Question

A single-line diagram of a four-bus system is shown in Figure 9.27. Equipment ratings and per-unit reactances are given as follows.

Machine and: $\begin{array}{l}\mathbf{100}\mathbf{MVA}\\ {\mathbf{X}}_0=\mathbf{0}\mathbf{.}\mathbf{04}\end{array}$ $\begin{array}{l}\mathbf{20}\mathbf{kV}\\ {\mathbf{X}}_n=\mathbf{0}\mathbf{.}\mathbf{05}\end{array}$ ${\mathbf{X}}_1={\mathbf{X}}_2=\mathbf{0}\mathbf{.}\mathbf{2}$

Transformers and: $\begin{array}{l}\mathbf{100}\mathbf{MVA}\\ {\mathbf{X}}_1={\mathbf{X}}_2={\mathbf{X}}_0=\mathbf{0}\mathbf{.}\mathbf{08}\end{array}$ $\mathbf{20}\mathbf{\Delta }/\mathbf{345}\mathbf{Y}\mathbf{kV}$

On a base of 100 MVA and 345 kV in the zone of the transmission line, the series reactances of the transmission line are${\mathbf{X}}_1={\mathbf{X}}_2=\mathbf{0}\mathbf{.}\mathbf{15}$and ${\mathbf{X}}_0=\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$. (a) Draw each of the sequence networks and determine the bus impedance matrix for each of them. (b) Assume the system to be operating at nominal system voltage without prefault currents when a bolted line-to-line fault occurs at bus 3. Compute the fault current, the line-to-line voltages at the faulted bus, and the line-to-line voltages at the terminals of machine 2. (c) Assume the system to be operating at nominal system voltage without prefault currents, when a bolted double line-to ground fault occurs at the terminals of machine 2. Compute the fault current and the line-to-line voltages at the faulted bus.

Short answer

(a)

Draw the zero-sequence network of the system.

Draw the positive-sequence network of the system.

Draw the negative-sequence network of the system.

(b)

The value of zero sequence bus impedance matrix is $\left[\begin{array}{cccc}j0.19& 0& 0& 0\\ 0& j0.08& j0.5& 0\\ 0& j0.5& j0.08& 0\\ 0& 0& 0& j0.19\end{array}\right]$.

The value of positive sequence bus impedance matrix is $\left[\begin{array}{cccc}j0.2& j0.08& j0.23& j0.31\\ j0.08& 0& j0.15& j0.23\\ j0.23& j0.15& 0& j0.08\\ j0.31& j0.23& j0.08& j0.2\end{array}\right]$.

The value of negative sequence bus impedance matrix is $\left[\begin{array}{cccc}j0.2& j0.08& j0.23& j0.31\\ j0.08& 0& j0.15& j0.23\\ j0.23& j0.15& 0& j0.08\\ j0.31& j0.23& j0.08& j0.2\end{array}\right]$

(b)

The value of positive, negative and zero sequence component of the bolted line-to-line fault occurs at bus-3 $I_1$, $I_2$and $I_0$is $-j2.556\text{\hspace{0.33em}pu}$, $j2.556\text{\hspace{0.33em}pu}$and 0 pu.

The value of line to line voltages at the faulted bus ${V^{\text{'}\text{'}}}_a$, ${V^{\text{'}\text{'}}}_{\text{b}}$and $V^{\text{'}\text{'}}\text{c}$are 1.732 pu, 0.886 pu and 0.866 pu.

The value of line to line voltages ${V^{\text{'}\text{'}}}_{\text{a}}$, ${V^{\text{'}\text{'}}}_b$and ${V^{\text{'}\text{'}}}_c$at the terminals of the machine are 1.732 pu, 4.26 pu and -0.798 pu.

(c)

The value of sub-transient fault currents $\left[\begin{array}{c}{I^{\text{'}\text{'}}}_{\text{a}}\\ {I^{\text{'}\text{'}}}_{\text{b}}\\ {I^{\text{'}\text{'}}}_{\text{c}}\end{array}\right]$at the faulted bus is $\left[\begin{array}{c}0\\ 5.038\angle 149.17°\\ 5.038\angle 30.83°\end{array}\right]$.

The value of line to line voltages, ${V^{\text{'}\text{'}}}_{\text{a}}$, ${V^{\text{'}\text{'}}}_b$and${V^{\text{'}\text{'}}}_c$ at the faulted bus are 1.704 pu, 0 pu and 0 pu.

