Answer:
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 %
