1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Power dissipation in parallel circuit

  1. May 21, 2012 #1


    User Avatar
    Homework Helper

    Right, so I was going over some old (basic) electromagnetism principles (to make sure it doesn't all fall out of my head). And I was thinking about two resistors in parallel, assuming some constant dc voltage across them. And I was thinking about the power dissipated by this simple circuit. And I came across something I thought is a bit odd. So anyway, I'll explain my calculation:

    For resistors 1 and 2, we know the power dissipated by each is [itex]I_1^2 R_1[/itex] and [itex]I_2^2 R_2[/itex] so the total power being dissipated by the circuit is:
    [tex]P= I_1^2 R_1 + I_2^2 R_2 [/tex]
    We also know that the sum of currents going into a node equal zero. So let's suppose we had current [itex]I[/itex] going into the circuit, which is then split into [itex]I_1[/itex] and [itex]I_2[/itex], so that:
    [tex]I = I_1 + I_2 [/tex]
    Let's suppose from here that [itex]I[/itex] is set. So how much current will go through one resistor and how much will go through the other resistor? We could work this out by using Ohm's law, but I tried out something else that comes up with the same answer. If we simply try to minimise the power lost by the circuit, then we will be given the correct answer. So here's my calculation:
    [tex]dP = 2I_1 R_1 dI_1 + 2 I_2 R_2 dI_2 [/tex]
    And we also know that [itex]dI_2 = - dI_1[/itex] since we are saying the current [itex]I[/itex] is set. So using this, we get:
    [tex]dP = 2(I_1 R_1 - I_2 R_2 )dI_1[/tex]
    So from this, we see there is a stationary point when [itex]I_1 R_1 = I_2 R_2 [/itex] So is this a minimum, or what? Well, we have
    [tex]\frac{dP}{dI_1} = 2(I_1 R_1 - I_2 R_2 )[/tex]
    So now, take the total differential of this with respect to [itex]dI_1[/itex], keeping in mind that [itex]dI_2 = - dI_1[/itex] and we get:
    [tex]\frac{d^2 P}{d I_1^2} = 2(R_1 + R_2) [/tex]
    So as long as the resistances are positive, the stationary point at [itex]I_1 R_1 = I_2 R_2 [/itex] will be a minimum!

    So what the heck does this all mean? I started with the equation for power dissipated by a resistor, and the rule saying that currents 'add up' at a node, and I said that the dissipated power must be minimised, and that told me that [itex]I_1 R_1 = I_2 R_2 [/itex] So the current flows through this circuit according to the principle of minimising the dissipated power!

    So what do you all think? Is this just a coincidence, or is there some simple explanation that I've overlooked, or is this an example of a principle that I am unaware of? Thanks in advance :)
  2. jcsd
  3. May 21, 2012 #2
    Not sure what you mean by this? since the voltage across circuits elements in parallel is the same by Ohm's law

    I1R1 always equals I2R2

    In fact it means that the currents are distributed in inverse proportion to the parallel resistors

    [tex]\left( {\frac{{{I_1}}}{{{I_2}}}} \right) = \left( {\frac{{{R_2}}}{{{R_1}}}} \right)[/tex]

    But yes your energy derivation is an example of the shakedown theorm.
  4. May 21, 2012 #3
    The fault here is that if the 2 currents I1 and I2 are changing then the resistances should also change (since I is constant) and can't be constant as at any instant I1*R1 must be = I2R2.
    And if I changes, then only ∂I1 will be = ∂I2.

  5. May 21, 2012 #4


    User Avatar
    Science Advisor

    ^ Oi, these guys totally missed the point.

    But yeah, it makes perfect sense. Lets say the circuit runs for some time τ. Some quantity of charge q has flown through it. The total energy dissipated is ∫Vdq = ∫Pdt. What would it mean for power to be unoptimized? It would mean that one of these elements dq would be able to find a path through your circuit with lower V. That would mean that different branches have different voltages. So the solution must certainly be a minimum in P in a steady state circuit.

    Now, for AC circuit, I expect things would be more interesting. Might still work out if you optimize RMS power, though.
  6. May 21, 2012 #5
    But how i am wrong in saying that the resistance should also change ?
    Here since I is constant so in order to change the 2 currents we need to change the resistances also.
    Please xplainn it to me sir K^2.
  7. May 21, 2012 #6
    I hope you weren't referring to my post.
  8. May 21, 2012 #7


    User Avatar
    Science Advisor

    Both of you are using the fact that V is the same in parallel circuit. That's why one of you is saying R must change, the other that I1R1=I2R2 is trivial. One or the other must be true if you keep both the Kirchhoff rules. But the whole point is that you keep the junction rule only and replace the loop rule with energy dissipation minimalization, which is an equivalent condition. In that case, condition I1R1=I2R2 must be derived.
  9. May 21, 2012 #8
    Indeed, according to Prigonine's theorem, energy dissipation minimalisation is the
    expected result in a steady-state process, at least as long as the system isn't too far
    from equilibrium(*) (for instance, if it can be considered to be in local thermodynamic

    (*) If I remember correctly, anyway.
  10. May 22, 2012 #9
    You entered this thread with a personally derogatory remark.