Step-by-step solution

Write the given data from the question.

Thevalue of base MVA is 100 MVA.

The value of series reactance of the line$X_1=X_2=0.15$ and $X_0=0.5\text{\hspace{0.33em}per\hspace{0.33em}unit}$.

The value of transmission line voltage is 345 kV.

Determine the formula of bolted line-to-line fault occurs at bus-3, line to line voltages at the faulted bus, line to line voltages at the terminals of the machine 2, sub-transient fault currents at the faulted bus and line to line voltages at the faulted bus.

Write the formula of positive-sequence component of the bolted line-to-line fault.

${\mathbf{I}}_1=\frac{{\mathbf{E}}_g}{{\mathbf{Z}}_1\mathbf{+}{\mathbf{Z}}_2}$ …… (1)

Here,${\mathbf{E}}_g$is generated voltage,${\mathbf{Z}}_1$is positive sequence impedance and${\mathbf{Z}}_2$is negative sequence impedance.

Write the formula of negative-sequence component of the bolted line-to-line fault.

${\mathbf{I}}_2=-{\mathbf{I}}_1$ …… (2)

Write the formula of zero-sequence component of the bolted line-to-line fault.

${\mathbf{I}}_0=\mathbf{0}$ …… (3)

Write the formula of line to line voltages at the faulted bus ${V^{\text{'}\text{'}}}_a$.

${V^{\text{'}\text{'}}}_a=\sqrt{3}{V^{\text{'}\text{'}}}_{\mathrm{ag}}$ …… (4)

Here,${V^{\text{'}\text{'}}}_{\mathrm{ag}}$is phase to ground voltages.

Write the formula of line to line voltages at the faulted bus ${V^{\text{'}\text{'}}}_b$.

${V^{\text{'}\text{'}}}_b=\sqrt{3}{V^{\text{'}\text{'}}}_{\mathrm{bg}}$ …… (5)

Here,${V^{\text{'}\text{'}}}_{\mathrm{bg}}$ is phase to ground voltages

Write the formula of line to line voltages at the faulted bus ${V^{\text{'}\text{'}}}_{\mathrm{cg}}$.

${V^{\text{'}\text{'}}}_c=\sqrt{3}{V^{\text{'}\text{'}}}_{\mathrm{cg}}$ …… (6)

Here,${V^{\text{'}\text{'}}}_c$ is phase to ground voltages.

Write the formula of line to line voltages,${V^{\text{'}\text{'}}}_a$ at the terminals of the machine 2.

${V^{\text{'}\text{'}}}_a=\sqrt{3}{V^{\text{'}\text{'}}}_{\mathrm{ag}}$ …… (7)

Here,${V^{\text{'}\text{'}}}_{\mathrm{ag}}$ is phase to ground voltages.

Write the formula of line to line voltages,${V^{\text{'}\text{'}}}_b$ at the terminals of the machine 2.

${V^{\text{'}\text{'}}}_b=\sqrt{3}{V^{\text{'}\text{'}}}_{\mathrm{bg}}$ …… (8)

Here, ${V^{\text{'}\text{'}}}_{\mathrm{bg}}$is phase to ground voltages.

Write the formula of line to line voltages, role="math" localid="1656588951604" ${V^{\text{'}\text{'}}}_c$at the terminals of the machine.

${V^{\text{'}\text{'}}}_c=\sqrt{3}{V^{\text{'}\text{'}}}_{\mathrm{cg}}$ …… (9)

Here,${V^{\text{'}\text{'}}}_{\mathrm{cg}}$ is phase to ground voltages.

Write the formula of sub-transient fault currents at the faulted bus.

$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathbf{a}}^2& a\\ 1& a& {\mathbf{a}}^2\end{array}\right]\left[\begin{array}{c}{\mathbf{I}}_0\\ {\mathbf{I}}_1\\ {\mathbf{I}}_2\end{array}\right]$ …… (10)

(a) Draw the sequence networks of the system and determine the sequence bus impedance matrix.

Refer to the Figure 9.27 in the textbook for the single-line diagram of a four bus system.

The zero-sequence network includes j0.15 ohm because machine 1 and machine 2 are grounded by a neutral reactance of j0.05 ohm, which is three times the neutral reactance.

