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Homework Help: Which resistor produces the most heat?

  1. Oct 11, 2015 #1
    1. The problem statement, all variables and given/known data
    fb2b563dc9.jpg
    Find the resistor that generates most heat.

    2. Relevant equations
    Principle of superposition.
    Voltage Div
    Current Div.
    Power = I^2R

    3. The attempt at a solution
    After aplying principle of super position and considering the voltage source in one scenario with current source open circuited and calculating current for each resistor and then short circuiting the voltage source and then again calculating each current. After that I added all currents for each individual resistor and caclulated the power output of each individual resistor by I^2R. The highest wattage I got was for the 5 ohm resistor (48.05W). So I concluded that the 5ohm resistor produces most heat. Is this correct? I have no solutions for this.
     
  2. jcsd
  3. Oct 11, 2015 #2

    BvU

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    I wonder how you can get so much current through this 5##\Omega## resistor. I do agree it's the one that generates the most heat,
     
  4. Oct 11, 2015 #3

    phinds

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    Off the cuff, that doesn't sound like the right thing to do, but then that's probably because I'm not used to using that technique. I just do loop analysis and find all the voltages and currents with both sources active. If your calculations are right then you should have the right answer.

    I don't understand the contradiction in your two statements, one of which gives an answer and the other of which says you don't have an answer.
     
  5. Oct 11, 2015 #4
    Sorry. I meant to say that I have no given solutions for this. The 5 ohm answer is my solution. My fault there
     
  6. Oct 11, 2015 #5

    phinds

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    Ah. Thanks for that clarification. That makes sense. With just a quick look, and no calculations, I'd say your answer is right.

    EDIT: well, that's embarrassing. I should have seen immediately that you don't even have to DO an calculations (that is, no circuit analysis). The power of each resistor is obvious at a glance and yes, the 5 ohm is the largest. Do you see why I say that? That is, do you see why I say no circuit analysis is required (mesh, loop, or any other kind).
     
  7. Oct 11, 2015 #6
    No. I don't see it. Keep in mind circuit analysis is really new for us. We've done only KCL, KVL, Superposition and Thevenins Theoram.
     
  8. Oct 11, 2015 #7

    gneill

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    The thing about ideal voltage sources is that they will produce or absorb ANY amount of current in order to maintain their specified potential difference. That potential difference cannot be changed, no matter what.

    The thing about ideal current sources is that they will present ANY amount of voltage, positive or negative, in order to maintain their specified current.

    Now consider the situations with components that are in parallel with, or in series with, such devices.
     
  9. Oct 11, 2015 #8

    CWatters

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    Then perhaps recall p=v^2/r or p=i^2r you don't need to calculate both I and v for each resistor (although you could).
     
  10. Oct 11, 2015 #9

    phinds

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    Yes, that's true. How does it apply to the question at hand?
     
  11. Oct 12, 2015 #10

    CWatters

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    For much the same reason gneil said in #7..

    If you have a resistor in parallel with a voltage source the voltage is known and it's a no-brainer to use p=v2/r. (eg on the 5R and 20R)
    If you have a resistor in series with a current source the current is known and it's a no-brainer to use p=I2r (eg on the 10R and 15R)
     
  12. Oct 12, 2015 #11

    phinds

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    Yes, *I* know that, I was trying to get the OP to see it.
     
  13. Oct 13, 2015 #12

    epenguin

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    Seems to me that minimum calculation is what is being looked for here. We have the same current going through two resistors so using I2R it cannot be the 10 Ω on that generates the most heat, it might be the 15 Ω one.

    Loking at the parallel branches what is their joint conductivity? It's 1/5 + 1/20 = 5/20 = 1/4 Ω-1. So the total conductivity of this parallel branch is more by a large margin than that of the 1/15 Ω-1 conductor in series with them so the latter is still the one that heats the most. There is a higher resistance (20 Ω) in there but not so much current goes through it. This however is not all that obvious just looking at it, I mean it is obvious it could be that way but not obvious that it must be, could it be different with different resistances? - so I am wondering if there is not a still more powerful approach to this with maximum power principle or something.
     
  14. Oct 13, 2015 #13

    BvU

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    Diamond is long gone: he/she has the answer (post #4) and moved on.
    I posted a synopsis/circuit analysis result of what gneill and CW pointed out but it was moderated to the black hole for reasons of PF rules violation.
     
