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Resistance with resistors between parallel resistors

  1. May 10, 2016 #1
    1. The problem statement, all variables and given/known data
    The overall resistance of the circuit is 12.5Ω. Calculate the value of the resistor labelled R in the circuit.
    14iertg.png

    2. Relevant equations
    1/R = 1/R1 + 1/R2
    R = V/I

    3. The attempt at a solution
    I understand that you use the first equation for parallel resistors and the second to calculate resistance from voltage and current. The bit I am having trouble with is the network of 4 resistors on the left hand side. The resistance on the right hand-side is 1/R = 1/2 + 1/6, so R = 1.5. So then 12.5 = 6 + 1.5 + the resistance of the left hand-side network.

    I just need guidance on how to calculate it because it puzzles me as there is 2 resistors between the other 2 in parallel.
     
    Last edited: May 10, 2016
  2. jcsd
  3. May 10, 2016 #2

    BvU

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    A matter of drawing; one could also say there are three in series with one in parallel ...
     
  4. May 10, 2016 #3

    cnh1995

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    Are you sure?
    Hint:Current through series resistors is same.
     
  5. May 10, 2016 #4

    Paul Colby

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    I find it helpful to follow the current around the circuit. Start at the battery. As the current flows where does it split? It's enough to start at the point in the circuit where you lost the bubble.
     
  6. May 10, 2016 #5
    Thanks guys! Just considered the other 3 (5, 1 and R) as a series and the 10 as parallel with those. :)
     
  7. May 11, 2016 #6

    CWatters

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    Yes that's ok.

    Yes that's ok for the "left hand side network".

    Just turn it all into an equation and solve for R
     
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