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I am trying to understand how this circuit works without burning or exploding.

When the inductor current starts to decrease, it acts in a way to increase it and if there is no path, then we see sparks or burning.

In the circuit below instead of DC, imagine we have a sinusoidal voltage source. The diode shuts off at some point when the inductor tries to push current into it and I am troubled to see how the circuit really works. Does the inductor keep the voltage across it if there is no path for discharge like a capacitor?

enter image description here

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    \$\begingroup\$ Assuming that the initial condition of the inductor is that its current is zero, and assuming that your diode is an ideal one (not a non-linear one) that holds a fixed voltage drop across it regardless of current through it, then don't you see that initially all of the voltage drop is across the inductor, giving the highest V(L)/L=dI/dt point at t=0+, but that as (I) increases R then drops more voltage gradually lowering V(L) and therefore lowering dI/dt while (I) still continues to increase over time? dI/dt doesn't change sign. It just gets smaller and smaller. \$\endgroup\$
    – jonk
    Aug 16, 2022 at 18:51
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    \$\begingroup\$ >> Does the inductor keep the voltage across it if there is no path for discharge like a capacitor? No. Soon as the current source goes away (you open the switch), the energy stored in the field around the inductor will try to escape. It does this by sending the voltage across that inductor up to (theoretically) infinity. Of course, before it hits infinity, the air will breakdown and there will be a big spark across it as the field collapses. This is how a car spark-plug works. It creates 5kV or so to make a spark from just a 12V battery. \$\endgroup\$
    – Kyle B
    Aug 16, 2022 at 20:18
  • \$\begingroup\$ > I am troubled to see how the circuit really works.< When the switch is open ... or closed? \$\endgroup\$
    – Antonio51
    Aug 17, 2022 at 10:17
  • \$\begingroup\$ OnurTR, if you haven't already been sufficiently helped and want to see something else, you will need to say something here. Are you here? Have you read what's been written? Do you still have a question? \$\endgroup\$
    – jonk
    Aug 18, 2022 at 4:18
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    \$\begingroup\$ @OnurTR If the diode is a non-linear (realistic) diode, then things get more complex. The voltage it drops will gradually increase as the current increases (not linearly so, though.) But I read your schematic as suggesting an 'ideal' diode in the sense that there is a fixed drop. If so, then all that happens is that the fixed drop subtracts from the voltage source, leaving a smaller applied voltage across the resistor/inductor series pair. I think you are taking things from different circumstances and imagining them here, perhaps? \$\endgroup\$
    – jonk
    Aug 20, 2022 at 3:40

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Here is an example of "similar" circuits showing what happens.
I used a switch to show behavior under random switching.

Voltages: one can see the "overvoltages" (very high) at switching OFF.

enter image description here

Currents

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And here is, what should be done, using a "protection" circuit (for example VDR and/or R serial C).
Overvoltage now limited as ~ 600 V. Varistor is B32K275.
Be aware that scales are not the same as above.

enter image description here

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