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This might sound a bit weird question but I’m having some confusion regarding some fundamentals on potentiometers used as rheostats.

Here I will try to explain by using the simple circuit below:

enter image description here

I simulated it so we can see the currents passing through each wire. R1, R2 values are 470 ohm and 10 ohm. Similarly R3 is 470 ohm and 10k Pot is set to 10 ohm.

So far so good, and we can see that the same amount of current passes through each component which is around 20mA.

R2 and Pot values are the same. Common sense says we can replace them safely since they are identical.

Let’s also assume their rated wattage is also the same i.e. 0.5 Watt. Now Iet’s examine if R2 can handle this circuit. The way to do that is to measure its power which is P = I^2*R. So for R2, P = 20mA*20ma*10 = 4mW roughly. Now the question to be asked here:

Does R2 have enough wattage? Since R2 is rated 0.5W which is 500mW, and since R2 in the circuit dissipates only 4mW we can say that the R2 is safe. Until recently I would use the same logic for the 10k Pot in the circuit which is set to 10 Ohm. So I would say it is safe.

But my previous question (A question on using potentiometer as a variable resistor) revealed the fact that when it is a potentiometer we examine the situation by the current passing through the pot not by the wattage.

Some explained it is done in the following way:

Wattage of the pot is given for its max resistance so in this case the max current of poti at 10k is I = sqrt(P/R). So I = sqrt(0.5W/10k) which is around 7mA. And this 7mA should never be exceeded whatever the pot value is set to. And since the current in my example circuit is 20mA, the pot is not safe since it exceeds 7mA limit.

So R2 is safe but not the pot?

My questions are:

Is the above logic is true? Why?

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  • \$\begingroup\$ So do the calculations of how much energy is dissipated by the pot when you have 7mA running through it in the 10k setting and you have your answer. \$\endgroup\$ Commented May 13, 2016 at 10:11
  • \$\begingroup\$ I dont get it. The pot is set to 10 ohm not 10k ohm \$\endgroup\$ Commented May 13, 2016 at 10:14
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    \$\begingroup\$ Imagine you have a 'pot' made up from 10 ohm resistors (all 1000 of them). The total wattage rating is the sum (I^2R1 + I^2R2 + ... I^2R1000) of the 1000 resistors is equal to the wattage rating of the 10k pot. That means that any individual 10 ohm resistor or 10ohm section of the pot only has a wattage rating of 1/1000 of the 10k pot. Andy gives the correct answer. \$\endgroup\$ Commented May 13, 2016 at 10:48

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If the 10k pot is rated at 500 mW then each 10 ohm section can only be rated at 0.5 mW. If you are only using a 10 ohm section then the current that can be put through that section is \$\sqrt{0.5 \space \text mW\text /10}\space \$ or 7.071 mA.

What if you used a 20 ohm section - the allowable power is 1mW and the allowable current is 7.07 mA. A 100 ohm section can dissipate 5 mW and the allowable current is therefore still 7.071 mA.

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  • \$\begingroup\$ So the wattage is directly proportional to resistor value of the pot? \$\endgroup\$ Commented May 13, 2016 at 10:21
  • \$\begingroup\$ Wattage is unrelated to the end-to-end resistance of the put but its physical dimensions. If you are using only 1% of the pot then the allowable power can only be 1% for that small section. \$\endgroup\$ Commented May 13, 2016 at 10:30

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