What is the percentage non-linearity error with a potentiometer?

If I understand your question, you have a 550 Ohm pot with cursor at center, and the lower half of it loaded with a 10 Ohm resistor: Let's call the top half of the pot R1 = half of 550 or 275. Calculate the parallel of 275 and 10, call this R2. Now calculate the voltage Vout across the 10 Ohms using the voltage divider equation: 10/(R1+R2)=Rout/R2.

With no 10 ohms load the Vout would be 5Volts. With the load it would be as you calculated it. The error is the ratio of the two voltages Vout/5 times 100, expressed as a percentage.

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