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 %
The non-linearity error will be "0.71 %".
An error seems to be the greatest discrepancy between the calibration curves as well as the "optimal simple or straight," which connects the output signal towards the applied pressure.
According to the question,
Resistance, [tex]R_p[/tex] = 1 kΩ
[tex]R_L[/tex] = 10KΩ
Displacement, x = [tex]\frac{1}{3}[/tex] or,
= 0.33
We know the relation,
Non-linearity error = [tex]\frac{\frac{R_p}{R_L} (x^2-x^3)}{\frac{R_p}{R_L} x(1-x)+1}[/tex] × 100
By substituting the values, we get
= [tex]\frac{\frac{1}{10} ((0.33)^2-(0.33)^3)}{\frac{1}{10} 0.33(1-0.33)+1}[/tex] × 100
= 0.71 %
Thus the above answer is appropriate.
Find out more information about non-linearity error here:
https://brainly.com/question/14888641
