GATE 2026 EE Q.41: solution and answer

GATE 2026 · EEQ.41 · 2 marks · MCQ

A three-phase two-winding transformer has a voltage transformation ratio $\frac{V _{P}}{V _{S}} = 0.866 + j 0.5$ , where $V_{P}$ is the primary side voltage in p.u., and $V_{S}$ is the secondary side voltage in p.u. $I_{P}$ and $I_{S}$ represent the currents injected into the primary and secondary sides of the transformer, respectively. The admittance corresponding to the leakage impedance of the transformer referred to the secondary is $y_{t}$ p.u. Neglect the magnetizing branch. The Y bus representation of this transformer is:

GATE 2026 EE question 41: Single-line diagram of a transformer symbol (two coupled circles) between two bus bars. — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
Figure — [I_P; I_S] = [[ y_t/(0.866+j0.5), -y_t/(0.866+j0.5) ], [ -y_t/(0.866+j0.5), y_t/(0.866+j0.5) ]] [V_P; V_S]
(B)
Figure — [I_P; I_S] = [[ y_t, -y_t ], [ -y_t, y_t ]] [V_P; V_S]
(C)
Figure — [I_P; I_S] = [[ y_t, -y_t/(0.866+j0.5) ], [ -y_t/(0.866+j0.5), y_t ]] [V_P; V_S]
(D)
Figure — [I_P; I_S] = [[ y_t, -y_t/(0.866-j0.5) ], [ -y_t/(0.866+j0.5), y_t ]] [V_P; V_S]

Answer and solution

ANSWER

The Y bus representation is $[y_{t} - y_{t} / (0.866 + j 0.5) - y_{t} / (0.866 - j 0.5) y_{t}]$ , which is option (D).

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

Let $a = \frac{V _{P}}{V _{S}} = 0.866 + j 0.5 = 1∠3 0^{\circ}$ . Then $∣ a ∣ = 1$ and $a^{*} = 0.866 - j 0.5$ . The leakage admittance $y_{t}$ is referred to the secondary, so model the transformer as an ideal complex-ratio transformer on the primary side, followed by the series admittance $y_{t}$ on the secondary side. All quantities are in p.u.
Let $V^{'}$ be the voltage on the secondary terminal of the ideal transformer, so that $\frac{V _{P}}{V ^{'}} = a$ and $V^{'} = \frac{V _{P}}{a}$ . The current $I_{S}$ injected at the secondary bus flows through $y_{t}$ into this terminal: $I_{S} = y_{t} (V_{S} - V^{'}) = - \frac{y _{t}}{a} V_{P} + y_{t} V_{S}$
An ideal transformer absorbs no net complex power. The current into its primary terminal is $I_{P}$ and the current into its secondary terminal is $I_{S}$ , so $V_{P} I_{P}^{*} + V^{'} I_{S}^{*} = 0 \Rightarrow I_{P}^{*} = - \frac{V ^{'}}{V _{P}} I_{S}^{*} = - \frac{I _{S}^{*}}{a}$ Taking the conjugate gives $I_{P} = - \frac{I _{S}}{a ^{*}}$ .
Substitute $I_{S}$ from step 2: $I_{P} = - \frac{1}{a ^{*}} (y_{t} V_{S} - \frac{y _{t}}{a} V_{P}) = \frac{y _{t}}{a a ^{*}} V_{P} - \frac{y _{t}}{a ^{*}} V_{S}$
Since $a a^{*} = ∣ a ∣^{2} = 1$ : $I_{P} = y_{t} V_{P} - \frac{y _{t}}{0.866 - j 0.5} V_{S}$
The secondary row, from step 2, is $I_{S} = - \frac{y _{t}}{0.866 + j 0.5} V_{P} + y_{t} V_{S}$
Collect the two rows: $[I_{P} I_{S}] = y_{t} - \frac{y _{t}}{0.866 + j 0.5} - \frac{y _{t}}{0.866 - j 0.5} y_{t} [V_{P} V_{S}]$ The matrix is not symmetric because the ratio is complex, so the transformer is a phase shifter. This is option (D). Options (A) to (C) are symmetric, so none of them can represent a phase-shifting transformer.

KEY IDEA

A transformer with complex tap ratio $a$ , with $y_{t}$ referred to the secondary, has $Y_{P P} = y_{t} /∣ a ∣^{2}$ , $Y_{P S} = - y_{t} / a^{*}$ , $Y_{S P} = - y_{t} / a$ and $Y_{S S} = y_{t}$ . Derive it from the series admittance plus an ideal transformer that conserves complex power. The $a$ and $a^{*}$ in the off-diagonal terms make the matrix non-symmetric for a phase shifter.

WATCH OUT FOR

Using the same factor $a$ in both off-diagonal terms, or treating the matrix as symmetric as for a real tap ratio. With a complex ratio the current relation is $I_{P} = - I_{S} / a^{*}$ , not $- I_{S} / a$ , so one off-diagonal term carries the conjugate.

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GATE 2026 EE Q.41: solution and answer

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