Hello guys.Sorry if i post my question in wrong section.How many current can handle a 50K potentiometer?I have a 50K potentiometer and i want to control a dc motor(650mah)and i am wondering if i can.Thank you very much.Have a nice day.
I think you need get acquainted with Mr Ohm and his eponymous law.
("mah" is not a recognised rating for motors)
What do you intend to do with this pot?
If it is simply putting it in seriese with the motor this is a poor way to control the speed.
Not only that bit you can get a pot in lots of different power ratings. The problem is that this is only relevant for the whole of the track. If you use the pot as a simple variable resistor then as their is less and less of the track dissipating the power. The result is that if the pot has not a very large power rating the end of the track vapourises. This will happen with all "film" pots so you would need a wire wound one. Even so it needs to be chunky.
JKO1992:
Hello guys.Sorry if i post my question in wrong section.How many current can handle a 50K potentiometer?I have a 50K potentiometer and i want to control a dc motor(650mah)and i am wondering if i can.Thank you very much.Have a nice day.
- Motor current rating is in mA or A, not mAh
- potentiometer is for signal, not power.
- A 50k potentiomer is going to see 0.25mA at 12V supply, I = V/R
You want to read the pot with analogRead, and control a motor driver or shield perhaps?
[u]Ohm's Law[/u] (Current = Voltage / Resistance) describes the relationship between voltage, current, and resistance, and it's the 1st thing you learn when you take an electronics class.
And, I assume you know that a pot is a kind of variable resistor. (So, the current will depend on the adjustment, and on whatever else is connected, and how it's wired.)
If you know the voltage rating on the motor, you can calculate it's resistance. But the 650mA rating is probably the "worst case" stall/start-up and it will draw less current (have more resistance) when it's up-and-running with no load.
If you put a variable resistor in series with the motor, you've created a [u]voltage divider[/u]. And if you do the calculations, you'll find that with a 50K series resistor and a much-lower resistance motor almost all of the voltage will be dropped across the resistor with very little voltage across the motor, and THE MOTOR WON'T RUN.
...Even with the pot adjusted to "90%" with 5K in series with the motor, the motor won't run. If you crank the pot all the way, there will still be some resistance but the motor may start to run. But, you'll probably also burn-out the pot.