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One for the weekend (3) - answer and follow-up question watch

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    Nearly everyone got right answer.
    The pd is 30V giving the resistance of the whole cube as 5 ohms.
    Use symmetry to work out the answer rather than attempt to figure out the series and parallel combinations.

    Each side has resistance 6 ohms.



    Here's a follow-up.

    What would be the resistance of the cube if you connected from A to C rather than A to D?

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    I have to confess to being full of self-doubt this time... is it 30 V?
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    Im going to go for 144V, probably completely wrong.

    Edit: I'm an idiot, it is 30V. Wasn't thinking it through properly.
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    I got 30V too by looking at it as 3 'stages' in series, with 3, 6, and 3 wires in parallel respectively.
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    I say 30V.
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    Agreed on 30V, interesting thought experiment!
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    I got 30 too!
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    fun fun fun!
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    I got 4.8 V. I think I'm wrong.

    I did it by considering this. (2 in series) parallel to (2 in series) parallel to (2 in series) and all of it parallel to 6 wires which are in turn parallel to each other.

    EDIT: I get it now. The above posters are correct.
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    I've posted the answer up in the 1st post.
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    For the second answer, here's my guess...

    EDIT: Misread question, but can't be bothered to go through working again :p: will post again later...
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    (Original post by Stonebridge)
    Here's a follow-up.

    What would be the resistance of the cube if you connected from A to C rather than A to D?
    My answer is R_{AC}=\tfrac{3}{4}R=4.5\ \Omega}.
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    (Original post by jaroc)
    My answer is R_{AC}=\tfrac{3}{4}R=4.5\ \Omega}.
    How did you do that anyway? Explain, please.
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    (Original post by Zishi)
    How did you do that anyway? Explain, please.
    If my result is correct, I won't do that until Stonebridge decides it's time to reveal the correct answer. I'd like to let other people work on the problem
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    (Original post by jaroc)
    My answer is R_{AC}=\tfrac{3}{4}R=4.5\ \Omega}.
    That's correct. Maybe you can answer Zishi's question now?
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    Sure

    Let's have a look at the cube (see attachments). By symmetry we can see that points B and D have equal potentials; the same can be said about F and G.

    If two points have the same potential, they are identical - in the sense, that you can draw them in the same point.

    Let's first consider paths ABC and ADC. Points B and D have equal potentials so we can connect them (it is probably not so easily seen why we would want to do that at the moment, but it will prove very useful soon). As a result we can draw this part of the circuit as in the second drawing in the first file.

    It's a bit more complicated with the other path, starting AE\ldots. With EFH and EGH it's exactly the same story as with ABC and ADC - because F and G have equal potentials. Then there is one resistor left between H and C. As a result, we can draw this part of the circuit as in the third sketch in the first file.

    If we sum up all resistors in the second and third drawing, we get 10. A cube has 12 edges - so we're missing two. Those are FB and GD. Now look at the second and first drawing. We see that the points B,D and F,G are in the middle of the first and second path respectively. This means that they have the same potential. If they have the same potential, no current goes between them, which means that we can forget about resistors FB and GD. If they have the same potential, we can connect them with a resistance-less wire, just like in the first sketch in the second file. This is obviously equivalent to the last graph.
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