Question

The venin equivalent sequence networks looking into the faulted bus of a power system are given with ${\mathbf{Z}}_{\mathbf{1}}\mathbf{=}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{15}$ ,${\mathbf{Z}}_{\mathbf{2}}\mathbf{=}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{15}$ ,${\mathbf{Z}}_{\mathbf{0}}\mathbf{=}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{2}$ and${\mathbf{E}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{\angle }\mathbf{0}\mathbf{,}$ per unit. Compute the fault currents and voltages for the following faults occurring at the faulted bus:

(a) Balanced three-phase fault

(b) Single line-to-ground fault

(c) Line-line fault

(d) Double line-to-ground fault. Which is the worst fault from the viewpoint of the fault current?

Short answer

(a) The fault current and voltage for three phase faults are $\left[\begin{array}{c}6.67\angle -90°\\ 6.67\angle 150°\\ 6.67\angle 30°\end{array}\right]\text{\hspace{0.17em}pu}$and $\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\text{\hspace{0.17em}pu}$respectively.

(b) The fault current and voltage for single line to ground faults are $\left[\begin{array}{c}6\angle -90°\\ 0\\ 0\end{array}\right]\text{\hspace{0.17em}pu}$and $\left[\begin{array}{c}0\\ 1.05\angle -124.7°\\ 1.05\angle 124.7°\end{array}\right]\text{\hspace{0.17em}pu}$ respectively.

(c) The fault current and voltage for single line to ground faults are $\left[\begin{array}{c}0\\ 5.76\angle 180\\ 5.76\end{array}\right]\text{\hspace{0.17em}pu}$and $\left[\begin{array}{c}1\\ 0.5\angle -179°\\ 0.5\angle 179°\end{array}\right]\text{\hspace{0.17em}pu}$ respectively.

(d) The fault current and voltage for double line to ground faults are$\left[\begin{array}{c}0\\ 6.38\angle 154.7°\\ 6.38\angle 25.3°\end{array}\right]\text{\hspace{0.17em}pu}$ and$\left[\begin{array}{c}1.092\\ 0\\ 0\end{array}\right]\text{\hspace{0.17em}pu}$ respectively.

Step-by-step solution

Write the given data from the question.

The voltage ${\mathrm{E}}_1=1\angle 0°\text{\hspace{0.17em}pu}$,

Write the equivalent sequence impedance of the system.

$\begin{array}{c}{Z}_1=j0.15\text{\hspace{0.17em}pu}\\ Z_2=j0.15\text{\hspace{0.17em}pu}\\ Z_0=j0.2\text{\hspace{0.17em}pu}\end{array}$

Determine the equation to calculate the fault current and voltages.  

The equation to calculate the sequence current for the three-phase fault is given as follows.

$\begin{array}{c}{\mathbf{I}}_{\mathbf{0}}\mathbf{=}{\mathbf{I}}_{\mathbf{2}}\\ {\mathbf{I}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{p}\mathbf{u}\\ {\mathbf{I}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{1}}}{{\mathbf{Z}}_{\mathbf{1}}}\end{array}$

…… (1)

Here,${\mathbf{V}}_{\mathbf{F}}$is the fault voltage,${\mathbf{Z}}_{\mathbf{1}}$is the positive sequence impedance, ${\mathit{I}}_{\mathbf{0}}$,$I_1$and${\mathit{I}}_{\mathbf{2}}$are the zero, positive and negative sequence component.

The equation to calculate the sub transient fault current is given as follows.

$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]$ …… (2)

The equation to calculate the fault voltages is given as follows.

$\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]$ …… (3)

The equations for single line to ground fault occur through an impedance.

The equation to calculate the fault current in the sequence component is given as follows.

$\begin{array}{c}{\mathrm{I}}_0={\mathrm{I}}_1\\ {\mathrm{I}}_0={\mathrm{I}}_2\\ {\mathrm{I}}_0=\frac{{\mathrm{E}}_1}{\mathrm{j}({\mathrm{Z}}_0+{\mathrm{Z}}_1+{\mathrm{Z}}_2+3{\mathrm{Z}}_{\mathrm{F}})}\end{array}$ …… (4)

The equation to calculate the sub transient fault current is given as follows.

role="math" localid="1655885026314" $\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]$ …… (5)

The equation to calculate the sequence voltage is given as follows.