Draw the circuit diagram of zero sequence network of the system.

Determine the zero impedance matrix.

$\begin{array}{c}{Z}_{\text{bus0}}=\left[\begin{array}{cccc}j0.15+j0.04& 0& 0& 0\\ 0& j0.08& j0.5& 0\\ 0& j0.5& j0.08& 0\\ 0& 0& 0& j0.15+j0.04\end{array}\right]\\ =\left[\begin{array}{cccc}j0.19& 0& 0& 0\\ 0& j0.08& j0.5& 0\\ 0& j0.5& j0.08& 0\\ 0& 0& 0& j0.19\end{array}\right]\end{array}$

Therefore, the value of zero sequence bus impedance matrixes is $\left[\begin{array}{cccc}j0.19& 0& 0& 0\\ 0& j0.08& j0.5& 0\\ 0& j0.5& j0.08& 0\\ 0& 0& 0& j0.19\end{array}\right]$.

Draw the circuit diagram of positive sequence network.

Determine the positive sequence impedance matrix.

$\begin{array}{c}{Z}_{\text{bus1}}=\left[\begin{array}{cccc}j0.2& j0.08& j0.08+j0.15& j0.08+j0.15+j0.08\\ j0.08& 0& j0.15& j0.15+j0.08\\ j0.08+j0.15& j0.15& 0& j0.08\\ j0.08+j0.15+j0.08& j0.08+j0.15& j0.08& j0.2\end{array}\right]\\ =\left[\begin{array}{cccc}j0.2& j0.08& j0.23& j0.31\\ j0.08& 0& j0.15& j0.23\\ j0.23& j0.15& 0& j0.08\\ j0.31& j0.23& j0.08& j0.2\end{array}\right]\end{array}$

Therefore, the value of positive sequence bus impedance matrix is $\left[\begin{array}{cccc}j0.2& j0.08& j0.23& j0.31\\ j0.08& 0& j0.15& j0.23\\ j0.23& j0.15& 0& j0.08\\ j0.31& j0.23& j0.08& j0.2\end{array}\right]$.

Draw the circuit diagram of negative sequence network.

Determine the negative sequence bus impedance matrix.

$\begin{array}{c}{Z}_{\text{bus2}}=\left[\begin{array}{cccc}j0.2& j0.08& j0.08+j0.15& j0.08+j0.15+j0.08\\ j0.08& 0& j0.15& j0.15+j0.08\\ j0.08+j0.15& j0.15& 0& j0.08\\ j0.08+j0.15+j0.08& j0.08+j0.15& j0.8& j0.2\end{array}\right]\\ =\left[\begin{array}{cccc}j0.2& j0.08& j0.23& j0.31\\ j0.08& 0& j0.15& j0.23\\ j0.23& j0.15& 0& j0.08\\ j0.31& j0.23& j0.08& j0.2\end{array}\right]\end{array}$

Therefore, the value of negative sequence bus impedance matrix is $\left[\begin{array}{cccc}j0.2& j0.08& j0.23& j0.31\\ j0.08& 0& j0.15& j0.23\\ j0.23& j0.15& 0& j0.08\\ j0.31& j0.23& j0.08& j0.2\end{array}\right]$.

(b) Draw the value of sequence component of the bolted line-to-line fault occurs at bus-3, value of line to line voltages at the faulted bus and value of line to line voltages at the terminals of the machine 2.

The line to line fault occurs at bus 3, determine the positive-sequence impedance.

$\begin{array}{c}{Z}_1=\left(j0.05+j0.2+j0.08+j0.15\right)\parallel \left(j0.08+j0.2+j0.05\right)\\ =\left(j0.48\right)\parallel \left(j0.33\right)\\ =\frac{\left(j0.48\right)\left(j0.33\right)}{j0.48+j0.33}\\ =j0.1955\text{\hspace{0.33em}pu}\end{array}$

Determine the negative-sequence impedance.

$\begin{array}{c}{Z}_2=\left(j0.05+j0.2+j0.08+j0.15\right)\parallel \left(j0.08+j0.2+j0.05\right)\\ =\left(j0.48\right)\parallel \left(j0.33\right)\\ =\frac{\left(j0.48\right)\left(j0.33\right)}{j0.48+j0.33}\\ =j0.1955\text{\hspace{0.33em}pu}\end{array}$

Determine the zero-sequence impedance.

