Finding Optimal Fixed Resistor for Potential Divider Circuit

[Jimmy87]

Homework Statement

Consider a potential divider circuit with an LDR and a fixed resistor. The circuit is powered by a battery which has an emf of 45V with negligible internal resistance.

(a) Describe and sketch a circuit that could use a voltmeter or an ammeter to act as a light meter such that an increase in light intensity will cause an increase in the light meter reading.
(b) The light intensity can cause a variation in the resistance of the LDR from 250-1500 Ohms. If the available options for the fixed resistor are 750 or 1000 Ohms. Use calculations to show which fixed resistor will give the biggest scale change on the light meter.
(c) How would you find the value for the fixed resistor that would give the largest possible scale change on the light meter.

Homework Equations

VT/(R1+R2) = Vo/R2 where R1 is the LDR, R2 is the fixed resistor, VT is the emf (45V), Vo is the potential difference across R2.

The Attempt at a Solution

(a) If choosing the ammeter as light meter it simply goes in series with the rest of the components. If choosing voltmeter must go across R2 in order to increase with increasing light intensity.
(b) From my calculations the 750 ohm resistor would give a voltage range of 18.75V across R2 when the LDR varies from 250-1500. The 1000 ohm resistor would give a range of 18V so the 750 ohm resistor is the best choice.
(c) I have no idea with this part! I can randomly select resistances and calculate the range they would give but I have no idea how you find the exact resistance that gives the largest range?

Thanks for any help given.

 

[gneill]

(c) I have no idea with this part! I can randomly select resistances and calculate the range they would give but I have no idea how you find the exact resistance that gives the largest range?

Can you write a single expression that would give you the range for a given value of fixed resistor R2?

 

[Jimmy87]

Can you write a single expression that would give you the range for a given value of fixed resistor R2?

Thanks. Sorry I don't understand if your asking me to write an expression (i.e. it can be done) or whether you are implying it can't be done? If it can, I'm not sure how to do this?

 

[gneill]

Thanks. Sorry I don't understand if your asking me to write an expression (i.e. it can be done) or whether you are implying it can't be done? If it can, I'm not sure how to do this?

You should be able to write an expression. You've already done it piece-wise when you found the range for the individual 750 and 1000 Ohm fixed resistors. What steps did you use to do that?

 

[Jimmy87]

You should be able to write an expression. You've already done it piece-wise when you found the range for the individual 750 and 1000 Ohm fixed resistors. What steps did you use to do that?

I did:

VTR2/(R1+R2) for R1 = 250 and 1500 when R2 = 750 and then R2 = 1000

The only thing I can think of along your lines is:

Vo1-Vo2 = VTR2/(R1+R2) - VTR2/(R3+R2) where R1 is when ldr is at 250 and R3 is when ldr is at 1500 and Vo1 and Vo2 are the output voltages across R2 therefore Vo1 -Vo2 is the range. However, trying to simply this expression makes it look more and more complicated the more I manipulate it :(

Is any of that right?

 

[gneill]

I did:

VTR2/(R1+R2) for R1 = 250 and 1500 when R2 = 750 and then R2 = 1000

The only thing I can think of along your lines is:

Vo1-Vo2 = VTR2/(R1+R2) - VTR2/(R3+R2) where R1 is when ldr is at 250 and R3 is when ldr is at 1500 and Vo1 and Vo2 are the output voltages across R2 therefore Vo1 -Vo2 is the range. However, trying to simply this expression makes it look more and more complicated the more I manipulate it :(

Is any of that right?

You're on the right track and you're nearly there. Let's clean up the math a bit by making the known fixed values (the LDR light and dark resistances) constants and letting the one unknown resistor be R.

$Range = VT \left[ \frac{R}{R + 250} - \frac{R}{R + 1500} \right]$

The source voltage VT just scales the problem since it multiplies everything by a constant value, so you can just concentrate on the terms involving resistor R as they will determine amount of VT that comprises the range. So as a function of R:

$f(R) = \frac{R}{R + 250} - \frac{R}{R + 1500}$

How do go about finding the minimum or maximum of a function?

 

[Jimmy87]

You're on the right track and you're nearly there. Let's clean up the math a bit by making the known fixed values (the LDR light and dark resistances) constants and letting the one unknown resistor be R.

$Range = VT \left[ \frac{R}{R + 250} - \frac{R}{R + 1500} \right]$

The source voltage VT just scales the problem since it multiplies everything by a constant value, so you can just concentrate on the terms involving resistor R as they will determine amount of VT that comprises the range. So as a function of R:

$f(R) = \frac{R}{R + 250} - \frac{R}{R + 1500}$

How do go about finding the minimum or maximum of a function?