$\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}0\\ E_1\\ 0\end{array}\right]-\left[\begin{array}{ccc}{Z}_0& 0& 0\\ 0& Z_1& 0\\ 0& 0& Z_2\end{array}\right]\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]$ …… (6)

The equation to calculate the voltage for three phase faults is given as follows.

$\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]$ …… (7)

The equations for line to line fault occur in the system.

The equation to calculate the fault current in the sequence component is given as follows.

$\begin{array}{l}{\mathbf{I}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\\ {\mathbf{I}}_{\mathbf{1}}\mathbf{=}\mathbf{-}{\mathbf{I}}_{\mathbf{2}}\\ {\mathbf{I}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{1}}}{{\mathbf{Z}}_{\mathbf{1}}\mathbf{+}{\mathbf{Z}}_{\mathbf{2}}\mathbf{+}{\mathbf{Z}}_{\mathbf{n}}}\end{array}$ …… (8)

The equation to calculate the sub transient fault current is given as follows.

$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]$ …… (9)

The equation to calculate the sequence voltage is given as follows.

$\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}0\\ E_1\\ 0\end{array}\right]-\left[\begin{array}{ccc}{Z}_0& 0& 0\\ 0& Z_1& 0\\ 0& 0& Z_2\end{array}\right]\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]$ …… (10)

The equation to calculate the voltage for three phase faults is given as follows.

$\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]$ …… (11)

The equations for double line to ground fault.

The equation to calculate the positive sequence fault current is given as follows.

${\mathbf{I}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{R}}}{{\mathbf{Z}}_{\mathbf{1}}\mathbf{+}\frac{\mathbf{(}{\mathbf{Z}}_{\mathbf{0}}\mathbf{+}\mathbf{3}{\mathbf{Z}}_{\mathbf{F}}\mathbf{)}{\mathbf{Z}}_{\mathbf{2}}}{{\mathbf{Z}}_{\mathbf{0}}\mathbf{+}{\mathbf{Z}}_{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{Z}}_{\mathbf{F}}}}$ …… (12)

The equation to calculate the negative sequence fault current is given as follows.

${\mathbf{I}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathbf{I}}_{\mathbf{1}}\left(\frac{Z_0+3Z_F}{Z_0+Z_2+3Z_F}\right)$ …… (13)

The equation to calculate the zero-sequence fault current is given as follows.

${\mathbf{I}}_{\mathbf{0}}\mathbf{=}\mathbf{-}{\mathbf{I}}_{\mathbf{1}}\left(\frac{Z_2}{Z_0+Z_2+3Z_F}\right)$ …… (14)

The equation to calculate the sub transient fault current is given as follows.

$\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]$ …… (15)

The equation to calculate the sequence voltage is given as follows.

$\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}0\\ E_1\\ 0\end{array}\right]-\left[\begin{array}{ccc}{Z}_0& 0& 0\\ 0& Z_1& 0\\ 0& 0& Z_2\end{array}\right]\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]$ …… (16)

The equation to calculate the voltage for three phase faults is given as follows.

$\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]$…… (17)

Calculate the fault current and voltages for balanced three phase fault.  

(a)

Draw the Thevenin equivalent circuit for three phase faults.

The zero and negative sequence current are zero for three phase faults.

Calculate the sequence current for the three-phase fault.

Substitute $1\angle 0°\text{\hspace{0.17em}pu}$for $E_1$and $j0.15\text{\hspace{0.17em}pu}$for $Z_1$into equation(1).

$\begin{array}{l}{I}_1=\frac{1\angle 0°}{j0.15}\\ I_1=-j6.67\text{\hspace{0.17em}pu}\end{array}$

Calculate the sub transient current.

Substitute $0\text{\hspace{0.17em}pu}$for $I_0$, $-j6.67\text{\hspace{0.17em}pu}$for $I_1$and $0\text{\hspace{0.17em}pu}$for role="math" localid="1655886583338" $I_2$into equation (2).

$\begin{array}{c}\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ -j6.67\\ 0\end{array}\right]\\ \left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{c}6.67\angle -90°\\ 6.67\angle 150°\\ 6.67\angle 30°\end{array}\right]\text{\hspace{0.17em}pu}\end{array}$

The sequence component of the voltages at the fault terminals are zero.

