Found a circuit where current is zero across an ohmic device (dV>0)

[DrBanana]

I know that a voltage difference doesn't necessarily imply current flow, but all the explanations I seem to find of that either say that that happens because the resistance is infinite, or there is nothing connecting the two points in question (the two are basically the same thing, as I understand it).

However in the above picture there is a battery, a wire and there is an ohmic device, so why is there no current between A and B? I understand that by virtue of Kirchhoff's current law, no current can really leave the loop on the right or the left, so physically it makes sense that no current flows between A and B. However how do I reconcile that with the fact that I've been taught that as long as there is a voltage difference and a favorable path, current must flow?

 

[phinds]

A to B is a single path (i.e. an open circuit), not a closed loop so of course no current flows.

 

[hutchphd]

However how do I reconcile that with the fact that I've been taught that as long as there is a voltage difference and a favorable path, current must flow?

By your reasoning, any battery on the shelf should be continuously discharging. There is no path in your circuit.

 

[kuruman]

However in the above picture there is a battery, a wire and there is an ohmic device, so why is there no current between A and B?

Imagine the dots in your drawing representing charge carriers like cars moving in traffic. Furthermore imagine your circuit to be a two-lane highway packed with cars moving at constant speed (drift velocity). What happens to the cars that go past point B and reach the upper left corner of the closed circuit on the right? They go around the cul-de-sac moving counterclockwise and exit on the other side at the upper left corner. You can see then that from the upper left corner of the left closed loop to point A, you have as much traffic going one way as you have going the other way which means no current. This is not the case with the closed loops where you get one-way circulation.

However how do I reconcile that with the fact that I've been taught that as long as there is a voltage difference and a favorable path, current must flow?

You reconcile it by modifying the take-home message to "as long as there is a voltage difference and closed path, current must flow. When you flip a switch to turn of a lamp, there is still a potential difference across the prongs of the plug but not a closed path to the resistor representing the light bulb.

 

[Averagesupernova]

The simple answer is you are misapplying what you've been taught.

 

[DrBanana]

For the future viewer I found this: https://physics.stackexchange.com/questions/410402/why-the-current-doesnt-flow-in-an-open-branch

Sorry I didn't find it before, it didn't come up in my googling.

Imagine the dots in your drawing representing charge carriers like cars moving in traffic. Furthermore imagine your circuit to be a two-lane highway packed with cars moving at constant speed (drift velocity). What happens to the cars that go past point B and reach the upper left corner of the closed circuit on the right? They go around the cul-de-sac moving counterclockwise and exit on the other side at the upper left corner. You can see then that from the upper left corner of the left closed loop to point A, you have as much traffic going one way as you have going the other way which means no current. This is not the case with the closed loops where you get one-way circulation.


You reconcile it by modifying the take-home message to "as long as there is a voltage difference and closed path, current must flow. When you flip a switch to turn of a lamp, there is still a potential difference across the prongs of the plug but not a closed path to the resistor representing the light bulb.

Does current not flow at all, or does it flow for a short period of time when this circuit is established?

 

[hutchphd]

There is no "circuit" and therefore no time when it is "established". There is a date of manufacture on the battery.....perhaps a time when wires were attached. See circuit. If one closes the circuit (switch in this hypothetical case) things will change to a new steady-state configuration over a short period of time which can be readily calculated by any proficient electrical practitioner. (For an idealized resistor battery circuit this happens instantly. Capacitance and Inductance will introduce time delays and perhaps oscillation. )

 

[sophiecentaur]

the fact that I've been taught that as long as there is a voltage difference and a favorable path, current must flow?

Who 'taught you' that?

 

[DrBanana]

Who 'taught you' that?

I think there was a line in my textbook that went something like "because these two charged spheres have different potential, current will flow if they are connected by a wire (until the potentials get equal again)." And there must've been a more general claim in the lower grades.

 

[sophiecentaur]

connected by a wire

exactly ; metal. The path has to be 'favourable', meaning a finite conductance so there's no 'division by infinity' to give you zero current.

 

[hutchphd]

This is absolutely true but only for transients. Please see #7 above.
All isolated conductors have a "self-capacitance" and will absorb or emit charge to match the local electrical potential. This is why one recieves a shock when discharging static buildup through a single finger. Such occurances are not usually part of lumped circuit analysis

 

[sophiecentaur]

Such occurances are not usually part of lumped circuit analysis

and, I would say, best left alone for a discussion at this level. People look for loopholes, rather than go mainstream. DC circuit theory is DC. AC next week perhaps?