Three single-phase \(11\text{kV}/3.3\text{kV}\) transformers are connected to form a three-phase transformer bank, with the HV and LV windings connected as shown.
Considering ABC phase sequence, the vector group of the transformer is:
Show Hint
Trace both deltas using the dot marks; the LV loop here closes in the opposite rotational sense to the HV loop, giving a clock number of 10, not 0.
Step 1: Read the dot convention on each single-phase unit.
Each of the three single-phase transformers has a dot on both its HV and LV winding. The dot marks the terminal that goes positive together on both windings at the same instant, so the HV winding EMF and the LV winding EMF of the same unit are always in phase with each other.
Step 2: See how the HV delta is wired.
On the HV side, each line terminal (A, B, C) is brought out at the dot end of its own winding, and the winding's far, undotted end is looped diagonally into the dot end of the next winding, closing A to B to C and back to A. So going around the HV delta, we always travel from a winding's dot end toward its undotted end.
Step 3: See how the LV delta is wired.
On the LV side, each line terminal (a, b, c) is also brought out at the dot end of its own winding, but here the undotted end of one winding feeds the dot end of the next, so going around the LV delta we travel the opposite way, from an undotted end toward the dot end of the following winding. The LV delta is closed in the reverse rotational sense compared to the HV delta.
Step 4: Write the HV line voltages in terms of the winding EMFs.
Taking \(E_A=E\angle0^{\circ}\), \(E_B=E\angle{-120^{\circ}}\), \(E_C=E\angle120^{\circ}\) (balanced ABC sequence, dot-positive), tracing the HV delta gives
\[
V_{AB}=E_A,\qquad V_{BC}=E_B,\qquad V_{CA}=E_C
\]
Step 5: Write the LV line voltages, tracing the reversed loop.
Because the LV loop is traced in the opposite sense relative to the dots, each LV line voltage comes out as the negative of the corresponding winding EMF one step ahead:
\[
V_{ab}=-e_b,\qquad V_{bc}=-e_c,\qquad V_{ca}=-e_a
\]
where \(e_a,e_b,e_c\) are in phase with \(E_A,E_B,E_C\) respectively, at the same angles with a scaled-down magnitude.
Step 6: Work out the phase shift.
\[
V_{ab}=-e_b=-E\angle{-120^{\circ}}=E\angle60^{\circ}
\]
Comparing to \(V_{AB}=E\angle0^{\circ}\), the LV line voltage \(V_{ab}\) leads the HV line voltage \(V_{AB}\) by \(60^{\circ}\), which is the same as saying it lags by \(360^{\circ}-60^{\circ}=300^{\circ}\).
Step 7: Convert the lag angle into a clock number.
Each clock hour represents \(30^{\circ}\) of lag.
\[
\text{Clock number}=\frac{300^{\circ}}{30^{\circ}}=10
\]
Step 8: Analyze the options.
(A) Dd0: Would need the LV delta wound the same rotational way as the HV delta, with dots aligned start-to-start on both sides, giving zero shift. Not the case here.
(B) Dd4: Would need a \(120^{\circ}\) lag, which does not match this connection. Incorrect.
(C) Dd6: Would need a full \(180^{\circ}\) reversal, meaning the LV dots reversed relative to HV on every leg, not just a reversed loop direction. Incorrect.
(D) Dd10: Matches the \(300^{\circ}\) lag found above. Correct.
