What is the non-linearity error, as a percentage of full range, produced when a 1K Ω potentiometer has a load of 10KΩ & is at one-third of its maximum displacement?

2.28%

Explanation:

Being at one third of its maximum range a potentiometer should output V0/3.

However if this 1kΩ potentiometer has a 10kΩ load:

(1) I1 = I2 + I3

(2) V0 = I1 * 2/3*Rp + I3 * 1/*Rp

(3) Vp = I2 * Rl

(4) Vp = I3 * 1/3 * Rp

Where

I1: current entering the potentiometer

I2: current going to the load

I3: current going to the other leg of the potentiometer

V0: supply voltage

Vp: output voltage of the potentiometer

Rp: total resistance of the potentiometer

Rl: load resistance

First we determine the intensity of I3 in function of supply power

I3 = 3 * Vp / Rp = 3 * Vp / 1000 = 0.003*Vp

Then the load current

I2 = Vp / Rl = Vp / 10000 = 0.0001*Vp

With these we determine I1

I1 = 0.003*Vp + 0.0001*Vp = 0.0031*Vp

Then

V0 = 0.0031 * Vp * 2/3*Rp + 0.003*Vp * 1/3*Rp

V0 = 0.00207 * Vp * Rp + 0.001 * Vp * Rp

V0 = 0.00307 * Vp * 1000

V0 = 3.07 * Vp

Vp = V0 / 3.07

Vp = 0.3257 * V0

Now the percentage error is:

(100 * (0.3333 - 0.3257)) / 0.3333 = 2.28 %