Thanks. That equation makes sense. However, I'm not sure how to find max and min of this function?

 

[gneill]

Thanks. That equation makes sense. However, I'm not sure how to find max and min of this function?

Have you studied calculus? If not you might use a graphical approach.

 

[Jimmy87]

Have you studied calculus? If not you might use a graphical approach.

Yes done A fair bit. Done power rule, chain rule and quotient rule. What do I need to use for this problem?

 

[gneill]

Yes done A fair bit. Done power rule, chain rule and quotient rule. What do I need to use for this problem?

You should have studied finding the extrema of a function. Say you have a function f(x) and you want to minimize or maximize it. What was the procedure?

 

[Jimmy87]

You should have studied finding the extrema of a function. Say you have a function f(x) and you want to minimize or maximize it. What was the procedure?

Ok, so I have looked through all my notes. We did this when we did suvat equations of motion. For example, we used derivatives to find the maximum height a ball can reach. We found the slope of the function by just differentiating the expression for height with respect to time. We then set this derivative equal to zero which would give the time when the slope is zero hence the height is maximum.

So do I differentiate your expression with respect to 'R' then set this derivative equal to zero and see what I get? How do you evaluate the maximum value though?

 

[gneill]

Ok, so I have looked through all my notes. We did this when we did suvat equations of motion. For example, we used derivatives to find the maximum height a ball can reach. We found the slope of the function by just differentiating the expression for height with respect to time. We then set this derivative equal to zero which would give the time when the slope is zero hence the height is maximum.

Right!

So do I differentiate your expression with respect to 'R' then set this derivative equal to zero and see what I get? How do you evaluate the maximum value though?

The value(s) of R you obtain from the above procedure will correspond to either maxima or minima for the function. You can either apply the second derivative test (not recommended here since the derivative will get pretty messy), or just plug in values close to but on either side of R to confirm that R gives you a maximum. Or, just plot the original function to show that there's a maximum around the value of R and so R is that maximum.

 

[Jimmy87]

Right!

The value(s) of R you obtain from the above procedure will correspond to either maxima or minima for the function. You can either apply the second derivative test (not recommended here since the derivative will get pretty messy), or just plug in values close to but on either side of R to confirm that R gives you a maximum. Or, just plot the original function to show that there's a maximum around the value of R and so R is that maximum.

So to differentiate your expression what rule do I use as R is on the numerator and denominator (which is quotient rule) but is also in two separate expressions?

 

[gneill]

The quotient rule applies.

 

[Jimmy87]

The quotient rule applies.

Ok so don't know if it's correct but I did the quotient rule on each term so I got:

(R+250) - R251/(R + 250)*2 - (R+1500) - R1501/(R+1500)*2

Where *2 is squared.

Have no idea if I have applied quotient rule correctly here?

 

[gneill]

The convention is to use "^" to signify exponents. Or you could use the edit panel x₂ and x² buttons to make actual subscripts and superscripts.

Your derivative doesn't look right to me. Maybe take one of the terms and show us your step by step work. Show us the derivative for:

R/(R + 250)

 

[Jimmy87]

The convention is to use "^" to signify exponents. Or you could use the edit panel x₂ and x² buttons to make actual subscripts and superscripts.

Your derivative doesn't look right to me. Maybe take one of the terms and show us your step by step work. Show us the derivative for:

R/(R + 250)

Ah I think I made a really silly mistake. Ok so for the above:

For quotient rule U = R and V = R+250.

So du/dR V = R+250
dv/dR U = R
So we get:

R+250 - R / (R+250)^2

Is that right?

 

[gneill]

Ah I think I made a really silly mistake. Ok so for the above:

For quotient rule U = R and V = R+250.

So du/dR V = R+250
dv/dR U = R
So we get:

R+250 - R / (R+250)^2

Is that right?

It would be right if you used parentheses to specify the correct order of operations. As it stands the standard interpretation of your expression would be wrong.

It should read:

(R + 250 - R)/(R + 250)^2

= 250/(R + 250)^2

 

[Jimmy87]

It would be right if you used parentheses to specify the correct order of operations. As it stands the standard interpretation of your expression would be wrong.

It should read:

(R + 250 - R)/(R + 250)^2

= 250/(R + 250)^2

Yes sorry I should have used brackets correctly. So using original expression you gave me the second term differentiated would be:

1500/(R+1500)^2

So the whole expression differentiated is:

250/(R+250)^2 - 1500/(R+1500)^2

So I set this equal to zero and solve for R? If so I get:

R = sqrt (250-1500)

So R=35

Is that right?

