Vic, what a relief that you have come along with some solid data on the DC part of this question. I've been silently gnawing my tongue here after reading some of the earlier posts on DC in inductors.
I've always viewed an inductor as a series device that always works to keep the current constant--if at zero, then zero, if at some value, then that value. It does this by building and collapsing its internal magnetic field. If it has an iron core, its capacity to build magnetic field is greatly increased but any coil, even air cores, work this same way.
Your first scope trace shows this as a voltage is applied to a coil at zero amps. The coil holds back the full level of current for a time as it builds its field. Once the circuit resistance becomes the current limiting factor, the field holds steady at a high level and the current holds steady at a high level. Under these steady state conditions, the circuit doesn't "see" the inductive part of the coil.
Your second trace isn't really applicable here because the iron core was moving generating a new voltage but the third trace is to be compared with the first except that there is now an iron core instead of an air core. The currents are similar with the time it takes to build the much larger field being longer. But, as before, once the field is fully built and the circuit resistance limits the current, it is as if the coil wasn't there.
In fact, as demonstrated, coils are used to avoid inrush in DC circuits.
The part that isn't shown by your traces, however, is what happens when the circuit opens and the current is supposed to fall immediately to zero. Remember that the nature of an inductor is to try to maintain the current the same, whatever it is. As soon as the circuit is opened, the coil uses the stored energy in its field to keep the current flowing at whatever steady state level it was at. The coil will exhaust all of its stored energy in this attempt.
Now, those of you with a calculator close by: if voltage = current times resistance and this coil is trying to push a finite amount of current across an open circuit (someone opened the switch, remember!), that would be a number times infinity which, in my world is very large! Since the inductor knows nothing but to keep the current flow constant and has no sense of voltage whatever, it will continue to increase the voltage across the gap until either it, or some insulation somewhere, or the switch contacts arc over. We've all been victimized by the stored energy in DC inductors this way, I would think. That is why, on DC coils, a flyback diode is normally installed so this current simply recirculates around the coil and no arc voltage surge occurs.
Vic, you might just want to reconnect your test rig and show us some traces of what happens when the circuit is opened. I would caution you to properly protect your test equipment from overvoltage.
Thanks again, Vic.
