Ohms law power problem
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From: John O'Flaherty on 23 Jul 2010 02:27 On Thu, 22 Jul 2010 18:34:37 -0700 (PDT), Bill Bowden <wrongaddress(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. You correctly figured the power at the two voltage levels, and subtracted the powers. >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. 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 p2 - p1 = 200 mW. -- John
From: Joe on 23 Jul 2010 03:10 In article <5b096e94-89b8-416a-9095-004aabb5868e(a)k19g2000yqc.googlegroups.com>, Bill Bowden <wrongaddress(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 ? > > I vote for A > > -Bill (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. --- Joe
From: Jon Kirwan on 23 Jul 2010 03:41 On Thu, 22 Jul 2010 22:22:08 -0700 (PDT), Bill Bowden <wrongaddress(a)att.net> wrote: >On Jul 22, 8:03�pm, "Shaun" <r...(a)nomail.com> wrote: >> "Bill Bowden" <wrongaddr...(a)att.net> wrote in message >> >> news:5b096e94-89b8-416a-9095-004aabb5868e(a)k19g2000yqc.googlegroups.com... >> >> > 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 ? >> >> > I vote for A >> >> > -Bill >> >> Your doing your math wrong. >> >> Learn to do math properly!! �If you do all three methods come up with the >> same answer, 200 mW >> >> Shaun > >Really? now would you mind explaining why an increase of 8.33mA in a >12 volt load amounts to 200mW? > >8.33mA times 12 volts is only 100mW. > >Where is the extra 100mW? > >-Bill 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): +----------------+---+ V+dV | (B) |(D)| +----------------+---+ V | / / / / / | | |/ / / / / /| | | / / (A) / / |(C)| | / / / / / | | |/ / / / / /| | 0 +----------------+---+ 0 V V+dV 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. Jon
From: Jon Kirwan on 23 Jul 2010 03:44 On Fri, 23 Jul 2010 00:10:38 -0700, none(a)given.now (Joe) wrote: ><snip> >Calculus is hardly necessary to explain either of the above errors. Completely agreed, as I've already evidenced with both algebra and simple areas. But calculus is still god! ;) Jon
From: Jasen Betts on 23 Jul 2010 07:19
On 2010-07-23, Bill Bowden <wrongaddress(a)att.net> wrote: > On Jul 22, 8:03 pm, "Shaun" <r...(a)nomail.com> wrote: >> Learn to do math properly!! If you do all three methods come up with the >> same answer, 200 mW > Really? now would you mind explaining why an increase of 8.33mA in a > 12 volt load amounts to 200mW? But, it's not a 12V load any more. --- news://freenews.netfront.net/ - complaints: news(a)netfront.net --- |