Using a 12 ohm load and 12 volt supply, what is the power gain when the voltage is raised to 12.1 volts?
Considering the formula P=E^2/R the power at 12 volts will be 144/12 =
12 watts. And, at 12.1 volts, the power will be 146.41/ 12 = 12.2 watts, for a gain of 200 milliwatts.
But if the current at 12 volts is 1 amp, and the current at 12.1 volts is 12.1/12= 1.00833 then the increase is 8.33mA and from the power formula of P=I*E we get .00833*12.1 = 101 milliwatts which is about half as much as the first number.
And,considering the increase in current, using the formula P=I^2 * R we get .00833^2 * 12 = 833 microwatts, which ain't much.
I don't like the use of 'gain' here. Might be dissipation, instead.
Yes. 12.1*(12.1/12).
Um. You missed the inclusion of a factor of 2, as you might surmise from this result. It's actually 200mW.
Here's the problem. You compute dV (I prefer V to E) and multipled it by the new V and divided by R. That's almost right, except you need a factor of 2. I'll explain a little by transposing your work.
Let's call V0=12 and V1=12.1. R=12. First, you computed V1/R and then subtracted from it V0/R to get the current difference. This is the same thing as computing (V1-V0)/R or another way of saying that is that since dV=V1-V0, you computed dV/R. Then you multipled this by V1. In short, what you computed was V*dV/R.
But look at the original power equation you used.
P = V^2/R
The derivative of that, assuming R is constant, is:
dP = 2*V*dV/R
The change in power is TWO TIMES V*dV/R and you only computed half that much when you computed V*dV/R. Double it and you are back in the saddle, again.
Another way of seeing this is to imagine the voltage as one side of a rectangle and the current as the other side. Power is then the rectangle area.
When you computed V1/R and subtracted V0/R from it, you basically took the width of a tiny strip of that area where I've kind of filled it with "." characters. Multiplying by V1 got you the area of that strip. But you completely missed that hollow part near the top there, from V0 to V1 and times V0/R.
To fix up your calculation, you'd need to add it back in. That would be (12.1-12)*12/12 or, as is obvious, 0.1. Which gets you right back to your 200mW.
Now you are computing dI^2*R.
Instead,
P = I^2*R
therefore,
dP = 2*I*dI*R
And if you used 2*1.008333*0.008333*12, you would once again get a much more sensible result.
You correctly figured the power at the two voltage levels, and subtracted the powers.
You subtracted the current, and then calculated the power based on the current difference. Do what you did for voltage: calculate the powers based on the two currents, and then subtract the powers. P = I^2R p1 = 1^2*12 = 12 W p2 = 1.00833^2 * 12 = 12.2 W
(B) The reason that you got the wrong result in (B) is that you took the
*increase* in current, .00833 amp and multiplied by 12.1, but you took the
1 amp and multiplied it by 12, when this current (1 amp) should also be multiplied by 12.1.
(C) The reason that you got the wrong result in (C) is quite analogous to an error in calculating the volume of concrete necessary to make a cylindrical conduit.
The true formula is (R^2 - r^2)*pi*L where R is the outside radius, r is the inside radius, and L is the length of the conduit. The fairly common mistake is doing (R-r)^2*pi*L, thus giving the whacky result that a 10' inside diameter conduit and a 1' inside diameter conduit having the same wall thickness would use the same amount of concrete.
Calculus is hardly necessary to explain either of the above errors.
Bill, we both agree that power is V^2/R, right? Let's look at this, term by term. It is V*V/R. Now, if you increase V by some small value... call it dV for now... then the new estimate is (V+dV)*(V+dV)/R. Right? Multiply it out. It becomes (V^2+2*V*dV+dV^2)/R, yes? Put another way, it is:
V^2/R + 2*V*dV/R + dV^2/R
Now take a deeper look at the above. There are three terms there. The first term is just the power computed in the first place. 12V*12V/12Ohms, or 12W. With dV=0.1V, the third term is barely noticeable. It is .01/12 or that 833uW you'd earlier computed (remember it from your 3rd calc?) The middle term is twice what you'd said in your 2nd calculation because of the 2 there. Note that this 2nd term is two times
12V times (12.1V-12V) divided by 12Ohms. But two times! Which accounts for earlier conceptual error. You failed to include the factor of 2 there.
