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RCL circuit alternating current, calculate current sum

  1. Mar 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider, in the circuit from the image,
    i1(t) = 5 cos(2t + 10º)
    v1(t) = 10 cos(2t - 60º).


    Find the value of the current ix(t).

    Options given:

    A: ix(t) = 9.9 cos(2t - 129.2º)
    B: ix(t) = 9 cos(2t - 29.2º)
    C: ix(t) = 99 cos(2t + 129.2º)
    D: ix(t) = 0.99 cos(2t + 130º)

    2. Relevant equations
    Ohm's law.

    3. The attempt at a solution
    My aproach was, first calculate the amount of current produced by the voltage source, combining the R + C + L in series and ignoring the current source, the result was 2<-59º(or 1.03 -1.7j)
    After I calculated the amount of current from the current source. I ignored the voltage source (short circuiting the circuit) and calculated L || (C + R) and the amount of current going to (C + R) is 11.86<-25.3º (or 10.7 -5.06j).

    Finally i summed the currents:

    1.03-1.7j + -(10.7-5.06j) ====> negative because it goes in the opposite direction from the one in the image.
    The result was -9.67+3.36j or 99.7<-19.16º [+180º];
    That is 99.7 < 160.84º or 99.7 cos(2t + 160.84º)
    That is near option C but not exactly! What am I doing wrong?

    Thanks for any help.
  2. jcsd
  3. Mar 11, 2014 #2


    User Avatar

    Staff: Mentor

    Can you show your calculations of the superposition currents in more detail? The values you're getting don't look right according to my own working, so I'd like to see how you're carrying out the operations.
  4. Mar 12, 2014 #3

    I reviewed all my calculations, I had huge mistakes all over!

    The current produced by the voltage source should be 2<-113º (or -0.78-1.84j).
    The amount of current from the current source (using a current divider) should be 8<47º (or 5.46+5.85j).

    -0.78-1.84j - (5.46+5.85j) = -6.24-7.69j

    -6.24-7.69j = 9.90<50.94º

    ix(t) = 9.9cos(2t + 50.94º) = 9.9cos(2t - 129.06º) [because cos(50.94º) = cos(-129.06º)]

    So... option A is correct.
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