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Equivalent resistance with possible short circuit

  1. Jan 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the current flowing through the voltage source in mA. Assume Vs = 3.9 V and R = 8.3 kohm. Round off your answer to two decimal points.

    2. Relevant equations

    1/R(equivalent) = 1/R1 + 1/R2 + ...


    Kirchoff's voltage law

    3. The attempt at a solution

    I considered the 3 resistors in the upper left of diagram (10k, R=8.3k, 10k) to be in parallel and calculated their equivalent resistance to be 3.12 kOhms. I basically ignored the two 10kOhm resistors in the bottom right because a) they didn't seem to be in series, or parallel so I didn't know what to do with them and b) it seemed that the current would avoid them by taking the path of least resistance after going through the 3 resistors in parallel in the upper left. I then wrote the KVL equation: Vs - I(3.12*10^3) = 0 and solved for I determining the current to be 1.25mA.

    Attached Files:

  2. jcsd
  3. Jan 27, 2012 #2


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    Gold Member

    Uh, dude, you REALLY need to rethink that analysis. How can the current through R possibly NOT flow through on or both of the resistors on the right & bottom?
  4. Jan 27, 2012 #3


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    Staff: Mentor

    Try redrawing the diagram, this time repositioning enough of the elements so that you can draw it so there are no sloping lines. I.e., arrange it so all elements and connecting wires are either vertical or horizontal, and with no ugly cross-overs.

    You may need to make a few attempts before you manage it. Just keep trying.
  5. Jan 27, 2012 #4
    Wow. Thank you both very much for the help. My first big problem was that I assumed there was a node where the wires crossed. For sure, I'll pay much closer attention to nodes from now on. Also, redrawing the diagram with the R=8.3k resistor going around the other circuit elements made the diagram much easier to understand.

    Thanks again!
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