Voltage across a potentiometer?

[bensm0]

The circuit of (attached file) 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.

I can't figure out how you would find the current first without knowing total resistance?
Any help would be greatly appreciated thanks.

 

[berkeman]

The circuit of (attached file) 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.

I can't figure out how you would find the current first without knowing total resistance?
Any help would be greatly appreciated thanks.

Write an equation for the output voltage in terms of the wiper position. Use some variable like "x" to represent the fraction of the pot resistance that is in parallel with the output resistor. Make x go from 0 to 1, to represent that fraction (as the wiper goes from the bottom to the top of the pot. "x" multiples the total resistance of the pot...

 

[bensm0]

Do i not need to find the currents? I have 0.6 mA going through the 5kΩ load resistor. I'm not sure about that equation? Please help?

 

[MATLABdude]

It may help to consider the potentiometer as two separate resistors, x*10k and (1-x)*10k. If you redraw the circuit, you now have three little resistors that may be a little easier to analyze.

You should then be able to put the bottom resistors in parallel, and form a voltage divider (so no currents required!) The method I used means that you'll have to eventually solve a quadratic equation, but only one of the roots makes sense.

 

[bensm0]

If we call the resistor in series R1 and the resistor in parallel R2 and the 5kΩ load resistor R3. Can i have R1+R2=10kΩ? And because R2 has a voltage of 3V and therefore R1 has a voltage of 6V can i say R2/R1=3/6 and R2=0.5R1? I'm still a little unsure?

 

[berkeman]

If we call the resistor in series R1 and the resistor in parallel R2 and the 5kΩ load resistor R3. Can i have R1+R2=10kΩ? And because R2 has a voltage of 3V and therefore R1 has a voltage of 6V can i say R2/R1=3/6 and R2=0.5R1? I'm still a little unsure?

You cannot ignore the parallel R3 in your calculation. You have a voltage divider that has 6V across the top resistance R1 and 3V across the bottom parallel resistance R2//R3. Write the voltage divider equation, and use the formulatio suggested by MATLABDUDE to define R1 and R2. Then solve away...

 

[bensm0]

so if i use the voltage divider rule:
V,out=V,in*R2/(R1+R2)
Could i say:

R1=10000x Ω
R2=10000(1-x) Ω
R3=5000Ω

and for the voltage divider rule R2 would be (5000 x 10000(1-x))/(5000 + 10000(1-x))??

 

[berkeman]

so if i use the voltage divider rule:
V,out=V,in*R2/(R1+R2)
Could i say:

R1=10000x Ω
R2=10000(1-x) Ω
R3=5000Ω

and for the voltage divider rule R2 would be (5000 x 10000(1-x))/(5000 + 10000(1-x))??

You are mixing your terms for R2. Keep R3 separate from R2 in your voltage divider equation (your first equation needs to have R3//R2 in it...)

 

[bensm0]

so if i use the voltage divider rule:
V,out=V,in*R2/(R1+R2)
Could i say:

R1=10000x Ω
R2=10000(1-x) Ω
R3=5000Ω

and for the voltage divider rule R2 would be (5000 x 10000(1-x))/(5000 + 10000(1-x))??

 

[berkeman]

so if i use the voltage divider rule:
V,out=V,in*R2/(R1+R2)
Could i say:

R1=10000x Ω
R2=10000(1-x) Ω
R3=5000Ω

and for the voltage divider rule R2 would be (5000 x 10000(1-x))/(5000 + 10000(1-x))??

This looks to be a duplicate post? You didn't include R3 in your first equation yet...

 

[bensm0]

Is it like this:

V,out=V,in*R3/(R1+R2)?

 

[bensm0]

Am i anywhere near?

 

[berkeman]

Is it like this:

V,out=V,in*R3/(R1+R2)?

Am i anywhere near?

No, sorry. Re-draw the circuit as suggested by MATLABDUDE. Show explicit resistors for R1 and R3, and divide up R2 into 2 separate resistors (divided at the wiper contact). Label the resistors as suggested, and write the voltage divider equation.

Please show us your new figure with 4 labeled resistors...

 

[bensm0]

Ive got it, phew.

Yes i used V,out = (V,in * (R2*R3/R2+R3))/(R1+(R2*R3/R2+R3))

After solving an quadratic equation i found x=0.5. Phew!

Thanks

 

[berkeman]

Ive got it, phew.

Yes i used V,out = (V,in * (R2*R3/R2+R3))/(R1+(R2*R3/R2+R3))

After solving an quadratic equation i found x=0.5. Phew!

Thanks

Good job!

