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SOLVING[Series-Parallel Circuits

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data
    R1-10
    R2-13
    r3-3
    R4-6
    What should be the total voltage so that I3 = 1A.
    2. Relevant equations
    n/a


    3. The attempt at a solution
    idk to solve it help me plss :(
     

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  3. Jan 17, 2012 #2

    gneill

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    Staff: Mentor

    1. You haven't identified where I3 is (Is it the current through R3?).
    2. There must be relevant equations! Otherwise, how would such problems get solved?
    3. You need to provide some attempt at a solution, or at least some discussion of what you've already tried.
     
  4. Jan 18, 2012 #3
    ohh sorry so this is my attempt.
    1. You haven't identified where I3 is (Is it the current through R3?).Yes
    2. There must be relevant equations! Otherwise, how would such problems get solved?
    http://en.wikipedia.org/wiki/Ohm's_law
    3. You need to provide some attempt at a solution, or at least some discussion of what you've already tried.

    Rt=R1+R2+R34
    Rt=10+13+2
    Rt=25

    I3=1
    V3=I3R3
    V3=1(6)
    V3=6

    I4=I2-I3
    I4=I2-1
    I2-I4=1
     
  5. Jan 18, 2012 #4

    gneill

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    Staff: Mentor

    Have you not learned about Kirchhoff's laws? KVL? KCL?
    Note that R1 is in parallel with the voltage source (let's call it V) . If the voltage source is ideal then the potential with respect to ground at the node where R1 and R2 meet is fixed by V regardless of the value of R1. So we can expect that the value of R1 will not play a role in determining V.

    It's not clear why you calculated Rt above, or what it represents, since R1 is not in series with R2 and R3||R4.
    Since you've determined the voltage that appears across R3 above, what can you say about the current through R4?
     
  6. Jan 19, 2012 #5

    Curious3141

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    Homework Helper

    The quickest way to do it is to use the voltage divider principle. But doing it with Kirchoff's laws is the most instructive way and teaches basic principles well. :smile:
     
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