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Very easy circuit question.

  1. Oct 3, 2006 #1
    I understand everything in the solution of this problem, except the expresson
    [tex]2I_{x}I_{2}=16W [/tex] supplied. can some one please tell me whats going on here?

    I thought that Power was the product of current and voltage. How can I get power out of the current source in the center of the mesh?


    [​IMG]
     

    Attached Files:

    Last edited: Oct 3, 2006
  2. jcsd
  3. Oct 3, 2006 #2
    I cannot see your document (approval usually takes a while), but I have a question: is the 2 referring to 2 ohms of resistance? Based on your description, it seems like the 2 is 2 ohms of internal battery resistance.

    Also, remember that:
    V = IR
    P = IV
    Therefore [tex]P = I^2R[/tex].
     
  4. Oct 3, 2006 #3
    No, sorry there are not resistors in the diagram.
     
  5. Oct 3, 2006 #4

    berkeman

    User Avatar

    Staff: Mentor

    Attachment is still pending approval, but you get power from a current source based on the voltage that is across the current source as it supplies the current. If you supply current into a 10V load, that's 10x the power compared to if you source that current into a 1V load.
     
  6. Oct 3, 2006 #5
    Can anyone see the image that I have linked to?
     
  7. Oct 3, 2006 #6
    Its approved now.
     
  8. Oct 3, 2006 #7
    any suggestins now that you can view the file?
     
  9. Oct 3, 2006 #8
    Are you doing mesh power(???) analysis?

    Anyway, to answer your question, the [tex] 2I_x I_2 = 16W[/tex] is a power because it uses the dependent voltage source, which has a voltage of [tex]2I_x[/tex].

    Any other questions?
     
  10. Oct 3, 2006 #9
    I see now. Thank you!!
     
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