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AC circuits and Kirchoff's laws

  1. Nov 11, 2005 #1
    I was trying to solve an exercise.

    We are given an AC circuit (see attachment). We had to find the current trough the resistor, the inductor and the capacitor.

    This can be solved using the impedance:
    The voltage by the source is sinusoidal: V=V0*sin(w*t)=(complex) V0*exp(j*w*t)
    The total (complex) impedance is: Z=R+(j*w*C+1/(j*w*L))^-1
    where j=sqrt(-1)

    The current trough the resistor is I0*exp(j*(w*t+phi))
    Since |Z|=V0/I0:

    For the current trough the inductor:

    For the current trough the capacitor:
    voltage(capacitor)=Q/C thus Q=C*voltage(capacitor)=C*(V0*exp(j*w*t)-I0*exp(j*(w*t+phi))*R)

    Now comes the problem. Kirchoff's laws should apply at any instant of time. Thus, using the junction rule, the algebraic sum of the three currents should be zero. (the current in the resistor + the one in the capacitor + the one in the inductor).
    However, using the results obtained above, this is not true.
    Can someone shed some light on the subject?

    Attached Files:

    Last edited: Nov 11, 2005
  2. jcsd
  3. Nov 11, 2005 #2
    I have managed to solve the problem. More calculations convinced me that the sum is, in fact, zero.
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