How would one go about using a 42V DC power source to power something which needs 36 V DC using resistors?
Not at all.
Resistors by definition cause a voltage drop proportional to the current draw.
So, you can't use resistors to make a supply at 36 V for varying current draw.
What you're looking for is a voltage regulator:
- a linear voltage regulator converts the voltage difference (10 V or 6 V, depending on whether I believe your title, or your question text) to heat, and acts like a "self-adjusting resistor". It's not made of resistors, but semiconductor devices (transistors, mainly).
- a switch mode supply instead charges the magnetic field in an inductor and releases just enough energy from that to its output that the voltage is stable at the target voltage.
Without knowing how much or how little current you'll draw, no definite recommendation can be made, but at 6 or 10 V drop, chances are you want a switch mode power supply, which is relatively complex in theory, but can in practice be built from a controller IC, inductor, a diode and a few capacitors.
This is a really bad idea, but can be done if you know exactly how much current the device draws, and if this is a fixed, constant value over time, temperature, and input voltage.
If the device draws 100 mA of current at 36 volts, you need a resistor that develops a voltage drop of 42-36=6 volts at 100 mA. Thanks to Ohm's law (R=U/I) this equals 6/0.1 = 60 Ohm in series with the device.
To calculate the power dissipation in this poor resistor you can get that by the voltage drop and the current directly: 6 * 0.1 = 0.6 W.
These calculations are pointless if your device changes the amount of current it draws, and completely impractical if it draws a lot of current in which case you need a beefy resistor.
After you figure out if it's actually 42, 46 or perhaps 48V PSU, you could make a string of diodes to drop off the voltage. Voltage drop needed divided by 0.7v = number of diodes needed. 42 => 36V, 6 volts off, 6V / 0.7V = 8 diodes needed. And so on. You could use a zener diode for a larger drop but depending on your current drain it'll get mighty hot. And speaking of hot, depending how many amps the charger pulls, those diodes will be piping hot. Don't use small dinky types. Or Schottky diodes.
It would also be possible to use a MOSFET to generate a fixed voltage drop but one might just as well get a standard LM317, TO-220 3-pin regulator and a heatsink if you're going in that direction. Hot.
I see a pattern:
36V is 3 x 12V
42V is 3 x "12V" lead acid batteries in series.
If the device was designed to be powered from the lead-acid batteries it will probably work just fine, if not it will probably break.
if you need 36V Then you need a voltage regulator. LM2596HVS is one example.
One usualy would not do it that way, using resistors that is. Special casses exist, but without knowing what is soupose to draw the current, it is impossible to answer your question. Here is a link to a wiki article that explains input impedance, link output impedance, and a link explaining voltage division with resistors. If you read these articels, it is clear that the resistor voltage dividers output voltage is dependant on the input impedance of the next stage, and that the current draw is limited with the resistor that connects to the input voltage node.