# Does a capacitor remove DC offset between grounds?

Should be a simple question but I've never seen a clear answer to this.

I know that capacitors remove DC bias on a signal, like with an AC signal being output from an amplifier; just couple the signal through an HPF and you're left with the AC signal oscillating around 0 V.

Does the same principle apply to two grounds that might have some DC offset? I know that a real capacitor will have some leakage current, but if we ignore that for a moment, conceptually does the capacitor remove DC offset between two grounds?

I ask because in a power supply I know that the isolated side and input side are sometimes bridged with a cap. If we assume for the moment that the output side is floating, wouldn't connecting the two sides with a capacitor set them to the same potential? Or does this just remove any DC offset for HF noise, so now the HF noise is oscillating around 0 V with respect to some other reference (like earth???)?

• DC current cannot flow through a capacitor, so putting one between two nodes at a different DC voltage will not being them to the same potential. Commented Nov 23, 2021 at 3:35
• The capacitor appears invisible to the DC, thus it will not make any difference. However, it will provide an impedance to AC currents. This is why you’ll commonly see cable shields tied to ground via a capacitor - high frequency AC currents get shunted to ground. If the shield was tied directly, DC currents might flow creating a ‘ground loop’ Commented Nov 23, 2021 at 3:36
• @Kartman well then in this case, it would remove an AC offset and not a DC offset, correct?
– user224284
Commented Nov 23, 2021 at 3:38
• It won’t necessarily remove, but it will attenuate the AC signal. Commented Nov 23, 2021 at 3:40
• @Kartman sure that makes sense, but going back to the DC offset for a moment, suppose there is some DC offset between two grounds... Connecting them with a cap would provide infinite impedance between these grounds, so any DC offset between them is entirely dropped across the cap. I suppose this means that the DC offset between them still exists, correct? EDIT: Meaning if we connect a voltmeter across them we would still measure a voltage... I suppose I'm answering my own question here...
– user224284
Commented Nov 23, 2021 at 3:44