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Find the Equivalent Resistance of a Network

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A network consists of a combination of parallel and series connections. What is the equivalent resistance of the network? (Hint: Use the series and parallel resistance formulas).

    The network looks like this:

    2. Relevant equations
    When a circuit is in series, R = R1 + R2 + R3...

    When a circuit is in parallel, R = 1/R1 + 1/R2 + 1/R3...

    3. The attempt at a solution
    First, c and b are in series, so I just add them up. 5 + 3 = 8.

    Now, a, d, and this new resistor of value 8 are in series. So add them up again.
    8 + 4 + 1 = 13.

    Did I do that correctly?
    Last edited by a moderator: Apr 14, 2017
  2. jcsd
  3. Feb 22, 2009 #2


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    Sorry. B and C are ||.

    Not in series.
  4. Feb 22, 2009 #3
    They look like they're in series to me... how can you tell?

    That changes my answer to 1/5 + 1/3 = 0.533. And 1/0.533 = 1.875.

    Then 1.875 + 4 + 1 = 6.875
  5. Feb 22, 2009 #4


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    That looks more like it.
  6. Feb 22, 2009 #5
    This is wrong. I think you meant to write:
    1/Req = 1/R1 + 1/R2 + 1/R3...

    Notice that the upper terminals of C and B share a common node. Similarly, the bottom terminals of C and B share a common node.
  7. Feb 22, 2009 #6
    Yeah, I meant to write 1/Req = 1/R1 + 1/R2...

    So a is in series with c because it only shares one common node? And b and c are in parallel because they share two nodes?
  8. Feb 22, 2009 #7
    Well, a better way to say it is this: A is in series with the combination C||B.

    If the same current flows through two resistive elements, the elements are in series.
    If the same voltage potential is across two elements, the elements are in parallel.

    Does that help?
  9. Feb 22, 2009 #8
    That helps a lot - I understand now. Thank you!
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