# Why are wires in simple circuits approximated as equipotentials?

Voltage in a circuit

by asserting that any two points joined only by a wire, but no other circuit elements, are at the same value of the electric potential. It is, of course, true that in the context of electrostatics, the values of the electric potential at any two points inside of a perfect conductor are the same, but the electrostatics argument seems irrelevant for circuits.

Why are we justified in using this same fact for circuits in which we are clearly no longer faced with electrostatics?

Edit. As prompted by user twistor59 in the comments below, let me include a bit more detail to clarify the question and hopefully indicate why I don't think the answer is obvious.

If we consider any segment of the wire as a small ohmic resistor, then as the resistance of any segment drops to zero, the voltage drop across that segment will be correspondingly small for a given current. In particular, zero resistance implies zero voltage drop. However, if I just have a piece of a perfect conductor sitting somewhere, then it is totally unclear to me why, or that it is even true, that the electric field will vanish in its interior if the situation is non-electrostatic. Perhaps, therefore, the magic lies in the microscopic physics that (in particular) lead to Ohm's Law?

• @twistor59 Yeah ok so current starts to flow, and once it does so, is it still the case that the statement about wires being equipotentials is approximately true? For example, imagine that I have a resistor-battery series circuit. When a real circuit of this type reaches its steady state with current flowing, is it still true that any two points on the wire are approximately at the same potential? Is this really simple? Am I just missing something totally obvious here? – joshphysics Jun 10 '13 at 18:41
• Hmm perhaps I am asking something a bit deeper. I understand that if we consider any segment of the wire as an ohmic resistor, then as the resistance of any segment drops to zero, the voltage drop across that segment will be correspondingly small for a given current. In particular, zero resistance implies zero voltage drop. However, if I just have a piece of a perfect conductor sitting somewhere, then it is totally unclear to me why, or that it is even true, that the electric field will vanish in its interior if the situation is non-electrostatic. Perhaps, therefore, the magic lies in – joshphysics Jun 10 '13 at 18:50
• (contd.) the microscopic physics that leads to Ohm's Law? – joshphysics Jun 10 '13 at 18:51
• Dear @joshphysics, Ohm's law $j=\sigma E$ holds even outside electrostatics, even in the presence of magnetic fields. An infinite conductivity $\sigma$ means that for any finite $j$, the electric field is simply zero (or infinitesimal) and the potential difference is meant to be the integral of $E$. Of course, in general, one may make a gauge transformation and completely redefine what we mean by a potential - different redefinition at each point in space - but that's not "the" potential we use in circuits. – Luboš Motl Jun 10 '13 at 19:16
• @Dilaton, it would be a shame to flag even the unedited question as too localized. This is an excellent conceptual question and goes to the heart of the common problem of forgetting the assumptions that must be made in order for a result to hold. – Alfred Centauri Jun 11 '13 at 22:18

Why are wires in simple circuits approximated as equipotentials?


Because one of the three assumptions of circuit theory is:

All electrical effects happen instantaneously throughout a circuit. If the circuit is small enough compared to the wave length of the signals applied, all electric signals travel through it so quickly, that we can assume that they affect every point in the network simultaneously.

For sinusoidal signals, we require that the wavelength of the signal be far larger than the length of the wire. If this isn't true, then we can no longer used the lumped element circuit model and must instead use the distributed element circuit model.

Here's a thought experiment. Consider a sinusoidal source with one end connected to ground and an ideal wire of length L. The "far" end end of the wire is not connected. At time $t=0$, the source is connected to the "near" end of the wire. Let's assume that, at that time, the sinusoidal source is just crossing zero and the voltage is increasing.

Since no signal can propagate faster than $c$, it is necessarily the case that the voltage, with respect to ground, at the far end of the wire must remain zero until at least $L/c$ seconds have elapsed.

Let's assume that the sinusoidal source has completed one-half cycle at the time that the voltage at the far end of the wire begins to change. It will be the case that standing waves of voltage and current will form on this wire and that the voltage and current waves will be in quadrature.

For circuits that operate at GHz frequencies, the physical circuits look far different from those in which the lumped element approximation is valid.

A circuit that operates in the 1MHz range:

Above, the wires and other components are very nearly lumped elements.

A distributed element circuit that operates in the 20GHz range:

In the above, some of the copper traces form microstrip filter elements. The signal voltage is not uniform along the length of these traces.

Just to add to Alfred's detailed answer. The rule of thumb among the electron pushers is if the the physical structure exceeds 1/10 of a wavelength at the maximum frequency of its usage then it can not be considered a "lumped element" and Maxwell's EM equations need to be employed in the analysis. Up until that point the lumped element model works "for all practical purposes".

Only in the lumped model are connected points considered 0V, otherwise they are modelled as transmission lines.