$\begin{array}{l}{V}_a=0\\ V_b=0\\ V_c=0\end{array}$

Calculate the fault voltages for three phase faults.

Substitute $0$for $V_0$, $0\text{\hspace{0.17em}pu}$ for $V_1$and $0\text{\hspace{0.17em}pu}$for $V_2$into equation (3).

Hence the fault current and voltage for three phase faults are$\left[\begin{array}{c}6.67\angle -90°\\ 6.67\angle 150°\\ 6.67\angle 30°\end{array}\right]\text{\hspace{0.17em}pu}$ and$\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\text{\hspace{0.17em}pu}$ respectively.

Calculate the fault current and voltages for single line to ground fault.  

(b)

Consider the sequence network for single line to ground fault.

Calculate the fault current is sequence component.

Substitute $1\angle 0°\text{\hspace{0.17em}pu}$for $E_1$, $j0.15\text{\hspace{0.17em}pu}$for $Z_1$, $j0.15\text{\hspace{0.17em}pu}$for $Z_2$, $j0.2\text{\hspace{0.17em}pu}$for $Z_0$and $0\text{\hspace{0.17em}pu}$for $Z_F$into equation (4).

$\begin{array}{l}{I}_0=\frac{1\angle 0°}{j(0.15+0.15+0.2+3\times 0)}\\ I_0=\frac{1\angle 0°}{j0.5}\\ I_0=-j2\text{\hspace{0.17em}pu}\end{array}$

Calculate the sub transient current.

Substitute $-j2\text{\hspace{0.17em}pu}$ for $I_0$, $-j2\text{\hspace{0.17em}pu}$for $I_1$and $-j2\text{\hspace{0.17em}pu}$for $I_2$into equation (5).

role="math" localid="1655887902109" $\begin{array}{l}\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}-j2\\ -j2\\ -j2\end{array}\right]\\ \left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{c}-j6\\ 0\\ 0\end{array}\right]\\ \left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{c}6\angle -90°\\ 0\\ 0\end{array}\right]\text{\hspace{0.17em}pu}\end{array}$

Calculate the sequence components of the voltage at fault terminal.

Substitute $1\angle 0°\text{\hspace{0.17em}pu}$ for $E_1$, $j0.2\text{\hspace{0.17em}pu}$ for $Z_0$, $j0.15\text{\hspace{0.17em}pu}$for $Z_1$, $j0.15\text{\hspace{0.17em}pu\hspace{0.17em}}$for $Z_2$, $-j2\text{\hspace{0.17em}pu}$for $I_0$,$-j2\text{\hspace{0.17em}pu}$for $I_1$ and $-j2\text{\hspace{0.17em}pu}$ for $I_2$into equation (6)

$\begin{array}{c}\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{ccc}j0.2& 0& 0\\ 0& j0.15& 0\\ 0& 0& j0.15\end{array}\right]\left[\begin{array}{c}-j2\\ -j2\\ -j2\end{array}\right]\\ \left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{c}0.4\\ 0.3\\ 0.3\end{array}\right]\\ \left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}-0.4\\ 0.7\\ -0.3\end{array}\right]\text{\hspace{0.17em}pu}\end{array}$

Calculate the fault voltage for single line to ground fault.

Substitute $-0.4\text{\hspace{0.17em}pu}$for$V_0$ , $0.7\text{\hspace{0.17em}pu}$ for $V_1$and$-0.3\text{\hspace{0.17em}pu}$ for$V_2$ into equation (7).

$\begin{array}{l}\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}-0.4\\ 0.7\\ -0.3\end{array}\right]\\ \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{c}0\\ 1.05\angle -124.7°\\ 1.05\angle 124.7°\end{array}\right]\text{\hspace{0.17em}pu}\end{array}$

Hence the fault current and voltage for single line to ground faults are$\left[\begin{array}{c}6\angle -90°\\ 0\\ 0\end{array}\right]\text{\hspace{0.17em}pu}$ and$\left[\begin{array}{c}0\\ 1.05\angle -124.7°\\ 1.05\angle 124.7°\end{array}\right]\text{\hspace{0.17em}pu}$ respectively.