  15. Oct 13, 2015 #14

    epenguin

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    If the too easily satisfied students go away once got the right answer, that's their loss. :oldsmile: We can still have a hopefully enlightening scientific discussion. In this case after a few minutes I decided now there are no great profundities. For more heating in the parallel branch each conductance less than half, I.e both resistances more than twice, of one in series is sufficient for more total heat in the parallel than the series resistor. Both conductances more than half, both resistances less than twice, sufficient for more heating in the series resistor. One smaller and one higher than those specifications then look at total conductance. That's for total heating in parallel circuit. General condition for more heating in just one of the parallel resistors is - left as an excercise for students. :oldbiggrin:
     
  16. Oct 13, 2015 #15

    BvU

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    Diamond doesn't loose anything, especially not if he/she has read gneill's post and connected it with phind's #5 : For the 5 and 20Ω, V is given and for the 10 and 15 I is given.

    Find ep reasoning hard to follow, except when I leave out the voltage source....
     
  17. Oct 13, 2015 #16

    epenguin

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    Because you can! Leave out the voltage source.

    If my answer is right, it remains so for the circuit independent of any voltage or current source and their nature. :oldsmile:
     
  18. Oct 13, 2015 #17

    BvU

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    Because I can what ? Leave out the voltage source ? I surely can not.
    Leaving out the voltage source causes the 15 Ω resistor to be the one that generates the most heat.
     
  19. Oct 13, 2015 #18

    epenguin

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    I say that's what you'll always get. Can you show me a way, a condition, to get any other result from this circuit?
     
  20. Oct 13, 2015 #19

    BvU

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    Leave in the voltage source and the 5 ohm resistor dissipates the most power.
     
  21. Oct 13, 2015 #20

    donpacino

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    are you saying the 15 ohm resistor generates the most heat???

    P=IV=V*V/R=I*I*R

    using those equations you can clearly see that the 5 ohm resistor will dissipate the most heat.

    you can't 'leave out' the voltage source you can only ZERO it for use in superposition analysis.
     
  22. Oct 13, 2015 #21

    epenguin

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    OK I have not been paying close enough attention. :redface:

    Isn't there trouble because we have been given a physically impossible situation? You cannot have the 10 V and 1 A shown in the illustration at the same time.

    I had just been looking at currents and you at voltages. Can we declare a draw? :oldbiggrin:
     
    Last edited: Oct 13, 2015
  23. Oct 13, 2015 #22

    CWatters

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    Yes you can.


    Power.jpg
     
  24. Oct 14, 2015 #23

    BvU

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    Dear CW, please be careful. I got cautioned and my post deported to the black hole for a less explicit account ! PF rules - when taken strictly - don't allow posting full solutions.

    I still consider that a bridge crossed a long time ago already and by now this thread is a private/public learning experience for ep, which is just fine
    (for a simple reason: I had the very same experience as ep - and to some extent Paul also - while trying to sort out how to look at this exercise in the proper way. gneill #7 was an eye opener or rather a fog lifter to show the easiest way right through. And it sort of rekindles my interest in more challenging circuit analysis exercises - if I only had time)

    So: ep, you happy and in control now ?
     
  25. Oct 14, 2015 #24

    CWatters

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    Thanks for the heads up.
     
  26. Oct 14, 2015 #25

    epenguin

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    Yes some learning experience, apologies if necessary, distractions or a speck of dust had surrounded my brain and for some reason I was considering the left side of the circuit as a capacitor or something essentially not there. Some things carry on through threads and one of a few days ago about how it is inaccurate to say "currents follow path of least resistance" had been working, so to say "the smallest resistance heats the most" is inaccurate if you have two parallel resistors that feed into a resistor in series with them. But that is in the case the same current flows through the parallel pair as through one in series with it. I think that holds up though no doubt obvious and familiar to you. In our case though it is not the same current through two sections, it does have somewhere else to go after the parallel pair.


    The reason I thought them incompatible and still do not understand how they can be compatible is that if 1 A flows in the right hand branch there is a total voltage drop of 25 V over the 25 Ω resistance between the points where it joins the rest of the circuit whereas looking at the left had side the voltage drop between the same points is supposed to be 10 V. Can you explain this?
     
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