Bottom line -- #10 for 3.6% voltage drop at last light.

Since I'm a student, I got out my trusty Ugly's book. Using the Ohm's Law wheel on the front I calculated that a 100w bulb on a 120v circuit has a resistance of 144 ohms. [E*E/W = 120*120/100]. Call the bulb resistance B. Since I'm a student, I'll ignore the fact that resistance of a light bulb varies with the applied voltage.

Next I looked up the AC resistance of of #14 through #6 wire. It is given for 1000ft so I divided it by 20 to get the resistance for 50ft:

14 - .16; 12 - .10; 10 - .06; 8 - .04; 6 - .02. Call the wire resistance C.

Now picture the circuit as a ladder with each rung a light. Label each light as R300 through R309, with R300 at the source and R309 farthest away. Label the left rail that connects R300 with R301 as R101. Number the rest of the left rails R102 through R109. Number the corresponding right rails as R201 through R209.

Now the fun begins: using the series and parallel resistance formulas calculate the circuit's total resistance.

First do the R109, R309, R209 group which is simply 2 * C + B, which for #10 is 144.12 ohms.

To calculate the resistance of the R108, R308 (in parallel with the R109, R309, R209 group), and R208 you need to first calculate R308 and the parallel circuit which is 72.03 [1/((1/Ra)+(1/Ry)) 1/((1/R308)+(1/144.12))] call this Rp and then add 2 * C which is 72.15 [R108+Rp+R208 2 * .06 + 72.03].

You repeat this calculation back to R101, R301, R201 which gives you a table of resistance values at the source side of the R101, R201 terminals to the R109, R209 terminals:

resistors resistance
Rxx1 16.42
Rxx2 18.38
Rxx3 20.91
Rxx4 24.30
Rxx5 29.06
Rxx6 36.22
Rxx7 48.19
Rxx8 72.15
Rxx9 144.12

Now that we know the resistances we can calculate the current, starting at the source. The source voltage is 120v, the resistance is 16.42 ohms, so the current is 7.31a [E/R = I 120/16.42]. The we can calculate the voltage drop in R101 and R201 to be .88v [R*I = E 2*C * 16.42] This leaves 119.12v across R301. Now we repeat the calculations and get the following table:

resistors current total volts drop
Rxx1 7.31a 120v .88v
Rxx2 6.48a 119.12v .78v
Rxx3 5.66a 118.35v .68v
Rxx4 4.84a 117.67v .58v
Rxx5 4.03a 117.09v .48v
Rxx6 3.22a 116.60v .39v
Rxx7 2.41a 116.22v .29v
Rxx8 1.61a 115.93v .19v
Rxx9 .80a 115.73v .10v

This gives a total voltage drop of 4.36v or 3.64%

JFW