$\begin{array}{c}{Z}_3=j0.08+j0.5\\ =j0.58\text{\hspace{0.17em}pu}\end{array}$

Determine the Bolted line-to-line fault occurs at bus-3.

Determine the positive-sequence component of the network.

Substitute 1 for $E_g$, $j0.1955$for $Z_1$and $j0.1955$for $Z_2$into equation (1).

$\begin{array}{c}{I}_1=\frac{1}{j0.1955+j0.1955}\\ =-j2.556\text{\hspace{0.33em}pu}\end{array}$

Determine the negative-sequence component of the network.

Substitute $j2.556\text{\hspace{0.17em}pu}$for $-I_1$into equation (2).

$I_2=j2.556\text{\hspace{0.33em}pu}$

Determine the zero-sequence component of the network.

$I_0=0$

Determine the sub-transient fault current in phase.

$\begin{array}{c}{I^{\text{'}\text{'}}}_{\text{a}}=I_0+I_1+I_2\\ =-j2.556+j2.556+0\\ =0\text{\hspace{0.33em}pu}\end{array}$

Determine the sub-transient fault current in phase.

$\begin{array}{c}{I^{\text{'}\text{'}}}_{\text{b}}=-j\sqrt{3}{I}_1\\ =-j\sqrt{3}\left(-j2.556\right)\\ =4.427\angle 180°\text{\hspace{0.33em}pu}\end{array}$

Determine the sub-transient fault current in phase.

$\begin{array}{c}{I^{\text{'}\text{'}}}_{\text{c}}=j\sqrt{3}{I}_1\\ =j\sqrt{3}\left(-j2.55\right)\\ =4.427\angle 0°\text{\hspace{0.33em}pu}\end{array}$

Therefore, the fault current ${I^{\text{'}\text{'}}}_{\text{a}},{I^{\text{'}\text{'}}}_b\&{I^{\text{'}\text{'}}}_c$are 0 pu, $4.427\angle 180°\text{\hspace{0.33em}pu}$and $4.427\angle 0°\text{\hspace{0.33em}pu}$

Determine the positive-sequence component voltage of the network.

$\begin{array}{c}{V}_1=1-I_1{Z}_1\\ =1-\left(-j2.556\right)\left(j0.1955\right)\\ =0.5\text{\hspace{0.17em}pu}\end{array}$

Determine the negative-sequence component voltage of the network.

$\begin{array}{c}{V}_2=V_1\\ =0.5\text{\hspace{0.33em}pu}\end{array}$

Determine the zero-sequence component voltage of the network.

$V_0=0$

Determine the phase to ground voltages.

$\begin{array}{c}{V^{\text{'}\text{'}}}_{\text{ag}}=V_0+V_1+V_2\\ =0+0.5+0.5\\ =1\text{\hspace{0.17em}pu}\end{array}$

Determine the phase to ground voltages.

$\begin{array}{c}{V^{\text{'}\text{'}}}_{\text{bg}}=V_0+a^2{V}_1+a{V}_2\\ =0+\left(1\angle 240°\right)\left(0.5\right)+\left(1\angle 120°\right)\left(0.5\right)\\ =0.5\angle 180°\text{\hspace{0.33em}pu}\end{array}$

Determine the phase to ground voltages.

$\begin{array}{c}{V^{\text{'}\text{'}}}_{\text{cg}}=V_0+a{V}_1+a^2{V}_2\\ =0+\left(1\angle 240°\right)\left(0.5\right)+\left(1\angle 120°\right)\left(0.5\right)\\ =0.5\angle 180°\text{\hspace{0.33em}pu}\end{array}$

Determine the line-to-line voltages for the system.

Substitute 1 for ${V^{\text{'}\text{'}}}_{\text{ag}}$into equation (4).

$\begin{array}{c}{V^{\text{'}\text{'}}}_{\text{a}}=\sqrt{3}\left(1\right)\\ =1.732\text{\hspace{0.33em}pu}\end{array}$

Determine the line-to-line voltages for the system.

Substitute 0.5 for ${V^{\text{'}\text{'}}}_{\text{bg}}$into equation (5).

$\begin{array}{c}{V^{\text{'}\text{'}}}_{\text{b}}=\sqrt{3}\left(0.5\right)\\ =0.886\text{\hspace{0.33em}pu}\end{array}$

Determine the line-to-line voltages for the system.

