GATE 2026 EE Q.19: solution and answer

Official answer: DSource: Official answer key, GATE 2026 EE (IIT Guwahati)

1. The zero sequence component is the average of the three phase voltages: ${\overline{V}}_0=\frac{1}{3}\left({\overline{V}}_A+{\overline{V}}_B+{\overline{V}}_C\right)$.

2. Case-1: ${\overline{V}}_{0,1}=\frac{1}{3}\left(V_A\angle {\theta }_A+V_B\angle {\theta }_B+V_C\angle {\theta }_C\right)$. From the figure, ${\overline{V}}_A$ points slightly above the real axis while ${\overline{V}}_B$ and ${\overline{V}}_C$ both point below it, and ${\overline{V}}_B$ is the longest phasor. The sum of the imaginary parts is therefore negative, so ${\overline{V}}_{0,1}\ne 0$. A nonzero ${\overline{V}}_{0,1}$ has a definite magnitude and a definite angle.

3. Case-2: every phasor keeps its magnitude and its angle is increased by the same $\Delta \theta$. Each phasor is therefore multiplied by $e^{j\Delta \theta }$.

4. Hence ${\overline{V}}_{0,2}=\frac{1}{3}{e}^{j\Delta \theta }\left(V_A\angle {\theta }_A+V_B\angle {\theta }_B+V_C\angle {\theta }_C\right)=e^{j\Delta \theta }{\overline{V}}_{0,1}$.

5. Multiplying by $e^{j\Delta \theta }$ rotates a phasor without changing its length, so $|{\overline{V}}_{0,2}|=|{\overline{V}}_{0,1}|$. The magnitudes are equal, which rules out options (B) and (C).

6. The angle of ${\overline{V}}_{0,2}$ is the angle of ${\overline{V}}_{0,1}$ plus $\Delta \theta$. Since $\Delta \theta \ne 0$ (and ${\overline{V}}_{0,1}\ne 0$), the phase angles differ. This rules out option (A).

7. The zero sequence components have the same magnitude but different phase angles, which is option (D).