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Superposition Theorem Circuit, need some help

  1. Oct 27, 2009 #1
    Just found this forum and found some great help here so far just reading some posts. But I am stuck on a question in my homework. I am asking you to help check my work and make sure im going in the right direction. I missed my lecture due to illness and have to figure this out on my own. Thank you in advance.

    The question asks for the 3 different currents through R2; With E1 as the source only, E2 as the source only, and both E1 and E2.

    Here are the variables:
    E1= 11V
    E2= 8V
    R1= 430ohms
    R2= 250ohms
    R3= 200ohms
    R4= 130ohms

    Here is the diagram

    image001.gif

    This is my work with E1 as the only source Voltage

    Rt= R1 + ((R2x(R3+R4)) / R2 + R3 + R4)
    Rt= 572.2414ohms
    Rt (of parallel circuit only is 142.2414ohms)

    It= E/Rt = 11/572...
    It= 19.223mA

    Based on KVL, the current of It = R1 = Ra (the parallel circuit of R2 and (R3 + R4 in series)

    Using the current divider, I got

    I2= Rt (of parallel) / Rt(prl) + R2 * It
    I2 = 6.971mA

    So far so good?

    With E2 as the only source voltage, I made R1 and R2 the parallel circuit and had the branch of R3, R4 and R2 as my series circuit

    Rt = R4 + R3 + (R1xR2 / (R1+R2))
    Rt= 488.0882ohms
    Rt of parallel = 158.0882ohms

    It = E/Rt
    It = 16.3905 mA

    Current divider rule

    I2 = 158.0882 / (158.0082 + 250) ohms * 16.3905mA
    I2 = 6.3495

    I do not have the answers to this question since its a online test and each section is worth 1% of my final mark. I would like you to help make sure I am doing this question right so I can move onto the other ones.

    Where I get stuck is how to find the answer with both voltage sources.

    I assume based on the fact there series adding that I would add them together for a total of 13.3205mA but this is where I am unsure.

    On top of that, for my future questions, how would I find Rt of the entire circuit with both sources? For the current total, since its a series adding circuit, could I add both to get 3X.XX mA for the It. If thats the case I could find Rt using ohms law, but any help is much appreciated.

    Thank you
     
  2. jcsd
  3. Oct 27, 2009 #2
    You got this right.

    You've got the current divider wrong. The divider fraction should be:

    (R3+R4)/(R2+R3+4)

    so that I2 = It * (R3+R4)/(R2+R3+4) = 10.937 mA

    You need to make the parallel circuit R1 in parallel with (R3+R4); this is then in series with R2. Redo your calculations. You don't need to use a current divider rule here because I2 is It when only E2 is active.

    With E2 only active, I get a current I2 = -18.319 mA. The minus sign means the current is directed upward.

    Yes, the final answer will be the sum of the two individual currents, which I get as -7.382 mA.
     
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