Question

Determine the symmetrical components of the following line currents:

(a) ${\mathit{l}}_{\mathbf{a}}\mathbf{=}\mathbf{6}\mathbf{\angle }\mathbf{90}\mathbf{°}\mathbf{,}\mathbf{}{\mathit{l}}_{\mathbf{b}}\mathbf{=}\mathbf{6}\mathbf{\angle }\mathbf{320}\mathbf{°}\mathbf{}{\mathit{l}}_{\mathbf{c}}\mathbf{=}\mathbf{6}\mathbf{\angle }\mathbf{220}\mathbf{°}\mathit{A}$and (b)${\mathit{l}}_{\mathbf{a}}\mathbf{=}\mathit{j}\mathbf{40}\mathbf{,}\mathbf{}{\mathit{l}}_{\mathbf{b}}\mathbf{=}\mathbf{40}\mathbf{,}\mathbf{}{\mathit{l}}_{\mathbf{c}}\mathbf{=}\mathbf{0}\mathit{A}$

Short answer

(a) Therefore, the symmetrical components of line currents are given as $l_{\mathrm{c}}=-j0.57A,l_1=j5.94A$and $l_2=j0.633A$

(b) Therefore, the symmetrical components of line currents are given as

$l_0=18.86\angle 45°,l_1=25.67\angle 105°A$and$l_2=6.91\angle 165°A$

Step-by-step solution

Write the given data from the question.

Assume that a line current of an un-balanced three-phase network are $l_0=6\angle 90°A,l_b=6\angle 320°A,and{l}_c=6\angle 220°A$

Assume that a line current of an un-balanced three-phase network are$l_a=40AA,l_b=40A,and{l}_c=0A$

Determine the formula of phase sequence current.

Write the formula of positive sequence current.

${\mathit{l}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left(l_a+a{l}_b+a^2{l}_c\right)$ …… (1)

Here,${\mathit{l}}_{\mathbf{a}}$, ${\mathit{l}}_{\mathbf{b}}$and ${\mathit{l}}_{\mathbf{c}}$are line current of three phase network.

Write the formula of negative sequence current.

${\mathit{l}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left(l_a+a^2{l}_b+a{l}_c\right)$ ..........(2)

Here, role="math" localid="1654594068297" ${\mathit{l}}_{\mathbf{a}}$, role="math" localid="1654594235866" ${\mathit{l}}_{\mathbf{b}}$and role="math" localid="1654594240237" ${\mathit{l}}_{\mathbf{c}}$are line current of three phase network.

Write the formula of zero sequence current.

${\mathit{l}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left(l_a+l_b+l_c\right)$ .........(3)

Here,${\mathit{l}}_{\mathbf{a}}$, role="math" localid="1654594272256" ${\mathit{l}}_{\mathbf{b}}$and ${\mathit{l}}_{\mathbf{c}}$are line current of three phase network.

Determine the line current of an un-balanced three-phase network.

First we convert the line voltage in polar form into rectangular form for further solving.

Write the value of line current$l_a$of three phase network

$$\begin{gathered} l_a=6\angle 90° \\ =j6 \end{gathered}$$

Write the value of line current $l_b$ of three phase network.

$$\begin{gathered} l_b=6\angle 320° \\ =4.60-j3.86 \end{gathered}$$

Write the value of line current $l_c$ of three phase network

$$\begin{gathered} l_c=6\angle 220° \\ =-4.60-j3.86 \end{gathered}$$

Determine the positive sequence current.

Substitute the value of $l_a$,$l_b$and $l_c$ into equation (1).

$$\begin{gathered} l_{1=}\frac{1}{3}\left(6\angle 90°+6\angle \left(320°+120°\right)+6\angle \left(220°+240°\right)\right) \\ =\frac{1}{3}\left(6\angle 90°+6\angle 440°+6\angle 460°\right) \\ =\frac{1}{3}\left(j6+1.04+j5.91-1.04+j5.91\right) \\ =\frac{1}{3}\left(j17.82\right) \end{gathered}$$

As further solve as

$l_1=j5.94A$

Determine the negative sequence current.

Substitute the value of $l_a$,$l_b$and $l_c$into equation (2).

$$\begin{gathered} l_2=\frac{1}{3}\left(6\angle 90°+6\angle \left(320°+240°\right)+6\angle \left(220°+120°\right)\right) \\ =\frac{1}{3}\left(6\angle 90°+6\angle 560°+6\angle 340°\right) \\ =\frac{1}{3}\left(j6+5.64-j2.05+5.64-j2.05\right) \\ =j0.633A \end{gathered}$$

Determine the zero sequence current.

Substitute the value of $l_a$,$l_b$ and $l_c$into equation (3).

$$\begin{gathered} l_0=\frac{1}{3}\left(6\angle 90°+6\angle 320°+6\angle 220°\right) \\ =\frac{1}{3}\left(j6+4.60-j3.86-4.60-j3.86\right) \\ =\frac{1}{3}\left(j6+4.60-j3.86-4.60-j3.86\right) \\ =\frac{1}{3}\left(-j1.72\right) \end{gathered}$$

As further solve as

$l_c=-j0.57A$

Therefore, the symmetrical components of line currents are given as , $l_c=-j0.57A,l_1=j5.94A$and $l_2=j0.633A$.

(b) Determine the line current of an un-balanced three-phase network.

First we convert the line voltage in polar form into rectangular form for further solving.

Write the value of line current$l_a$of three phase network.

$$\begin{gathered} l_a=j40 \\ =40\angle 90° \end{gathered}$$

Write the value of line current $l_b$of three phase network.

$$\begin{gathered} l_b=40 \\ =40\angle 0° \end{gathered}$$

Write the value of line current$l_c$of three phase network.

$l_c=0$

Determine the positive sequence current.

Substitute the value of $l_a$,$l_b$and$l_c$into equation (1).

$$\begin{gathered} l_1=\frac{1}{3}\left(40\angle 90°+40\angle \left(0°+120°\right)+0\right) \\ =\frac{1}{3}\left(40\angle 90°+40\angle 120°\right) \\ =\frac{1}{3}\left(j40-20+j34.64\right) \\ =\frac{1}{3}\left(-20+j74.64\right) \end{gathered}$$

Further solve as

$l_1=-6.67+j24.88$or$25.67\angle 105°A$

Determine the negative sequence current.

Substitute the value of $l_a$,$l_b$and$l_c$into equation (2).

$$\begin{gathered} l_2=\frac{1}{3}\left(40\angle 90°+40\angle \left(0°+240°\right)+0\right) \\ =\frac{1}{3}\left(40\angle 90°+40\angle 240°\right) \\ =\frac{1}{3}\left(j40-20-j34.64\right) \\ =\frac{1}{3}\left(-20+j5.36\right) \end{gathered}$$

Further solve as

$l_2=-6.67+j1.79$or$6.91\angle 165°A$

Determine the zero sequence current.

Substitute the value of $l_a$, $l_b$and $l_c$ into equation (3).

localid="1654597637264" $l_{0=}\frac{1}{3}\left(j40+40+0\right)$

$=13.33+j13.33$or$18.86\angle 45°A$

Therefore, the symmetrical components of line currents are given as localid="1654597798226" $l_0=18.86\angle 45°,$

$l_1=25.76\angle 105°A$and $l_2=6.91\angle 165°A$.