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Superposition: Find V(t) in complex (AC) circuit

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data

    upload_2015-4-16_15-42-26.png
    2. Relevant equations
    phasor forms
    voltage division
    current division

    3. The attempt at a solution

    Using superposition, considering only the varying voltage source.

    Z (L) = 4j
    Z (C) = 5j

    Total impedance:
    4 is parallel with 5 = 2.44 + 1.95j
    series with 1 + 4j
    Total impedance: 3.44 + 5.95j = 6.87 <60°

    Voltage source: 10<0°

    Voltage division Vo(t) = 10<0° * (1/(6.87 <60°)) = 1.45<-60°
    = 1.45cos (2t - 60°)

    However this is off from the cos part of the correct answer.

    What's wrong with my working?

    Thank you very much.
     
  2. jcsd
  3. Apr 16, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    Check the signs of your reactive component impedances.
     
  4. Apr 16, 2015 #3
    Thank you very much for your response.

    With the capacitor's impedance set to negative and the correct answer pops up.

    What is the rule to determine whether an impedance is positive or negative?
     
  5. Apr 16, 2015 #4

    gneill

    User Avatar

    Staff: Mentor

    In general inductive impedances are positive while capacitive impedances are negative.

    This comes from the formulas for impedances of inductors and capacitors:

    ##Z_L = j \omega L##
    ##Z_C = \frac{1}{j \omega C}##

    When the j in the denominator of the capacitive impedance is "moved" to the numerator, its sign changes.
     
  6. Apr 16, 2015 #5
    I see, thank you!
     
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