• @Voitcus the only reason 50/60hz needs "special" theory for long lines is because the physical structure is greater than 1/10 the wavelength. It's less about approximations because in those cases lumped element models won't work. – user6972 Jun 11 '13 at 8:49

This is a simple consequence of Ohm's law, V=IR (note that this is by its very nature, an equation outside the realm of electro-statics, and that it is exact regardless of the drive frequency for a purely resistive element like a wire). A wire, by definition, has a very small resistance (precisely zero in the "ideal" case, some tiny number in most practical circuits), and so V, the voltage drop between two points on the same wire, is correspondingly very small. It really is just that simple; I think some of the answers here are extemely over-complicated and miss the point. FYI, I am a physics Ph.D. here with 10+ years teaching and research experience.

• +1: I am inclined to agree. I somehow over-complicated this in my own mind when I asked the original question. There is the additional question of the domain of validity of Ohm's law, but it wasn't really the original aim of this question to address that. – joshphysics Jun 14 '13 at 4:46
• This is true if the current is finite. In theory you can have an infinite current and a finite voltage which still makes ohms law correct. – Larry Harson Jun 14 '13 at 15:44

This is a difference between theory and practice.

I remember when I was beginning my studies I had lots of problems to understand why every teacher takes $\pi$ as 3.14 and not 3.1415926..., as I learned in school. In algebra $\pi$ was never calculated and the results were something like $2\sqrt2 \pi$. That was because in engineer calculations we don't care about so much precision.

In simple theoretical circuits we make assumptions as well. Because voltage drop coming from Ohm's law is very small, we consider it zero. We are not interested if the current in main circuit is 1 A or 1.00089 A, 1% accuracy is enough for almost all engineering purposes.

For practical calculations we are even not able to consider all impacting factors, while some (with different impact) are:

• influence of temperature on resistance,
• the fact, that resistance is not distributed regularly along the line length, so $dR / dx \neq const$,
• influence of humidity, insulation resistance which leads to current flows from the wire to earth,
• magnetic fluxes from other sources,
• internal and external capacitances,
• photo-voltaic,
• magnetic field of human brain of the scientist (why not include this?),
• effects of gravity of Jupiter (theoretical physics says there is such impact),
• etc.

Usually calculations are made to know what current (eg. short-circuit current) will flow. If we do calculations to find what switchgear or fuse we need, it doesn't matter if the current should be 15.23213121 A or 15.23943 A, because we choose 16 A or 25 A.

The real element (a wire) could be considered as being something like this:

This is an equivalent scheme of wire. The resistance is not the only parameter here, so is inductance of the wire (because it forms some kind of loop and is producing magnetic flux) and there is internal capacitance between both ends. You may say that for DC circuits L and C do not matter, but it is not true. In transients states they have quite large impact on the current / voltage, especially for high frequencies. In electronics even short wires can create many bad phenomena (that's why it is now not possible to make a microprocessor with frequencies larger than some GHz), that cannot be easily handled.

But in "normal" frequencies like 50/60 Hz or DC (which we assume to be constant) this really doesn't matter. L and C (and R) can be safely omitted and we still get almost perfect results, and exact for the purposes we are doing calculations for.

However, if the line is quite long, it is not possible to omit these values. By "long" I mean hundreds of kilometres/miles and there is a special theory of dealing with long lines. The higher frequency we use, the shorter "long line" is and for electronics a long line can be even some millimetres.

So, finally, answering to your question: we are justified to say that a wire is $0 \Omega$ resistance (or rather: impedance) if and only if the error made by this assumption is not disturbing required accuracy of calculations we do.

We are never justified if the impact of this assumption is violating our desired accuracy.

In theoretical calculations (eg. in school) you are to learn that $U = I R$, so you should pay attention to this and nothing else; this is a hypothetical example, however as all the physics is. The real world "violates" theoretical physics: there are no elastic collisions, there is friction, and there are no ideal conductors.

Wires can be approximated as equipotentials in circuits if boundary conditions force the electric field inside the wire to be maintained at $0 vm^{-1}$ through the creation of a cancelling internal static electric field.

Suppose we take an ideal battery and connect an ideal wire across it. Since charge is a Lorentz scalar with its speed limited to $c$, the current will approach a constant finite value, while it's energy will continue to increase. There isn't a boundary condition to force the creation of an internal electric field to cancel the applied electric field of the battery, and so the wire isn't an equipotential.

Now place a resistance in series with the wire. This time the energy of the current is fixed by the finite resistance and battery voltage, forcing charge to flow at a constrained rate across both ends of the wire, and creating a static charge distribution on the wire's surface and connection points to the resistance. So the boundary condition of current flow for the wire, forces the creation of a cancelling static electric field, making the wire an equipotential in this case.