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Currents and resistance

  • Thread starter cemar.
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
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okay im technically challenged.
Take two on uploading this badboy.
 

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  • #3
rock.freak667
Homework Helper
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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.
 
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