Intro to three phase systems - 4

Electricity and Magnetism Level 5

Consider the following three phase symmetrical system.

Let the line voltage be $V = 220\angle{0^{\circ}} \begin{bmatrix} 1 \\ \alpha^2 \\ \alpha \end{bmatrix}$ , and consider the following impedances:

$\begin{cases} Z_a = 3 + 4j \\ Z_b = 10j\\ Z_c = -10j \end{cases}$

What is the magnitude of the voltage difference from the node $N'$ to the ground? In other words, what is $||V_{NN'}||$ ?

Assumptions and Clarifications

$j = \sqrt{-1}$
$\alpha = 1\angle{120^{\circ}} = 1\cdot e^{j\frac{2\pi}{3}} = (\frac{-1}{2} + \frac{j\sqrt{3}}{2})$

1 solution

Mateus Lucas
Sep 14, 2020

The phase voltages are:

$\begin{bmatrix} V_a\\V_b\\V_c\end{bmatrix}=\dfrac{220}{\sqrt{3}}\angle{-30^\circ}\begin{bmatrix} 1\\a^2\\a\end{bmatrix}$

Let $g_{ij}$ and $Z_{ij}$ be, respectivelly, the admittance and the impedance of the parallel association of $Z_i$ and $Z_j$ :

$g_{ab}=\dfrac{1}{Z_a}+\dfrac{1}{Z_b}=0.12+0.26j\implies Z_{ab}\cong1.46+3.17j$

$g_{bc}=\dfrac{1}{Z_b}+\dfrac{1}{Z_c}=0\implies Z_{bc}\to\infty$

$g_{ca}=\dfrac{1}{Z_c}+\dfrac{1}{Z_a}=0.12-0.06j\implies Z_{ca}\cong6.67+3.33j$

Using superposition and voltage divider:

Case 1: $V_b=V_c=0 \implies V_{N'N1}=V_a\cdot \lim_{Z_{bc}\to+\infty}\dfrac{Z_{bc}}{Z_a+Z_{bc}}=V_a\cong127.017\angle{-30^\circ}$
Case 2: $V_a=V_c=0 \implies V_{N'N2}=V_b\cdot \dfrac{Z_{ca}}{Z_b+Z_{ca}}\cong63.509\angle{173.13^\circ}$
Case 3: $V_a=V_b=0 \implies V_{N'N3}=V_c\cdot \dfrac{Z_{ab}}{Z_c+Z_{ab}}\cong63.509\angle{-126.87^\circ}$

$V_{N'N}=V_{N'N1}+V_{N'N2}+V_{N'N3}\cong107.084\angle{-85.26^\circ}$

$||V_{NN'}||\cong107.084$