    When asked specifically about this you confirmed you intended it to apply to me.

    You then repeated in 6 lines what I had said in one line.

    I did not mention Kirchoff, you did, and his rule is not needed.

    First take the definition of parallel - both resistors are connected to the same pair of nodes or across the same voltage.

    Consider the voltage across these nodes and apply Ohm's law, as I did say leads to the expression I1R1=I2R2.

    If you wish to look up energy methods you will find reference to the equivalence in mechanical and electrical systems (the shakedown theorem) in the works of Horace Lamb and Arthur Morley, well before Prigogine.

    I also consider the lack of response to Zubeen, who asked politely where he was wrong, insulting.

    So here is an explanation for him.

    The value of the resistors was set as part of the original statement, as was the configuration, and cannot be varied.
    However the voltage applied was not set and therefore neither were the currents, so Bruce was quite correct in varying these.
  11. May 22, 2012 #10
    @ Sir Studiot,
    OK then let R(1)/R(2) = k (constant)
    then since I(1)R(1)=I(2)R(2) then it applies that,
    I(2)/I(1)=k , differentiating w.r.t I(1) we get
    is this the point where Bruce is missing ?
    and also since Io is constant (in derivation of Bruce) Vo must be constant and hence, the only mechanism to change currents in the 2 wires is by changing there resistance.
    in short, this making resistance constant while changing the current is disturbing me....
    please sir i beg your help to understand this, please ... :cry:
  12. May 22, 2012 #11
    Bruce's derivation preserves Kirchhoff's current law but violates Kirchhoff's voltage law because if you vary the current in the two parallel branches, the sum of voltages in all the three loops don't sum up to zero. You're not allowed to do this in the first place, so I'd say your derivation is meaningless and potentially confusing.
  13. May 22, 2012 #12
    Hello zubeen,

    Firstly I don't think anyone (Bruce and K2 included) is missing anything.
    What I think is happening is that we all, in general, are showing the self consistency of the set of equations that apply to electric circuits.
    For simple circuits you do not need to employ all of them to 'solve' the circuit so there is more than one way to calculate.

    As to the question of resistors:

    A good way to proceed in technical lines of thinking is to start with definitions and then apply the rule or process. Of course these must be applied within the boundary conditions.

    We have here the definition that the resistors are in parallel.

    This definition requires that the voltage across each is the same.

    So by simple application of Ohm's law we can determine the current in each resistor.

    We could also determine it from the power equation V2/R.

    If we applied a different voltage the currents would be different.

    Bruce did not specify a set voltage, he did specify (I think) two set resistors R1 and R2.

    So we can vary the current by varying the voltage. This also varies the power by the above formula.

    So I am not saying that you could not, in general, vary the resistance. Yes you could, and that would indeed vary the current for a fixed voltage.
    I am saying that I think that Bruce fixed the resistance, not the voltage.
    I am saying this because we are considering a fixed system (set of parallel resistors) and varying the impressed load (which is a voltage).
    So we are studying the effect of varying the load on a (fixed or given) system to arrive at the conclusion that the response of the system is to minimise the energy or power in some way. This is a common and important conclusion that covers a wide variety of physical systems. It just happens that it is not used so often in electrical engineeering because other methods are simpler.

    Either way we take the independent variable as voltage or resistance and the dependent ones as current or power.

    Does this help?
    Last edited: May 22, 2012
  14. May 22, 2012 #13
    He fixed both the resistance and the voltage(he said that the current I is set). In that case, varying the current I1 is simply wrong
  15. May 22, 2012 #14
    But sir in bruce's derivation -
    since I is set, so V must also be Set (this must be because we are saying that R1 and R2 are set so net resistance R0 must be set and since I is set so V=IR0 must be set ) ?????
  16. May 22, 2012 #15
    Vin300 and Zubeen, good points.
    Too many parameters are fixed here and there is little valour in proving that

    0 = 0

    So instead of squabbling over presentation the thing to take away from this thread is the minimum energy response principle exhibited by many systems in many situations.
  17. May 22, 2012 #16
    I am still a bit confused about what is happening here.
    Can you please suggest me the waty i should think, or some questions i should ask myself or can you tell me the point i am wrong it.
    This derivation is nothing important to me, but what is important here is my understandings about this topic, since i am not understanding the thing you all are trying to say, i must be wrong at some point, i just want to correct that point.
    Please help me on it .......
  18. May 22, 2012 #17
    Why should this be so?

    You could set R1 at 10 and R2 at 100, 1000 or whatever.

    It would still be true that

    I1R1 = I2R2

    Why do you think Bruce was wrong?
  19. May 23, 2012 #18
    Really neat observation.
    Too bad there is so much QRM on this thread.
  20. May 23, 2012 #19
    My point was that the derivation requires varying the current in the branch while maintaining the voltage constant, which is not the right thing to do.
  21. May 23, 2012 #20
    Not quite sure what you mean.

    Differentiating a constant is a perfectly respectable operation.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook