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KVL and KCL

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data
    In the circuit shown in the figure, find (a) the current in resistor R; (b) the resistance R; (c) the unknown emf E. (d) If the circuit is broken at point x, what is the current in the 28v battery?


    2. Relevant equations
    KVL and KCL


    3. The attempt at a solution
    I don't know if these are correct :S
    KVL:
    3(6) + 6(4) - E = 0
    E = 42v
    So my E will be 42 volts?

    And in question (a) since 6A will pass in node a and the current in I2 is 4A, then logically, the remaining 2A will flow through R. So the current in R will be 2 Amperes.

    So now I can solve for the Resistance in R (question B)
    4(6) - 48 - 28 + 2(R) = 0

    24 - 48 - 28 = -2(R)

    R = 26 ohms

    Are my solutions correct? In question D, the current will be 2A if the circuit is broken at point x. Correct me if I'm wrong. Thanks in advance!
     

    Attached Files:

    Last edited: Sep 21, 2008
  2. jcsd
  3. Sep 21, 2008 #2

    Defennder

    User Avatar
    Homework Helper

    Correct so far.

    Where did -48 come from? Also check your sign convention for the 28V voltage source. If you're going in a clockwise loop, and you enter the 28V voltage source through the +ve terminal, then you add, not subtract the potential drop.

    Why would it be 2A? Remember that once x is cut, E doesn't factor into the calculations at all. You would then have a single loop consisting of currents through top and bottom wires.
     
  4. Sep 22, 2008 #3
    Okay I already get it now. Thanks Defennder! :approve:
     
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