I have a 10 kilohm potentiometer being powered by a twelve volt battery. After I run the circuit, the potentiometer begins to overheat. I thought that a 10 kilohm potentiometer could take 12 volts. Was I wrong or was it an issue with the potentiometer?


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Draw a schematic of how you have misconnected it using the CircuitLab button on the editor toolbar. \$\endgroup\$
    – Transistor
    Feb 26, 2019 at 16:24
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    \$\begingroup\$ "kiloohm" or "kΩ", but, please, not "kilohm". \$\endgroup\$ Feb 26, 2019 at 16:28
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    \$\begingroup\$ Question: do you know what the power dissipated over a resistor of given resistance under given voltage is? \$\endgroup\$ Feb 26, 2019 at 16:29
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    \$\begingroup\$ @MarcusMüller No, kilohm is correct. Also, megohm is correct but "megaohm" is not. There are three combinations of prefix and unit where a vowel is dropped, according to the SI, and these are two of them. \$\endgroup\$ Feb 26, 2019 at 16:52
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    \$\begingroup\$ @ElliotAlderson I'm in awe. \$\endgroup\$ Feb 26, 2019 at 17:05

2 Answers 2


Power is equal to the square of the voltage divided by the resistance, or $$P=\frac{V^2}{R} $$

For 12 volts and 10k, P = .0144 watts, which certainly will not overheat.

The thing is, you are not applying the voltage across the fixed leads of the pot. Instead, you are applying voltage to the wiper, and as you change the wiper position you change the resistance. As the resisistance goes down (the shaft approaches one limit) the power will go up. In the worst case, if you turn the pot all the way to its limit, the resistance will nominally become zero, and the power (nominally) infinite.

So don't do it. Or, if you must, put a fixed resistor in series with the pot so that the minimum resistance produces an acceptably low power.

  • \$\begingroup\$ Not only will the power go up, but it will have to be dissipated from a much smaller area because of the potentiometer construction. \$\endgroup\$
    – pipe
    Feb 26, 2019 at 17:07
  • \$\begingroup\$ would it be alright if I used a 100k ohm pot? \$\endgroup\$ Feb 26, 2019 at 20:51
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    \$\begingroup\$ @StrawberryMilk - I'm afraid you've missed the point. At some point the variable pin (the one shown as being an arrow) will make electrical contact with one of the fixed pins, and the resistance between them will go to zero. Turn the shaft all the way in the other direction and you contact the other. So the pot value does not make any difference. Any value pot will burn out if you turn it to zero ohms with a decent voltage source attached. As in your schematic. About the only voltage which would be safe in this situation is a coin battery. \$\endgroup\$ Feb 26, 2019 at 20:57
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    \$\begingroup\$ @StrawberryMilk A 10 ohm resistor will overheat. A 10k ohm pot set to 10 ohms will overheat. A 100k ohm pot set to 10 ohms will overheat, because all of them are 10 ohms. \$\endgroup\$
    – user253751
    Feb 26, 2019 at 22:45
  • \$\begingroup\$ @WhatRoughBeast Is there any other way that I could moderate the electricity I give out then? I am trying to produce heat, so I don't think that a coin battery would be an option. \$\endgroup\$ Feb 27, 2019 at 15:07

Your circuit shows that as you move the wiper towards the bottom position that it will progressively decrease the resistance until it is a short circuit at the bottom. In that case the current through the wiper will be limited only by the battery's internal resistance and the wiper contact resistance. Most likely this current will be high enough to destroy the potentiometer.

If you edit your question to explain what you are trying to make we can help further.

enter image description here

Figure 1. Uxcell Ceramic Tube Adjustable Rheostat Variable Resistor.

If you really want to use a variable resistor for your circuit you may need a rheostat. You need to work out what the maximum current your "custom wire" will draw on a 9 V power supply and get a rheostat that can handle that current and has a resistance high enough for your application.

  • \$\begingroup\$ this circuit also has extra parts but unfortunately, they were not included with the schematic making system \$\endgroup\$ Feb 26, 2019 at 16:37
  • \$\begingroup\$ There is a Custom Component symbol. Double-click to edit any component. You can add pins and name them. \$\endgroup\$
    – Transistor
    Feb 26, 2019 at 16:39

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