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How would you approach this problem: enter image description here

I went about it by first converting everything to its phasor values, then applying the principle of superposition. According to my calculations:

$\omega = 2, Z_L=j\omega L = 2j, Z_C=\frac{-j}{\omega C}=-j$

I ran into an issue when I tried to calculate the current from the voltage source. This is my working:

enter image description here

Is there something wrong in my approach? I was going to then get the current from the fact that the voltage drop across the two branches should be the same, and so since the impedance of the capacitor is half that of the inductors, to balance it would draw double the current. And, of course, the sum of those two currents is the total current that I calculated by dividing 2.5 by the total impedance. Is this a valid approach?

Edit: Holy crap, I've lost my mind. Totally forgot that with current sources for superposition they act as an open circuit. If that was the only problem, then the solution should be as follows, correct?:

enter image description here

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  • $\begingroup$ Just one note. Things can have 0 impedance if the frequency of the load is such, that the reactances of reactive elements like ideal coils and ideal capacitors are opposite in value. $\endgroup$
    – MaDrung
    Jan 17 '18 at 7:19
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$$\text {source 1: }I_g=0\quad\Longrightarrow\quad I'_c=\frac {V_g}{Z_C+Z_L}$$

$$\text {source 2: }V_g=0\quad\Longrightarrow\quad I''_c=I_g\frac {Z_L}{Z_C+Z_L}$$

Thus $$I_c=I_c'+I_c''=\frac {I_gZ_L+V_g}{Z_C+Z_L}=\frac {3j+2.5}{j}=3-2.5j$$

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