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I know this is a very basic concept, and has been described in many places, but I still can't 100% understand inductive kickback.

schematic

simulate this circuit – Schematic created using CircuitLab

If the switch in the circuit above is initially closed, and then immediately opened, we will have an instantaneous drop in current. Using the following equation describing voltage across an inductor: $$V = L\frac{dI}{dt},$$ we would get the voltage at point A to be a high positive number, to induce a negative voltage across the inductor. Since inductors are suppose to resist changes in current, how would a very high voltage at point A maintain the flow of current? Wouldn't it result in current in the opposite direction?

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Once you close the switch, the inductor will face a really high $$\frac{Di}{Dt}$$, and from that the inductor will try to maintain the current flowing to the same direction yielding a high voltage(which can be understood from the inductive voltage formula you provided). This kickback can be understood from Lenz-Law since the change in current causes a change in magnetic field and as the inductor faces a change in the opposite direction, this field will force a current in the same direction it was flowing in the beginning, which is entering point A.

So answering your question: No, the current will still flow to the same direction.

Once the inductor tries to maintain this current flowing, the switch is now going to behave as a high resistance path, and for the current to flow there should be a really high voltage produced across the switch, which causes the famous arc, a phenomenon when the voltage is too big to the point of breaking that little gap insulation, causing dielectric rupture.

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  • \$\begingroup\$ Got it. Having a high voltage across the switch makes sense. But what prevents current from flowing in the reverse direction? Since there is a voltage drop in that direction as well? Also, let's say we add a kickback diode, that will clamp point A at about V+ + 0.6V. What's forcing the current to specifically flow up through the diode and down the inductor, rather than up through the inductor? \$\endgroup\$ – user198450 Oct 2 '18 at 1:29
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    \$\begingroup\$ @user198450, the current starts out flowing down the inductor when the switch is closed. It can't change instantly because \$\frac{dI}{dt}=V/L\$ and \$V\$ is finite. So it keeps flowing downward when the switch opens, either by creating an arc across the switch or by flowing up through a diode that you added to the circuit. \$\endgroup\$ – The Photon Oct 2 '18 at 1:36
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If the switch in the circuit above is initially closed, and then immediately opened, we will have an instantaneous drop in current

No, you might have a high rate of change of current but it won't be instantaneous due to there always being a practical current flow through the coils parasitic capacitance (causing a ringing waveform) and the voltage breakdown of the switch as it opens will produce a spark that continues the current.

how would a very high voltage at point A maintain the flow of current? > Wouldn't it result in current in the opposite direction?

The coil turns into a generator and forces current in the same direction it was previously travelling (albeit decaying quite rapidly). This means that point 'A' produces a big positive voltage so that current is forced in the same magnitude and direction through the external circuitry connected around the coil.

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