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A strange electrical circuit - resistance calculation

  1. Feb 26, 2013 #1
    Hello. So I've got an electrical circuit on page 4 of this: http://www.iuventa.sk/files/documents/2_olympiady/fo/54.%20rocnik%202012-2013/ulohy%20a%20riesenia/skolske%20kolo/f54skcul13.pdf and I am to calculate the total resistance of the circuit, but I have no idea on how to solve that strange, circular group of resistors. Could anyone help me? I don't want the solution, just a little clue on what to do when the resistors are grouped like this. Thanks a lot.
     
  2. jcsd
  3. Feb 26, 2013 #2

    gneill

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    Look up Y-Delta transformations. Otherwise, use Kirchhoff's laws or mesh or nodal analysis to find the current that the battery supplies and from that the resistance of its load.

    EDIT: I just noticed the ratios of the resistors in the group between nodes A and B. What can you say about the potentials at the nodes at either end of the 5.0Ω resistor?
     
  4. Feb 26, 2013 #3

    NascentOxygen

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    Well spotted. That's an observation which can only come from experience! (it's the sort of exam question which I hated.)
     
  5. Feb 26, 2013 #4

    gneill

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    Thanks! :smile: I find it's usually worth looking for an "out" when there's a Δ-Y or Y-Δ "opportunity" in the offing...
     
  6. Feb 27, 2013 #5
    Thanks, that's interesting. I'm just still not sure how to use that to solve the circuit…
     
  7. Feb 27, 2013 #6

    NascentOxygen

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    The sub-circuit between points A and B includes two potential dividers. The resistors make them practically equivalent. As gneill suggested, look at the voltages at each end of that 5 Ohm resistor....
     
  8. Feb 27, 2013 #7
    So does that mean that resistance on both branches of the A-B part of the circuit will be the same?
     
  9. Feb 27, 2013 #8

    gneill

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    No, not that. What do you notice about the resistor ratios in both branches?
     
  10. Feb 27, 2013 #9

    NascentOxygen

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    Name the resistor values that you see comprising the two potential dividers we refer to.
     
  11. Feb 27, 2013 #10
    Yeah, I noticed they are 1:2.

    However, following a consultation with a friend who studied electrotechnics, I managed to do the delta-y transformation and calculate the total resistance, which appears to be 10/3Ω. Now, the next part of the task is to calculate the currents I1 through I6. So my current question is: how do I calculate the currents of the original resistors using the delta-y transformation? Thanks.
     
  12. Feb 27, 2013 #11

    gneill

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    I think that perhaps something went awry with your Delta-Y transformation. I'm not seeing 10/3 Ω as the equivalent resistance.

    So you've noticed that the ratio of the resistors in the two parallel branches are 1:2. Suppose that there is some potential across the nodes AB. What can you say about the potentials at either end of the 5Ω resistor?
     
  13. Feb 27, 2013 #12
    Well... do I conclude right that they are the same?

    Gotta look to my physics notebook from last year as I don't remember the relation between electrical potential, current and voltage. Thanks for help so far.

    By the way, if I wanted to solve it using the delta-y transformation (provided I do it right, which I didn't the first time, apparently), how do I calculate the current on each of the original resistors that I delta-y transformed?
     
  14. Feb 27, 2013 #13

    gneill

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    Yes, they would be the same. If both ends of the 5Ω resistor are at the same potential, what is the current though it?
    You can't calculate the currents through resistors that you've transformed away, so you have to return to the original configuration.

    You would first find the current supplied by the battery into the equivalent resistance, then the potential drop across the 2.4Ω resistor (potential BC). That in turn would give you the potential drop from A to B. Then you'd have to analyze the sub-circuit between A and B by other means (maybe mesh or nodal analysis).

    Trust me, going by way of the fortuitous resistor ratio will be simpler overall :wink:
     
  15. Feb 27, 2013 #14
    Well, zero? So should I just calculate as the 5Ω resistor wasn't there and it's just a normal two-branch circuit?
     
  16. Feb 27, 2013 #15

    gneill

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    Yes, exactly. Much simpler, no?

    When two nodes are at the same potential you are allowed to remove what's between, or connect them together with a wire (short circuit). This can come in very handy in certain circumstances (such as this), and often occurs when a circuit exhibits symmetry.
     
  17. Feb 27, 2013 #16
    Yeah, it surely is much easier. Physics never stops to amaze me… :)

    Thanks a lot for all your help, I really appreciate that :)
     
  18. Feb 27, 2013 #17

    gneill

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    Glad I could help. Good luck with your studies.
     
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