Again, but slightly differently than I did before, let's draw out a square of sorts. I'm going to remove the 1/R part of the power equation so that it just leaves us with V*V and also (V+dV)*(V+dV). Drawing that geometrically instead of algebraically provides something like this (use a fixed spaced font, of course, to "see" it):
Region A represents V^2 at V=12V. Region A+B+C+D represents (V+dV)^2 at V+dV=12.1V. If you then divide A by R=12 you get the power before changing the voltage by dV. If you then divide A+B+C+D by R you get the power afterwards. Note that the difference in area between A and A+B+C+D is B+C+D. B is just dV*V; C is just dV*V; and D is just dV*dV or dV^2. If you add B and C together, you get 2*V*dV, right? Once again, there is that pesky 2 factor. It's there because there are two areas that are the same shape and size.
The point here is that when you did your 2nd calculation, you either neglected to take into account B, or else C, and just calculated one of them. That's why you were basically off by a factor of two. You only counted one region instead of two. And there are two of them. Even then, you missed a tiny bit, namely D. Which as you calculated elsewhere, was only 833uW.
After dividing each by R=12, A=12W, B=0.1W, C=0.1W, and D=833uW. The total difference is B+C+D or 0.200833W, no matter how you cut it.
Your 3rd calculation came up with the area of D because what you did was to compute (dV/R)*(dV/R)*R. This is just dV^2/R. And since region D is just dV^2 (with an implicit /R to it), it's what you were actually getting.
I don't know if any of that helps. But there it is. Algebra AND geometry. Either way, it gets you to the same place.
I will stick my version of an explanation for this.
I think that you already know that A is the correct answer.
Power is current time voltage.
Increasing the voltage increases the current. If we call the the original voltage V0 and the initial current I0 and we call the increase in voltage dV and the increase in current dI.
Then the original power (at 12 volts) is given by.
P0 = V0 * I0 = 12 volts * 1 amp = 12 watts.
The power after the current is increased is given by
P1 = (V0 + dV) * (I0 + dI)
multiplying out the terms gives us:
P1 = V0 * I0 + dV * I0 + v0 * dI + dV * dI
Now let us look at the terms in this equation.
First we have V0 * I0. This is the original power before the voltage was increased. It is 12 volts times 1 amp = 12 watts.
Second we have dV * I0. I am going to save this term for a little later since this is the real cause of your confusion.
The third and fourth terms are V0 * dI plus dV * dI. This is the power increase that you have mentioned. I.e. they are 12 volts *
0.00833 amps = 0.1 watts and 0.1 volts * 0.00833 amps = 0.000833 watts. Or you can (and you did) combine these together to be 12.1 volts times 0.0833 amps. Either way you get 0.100833 watts.
Now lets go back to the dV * I0 term. This power is 0.1 volts times
1 amp which is 0.100 watts. This term represents the increase in power due to the increase in the voltage combined with the original current. This is the amount that you are missing.
Well, as another said, you don't need calculus. It just adds another perspective here that gets you to the same place as all the other approaches also do. And it adds insight where other methods fail. As a counter to this, some methods (such as hodographs, for example) provide insights where calculus fails to help nearly as well. So although it is god, it isn't a panacea.
Anyway, I hope the pictures I drew (and the algebra) helped. There was no calculus needed for some of what I wrote.
Well, that's a little off-putting. I think most 6th graders I've been exposed to (and I volunteered 300 hours a year for about 5 years running at one grade school, some years back, as an in-class aide) wouldn't be able to properly help Bill through the problem. A rare one, maybe.
6th graders are supposed to know how to represent rationals either as fractions or decimals, can use and apply ratios, solve percentage problems, and hopefully have been exposed to talking about things as x grams, y apples, and z frogs. Solve a few simple things like 3X=5, 2+Y=16, maybe. A very few go much beyond that.
