Calculus for puzzlers, plug and chug for EEs?

There are 2 legs, so just analyzing 1 at first.

The idea is to determine the average current (conceptual) through the 28 resistors.

This is not avoiding calculus, rather simplifying it. Because all the loads are equal the average is 1/2 the total.

Try it with 4 loads of 1A to understand. Make each section 1 ohm too. Draw it out, write the currents near each resistor...

The average current is the sum of all the load currents divided by the number of loads.

That would be (4A + 3A + 2A + 1A ) / 4 loads = 2.5A

Easy with only 4 loads, but with 28 loads tedious, so the average current is:

Iavg = 0.5 * ( (Number of loads * current per load) + ( 1 * current per load) ) = 2.5A

Next, 4 loads has 4 resistor sections, so 4 ohms (one leg).

Vdrop = I * R = 2.5A * 4 ohms = 10V.

Total is double that for 2 legs.

In your example:

With 28 loads of 0.5A

The average current for all the resistors is:

Iavg = 0.5 * ( (28 * 0.5) + 0.5) = 7.25A

The total resistance of one leg is 28 * 0.0104 = 0.2912 ohms

The voltage drop in one leg:

V = IR = 7.25A * 0.2912 ohms = 2.1112V

Double it for both legs = 4.2224V =====> answer

The voltage across the final load is 24V - 4.2224 = 19.7776V

resistance of first connection is .....

4 x 5.2 = 20.8 milli-ohm and voltage drop = 20.8 x 0.5 = 10.4 milli-volt

if you add the second load ...voltage drop = 2 x 10.4 = 20.8 mV

and on 14th load >>> 14 x 10.4 = 145.6 mV