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Delta/star 120 kVA transformer values

Short answer

Primary (delta): VL=11 kV, Vph=11 kV, IL=6.30 A, Iph=3.64 A. Secondary (star): VL=1905 V, Vph=1100 V, IL=36.4 A, Iph=36.4 A.

An ideal 3-phase delta/star step-down transformer delivers power to a balanced 3-phase load of 120 kVA at 0.8 power factor. The input line voltage is 11 kV and the turns ratio of the transformer, phase to phase is 10. Determine the line voltage, line currents, phase voltages and phase currents on both the primary and secondary sides.

A balanced three-phase load of 120 kVA is supplied. For a delta/star step-down transformer, the primary is delta-connected and the secondary is star-connected.

Given: S = 120 kVA, cos φ = 0.8 lag; primary line voltage VL1 = 11 kV; phase turns ratio N1/N2 = 10.

Assumption: the transformer is ideal, so input kVA = output kVA = 120 kVA.

Primary side (delta)

Line voltage       VL1 = 11 000 V
Phase voltage      Vph1 = VL1 = 11 000 V   (delta)
Line current       IL1 = S/(√3 VL1)
                       = 120 000/(√3 × 11 000) = 6.30 A
Phase current      Iph1 = IL1/√3 = 6.30/1.732 = 3.64 A

Secondary side (star)

Phase voltage      Vph2 = Vph1/10 = 1100 V
Line voltage       VL2 = √3 Vph2 = 1.732 × 1100 = 1905 V
Line current       IL2 = S/(√3 VL2)
                       = 120 000/(√3 × 1905) = 36.4 A
Phase current      Iph2 = IL2 = 36.4 A   (star)

Check: Iph2/Iph1 = 36.4/3.64 = 10, as required by the turns ratio.

QuantityPrimary (delta)Secondary (star)
Line voltage11 000 V1905 V
Phase voltage11 000 V1100 V
Line current6.30 A36.4 A
Phase current3.64 A36.4 A

The power factor of 0.8 gives the load power of 96 kW, which does not change the above currents, because they are determined from the kVA.

✓ Answer checked by Claude Opus 5.5 (high effort)

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