Power Division

Yes, simple algebra. power = v^2 / R or I * V or I^2 * R Solve for the current You do it.

If you need another hint:

1) Write the equation for total power in terms of total resistance and current. 2) write the equation for R1 power in terms of R1 and current. 3) Use the two equations above to get the ratio you want.

Ptotal =I^2(R1 +R2) =I^2R1 +I^2R2 =P1 +P2 so P1/Ptotal =R1/R1+R2 Without knowing I (=V/R1+R2) at least you get a ratio Is this what you want?

Thank you. I was looking for a simple equation using voltage and resistance terms only. What do you think of the following?

R1=5 ohms R2=10 ohms |--------\/\/\/-------\/\/\/---- | | E = 30 volts | |-------------------------------

In case ASKI gets screwed up, two resistors (5 and 10 ohm) and

30 volts all in series.

I=E/R1+R2 = 2 amps P total = I^2 R1+R2 = 60 watts PI = I^2 R1 = 20 watts P2 = I^2 R2 = 40 watts Check: Total power being dissipated = P1+P2=20+40=60 watts

Now the equation I came up with using E and R only:

P1 = (E/R1+R2)^2 * R1 = 20 watts P2 = (E/R1+R2)^2 * R2 = 40 watts

Note the first term is just the current squared. No monkeying around "hunting" current and working it in. How does this look to you? TIA.

JC the elder

Thanks.

Thanks. All hints appreciated.

In fact, you have "hunted" for and found the current as E/(R1+R2)=2 Amps =I no matter what you call it so you have done exactly the same here as done above. What are you gaining? By explicitly using I you will have a better conceptual understanding rather than simply plug in and turn the crank

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But I ask for a simple equation and your answer was convoluted. I would have thought that you would have picked up on the following:

You said: P1/Ptotal=R1/R1+R2 then P1=Ptotal*(R1/R1+R2) or P1=(E/R1+R2)^2*(R1+R2)*(R1/R1+R2) (messy route) Cancelling (R1+R2) P1=(E/R1+R2)^2 R1 (note you did not mention this)

Einstein said things should be simple as possible, but no simpler. :-)

Anyway.. If one doesn't know that I=E/Rtotal, he probably shouldn't be trying to apportion power anyway. Might get shocked.

JC the elder

Least convoluted: I=V/(Rtotal) (1) Pn =(I^2)*Rn for resistor n (2)

Can it be simpler?

------

Least convoluted? Your original response was

"Ptotal =I^2(R1 +R2) =I^2R1 +I^2R2 =P1 +P2 so P1/Ptotal =R1/R1+R2"

And still no provision for current. Arghhhh.......

Also yout last response above is misleading. You left out the step to sum R1 and R2. When this step is added, your method is:

P1=V/(R1+R2)^2 R1 My method was:

P1=(E/R1+R2)^2 R1

You tried a bit of artfulness by leaving out the summing step, and I had to lead you to your arrangement of terms, therefore I win. :-)

JC the elder

I believe you wanted to make a very brief equation, at least that's what you did, but you "cheated" leaving out provision for current.

Are you kidding? You would have had to remember the equation for current plus derive it in your original equation:

"Ptotal =I^2(R1 +R2) =I^2R1 +I^2R2 =P1 +P2 so P1/Ptotal =R1/R1+R2"

Sorry but this was strung out, messy, and no provision for current, whereas my equation gave everything needed:

"P1 = (E/R1+R2)^2 * R1 = 20 watts. Note the first term is just the current squared."

Please explain how your original equation would show a newcomer to "know what you are doing" better than mine, which *does* give current and even briefly noted how by adding " Note the first term is just the current squared."

Now you parrot *exactly* what I did, I reduced the baggage with nothing pertinent left out. You should have taken your own advice.

Sorry, but "Ohms law" type of equations are not trivial. In fact, they are so basic there is no need to confuse a situation with long and strung out expressions that can be made simple as Einstein suggested. Therefore I still suggest that I won. :-)

JC the elder.

The relationship for resistors in series is not Ohm's law . It depends on Kirchoff's Laws and the concept of linearity- these are not trivial but not exactly difficult. Would it not be better to concentrate on Kirchoff's laws and also the rather intellectually trivial "E=IR" (which is not ohms law unless R is constant)?. Learning these would eliminate the necessity for a long list of formulae- replacing rote memory by some thinking/ understanding.