The algebra is this:
V is some voltage R is some resistance P is some power P = V^2/R
This last expression is the power difference, of course. And it says that for small dV, it is mostly determined by the first part, or 2*V0*dV/Rx. But I wouldn't expect a 6th grader or even an 8th grader to get this far with the problem or to exhibit much insight.
I think Bill attempted to use some intuitive ideas to examine the problem from different angles and I'm impressed that he struggled with two additional approaches that otherwise might have been _powerful_, had he understood their meaning fully. In fact, the values he computed were dV*V/Rx and dV^2/Rx, which are important parts had he visualized the picture they were part of, correctly. I'm also impressed that he exposed himself to criticism, as that a good way to learn. I hope he won't take your kicking sand in his face as teaching him the wrong lesson.
The point that Joe made, that the error Bill made was similar to the cylindrical conduit problem is apt, but I'm not sure any light was shed for Bill in saying so. Bill's problem is more easily seen, not with a small margin perimeter where a smaller square is centered inside the larger -- which is more as the conduit case -- but instead with two sides and a corner of the two squares superimposed, I think. All ways work, but mimicking the conduit by centering squares would deviate more from Bill's initial insights, I think.
For someone wanting to "understand" and not just work some recipe that gets right results, I'd recommend looking again at P=V^2/R. The changing part, per the stated problem, is V (and consequently P.)
Ignore the R part for now. P is proportional to V^2, where V is subjected to change. V^2 is a square, with sides of length V. Changing V is, in effect, changing the length of both sides of the square. The difference between one square and a slightly larger square (aligned at one lower-left corner) imposed on top of it is a rectangular margin on the right and a similarly rectangular margin at the top. There are TWO of these margins, so the change in area is 2 times the area of one rectangle by itself. And the area of one rectangle is the change in V, the tiny width from V0 to V1 for example, times V. Which is where the 2*dV*V comes from. That represents the two rectangles, one on the right and one at the top, each dV*V in size. The last term comes from the tiny square in the upper right corner that isn't covered by either of the two marginal rectangles. And that tiny piece is dV*dV in size. Which explains the entire algebra equation using entirely visual, geometric and non-algebra means to yield the same resulting concept.
Bill tried to use finite differences, which is the beginning of moving towards very powerful concepts. He wrote, elsewhere:
And I think it means he is grasping very close to calculus thinking. He may not realize just how close he is to "getting it" and I'd like to encourage him to take on more, not make him feel badly about failing at "sixth grade arithmetic." He's close. Very close.
Since I know Ohm's law to be a law (and not a suggestion) and I got an answer _different_ from P=3DE*I, I would assume I made some kind of mistake and then hunt it down to maintain my motto : Always make _new_ mistakes.
It wasn't meant as a put-down, more of a don't make things more complex than they need to be and don't be afraid of it.
Hehe. Okay. I think that may actually be part of why Bill wrote. He got a group of "different answers" and wanted to understand why, when looking at this from different angles, he didn't come up with the same result. I think it is valid to think about the world using finite differences (in short, moving towards a differential viewpoint) and to try and make sure that works as well as using traditional finite averages in equations deduced from basic laws.
I think Bill _did_ assume he'd made some kind of mistake. He just didn't know how to 'right' himself while hanging onto the differential mindset at the same time.
I have mixed feelings about that approach. If all one wants is a correct answer for a particular situation and has no further interest (or ability) in developing a deeper understanding, I agree with your comment.
But if someone _is_ interested in precise boundaries and profound meaning to such laws, and not just one of the many expressions (facets) of them that just happens to be in front of them at the time, then I think it is wonderful to try and take something concrete and immediate and play with it more in order to help develop a deeper insight into their nature. That may mean doing what Bill attempted (and slightly failed at.) It is through such questions and struggles that one may, at times, gain a higher ground from which to see.
In such cases, saying to "not make things more complex than needed" is not entirely unlike telling a 5th grader who _may_ be asking a profound question like "Why is the moon round?" and rather than exploring that with them while they show some interest, instead answering it with a put down, "What? Do you think it should be square?" as though they were stupid for asking in the first place. They will get the clue, of course. And stop asking. But for exploring minds, that is destructive.
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