Voltage Increase at the Induction

Perhaps it is:Inrush current - Wikipedia, the free encyclopedia

Better to put my qwestion so:

Why instantaneous voltage rise on an disconnected coil can exceed many times the nominal supply level ?

Why not make it easy on us all?

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dI/dt.

That, too.

In the text book under inductors you will find an equation for voltage that looks like this: v=di/dt

This says that the voltage on the inductor = rate of change of the current.

So if the current was flowing along, just minding its business and you come along and change it by disconnecting it, the rate of change is very very small.

Now, what happens when you divide by something very very close to zero?

So, it is also true for an active load too ?

#8↓ is true for all loads.

V=di/dt

V=L.di/dt

Thanks ! - now I am aware of this.