Finding Optimal Fixed Resistor for Potential Divider Circuit

[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!