Electronic – Inductive kickback


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


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?

Best Answer

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.