Vacuum interrupter contacts and displacement current

[kenneth edmiston]

TL;DR Summary

Why can displacement current travel through dielectric medium in capacitors but has no effect between interrupted contacts.

I’m self taught so I have a lot of holes in my understanding. I also have little to no mathematical understanding. Even though ironically, I tend to prefer to picture electricity as numbers rather than “water” in a pipe, etc.

I’ve been studying displacement current recently and have a question.

Displacement current can exist in a vacuum, and the contact gap in a MV/ HV vacuum interrupter is relatively small. Why do you not see current flow through the circuit with contacts open?

My theories are that;
1) if Displacement current = conduction current, you need a closed circuit conducting current to obtain displacement current. No conduction, no displacement.
2) the gap on a vacuum interrupter is too far apart

Help understanding this would be much appreciated.

 

[zoki85]

Displacement current can exist in a vacuum, and the contact gap in a MV/ HV vacuum interrupter is relatively small. Why do you not see current flow through the circuit with contacts open?

AC circuit ⇒ there is displacement current. But it is very small for ordinary power frequencies since the capacitance involved is very small

 

[kenneth edmiston]

What allows it to pass through a capacitor dielectric at that frequency? The voltage at the capacitor plates would have the same frequency as an interrupter contact, right? Just curious. Thanks for the quick reply.

 

[zoki85]

Very contacts separated by vacuum gap form a small capacitor and change of electrical field there must be accompanied by change of charges on the electrode contacts which in turn must be accompanied by flow of current in an external circuit. Yes, the same frequency

 

[Baluncore]

When you examine a capacitor closely, the charge that flows into one terminal of the capacitor is equal to the charge that flows out of the other terminal. That must be true because the terminal currents are actually the same current, flowing through the unseen circuit behind you.

Since the two terminal currents are identical, they can be numerically equated and referred to by the same name, capacitor displacement current. Like double entry book keeping, the hypothetical displacement current is simply there to close the circuit, so as to reconcile Kirchhoff's current law.

So, electron current cannot flow through the insulation, yet the hypothetical displacement current does flow through the capacitor.
C = Q / V
E = ½ · C · V²

 

[kenneth edmiston]

And the same applies to both ac and dc correct? A charging or discharging dc capacitor will have a changing electric field which would cause the displacement? I’m an electrician, no one explains capacitors in this way. “A charge on one terminal pulls a charge from the other side” etc. (I’m referring to charge in the post and not the movement of particles. People seem to forget the charge flow when explaining capacitors) So I would be correct in saying displacement current is the way current continues through a dielectric regardless if it’s ac or dc? In an ac circuit it is a constant property, in dc it applies only to a change in magnitude of the charge? Thank you Edit: also why do you refer to it as hypothetical? Is it unclear exactly what is happening even though it is know that the effect is a current through a dielectric.

 

[zoki85]

When you turn on/off power in DC circuit too ( first a current spike, than in steady state displacement current is 0).That current is possible due to "self"-capacity of the circuit. In AC circuit capacitive current is present all the time in steady state. Generally, the current is small

 

[kenneth edmiston]

Zoki the steady state capacitive current in AC is small and inherent in the circuit, but the displacement current through a capacitor equals the conduction current at the capacitor, correct?

 

[zoki85]

Correct