Please also note that my original reply was to your original question and was not intended to produce a neat canned formula but simply to point out a relationship which you apparently could not figure out at the time. . I could have given your result directly but chose not to, on the assumption that you knew how to find the current I. I may have misread your query- if so- I am sorry - but my point still stands. If you "have won" it is because you appear to have learned something. Fair enough.

Bye, have fun

And where did I say it was?

??? Ohms law states: R=V/I. R is the variable, certainly not constant.

My solution was not a "long list of formulae" is was simply: P1=(E/R1+R2)^2*R1

Sorry but I did ask "Is there a simple power divider rule for two resistors in series?" Note the word simple. You gave a rather convoluted reply.

Did you not note the smiley, i.e. :-) ? In any event, please state specifically what you taught me. Should be interesting. :-)

JC the elder.

------ You didn't and I may have read this into what you said when you went on about the complexity of Ohms law.>

Strictly speaking "Ohm's Law" as originally given and still valid holds for R constant. It described a relationship, found by experiment (at the time and for the materials considered) which was a linear relationship between E and I --that is R independent of either E or I. Nonlinear materials do exist but for such materials and then R is dependent on current or voltage. In such a case, the standard circuit analysis theorems are ratshit except for Kirchoff's Laws as they are based on linearity. In the simple series circuit it is not a big problem.

Example: take R1= 2 ohms and R2 =2*(I*0.5) and what is R1 + R2 -- You can say that E =2*(1+I*0.5)*I - so Rtotal =2(1+I*0.5). Fine- Now, given E=10V, what is I? , what Rtotal do you use to find I? In fact the concept of "R" is rather useless,except for visualisation, in this case Now consider the case where a "real" circuit with a nonlinear element exists- it gets messy.

No. I "went on" about the complex way you responded.

Halliday (6th ed.) says Ohms law states that current is proportional to the potential applied. The example I gave (which you snipped out) clearly follows this rule. Sorry, but there is no need for all the "complexity" you are into in ***the example I gave***.

Thanks, but I had it worked out 50+ years ago. Had intended the post as a ***lead-in*** for more complicated stuff - just building up to such.

Mr. Kelly, if I may........ I have respect for your abilities. Seriously, you are damned sharp. However, you tend to be a bit condescending at times, with assumption the other person is unlearned at best... Plus.. (my opinion follows)... your store of knowledge gets in the way of Mr. Einstein's credo that things should be simple as possible. In any event, thank you for your time and response.

JC the elder

In spite of what I said before, I have to respond to this. I have no problem with Halliday but simply pointed out that Ohm actually came up with a "Law" which was that E is directly proportional to current- Halliday simply left out the "directly' as many others have done. I also attempted, unsuccessfully that generalising R to a non-linear function of current or voltage leads to the elimination of the basis of most of circuit analysis- linearity.

If I have pontificated, I am sorry. I tend to do so.

If I disagree with you regarding what is simple - that's life. Nothing that I have said, in this situation, is complex, but I simply don't believe in canned formulae unless I know just how they are obtained (and if I know that- why bother?).

If you want to discuss deeper aspects- fine by me- but put in a stronger lead in ( I also have had over 50 years experience-on both the giving and recieving end- so that makes us both old farts:).

Possibly under a different thread title. Cheers,

Agreed, but you apparently miss my point. What you wrote was more complex then needed. Sorry, but I will not indulge in this roundabout game of math. My goal is to obtain the result with the path to it made short and clear, and Ohms law is more instrumental in the process than any light I can shine on my math capabilities.

Exactly how strong would you like it to be? :-)

I try to hold farting to a minimum, note. It disturbs the cat.

JC the elder

As to complexity - yes - i did give a long answer and pointed out a way for you to go assuming your original query was genuine. In fact, the approach that I gave later, which you found complex required exactly the same numerical steps and effort as your canned formula and was just as explicit, with, in fact, for most, less memorisation. However, there is likely no point going after that any more.

As to Ohm's Law- please recognise that you were, in fact , using something a bit more fundamental than Ohm's Law which is simply the expression of a

• relationship between the voltage across an energy dissipating element and the current through it - i.e valid for what is called an "ohmic" resistance. It is, in practical terms, no more than a useful mathematical idealisation (as is all circuit theory). Math is involved, whether or not you like it - particularly when you go beyond elementary DC analysis.

As for deeper aspects- Whatever strength you want. :)

• references on hand- can be quoted if you wish.