> > 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
> After reading all the responses I got dizzy and I have been teaching this > for 40 years. > You made a math error. You seem capable of catching it. > Check your work.
> Tom
I checked the work, but it doesn't add up.
a small change of 8.33mA and 100mV is only 833 microwatts when it should be 200 milliwatts.
I think an analogy might be a freeway traffic jam where you add one more car and everybody stops?
>> > 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
>> After reading all the responses I got dizzy and I have been teaching this >> for 40 years. >> You made a math error. You seem capable of catching it. >> Check your work.
>> Tom
>I checked the work, but it doesn't add up.
>a small change of 8.33mA and 100mV is only 833 microwatts when it >should be 200 milliwatts.
--- Algebraically, you can't just use the change in quantity, you have to use the whole new quantity.
> > 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
> After reading all the responses I got dizzy and I have been teaching this > for 40 years. > You made a math error. You seem capable of catching it. > Check your work.
> Tom
I checked the work, but it doesn't add up.
a small change of 8.33mA and 100mV is only 833 microwatts when it should be 200 milliwatts.
I think an analogy might be a freeway traffic jam where you add one more car and everybody stops?
-Bill
You can't just take pieces of what you want. The current produced was 1.00833 Amps not .00833 Amps. The 8.33mA was in addition to the one Amp. Concept of math error. Keep pluggin' its fun.
> a small change of 8.33mA and 100mV is only 833 microwatts when it > should be 200 milliwatts.
Okay, let's do this as I would in a decent algebra class:
Power is current x voltage.
Let's make this as a formula P = I*V
Better, from an algebraic understanding, is to write this in FUNCTION NOTATION (dammit!).
P(I, V) = I*V
Where you are confused is thinking that this FUNCTION has the property
<BOGUS> P(I+i, V+v) = P(I,V) + P(i,v) </BOGUS>
<CORRECT> P(I+i, V+v) = (I+i)*(V+v) </CORRECT>
Now, compare <BOGUS> with <CORRECT>:
<CORRECT> Gives P(I+i, V+v) = (I+i)*(V+v) = (remember "FOIL") IV + Iv + iV + iv
<BOGUS> Gives P(I+i, V+v) = (BOGUS) P(I,V) + P(i,v) = IV + iv which is WRONG!
<CORRECT> gives two additional terms: Iv and iV
** NOTE: I never heard of "FOIL" as an acronym until a younger relative told me about it. Some of my students seem to cling to a rule rather than an understanding of what is happening (the distributive law). I sometimes mocked mindless rules by referring to the "FOIL" situation as "Leo Rio". Which actually stood for "LIORIO" (my own "Left Inner Outer Right Inner Outer" Roole). :)
Joe wrote: > In article > <606dd3ae-7b34-4da6-a306-32caba98a...@t2g2000yqe.googlegroups.com>, Bill > Bowden <wrongaddr...@att.net> wrote:
>>I checked the work, but it doesn't add up.
>>a small change of 8.33mA and 100mV is only 833 microwatts when it >>should be 200 milliwatts.
> Okay, let's do this as I would in a decent algebra class:
> Power is current x voltage.
> Let's make this as a formula P = I*V
> Better, from an algebraic understanding, is to write this in FUNCTION > NOTATION (dammit!).
> P(I, V) = I*V
> Where you are confused is thinking that this FUNCTION has the property
> <BOGUS> P(I+i, V+v) = P(I,V) + P(i,v) </BOGUS>
> <CORRECT> P(I+i, V+v) = (I+i)*(V+v) </CORRECT>
> Now, compare <BOGUS> with <CORRECT>:
> <CORRECT> Gives P(I+i, V+v) = (I+i)*(V+v) = (remember "FOIL") IV + Iv + iV + iv
> <BOGUS> Gives P(I+i, V+v) = (BOGUS) P(I,V) + P(i,v) = IV + iv which is WRONG!
> <CORRECT> gives two additional terms: Iv and iV
> ** NOTE: I never heard of "FOIL" as an acronym until a younger relative > told me about it. Some of my students seem to cling to a rule rather than > an understanding of what is happening (the distributive law). I sometimes > mocked mindless rules by referring to the "FOIL" situation as "Leo Rio". > Which actually stood for "LIORIO" (my own "Left Inner Outer Right Inner > Outer" Roole). :)
> --- Joe
In other words, what you are saying is that the increase in power cannot be determined by multiplying the increase in current by the increase in voltage. True.
And, even if you don't know that, when you write the equation using functional notation (correctly) the correct answer comes out.
I screw up like that a lot. My Dad used to tell me that mistakes are harder to spot when your nose is too close to the work. Taking a step back gives a wider view.
We're all striving to be better. I note with some interest that one or two denizens of this fine group have already attained perfection. A weird thing though..perfection seems to bring with it uncontrolled profanity.
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