Star-Connected System Phase Angle Difference Explained

In a three-phase star-connected system, the relationship between the line voltage ($V_L$) and the phase voltage ($V_p$) is a fundamental concept. Understanding the phase angle difference between these voltages is crucial for analyzing such systems.

In a star connection, the ends of the three coils or windings are connected together at a common point, called the neutral point. The other ends of the three coils are connected to the three line terminals (R, Y, B). The phase voltages are the voltages measured between each line terminal and the neutral point ($V_R$, $V_Y$, $V_B$). The line voltages are the voltages measured between any two line terminals ($V_{RY}$, $V_{YB}$, $V_{BR}$).

Let's consider the phase voltages $V_R$, $V_Y$, and $V_B$. In a balanced three-phase system, these phase voltages are equal in magnitude and are displaced from each other by $120^\circ$ electrical degrees. We can represent these phase voltages using phasors:

• $V_R = V_p \angle 0^\circ$
• $V_Y = V_p \angle -120^\circ$
• $V_B = V_p \angle -240^\circ$ or $V_p \angle 120^\circ$

The line voltage $V_{RY}$ is the vector difference between the phase voltages $V_R$ and $V_Y$.

$\vec{V}_{RY} = \vec{V}_R - \vec{V}_Y$

To perform this vector subtraction, we can represent $-\vec{V}_Y$. The vector $-\vec{V}_Y$ has the same magnitude as $\vec{V}_Y$ but is at an angle $180^\circ$ away from $\vec{V}_Y$. Since $\vec{V}_Y$ is at $-120^\circ$, $-\vec{V}_Y$ is at $-120^\circ + 180^\circ = 60^\circ$.

Now, we can find $\vec{V}_{RY}$ by adding the phasors $\vec{V}_R$ ($V_p \angle 0^\circ$) and $-\vec{V}_Y$ ($V_p \angle 60^\circ$). Using the parallelogram law of vector addition, or by resolving components:

$V_{RY} = \sqrt{V_p^2 + V_p^2 + 2 V_p V_p \cos(60^\circ)}$

$V_{RY} = \sqrt{2V_p^2 + 2V_p^2 (0.5)}$

$V_{RY} = \sqrt{2V_p^2 + V_p^2}$

$V_{RY} = \sqrt{3V_p^2}$

$V_{RY} = \sqrt{3} V_p$

This confirms the magnitude relationship between line and phase voltage in a star connection: $V_L = \sqrt{3} V_p$.

Now let's consider the phase angle of $V_{RY}$. The phasor $V_{RY}$ is the resultant of adding $V_R$ (at $0^\circ$) and $-V_Y$ (at $60^\circ$). In a phasor diagram, the resultant vector $V_{RY}$ bisects the angle between $V_R$ and $-V_Y$. The angle between $V_R$ and $-V_Y$ is $60^\circ$. Therefore, $V_{RY}$ is at an angle of $0^\circ + (60^\circ / 2) = 30^\circ$ with respect to $V_R$. Similarly, considering the vector addition of $V_R$ and $-V_Y$, the phase angle of $V_{RY}$ is $30^\circ$ with respect to the reference phasor $V_R$ (which is at $0^\circ$).

So, the line voltage $V_{RY}$ leads the phase voltage $V_R$ by $30^\circ$. This phase difference of $30^\circ$ is consistent for all line voltages relative to their corresponding phase voltages (e.g., $V_{YB}$ leads $V_Y$ by $30^\circ$, $V_{BR}$ leads $V_B$ by $30^\circ$).

Therefore, in a star-connected system, the phase angle difference between any line voltage and its corresponding phase voltage is $30^\circ$.

Let's verify with the options:

Based on our analysis, the phase angle difference is $30^\circ$.