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Electric Circuit - Equivalent Resistance

  1. Aug 17, 2011 #1
    1. The problem statement, all variables and given/known data

    For 1 and 2 : Find the equivalent resistance for the resistor network provided


    2. Relevant equations

    For resistor in series : Req = R[itex]_{1}[/itex]+R[itex]_{2}[/itex]+...R[itex]_{n}[/itex]

    For resistors in parallel : Req = (1/R[itex]_{1}[/itex]+1/R[itex]_{2}[/itex]+... 1/R[itex]_{n}[/itex])[itex]^{-1}[/itex]

    3. The attempt at a solution

    For question 1 :

    I have considered the two paths shown :

    1st path (blue) : total resistance = 2 ohm
    2nd path (orange) : total resistance = 2 ohm

    so since these two paths are in parallel , Req = 1 ohm

    But answer provided = 2/3 ohm

    For Question 2 :

    Since no current passes through the middle wire, upper path = 10 ohm
    lower path = 10 ohm

    So, total resistance = 5 ohm

    However , the answer used the method of (1/4+1/8)^-1 + (1/2+1/6)^-1 = 4.17 ohm

    But I don't understand this part, if there is current flowing in the middle wire, there would be one current flowing down and one current flowing up. Won't they cancel out each other? Thanks !!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Aug 17, 2011 #2

    PeterO

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    For question 1 there is a third path you have not considered.
     
  4. Aug 17, 2011 #3
    I have thought of two : One through the middle through all 3 resistors, and another one , turning left between middle and rightmost resistor to pass through the middle resistor.

    Which one is correct?
     
  5. Aug 17, 2011 #4

    PeterO

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    For the second part, there may well be current flowing through the middle wire - but since we consider a wire to have zero resistance, the potential at each end will be the same.
    re-draw the circuit with the centre wire much, much shorter [like zero length] and see what the circuit looks like. you should then see where their answer comes from.

    EDIT: Sorry but i don't know how to load pictures
     
  6. Aug 17, 2011 #5

    PeterO

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    The second - as it offers another path back to the battery having passed through only one of the resistors - the same as the two paths you identified at first.
     
  7. Aug 17, 2011 #6

    PeterO

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    Try this: use a high-lite pen to high-lite the wire from the positive terminal of the battery. continue shading as far as you can along any wires without passing through a resistor, then have a look at what you have shaded.

    Preferably with a different colour, start at the negative terminal and shade back along wires - but again don't go through any resistor.

    What do you notice?

    EDIT: It is 2:30 am here, I am off to bed!!
     
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