I was solving some problems with RL circuits today and a thought occurred to me that I couldn't answer. Say you have a Vsource Resistor and Inductor all nicely in a loop. You let the voltage source run until everything reaches a happy steady state. The Inductor should have energy

$$E = \frac{1}{2}L \frac{V^2}{R^2}$$

Now, what happens if I cut the wire? Where does the energy go?

Sorry for the needless schematic... I just discovered the feature and thought it was cool. :)


simulate this circuit – Schematic created using CircuitLab


1 Answer 1


In a real circuit there would be a spark generated as you cut the wire because the inductor tries to maintain that current. The stored energy in the inductor produces the spark in order to maintain this current as best it can. The energy in the spark will be part of the energy initially stored in the inductor and there will be EM transmissions and light and heat and, you may get a jolt through your fingers. That's where the energy goes.

Your simulator may have default values for leakage components such as stray capacitance and this will show up in your transient analysis as an initial high peak of voltage and a rapidly decaying sine-wave probably finishing in less than a fraction of a micro second.

  • 1
    \$\begingroup\$ Thanks! I guessed that would be the case, but i wasnt sure! \$\endgroup\$ Mar 20, 2013 at 14:47
  • \$\begingroup\$ See Lenz's Law and Faraday's Law of Induction for further details and explanations. \$\endgroup\$
    – sherrellbc
    Jun 27, 2014 at 17:02
  • \$\begingroup\$ Suddenly removing the current path in an energized inductor is the primary mechanism to increase voltage in a switching boost regulator. When a transistor in series with the inductor is switched off, the voltage across the inductor spikes higher than the input voltage. \$\endgroup\$
    – Dan Laks
    Jun 27, 2014 at 17:11

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