Question

For a completely transposed three-phase line that has balanced positive sequence voltages, the total reactive power supplied by the three-phase line, in var, is given by ${\mathbf{Q}}_{C3}=$ ______ in terms of frequency role="math" localid="1655303157282" $\mathbf{\omega }$, line-to-neutral capacitance Can, and line-to-line voltage V_LL.

Short answer

The correct answer is ${\mathrm{\omega C}}_{\mathrm{an}}{\mathrm{V}}_{\mathrm{LL}}^2$ .

Step-by-step solution

Write the expression for charging current 

For a single-phase circuit operating at line-to-line voltage ${\mathit{V}}_{\mathbf{x}\mathbf{y}}\mathbf{=}{\mathit{V}}_{\mathbf{x}\mathbf{y}}\mathbf{\angle }\mathbf{0}\mathbf{°}$, the charging current can be written as,

$\begin{array}{c}{\mathbf{I}}_{\mathbf{chg}}\mathbf{=}{\mathbf{Y}}_{\mathbf{xy}}{\mathbf{V}}_{\mathbf{xy}}\\ \mathbf{=}{\mathbf{j\omega C}}_{\mathbf{xy}}{\mathbf{V}}_{\mathbf{xy}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}A}}\end{array}$

Here, ${\mathrm{Y}}_{\mathrm{xy}}$ is the line admiitance, $V_{xy}$ is the line voltage, $C_{xy}$ is the line capacitance, $\omega$ is the frequency.

For a completely transposed three-phase line that has balanced positive-sequence voltages with role="math" localid="1655303405694" $V_{an}=V_{LN}\angle 0°$ , the phase $a$ charging current is

Here, $V_{an}$ is neutral voltage , $C_{an}$is the neutral capacitance, $V_{LN}^{}$ is the line to neutal voltage.

write the expression of total reactive power

The total reactive power supplied by the three-phase line is written as

Here, $V_{LL}^{}$ is the line to line voltage.

Thus the correct answer is ${\mathrm{\omega C}}_{\mathrm{an}}{\mathrm{V}}_{\mathrm{LL}}^2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}var}$.