Question

A three-phase balanced voltage source is applied to a balanced Y-connected load with ungrounded neutral. The Y-connected load consists of three mutually coupled reactances, where the reactance of each phase is $\mathbf{j}\mathbf{12}\mathbf{\Omega }$, and the mutual coupling between any two phases is $\mathbf{j}\mathbf{4}\mathbf{\Omega }$. The line-to-line source voltage is $\mathbf{100}\sqrt{3}\mathbf{}\mathbf{V}$. Determine the line currents (a) by mesh analysis without using symmetrical components and (b) using symmetrical components.

Short answer

(a) The line currents by mesh analysis without using symmetrical components is$I_a=12.5\angle -90°\text{\hspace{0.33em}A}$,$I_b=12.5\angle 150°\text{\hspace{0.33em}A}$and$I_c=12.5\angle 30°\text{\hspace{0.33em}A}$.

(b) The line currents by mesh analysis using symmetrical components is $I_a=12.5\angle -90°\text{\hspace{0.33em}A}$,$I_b=12.5\angle 150°\text{\hspace{0.33em}A}$ and $I_c=12.5\angle 30°\text{\hspace{0.33em}A}$.

Step-by-step solution

Write the given data from the question.

The reactance of each phase, $Z_s=j12\Omega$.

The mutual coupling between two phases, $Z_m=j4\Omega$.

Line to line source voltage, role="math" localid="1655883640124" $V_{LL}=100\sqrt{3}\text{\hspace{0.33em}V}$.

Determine the formula of line currents.

Write the formula of line currentsby mesh analysis without using symmetrical components.

${\mathbf{I}}_a\mathbf{+}{\mathbf{I}}_b\mathbf{+}{\mathbf{I}}_c\mathbf{=}\mathbf{0}$ …… (1)

Here ${\mathbf{I}}_a\mathbf{,}\mathbf{}{\mathbf{I}}_b\mathbf{}\&\mathbf{}{\mathbf{I}}_c$are the line current.

Write the formula of line currentby mesh analysis using symmetrical components.

$\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]\mathbf{=}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[I_s\right]$ …… (2)

Here, ${\mathbf{I}}_{\mathbf{s}}$is phase sequence current and a is an identity.

(a) Determine the line currents by mesh analysis without using symmetrical components.

Draw the following circuit diagram.

Apply Kirchhoff’s voltage law for the loop$V_a$to $V_b$.

$\begin{array}{c}{V}_a-V_b=j12I_a+j4I_b-j12I_b-j4I_a\\ V_a-V_a\angle -120°=j8I_a-j8I_b\\ V_a\left(1-1\angle -120°\right)=j8I_a-j8I_b\\ V_a\sqrt{3}\angle 30°=j8I_a-j8I_b\end{array}$

Solve further as,

role="math" localid="1655884120091" $\begin{array}{c}100\sqrt{3}\angle 30°=j8I_a-j8I_b\\ \frac{100\sqrt{3}\angle 30°}{8j}=I_a-I_b\\ \end{array}$

$I_a-I_b=21.65\angle -60°\text{\hspace{0.33em}A}$ …… (3)

Here,$I_a$and $I_b$are line current.

Apply Kirchhoff’s voltage law for the loop$V_b$to $V_c$.

$\begin{array}{c}{V}_b-V_c=12j{I}_b+4j{I}_c-12j{I}_c-4j{I}_b\\ V_a\angle -120°-V_a\angle 120°=j8I_b-j8I_c\\ V_a\left(1\angle -120°-1\angle 120°\right)=j8I_b-j8I_c\\ V_a\sqrt{3}\angle -90°=j8I_b-j8I_c\end{array}$

Solve further as,

$\begin{array}{c}100\sqrt{3}\angle -90°=j8I_b-j8I_c\\ \frac{100\sqrt{3\angle }-90°}{j8}=I_b-I_c\end{array}$

$I_b-I_c=21.65\angle 180°\text{\hspace{0.33em}A}$ …… (4)

Here,$I_b$ and $I_c$are line current.

Determine the line current.

Subtract equation (4) from (3) and subtract it from to equation (3).

$\begin{array}{c}\left(I_a+I_b+I_c\right)-\left[I_a-I_b-\left(I_b-I_c\right)\right]=0-\left[\left(21.65\angle -60°\right)-\left(21.65\angle 180°\right)\right]\\ \left(I_a+I_b+I_c\right)-\left[I_a-2I_b+I_c\right]=-\left(32.475-j18.75\right)\\ I_b=\frac{\left(37.5\angle 150°\right)}{3}\\ =12.5\angle 150°\text{\hspace{0.33em}A}\end{array}$

Substitute$12.5\angle 150°$ for$I_b$ into equation (3).

