# Potentiometer electrical engineering question

1. Aug 15, 2017

### james123

1. The problem statement, all variables and given/known data
The circuit of FIGURE 2 shows 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.

2. Relevant equations

3. The attempt at a solution

I really don't know the right way to start this.

If I'm being honest, I've seen the answer posted elsewhere on this website but I don't just want to copy it, I want to learn it but the way it's explained elsewhere isn't clear to me.

If anyone could provide a starting point/simpler explanation on how to approach this, I'd massively appreciate it

2. Aug 15, 2017

### cnh1995

There is no image in your post.

3. Aug 15, 2017

### james123

Apologies, this is the image

4. Aug 15, 2017

### cnh1995

You need to show your attempt.
Have you studied Kirchhoff's laws?

5. Aug 15, 2017

### james123

I have, but like I said, I've seen the answer elsewhere I just don't want to copy it though
I understand that the 10kΩ resistor needs to be split into 2 resistors and I understand that it will end with a quadratic.
Just not sure how to begin thats all.

6. Aug 15, 2017

### Staff: Mentor

Draw the circuit as 3 resistors, the right one is 5kOhm, the left 2 are R and 10kOhms - R. Then write the voltage divider equation and solve. Please show your work. Thanks.

7. Aug 15, 2017

### cnh1995

Draw a diagram as berkeman described and apply KCL (with Ohm's law) at the junction of the three resistors.

8. Aug 15, 2017

### james123

Ok, so I re-drew and instead of the 10kΩ resistor I have the top one 'R', and the bottom one '10kΩ-R'

The used product over sum rule to deal with the bottom resistor and the 5kΩ resistor and got:

10kΩ-R x 5kΩ/10kΩ-R + 5kΩ
= 50kΩ-R/15kΩ-R
= 3.3kΩ-R

Now I'm confused again, do I put put this value into:

with R1 being 'R' and R2 being '3.3kΩ'?

9. Aug 15, 2017

### cnh1995

This is not correct, both algebraically and dimensionally.

Try the KCL method.

10. Aug 15, 2017

### james123

I have KCL as this:

'at any electrical junction the sum of the currents flowing towards the junction equals the sum of the currents flowing away from the junction.'

I still don't get how to apply this with the info I'm given? Maybe I'm punching above my weight with this course

11. Aug 15, 2017

### cnh1995

You have it right. Mark three currents i.e. current through R, current through 10k-R and current through 5k resistance. Which one(s) is/are entering the junction? Which one(s) is/are leaving the junction? What is the KCL equation for this junction?
What is the voltage across R? What is the voltage across 10k-R? What is the current through the 5k resistor?

12. Aug 15, 2017

### james123

The voltage across R is 6V
The voltage across 10k-R is 3V?
The voltage across 5k is 3V?

13. Aug 15, 2017

### cnh1995

Right.

14. Aug 15, 2017

### james123

Current running through R is going to the junction,
current going toward 10k-R is going away from the junction
current going toward 5k is going away from the junction

how does this relate to the formula though?

15. Aug 15, 2017

### cnh1995

Correct.

You know the voltages across all these resistances. Can you write the KCL equation at their junction using the resistances and their respective voltages?

16. Aug 15, 2017

### james123

But we don't know their resitances?

and is it this equation?

17. Aug 15, 2017

### cnh1995

Yes, but you know all the voltages and one of the three currents. You want to find R, so that should be the unknown in the equation.
No.

You have already stated KCL in an earlier post.
How will you write the current through R using Ohm's law if voltage across R is 6V?

18. Aug 15, 2017

### james123

If R is the unknown then it'll be R=V/I

But we don't know the current??

19. Aug 15, 2017

### cnh1995

Or I=V/R.

You don't know the currents, but you can form a KCL equation with only one unknown.

So what's the current through R?
So what's the current through 10k-R?
What is the current through the 5k?

20. Aug 15, 2017

### james123

So,

Current for R is I=6/R

Current for 10k-R is I=3/10k-R

Current for 5k is I=3/5= 0.6A??