Ohms law power problem
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From: John Fields on 23 Jul 2010 11:40 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. > >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: E� 12V� 1441 P1 = ---- = ----- = ----- = 12W watts R 12V 12 12.1V� 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 = I�R = 1.00833...A� * 12R = 1.0167 * 12R = 12.2 watts DP3 = P3 - P2 = 0.2 watts.
From: Dan Coby on 23 Jul 2010 19:00 On 7/22/2010 6:34 PM, Bill Bowden 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 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.
From: Bill Bowden on 23 Jul 2010 22:20 On Jul 23, 12:10 am, n...(a)given.now (Joe) wrote: > In article > <5b096e94-89b8-416a-9095-004aabb58...(a)k19g2000yqc.googlegroups.com>, 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 ? > > > 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. > Yes, that looks right. I missed the 1 amp at 12.1V for the extra 100mW. > (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: Bill Bowden on 23 Jul 2010 22:33 On Jul 22, 7:12 pm, Jon Kirwan <j...(a)infinitefactors.org> 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? > > I don't like the use of 'gain' here. Might be dissipation, > instead. > > >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. > > Yes. 12.1*(12.1/12). > > >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. > > 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. > > V1 +----------------------------+---+ > | |. .| > V0 +----------------------------+---+ > | / / / / / / / / / | . | > |/ / / / / / / / / /|. .| > | / / / / / / / / / | . | > | / / / / / / / / / |. .| > |/ / / / / / / / / /| . | > 0 +----------------------------+---+ > 0 V0/R V1/R > > 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. > > >And,considering the increase in current, using the formula P=I^2 * R > >we get .00833^2 * 12 = 833 microwatts, which ain't much. > > 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. > > >So, which is correct A,B C ? > > >I vote for A > > Calculus is god, though. Best to use that. > > Jon 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. -Bill
From: stratus46 on 23 Jul 2010 22:53
On Jul 23, 7:33 pm, Bill Bowden <wrongaddr...(a)att.net> wrote: > On Jul 22, 7:12 pm, Jon Kirwan <j...(a)infinitefactors.org> 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? > > > I don't like the use of 'gain' here. Might be dissipation, > > instead. > > > >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. > > > Yes. 12.1*(12.1/12). > > > >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. > > > 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. > > > V1 +----------------------------+---+ > > | |. .| > > V0 +----------------------------+---+ > > | / / / / / / / / / | . | > > |/ / / / / / / / / /|. .| > > | / / / / / / / / / | . | > > | / / / / / / / / / |. .| > > |/ / / / / / / / / /| . | > > 0 +----------------------------+---+ > > 0 V0/R V1/R > > > 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. > > > >And,considering the increase in current, using the formula P=I^2 * R > > >we get .00833^2 * 12 = 833 microwatts, which ain't much. > > > 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. > > > >So, which is correct A,B C ? > > > >I vote for A > > > Calculus is god, though. Best to use that. > > > Jon > > 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. > > -Bill You don't need calculus for this. As my high school physics teacher would say, "it's just sixth grade arithmetic". G² |