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Simple Circuit Question (KVL/KCL)

  1. Oct 16, 2009 #1
    [Solved]

    1. The problem statement, all variables and given/known data

    4017773043_4297587691_o.jpg

    2. Relevant equations

    KCL, KVL, Ohms Law

    3. The attempt at a solution

    6 = i + (i/4) + (v/8)

    I just can't get another equation that relates v & i.

    Any suggestions? Am I going about this the right way?
     
    Last edited: Oct 17, 2009
  2. jcsd
  3. Oct 16, 2009 #2

    cepheid

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    The branch with the 2 ohm resistor tells you that the voltage drop across it is given by i*(2 ohms), which, from the way the circuit is laid out, must also be equal to v.
     
  4. Oct 16, 2009 #3
    we know that the equivalent resistance of the io/4 branch is 4 times that in the 2 ohm resistor, because 1/4 the current flows. So it's easy to redraw the circuit using an 8 ohm resistor in place of that branch, and that makes it all nice and easy.
     
  5. Oct 16, 2009 #4
    If that's the case, then won't I get:

    6 = i + (i/4) + (2i/8) = (5i/2)
    Which would give: i = 12/5.

    Am I missing something?
     
  6. Oct 16, 2009 #5

    cepheid

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    Bad arithmetic:

    i = 4i/4

    i/4 = i/4

    2i/8 = i/4

    so the sum is:

    4i/4 + i/4 + i/4 = 6i/4 = 3i/2

    if 3i/2 = 6, then 3i = 12, and i = 4.
     
  7. Oct 16, 2009 #6
    Ah, that's right. Not sure how I made such a silly mistake.
    Thank you.
     
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