Ohms law power problem
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From: Bill Bowden on 22 Jul 2010 21:34 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
From: Jon Kirwan on 22 Jul 2010 22:12 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? 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
From: Shaun on 22 Jul 2010 23:03 "Bill Bowden" <wrongaddress(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
From: Bill Bowden on 23 Jul 2010 01:22 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
From: ehsjr on 23 Jul 2010 01:48
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. Correct. > > 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. The error above is that you used P=I*E to compute based on the _change_ in current (dI). P <> dI*E Same thing, below. You used P=I^2 * R to compute based on the _change_ in current (dI). P <> dI^2 * R Ed > > 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 > |