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Potentiometer position

  • #1
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)
 
Last edited:

Answers and Replies

  • #2
  • #3
berkeman
Mentor
56,455
6,369
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)
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
1,506
17
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:)
 

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