EDIT: I realized I didn't draw the schematic properly. Previously, the 100 Ohm resistor wasn't there.

I thought I'd done the math correctly for this, but clearly something conceptually isn't straight for me. I need a 5K potentiometer, but the sources I've looked through didn't have it in the package I'd like. I did see a 10K pot in the package though. So I thought by shorting the wiper terminal with one of the ends of the trim pot I'd be able to half the maximum possible resistance offerable by the pot.

$$\frac{1}{R_{total}} = \frac{1}{R_{wiper}} + \frac{1}{10000}$$ $$\frac{1}{R_{total}} = \frac{R_{wiper} + 10000}{10000 \cdot R_{wiper}}$$ $$R_{total}=\frac{10000 \cdot R_{wiper}}{R_{wiper} + 10000}$$

Since the highest resistance Rwiper can reach is 10K, the resulting highest value would be

$$R_{total}=\frac{10K \cdot 10K}{10K + 10K} = \frac{10K \cdot 10K}{2 \cdot 10K} = 5K$$

I could also make a graph by plotting the function of Rtotal.

By this point I was convinced. I tried to measure the resistance on a spare 10K pot lying around, but I saw that maximum value remained 10K. Why is this?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Adding R2 just increases the total resistance by 100 Ohms, regardless of the position of the pot. \$\endgroup\$ Commented Jul 30, 2020 at 16:27
  • \$\begingroup\$ @PeterBennett Not 'just', it prevents shorting V1. That's quite important. \$\endgroup\$
    – Mast
    Commented Jul 31, 2020 at 12:40
  • \$\begingroup\$ If you paralleled a 10K resistor with your pot in the diagram, the math would work as you have written, and you've have a (nonlinear) 5K pot. \$\endgroup\$
    – Dave X
    Commented Jul 31, 2020 at 18:52

3 Answers 3


The formulas do not match the picture. In the picture, the resistance between ground and 5V can be set between 0R where the wiper bypasses the resistance and 10k where the wiper selects full resistance.

To whatever position you set the wiper, it will short out the unused portion of the resistance when it is connected like in the picture, so it will have no effect.


If you connect to the top and wiper, you can use the potentiometer as a simple variable resistor.

If you connect to the top and wiper, AND short wiper to bottom, you are still using it as a variable resistor, but with one important difference : if the wiper goes open circuit, the resistance value is 10K rather than infinity.

That can:

  • reduce crackling noises in an audio circuit when the track is old or dirty
  • prevent loss of signal or things blowing up due to wiper failure
  • prevent gain (and volume) attempting to reach infinity in circuits where gain is proportional to resistance

If you have to design circuits to tolerate component failure, this is worth remembering.

  • 2
    \$\begingroup\$ It may be helpful to note that tying the wiper to the end of the pot is most important in circuits where gain is essentially proportional to resistance, and thus an open resistor may set gain not to zero, but to infinity. \$\endgroup\$
    – supercat
    Commented Jul 30, 2020 at 19:38
  • \$\begingroup\$ I also have this idea in my head that if you leave the terminal open, in certain high impedance ckts it is easier for it to pick up noise - but I am not so sure this is entirely accurate. \$\endgroup\$ Commented Jul 31, 2020 at 7:44

If the wiper is at the bottom of the resistance element in your schematic, the whole 10K resistance element is in circuit.

At any point in the wiper's travel, you will only have one part of the resistance element in the circuit. There is no way to get two parts of the resistance element in parallel, as your calculation would require.

  • 2
    \$\begingroup\$ I'm not saying that this has ANY sense, but certainly there is a way to get two parts of potentiometer's resistive element in parallel :D -> pasteboard.co/Jk7amzS.png \$\endgroup\$ Commented Jul 30, 2020 at 23:38

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