Substitute 0.5 for ${V^{\text{'}\text{'}}}_{\text{cg}}$into equation (6).

$\begin{array}{c}{V^{\text{'}\text{'}}}_{\text{c}}=\sqrt{3}\left(0.5\right)\\ =0.886\text{\hspace{0.17em}pu}\end{array}$

Therefore, The value of line to line voltages at the faulted bus, ${V^{\text{'}\text{'}}}_{\text{a}},{V^{\text{'}\text{'}}}_b\&{V^{\text{'}\text{'}}}_c$are 1.732 pu, 0.886 pu and 0.866 pu.

Determine the voltage between the phase a and ground at terminals of machine 2.

$\begin{array}{c}{V^{\text{'}}}_{\text{ag}}={E^{\text{'}}}_{\text{g}}+j{I}_{\text{ag}}{X^{\text{'}}}_{\text{d}}\\ =1+0\\ =1\text{\hspace{0.33em}pu}\end{array}$

Determine the voltage between the phase b and ground at terminals of machine 2.

$\begin{array}{c}{V^{\text{'}}}_{\text{bg}}={E^{\text{'}}}_{\text{g}}+j{I}_{\text{bg}}{X}_{\text{d}}\\ =1+j\left(4.427\angle 180°\right)\left(0.33j\right)\\ =2.461\text{\hspace{0.33em}pu}\end{array}$

Determine the voltage between the phase c and ground at terminals of machine 2.

$\begin{array}{c}{V^{\text{'}}}_{\text{cg}}={E^{\text{'}}}_{\text{g}}+j{I}_{\text{cg}}{X^{\text{'}\text{'}}}_{\text{d}}\\ =1+j\left(4.427\angle 0°\right)\left(0.33j\right)\\ =-0.461\text{\hspace{0.33em}pu}\end{array}$

Determine the line-to-line voltages at the terminals of the machine 2 for the system.

Substitute 2.461 for ${V^{\text{'}\text{'}}}_{\text{bg}}$into equation (7).

$\begin{array}{c}{V^{\text{'}\text{'}}}_{\text{b}}=\sqrt{3}\times 2.461\\ =4.26\text{\hspace{0.33em}pu}\end{array}$

Determine the line-to-line voltages at the terminals of the machine 2 for the system.

Substitute -0.461 for ${V^{\text{'}\text{'}}}_{\text{cg}}$into equation (9).

$\begin{array}{c}{V^{\text{'}\text{'}}}_{\text{c}}=\sqrt{3}\left(-0.461\right)\\ =-0.798\text{\hspace{0.33em}pu}\end{array}$

Therefore, the value of line to line voltages${V^{\text{'}\text{'}}}_{\text{a}},{V^{\text{'}\text{'}}}_b\&{V^{\text{'}\text{'}}}_c$at the terminals of the machine 2 are 1.732 pu, 4.26 pu and -0.798 pu.

(c) Draw the value of sub-transient fault currents at the faulted bus and value of line to line voltages at the faulted bus.

Assume that Bolted double line-to-ground fault occurs at terminals of machine 2.

Determine the positive-sequence current component of the system.

$\begin{array}{c}{I}_1=\frac{V_{\text{F}}}{Z_1+\left(0.2\parallel \left(0.04+3\left(0.05\right)\right)\right)}\\ =\frac{V_{\text{F}}}{0.2+\left(0.2\parallel \left(0.04+3\left(0.05\right)\right)\right)}\\ =\frac{V_{\text{F}}}{0.2+\left(0.2\parallel 0.19\right)}\\ =\frac{1}{j\left(0.2+\left(\frac{0.2\times 0.19}{0.2+0.19}\right)\right)}\end{array}$

Solve further as,

$\begin{array}{c}{I}_1=\frac{1}{j\left(0.2+0.0974\right)}\\ =-j3.36\text{\hspace{0.33em}pu}\end{array}$

Determine the negative-sequence current component of the system.

$\begin{array}{c}{I}_2=\left(-I_1\right)\left(\frac{Z_0+3Z_{\text{F}}}{Z_2+Z_0+3Z_{\text{n}}}\right)\\ =\left(j3.36\right)\left(\frac{0.04+\left(3\right)\left(0.05\right)}{0.2+0.04+\left(3\right)\left(0.05\right)}\right)\\ =\left(j3.36\right)\left(0.487\right)\\ =j1.636\text{\hspace{0.33em}pu}\end{array}$

Determine the zero-sequence current component of the system.

