Question

The initial three-phase voltage phasors (\(\bar V_A\), \(\bar V_B\), and \(\bar V_C\)) at a bus of a power network are as shown in Case-1. Due to a disturbance, the bus voltage phasors changed in phase by a small angle \(\Delta\theta\), and the magnitudes remained the same as depicted in Case-2.

Which one of the following statements is correct about the zero sequence components?

• A. The zero sequence components in Case-1 and Case-2 have the same phase angle and magnitude
• B. The magnitude of the zero sequence component in Case-1 is greater than that in Case-2
• C. The magnitude of the zero sequence component in Case-2 is greater than that in Case-1
• D. The zero sequence components in Case-1 and Case-2 have the same magnitude but different phase angles

Hint:

Rotating all three phase voltages by the same angle rotates their zero-sequence average by that same angle too, leaving its magnitude unchanged.

Answer 1

The Correct Option is D

Step 1: Write the zero sequence component as a phasor sum.
For any three-phase set, the zero sequence component is
\[ \bar V_0=\frac{1}{3}\left(\bar V_A+\bar V_B+\bar V_C\right) \]
This is defined even when the three phasors are not balanced (not equal in magnitude or not exactly \(120^{\circ}\) apart), which is important here since Case-1's phasors, as drawn, are not necessarily a perfectly balanced set.
Step 2: Express Case-2's phasors in terms of Case-1's.
Each Case-2 phasor keeps the same magnitude as its Case-1 counterpart but is rotated by the same angle \(\Delta\theta\):
\[ \bar V_A'=\bar V_Ae^{j\Delta\theta},\qquad \bar V_B'=\bar V_Be^{j\Delta\theta},\qquad \bar V_C'=\bar V_Ce^{j\Delta\theta} \]
since multiplying a phasor by \(e^{j\Delta\theta}\) rotates it by \(\Delta\theta\) while leaving its magnitude unchanged.
Step 3: Compute the zero sequence component in Case-2.
\[ \bar V_0'=\frac{1}{3}\left(\bar V_A'+\bar V_B'+\bar V_C'\right)=\frac{1}{3}\left(\bar V_Ae^{j\Delta\theta}+\bar V_Be^{j\Delta\theta}+\bar V_Ce^{j\Delta\theta}\right) \]
Factor out the common term \(e^{j\Delta\theta}\):
\[ \bar V_0'=e^{j\Delta\theta}\cdot\frac{1}{3}\left(\bar V_A+\bar V_B+\bar V_C\right)=e^{j\Delta\theta}\,\bar V_0 \]
Step 4: Interpret this result.
The Case-2 zero sequence phasor is simply the Case-1 zero sequence phasor multiplied by \(e^{j\Delta\theta}\). Multiplying by \(e^{j\Delta\theta}\) rotates a phasor by angle \(\Delta\theta\) without changing its magnitude at all, since \(|e^{j\Delta\theta}|=1\). So
\[ |\bar V_0'|=|\bar V_0| \]
while the phase angle of \(\bar V_0'\) is that of \(\bar V_0\) shifted by \(\Delta\theta\).
Step 5: Rule out the other options.
Option (A) claims the phase angle also stays the same, but Step 4 shows it shifts by \(\Delta\theta\) (which is nonzero, since the disturbance genuinely changes the phase). Options (B) and (C) both claim the magnitudes differ between the two cases, but Step 4 shows they are exactly equal.
Step 6: Final Answer.
\[ \boxed{\text{Same magnitude, different phase angles}} \]