# Where does all the inductor stored energy go?

If I apply a voltage V to an inductor's terminals and wait for all transients to cease it will store an energy given by the formula: $L \frac {i^2}{2}$ If i remove both terminals rapidly enough to avoid any arc formation, where does all the inductor stored energy go?

• – MaximGi Mar 17 '16 at 12:40
• You can't avoid arc formation, even if you can, the insulation will break and the inductor will discharge on itself, or if you are lucky the inductor will discharge into the self capacitance. – Marko Buršič Mar 17 '16 at 13:14
• @MarkoBuršič And where does the energy in self capacitance go? – PDuarte Mar 17 '16 at 20:18
• @PDuarte Back and forth L->C->L->C....until it is consumed in resistance of the wire. – Marko Buršič Mar 17 '16 at 21:28
• @Marko Buršič: And carried away by electromagnetic radiation. EMI Yay! – DerManu Mar 18 '16 at 0:01

If i remove both terminals rapidly enough to avoid any arc formation

The more rapidly you remove the terminals, the more an arc will want to form.

Ideal inductance is defined by:

$$v(t)= L\frac{\mathrm di}{\mathrm dt}$$

If you remove the terminals "instantly", then the current must stop "instantly". That means the di/dt term will approach infinity, and consequently the voltage term will, also.

The faster you stop the current, the higher the voltage generated. That energy has to go somewhere, and the harder you try to stop it, the more it will try to get out. If considering this situation in a purely theoretical, ideal situation where you posit the energy can't go anywhere, it's simply not possible. There is no real mathematical solution.

In practice, there are places the energy can go besides arcing. If you could somehow magically eliminate arcing in an otherwise real circuit, you'd have to consider:

• capacitance in parallel with the inductor, including the parasitic capacitance of the inductor and the wires connecting to it, which makes a resonant circuit where the energy oscillates between the capacitance and inductance
• resistive, magnetic, and dielectric losses, creating heat
• corona discharge

In practice several of these will be in play, in addition to arcing. Depending on the particular design of the circuit some may be more significant than others.

• Crystal clear.. – PDuarte Mar 17 '16 at 14:17
• I suspect if you make it ideal enough you will blow the vacuum breakdown voltage and have a nice spark to somewhere else. – Joshua Mar 18 '16 at 3:06

This question covers most of what you're asking.

However, if you could hypothetically avoid making an arc, then you need to consider the parasitic capacitance between the ends of the inductor. (All circuits of nonzero size have nonzero inductance and nonzero capacitance.) The energy stored in the magnetic field would end up charging this capacitance to a very high voltage, after which the current flow would reverse.

This would repeat indefinitely — back and forth — gradually decaying in amplitude as the energy is radiated away as an electromagnetic wave.

The interwinding capacitance of most inductors is typically in the pF range. When the terminals are removed the inductor voltage will increase rapidly as the interwinding capacitance is charged up. But since the capacitance is so small it is likely that (neglecting other factors) the peak voltage would be several tens of kV.

At some point the voltage will probably get high enough that one of two things will happen.

1) The insulation on the wire will break down.
2) Based on the terminal spacing you will reach the breakdown voltage of the surrounding air.

Lets take as an example an inductor that has terminals spaced 1cm apart and is wound in a single layer around a cylindrical core, and there are 10 windings.

The breakdown voltage of air is about 30kV/cm depending on humidity. If the voltage across the terminals, spaced 1cm apart, passed 30kV then an arc would form in air between the terminals.

Additionally, since there are 10 windings then 1/10th of the voltage is across each winding. If the breakdown voltage of the wire enamel was say 1kV then arcs would begin to form between the windings due to insulation breakdown if the terminal voltage exceeded 10kV.

For inductors wound as toroids, in multiple layers, or on partially-conductive cores, the breakdown voltage may be much lower (as low as the breakdown voltage of one layer of enamel).

So no matter how fast you remove the connections you will probably still make an arc one way or another.

But assuming you don't make an arc, the combination of the interwinding capacitance the inductance and the wire resistance will form a damped oscillator. The energy in that scenario will eventually dissipate from three sources. The first being heat due to the wire resistance and instantaneous current through it (w=I^2*R). The second being heat from core losses (which could be zero for an air core). The third is radiated energy as EM waves at the frequency of oscillation.