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I have a question about how to intuitively think about the polarity reversal (flyback voltage) that occurs in an inductor when removed from a supply.

Here is my understanding: When the switch is closed, current flows through the inductor to ground, with a positive voltage across the inductor.

Then, when the switch is opened, I understand that the polarity reverses, but I'm trying to make sense of this fact.

I know that the average value of the voltage across an inductor must be 0. From this perspective, it makes sense that the polarity switches, it was previously positive, and now must be negative to have an average value of 0. However, when thinking about it in terms of maintaining current direction, I can't wrap my head around it.

I know that the inductor will try to resist the change in current flow. So when the switch is closed, it is storing up a magnetic field with the original positive polarity. Then, when the switch opens, the inductor should act in a way to keep the current flowing in the same direction as when the switch was closed. Knowing that current flows from high potential to low potential, it would make sense that the inductor would force a positive voltage across it, to keep the current flowing in the same direction. However, what actually happens is it forces a negative voltage across itself. enter image description here

Wouldn't flipping the polarity mean that current goes in the opposite direction as before, since it must go from high potential to low potential?

Can someone help clear the confusion for me?

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  • \$\begingroup\$ Think about the voltage across SW1. If the voltage of V1 and L1 was the same and of the same polarity, then there would be 0V across the switch. Reverse the polarity of L1 and you’d get 2V. This voltage would rise as the inductor attempts to maintain the current. \$\endgroup\$
    – Kartman
    Mar 15, 2021 at 3:10

3 Answers 3

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Wouldn't flipping the polarity mean that current goes in the opposite direction as before, since it must go from high potential to low potential?

There is no rule for an inductor that says "[current] must go from high potential to low potential". That is a rule for a resistor.

The rule for an inductor is "the current increases (becomes more positive) with a finite rate of change when the voltage across the terminals is positive and decreases (or becomes more negative) with a finite rate of change when the voltage across the terminals is negative".

Or, mathematically,

$$\frac{dI}{dt}=\frac{V}{L}.$$

If you think about it a bit, this rule explains exactly why the voltage must be negative to get the current in the inductor to change from some positive to zero.

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  • \$\begingroup\$ Thanks. If you had to imagine this without math, as a physical phenomena going on around the inductor in slow motion. How would you describe it? if you feel inspired, how would you explain it? (note descriptions < explanations) \$\endgroup\$
    – juanmf
    Mar 19 at 14:50
  • \$\begingroup\$ @juanmf, P = VI. Using the passive current convention for the sign of I, this means that when P is positive, power is being delivered to the device. When P is negative the device is delivering power to the rest of the circuit. When you deliver power to an inductor you build up its magnetic field. When the inductor delivers power to the circuit its magnetic field decays or collapses. But power can't be infinite (energy can't be transferred instantly), these processes (either building up or collapsing the magnetic field) take time. \$\endgroup\$
    – The Photon
    Mar 19 at 15:38
  • \$\begingroup\$ Thanks, it's interesting, that what takes time has a magnitude, or is magnetism. Literally a vortex (shown experimentally). But electrostatic discharges are much faster, a minimum of 10x faster, when no vortex/magnetic field is involved, and is still energy transfer, with no Eddy currents. Makes me wonder, formulas are useful but visualizing what's actually happening satisfies me. \$\endgroup\$
    – juanmf
    Mar 21 at 15:08
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Looking at it another way, the state changes from the circuit pushing current through the inductor (current flows in to the terminal with the higher voltage applied) to the inductor pushing current through the circuit (current flows out of the terminal with the higher voltage generated).

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I recently had an opportunity to think this through, while answering a similar question. The situation is that the polarity reverses, while the current continues in the same direction. The key is that with the switch closed, the inductor is "resisting" (impeding?) the current flow, but after the switch opens, the inductor itself becomes the voltage source!

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