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    Sorry I'm posting quite a few questions but I'm only trying to learn and see if I'm doing this right! Would anyone be able to tell me if this is right? And would (a) (ii) X and Z be 8 Ohms? Name:  ImageUploadedByStudent Room1421184629.941095.jpg
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    (Original post by ohnanailikenanas)
    Sorry I'm posting quite a few questions but I'm only trying to learn and see if I'm doing this right! Would anyone be able to tell me if this is right? And would (a) (ii) X and Z be 8 Ohms? Name:  ImageUploadedByStudent Room1421184629.941095.jpg
Views: 94
Size:  91.7 KB


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    How have you worked these out? I don't think so. What you want to do in both cases is:

    1) Find all paths between the two points (there will be two routes between each pair, one going clockwise and the other counterclockwise)
    2) Find the resistance for each path
    3) Work out the total resistance
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    (Original post by ohnanailikenanas)
    Sorry I'm posting quite a few questions but I'm only trying to learn and see if I'm doing this right! Would anyone be able to tell me if this is right? And would (a) (ii) X and Z be 8 Ohms? Name:  ImageUploadedByStudent Room1421184629.941095.jpg
Views: 94
Size:  91.7 KB


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    Two rules:

    1) Series resistors are summed: R1 + R2 + R3 etc.

    2) Parallel resistors are: \frac{1}{R_{total}} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} \mathrm \ {etc}.

    In part a) think of the paths current will take flowing between the two points in question.

    In the first there are two paths which means these paths are in parallel:

    1) The first is directly between X and Y flowing through the 6 ohms resistor on its own.

    2) The second flows via 3 resistors passing through each in turn (series) 6, 3 and 12 ohms.

    To calculate the resistance between the terminals, calculate the resistance between each path independently (the series paths). Replace each of the series paths with a single equivalent resistance.

    You then have only two resistances to deal with and both are in parallel.

    The final resistance can now be calculated.
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    Hm, I thought that's what I was doing Since I have to find the resistance b/w X and Y.. It's in parallel with the 3 Ohms so do I not do 1/3 + 1/6 = 4/12 + 2/12 = 6/12 then flip 12/6 = 2 Ohms??? And then between X and Z is it not 6 Ohms because the 6 Ohms is parallel to the 12 Ohms which equals to be 4 Ohms then if the other is 2 Ohms add it together to make 6 Ohms? :/


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    (Original post by ohnanailikenanas)
    Hm, I thought that's what I was doing Since I have to find the resistance b/w X and Y.. It's in parallel with the 3 Ohms so do I not do 1/3 + 1/6 = 4/12 + 2/12 = 6/12 then flip 12/6 = 2 Ohms??? And then between X and Z is it not 6 Ohms because the 6 Ohms is parallel to the 12 Ohms which equals to be 4 Ohms then if the other is 2 Ohms add it together to make 6 Ohms? :/


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    It's not in parallel with the 3Ohms only though- it's in parallel with the entire anti-clockwise route(6,3,12). You can find the resistance of that by combining them using the rules for series circuits.
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    Wait.. Are you trying to explain to me that b/w X and Y you count it as if 6, 3, 12 Ohms then the shorter way 6 Ohms? Do I still use the parallel equation thing? Eg 1/6 + 1/12 etc


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    (Original post by ohnanailikenanas)
    Wait.. Are you trying to explain to me that b/w X and Y you count it as if 6, 3, 12 Ohms then the shorter way 6 Ohms? Do I still use the parallel equation thing? Eg 1/6 + 1/12 etc


    Posted from TSR Mobile
    yes. the longer path is 6,3 and 12 ohms. These resistors are in series.

    The equivalent resistance is 6 + 3 + 12 = 21 ohms.

    This equivalent is in parallel with the 6 ohms resistor directly connected between X and Y.

    21 ohms equivalent is in parallel with 6 ohms.
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    So the resistance b/w X and Y is 1/21 + 1/6 = 2/42 + 7/42 = 9/42 = 42/9 = 4.66 Ohms?


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    (Original post by ohnanailikenanas)
    So the resistance b/w X and Y is 1/21 + 1/6 = 2/42 + 7/42 = 9/42 = 42/9 = 4.66 Ohms?


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    Correct.

    You should round your answer to 4.67 ohms or state it exactly as 4\frac{2}{3} \Omega
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    Oh okay, great! Thank you so much! I'll try and work out the second question now.


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    (Original post by ohnanailikenanas)
    So the resistance b/w X and Y is 1/21 + 1/6 = 2/42 + 7/42 = 9/42 = 42/9 = 4.66 Ohms?


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    Now you know how to do it, what did you make the answer between X and Z?
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    I did 6 + 3 = 9 then 6 + 12 = 18. 1/9 + 1/18 = 4/36 + 2/36 = 6/36 = 36/6 = 6 Ohms?


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    (Original post by ohnanailikenanas)
    I did 6 + 3 = 9 then 6 + 12 = 18. 1/9 + 1/18 = 4/36 + 2/36 = 6/36 = 36/6 = 6 Ohms?


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    You got it!
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    Ah thank you ever so much! You don't realise how much I appreciate this!


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    (Original post by ohnanailikenanas)
    Ah thank you ever so much! You don't realise how much I appreciate this!


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    You are very welcome.
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    Sorry, again. Is part (b) right? Name:  ImageUploadedByStudent Room1421189978.684356.jpg
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    (Original post by ohnanailikenanas)
    Sorry, again. Is part (b) right? Name:  ImageUploadedByStudent Room1421189978.684356.jpg
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    Posted from TSR Mobile
    No.

    As before, there are two current paths between X and Z and we must consider the path with the 3 ohms resistor in it.

    I3ohms = VXZ / RXZ

    That is because all of the series current flowing in that path must flow through the 3 ohms resistor.

    So

    VXZ = 12V

    RXZ = total resistance between XZ (current path with 3 ohms resistor in it) = (3 + 6)ohms = 9 ohms

    I3ohms = 12 / 9 = 1.33 amps.
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    Ah okay I see, thank you once again!


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