$\begin{array}{c}{I}_a-12.5\angle 150°=21.65\angle 180°\\ I_a=\left(21.65\angle -60°\right)+\left(12.5\angle 150°\right)\\ I_a=12.5\angle -90°\text{\hspace{0.33em}A}\end{array}$

Substitute $12.5\angle 150°$for $I_b$into equation (4).

$\begin{array}{c}12.5\angle 150°-I_c=21.65\angle 180°\\ I_c=\left(12.5\angle 150°\right)-\left(21.65\angle 180°\right)\\ I_c=12.5\angle 30°\end{array}$

Therefore, the line current is

$\begin{array}{c}{I}_a=12.5\angle -90°\text{\hspace{0.33em}A}\\ I_b=12.5\angle 150°\text{\hspace{0.33em}A}\\ I_c=12.5\angle 30°\text{\hspace{0.33em}A}\end{array}$

(b) Determine the line currents by mesh analysis using symmetrical components.

Determine the sequence components of line to ground voltages using the relation.

$\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{V}_{ag}\\ V_{bg}\\ V_{cg}\end{array}\right]$ …… (5)

Here, $V_{ag},V_{bg}\&V_{cg}$are line to ground voltages.

Substitute $100$for $V_{ag}$, $100\angle -120°$for $V_{bg}$, $100\angle 120°$for $V_{cg}$and $1\angle 120°$for a into equation (5).

$\begin{array}{c}\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}100\\ 100\angle -120\\ 100\angle 120°\end{array}\right]\\ =\frac{1}{3}\left[\begin{array}{c}100+100\angle -120°+100\angle 120°\\ 100+100\angle \left(-120+120\right)°+100\angle \left(120+240\right)°\\ 100+100\angle \left(-120+240\right)°+100\angle \left(120+120\right)°\end{array}\right]\\ V_s=\left[\begin{array}{c}0\\ 100\\ 0\end{array}\right]\end{array}$

Determine the sequence impedance matrix.

$\begin{array}{c}\left[Z_s\right]=\left[\begin{array}{ccc}{Z}_s+3Z_m+2Z_m& 0& 0\\ 0& Z_s-Z_m& 0\\ 0& 0& Z_s-Z_m\end{array}\right]\\ =\left[\begin{array}{ccc}j12+3\left(0\right)+2\left(j4\right)& 0& 0\\ 0& j12-j14& 0\\ 0& 0& j12-j4\end{array}\right]\\ =\left[\begin{array}{ccc}j20& 0& 0\\ 0& j8& 0\\ 0& 0& j8\end{array}\right]\end{array}$

The relationship between sequence voltages and currents as follows:

$\left[I_s\right]=Z_s^{-1}\left[V_s\right]$ …… (6)

Here,$Z_s^{-1}$ is inverse sequence impedance matrix and$V_s$ is a sequence voltage.

Substitute equation (6) into equation (2).

$\begin{array}{c}\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]Z_s^{-1}\left[V_s\right]\\ =\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]{\left[\begin{array}{ccc}j20& 0& 0\\ 0& j8& 0\\ 0& 0& j8\end{array}\right]}^{-1}\left[\begin{array}{c}0\\ 100\\ 0\end{array}\right]\\ =\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{ccc}-0.05j& 0& 0\\ 0& -0.125j& 0\\ 0& 0& -0.125j\end{array}\right]\left[\begin{array}{c}0\\ 100\\ 0\end{array}\right]\\ =\left[\begin{array}{ccc}-0.05j& -0.125j& -0.125j\\ -0.05j& 0.125\angle 150°& 0.125\angle 30°\\ -0.05j& 0.125\angle 30°& 0.125\angle 150°\end{array}\right]\left[\begin{array}{c}0\\ 100\\ 0\end{array}\right]\end{array}$

As further solve as

$\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{c}12.5\angle -90°\\ 12.5\angle 150°\\ 12.5\angle 30°\end{array}\right]$

Therefore, the values of line currents are,

$\begin{array}{c}{I}_a=12.5\angle -90°\text{\hspace{0.33em}A}\\ I_b=12.5\angle 150°\text{\hspace{0.33em}A}\\ I_c=12.5\angle 30°\text{\hspace{0.33em}A}\end{array}$