The humidity sensor in the question operates on an AC signal of up to 1 Volt RMS. The datasheet specifically mentions an operating frequency range of 0.5 to 2 KHz, as well.
If the sensor is operated with DC supply, one electrode (the negative one, if I remember correctly) will deteriorate rapidly due to ion migration towards one plate in preference to the other, rendering the part inoperative.
Now, regarding a suitable operating mechanism:
The impedance curve of the device spans a range of anywhere from 1 to 10 Megaohm at 20% relative humidity, down to between 1 and 5 kΩ at 90% RH. The impedance table in the datasheet specifies values from 1.1 kΩ to 7.2 mΩ, too broad a span for a voltage divider to work.
Calculating current through the device for 1 Volt across it, spanning this impedance range to worst-case limits:
At 90% RH, for 1 kΩ, I = 100 μA
At 20% RH, for 10 MΩ, I = 100 nA
Thus, a very low impedance (100 Ohm or less) AC voltage source would be needed, to drive this sensor suitably, if it is to be operated in voltage driven mode. This shows that a voltage divider would be a very inefficient, and somewhat ineffective, way of driving the sensor.
Instead, a more viable approach would be to drive the sensor using a current source, with the DC blocked using a suitably large capacitor.
There are several current source circuits out there, using bijunction transistors, FETs, or op-amps. Pick one that suits your purpose and budget, gate the current with an input from one pin of your microcontroller being toggled at say 1 KHz, and read the voltage across the sensor using an ADC pin of the MCU.
Note that this will not give very precise results, as such electrode-based humidity sensors are characterized using a bipolar sine wave. Improvements to the solution could include using an RC or LC filter to bypass the higher harmonics of the 1 KHz signal, leaving an approximation of a 1 KHz sine wave.
Actually designing such an AC (near)sine wave, stiff current source is left as an exercise to others less preoccupied than me.
You're on the right track in two regards; your calculations headed in the right direction, and the Radio Shack guy doesn't know what he's talking about. A general good start in thinking about electrics is to consider vague magnitude. You have a project dealing with 40W here. Does 1/8 of a Watt sound like the right sort of answer?
The bit of calculation you need here is the power dissipation in the resistor. A series resistance circut (when resistors are connected end to end, "in series") is pretty simple. In this case, we know that the current flowing in the circuit is 0.335A or so. We also know that the resistor you are using is 100 Ohms.
The power dissipation in a component is Volts X Amps. Combining this with Ohms Law, we can show that this power is also Amps squared X the resistance in Ohms.
This gives us 0.335 X 0.335 X 100. Which gives an answer of 10.77 Watts.
So the vague magnitude speculation actually gave you a pretty good answer, calculating it we find a 20W rated resistor will comfortably handle this load. And not to trust the guy in Radio Shack.
edit Oh poo, I didn't realise you have two parallel strings. And now I'm typing this retraction I can't see your picture. BRB.
edit 2.
Okay, we now have 50 lamps across 120V. The total resistance of all the lamps is V*V/P, which is 120*120/20, or 120*6, which is 720 Ohms. Divide that by 50 to get the resistance of 1 lamp, which is 14.4 Ohms.
Now the resistor needs to be the same resistance as 27 lamps, so it's 27*14.4 which is about 388.8 Ohms. So that's our resistor.
The current flowing is I/R, which is 120/720, about 0.166 Amps. And the power dissipated by the resistor is 0.166*0.166*388. Which comes back to 10.77 Watts again; which it ought to because we're dissipating about (rule of thumb) 1/4 of the entire string's rated power as useless heat in the resistor.
Best Answer
(1) So called "Auto transformers" which are essentially a single winding with a tap part way down the winding, are available which "buck" or boost the input voltage by a small amount. These are substantially less costly than a normal transformer of the same input power as they are effectively transforming only the voltage difference.
(2) A resistor can be wound with suitably heavy Nichrome wire. This is usually available for electrical suppliers. A self supporting air cooled coil or coils is easy enough to make.
If you are converting from 120 VAC to 100 VAC at 100A the dissipation required is about 2 kW. That is 100 to 200% of a typical toaster, giving you some idea of what can be achieved. It may be possible to use toaster elements with rearranged wiring to parallel several sections to lower resistance.
(3) Incandescent light bulbs may be useful. Their wide cold to hot resistance change may make their use hard.
(4) A series inductor may work depending on the load characteristics. An old microwave over transformer (or several) with added airgap to lower inductance and increase current handling. These often have welded core laminations. An angle grinder + hacksaw may be needed.