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Equivalent Resistance

  1. Jul 1, 2010 #1
    1. The problem statement, all variables and given/known data
    This should be quite simple for most of you although I can't understand it.
    Circuit is attached below

    2. Relevant equations

    3. The attempt at a solution
    The answer is 20ohms, although I would have done R1+R3 + R4//R5 which would get 25ohms.
    The previous question was to find Rab and Rab = 25ohms.

    How do i view the circuit to get 20ohms? I'm aware it would just be R1+R3, but why can't you do it my way?
    Last edited: Jul 1, 2010
  2. jcsd
  3. Jul 1, 2010 #2


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    Do you agree with the way I have redrawn the circuit?

    Attached Files:

  4. Jul 1, 2010 #3
    Yeah that makes sense. I just have trouble redrawing circuits. I need to work on that. Do you have any pointers or suggestions on how I need to approach it?

    Should I just imagine a current starting from A and redrawing it according to where the current branches off etc?
  5. Jul 1, 2010 #4
    My approach to redraw a circuit is to simply reduce the lengthy wires (because they have zero resistance) and bring resistors together as close as possible. Now it should be more clear to see whether the resistors are in parallel or series.
    As Zryn has redrawn the circuit, it will quickly indicate you that R4 and R5 are parallel. Find equivalent resistance of R4 and R5 and add it with R1 and R3.
  6. Jul 1, 2010 #5


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    Keep in mind that circuits that test your knowledge of how resistors can be combined will be drawn in such a fashion as to confuse you.

    If you have some time, you can practice redrawing them by getting several different color pencil/pen/highlighter/crayon/whatever and color over each different node in a different color, then redraw the circuit in a linear fashion from top to bottom or left to right so that all the resistors point the same direction, and double check that each resistor is connected to the right color nodes.
  7. Jul 1, 2010 #6
    That last tip with the labeling each node thing is a nice tip for me. Thanks for that, I understand it better now. I just need to find some more circuits to practice on to see if I really have it.
  8. Jul 1, 2010 #7


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    In this circuit, the path of current goes from "a" through R1 and R3, to "d".

    R2 goes nowhere, so you can ignore it.

    Although R4 and R5 are certainly in parallel, they are not in the path of the current from "a" to "d".
    They are shorted out by the wire going from R3 to "d".
    So you can ignore them too.
  9. Jul 2, 2010 #8
    Yeah, it's just hard for me to see in the picture unless it's redrawn like Zryn did. That's just for me though.
  10. Jul 2, 2010 #9


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    [PLAIN]http://dl.dropbox.com/u/4222062/series%20R2.PNG [Broken]

    Maybe this will make it clearer why R2, R4 and R5 are not involved. You could remove all of them without having any effect on the current in the meter marked "A" in the diagram.

    Current only flows in the parts marked in red.
    Last edited by a moderator: May 4, 2017
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