I have to calculate the voltage on nodes A,B,C using KCL nodal analysis method. (The two sources work at 60 Hz)

enter image description here

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?

  • \$\begingroup\$ Is this for 60 Hz? \$\endgroup\$
    – jonk
    Commented Sep 25, 2017 at 18:31

1 Answer 1


Assuming \$f=60\:\textrm{Hz}\$ and using the included schematic editor:


simulate this circuit – Schematic created using CircuitLab

From this, I get:

$$\begin{align*} V_B &= V_C + V_2\\\\ \frac{V_A}{4}+\frac{V_A}{3-2j}+\frac{V_A}{4j}&=\frac{V_1}{4}+\frac{V_C}{3-2j}+\frac{V_B}{4j}\\\\ \frac{V_B}{4j}+\frac{V_B}{2j}&=I_2+\frac{V_A}{4j}\\\\ \frac{V_C}{14}+\frac{V_C}{3-2j}+I_2&=\frac{V_A}{3-2j} \end{align*}$$

With \$V_1=100\angle 0^\circ\$ and \$V_2=220\angle 0^\circ\$, These solve out simultaneously as:

$$\begin{align*} V_A&=-28.9859544 - 29.8351871j\\ V_B&=-1.19022489 + 96.0915807j\\ V_C&=-221.190225 + 96.0915807j\\ I_2&=79.5274823 - 6.35381992j \end{align*}$$ or, $$\begin{align*} V_A&\approx &41&.6\:\textrm{V}& \angle -134.2&^\circ\\ V_B&\approx &96&.1\:\textrm{V}& \angle +90.7&^\circ\\ V_C&\approx &241&.2\:\textrm{V}& \angle +156.5&^\circ\\ I_2&\approx &79&.8\:\textrm{A}& \angle -4.6&^\circ \end{align*}$$

Computing \$V_B-V_A\$ (the voltage across \$L_1\$) I get \$V_B-V_A\approx 129\:\textrm{V} \angle +77.553^\circ\$. If all these were RMS voltages, then this would in fact mean that the peak voltage across \$L_1\$ would be roughly \$182\:\textrm{V}\$.


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