Question

In a balanced three-phase Y-connected system with a positive-sequence source, the line-to-line voltages are $\sqrt{\mathbf{3}}$ times the line-to-neutral voltages and lend by $\mathbf{30}\mathbf{°}$.

(a) True

(b) False

Short answer

Answer

Therefore, the statement is true.

Step-by-step solution

Define equations for line to line and line to neutral voltages.

Consider the three phases${\mathit{E}}_{\mathbf{a}\mathbf{h}\mathbf{,}}{\mathit{E}}_{\mathbf{b}\mathbf{h}}$and ${\mathit{E}}_{\mathbf{c}\mathbf{h}}$. The voltage between these phases is called the line-to-line voltage.

Consider the KVL equation for these phases for the buses a, b and h.

$E_{ab}=E_{ah}-E_{bh}$

Consider the magnitude and the phase of the voltages as,

role="math" localid="1652691971726" $$\begin{gathered} E_{ah}=10\angle 0° \\ {E}_{bh}=10\angle -120° \\ {E}_{ch}=10\angle +120° \end{gathered}$$

Rewrite the KVL equation as,

$$\begin{gathered} E_{ab}=10\angle 0°-10\angle -120° \\ =10-10\left[\frac{-1-j\sqrt{3}}{2}\right] \\ =\sqrt{3}\left(10\right)\left[\frac{\sqrt{3}+j1}{2}\right] \\ =\sqrt{3}\left(10\angle 30°\right) \end{gathered}$$

Consider the line to line voltage $E_{bc}$.

$$\begin{gathered} E_{bc}=E_{bh}-E_{ch} \\ =10\angle -120°-10\angle 120° \\ =\sqrt{3}\left(10\angle 90°\right)V \end{gathered}$$

Consider the line to line voltage $E_{ca}$.

$$\begin{gathered} E_{ca}=E_{ch}-E_{ah} \\ =10\angle 120°-10\angle 0° \\ =\sqrt{3}\left(10\angle 150°\right)V \end{gathered}$$

Compare line to line and line to neutral voltages.

Compare the line to line voltages and line to neutral voltages.

$$\begin{gathered} E_{ab}=\sqrt{3}{E}_{an}\angle 30° \\ {E}_{bc}=\sqrt{3}{E}_{bn}\angle +30° \\ {E}_{ca}=\sqrt{3}{E}_{cn}\angle +30° \end{gathered}$$

The comparison shows that line to line voltage is $\sqrt{3}$ times line to neutral voltage and it lead by $30°$.

Thus, the statement is true.