AC circuits and Kirchoff's laws

AI Thread Summary
The discussion revolves around solving an AC circuit exercise using Kirchhoff's laws and impedance. The circuit analysis involves calculating the current through a resistor, inductor, and capacitor using complex impedance and sinusoidal voltage. Initially, there was confusion regarding Kirchhoff's junction rule, as the sum of the currents appeared not to equal zero. However, further calculations confirmed that the algebraic sum of the three currents does indeed equal zero, aligning with Kirchhoff's laws. The resolution emphasizes the importance of careful calculations in AC circuit analysis.
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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:
I0=V0/|Z|
phi=arg(Z)=arctan(w*L/(R*(1-w^2*L*C)))

For the current trough the inductor:
Voltage(inductor)=L*dI/dt
thus
I=1/L*intrgral(V0*exp(j*w*t)-I0*exp(j*(w*t+phi))*R)

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)
and
I=dQ/dt
thus
I=d(C*(V0*exp(j*w*t)-I0*exp(j*(w*t+phi))*R))/dt

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?
 

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I have managed to solve the problem. More calculations convinced me that the sum is, in fact, zero.
 
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