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Calculating Power - Simple DC Circuit

  1. Mar 2, 2014 #1
    1. The problem statement, all variables and given/known data

    http://puu.sh/7gYwk.png [Broken]

    Consider the circuit shown above. Find the current Ia flowing through the resistor and the power for each element in the circuit. Which elements are absorbing power?

    2. Relevant equations

    P = V x I = I^2 x R = V^2 / R

    3. The attempt at a solution

    I tried to find Ia first, the voltage source with the load produce 2A going the opposite direction from the current source, so I got Ia = 0A and makes all the power = 0W

    Is this correct?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 2, 2014 #2


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    Staff: Mentor

    Hi JosefMTD! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    Is that a constant current source in your circuit? What do you understand is the behaviour of a "constant current source"?
    Last edited by a moderator: May 6, 2017
  4. Mar 2, 2014 #3
    I think it's a constant current source.

    It's always of the same value and stays on the same direction, does it override all other current source though? I have no idea, I may have missed this during class :-\
    Last edited: Mar 2, 2014
  5. Mar 2, 2014 #4


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    Staff: Mentor

    Yes, the constant current source always wins! Everything else must accommodate the dictates of a constant current source.

    So you never arrange two where there would be conflict, or there will be fireworks!
  6. Mar 2, 2014 #5
    Well, thank you very much. I must have fail to listen carefully during class while this happens :P

    But then,
    Power on resistance = (2A)(2A)(5Ω) = (10 V)(10 V)/(5Ω) = 20 W
    Power on voltage source = (10V)(2A) = 20 W
    Power on current source = (-10V)(2A) = -20 W

    It adds up to 20W, but according to conservation of energy: ∑p = 0 for a closed circuit. My "gut" feeling says the voltage source should have 0 W power, dunno why. So anyways, where did I do wrong :eek:?
  7. Mar 2, 2014 #6


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    Why did you decide to throw in that negative sign when calculating the current source's power output?
  8. Mar 2, 2014 #7
    Because of the active sign convention.

    I did it like I did on the previous question:
    http://puu.sh/7h4PK.png [Broken]
    Last edited by a moderator: May 6, 2017
  9. Mar 2, 2014 #8
    By the way, can you offer an explaintion of why you believe the current source has a voltage of 10v?
  10. Mar 2, 2014 #9
    I'm taking reference from the previous question which is shown up there, dunno, I figured it should be the same, or it shouldn't? :eek:
  11. Mar 2, 2014 #10
    Well this one has a resistor in the circuit, so that makes it a bit different.

    If you take the voltage at the bottom of the voltage source as V0 = 0v, then work around the circuit and calculate the voltage at each node.
    At the node bewtween the voltage source and the resistor, V1 = ?
    At the node between the resistor and the current source, V2=?
    At the node between the current source and the voltage source V3=?
  12. Mar 2, 2014 #11
    Oh, I see.
    V1 = 10V
    V2 according to KCL
    ∑I = 0, V2-V1/5Ω - 2A = 0
    V2 - 10V = 10V, V2 = 20V
    V3 according to KVL
    V2-V3 = V3-V0
    20V = 2V3
    V3 = 10V

    Damn, I'm confused :confused:
  13. Mar 3, 2014 #12


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    Staff: Mentor

    How many volts across the constant voltage source?
    How many volts across the resistor itself?
  14. Mar 3, 2014 #13
    I guess that is confusing, when you put it that way.

    I thought you would have recognized the following:
    V3 = V0 = 0 since they are the same node.
    You really do not have to do a KVL calculation on it.

    With KVL you can do:
    V2-V3 = V2-V0 ie summation of voltages going left from V2 = going right from V2
  15. Mar 3, 2014 #14
    1. 10V
    2. With using KCL on the node between current source and resistor, I got 10 V across the resistor

    I don't know why, but I never had problems in solving these kinds of problems, I think I forgot all of the basics because I had a too long vacation :P
  16. Mar 3, 2014 #15

    V2 = 0 V
    (V1 - V2)/5 ohm = 2A [KCL]
    (10 V - V2) = 10 V
    V2 = 10 V - 10 V = 0 V

    No voltage across current source.

    Do we use the generator-load convention to make the 20 W on voltage source to be generating power not absorbing because it's positive?

    Thank you guys so much :D
    Finally, I think I'm getting to know the basics again XD
  17. Mar 3, 2014 #16


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    Gold Member

    Really? I think maybe you still have a little work to do on the basics.

    Also, as nascentoxygen alluded to in post #4, if this were a real world situation and the voltage source was a battery, it would most likely explode. Do you understand why?
  18. Mar 3, 2014 #17
    Because the current generated from the constant voltage source would always make it 2A to the opposite side of constant current source. Ka boom.
  19. Mar 3, 2014 #18
    You already had across the voltage-source 10v, and across the resistor 10v.
    so how do you now get 0v????

    (V1 - V2)/5 ohm = 2A [KCL]
    (V2- V1)/5 ohm = 2A [KCL]
    in the direction of the current.
  20. Mar 3, 2014 #19

    Yeah, I messed up XD
    I realized where I did wrong XD
    Thank you guys for bearing with me ;D

    (V2 - V1)/5 ohm = 2 A
    V2 - V1 = 10 V
    V2 = 20 V

    V3 = 0 V

    Voltage across the constant current source = 20 V
    with the active sign convention, the power on constant current source is - 40 W.
    with the passive sign convention, the power on constant voltage source is 20 W
    the power on resistor is 20 W

    Summation of all the power in components is 0. I'm a bit sure that this one is right, now ;D
  21. Mar 3, 2014 #20
    OK. That's fine.
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