 

[gneill]

Yes sorry I should have used brackets correctly. So using original expression you gave me the second term differentiated would be:

1500/(R+1500)^2

So the whole expression differentiated is:

250/(R+250)^2 - 1500/(R+1500)^2

So I set this equal to zero and solve for R?

Right.

If so I get:

R = sqrt (250-1500)

So R=35

Is that right?

Nope. Show your work.

 

[Jimmy87]

The quotient rule applies.

Right.

Nope. Show your work.

I put denominators as R^2 + 250^2 and R^2 + 1500^2. Is this my mistake do I need to expand the brackets like this:

(R+250) (R+250)?

 

[gneill]

I put denominators as R^2 + 250^2 and R^2 + 1500^2. Is this my mistake do I need to expand the brackets like this:

(R+250) (R+250)?

Yes!

 

[Jimmy87]

Yes!

Great thanks. So I have got this far:

1/(R^2+500R+500) - 1/(R^2+3000R+1500) = 0

So do I now solve both denominators as quadratic equations?

 

[gneill]

Great thanks. So I have got this far:

1/(R^2+500R+500) - 1/(R^2+3000R+1500) = 0

So do I now solve both denominators as quadratic equations?

How come it's 1/(stuff) for each term? What happened to the original numerators?

You need to form a single expression by putting everything over a common denominator. Hint: in the end the denominator itself won't matter because zeroes can only happen when the numerator is zero. Concentrate on finding the combined numerator.

 

[Jimmy87]

How come it's 1/(stuff) for each term? What happened to the original numerators?

You need to form a single expression by putting everything over a common denominator. Hint: in the end the denominator itself won't matter because zeroes can only happen when the numerator is zero. Concentrate on finding the combined numerator.

The original numerators were 250 and 1500 but each expanded term in the denominator had a 250^2 term and 1500^2 term so would the numerators cancel?

 

[gneill]

The original numerators were 250 and 1500 but each expanded term in the denominator had a 250^2 term and 1500^2 term so would the numerators cancel?

That may well be, but you still need to form a single fraction out of the two separate terms.

 

[Jimmy87]

That may well be, but you still need to form a single fraction out of the two separate terms.

Oh I see. So if we expand denominators we get:

250\(R^2+500R+250^2) - 1500/(R^2+3000R+1500^2)
= 0

That seems impossible to make a common denominator unless I multiply each denominator by the others denominator?

 

[gneill]

Oh I see. So if we expand denominators we get:

250\(R^2+500R+250^2) - 1500/(R^2+3000R+1500^2)
= 0

That seems impossible to make a common denominator unless I multiply each denominator by the others denominator?

Yeah. You've made things overly complicated by expanding the denominators first. Don't do that. Leave them alone and treat them as individual terms until you have the numerator fleshed out.

 

[Jimmy87]

Yeah. You've made things overly complicated by expanding the denominators first. Don't do that. Leave them alone and treat them as individual terms until you have the numerator fleshed out.

Oh ok. So I would still multiply each expression by the others denominator:

250 (R+1500)^2/(R+250)^2 (R+1500)^2 - 1500 (R+250)^2/(R+250)^2 (R+1500)^2

Then combined we now get:

250(R+1500)^2 - 1500(R+250)*2 / (R+250)^2 (R+1500)^2

Am I now on the right lines?

 

[gneill]

Yes. Again, you need to use more parentheses to ensure the correct interpretation of what you write. A divide operator "/" generally applies only to the immediately adjacent terms!

Expand the numerator and find R that makes it zero.

 

[Jimmy87]

J

Yes. Again, you need to use more parentheses to ensure the correct interpretation of what you write. A divide operator "/" generally applies only to the immediately adjacent terms!

Expand the numerator and find R that makes it zero.

just to check with you first. When you expand the numerator brackets do you multiply by the coefficient first or after? I'm thinking after as per BODMAS?

 

[gneill]

Definitely after. Unless you want to bring the square root of the coefficients into the brackets and distribute it before expanding the brackets. Too messy in my opinion. Expand the bracketed terms, then apply the coefficients, then collect terms.

 

[Jimmy87]

Definitely after. Unless you want to bring the square root of the coefficients into the brackets and distribute it before expanding the brackets. Too messy in my opinion. Expand the bracketed terms, then apply the coefficients, then collect terms.

Ok so for numerator I ended up with:

250R^2 - 1500R^2 = -1250R^2

With denominator I'm a bit stuck. I multiplied each bracket separately and now have:

(R^2+500R+250^2) x (R^2+300R+1500^2)

How do I now multiply these?

