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In previous courses, I remember we were told that capacitors act as an open circuit at DC and inductors act as a short circuit.

I came across this video: https://www.youtube.com/watch?v=WR6qVvnDnI4 and in this, it has a DC source.

Now with my previous understanding, does this mean that all the voltage is actually dropped by the non-ideal series resistance of the inductor?

I'm just trying to make connections between different courses I took/am taking.

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Yes, the ideal inductor has zero DC resistance. To model a real world inductor, we often add a series resistance, which may sustain a DC voltage drop. There is no way to measure this voltage drop independent of the inductor, however.

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    \$\begingroup\$ What do you mean by "measuring the voltage drop independant of the inductor"? The real world inductor is modelled by an ideal inductor and a series resistance, so any voltage drop across the real inductor should be purely due to the drop across the series resistance of the real inductor, correct? \$\endgroup\$ – AlfroJang80 Sep 18 '19 at 0:32
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    \$\begingroup\$ @AlfroJang80 This phrase means that for a real inductor you can't separate the voltage drop across the inductance from the voltage drop across the parasitic resistance. This is true at dc, but you can apply ac signals of different frequencies and thereby determine the actual inductance and actual resistance. \$\endgroup\$ – Elliot Alderson Sep 18 '19 at 12:04
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There will be a voltage across an inductor as the current in the inductor changes. Once the current reaches its steady-state value it will have zero voltage drop, because the current will not be changing. When the switch first closes there will be zero current and the resistor will have no voltage drop across it, thus the full voltage drop will be from the inductor.

I found my understanding of inductor circuits improved a lot when I learned about volt-second balance and was able to solve a simple Buck converter.

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