I'm using 0.2 W, 500 Ω potentiometer with 300mA safely (there is no heat or abnormal behavior). But according to calculation, max current of this potentiometer is about 20mA. Is my calculation wrong?

$$P = VI = I^2 R$$ $$0.2W = I^2 500 \Omega$$ $$I = \sqrt{\frac{0.2W}{500\Omega}} = 0.02A$$


I'm confused about power rating of potentiometer. It's max value is 500 Ω and power rating is 0.2 W, so max current is 20 mA. But when I use only part of the potentiometer, 20 mA limit can be applied? I'm using 3 Ω part with 300 mA.


Entire circuit is like this:


simulate this circuit – Schematic created using CircuitLab

There are 40 set of R - LED - LED totally. The current of each set is about 7 mA. (I measured voltage across the 100 Ω resistors and it's 0.7 V) And I measured current between P1 and 40 (R - LED - LED) s and it was about 280 mA. I used pot for brightness control.


enter image description here

I used only 2 part of the pot. Is this not recommended? Why? (I'm just hobbyist and forgive me if this is too basic question)

This is test video: https://www.youtube.com/watch?v=Su1ZFqXYfIE The voltage seen on the multimeter is the voltage across a 100 Ω resistor, as shown on the circuit diagram.

  • \$\begingroup\$ So to get 300mA you must be applying 150 V across the pot. (V= IR = 300 x 500 /1000 = 150 V ) \$\endgroup\$ Commented Jul 27, 2016 at 9:37
  • 1
    \$\begingroup\$ Your calculation is correct; P=I^2R, so current at max power dissipation is 20mA. As Jim says though, you're apply 150 V across the potentiometer? How are you measuring the current? What voltage do you think you're putting across it? Is this AC or DC? \$\endgroup\$
    – Puffafish
    Commented Jul 27, 2016 at 9:46
  • \$\begingroup\$ I'm using part of potentiometer(about 3ohm). So voltage across potentiometer is about 0.3v(0.3^2 * 3). I'm confusing about the meaning of power rating. Power source is DC and current is measured with multimeter(serial connection) \$\endgroup\$
    – firia2000
    Commented Jul 27, 2016 at 9:54
  • 1
    \$\begingroup\$ "I'm using 0.2W, 500ohm potentiometer with 300mA safely ..." - No, you're not. \$\endgroup\$
    – marcelm
    Commented Jul 27, 2016 at 10:03
  • 1
    \$\begingroup\$ If you can edit your question and add a little schematic including how you are measuring the current then it will save everyone a lot of time \$\endgroup\$
    – user103993
    Commented Jul 27, 2016 at 10:56

2 Answers 2


I'm using 0.2W, 500ohm potentiometer ... But according to calculation, max current of this potentiometer is about 20mA.

$$ I_{MAX} = \sqrt {\frac {P}{R}} = \sqrt {\frac {0.2}{500}} = 20~mA $$

Your calculation is correct.

You need to remember that this is the maximum current that can run through any portion of the potentiometer track. A common mistake is to think that if the pot was turned down to 20% that we could run five times the current through it and remain within the power limit.

enter image description here

Figure 1. Single-turn potentiometer with metal casing removed to expose wiper contacts and resistive track. Source: Photo by Junkyardsparkle, Wikipedia Commons.

It should be obvious from the photo that we can't expect to dissipate 100% of the power rating in a fraction of the potentiometer. That portion of the track would overheat.

The power rating is a useful number for the manufacturer as they can make a range of values sharing the common design. For the user the maximum current may be a more relevant parameter.

I'm using 3 ohm part with 300 mA.

We're all surprised the potentiometer still works. You may be fortunate that at minimum setting the metal contacts are touching the tab and giving you a few ohms resistance.

You are using the potentiometer as a "rheostat" - a variable resistor rather than varying the potential or voltage along the track. That's OK but not at the current you are putting through that potentiometer.

This is test video. The voltage of multimeter is voltage aroung 100 ohm resistor.

Presuming you mean the "voltage across" one of the 100 Ω resistors. In the video this varies from about 13 mV to 400 mV which means your current through that leg of the circuit (one pair of LEDs) is varying (using I = V/R) from 0.13 mA to 4 mA.

  • \$\begingroup\$ @firia2000: See update. \$\endgroup\$
    – Transistor
    Commented Jul 27, 2016 at 11:47
  • \$\begingroup\$ Thank you for kind explanation. The video is vcc downed version so current through pot is 4*40=160mA. But this value is still 8 times of max current. Am I just fortunate? Is this really possible? \$\endgroup\$
    – firia2000
    Commented Jul 27, 2016 at 12:21
  • \$\begingroup\$ You might get away with it for a while. \$\endgroup\$
    – Transistor
    Commented Jul 27, 2016 at 12:23

The first tab on the potentiometer that you did not connect is the ground and allows you to put the resistance all the way to zero.


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