Position Potentiometer for 3V Across XX

In summary, a 10 kΩ potentiometer with a 5 kΩ load is set up in a circuit with resistors in series and parallel. By using the voltage divider formula and solving for the slider position, it is determined that the slider should be set at the halfway point, or 0.5, to achieve a voltage of 3 V across points XX. Checking the solution with a final voltage divider confirms the accuracy of the result.
  • #1
charger9198
60
0
A circuit has a 10 kΩ potentiometer with a 5 kΩ load. Determine the position of the slider on the ‘pot’ when the voltage across points ‘XX' is 3 V.

Think I'm there but not sure of my working out is excessive or I've taken the long route


R1 in series with 2 resistors in parallel (R2 and 5k ohm)

R1=10x, R2=10(1-x)

R2 + 5k ohm = RC

1/RC = 1/(10(1-x)) + 1/5.
1/RC = (5+10(1-x))/(50(1-x))
RC = 50(1-x) / (5+10(1-x))
= 10(1-x) / (1+2(1-x)

Voltage drop will be 9-3=6v
I = V/R
I = 6/R1 = 3/RC
R1 = 2RC

10x= 2x10(1-x)/ (1+2(1-x))
x = 2(1-x)/(1+2(1-x))
x(1+2(1-x))=2(1-x)
x+2x-2x^2 =2- 2x
2x^2-5x+2 = 0

Quadratic gives
x= 2 or x= .5

x=0.5 (halfway point)
 
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  • #2
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  • #3
charger9198 said:
A circuit has a 10 kΩ potentiometer with a 5 kΩ load. Determine the position of the slider on the ‘pot’ when the voltage across points ‘XX' is 3 V.

Think I'm there but not sure of my working out is excessive or I've taken the long route


R1 in series with 2 resistors in parallel (R2 and 5k ohm)

R1=10x, R2=10(1-x)

R2 + 5k ohm = RC

1/RC = 1/(10(1-x)) + 1/5.
1/RC = (5+10(1-x))/(50(1-x))
RC = 50(1-x) / (5+10(1-x))
= 10(1-x) / (1+2(1-x)

Voltage drop will be 9-3=6v
I = V/R
I = 6/R1 = 3/RC
R1 = 2RC

10x= 2x10(1-x)/ (1+2(1-x))
x = 2(1-x)/(1+2(1-x))
x(1+2(1-x))=2(1-x)
x+2x-2x^2 =2- 2x
2x^2-5x+2 = 0

Quadratic gives
x= 2 or x= .5

x=0.5 (halfway point)

charger9198 said:

Your solution looks good. You can do a final check of your work by checking what voltage divider you get with your final configuration (5k on top, and 5//5k on the bottom)...
 
  • #4
I split the potentiometer into 2 parts , 10-R at the top and R at the bottom giving R in parallel with the 5k
Resistance of parallel combination is then 5R/(5+R)
The voltage across this combination = 3V and the voltage across the toppart (10-R) =
6V
This means that (10-R) = 2 x 5R/(5+R)
This gives R^2 + 5R - 50 =0 with R=5 being the +ve solution...half way
I think this is more or less the procedure you used.
 
  • #5
Thanks for both your help guys; technician: I like your method, thanks for the input:)
 

What is a Position Potentiometer?

A Position Potentiometer is a type of electrical resistor that changes its resistance based on the position of a sliding contact. It is commonly used to measure the position or movement of an object.

How does a Position Potentiometer work?

A Position Potentiometer consists of a resistive element, a sliding contact (also known as a wiper), and two terminals. The resistive element is connected to a voltage source, and the wiper moves along the resistive element as the object being measured moves. This changes the resistance of the potentiometer, which can then be measured to determine the position of the object.

What is the voltage rating for a Position Potentiometer?

The voltage rating for a Position Potentiometer can vary depending on the specific model and manufacturer. However, for a 3V application, a potentiometer with a voltage rating of at least 3V would be suitable.

What does XX refer to in "3V Across XX"?

The XX in "3V Across XX" refers to the specific unit of measurement being used for the position or movement being measured. This could be inches, centimeters, degrees, etc.

What are some common uses for a Position Potentiometer?

Position Potentiometers are commonly used in various industrial and scientific applications for position sensing and control. They can also be found in consumer electronics such as volume knobs on audio equipment and joysticks on gaming controllers. Other uses include robotic and automation systems, motion control systems, and measurement and testing equipment.

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