As in figure (b), a balanced three-phase load is Δ-connected with an impedance of Z = 30 Ω per phase. If it is replaced by an equivalent Y connection as in figure (a) that draws the same line currents, which of the following is the impedance per phase?

Answer: A. 10 Ω

Method 1 (Δ-Y conversion formula): if the sides of the Δ are Zab, Zbc and Zca, phase a of the Y is Za = Zab·Zca/(Zab + Zbc + Zca); for a balanced load this is 30 × 30/90 = 10 Ω, that is, Z/3. Method 2 (check with the line current): with a line voltage VL, the Δ connection has a phase current of VL/30 and a line current √3 times larger, √3 VL/30. In the Y connection the line current is the phase voltage VL/√3 divided by ZY, so (VL/√3)/ZY = √3 VL/30 gives ZY = 10 Ω.

• A. Correct. For a balanced load, ZY = ZΔ/3 = 10 Ω.
• B. 30/√3 ≈ 17 Ω wrongly applies the factor √3 between voltages or currents directly to the impedance. For impedance the factor √3 appears twice, giving one third.
• C. Multiplying by 3 is the opposite conversion, used when turning a Y-connected Z into an equivalent Δ connection.