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Currents and resistance

  1. Feb 3, 2008 #1
    1. There is a current of 0.25 A in the circuit of the figure
    What is the value of the resistance R?


    Hockay so. I know the change of voltage throughout the loop is 0.
    therefore
    0 = E1 - I(R1+R2) - E2 - 1R(unknown)
    0 = 6V - 0.25A*18ohm - 12V - 0.25A*R
    through algebra i get R=42.
    And thats wrong. So i dontk now what im doing wrong.

    And it also asks for the power dissipated by R and i dont know whether to use the energy of both batteries or do a long equation with info from the whole circuit or what. I do know how ever that i should use the PR = IV or I^2R or (V^2/R)

    thanks a bunch.
     
  2. jcsd
  3. Feb 3, 2008 #2
    okay im technically challenged.
    Take two on uploading this badboy.
     

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  4. Feb 3, 2008 #3

    rock.freak667

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    Attachments take long to approve. Upload the image on http://imagshack.us and use one of the links in it so that is shows up.
     
  5. Feb 3, 2008 #4
    Last edited by a moderator: May 3, 2017
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