If the supply voltage is constant, increasing the length of the conductor increases the total resistance. With the same current flowing through the circuit, this should produce a larger voltage drop according to Ohm's law.

What I'm less clear about is how this develops along the length of a circuit rather than simply appearing as a single voltage loss. Does the potential decrease continuously along the conductor, and can this be thought of as an electric field driving the current through the resistance of the wire?

I'm also interested in how the wire's cross-sectional area changes the situation. A thicker conductor has lower resistance, but is there a useful way to relate this directly to the potential gradient along the conductor?

I'd appreciate an explanation from the physics side rather than just a practical wiring rule.

The supply voltage is fixed, Vs.
The LED voltage is fixed, Vd, (by the band gap of the material).
The resistance, R, of the circuit wires is proportional to length.
The current that flows in the circuit will be, I = ( Vs – Vd ) / R .

If you increase R, then I will fall.

If each LED has its own switching current source that draws power from the string wires, then the LEDs will be the same brightness even though the string wiring voltage decreases with distance. Each LED circuit will draw the same

Khan academy has some really good tutorials on electronic circuits.

One way to reduce that variation is to feed the strip from both ends. Connect the positive supply at one end, with the negative supply at the other end. That places the same series supply resistance in all LED current paths, so the voltage variation should not result in a current or brightness variation along the strip.

All LEDs will then be equally bright, the same brightness as the LED in the middle of the strip would be, if fed with both supplies from one end only.

LEDs, or loads, on an intact UK ring main configuration would be brightest at the ends, with the dimmest LED being in the middle of the ring. The UK ring main configuration is actually used for most LED strips, because it provides more light, but it only halves the brightness variation, it does not cancel the variation completely.

One pair of conductors, in a ring, with only one conductor fed at each end, equalises the line resistance to all nodes on the ring. Unlike in the UK ring main, any break in either conductor will extinguish all nodes between the break and the open end of that conductor. For that reason, it is easier to diagnose the presence and position of a break, than it is in the UK ring main system, but it doubles the cost of the copper needed in the wires.

With just one ring main conductor cut, the system works 'fine' except under heavy loads. Checking for a break is not easy without testing for continuity with the whole ring disconnected from the board. I think it will be a matter of differences between power dissipation in the power and the lighting applications.

Assuming the buck switching circuit has high efficiency, then the power drawn from the distribution wires is just the LED current multiplied by the voltage drop across the LED and the current-sensing resistor, which is the constant power that I mentioned. The LED circuits at the end of the distribution wiring will see an input voltage less than the source 12Vdc, but they will still draw the same power determined by the LED current and the voltage drop of the LED+resistor. Since they are constant power consumption devices, the input current will be proportionally higher since the input voltage is lower.

Does that help?

When all LEDs are on, the variation in the ring-main easily beats the variation in the single-ended, and comes close, but not equal to, the variation in the equal-brightness circuit.

I agree that, the ring-main is a clear win for the economy.