find the energy stored in coupled inductors

Discussion in 'Homework Help' started by rogerloh4.0, Nov 9, 2012.

  1. rogerloh4.0

    Thread Starter New Member

    Sep 26, 2012
    10
    0
    Q: For the circuit in the figure, determine the energy stored in the coupled inductors at t = 1.5s.

    A: 39.4J

    Following is my approach on this problem:

    1/8 F => -j4Ω
    2H => j4Ω
    1H => j2Ω

    I determine I1 to be the current in the mesh on the primary side going clockwise, I2 to be the current in the mesh on the secondary side going clockwise.

    40 = 4I1-j2I2
    0 = (2+j2)I2-j2I1

    I1 = (1-j)I2

    (4-j6)I2 = 40

    I2 = 5.547\angle56.3^{\circ}
    I1 = 7.845\angle11.3^{\circ}

    i2 = 5.547cos(2t + \56.3^{\circ})
    i1 = 7.845cos(2t + \11.3^{\circ})

    at t = 1.5s:

    i1 = 7.845cos(\183.2^{\circ}) = -7.833A
    i2 = 5.547cos(\228.2^{\circ}) = -3.697A

    w = \\frac{1}{2}*2*(-7.833)^{2}+\frac{1}{2}*(-3.697)^{2}+(-3.697)(-7.833) = 97.148J

    Please tell me where have I got wrong, thanks!
     
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