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Combining parallel resistances

  1. Apr 25, 2010 #1
    Hi,

    I have the following circuit:
    [PLAIN]http://img689.imageshack.us/img689/9694/parallel.png [Broken]

    I need to find the equivalent resistance between the points A and B.

    I have tried several ways, but no matter how i try to combine them i always end up with the wrong answer.

    /James
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 25, 2010 #2

    Filip Larsen

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    You can step by step reduce the network by combining two resistors at a time into one replacement resistance using either a serial or parallel connection. As a start you can for instance reduce the two top right resistances with one using a parallel connection. After three more reductions you are left with only one resistance between A and B, which is the equivalent resistance of the network.
     
  4. Apr 26, 2010 #3
    Are the top two resistors then parallel with the bottom two resistors?
     
  5. Apr 26, 2010 #4

    Filip Larsen

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    Indirectly, yes. If you have trouble visualizing the structure of the network, then try make a sketch of the same network but with A to the left and B to the right (or top and bottom). It should hopefully then be more obvious that you initially can make two independent parallel reductions, and after those a serial reduction and then finally a parallel reduction.
     
  6. Apr 26, 2010 #5
    Hi,

    So i tried
    [tex]\frac{35\cdot42}{35+42} = 19.09[/tex]

    [tex]\frac{70*30}{70+30} = 21[/tex]

    [tex]\frac{21*19.09}{21+19.09} = 9.999[/tex]

    And finally [tex]\frac{9.999*7}{9.999+7} = 4.11[/tex]

    But the correct answer is supposed to be 11.635 ohm

    Any ideas?
     
  7. Apr 26, 2010 #6
    You need 19.09 in series with 7, then that combination in parallel with 21
     
  8. Apr 26, 2010 #7
    ok, to me it looked like the 7 ohm resistor was in parallel with the rest.
     
  9. Apr 26, 2010 #8

    vela

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    When two elements are in parallel, one element is connected to the same two nodes as the other element. In your circuit, the 35-ohm and 42-ohm resistors are both connected to node A, and their other ends are connected together at the top. Therefore, they are in parallel. Similarly, the 70-ohm and 30-ohm resistors are both connected to node A on one end and node B on the other, so they are also in parallel.

    This is not the case for the resulting 21-ohm and 19.09-ohm resistors, however. While both are connected to node A, one is connected to the top of the 7-ohm resistor, and the other, to the bottom. They're not connected to the same two nodes, so they are not in parallel.
     
  10. May 29, 2010 #9
    I think that the key to this question is "art" over "maths": you must redraw the circuit in a manner that is less confusing.

    [PLAIN]http://www.xphysics.co.uk/x/R1 [Broken]

    Then redraw...

    [PLAIN]http://www.xphysics.co.uk/x/R2 [Broken]

    and again...

    [PLAIN]http://www.xphysics.co.uk/x/R3 [Broken]

    Simples. :)
     
    Last edited by a moderator: May 4, 2017
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