Calculating Resistor Values for Voltage Reduction from 12V to 4-5V for Fans and LEDs

What is the formula for the voltage drop across a resistor? I mean the formula by which I can count, for example, the value of a resistor that allows to lower the fan voltage from 12V to about 4-5V (or a potentiometer) or to power the led from 12V.

Ohm's Law: The ratio of voltage to current is constant. (R = U / I) and is called the electrical resistance. You need to know the electricity.

Added after 16 [seconds]:

Ohm's Law: The ratio of voltage to current is constant. (R = U / I) and is called the electrical resistance. You need to know the electricity.

And how to calculate the current for a windmill for example. Do you follow its rated current stated on the label? This is probably the current for the rated voltage (12V). How to calculate the current for a lower voltage, e.g. 4-5V?

If you have given the current for 12V, the current for the lower voltage is proportionally lower (e.g. for 6V it will also be half the value).

It has to be measured, it cannot be counted directly.

In addition, the starting current must be taken into account. It cannot be too small because the fan will not start.

It is good to have a regulated power supply, choose the voltage so that the fan rotates at the desired speed and measure the current consumed.

We have:

Uz - measured voltage from the regulated power supply
Iz - current consumed at this voltage

Ucc - target supply voltage

In steady state (i.e. when the fan is rotating at a constant speed)

Ucc = Uz + Iz * Rx

Rx - series resistor

that is

Rx = (Ucc-Uz) / Iz

The power of the series resistor must be greater than

Prx> (Ucc-Uz) * Iz in Watts

For such a circuit to start reliably, an electrolytic capacitor of 470 uF or more can be placed in parallel to the series resistor.

At the moment of starting, the charging current of this capacitor will give a current pulse to facilitate starting. After charging to voltage (Ucc-Uz), the current stops flowing and we have a voltage on the fan limited by a resistor.
When the power is turned off, the capacitor discharges through the limiting resistor.

As for the rated current written on the housing, I measured myself an Adda Corp brushless motor. AD0812MS-A70 a little used already (plain bearings)

According to what is on the housing, it consumes 120 mA at 12V

According to tables from ADDA Corp. this type draws 150 mA at 12V
And according to measurements:

[code:1:41dfb82822]
Uz Iz
12V 190 mA
11V 171 mA
10V 151 mA
9V 133 mA
8V 116 mA
7V 99 mA
6V 81 mA
5V 63 mA
4V 45 mA (niestabilnie)
Rx = (Ucc-N * Ud) / Id

N - number of diodes connected in series
Id - diode current 10-20 mA
Ud - voltage drop on one diode (from 1.8 to 4.1V depending on the color and Id current)

Ucc supply voltage

The power of the resistor must be greater than:

Prx> (Ucc-N * Ud) * Id