I want to use a 7.2V (4200mAh) battery with my raspberry pi but I need to convert the output to 5V (2000mAh), I tried the following formula:
Ω = V / A
but when I try to find the ohms the result is 0.78
|7.2/4.2 - 5/2| = 0.78
How can I find the correct ohm (to use a resistor) to get an output of 5V and 2A?
This will NOT work.
You cannot just use a resistor. If you try you will destroy your Pi.
You will need a regulator. A switching regulator is recommended.
Some ready modules do the task. Essentially, same circuit inside as UBEC switch mode DC/DC regulator but added USB socket.
Example, this one max 3A and ok as RPi max demand is less than 3A
Note other similar looking modules boost voltage (instead of dropping) from 3.7V Lipo cell to 5V
As the load is RPi and it varies, I recommend using a voltage regulator. For 5V it is L7805. Connect as shown here.
If you want to use resistors only, try using a voltage divider network.
Vout = (Vin * R1) / (R1 + R2)
If you dont want theory use this calculator. As rule of thump use R2 = 2 * R1 so you get a 4.6V.
EDIT: The best option is using a UBEC switch mode DC/DC regulator
Would this be a sufficient solution to this problem ? When the Ni-Mh battery is connected to the power port of this board:
These regulators also have internal diode protection and can handle transient surge currents from 50 A to 100 A.
The operating instructions state that the input voltage through the barrel socket must be between 6.5 V and 12 V. Hence, if you wish to use it to its maximum capability you will need to remain in that range. This is a non-adjustable fixed power supply model, which is good enough for most applications. The manual states maximum output current to be 700 mA. However, it is probably better to use much lower voltages and current to be on the safe side in case you make a mistake on your breadboard circuit.
