Question

Determine the symmetrical components of the following line currents:

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

Short answer

Answer

(a) Therefore, the symmetrical components of line currents are given as $I_0=-j0.57A,I_1=j5.94A$and $I_2=j0.633A$.

(b) Therefore, the symmetrical components of line currents are given as$I_0=18.86\angle {45}^{°},I_1=25.76\angle {105}^{°}A$and $I_2=6.91\angle {165}^{°}A$.

Step-by-step solution

Write the given data from the question.

Assume that a line current of an un-balanced three-phase network are$I_a=6\angle {90}^{°}A,I_b=6\angle {320}^{°}A$, and $I_c=6\angle {220}^{°}A$.

Assume that a line current of an un-balanced three-phase network are$I_a=j40AA,I_b=40A,$and $I_c=0A$.

Determine the formula of phase sequence current.

Write the formula of positive sequence current.

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

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

Write the formula of negative sequence current.

${\mathit{I}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{I}}_{\mathbf{a}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}{\mathbf{I}}_{\mathbf{b}}\mathbf{+}{\mathbf{aI}}_{\mathbf{c}}\mathbf{)}$ …… (2)

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

Write the formula of zero sequence current.

${\mathit{I}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{I}}_{\mathbf{a}}\mathbf{+}{\mathbf{I}}_{\mathbf{b}}\mathbf{+}{\mathbf{I}}_{\mathbf{c}}\mathbf{)}$ …… (3)

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

(a) 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$I_a$of three phase network.

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

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

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

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

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

Determine the positive sequence current.

Substitute the value ofrole="math" localid="1656317196402" $I_a$,width="14" height="24" role="math">Iband $I_c$into equation (1).

$$\begin{gathered} I_1=\frac{1}{3}\left(6\angle {90}^{°}+6\angle \left({220}^{°}+{240}^{°}\right)\right) \\ =\frac{1}{3}\left(6\angle {90}^{°}+6\angle {440}^{°}+6\angle {240}^{°}\right) \\ =\frac{1}{3}\left(j6+1.04+j5.91-1.04+j5.91\right) \\ =\frac{1}{3}\left(j17.82\right) \end{gathered}$$

As further solve as

$I_1=j5.94A$

Determine the negative sequence current.

Substitute the value ofrole="math" localid="1656317543125" $I_a$,$I_b$and $I_c$into equation (2).

$$\begin{gathered} I_2=\frac{1}{3}\left(6\angle {90}^{°}+6\angle \left({320}^{°}+{240}^{°}\right)+6\angle \left(220°+120°\right)\right) \\ =\frac{1}{3}\left(6\angle {90}^{°}+6\angle 560°+6\angle 340°\right) \\ =\frac{1}{3}\left(j6-5.64-j2.05+5.64-j2.05\right) \\ =j0.633A \end{gathered}$$

Determine the zero sequence current.

Substitute the value of$I_a$,$I_b$and $I_c$into equation (3).

$$\begin{gathered} I_0=\frac{1}{3}\left(6\angle {90}^{°}+6\angle 320°+6\angle 220°\right) \\ =\frac{1}{3}\left(j6+4.60-j3.86-4.60-j3.86\right) \\ =\frac{1}{3}\left(j6+4.60-j3.86-4.60-j3.86\right) \\ =\frac{1}{3}\left(-j1.72\right) \end{gathered}$$

As further solve as

$I_0=-j0.57A$

Therefore, the symmetrical components of line currents are given as $I_0=-j0.57A,I_1=j5.94A$and $I_2=j0.633A$.

(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$I_a$of three phase network.

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

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

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

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

$I_c=0$

Determine the positive sequence current.

Substitute the value of$I_a$,$I_b$and $I_c$into equation (1).

$$\begin{gathered} I_1=\frac{1}{3}\left(40\angle 90°+40\angle \left(0°+120°\right)+0\right) \\ =\frac{1}{3}\left(40\angle 90°+40\angle 120°\right) \\ =\frac{1}{3}\left(j40-20+j34.64\right) \\ =\frac{1}{3}\left(-20+j74.64\right) \end{gathered}$$

Further solve as,

$I_1=-6.67+j24.88or25.76\angle 105°A$

Determine the negative sequence current.

Substitute the value of$I_a$,$I_b$and $I_c$into equation (2).

$$\begin{gathered} I_2=\frac{1}{3}\left(40\angle 90°+40\angle \left(0°+240°\right)+0\right) \\ =\frac{1}{3}\left(40\angle 90°+40\angle 240°\right) \\ =\frac{1}{3}\left(j40-20-j34.64\right) \\ =\frac{1}{3}\left(-20+j5.36\right) \end{gathered}$$

Further solve as

$I_2=-6.67+j1.79or6.91\angle 165°A$

Determine the zero sequence current.

Substitute the value of$I_a$,$I_b$and $I_c$into equation (3).

$$\begin{gathered} I_0=\frac{1}{3}\left(j40+40+0\right) \\ =13.33+j13.33or18.86\angle 45°A \end{gathered}$$

Therefore, the symmetrical components of line currents are given as$I_0=18.86\angle 45°,I_1=25.76\angle 105°$and $I_2=6.91\angle {165}^{0}A$.