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Homework Help: Circuit Analysis

  1. Oct 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Image: http://puu.sh/1ejYb [Broken]
    http://puu.sh/1ejYb [Broken]

    2. Relevant equations
    Req of n resistors in series = R1 + R2 + ... + Rn
    Req of n resistors in parallel = 1/R1 + 1/R2 + ... + 1/Rn

    3. The attempt at a solution

    The answer is supposed to be 10 ohms. However, I don't know how they got that.

    What I was thinking that some of the resistors short circuit and the current does not pass through them. So, it would be like: 3 ohms, then 6 ohms and finally 5 ohms. However, that adds up to 14 ohms and also I'm very sure that's wrong.

    Thanks a lot!
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 13, 2012 #2
    It is clear that the 12 ohm resistor short circuit. Now try to see whether the 6 ohm resistor on the right in the upper line and the 6 ohm resistor in the bottom line are in series or parallel.
  4. Oct 13, 2012 #3


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    Gold Member

    The 6Ω resistor in the right at top can be ruled out i think.. bcuz even that is shorted by a wire from the junction of 5Ω and 6Ω resistors at the bottom.
  5. Oct 13, 2012 #4


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    Science Advisor
    Homework Helper
    Gold Member

    PhysicoRaj - I don't think that's correct.

    After removing the 12 Ohm..

    The 6 Ohm at the top right is in parallel with the 6 Ohm at the bottom. So delete the 6 Ohm at top right and change the 6 Ohm at the bottom to 3 Ohm.

    Then the new 3 Ohm just created is in parallel with the remaining 6 Ohm making 2 Ohms.

    At this point you have 3, 2 and 5 in series.
  6. Oct 13, 2012 #5


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    Staff: Mentor

    This is a creative problem. Under examination pressures, where it is often difficult to get ones thoughts clear, there may not be a high score rate on this question.
  7. Oct 13, 2012 #6
    12Ω resistor is short cirrcuit. three 6Ω are parallel.
    So, equivalent resistor of circuit is 3+6/3+5=10Ω
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