Calculate the fault current and voltages for line to line fault.  

(c)

Calculate the fault current is sequence component.

Substitute$1\angle 0°\text{\hspace{0.17em}pu}$for$E_1$,$j0.15\text{\hspace{0.17em}pu}$ for$Z_1$, $j0.15\text{\hspace{0.17em}pu}$for $Z_2$, and$0\text{\hspace{0.17em}pu}$for$Z_F$into equation (8).

$\begin{array}{l}{I}_1=\frac{1\angle 0°}{j(0.15+0.15+3\times 0)}\\ I_1=\frac{1\angle 0°}{j0.3}\\ I_1=-j3.33\text{\hspace{0.17em}pu}\end{array}$

Calculate the sub transient current.

Substitute $0\text{\hspace{0.17em}pu}$for $I_0$, $-j3.33\text{\hspace{0.17em}pu}$for $I_1$and $j3.33\text{\hspace{0.17em}pu}$ for $I_2$into equation (9).

role="math" localid="1655889482738"

Calculate the sequence components of the voltage at fault terminal.

Substitute $1\angle 0°\text{\hspace{0.17em}pu}$for $E_1$, $j0.2\text{\hspace{0.17em}pu}$for $Z_0$, $j0.15\text{\hspace{0.17em}pu}$ for $Z_1$, $j0.15\text{\hspace{0.17em}pu\hspace{0.17em}}$for $Z_2$, $0\text{\hspace{0.17em}pu}$for $I_0$, $-j3.33\text{\hspace{0.17em}pu}$for $I_1$and $j3.33\text{\hspace{0.17em}pu}$ for $I_2$into equation (10)

$\begin{array}{l}\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{ccc}j0.2& 0& 0\\ 0& j0.15& 0\\ 0& 0& j0.15\end{array}\right]\left[\begin{array}{c}0\\ -j3.33\\ j3.33\end{array}\right]\\ \left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ 0.4955\\ -0.4955\end{array}\right]\\ \left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}0\\ 0.5045\\ 0.4955\end{array}\right]\text{\hspace{0.17em}pu}\end{array}$

Calculate the fault voltage for line to line fault.

Substitute $0\text{\hspace{0.17em}pu}$for $V_0$, $0.5045\text{\hspace{0.17em}pu}$ for $V_1$and $0.4955\text{\hspace{0.17em}pu}$for $V_2$into equation (11).

$\begin{array}{l}\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0\\ 0.5045\\ 0.4955\end{array}\right]\\ \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{c}1\\ 0.5\angle -179°\\ 0.5\angle 179°\end{array}\right]\text{\hspace{0.17em}pu}\end{array}$

Hence the fault current and voltage for line to line faults are$\left[\begin{array}{c}0\\ 5.76\angle 180\\ 5.76\end{array}\right]\text{\hspace{0.17em}pu}$ and $\left[\begin{array}{c}1\\ 0.5\angle -179°\\ 0.5\angle 179°\end{array}\right]\text{\hspace{0.17em}pu}$respectively.

Calculate the fault current and voltages for double line to ground fault.  

(d)

Calculate the positive sequence component of fault current.

Substitute $1\angle 0°\text{\hspace{0.17em}pu}$for $E_1$, $j0.15\text{\hspace{0.17em}pu}$for $Z_1$, $j0.15\text{\hspace{0.17em}pu}$for $Z_2$, $j0.2\text{\hspace{0.17em}pu}$for $Z_0$and $0\text{\hspace{0.17em}pu}$for $Z_F$into equation (12).

$\begin{array}{c}{I}_1=\frac{1\angle 0°}{j\left(0.15+\frac{\left(0.15\right)(0.2+3\left(0\right))}{0.15+0.2+3\left(0\right)}\right)}\\ I_1=\frac{1\angle 0°}{j0.2356}\\ I_1=-j4.24\text{\hspace{0.17em}pu}\\ I_1=4.24\angle -90°\text{\hspace{0.17em}pu}\end{array}$

Calculate the negative sequence component of fault current.

Substitute $4.24\angle -90°\text{\hspace{0.17em}pu}$ for $I_1$, $j0.15\text{\hspace{0.17em}pu}$for $Z_2$, $0\text{\hspace{0.17em}pu}$ for $Z_F$ and $j0.2\text{\hspace{0.17em}pu}$for $Z_0$into equation (13).