 

[spkevin]

i know this is an old thread but i was using it as a guide and I am struggling to get down to a quadratic equation. i understand up to v,out = v,in*(R2*R3/R2+R3))/(R1 + (R2*R3/R2+R3)) but when simplifying this i get nowhere.

 

[MATLABdude]

If you're just "following along", start from the circuit diagram and derive (note that what was in posts 6 and 7 aren't correct). When I was first learning, it took me a while to get it, but always, ALWAYS draw the circuit diagram and start from there. If you post an intermediate step or two, we can help you along.

In this specific instance, R1 and R2 can be expressed in terms of something involving x (the wiper position of the 10k pot) and R3 is known.

 

[spkevin]

i have since tried this again i got x=0.54 i see earlier someone had got 0.5. not sure where i went wrong. am i right i saying i can simplify the 9(R2R3/R2+R3)/R1 +(R2R3/R2+R3) to 8(R2R3/R2+R3)/R1+1. i went on to multiply out and got to 3R1R2+15R1-37R2+15=0 before substituting R1=10x and R2=10(1-x). Sorry if this is all a bit confusing, could you tell me if this seems right or possibly where I've went wrong.Thanks

 

[tufty1980]

Hi everyone I know this is an old thread but I have the same question on my HNC I got to the formulae Vp=Vs(R2R3)/(R2+R3)/R1+(R2R3)/(R2+R3) using R1 is 10x and R2 10(1-x) could someone show me the steps to find x I wouldn't normally ask for a direct answer but I'm going out of my mind and A4 sheets of paper trying to get the answer thanks

 

[gneill]

Hi everyone I know this is an old thread but I have the same question on my HNC I got to the formulae Vp=Vs(R2R3)/(R2+R3)/(R1+(R2R3)/(R2+R3)) using R1 is 10x and R2 10(1-x) could someone show me the steps to find x I wouldn't normally ask for a direct answer but I'm going out of my mind and A4 sheets of paper trying to get the answer thanks

Note: You need to use more parentheses in order to make the mathematical order of operations clear. Otherwise, strict left to right interpretation of your equation would make it incorrect. I've added a pair (in red) to your equation above.

Note also that we cannot solve the problem for you here. That's against the rules, so you'll have to show an attempt that we can comment on and help you to sort out.

Really, once you've got the equation and the definitions of the potentiometer R's in terms of x the rest is just algebra. A bit tedious, perhaps, but just algebra. As a suggestion I would plug in some of the hard numbers at this point to leave you with an equation in x alone. Since the resistances are all given in kΩ, just drop the units for the algebra. Same goes for the voltages. So the potentiometer resistance is 10 and the load resistance is 5. The input is 9 and the output is 3.

Can you show the result after plugging in the numbers?

 

[tufty1980]

From 3=9(10(1-x))5)/(10(1-x))+5)
I get to 3=9(50(1-x))/(10(1-x))+5)/
10x+(50(1-x))/(10(1-x))+5) it's around this point that I get confused I think it's how I'm treating the (1-x) in the equation

 

[gneill]

From 3=9(10(1-x))5)/(10(1-x))+5)
I get to 3=9(50(1-x))/(10(1-x))+5)/
10x+(50(1-x))/(10(1-x))+5) it's around this point that I get confused I think it's how I'm treating the (1-x) in the equation

There should be more terms in your first equation. Start with your complete equation in symbolic form:

