0
\$\begingroup\$

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?

\$\endgroup\$
  • \$\begingroup\$ Is this for 60 Hz? \$\endgroup\$ – jonk Sep 25 '17 at 18:31
1
\$\begingroup\$

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

schematic

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}\$.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.