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Resistors in Series, in Parallel

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    See the attached image.

    2. Relevant equations

    V=IR, Kirchoff's Law

    3. The attempt at a solution

    The voltage drop across resistors in parallel are equivalent due to Kirchoffs law (at least I think so?), so V=R(eq,top)I(top)=R(eq,bottom)I(bottom)--> I(top)(52)=I(bottom)(7+Rx)

    For each resistor, an equation using Ohm's Law:
    V1=I(top)(14)
    V2=I(top)(38)
    V3=I(bottom)(7)
    V4=I(bottom)(Rx)

    As the ammeter reads zero, I thought the voltage drop across resistor 1 and 3 were equal, so V1=V3. Rearranging the equation and substituting in the top equation-->
    (V/14)(52)=(V/7)(7+Rx), which yields Rx=19.

    I don't think this is right though. Please help! Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Feb 13, 2012 #2

    PeterO

    User Avatar
    Homework Helper

    19Ω is a perfect answer - so probably your reasoning is correct too as answers like 19 don't usually appear by co-incidence.
     
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