# Potentiometer Power Handling

I have a circuit with a voltage of about 150 volts but with just about 1.75 mA current draw. Will I be alright with a 1K, 3K or 5K 1/2 watt or a 1 watt potentiometer to adjust and dial in the voltage? This is sometimes nice for a microphone in order to get the desired sound. I would of course rather use a 5 watt potentiometer but I will need to order one.

What kind of power rating do I need for a potentiometer if I run 150V across it?

• Can you draw a circuit for that? Something sounds fishy.. – Trevor_G Mar 24 '17 at 18:23
• I don't see why you can't calculate the power dissipation in the potentiometer the same way as any other resistor. If the current in the potentiometer really is 1.75mA, then I would think a half watt will be fine for 5k. Using I^2 * R, the dissipation would only be 15mW. There could possibly be a shock hazard associated with turning the knob on the poteniometer. But I guess that depends on the details. – mkeith Mar 24 '17 at 18:38
• @mkeith, calculating it that way yes, but if that 5K pot is across 150V it's 4.5W... assuming whatever is supplying it can deliver the 30mA. Like I said.... something isn't right here. – Trevor_G Mar 24 '17 at 18:43
• That is why I said "if the current... really is 1.75 mA". I think the idea is that there is a main divider dissipating most of the power, and the pot is just to trim the voltage. But, with no schematic, I am just guessing. – mkeith Mar 24 '17 at 18:57
• One other piece of information for the OP. There is actually a built-in tool for drawing schematics in questions. You can edit your question by clicking on "edit". After you do that, there is an icon that looks like schematic symbols. You can click on the icon to draw a schematic. – mkeith Mar 26 '17 at 8:15

If you put V volts across a potentiometer of R ohms, and draw no current from the wiper, then you will be dissipating $\frac{V^2}{R}$ watts. So with 150V and a 5k pot, you need a 5 watt pot.