$\begin{array}{l}{I}_2=-(4.24\angle -90)\times \frac{j0.2+3\left(0\right)}{j0.2+j0.15+3\left(0\right)}\\ I_2=-(4.24\angle -90)\times \frac{j0.2}{j0.35}\\ I_2=2.42\angle 90°\text{\hspace{0.17em}pu}\end{array}$

Calculate the zero-sequence component of fault current.

Substitute $4.24\angle -90°\text{\hspace{0.17em}pu}$for $I_1$, $j0.15\text{\hspace{0.17em}pu}$ for $Z_2$, $0\text{\hspace{0.17em}pu}$for $Z_F$ and $j0.2\text{\hspace{0.17em}pu}$for $Z_0$into equation (14).

$\begin{array}{l}{I}_0=-(4.24\angle -90°)\times \frac{j0.15}{j0.2+j0.15}\\ I_0=(4.24\angle 90°)\times \frac{j0.15}{j0.35}\\ I_0=1.81\angle 90°\text{\hspace{0.17em}pu}\end{array}$

Calculate the sub transient current.

Substitute $1.81\angle 90°\text{\hspace{0.17em}pu}$ for $I_0$, $2.42\angle 90°\text{\hspace{0.17em}pu}$for $I_L$and $4.24\angle 90°\text{\hspace{0.17em}pu}$ for $I_2$into equation (15).

$\begin{array}{l}\left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}1.81\angle 90°\\ 2.42\angle 90°\\ 4.24\angle 90°\end{array}\right]\\ \left[\begin{array}{c}{I^{\text{'}\text{'}}}_a\\ {I^{\text{'}\text{'}}}_b\\ {I^{\text{'}\text{'}}}_c\end{array}\right]=\left[\begin{array}{c}0\\ 6.38\angle 154.7°\\ 6.38\angle 25.3°\end{array}\right]\text{\hspace{0.17em}pu}\end{array}$

Calculate the sequence components of the voltage at fault terminal.

Substitute $1\angle 0°\text{\hspace{0.17em}pu}$for $E_1$, $j0.2\text{\hspace{0.17em}pu}$for $Z_0$, $j0.15\text{\hspace{0.17em}pu}$ for $Z_1$, $j0.15\text{\hspace{0.17em}pu\hspace{0.17em}}$for $Z_2$, $0\text{\hspace{0.17em}pu}$for $I_O$, $6.38\angle 154.7°\text{\hspace{0.17em}pu}$for $I_1$and $6.38\angle 25.3°\text{\hspace{0.17em}pu}$for $I_2$into equation (16)

$\begin{array}{l}\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{ccc}j0.2& 0& 0\\ 0& j0.15& 0\\ 0& 0& j0.15\end{array}\right]\left[\begin{array}{c}0\\ 6.38\angle 154.7°\\ 6.38\angle 25.3°\end{array}\right]\\ \left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}0\\ 1\angle 0°\\ 0\end{array}\right]-\left[\begin{array}{c}-0.364\\ 0.636\\ -0.364\end{array}\right]\\ \left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\left[\begin{array}{c}0.364\\ 0.364\\ 0.364\end{array}\right]\text{\hspace{0.17em}pu}\end{array}$

Calculate the fault voltage for double line to ground fault.

Substitute $0.364\text{\hspace{0.17em}pu}$for $V_0$, $0.364\text{\hspace{0.17em}pu}$for $V_1$and $0.364\text{\hspace{0.17em}pu}$for $V_2$into equation (17).

$\begin{array}{l}\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}0.364\\ 0.364\\ 0.364\end{array}\right]\\ \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{c}1.092\\ 0\\ 0\end{array}\right]\text{\hspace{0.17em}pu}\end{array}$

Hence the fault current and voltage for double line to ground faults are

$\left[\begin{array}{c}0\\ 6.38\angle 154.7°\\ 6.38\angle 25.3°\end{array}\right]\text{\hspace{0.17em}pu}$and $\left[\begin{array}{c}1.092\\ 0\\ 0\end{array}\text{\hspace{0.17em}p}\right]\text{u}$respectively.

The worst fault is three phase balanced fault because magnitude of current in this fault is highest.