What happens if voltmeter is connected across arbitrary points on two separate circuits?

Question

My question is pretty much what the title says. If we have two separate circuits not connected to each other and we connect a voltmeter across any two arbitrary points one on each of them, would we get a reading?

Theoretically there should be a potential difference across the two points.

But I've studied that voltmeters are really just galvanometers with a high resistance in series and calibrated to directly display potential difference so they practically shouldn't function without current flow through them so I'm confused.

And another scenario I thought of was that if we grounded some point in a circuit we can clearly define that as the point of zero potential in both circuits so we can even calculate the numerical value of potential difference between these two points.

So do we get a reading or not? Any help would be appreciated.

Answer 1

Circuits are approximately devoid of node capacitance (i.e. unlike the sphere atop a van de Graaff generator, they are presumed to hold negligible net charge), so the voltmeter (in circuit-theory approximation) would read zero. If the 'isolated' circuits have a voltmeter wired in, of course, they aren't isolated from each other any more.

Even without isolation, in circuit-theory approximation, there would be no current through the voltmeter because that would create or destroy net charge of the two separate entities' nodes, because the node capacitance is presumed negligible and thus can hold no charge. That's the equivalent of Kirchoff's first circuit law.

The only electrical theories that allow for net charge on the two circuits, and would (with the proper type of electrostatic difference voltmeter) show a voltage difference, are those of electrostatics, which do not require complete circuits of any sort.

Answer 2

Great question. An ideal voltmeter with infinite series resistance would probably measure a non-zero potential difference. However, a real voltmeter will have a large (but finite) series resistance, so what you would actually see is a potential difference that is probably non-zero initially, but that decays to zero roughly exponentially.

First, picture the circuits without the voltmeter. The two circuits are coupled via parasitic capacitances. In general, depending on a number of factors including the net charge on each circuit, the battery voltages, the resistor values and the geometry of the circuits, there would probably be a non-zero potential difference $V_0$ between the two wires you would hook up the voltmeter to. It is not trivial to estimate what this potential difference might be: you would need to solve this electrostatics problem considering all the factors mentioned above.

Now consider what happens when you connect the voltmeter. You can model the effect of the parasitic capacitances between the two wires as a lumped capacitance $C$ in parallel with the voltmeter, where the voltmeter has a series resistance of $R_V$. As soon as as you connect the voltmeter at time $t=0$, the capacitance $C$ will begin to discharge through the voltmeter. The time constant associated with this discharge is $\tau=R_VC$, so the voltmeter will measure a voltage given roughly by $$V(t)=V_0e^{-\frac{t}{\tau}}=V_0e^{-\frac{t}{R_VC}}.$$

Note that after you wait long enough, the current through the voltmeter must be zero, because otherwise there would have to be a steady non-zero current through the voltmeter, which violates Kirchhoff's current law (charges would pile up in both circuits). So eventually the voltmeter will measure zero.

Answer 3

There will be a non zero reading in it since there will be current a flowing through the voltmeter (unless $e^1/r^1=e^2/r^2$ in which the reading would be $0$). Also after you connect the voltmeter the two circuits are no longer be independent so you cant define two zero potential points as it becomes a single circuit.