Ohms law power problem
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From: Bill Bowden on 26 Jul 2010 02:37 On Jul 25, 9:03 am, "Tom Biasi" <tombi...(a)optonline.net> 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 > > 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
From: John Fields on 26 Jul 2010 09:55 On Sun, 25 Jul 2010 23:37:03 -0700 (PDT), Bill Bowden <wrongaddress(a)att.net> wrote: >On Jul 25, 9:03�am, "Tom Biasi" <tombi...(a)optonline.net> 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 >> >> 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. JF
From: Tom Biasi on 26 Jul 2010 14:54 "Bill Bowden" <wrongaddress(a)att.net> wrote in message news:606dd3ae-7b34-4da6-a306-32caba98a2c3(a)t2g2000yqe.googlegroups.com... On Jul 25, 9:03 am, "Tom Biasi" <tombi...(a)optonline.net> 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 > > 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. Tom
From: Joe on 26 Jul 2010 16:02 In article <606dd3ae-7b34-4da6-a306-32caba98a2c3(a)t2g2000yqe.googlegroups.com>, Bill Bowden <wrongaddress(a)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
From: ehsjr on 26 Jul 2010 23:37
Joe wrote: > In article > <606dd3ae-7b34-4da6-a306-32caba98a2c3(a)t2g2000yqe.googlegroups.com>, Bill > Bowden <wrongaddress(a)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. Nice. Ed |