I have to calculate the voltage on nodes A,B,C using KCL nodal analysis method. (The two sources work at 60 Hz)
I set the next equations system:
on node A: $$\frac{100\angle 0}{4\angle0} +\frac{VA -VC}{4\angle90°}+\frac{VA - VB}{3.6\angle-33.7°} = 0$$ on node B: $$\frac{VB}{14\angle0}+\frac{VB-VA}{3.6\angle-33.7°}=0$$ on node C: $$\frac{VC}{2\angle90°}+\frac{VC-VA}{4\angle90°}=0$$
I suppose that voltage on Node C is the same of the source connected to it (220V), so the third equations results: $$200\angle0(\frac{1}{2\angle90°}+\frac{1}{4\angle90°})=\frac{VA}{4\angle90°}$$ then: $$220\angle0°*0.75\angle-90°=\frac{VA}{4\angle90°}$$ hereby: $$VA = 165\angle-90° * 4\angle90° = 660\angle0 V$$
I think this value is a bit too high. I simulated the circuit with an online page, and the branch of the 10.6mH inductor (wich produces a inductive reactance of j4 Ohm) should have a peak voltage of around 172 V. So I think Im doing something wrong, but I can not make it out. Could you help me?