 

[gneill]

Your numerator result doesn't look right. Why not present your work step by step?

You can ignore the denominator entirely. The denominator cannot produce a zero for the function (for finite values of R). Only the numerator can produce a zero by itself becoming zero.

So you're starting with:

250(R+1500)^2 - 1500(R+250)^2 = 0

 

[Jimmy87]

Your numerator result doesn't look right. Why not present your work step by step?

You can ignore the denominator entirely. The denominator cannot produce a zero for the function (for finite values of R). Only the numerator can produce a zero by itself becoming zero.

So you're starting with:

250(R+1500)^2 - 1500(R+250)^2 = 0

Thanks. Now I actually have done it systematically I got a different answer. So, I did it in these steps:

(R+1500) (R+1500) = R^2 + 3000R +1500^2
Then multiplied by the coefficient: 250 (R^2 + 3000R + 1500^2) = 250R^2 + 750,000R + 562500000
Now onto the next bracket:
(R+250)(R+250) = R^2 + 500R + 250^2
1500(R^2+500R+250^2) = 1500R^2 + 750000R + 937500000

Finally we have:

250R^2 + 750,000R + 562500000 - 1500R^2 + 750,000R + 937500000

Which simplifies to:

1250R^2 - 375,000,000 = 0

1250R^2 = 375,000,000

R^2 = 300,000

R = 548 (to 3 s.f.)

So if this is correct what have we just worked out? Is this the resistance that will give the largest range or is this the resistance at the minimum?

Btw thanks for all your help and patience. I have learned more from this single thread than I have from a month in class!

 

[gneill]

You're definitely getting closer!

Finally we have:

250R^2 + 750,000R + 562500000 - 1500R^2 + 750,000R + 937500000

Check the signs on the last two terms. When you expanded 1500(R + 250)^2 there was a "-" associated with the 1500. That must distribute over the whole thing when you distribute the 1500.

 

[Jimmy87]

You're definitely getting closer!

Check the signs on the last two terms. When you expanded 1500(R + 250)^2 there was a "-" associated with the 1500. That must distribute over the whole thing when you distribute the 1500.

Whoops forgot about that. I have copied and pasted and corrected (hopefully):

(R+1500) (R+1500) = R^2 + 3000R +1500^2
Then multiplied by the coefficient: 250 (R^2 + 3000R + 1500^2) = 250R^2 + 750,000R + 562500000
Now onto the next bracket:
(R+250)(R+250) = R^2 + 500R + 250^2
-1500(R^2+500R+250^2) = -1500R^2 - 750000R - 937500000

Finally we have:

250R^2 + 750,000R + 562500000 - 1500R^2 - 750,000R - 937500000

Which simplifies to:

-1250R^2 - 375,000,000 = 0

-1250R^2 = 375,000,000

-R^2 = 300,000

R = ?

Since we can't sqrt a negative number what do I do now?

 

[gneill]

Now onto the next bracket:
(R+250)(R+250) = R^2 + 500R + 250^2
-1500(R^2+500R+250^2) = -1500R^2 - 750000R - 937500000

I think you've added one too many zeros to the constant term. Check that multiplication again.

 

[Jimmy87]

I think you've added one too many zeros to the constant term. Check that multiplication again.

So should have one less zero on last constant term:

250R^2 + 750,000R + 562500000 - 1500R^2 - 750,000R - 93750000

Which simplifies to:

-1250R^2 - 468,750,000 = 0

-1250R^2 = 468,750,000

-R^2 = 375,000

Do we still not have a negative though which we can't square root?

 

[gneill]

562500000 - 93750000 = ?

Is the result positive or negative?

 

[Jimmy87]

562500000 - 93750000 = ?

Is the result positive or negative?

I feel like a complete fool - sorry. It should have been:

-1250R^2 + 468,750,000 = 0

468,750,000 = 1250R^2

R^2 = 375,000

R = 612 (to 3s.f.)

You said this could be a max or min. If we use this resistance in the original post in part (b) of the question we get a bigger range than the 750 Ohms so it must be a maximum? Is that right? Is it random whether or not you get a max or min because there must also be some other point in the function where you have a resistance that would give the smallest range (i.e. minima). What determines if you end up with a min or max at the end?

 

[gneill]

That's correct! (*whew* )

You said this could be a max or min. If we use this resistance in the original post in part (b) of the question we get a bigger range than the 750 Ohms so it must be a maximum? Is that right? Is it random whether or not you get a max or min because there must also be some other point in the function where you have a resistance that would give the smallest range (i.e. minima). What determines if you end up with a min or max at the end?

