Ohms law power problem
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From: Bill Bowden on 23 Jul 2010 22:56 On Jul 23, 8:40 am, John Fields <jfie...(a)austininstruments.com> wrote: > On Thu, 22 Jul 2010 18:34:37 -0700 (PDT), Bill Bowden > > > > <wrongaddr...(a)att.net> wrote: > >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. > > >So, which is correct A,B C ? > > --- > All three, when they're worked out properly. :-) > > A: > E² 12V² 1441 > P1 = ---- = ----- = ----- = 12W watts > R 12V 12 > > 12.1V² 146.41 > P2 = ------- = -------- = 12.2 watts > 12R 12R > > Dp1 = P2 - P1 = 0.2 watts > > B: > E 12V > I1 = --- = ----- = 1 ampere > R 12R > > P1 = EI = 12V * 1A = 12 watts. > > 12.1V > I2 = ------- = 1.00833... ampere > 12R > > P2 = 12.1V * 1.00833A = 12.2 watts > > DP2 = P2 - P1 = 0.2 watts > > C: > > P3 = I²R > > = 1.00833...A² * 12R > > = 1.0167 * 12R = 12.2 watts > > DP3 = P3 - P2 = 0.2 watts. Yes, that makes sense using hard numbers, I was trying to do it using small changes, thinking I could get the same result, but it didn't work. -Bill
From: Jon Kirwan on 24 Jul 2010 02:31 On Fri, 23 Jul 2010 19:33:26 -0700 (PDT), Bill Bowden <wrongaddress(a)att.net> wrote: ><snip> >I can't do calculus Jon. I took it one semester in school and got a >"D" >All I remember is the derivative of x^2 is 2x. 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. Jon
From: Jon Kirwan on 24 Jul 2010 03:50 On Fri, 23 Jul 2010 19:53:53 -0700 (PDT), stratus46(a)yahoo.com wrote: ><snip> >You don't need calculus for this. As my high school physics teacher >would say, "it's just sixth grade arithmetic". 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 dV = 0.1 V0 = 12 V1 = V0 + dV = 12.1 Rx = 12 P0 = V0^2/Rx P1 = V1^2/Rx = (V0+dV)^2/Rx = (V0^2+2*V0*dV+dV^2)/Rx = V0^2/Rx + 2*V0*dV/Rx + dV^2/Rx = P0 + 2*V0*dV/Rx + dV^2/Rx or, moving P0 to the left side, P1 - P0 = 2*V0*dV/Rx + dV^2/Rx 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: >: Yes, that makes sense using hard numbers, I was trying >: to do it using small changes, thinking I could get the >: same result, but it didn't work. 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. Jon
From: stratus46 on 24 Jul 2010 15:26 On Jul 24, 12:50 am, Jon Kirwan <j...(a)infinitefactors.org> wrote: > On Fri, 23 Jul 2010 19:53:53 -0700 (PDT), stratu...(a)yahoo.com > wrote: > > ><snip> > >You don't need calculus for this. As my high school physics teacher > >would say, "it's just sixth grade arithmetic". > > 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 > > dV = 0.1 > V0 = 12 > V1 = V0 + dV = 12.1 > Rx = 12 > P0 = V0^2/Rx > P1 = V1^2/Rx = (V0+dV)^2/Rx = (V0^2+2*V0*dV+dV^2)/Rx > = V0^2/Rx + 2*V0*dV/Rx + dV^2/Rx > = P0 + 2*V0*dV/Rx + dV^2/Rx > > or, moving P0 to the left side, > > P1 - P0 = 2*V0*dV/Rx + dV^2/Rx > > 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: > > >: Yes, that makes sense using hard numbers, I was trying > >: to do it using small changes, thinking I could get the > >: same result, but it didn't work. > > 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. > > Jon Since I know Ohm's law to be a law (and not a suggestion) and I got an answer _different_ from P=E*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. G²
From: Jon Kirwan on 24 Jul 2010 16:22
On Sat, 24 Jul 2010 12:26:07 -0700 (PDT), stratus46(a)yahoo.com wrote: ><snip> >Since I know Ohm's law to be a law (and not a suggestion) and I got an >answer _different_ from P=E*I, I would assume I made some kind of >mistake and then hunt it down to maintain my motto : Always make _new_ >mistakes. 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. >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. 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. Jon |