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Homework Help: Power in voltage/current sources, passive sign convention

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine which of the five sources in Fig. (attached) are being charged (absorbing positive power), and show that the algebraic sum of the five absorbed power values is zero.

    I've labelled how I'm referring to the sources in red

    http://img17.imageshack.us/img17/498/circuitt.png [Broken]

    2. Relevant equations
    p = vi, passive sign convention

    3. The attempt at a solution

    source one: p = (2v)(2A) = 4W generated (not PSC)
    source two: p = (8V)(2A) = 16W generated (not PSC)
    source three: p = (2V)(-4A) = -8W generated (not PSC) --> 8W absorbed
    source four: p = (10V)(5A) = 50W generated (not PSC)
    source five: p = (10V)(-3A) = -30W generated (not PSC) --> 30W absorbed


    So I know I'm wrong because the question wants to know which ONE is absorbing power. And the sums don't add up in the second part.

    I'm clueless, been looking at this problem for a long time now, what have I missed?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 4, 2009 #2
    Edit: caesius, ignore my posts. mplayer seems to have gotten it right...
    Last edited: Mar 4, 2009
  4. Mar 4, 2009 #3
    You have the wrong voltage for Source 3.
    Source 5 seems fine to me unless I'm missing something.

    The rest are correct, just correct source 3 and then:
    (power generated) = (power absorbed)
    (power generated + power absorbed) = 0 W

    If the problem is asking for just one power absorbing element, then I'm not sure what to tell you...I'm getting 2 elements absorbing, and the numbers add up to zero. Hopefully that can help, though. :uhh:
  5. Mar 4, 2009 #4
    Maybe, I should shut my mouth now. :redface: My method seems to be wrong...
    Last edited: Mar 4, 2009
  6. Mar 4, 2009 #5
    Source 2 seems OK by me. Positive current flowing from the negative terminal to positive terminal on the voltage source. That generating power P = (2A)(8V) = 16W.

    I don't think there's a problem with Source 5, it looks like the current and voltage parameters are explicitly defined there. You may be seeing a problem I'm not though, what do you think?
  7. Mar 4, 2009 #6
    All currents flowing into that top node are summing to zero so that's good:

    2A - 4A + 5A - 3A = 0A

    I don't think there are any other effects on Source 5.
  8. Mar 4, 2009 #7
    Source 3 is a current source in parallel with a voltage source, Source 4. The specified current is moving a negative terminal, towards a positive one. That satisfies the passive sign convention. Now you can calculate its power. You have the current, (-4A), and you have the voltage across the current source, since it is connected in parallel with voltage source 4 at 10V.

    P = (-4A)(10V) = -40W = 40W absorbed

    Source 5 is a current source of (-3A) moving from negative to positive terminals through the element. This satisfies the passive sign convention. The voltage across that device is given, 10V.

    P = (-3A)(10V) = -30W = 30W absorbed

    Did that help?
  9. Mar 4, 2009 #8
    Nope. I can't understand, why the voltage across the supplies 3 and 5 is 10V. I understand, that the supply 4 creates 10V across the both of them, but then there's supplies 1 and 2, which should create 10V across both of them. So their voltage ought to be 20V, but that doesn't make [tex] \sum P=0[/tex]
    Last edited: Mar 4, 2009
  10. Mar 4, 2009 #9
    It might be better for you to redraw the circuit so yo have two vertical lines with the current source and 10V voltage source for clarity. But the question is only about the power at the particular sources, so all we are considering here are voltages across a single source, and currents through a single source.

    Source 3 is in paralell with 10 V Source 4.
    Source 5 has a 10V across the current souce becasue it was labeled that way.
  11. Mar 4, 2009 #10
    But it's in parallel with supplie 1 and 2 as well. So using superposition, the voltage should be 20V.

    Then it's obviously wrong. :biggrin:

    I guess the same methods don't apply with supplies and resistors.
    Last edited: Mar 4, 2009
  12. Mar 4, 2009 #11


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    I don't think it asks for which one is absorbing, it asks which elements are absorbing.
  13. Mar 4, 2009 #12
    Okay, just to let you know, I had a total brain freeze there. Obviously you can't use the superposition method here, since all the elements are supplies. You just need to make sure each loop has [tex] \sum V=0[/tex].
  14. Mar 4, 2009 #13
    Thanks mplayer you're great, I'm just skimming over what you've written but got confused by the above quote, we've been taught in class that the PSC convention is current entering the positive terminal, not the other way around...?
  15. Mar 4, 2009 #14
    What I said was confusing, sorry about that. I was trying to say that it was the correct convention for the assumption that the particular source was going to generate power. If the answer came out positive, then the assumption was correct and the source is generating, if the answer came out negative, then the assumption was incorrect and the source is absorbing.

    For example:
    In Source 3's case, my assumption was the source was generating power. The convention for power generation is current flow from - to + terminal. I calculated power by P = (-4A)(10V) = -40W. The answer came out negative, so my assumption of power generation was incorrect. Therefore, the source is absorbing.

    I hope that's clear and am not confusing you more. :yuck:
  16. Mar 4, 2009 #15
    Thanks, that makes sense. Another question though, looking at source three, it looks to me like there are two voltages over it, 2V from the very left hand side and 10V from the voltage source. Why do you pick 10V over the 2V?

    As far as I understood everything in parallel has the same voltage drop, how can there be two DIFFERENT voltage drops I guess is what I'm asking...

    Thanks again
  17. Mar 4, 2009 #16
    Source 3 is not in parallel with Source 1. But, notice that Source 3 is in parallel with the combination of Source 1 and Source 2. If Source 1 and Source 2 were combined, it would have a potential difference of 10 V, just like Source 3 and Source 4. Does that make more sense?
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