If this was an exam you might be asked to show mathematically that it is the requested maximum. In that event you might resort to the second derivative test for local extrema (look up the second derivative test). That would tell you whether it is a maximum or a minimum.

For this type of work though, it would suffice to show that the given solution is a maximum by testing nearby values and showing that they yield smaller results, or by providing a plot.

The first derivative method will find ALL the extrema of the function. It will return values for R for every one of them. In this case you ended up with a quadratic so there were potentially two values of R that produce extrema for the function. In this case the roots happened to be (approximately) +612 and -612 Ohms. Since a negative resistance is un-physical for real components (at least the type of components we're using here), the negative value is discarded and you're left with one possible solution. It could be either a minimum or a maximum, and you should verify that it is a maximum by testing it in the function as mentioned.

Note that when you solved this problem the value of R was not constrained in any way. So every possible value of R is included in the "test", and there are no boundaries to check.

If you had a function that returned several plausible values for R then you would have to check each one to see if it corresponded to a maximum or a minimum.

 

[Jimmy87]

That's correct! (*whew* )
If this was an exam you might be asked to show mathematically that it is the requested maximum. In that event you might resort to the second derivative test for local extrema (look up the second derivative test). That would tell you whether it is a maximum or a minimum.

For this type of work though, it would suffice to show that the given solution is a maximum by testing nearby values and showing that they yield smaller results, or by providing a plot.

The first derivative method will find ALL the extrema of the function. It will return values for R for every one of them. In this case you ended up with a quadratic so there were potentially two values of R that produce extrema for the function. In this case the roots happened to be (approximately) +612 and -612 Ohms. Since a negative resistance is un-physical for real components (at least the type of components we're using here), the negative value is discarded and you're left with one possible solution. It could be either a minimum or a maximum, and you should verify that it is a maximum by testing it in the function as mentioned.

Note that when you solved this problem the value of R was not constrained in any way. So every possible value of R is included in the "test", and there are no boundaries to check.

If you had a function that returned several plausible values for R then you would have to check each one to see if it corresponded to a maximum or a minimum.

We didn't end up with a quadratic did we? The final solution yielded one value for R didn't it? Am I right that the minima would be the resistance that gives the smallest range? What would happen if the question asked for this in part (c) instead of the maximum range? Is it easy to go from the maxima to the minima? Final question (I promise) - is it difficult to plot the original function we had on a graph to show the maxima?

 

[Jimmy87]

Ignore my first comment. Of course it was quadratic as the solution can either be -612 or +612! Sorry!

 

[gneill]

We didn't end up with a quadratic did we? The final solution yielded one value for R didn't it? Am I right that the minima would be the resistance that gives the smallest range? What would happen if the question asked for this in part (c) instead of the maximum range?

You would solve as before, find out that the plausible values of R returned gave only a maxima, and conclude that there's no fixed value of R greater than zero that yields a minima. You could then note that by inspection of the function the range is zero when R is zero. Since presumably negative ranges are not a valid option, R = 0 and a range of zero would be your solution.

Is it easy to go from the maxima to the minima?

Not sure what that means. The values of R returned by the derivative method will correspond to either maxima or minima. So finding either, when they exist, takes the same mount of effort.

Final question (I promise) - is it difficult to plot the original function we had on a graph to show the maxima?

Depends on what graphing software you have on hand. Some people use Excel. Others use online apps or graphing calculators. Still others have dedicated software packages such as Matlab. Personally I like to use an ancient copy of MathCad

Just define the function: $f(R) = \frac{R}{(R + 250)} - \frac{R}{(R + 1500)}$ and feed it to your software. You might have to choose a suitable domain for R for the plot.

 

[Jimmy87]

You would solve as before, find out that the plausible values of R returned gave only a maxima, and conclude that there's no fixed value of R greater than zero that yields a minima. You could then note that by inspection of the function the range is zero when R is zero. Since presumably negative ranges are not a valid option, R = 0 and a range of zero would be your solution.

Not sure what that means. The values of R returned by the derivative method will correspond to either maxima or minima. So finding either, when they exist, takes the same mount of effort.

Depends on what graphing software you have on hand. Some people use Excel. Others use online apps or graphing calculators. Still others have dedicated software packages such as Matlab. Personally I like to use an ancient copy of MathCad

Just define the function: $f(R) = \frac{R}{(R + 250)} - \frac{R}{(R + 1500)}$ and feed it to your software. You might have to choose a suitable domain for R for the plot.

Amazing. I can't thank you enough for all your help!