$\begin{array}{c}{I}_0=\left(-I_1\right)\left(\frac{Z_2}{Z_2+Z_0+3Z_{\text{n}}}\right)\\ =\left(j3.36\right)\left(\frac{0.2}{0.2+0.04+\left(3\right)\left(0.05\right)}\right)\\ =\left(j3.36\right)\left(0.512\right)\\ =j1.72\text{\hspace{0.33em}pu}\end{array}$

Determine the sub-transient fault current at the faulted bus.

Substitute $j1.72$for $I_0$, $-j3.36$for $I_1$and $j1.636$for $I_2$into equation (10).

$\begin{array}{c}\left[\begin{array}{c}{I^{\text{'}\text{'}}}_{\text{a}}\\ {I^{\text{'}\text{'}}}_{\text{b}}\\ {I^{\text{'}\text{'}}}_{\text{c}}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}j1.72\\ -j3.36\\ j1.636\end{array}\right]\\ =\left[\begin{array}{c}0\\ 5.038\angle 149.17°\\ 5.038\angle 30.83°\end{array}\right]\end{array}$

Therefore, the value of sub-transient fault currents $\left[\begin{array}{c}{I^{\text{'}\text{'}}}_{\text{a}}\\ {I^{\text{'}\text{'}}}_{\text{b}}\\ {I^{\text{'}\text{'}}}_{\text{c}}\end{array}\right]$at the faulted bus is $\left[\begin{array}{c}0\\ 5.038\angle 149.17°\\ 5.038\angle 30.83°\end{array}\right]$.

Determine the sequence voltage components of the system.

$\begin{array}{c}{V}_0=V_1\\ =V_2\\ =1-I_1{Z}_1\\ =1-\left(3.36\right)\left(0.2\right)\end{array}$

Solve further as,

$V_0=0.328\text{\hspace{0.17em}pu}$

Determine the phase-to-ground voltages of the system.

$\begin{array}{c}{V^{\text{'}\text{'}}}_{\text{ag}}=V_0+V_1+V_2\\ =0.328+0.328+0.328\\ =0.984\text{\hspace{0.33em}pu}\end{array}$

Determine the phase-to-ground voltages of the system.

$\begin{array}{c}{V^{\text{'}\text{'}}}_{\text{bg}}=V_0+a^2{V}_1+a{V}_2\\ =0.328+\left(0.328\right)\left(1\angle 240°\right)+\left(0.328\right)\left(1\angle 120°\right)\\ =0\end{array}$

Determine the phase-to-ground voltages of the system.

$\begin{array}{c}{V^{\text{'}\text{'}}}_{\text{cg}}=V_0+a{V}_1+a^2{V}_2\\ =0.328+\left(0.328\right)\left(1\angle 120°\right)+\left(0.328\right)\left(1\angle 240°\right)\\ =0\end{array}$

Determine the line-to-line voltages at the faulted bus.

Substitute 0.984 for ${V^{\text{'}\text{'}}}_{\text{ag}}$into equation (4).

$\begin{array}{c}{V^{\text{'}\text{'}}}_{\text{a}}=\sqrt{3}\times 0.984\\ =1.704\text{\hspace{0.17em}pu}\end{array}$

Determine the line-to-line voltages at the faulted bus.

Substitute 0 for ${V^{\text{'}\text{'}}}_{\text{bg}}$into equation (5).

$\begin{array}{c}{V^{\text{'}}}_{\text{b}}=\sqrt{3}\left(0\right)\\ =0\end{array}$

Determine the line-to-line voltages at the faulted bus.

Substitute 0 for ${V^{\text{'}}}_{\text{cg}}$into equation (6).

$\begin{array}{c}{V^{\text{'}}}_{\text{c}}=\sqrt{3}\left(0\right)\\ =0\end{array}$

Therefore, the value of line to line voltages, ${V^{\text{'}\text{'}}}_{\text{a}},{V^{\text{'}\text{'}}}_b\&{V^{\text{'}\text{'}}}_c$at the faulted bus are 1.704 pu, 0 pu and 0 pu.