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?

 

[cnh1995]

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

There is.
Only R is unknown.

Write the equation and simplify.

 

[james123]

So for R it's 6/0.6=10Ω
& for 10k-R it's 3/0.6=5Ω

??

 

[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

I think I misread your post.

The "sum" of the blue currents is equal to the red current. Write the equation and solve for R.

 

[james123]

but how can I find the current for 10k-R resistor?

 

[cnh1995]

but how can I find the current for 10k-R resistor?

Isn't it 3/(10k-R)?

 

[james123]

you don't mean 3/10=0.3A ??

 

[berkeman]

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

I could have missed it, but I didn't see you writing the KCL equation for the node at the wiper of the pot. Did you ever take that approach, or are you saying that KCL has not been covered in your class yet?

If not, this problem is simple enough that you can just use a voltage divider equation. I've seen that posted in the thread so far, but you seem mired in the details. What is the overall equation you are working with at this point?

 

[james123]

Hi there, I believe this is a voltage divider equation question. However this thread has confused me I must admit.

My problem is I know the answer will end in a quadratic equation with x=0.5 as the final answer, but I can't figure out how to get there! People have used different methods(none of which are shown in my course materials) and arrived at the same answer.

I just need a clear outlined method on how to arrive at the answer..

 

[Numbskull]

I did this some time ago, and I feel your pain with the learning materials, they are woefully inadequate!

I called the parallel resistors $R_o$ and followed with $R_o = \frac{10 \left(1 - x\right) \text{ x } 5}{10 \left(1 - x\right) + 5}$

Does that help?

 

[james123]

Hi there, yeh they're a struggle to follow, given you're expected to essentially teach yourself!

I have seen that method before but never understood it.

Does this make sense to you? Using this diagram:

As the question says, 'the voltage across points ‘XX' is 3 V.' Meaning 'R' must be 5kohm as it has the same voltage across it as its parallel 5kohm resistor does, right?

And as you said, 'total resistance in the parallel section is 1/2 the resistance of the section in series ...'

So because the voltage drops from 6V to 3V (50%) between '10kohm-R' and the 'R' & '5kohm' parallel resistors.

Can we conclude that the slider position on the 'pot' will be 50% of the way down (0.5) ?

 

[Numbskull]

Hi there, yeh they're a struggle to follow, given you're expected to essentially teach yourself!

Can we conclude that the slider position on the 'pot' will be 50% of the way down (0.5) ?

Yes, the correct final position of the slider is 0.5. I then fiddled with the equation I posted to get $-50x^2 + 125x - 50 = 0$ and popped that into the quadratic formula for which there is only one credible answer.

 

[cnh1995]

Can we conclude that the slider position on the 'pot' will be 50% of the way down (0.5) ?

Although the final answer is correct, your analysis is not.

Meaning 'R' must be 5kohm as it has the same voltage across it as its parallel 5kohm resistor does, right?

No. The reason R=5k is different. Parallel components always have the same voltage across them, whether or not they are equal.

And as you said, 'total resistance in the parallel section is 1/2 the resistance of the section in series ...'

What does this mean?

 

[james123]

Okay I have this, using your equation but ended with slightly different numbers:

Ro=10(1-x) x 5/10(1-x) + 5
=50(1-x)/10+5(1-x)

So, we know that the parallels resistors are half the value of the series resistor:

50x=2x50(1-x)/10+5(1-x)
x=2(1-x)/10+5(1-x)
x(10+5(1-x)) = 2(1-x)
10x+5x-5x^2 = 2-2x
0 = 5x^2-13x+2

By the way, I know this doesn't pull up 0.5 as the answer but could some direction as to where I've gone wrong please??

 

[Diegor]

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

Hi james123. Maybe you should go back to really basic cicuit examples like two different resistors in series or two resistors in parallel and try to understand how KVL, KCl and Ohms law work there to get a better understanding before you jump to a more complex situation.

 

[MaRkR78]

I have the exact same question to answer and have been using KCL/KVL/Ohms law to attempt it. I'd love to pick this up for myself from post #23 with cnh1995 if at all possible as he was providing insight into the use of those laws to determine the solution.

As I see it the KCL equation would be I1=I2+I3 where I1 is the current entering the junction after resistor R, I2 is the current leaving the junction towards resistor 10k-R and I3 is the current leaving the junction towards resistor 5k (load).

I already have current I3 as 3/5000 = 0.0006A or 6x10^-4 or 0.6m/A

Therefore to determine I1 we also need I2. In a slightly earlier post #35 cnh1995 suggested that I2 would be calculated with ohms law I=V/R which would give I2=3/10000-R this is where it becomes a little unclear to me. The reason is that although I can calculate I2=3/10000-R it still leaves me with 0.0003-R or 3x10^-4-R or 0.3m/A-R. We still don't have a value for R...

Even if I state I1=I2+I3 and use I1=0.0003-R+0.0006 that gives me 0.0009-R for I1.

You state to solve for R and this is where I'm hitting a wall. Do I just need to apply some more algebra to make R the subject?
I have seen others use product over sum for the parallel resistors but I'm keen to carry on with Kirchhoff/Ohms if it is doable.

Regards Mark.

 

[cnh1995]

Hi Mark,
Welcome to PF!

I'm keen to carry on with Kirchhoff/Ohms if it is doable.

It certainly is.
Refer the diagram in #40 (Ignore the reasoning).
You know the current through the 5k resistance.
How would you write KCL equation for the junction of the three resistors?
It would be an equation with R as the only unknown and you'll be able to solve it with some algebraic manipulations.

 

[MaRkR78]

Many thanks for your reply and help.
I think I get where you're coming from. I'm going away for the Easter break so won't get to look at this for three days which is probably a good thing as my head is frazzled. I've had little rest from study for a while now.
Before I go perhaps you can just confirm something for me?
The diagram you refer to in #40. I had already redrawn the circuit expanding the 'Pot' out to give three resistors and of course the necessary junction from which to work out KCL. However I have resistor 10k-R in parallel with the 5k resistor where as diagram #40 has it the other way round. Will this matter?

Again appreciate the input and hoping a fresh mind will get me through this in a couple of days time :)
Mark.

 

[cnh1995]

Will this matter?

No. Both the diagrams will give the same answer, but in different forms.
If using one diagram you get the slider position as "4k from top", the other diagram will give "6k from the bottom"(both of them give the same position).

 

[MaRkR78]

Ah.
I'll go back through my scribbles later as I'm convinced I had an answer very similar to those results but made the decision they were wrong. Reason being was that a friend has used a different method and had the slider at halfway. I've seen other stuff online giving the same result so assumed I must have been wrong with so many examples showing the slider at 5k or halfway. Worrying to think so many people have this incorrect??
Thanks again and I'll post back after Easter once I have it nailed down. :)

 

[Jason-Li]

Hi,

Sorry to have to ask, you replied to James saying that 5k was correct however to Mark you have said that the 4&6 options are correct. I have completed the question as per how Mark completed it (paralleled resistor = 10000 - R1). I have attached my workings, would you be able to explain why it is correct / incorrect? Reading this post seems to have re-confused me!

Thanks and much appreciated,
Jason