Question

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:

• A. \[ \begin{bmatrix} I_P \\ I_S \end{bmatrix} = \begin{bmatrix} \dfrac{y_t}{0.866+j0.5} & -\dfrac{y_t}{0.866+j0.5} \\ -\dfrac{y_t}{0.866+j0.5} & \dfrac{y_t}{0.866+j0.5} \end{bmatrix} \begin{bmatrix} V_P \\ V_S \end{bmatrix} \]
• B. \[ \begin{bmatrix} I_P \\ I_S \end{bmatrix} = \begin{bmatrix} y_t & -y_t \\ -y_t & y_t \end{bmatrix} \begin{bmatrix} V_P \\ V_S \end{bmatrix} \]
• C. \[ \begin{bmatrix} I_P \\ I_S \end{bmatrix} = \begin{bmatrix} y_t & -\dfrac{y_t}{0.866+j0.5} \\ -\dfrac{y_t}{0.866+j0.5} & y_t \end{bmatrix} \begin{bmatrix} V_P \\ V_S \end{bmatrix} \]
• D. \[ \begin{bmatrix} I_P \\ I_S \end{bmatrix} = \begin{bmatrix} y_t & -\dfrac{y_t}{0.866-j0.5} \\ -\dfrac{y_t}{0.866+j0.5} & y_t \end{bmatrix} \begin{bmatrix} V_P \\ V_S \end{bmatrix} \]

Hint:

Write the transformer as an ideal complex-ratio transformer in series with \(y_t\), then eliminate the internal node voltage; check that \(|a|^2=1\) before simplifying.

Answer 1

The Correct Option is D

Step 1: Recall the general model.
Any two-winding transformer with a possibly complex turns ratio $a=\dfrac{V_P}{V_S}$ and a series admittance $y_t$ placed on the secondary side can be treated as an ideal $a:1$ transformer followed by $y_t$ in the secondary circuit.

Step 2: Write down the standard admittance matrix.
For this kind of transformer, the injected currents work out to
\[ I_P = \frac{y_t}{|a|^2}V_P-\frac{y_t}{a^{*}}V_S \]
\[ I_S = -\frac{y_t}{a}V_P+y_t V_S \]
This form comes from writing the ideal-transformer voltage and current relations and eliminating the internal node voltage, and it is the standard result used for phase-shifting/off-nominal-tap transformers in load flow studies.

Step 3: Plug in the numbers.
Here $a=0.866+j0.5$, so $a^{*}=0.866-j0.5$. Check the magnitude:
\[ |a|^2=(0.866)^2+(0.5)^2=0.75+0.25=1 \]
Since $|a|^2=1$, the first coefficient becomes simply $y_t/1=y_t$.

Step 4: Write the two rows separately.
\[ I_P=y_t V_P-\frac{y_t}{0.866-j0.5}V_S \]
\[ I_S=-\frac{y_t}{0.866+j0.5}V_P+y_t V_S \]

Step 5: Sanity check with reciprocity.
Notice the mutual terms are not equal to each other, $-y_t/a^{*}\neq -y_t/a$ unless $a$ is real. This is expected: a transformer with a complex (phase-shifting) ratio gives a non-symmetric Y bus, which already rules out any option that uses the same expression for both off-diagonal entries or keeps them numerically equal.

Step 6: Match with the given choices.
Only the option that keeps the diagonal terms as plain $y_t$ (since $|a|=1$) and uses $a$ in the $I_S$ row but $a^{*}$ in the $I_P$ row fits both requirements.
\[ \boxed{I_P=y_tV_P-\frac{y_t}{0.866-j0.5}V_S,\ \ I_S=-\frac{y_t}{0.866+j0.5}V_P+y_tV_S} \]