Question

A balanced $\mathbf{∆}$-connected load is fed by a three-phase supply for which phase C is open and phase A is carrying a current of $\mathbf{10}\mathbf{\angle }\mathbf{90}\mathbf{°}\mathbf{A}$. Find the symmetrical components of the line currents. (Note that zero-sequence currents are not present for any three-wire system.)

Short answer

The sequence current for phase A, B and C are

$$\begin{gathered} l_{a0}=0\mathrm{A} \\ {l}_{a1}=5.774\angle -30°\mathrm{A} \\ {l}_{a2}=5.774\angle -30°\mathrm{A} \end{gathered}$$,$$\begin{gathered} l_{b0}=0\mathrm{A} \\ {l}_{b1}=5.774\angle -150°\mathrm{A} \\ {l}_{b2}=5.774\angle -150°\mathrm{A} \end{gathered}$$and$$\begin{gathered} l_{c0}=0\mathrm{A} \\ {l}_{c1}=5.774\angle -90°\mathrm{A} \\ {l}_{c2}=5.774\angle -90°\mathrm{A} \end{gathered}$$

Step-by-step solution

Write the given data from the question.

The current in the in-phase C$l_c=0$.

The current in phase A$l_a=10\angle 0°A$.

Determine the symmetrical component of the line current.

The sum of the all three phases are equal to zero for the $∆$-connected load.

${\mathbf{l}}_{\mathbf{a}}\mathbf{+}{\mathbf{l}}_{\mathbf{b}}\mathbf{+}{\mathbf{l}}_{\mathbf{c}}\mathbf{=}\mathbf{0}$ …… (1)

The equation to calculate the positive sequence current for $a$is given as follows.

${\mathbf{l}}_{\mathbf{a}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left(l_a+{\mathrm{al}}_b+l_c\right)$ …… (2)

Here

$$\begin{gathered} \mathbf{a}\mathbf{=}\mathbf{1}\mathbf{\angle }\mathbf{120}\mathbf{°} \\ {\mathbf{a}}^{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{\angle }\mathbf{240}\mathbf{°} \end{gathered}$$

The equation to calculate the Negative sequence current for is given as follows.

${\mathbf{l}}_{\mathbf{a}\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}{\mathbf{l}}_{\mathbf{a}}\mathbf{+}{\mathbf{al}}_{\mathbf{b}}\mathbf{+}{\mathbf{l}}_{\mathbf{c}}\mathbf{)}$ …… (3)

The equation to calculate the positive sequence current for $b$is given as follows.

${\mathbf{l}}_{\mathbf{b}\mathbf{1}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{a}\mathbf{1}}$ …… (4)

The equation to calculate the Negative sequence current for $b$is given as follows.

${\mathbf{l}}_{\mathbf{b}\mathbf{2}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{a}\mathbf{2}}$ …… (5)

The equation to calculate the positive sequence current for $c$is given as follows.

${\mathbf{l}}_{\mathbf{c}\mathbf{1}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{a}\mathbf{1}}$ …… (6)

The equation to calculate the Negative sequence current for $c$is given as follows.

${\mathbf{l}}_{\mathbf{c}\mathbf{2}}\mathbf{=}{\mathbf{al}}_{\mathrm{b}2}$ …… (7)

Calculate the symmetrical component of the line current.

In a $∆$-connected three-phase system, line currents have no zero-sequence component.

$$\begin{gathered} l_{a0}=l_{b0} \\ {l}_{a0}=l_{c0} \\ {l}_{a0}=0\mathrm{A} \end{gathered}$$

Calculate the current in phase $b$.

Substitute $0\mathrm{A}$for $l_c\mathrm{and}10\angle 0°\mathrm{A}\mathrm{for}{l}_a$into equation (1).

$$\begin{gathered} 10\angle 0°+l_b+0=0 \\ {l}_b=-10\angle 0°A \\ {l}_b=-10\angle 180°A \end{gathered}$$

Calculate the positive sequence current for $a$.

Substitute $10\angle 180°A\mathrm{for}{l}_{b2}\mathrm{oA}\mathrm{for}{l}_c$and $10\angle 0°A\mathrm{for}{l}_a$into equation (2).

$$\begin{gathered} l_{a1}=\frac{1}{3}\left(10\angle 0°+10\angle \left(180°+120°\right)+0\right) \\ {l}_{a1}=\frac{1}{3}\left(15-j5\sqrt{3}\right) \\ {l}_{a1}=5-j2.887 \\ {l}_{a1}=5.774\angle -30°A \end{gathered}$$

Calculate the negative sequence current for $a$.

Substitute $10\angle 180°A\mathrm{for}{l}_{b2}\mathrm{oA}\mathrm{for}{l}_c$and $10\angle 0°A\mathrm{for}{l}_a$into equation (3).

$$\begin{gathered} l_{a2}=\frac{1}{3}\left(10\angle 0°+10\angle \left(180°+24°\right)+0\right) \\ {l}_{a2}=\frac{1}{3}\left(15-j5\sqrt{3}\right) \\ {l}_{a2}=5-j2.887 \\ {l}_{a2}=5.774\angle -30°A \end{gathered}$$

Calculate the positive sequence current for $b$.

Substitute $5.774\angle -30°\mathrm{A}\mathrm{for}{l}_{a1}$into equation (4).

$$\begin{gathered} l_{b1=}5.774\angle \left(240-30\right)° \\ {l}_{b1=}5.774\angle 210°A \\ {l}_{b1}=5.774\angle 150°A \end{gathered}$$

Calculate the negative sequence current for $b$.

Substitute $5.774\angle -30°\mathrm{A}\mathrm{for}{l}_{a2}$into equation (5).

$$\begin{gathered} l_{b2}=5.774\angle \left(240+30\right)° \\ {l}_{b2}=5.774\angle 270°A \\ {l}_{b2}=5.774\angle 150°A \end{gathered}$$

Calculate the positive sequence current for $c$.

Substitute $5.774\angle 150°A\mathrm{for}{l}_{b1}$into equation (6).

$\begin{array}{l}{l}_{c1}=5.774\angle \left(210+240\right)°\\ l_{c1}=5.774\angle 90°A\end{array}$

Calculate the negative sequence current for $c$.

Substitute $5.774\angle 150°A\mathrm{for}{l}_{b2}$into equation (7).

$$\begin{gathered} l_{c2}=5.774\angle \left(150+120\right)° \\ {l}_{c2}=5.774\angle 270°A \\ {l}_{c2}=5.774\angle -90°A \end{gathered}$$

Hence the sequence current for phase A, B and C are,

$$\begin{gathered} l_{a0}=0A \\ {l}_{a1}=5.774\angle -30°A \\ {l}_{a2}=5.774\angle -30°A \end{gathered}$$

Also,

$$\begin{gathered} l_{b0}=0A \\ {l}_{b1}=5.774\angle -150°A \\ {l}_{b2}=5.774\angle 150°A \end{gathered}$$

And,

$$\begin{gathered} l_{c0}=0A \\ {l}_{c1}=5.774\angle 90°A \\ {l}_{c2}=5.774\angle -90°A \end{gathered}$$