# Voltage drop in a conductor

My textbook, Practical Electronics for Inventors, Fourth Edition, by Scherz and Monk, says the following in section 2.3.1 The Mechanisms of Voltage:

In regard to potential energies of free electrons within the conductors leading to and from the battery, we assume all electrons within the same conductor have the same potential energy. This assumes that there is no voltage difference between points in the same conductor. For example, if you take a voltmeter and place it between any two points of a single conductor, it will measure 0 V. (See Fig. 2.8.) For practical purposes, we accept this as true. However, in reality it isn't. There is a slight voltage drop through a conductor, and if we had a voltmeter that was extremely accurate we might measure a voltage drop of 0.00001 V or so, depending on the length of the conductor, current flow, and conductor material type. This is attributed to internal resistance within conductors - a topic we'll cover in a moment.

I can't help but wonder if the authors are understating the potential for voltage drop through a conductor. Their estimation of a 0.00001 V drop seems reasonable for a relatively small electrical circuit, but what about for large-scale power systems, where the conductor may span many miles (say, an undersea cable, an electrical grid (although, I assume that electrical grids have hardware in place to "boost" the voltage between power plants and delivery destinations), a solar farm in the desert, etc.)? In my, perhaps naive, mind, I wonder if the voltage drop would not just be minor in such situations, but significant enough to cause practical problems?

I would appreciate clarification on this. Thank you.

• thats why high voltage AC transmission. – Mitu Raj Nov 27 '19 at 4:02
• I think the author is talking about the voltage difference across a very small size conductor. In power grid the voltage drop usually ranges about 2~4%. – ab29007 Nov 27 '19 at 4:23

The conductivity of these materials is well known. Copper is $$\5.96\times 10^7 {\rm \frac{S}{m}}\$$. It varies a little depending on temperature and how the metal was worked.
The resistance of a round wire is easy to calculate from the conductivity of the material, $$\R=\frac{l}{A\sigma}\$$, where $$\A\$$ is the cross-sectional area of the wire, $$\l\$$ is the length of the wire, and $$\\sigma\$$ is the conductivity of the material. So for any given length and diameter of round copper wire, you can calculate the resistance. Let's say 1 km of 10 mm diameter. Then you have $$R=\frac{(1000\ {\rm m})}{\pi (0.005 {\rm m})^2(5.96\times 10^7\ {\rm\frac{S}{m}})}=0.21\ \Omega$$
And the voltage drop along the wire is given by Ohm's law: $$\V=IR\$$. So if you have 1 A flowing through this wire, 0.21 V drop. If you have 100 A flowing through this wire, 21 V drop. Whether that's significant or not depends on the voltage being used. 21 V drop on a 50 kV transmission line is pretty small potatoes.