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Simultaneous Equation

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data

    [eq1]: (Va-Vb)/(2i) = {[Vb-(10<0)]/(4-8i)} + {(Vb)/(6i)}

    [eq2]: (Va-Vb)/(2i) = (8<20) - Va

    2. Relevant equations

    Note: < is angle

    3. The attempt at a solution

    Solving [eq2]

    (Va-Vb) = 2i[(8<20) - Va]
    (Va-Vb) = (2i)(8<20) - (2i)(Va)
    -Vb = (2i)(8<20) - (2i)(Va) - Va
    Vb = -(2i)(8<20) + (2i)(Va) + Va
    Vb = -(16<110) + (2i)(Va) + Va

    Then plugging into [eq1]

    [Va + (16<110) + (2i)(Va) + Va]/[2i] = {[-(16<110) + (2i)(Va) + Va - (10<0)]/[4-8i]} + {[-(16<110) + (2i)(Va) + Va]/[6i]}

    [2Va + (16<110) + (2i)(Va)]/[2i] = {[-(16<110) + (2i)(Va) + Va - (10<0)]/[4-8i]} + {[-(16<110) + (2i)(Va) + Va]/[6i]}

    I am stuck at this point. If I could find the LCM of the the 2 fractions on the right and add them together I would be fine, but since there are imaginary numbers I'm not sure if that's even possible. (I tried using the lcm feature on my TI-89, didn't work).

    Or if there's a better way to solve this please let me know. Thanks.
     
  2. jcsd
  3. Nov 10, 2009 #2

    berkeman

    User Avatar

    Staff: Mentor

    Perhaps it can be solved in that polar format for the complex numbers, but I would find it easier to solve it in rectangular format. That is, convert each complex number from "magnitude>angle" format into A+jB format. Then to solve, the real parts have to be equal, and the imaginary parts have to be equal. You can convert it back to polar form in the end, if that's the format that the complex number answer is supposed to be in.
     
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