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Resistors Question

  1. Jul 11, 2014 #1
    1. The problem statement, all variables and given/known data


    I'm not sure how to do these. Other resistor questions are relatively easy but these had me stumped.
    An explanation for them would be helpful since all I got are the answers.

    The first one is 9.6ohms, second is 0.55ohms.

    Not sure if this is advanced but I'm seeking an explanation for these guys. Much thanks!
  2. jcsd
  3. Jul 12, 2014 #2

    Simon Bridge

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    It can help to redraw the circuits.
  4. Jul 23, 2014 #3
    Are you familiar with Wye-Delta transformations?
  5. Jul 24, 2014 #4


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    When given structures like that in (a), ALWAYS first look closely to see whether it has a particular symmetry. If present, this can make calculations SO MUCH EASIER!! :wink: :wink:

    Do you see a symmetry in the resistance values in that figure?
  6. Jul 25, 2014 #5


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    Don't think they are needed for these two problems. There are more obvious ways to simplify them.
  7. Jul 25, 2014 #6


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    Hello pandaboi,

    Welcome to Physics Forums! :smile:

    For part b), it might help to draw a "virtual" voltage source connected between X and Y (by that I mean connect the positive terminal to X and the negative terminal to Y). Ultimately, you want to find the current though that voltage source, so that you can divide its voltage by the current through it, and then you've found the overall resistance.

    Once you've included the voltage source, stop and think about what the voltage drop must be for each of the resistors in the circuit. Sure, you could use Kirchhoff's circuit laws and solve a set of simultaneous equations (which will work -- it's a perfectly valid way to solve this), but there may be a simpler way than that for this particular problem. If you find each resistor's voltage drop, you can find the currents through each resistor, and then ultimately the current though the voltage source. You'll still need to heed Kirchhoff's laws, but you might be able to skip the tedium of solving simultaneous equations (for this particular problem).

    Btw, part a) is easier because it has that symmetry that NascentOxygen mentioned. But if it didn't have that symmetry, this process I describe of adding a virtual voltage source and solving the simultaneous equations is a sure-fire way to solve these (even problems that don't have shortcuts like these). :wink:
  8. Jul 25, 2014 #7


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    I think part b) is easier than a).

    The circuit in b) is trivial. It can be very easily simplified. It's drawn to make it look hard.
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