure that you know this, but someone stumbling on this answer in 10 years might be looking to fix a problem and I believe this is worth adding. \$\endgroup\$ Apr 9, 2016 at 0:43
  • 1
    \$\begingroup\$ The answer has nothing to do with Ohm, and everything to do with Faraday, Lenz, and Henry. Read this: \$\endgroup\$
    – EM Fields
    Apr 9, 2016 at 11:30
  • \$\begingroup\$ Ohm's law has nothing to do with the inductor which is true, but it does apply to the switch. \$\endgroup\$
    – Krentin018
    Apr 9, 2016 at 11:33

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