For differential protection of a Delta–Wye transformer, the current transformers on the two sides are connected the opposite way round from the transformer itself: the CTs on the Delta side are connected in Wye, and the CTs on the Wye side are connected in Delta. It looks backwards — until you remember that the transformer shifts the phase of the currents by 30°. This post explains why the crossed connection is necessary, and works through the numerical example from the exam.

The problem: a 30° phase shift

A Delta–Wye transformer (vector group Dyn11) displaces the primary and secondary line currents by a 30° phase shift. There is a second problem too: a zero-sequence ground-fault current I₀ on the grounded Wye side cannot appear in the Delta primary lines at all — it circulates inside the delta winding instead.

If both sets of CTs were connected in Wye, the differential relay would see a false difference current on perfectly normal load:

False differential current = 2I sin 15° = 0.518 I

— more than half the load current, flowing through the relay coils on every normal day. The same false difference would appear during an external earth fault on the Wye side. The relay would trip when nothing is wrong.

Figure 7(b): Why Δ–Y transformer CTs are connected Y–Δ. The crossed CT connection cancels the transformer's 30° shift, and the pilot currents match at 5 A with 0° phase shift.

The solution: cross the CT connections

Connecting the Delta-side CTs in Wye (Y) and the Wye-side CTs in Delta (Δ) does two things at once:

• It cancels the 30° phase shift, so the two pilot currents arrive at the differential relay in phase with each other and cancel exactly on normal load and external faults.
• It traps zero-sequence current inside the Δ CT loop, so a Wye-side earth fault cannot send a false differential current into the relay.

The rule is easy to remember: the CTs are always connected in the opposite configuration to the winding they protect — Delta winding, Wye CTs; Wye winding, Delta CTs.

Numerical example: 10 MVA, 33 kV / 11 kV

For a 10 MVA, 33 kV (Δ) / 11 kV (Y) transformer, the full-load line currents are:

I_L1 = 10×10⁶ / (√3 × 33×10³) = 174.95 A   (33 kV Delta side)
I_L2 = 10×10⁶ / (√3 × 11×10³) = 524.86 A   (11 kV Wye side)

(1) 11 kV Wye side, CTs in Delta: the pilot-wire current must be 5 A, and for a delta connection I_pilot = √3 × I_CT2, so the CT secondary phase current is I_CT2 = 5/√3 = 2.887 A. The Wye-side CT ratio is therefore 525/2.89 A (also written 909/5 A).

(2) 33 kV Delta side, CTs in Wye: for a wye connection the pilot current equals the CT secondary current, I_pilot = I_CT1 = 5 A, so the Delta-side CT ratio is 175/5 A.

With these ratios the pilot currents match exactly — 5 A = 5 A, with 0° phase shift — so the differential relay sees zero difference current on normal load and operates only for a genuine internal fault.

Key points to remember

• Δ–Y transformer (Dyn11): 30° phase shift between primary and secondary line currents; zero-sequence current on the Wye side cannot appear in the Delta lines.
• Both CTs in Wye would give a false differential current of 2I sin 15° = 0.518 I on normal load.
• Solution: Delta-side CTs in Wye, Wye-side CTs in Delta — cancels the 30° shift and traps zero-sequence current in the Δ CT loop.
• Example (10 MVA, 33/11 kV): I_L1 = 174.95 A, I_L2 = 524.86 A.
• CT ratios: 175/5 A on the 33 kV Delta side (CTs in Wye); 525/2.89 A (909/5 A) on the 11 kV Wye side (CTs in Delta). Pilot currents match: 5 A = 5 A, 0° shift.

Based on Power System Protection & Switchgear, Paper II, Q7(b), Exam 11 June 2026.