Previous Year Question
Correct answer: A
Explanation
A — Dd1. A conventional delta–delta transformer connection does not provide the 30° displacement required by clock number 1.
Formula
Step-by-step solution
Correct answer: A — Dd1
Decode the vector-group notation
A transformer vector group identifies the winding connections and the phase displacement between the high-voltage and low-voltage voltage systems. The capital letter identifies the HV winding, the lowercase letter identifies the LV winding, and the clock number expresses displacement in multiples of 30°.
Step 1: Interpret Dd. Both windings are delta connected. Each delta line voltage is directly associated with a winding phase voltage, so the connection does not introduce the star–delta 30° conversion.
Step 2: Interpret clock number 1. With the HV reference at 12 o’clock, clock 1 represents the LV reference lagging by 30°:
Step 3: Compare the required and available displacement. The common delta–delta groups are Dd0 and Dd6, corresponding to 0° and 180°. Conventional delta–delta winding polarity and phase arrangements do not produce the odd 30° displacement demanded by Dd1.
Therefore Dd1 is the invalid connection intended by the question. The clock number must agree with the actual winding phasors; it cannot be attached arbitrarily to any letter combination. Zigzag connections require their own split-winding phasor construction, so the delta–delta argument should not be applied to a group containing z.
Other options
Yy0 is a standard zero-displacement star–star group. Yd11 is a standard star–delta group with 330°/−30° clock displacement. Dz6 is a recognized delta–zigzag form with 180° displacement. Dd1 is the nonstandard conventional Dd combination.
Common mistake
Assuming any clock number 0–11 can be paired with any winding-connection letters.
Exam tip
Clock number ×30° gives displacement. Same-connection groups such as Dd do not normally give 30° displacement.
Quick method
Dd + clock 1 (30°) is the mismatch.