# Homework Help: Potentetiometer - position of slider

1. Jul 29, 2013

### damon669

1. The problem statement, all variables and given/known data
https://www.physicsforums.com/attachment.php?attachmentid=60570&stc=1&d=1375128608
Thew circuit shows a 10KΩ potentiometer with a 5KΩ load. determine the position of the slider on the 'pot' when tthe voltage across points 'XX' is 3V

2. Relevant equations

3. The attempt at a solution
Vi=IR
I=V/R = 12/10x10^3 = 0.0012A = 1.2mA
Vo=IR
I=Vo/Io = 3/5x10^3 = 0.0006A = 0.6mA
Vi/Ri+Ro = 9/10x10^3 + 5x10^3 = 0.6mA
VD Across Ri =Vi =RiVi/Ri+Ro = RiVi/R

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2. Jul 29, 2013

### rude man

?
Let x = fraction of distance from ground to wiper.
Call the potentiometer resistance (10K) = R1 and the external resistor (5K) = R2.
Call 9V = Vi and wiper voltage = Vo = 3V.
Then redraw your circuit so that Vo/Vi consists of a voltage divider between two equivalent resistors.
What is the equivalent resistor going from Vi to Vo?
What is the equivalent resistor going from Vo to ground?
Solve for x.

3. Aug 3, 2013

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4. Aug 3, 2013

5. Aug 4, 2013

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6. Aug 4, 2013

### Staff: Mentor

That is right.
I would align the two resistors to be in the same column, but that is just a graphical thing and does not influence the electric properties.

7. Aug 4, 2013

### damon669

so now to answer the question

I need to find the new parallel resistence which is (R_1 R_2)/(R_1+R_2 )
=33x10^3 ohms

Then I think I have to put 3 V in the equation?

8. Aug 4, 2013

### Staff: Mentor

The parallel resistance should depend on the position of the slider, and don't forget the third resistor in your analysis.
What do you mean with "put 3 V in the equation"? Just do it.

9. Aug 4, 2013

### rude man

Your diagram is correct. But the parallel resistance is obviously a function of p.

yes, put 3V at the output and solve the whole thing for p.

10. Aug 4, 2013

### damon669

I think somebody needs to get the glove puppets out :)

11. Aug 6, 2013

### damon669

Vl=RoVi/R1+R2

3=Ro•9/(10•10^3)+(5•10^3)

= 5k ohms

12. Aug 6, 2013

### Staff: Mentor

I don't understand where that formula comes from. And where did you define all those parameters?

13. Aug 6, 2013

### damon669

Vl = 3 volts across the load
Ro = what I'm trying to find
Vi = voltage on the circuit
R1 = 10k ohms
R2 = 5k ohms

14. Aug 6, 2013

### damon669

That formula came from my notes

15. Aug 6, 2013

### damon669

voltage division.

is it correct then?

16. Aug 6, 2013

### Staff: Mentor

There is no 10kOhm-resistor in your network.
It should come together with a circuit where you can apply it. You cannot just use a formula for some circuit here.

17. Aug 6, 2013

### damon669

10k ohm is the max position of the slider @ 9V

18. Aug 6, 2013

### Staff: Mentor

Sure, but 10kOhm is not a single resistor in your network.
The maximal position of your slider is not the solution to this problem.

19. Aug 6, 2013

### damon669

If the slider was at 5k ohm would that not give 3 volts across the load?

20. Aug 6, 2013

### Staff: Mentor

It would, but I think it is an accident that the formula gives the right result.
Try to do the same calculation for 0V, 9V and 4V, and see what you get (and show what you plugged in where please). Are the results reasonable?

21. Aug 7, 2013

### damon669

Can this be done without using a Quadratic equation?

22. Aug 7, 2013

### rude man

No, but it's a piece of cake - especially if you use Wolfram Alpha!

23. Aug 7, 2013

### damon669

so you DO have to use a Quadratic to solve then

24. Aug 7, 2013

### rude man

Yes.

25. Aug 7, 2013

### damon669

OK this is what i got then

Let p represent the position of the ‘pot’ slider where 0≤p≤1 (i.e. the proportion of the ‘pot’ resistance that is in parallel with the load resistor). For p=0 the slider is at the bottom (minimum resistance) and for p=1 the slider is at the top (maximum resistance). The equivalent circuit is shown below.

The parallel resistance R_P is given by

R_P=(10p×5)/(10p+5) kΩ

=50p/(10p+5) kΩ

=10p/(2p+1) kΩ

and the current I is given by

I=9/〖10(1-p)+R〗_P
=9/(10(1-p)+R_P ) mA

The voltage V_L across R_P is

V_L=I×R_P

=(9R_P)/(10(1-p)+R_P )

=(9×10p)/(10(1-p)(2p+1)+10p)

Now V_L=3 so

3=(9×10p)/(10(1-p)(2p+1)+10p)

3=9p/((1-p)(2p+1)+p)

(1-p)(2p+1)+p=3p

2p+1-2p^2-p+p=3p

0=2p^2+p-1

Therefore,
p=(-1±√(1-2(2)(-1) ))/(2×2)

=(-1±√9)/4

=(-1±3)/4

Since 0≤p≤1 then p=0.5. Thus the slider is at the midpoint position of the ‘pot’.

Check: R_P=2.5 kΩ which is in series with a resistance of 5 kΩ so the voltage V_L across the load is

V_L=(9×2.5)/(2.5+5)

=3 V