How Do You Calculate Slider Position on a Potentiometer?

[james123]

Homework Statement

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.

Homework Equations

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

 

[cnh1995]

Homework Statement

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.

Homework Equations

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

There is no image in your post.

 

[james123]

Apologies, this is the image

 

[cnh1995]

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

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

 

[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 that's all.

 

[berkeman]

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.

 

[cnh1995]

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 that's all.

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

 

[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'?

 

[cnh1995]

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

This is not correct, both algebraically and dimensionally.

Try the KCL method.

apply KCL (with Ohm's law) at the junction of the three resistors.

 

[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

 

[cnh1995]

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.'

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?

I have the top one 'R', and the bottom one '10kΩ-R'

What is the voltage across R? What is the voltage across 10k-R? What is the current through the 5k resistor?

 

[james123]

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

 

[cnh1995]

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

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?

 

[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?

 

[cnh1995]

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

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?

 

[james123]

But we don't know their resitances?

and is it this equation?

 

[cnh1995]

But we don't know their resitances?

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.

and is it this 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?

 

[james123]

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

But we don't know the current??

 

[cnh1995]

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

Or I=V/R.

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

The voltage across R is 6V

So what's the current through R?

The voltage across 10k-R is 3V?

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

The voltage across 5k is 3V?

What is the current through the 5k?

 

[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??

 

[cnh1995]

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??

Correct.

So what is the KCL equation from this information?

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

 

[james123]

I honestly don't know, this is what I don't get, how can you find anything without at least two values?

Nowhere in my learning materials does it give an equation for KCL

 

[cnh1995]

Current for R is I=6/R

This current enters the junction.

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

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

These two leave the junction.

And this is KCL...

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.'

Now write the KCL equation and solve for R.

 

[james123]

Are you saying the current for the other two equations is 0.6A?

So, for R it's R=6/0.6=10
for 10k-R it's R=3/0.6=5A

??

 

[cnh1995]

Are you saying the current for the other two equations is 0.6A?

So, for R it's R=6/0.6=10
for 10k-R it's R=3/0.6=5A

??

No.

You have to find R.

You have three currents. You know only one of them, but you also know that sum of two of them is equal to the third (KCL).

Think on #23.

 

[james123]

so you're saying the sum of the two unknown current values is equal to 0.6A?

 

[cnh1995]

so you're saying the sum of the two unknown current values is equal to 0.6A?

Not exactly, but you are close.

Read KCL again. Which two currents would you add? Which current splits off at the junction?

 

[james123]

So the current for these two:

10k-R is I=3/10k-R
5k is I=3/5= 0.6A

Is equal to:
R is I=6/R

 

[cnh1995]

So the current for these two:

10k-R is I=3/10k-R
5k is I=3/5= 0.6A

Is equal to:
R is I=6/R

Yes. Solve for R.

 

[james123]

How? there's not enough info there to do it?