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Baffling Circut problem

  1. Mar 14, 2005 #1
    In the figure provided, all the resistors have a resistance of 4 ohms and all the ideal batteries have an emf of 4 V. What is the current through resistor R?

    This question had me baffled for a while, I can't quite trace through the circut. Even my physics teacher got lost with the unbelievable loops in the circut. Any help would be greatly appreciated!
     

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  2. jcsd
  3. Mar 16, 2005 #2
    ANYONE? Come on, this can't be that hard.
     
  4. Mar 16, 2005 #3

    robphy

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    I think it's 2A.
    Use Kirchoff's Voltage Law.
    Start at the junction below the lower-left battery.
    Travel up to the upper left corner, passing along three batteries along the way.
    Then travel right to the battery near the right-side, passing it "backwards".
    Then travel to your resistor, then down to the bottom wire, which is
    equivalent to your starting point.

    (+4V) + (+4V) + (+4V) + (-4V) + (-IR) = 0
    or
    (+8V)=IR... so I=(8 V)/R = (8 V)/(4 ohms)= 2A.
     
  5. Mar 17, 2005 #4

    Chi Meson

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    Aha! I knew there had to be a trick! Here I was starting to set up a 19 x 19 matrix. After bogging down in all the junctions I thought I'd just wait for the answer to show up. So, to summarize: find a loop that only passes through the batteries and the resistor that you want, eh? Brilliant! I love it!
     
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