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Find the Electric power of the Current source

  1. Nov 29, 2011 #1
    in the picture you can see the electrical network.
    in blue is the Current of each side of the electrical network. (did it with loop equations)
    i asked to find the Electric power of E and the Electric power of the Current source.

    so for E i found it and i got pe=11.4w.

    i also know what is the Current in each resistor.

    i have no problem to find the electric power of voltage source, but for current source i need some help.

    thanks.
     

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  3. Nov 29, 2011 #2

    Redbelly98

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    That's good. What is the current in R4? And therefore, what is the voltage across R4?
     
  4. Nov 29, 2011 #3
    the current in R4 is 0.2A and the voltage across R4 is 0.2*8=1.6v

    do i correct ?

    thanks.
     
  5. Nov 29, 2011 #4

    Redbelly98

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    I don't know if it's correct (unless you want to show all your work so far), but knowing the voltage across R4 should help you with finding the power produced by the current source.
     
  6. Nov 29, 2011 #5
    ok so just tell me how is help me to find the current source?

    the final answer need to be P( of the current source)=36.9w.

    thanks
     
  7. Nov 29, 2011 #6

    gneill

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    You need to find the voltage that appears across the current source. A path can be drawn from one terminal of the current source through RL and R3 and back to the current source. Use KVL around that path.
     
  8. Nov 29, 2011 #7
    the voltage that appears across the current source, i get 18v (if i include R4)

    but if i take only RL and R3 i get 16.4v and then multiply it by 2.25A i get that the Plo=36.9w and for real this is the correct answer.

    can you explain me why i not need to include R4?

    thank you.
     
  9. Nov 29, 2011 #8

    gneill

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    R4 is connected to the same nodes as the current source (they're in parallel). The path that you want should have ends that start at one node (B) and end at the other (A), and not 'close the gap'.

    attachment.php?attachmentid=41354&stc=1&d=1322587471.jpg
     

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  10. Nov 29, 2011 #9
    great answer thank you :)
     
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