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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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.
Step 1: Set up the transformer model.
Model the transformer as an ideal transformer with complex turns ratio \(a = \dfrac{V_P}{V_S} = 0.866+j0.5\), followed by (referred to the secondary) a leakage admittance \(y_t\) that connects the internal secondary-side node to the actual secondary bus. Let \(V_1\) be the voltage at this internal node, right at the ideal transformer's secondary terminal, before the admittance \(y_t\).
Step 2: Find the magnitude of the ratio.
\[ |a|^2 = (0.866)^2+(0.5)^2 = 0.75+0.25 = 1 \]
So \(a\) has unit magnitude, meaning this transformer only shifts phase (by \(30^{\circ}\)) and does not change voltage magnitude. This fact will simplify the diagonal terms of the Y bus.
Step 3: Write the ideal transformer voltage relation.
For the ideal transformer,
\[ V_P = a V_1 \implies V_1 = \frac{V_P}{a} \]
Step 4: Write the ideal transformer current relation.
An ideal transformer carries no real or reactive power itself, so the complex power on both sides is equal:
\[ V_P I_P^{*} = V_1 I_1^{*} \]
where \(I_1\) is the current leaving the ideal transformer's secondary winding into the admittance branch. Substituting \(V_1 = V_P/a\) and taking the conjugate of the resulting relation gives
\[ I_P = \frac{I_1}{a^{*}} \]
Step 5: Write the current through the admittance branch.
The admittance \(y_t\) connects node \(1\) to bus \(S\), so
\[ I_1 = y_t(V_1-V_S) = y_t\left(\frac{V_P}{a}-V_S\right) \]
Step 6: Combine to get \(I_P\).
\[ I_P = \frac{I_1}{a^{*}} = \frac{y_t}{a^{*}}\left(\frac{V_P}{a}-V_S\right) = \frac{y_t}{aa^{*}}V_P-\frac{y_t}{a^{*}}V_S = \frac{y_t}{|a|^2}V_P-\frac{y_t}{a^{*}}V_S \]
Since \(|a|^2=1\),
\[ I_P = y_t V_P-\frac{y_t}{a^{*}}V_S = y_t V_P-\frac{y_t}{0.866-j0.5}V_S \]
because the conjugate of \(0.866+j0.5\) is \(0.866-j0.5\).
Step 7: Find \(I_S\) using current continuity.
The current \(I_1\) flows out of node \(1\) toward bus \(S\); with \(I_S\) defined as the current entering the transformer branch from bus \(S\), \(I_S = -I_1\):
\[ I_S = -I_1 = -y_t\left(\frac{V_P}{a}-V_S\right) = -\frac{y_t}{a}V_P+y_t V_S = -\frac{y_t}{0.866+j0.5}V_P+y_t V_S \]
Step 9: Rule out the other options.
Option (B) is correct only if \(a=1\) (an ordinary transformer with no phase shift); it ignores the ratio entirely. Option (A) divides every entry of the matrix by \(a\), which wrongly changes the diagonal terms too, when in fact \(|a|^2=1\) keeps the diagonal terms equal to plain \(y_t\). Option (C) uses the same denominator \(0.866+j0.5\) for both off-diagonal terms, but the two off-diagonal terms must use \(a\) and its conjugate \(a^{*}\) separately, since a complex (phase-shifting) ratio makes the bus admittance matrix non-symmetric.
Final Answer:
\[ \boxed{I_P = y_t V_P-\dfrac{y_t}{0.866-j0.5}V_S,\qquad I_S = -\dfrac{y_t}{0.866+j0.5}V_P+y_t V_S} \]
This matches option (D).
