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    Hi guys,




    For a), apparently the answer is 2A. Could any 1 help me out with this plz?


    Thanks!
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    Does Kirchhoff go ding-a-ling?
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    (Original post by K the Failure)
    Does Kirchhoff go ding-a-ling?

    Kirchhoff! Cant belive i've forgotten so much in my gap year lol! So its current in = current out init? The thing i'm not too sure about is how the elec flows to get to H (what direction)?

    Thanks.
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    Kirchhoff is probably a good thing to look at. Failing that look at it as a maths/routing problem.

    How long is the shortest path from A to H and what routes are that length?
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    (Original post by mfaxford)
    Kirchhoff is probably a good thing to look at. Failing that look at it as a maths/routing problem.

    How long is the shortest path from A to H and what routes are that length?

    Hmmm.. could u help me get started plz?

    Thanks.
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    (Original post by mfaxford)
    Kirchhoff is probably a good thing to look at. Failing that look at it as a maths/routing problem.

    How long is the shortest path from A to H and what routes are that length?

    Hmmm.. could u help me get started plz? (with the current bit)

    Thanks.
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    (Original post by wizz_kid)
    Hmmm.. could u help me get started plz?

    Thanks.
    Try answering these questions

    How long is the shortest path from A to H
    What routes are that length?
    the aim of those questions is to help determine what proportion of the current goes down each route.
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    (Original post by wizz_kid)
    Kirchhoff! Cant belive i've forgotten so much in my gap year lol! So its current in = current out init? The thing i'm not too sure about is how the elec flows to get to H (what direction)?

    Thanks.
    Yes, the net current = 0.

    Regarding the direction of the current: current always flows in the path of least resistance (hint: not backwards.)
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    (Original post by wizz_kid)
    Hmmm.. could u help me get started plz? (with the current bit)
    Thanks.
    The thing to remember with this question is that there is symmetry here.
    This diagram is similar to the one in your book.

    When the current enters the cube at A it has 3 options. Each option AB (red), AC (blue) and AD(green) has the same resistance, so the current divides equally between the 3.
    So you know the current through those 3.
    Next.
    The current blue current, AC, when it gets to C, has two options. Again the resistance is the same through CF and CG, so it divides into 2 equal parts.
    So you know the current through CF and CG.
    Exactly the same happens for the green and the red currents.
    These currents then recombine (add) to form the black currents through EH, FH and GH.
    (Which finally add to give 6A out the other end.)
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    (Original post by Stonebridge)
    The thing to remember with this question is that there is symmetry here.
    This diagram is similar to the one in your book.

    When the current enters the cube at A it has 3 options. Each option AB (red), AC (blue) and AD(green) has the same resistance, so the current divides equally between the 3.
    So you know the current through those 3.
    Next.
    The current blue current, AC, when it gets to C, has two options. Again the resistance is the same through CF and CG, so it divides into 2 equal parts.
    So you know the current through CF and CG.
    Exactly the same happens for the green and the red currents.
    These currents then recombine (add) to form the black currents through EH, FH and GH.
    (Which finally add to give 6A out the other end.)

    Veeery well explained! Your a star! Thanks a million matey! Rep to follow!
 
 
 
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