I am very inexperienced with regards to electronics and solderings. It seems most if not all voltage reduction cables just have a single resistor on the wire for the yellow (+V) or red (+V), depending on which you have; the +V cable has a resistor to it.

If you're talking about 2 resistors Ohm's law applies and the voltage drop is the ratio between the resistors. So a resistor of 1 Ω and one of 2 Ω will see 1/3 of the voltage being consumed by the first, and 2/3 by the second.

This also means that the current going through the circuit is going to be e.g. 12V / 3 = 4 amps? That means that this circuit would draw 4 x 12 = 48W ??.

But Ohms law does not apply to dynamic materials. A fan rated at e.g. 0.33 A on 12V, would, according to Ohms law, have a resistance of 36 Ω. Then, if you wanted to drop the voltage by 1/3, you would ensure that the preceding resistor has a resistance of 18 Ω and then it would consume 4V, leaving 8V for the Fan. If the Fan is still drawing 0.33 A, the resistor would consume 4V * 0.33 = 1.33 W.

When I look online at selections of resistors they are all rated for 0.25W, they are those tiny things. But you can also easily find .e.g 3W online. Looking at some of the loudest fans they draw at most 4W at 92x92mm. Yes, this is for a graphics card and I haven't checked but its fan might go up to 3600 RPM. It's 80 degrees value was 2250 RPM I believe, and its 100 degrees value was 3600.

Other fans that can do at most 2500 RPM draw at most 2.1W. So for a resistor I assume for a 1/3 voltage drop you will be looking for some 1-2W maximum for whatever fan you hook up. A 2W resistor is simply going to be smaller.

However I have no clue what would happen with PWM. If resistance of the 2nd component (the load) goes up and down, then the voltage drop will differ from time to time. If the thing uses a lot of current the resistance will be low and the voltage drop of the resistor will be higher, which means that when it draws more current, it will get less voltage, and that might cancel it out in some way.

Then, to get a fixed voltage drop, I would have to employ

this thing. To create a voltage divider I must employ two resistors in parallel where... I mean. They have to be in sequence but the gap between them must feed to the fan. So now I need two resistors where one goes back to the black wire (ground) and the other one is in sequence to the fan, that also feeds back on the black wire.

However the second resistor (that feeds back to ground) is in effect part of a parallelel circuit that also contains the load:

**Code:**

PSU ---> resistor 1 -------> resistor 2 -------> ground

\---> load ----------/

Since the voltage drop for resistor 2 and load is going to be identical its size (magnitude) is going to depend on the resistence of the parallel block with respect to resistor 1. If the maximum power draw (current draw) for the load is say 0.30 A with accompanying resisance of perhaps 40 Ω, then at maximum draw if you put a 40 Ω resistor at the 2nd spot and a 20 Ω resistor at the first spot, you will have 8V for the fan.

But if the amperage goes down and resistance goes up then the weight of resistor 1 becomes heavier and ... the voltage drop will be more pronounced. But I am already mistaken. Two identical ohms will see the resistance of the second block go down to 20 Ω in our example. Thus, the first resistance must be 10 Ω. The resulting resistance of the 2nd block (the parallel block) will always be less than its components. But if the resistance goes up.... let's say it goes up to 80 Ω for the fan.

1/40 + 1/80 = 3/80 --> 80/ 3 = 26.667 Ω for the second block with the first resistor at 10 Ω gives a 10/36 * 12 = 3.3V voltage drop, leaving 8.7V for the fan, so I am mistaken, the voltage drop is less pronounced. The weight of resistor 1 does not become heavier obviously, it becomes lighter. My mistake. With infinite resistance the voltage drop approaches 1/2, or 6V.

So we'd end up with a voltage of 8V (when the fan is spinning fastest) down to 6V (when the fan is spinning very slowly). That means the speed reduction will be more pronounced in the lower speed range. Building on this, you could reduce the voltage reduction by placing e.g. some 5 Ω resistor at first spot at which point the maximum voltage drop will only be 1/5.