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    ??
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    (Original post by Dynamo123)
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    ??
    I'm not sure but they may be saying that the 3 resistors in the top-left look like a 2/3R resistor from the point of view of both nodes to which they are connected (similar to the replacement in step 1) so they have simply arranged for both nodes to see 2/3R.

    If that is the argument, then I'm not sure it's valid, since the 3 resistors do not form a stand-alone circuit element in this case, whereas in step 1 the replacement looks fine.
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    (Original post by atsruser)
    I'm not sure but they may be saying that the 3 resistors in the top-left look like a 2/3R resistor from the point of view of both nodes to which they are connected (similar to the replacement in step 1) so they have simply arranged for both nodes to see 2/3R.

    If that is the argument, then I'm not sure it's valid, since the 3 resistors do not form a stand-alone circuit element in this case, whereas in step 1 the replacement looks fine.
    what do you do to solve 1 i've asked my teacher this before and she couldnt do this,so annoying
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    Apply Kirchhoff's rules. Just 2 simply stated rules that will solve any of these type of problems for you.
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    I figured that they had used a delta-wye conversion there. Kirchoff's rules are good, but I can't use them rapidly to solve these kinda questions

    Oh, and I guess this thread's resolved
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    (Original post by physics4ever)
    what do you do to solve 1 i've asked my teacher this before and she couldnt do this,so annoying
    The three resistors at the top right make a triangle. In a triangle combination (or a delta combination), two of the resistors are in series, while one is in parallel. So the REq= R*(R+R)/R+R+R = 2/3R as shown in the figure.

    I hope that helps :P
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    (Original post by Dynamo123)
    The three resistors at the top right make a triangle. In a triangle combination (or a delta combination), two of the resistors are in series, while one is in parallel. So the REq= R*(R+R)/R+R+R = 2/3R as shown in the figure.

    I hope that helps :P
    i made the top right deltacombination equivalent a 2R/3 resistor, i did the same thing to the other 2 deltacombinations,from A to B does the circuit go, one resistor in series,1 resistor in parallel but with just a wire,and then another resistor in series? like this
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    so that should be 2R?
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    No, for the other two, you have to apply delta-wye conversions. The two delta circuits (top and bottom left) are converted into wye circuits as shown in step 2. Here is a simple worksheet how to do it.
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    (Original post by Dynamo123)
    No, for the other two, you have to apply delta-wye conversions. The two delta circuits (top and bottom left) are converted into wye circuits as shown in step 2. Here is a simple worksheet how to do it.
    i dont know what delta wye is,is this A level stuff?
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    (Original post by physics4ever)
    i dont know what delta wye is,is this A level stuff?

    No, undergraduate stuff. You won't get such questions in A-levels.
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    (Original post by Dynamo123)
    No, undergraduate stuff. You won't get such questions in A-levels.
    oh right thats good,i'll try learning these then thanks for the worksheet ,why did you put it sixth form if its undergrad stuff?
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    (Original post by physics4ever)
    oh right thats good,i'll try learning these then thanks for the worksheet ,why did you put it sixth form if its undergrad stuff?
    I was a bit..whatchamacallit...confuzzle d as to where to place it.
 
 
 
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