- Open Delta Connection Definition: An open delta connection transformer uses two single-phase transformers to create a three-phase supply, typically used in emergencies.
- Efficiency: Open delta systems are less efficient than closed delta systems because they provide less power output while operating at full transformer capacity.
- Calculation Formula: The capacity of an open delta system is found by multiplying the square root of three with the rating of one transformer, resulting in a lower total power output compared to a closed delta system.
- Diagram: The connection diagram shows how two transformers supply a three-phase load with a unity power factor, illustrating the system’s operation.
- Load Distribution: In an open delta system, each transformer supplies 10 kVA, totaling 17.32 kVA, demonstrating how power is distributed and why efficiency drops.
What is an Open Delta Connection Transformer?
An open-delta transformer bank uses two single-phase units to supply a three-phase, three-wire load. The V-V connection can keep service available after one unit is removed from a closed-delta bank. It can also be a planned arrangement for smaller loads or later expansion. Its main limitation is reduced bank capacity, not an automatic loss of efficiency.
The example below separates transformer nameplate capacity, bank output and conversion efficiency.
Start with three identical single-phase transformers, each rated 10 kVA. Connecting both sides as a delta connection forms a closed delta-delta bank.
For an ideal balanced load, what is the bank’s maximum three-phase apparent-power rating?
The closed bank is rated 30 kVA because each of its three units can carry 10 kVA at the same time.
If one unit is isolated, the two remaining units form an open-delta bank. Protection, isolation and reconnection must follow the transformer ratings, utility rules and an engineered switching procedure.
The remaining bank still supplies three line-to-line voltages, but at what balanced three-phase rating?
The two nameplates add to 20 kVA, but the bank cannot deliver 20 kVA to a balanced three-phase load without exceeding a transformer’s rating.
The maximum balanced bank rating is 17.32 kVA when each transformer carries its full 10 kVA.
This 17.32 kVA value is 86.6% of the two transformers’ combined 20 kVA nameplate rating. That ratio is a bank-utilisation factor. Efficiency is output power divided by input power and depends on core loss, winding loss, power factor, temperature and loading. The connection alone does not prove that open delta is less efficient.
The 20 kVA sum cannot be used as balanced three-phase output because each transformer’s current is displaced from its voltage. The phasor relationship limits the bank to √3 times one unit’s rating.
Open delta therefore maintains three-phase service at reduced capacity. A real installation may need further derating for unbalanced loads, harmonics, ambient conditions or equipment-specific limits.
The next section derives the 17.32 kVA bank rating.
Open Delta Transformer Calculations (Example Problem)
Let S_u be the kVA rating of either identical transformer.
For a balanced load, open-delta bank capacity = 
x rating of one transformer = x 10 kVA= 17.32 kVA.
Equivalently:
Open-delta bank capacity = 0.577 x the rating of the original closed-delta bank = 0.577 x 30 kVA = 17.32 kVA.
The two-unit bank therefore carries 57.7% of the original three-unit closed bank’s balanced kVA rating. It also carries 86.6% of the two remaining nameplates’ 20 kVA sum.
The circuit diagram shows why line current and transformer winding current produce this rating:
The diagram uses a balanced unity PF load. Each transformer is rated 10 kVA, with 1000 V across its winding and a 10 A rated winding current. Current at node A follows Kirchoff’s Current Law. For the balanced three-phase load, apparent power is 
x1000x10=17320 VA=17.32 kVA
Each transformer carries 1000 V x 10 A = 10 kVA. Both units are fully loaded, while their phasor sum supplies a 17.32 kVA balanced bank load rather than 20 kVA.
To calculate load as complex power, use the voltage phasor multiplied by the conjugate current phasor. For transformer 1: VI*= 
=8660.2-j5000 VA.
For transformer 2, the chosen current reference gives:
VI*= 
=8660.2+j5000
The negative current sign follows the reference direction in the diagram. Reversing that reference changes the phasor angle by 180°, but not the physical result.
The real-power contribution of each transformer is 8660.2 Watt, or about 8.66 kW. Together they supply the unity-power-factor three-phase load with about 17.32 kW.
The transformer complex powers have equal and opposite 5 kVAR components, so their net reactive power is zero for the resistive load. Each unit still carries 10 kVA. This internal phasor relationship explains why the combined kVA rating of transformers exceeds the 17.32 kVA bank output.
For a balanced load with power-factor angle ø, the individual transformer power factors follow cos(30°+ø) and cos(30°-ø), with assignment depending on phase sequence and reference direction. This can make one unit carry a less favourable power factor even when their kVA magnitudes match.
In this unity-power-factor example, ø=0. Each transformer therefore supplies 10 kVA x cos30° = 8.66 kW of real power. Actual bank design must also check voltage regulation, unbalance, protection, grounding and the transformer’s permitted loading.
