Equivalent Resistance

Discussion in 'Homework Help' started by Tmm46, Mar 25, 2013.

  1. Tmm46

    Thread Starter New Member

    Mar 25, 2013
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    0
    I'm having major trouble with this circuit, I can get the series resistors easy, but the rest is puzzling.
     
  2. panic mode

    Senior Member

    Oct 10, 2011
    1,321
    304
    show your work:

    do the series resistors, then redraw the circuit. if that is not clearing the picture up, we'll help
     
  3. WBahn

    Moderator

    Mar 31, 2012
    17,768
    4,802
    The problem is that some of the relationships are not obvious because they are draw in a screwy way (which was the intent).

    You can always apply a test voltage and determine the current that would flow and take the ratio to find the equivalent resistance.

    But in this case you can just redraw the circuit an a more usual way, preferably with no lines crossing. Once you do that (and you may have to redraw it a couple of times since each redrawing will reveal some more of the traditional structure to you) you will be able to get the equivalent resistance by hand.

    As a check, assuming I haven't done anything wrong, the result should be between 35Ω and 40Ω and is much closer to one end of the range than the other.
     
  4. justtrying

    Active Member

    Mar 9, 2011
    329
    352
    label the nodes when redrawing, so that you do not make mistakes
     
  5. nomad1003

    New Member

    Apr 17, 2013
    4
    0
    we can numarete all current values for junction (rule) in points that accrossed. (conversation of electric charge etc.) and that helps us to make the loop rule (which it's an electrostatic force is conservative.) Suppose we go around a loop, measuring the potential differences across successive circuits elements as we go. (the head-bang is about the direction of loops that you choose to get signs) Rest part is just about the Kirchhoff Rule's Applications.
     
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