$$Vp = Vs \left( \frac{\frac{R2 R3}{R2 + R3}}{R1 + \frac{R2 R3}{R2 + R3}} \right) \\
~~~Where: R1 = 10x ~~~~ R2 = 10(1 - x) ~~~~ R3 = 5$$
Then copy it out again but substitute your numbers and replacement expressions. You can move the Vs to the left side at the same time:

$$\frac{3}{9} = \frac{\frac{10(1 - x) 5}{10(1 - x) + 5}}{10x + \frac{10(1 - x) 5}{10(1 - x) + 5}} $$
and hammer away at simplifying it.

As a suggestion, since R2 is in parallel with R3 and you'll need to expand that twice, why not make R2 = 10x and R1 = 10(1-x)? That will cut down the amount of work you need to do when you expand the parallel combinations. It just means that x will be the "distance" from the bottom end of the potentiometer rather than the top end.

 

[tufty1980]

I've done it thank you very much needed someone to set my mind straight after 2 days on this i was going mad thanks again

 

[Electest]

Interesting thread. I found the answer using a different approach but am keen to use this alternative method but I'm having trouble simplifying the complex fraction. Any help would be greatly appreciated.

Regards.

 

[gneill]

Interesting thread. I found the answer using a different approach but am keen to use this alternative method but I'm having trouble simplifying the complex fraction. Any help would be greatly appreciated.

Regards.

Try using the suggestion in post #22 regarding the assignment of x and (1-x) portions of the potentiometer, and then show us your effort and point out where you're getting stuck.

 

[Electest]

Hi Gneill

I already understand the assignment of the x and x-1 as I used this to gain my answer initially. Then I read this thread and thought I would check my answer using the complex fraction method.

Ok this is where I am at (using the top expression in your post 22):On the bottom part of the fraction I cross multiply so I can add the fractions together, giving:

R1(R2+R3)+R2R3/R2+R3

Following this I flip the bottom half of the fraction and multiple out with the top half. (cancelling out R2+R3) giving:

R2R3/R1(R2+R3)+R2R3

Multiplying out gives:

R2R3/R1R2+R1R3+R2R3

Substituting in R values and expanding brackets I get

R1= 10X
R2= 10(1-X)
R3= 5

3/9=50-50X/100X-100X^2+50X+50-50X

therefore

3/9=50-50X/-100x^2+100X+50

I then move -100x^2+100X+50 to left hand side of = and the 9 to the right hand side by multiplying through. Giving me:

3(-100x^2+100X+50)=9(50-50X)

Therefore:

-300X^2+300X+150=450-450X

Giving:

-300x^2+750X-300=0

Using the Quadratic Formulas

A=-300 B=750 C=-300

Giving:

-750+ or - 450/-600 = 2 or 0.5

Must be 0.5 as cannot be greater than 1

-300(0.5^2)+750(0.5)-300=0

-75+375-300=0

I can seem to get the answer if I plug in the R values after re-arranging the equation. But would like to do it earlier as shown in post 22 bottom complex fraction. Driving me nuts!

Regards

 

[gneill]

It really is just a lot of algebraic slogging, and everyone seems to have their own favorite techniques and paths to the finish. If you show your steps we might be able to point out an error.

My usual first step is to try to concentrate as much of the nastiness into as few terms as possible, so I'd divide through by the numerator term first, since it appears in both the top and bottom:

$$\frac{3}{9} = \frac{1}{10x \left[ \frac{10(1-x) + 5}{10(1-x)5 }\right] +1}$$
If you invert both sides it'll already start looking a bit better...

 

[Electest]

Thanks again.

How do you post fractions like the ones you show? That would be easier for me to show you my workings out.

Regards

 

[gneill]

Thanks again.

How do you post fractions like the ones you show? That would be easier for me to show you my workings out.

Regards

LaTeX. Physics Forums takes advantage of software which will interprets embedded LaTeX commands to render mathematical formulas. If you go to a post showing such math rendered and hit the "Quote" button as though you wanted to quote the material to reply, then you'll see the LaTeX commands that resulted in given math formulas.

 

[Electest]

Ok so its finally come together as I realized I missed off 2 terms with I had initially put in the substituted values.

should have read:

Just trying Latex so may need some patience

$$\frac{3}{9}=\frac{\frac{10(1-x)5}{10(1-x)+5}}{\frac{10x(10(1-x)+5)+10(1-x)5}{10(1-x)+5}}$$

 

[Electest]

This expands to

$$\frac{3}{9}=\frac{50-50x}{100x-100x^2+50x+50-50x}$$

Which then matches my original answer further up in this thread. Funny how something so simple was overlooked on my part.

Could you explain a little further how you managed to get to that stage in your last post. Always keen to learn new methods.

Regards

Electest

 

[gneill]

Could you explain a little further how you managed to get to that stage in your last post. Always keen to learn new methods.

I mentioned it in the post. Beginning with the full "complex fraction" of post 22, I divided numerator and denominator by the numerator expression. Naturally that leaves a "1" for the numerator and "<mess> + 1" for the denominator.

 

[agreaves]

As a suggestion, since R2 is in parallel with R3 and you'll need to expand that twice, why not make R2 = 10x and R1 = 10(1-x)? That will cut down the amount of work you need to do when you expand the parallel combinations. It just means that x will be the "distance" from the bottom end of the potentiometer rather than the top end.

Hi, I've just tried this method and couldn't get it to work, can you show me this other method, when i tried i canceled it down to 2x² -x -1 =0
not sure where I made my mistake, I got the other way to work fine, can't see why it didn't work swapping them round

 

[agreaves]

also, wouldn't it work if the slider was at the bottom, which would put the resistors in series, as 5/(5+10) is still a third which is what we want

 

[gneill]

Hi, I've just tried this method and couldn't get it to work, can you show me this other method, when i tried i canceled it down to 2x² -x -1 =0
not sure where I made my mistake, I got the other way to work fine, can't see why it didn't work swapping them round

You'll have to show us your workings, not just your conclusion, in order for us to see where your attempt has gone awry.