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Resistors connected in parallel and series

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the equivalent resistance between the points A and B of the network shown in figure ( file has been attached )

    2. Relevant equations
    Req( series) = R1+ R2 + R3 .. + Rn
    1/Req ( parallel) = 1/R1 + 1/R2 + 1/R3 .. + 1/Rn

    3. The attempt at a solution
    We have to find the equivalent resistance between a and b.
    The way I solved it was :
    1) the second ( from left )4 ohm and the 3 ohm are connected in series, so Req = 7 ohm.
    2) now the 7 ohm and 2 ohm are connected in parallel, so Req = 1.56 ohm
    3) the remaining 4 ohm and 1.56 are connected in series, so Req = 4.56 ohm.

    But this answer is wrong..According to the book,
    The 2 ohm resistance is connected in parallel with the 3 ohm resistor. Their equivalent resistance is 2 ohm which is connected in series with the first 4 ohm resistor from left. Thus equivalent resistance between A and B is 6 ohm.
    Please help :/
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Mar 23, 2012 #2

    gneill

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    Staff: Mentor

    No, those resistors are not connected in series; The first 4 Ω resistor also connects at the node where they meet. For two components to be in series, no other component can share their connection point.

    I suggest that you begin from the 'back end" of the circuit, with the 2 Ω resistor, and work "forward" from there :wink:
     
  4. Mar 23, 2012 #3

    pcm

    User Avatar

    4 and 3 are not in series.think...
     
  5. Mar 23, 2012 #4
    the problem with these questions is that, you think that some of the resistors are in series with the other, where in reality they are not. Instead of drawing the R3 resistor on the same horizontal line as the other 4 ohm resistor(R1), you could move it closer to the 2ohm resistor and draw them on the same vertical line.

    You will see that R3 and R4 are actually in series, which means you could add it up together, which gives 6ohms. We'll name this equivalent resistor R34. You will also notice that R34 is parallel to R2. So 1/(1/6 + 1/3) = 2 ohms. Now we name this simplified resistances R234.

    R234 is in series with R1. 2+4 = 6ohms.
     
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