# Potentiometer and Errors

#### Potentiometer and Resistor

The resolution of a potentiometer is 0.1 mm, its length is 50 mm and its diameter is 5 mm. Please find the total resistance if the wire resistivity is $\rho=4 \mu ohms*cm$. What is its power if the power supply has 5 V? Please find the percentage error if the resistance in the external circuit is 5 times the device resistance.

[A potentiometer is a variable resistor, that is made up from a wire of high resistanceÂ wrapped on an insulator support. An cursor can slide over this wire making electical contact with it. Thus a potentiometer has three terminals, two for the fixed resistance and one for the cursor, where a variable resistance is available. Usually the power of a potentiometer (between 1-10 W) is bigger than the power of a fixed resistor(betwee 0.25-1 W)]

#### Answer

The distance between 2 consecutive turns of the wire gives the linear resolution. Thus the total number of turns of the winding is

$N = 50/0.1 =500 turns$

We can now find the length of the wire used in this device winding

$L =N*\pi*d =500*\pi*0.005 =7.854 m =785.4 cm$

The section of the wire is (its diameter is $D =0.0001 m =0.1 mm$)

$S = \pi*D^2/4 = \pi*0.0001^2/4 =7.854*10^{-9} m^2 =7.854*10^{-5} cm^2$

Therefore the total resistance of the wire that makes up this device is

$R = \rho*L/S =4*10^{-6} *785.4/7.854*10^{-5} =40 ohm$

The power consumption is

$P =U^2/R = 5*5/40 =0.625 W$

#### Error

The percentage error is defined for the position of the cursor at the end of the potentiometer (R =40 ohm) when the error is maximum. The explanation is that at lower values of the device resistance, same load resistor value counts less in the error (resistances are in parallel).

$(R(pot) || RL) =error*R(pot)$

$error = (R(pot) || RL)/R(pot) = (RL*R(pot)) /(RL+R(pot)) *1/R(pot)=$

$=RL/(RL+R(pot)) = RL/(6RL)$

$error =0.1667 =16.67 %$

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