# Potential drop in a circuit

• hokhani
In summary, the electric field in the wire and the resistance is different. Moving in the wire the potential should change point by point while this way, we neglect the electric filed in the wire.f

#### hokhani

A battery induces an electric field throughout a circuit in such a way that the electric field exists in the wire and the resistance. By moving in the direction of electric field, we expect that the potential decreases no matter we move in the wire or in the resistance. However, we only consider the potential drop in the resistance! Why we don't consider the potential drop in the wire?

Usually because the potential drop in the wire is so small that it can be neglected. So to a certain level of accuracy, we can assume that all points on the wire are at the same potential.

davenn, Asymptotic, jim hardy and 1 other person
Usually because the potential drop in the wire is so small that it can be neglected. So to a certain level of accuracy, we can assume that all points on the wire are at the same potential.
Thanks, we have electric field in both the wire and the resistance. Therefore, by moving in the wire the potential should change point by point while this way, we neglect the electric filed in the wire. Could you please guide me if I am wrong?

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This Insights article discusses the assumptions of circuit analysis, including those things that we deliberately neglect. You are not forced to use circuit analysis (Ohm's Law, Kirchoff's Laws). You can do it all using Maxwell's equations where the 3D location of each bit of wire matters.

https://www.physicsforums.com/insights/circuit-analysis-assumptions/

But it you try to mix Circuits and Maxwells in your head, you'll be terribly confused. It sounds like that is what you are doing in this thread.

Fisherman199 and sophiecentaur
Therefore, by moving in the wire the potential should change point by point
If the wire resistance is low enough (ideal wire) there is no change in potential over the wires joined to one node of the circuit. Charge can flow anywhere without dissipation any energy. Under DC conditions, there is no difference in potential so there is no field. Trying to do this by working out the Fields over the circuit is very complicated for anything but the simplest and most symmetrical of circuits - which is why Electrical Engineers stick to Potential Difference in nearly all circumstances.
Take a short (1mm) 100Ω resistor and a long 100Ω (1km) resistor and apply 10V. The current will be the same (0.1A) for both but the Fields will be different by a factor of 106 if you treat the circuits as 'ideal'. Life's too short to worry too much about it, imo.

davenn, jim hardy, Dale and 1 other person
Thanks, we have electric field in both the wire and the resistance.

When you're just starting out and struggling with concepts
it's usually helpful to look at the units used in whatever phenomenon you're studying.

Electric field is volts per meter.
Because copper conducts so well it takes only a miniscule field to make the sort of current density one is likely to encounter in everyday experience.
That's why we're telling you that there's no electric field inside the wire ---
-------- because it's so small it's almost zero volts per meter,
------------- and if it weren't that small preposterously impractical amounts of current would flow..
The electric field in your resistor is many times more volts per meter,
and will in fact consume very nearly all the volts your circuit has availabe. .

To keep it reasonably easy for students to learn fundamentals we tell the small lie that the field inside a wire is zero,
and because of that there's no voltage drop along a wire..
That's so close to the truth that it will cause you no difficulty in 95% of the work you do as an engineer.

Where it will fail you is in power distribution.
We have to account for voltage drop in long wires carrying a lot of power to let's say a motor or electric furnace..
A rule of thumb is design for voltage drop in the feeder line not more than 3% of supply.

But in beginning circuit analysis we use the simplification that wires are perfect.
,,,, and no E-field can exist inside a perfect conductor.

Fisherman199, dlgoff, sophiecentaur and 2 others
Because copper conducts so well it takes only a miniscule field to make the sort of current density one is likely to encounter in everyday experience.
Imagine you have two large parallel metal plates, 1m apart and with 100V across them. You will get a 'uniform' Electric field of 100V/m between them (the field lines will all be parallel. Then take a 0.9m piece of wire and hang it between the two plates (at right angles to them. Because of its low resistance, the field will now be distorted by the wire. You could say that the wire "shorts out" the local field. What's left of the 100V will be shared between the gaps at each end of the wire and there will be a high field across the two gaps (50/0.05 V/m = 1000V/m) and zero field right next to the wire. Way out to the side, where the wire is far enough away, the field will settle down to 100V/m.

Fisherman199, Tom.G, Asymptotic and 1 other person
Imagine you have two large parallel metal plates, 1m apart.....

Wow ! What a word-picture , and what a masterful lead-in to field thinking.

Keep in mind, too, that the major flow of power is not through the wires but just outside, via the Poynting vector. Lot's of papers on this but one to suggest is "Electrodynamics by John Krauss. The resistance does cause some power to intercept the wires and cause the heating seen.

sophiecentaur