GATE 2026 EE – Question 41
A three-phase two-winding transformer has a voltage transformation ratio $\dfrac{V_P}{V_S}=0.866+j0.5$, 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:

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Correct answer: (D) $\begin{bmatrix}I_P\\I_S\end{bmatrix}=\begin{bmatrix}y_t&-\dfrac{y_t}{0.866-j0.5}\\-\dfrac{y_t}{0.866+j0.5}&y_t\end{bmatrix}\begin{bmatrix}V_P\\V_S\end{bmatrix}$
Explanation
Let $a=\dfrac{V_P}{V_S}=0.866+j0.5$, with $|a|=1$. The secondary current is $I_S=y_t\left(V_S-\dfrac{V_P}{a}\right)$. An ideal transformer with a complex ratio conserves complex power, so $I_P=-\dfrac{I_S}{a^*}=\dfrac{y_t}{|a|^2}V_P-\dfrac{y_t}{a^*}V_S$. With $|a|=1$: $Y_{PP}=y_t$, $Y_{PS}=-\dfrac{y_t}{a^*}=-\dfrac{y_t}{0.866-j0.5}$, $Y_{SP}=-\dfrac{y_t}{a}=-\dfrac{y_t}{0.866+j0.5}$, $Y_{SS}=y_t$. The matrix is not symmetric, which matches option D.