Three-phase circuits, delta-wye and the 30° shift on the PE Power exam

Three-phase questions on the PE Power exam ask for a line current, a winding voltage, a power, or the angle of one phasor against another. The NCEES handbook holds the relations under 3-Phase Circuits in its Circuits chapter: the delta and wye line and phase relations, balanced three-phase power, the positive-sequence phasor sets, and the delta-wye load conversion. Transformer vector groups, open-delta capacity and the two-wattmeter method are not printed there.

What you actually need to own

• Line and phase relations. In a wye, V_L = √3 · V_P and I_L = I_P. In a delta, V_L = V_P and I_L = √3 · I_P. A delta element sees the full line-to-line voltage and carries line current divided by √3.
• Balanced power. |S| = √3 · V_L · I_L, with P = |S| cos θ and Q = |S| sin θ. P and Q combine as |S| = √(P² + Q²).
• Phase sequence and the 30° shift. For positive (abc) sequence with V_an at 0°, V_bn sits at −120° and V_cn at +120°. The line-to-line set is V_ab = √3 · V_P at +30°, V_bc at −90° and V_ca at +150°. The 30° comes out of the subtraction V_ab = V_an − V_bn. On acb sequence the same subtraction puts V_ab at −30°. The handbook prints the abc set only.
• Delta-wye load conversion. For three equal impedances, Z_Δ = 3 · Z_Y. The handbook prints the general conversion for unequal branches.
• Unbalanced loads. The √3 relations hold only for a balanced set. With unequal branches, find each branch current as its own voltage phasor divided by its own impedance, then add the phasors at the node. A resistive branch keeps the angle of its own voltage. Take three resistive line-to-neutral loads drawing 30 A, 20 A and 10 A on abc sequence. The neutral carries 30∠0° + 20∠−120° + 10∠120° = 15 − j8.66 A, which is 17.3 A.
• Transformer banks. A delta winding is rated at line-to-line voltage. A wye winding is rated at line-to-neutral voltage. The winding turns ratio of a delta-wye bank therefore differs from the ratio of the two line voltages by √3. Under the ANSI/IEEE standard connection, the low-voltage line voltages of a delta-wye or wye-delta bank lag the high-voltage ones by 30° on abc sequence. The same bank on acb sequence shows a 30° lead. Clock notation counts that lag in 30° steps, so hour 1 is a 30° lag and hour 11 is 330° of lag, which is a 30° lead. A field-assembled bank can go either way. Read the polarity marks: the two windings of one unit are in phase, marked end to unmarked end.
• Open delta. Two single-phase units carry √3 times one unit's rating. That is 57.7% of the original three-unit bank and 86.6% of the two units still in service.

Where people lose points

• Dropping the √3, or using it twice. √3 · V_L · I_L takes the line-to-line voltage. Putting the line-to-neutral voltage into it leaves the answer √3 low. Writing 3 · V_L · I_L leaves it √3 high. The 3 goes only with phase values, as 3 · V_P · I_P.
• Ignoring the stated connection. The same impedances draw three times the line current in delta as in wye. Read the drawing.
• Running the conversion backwards. Z_Y = Z_Δ/3. Multiplying by 3 when dividing is needed puts the current off by a factor of nine.
• Using I_L = √3 · I_P on an unbalanced delta. With an open element or a blown fuse, redraw the circuit and solve the paths that remain.
• Reconnecting wye to delta. Each element's voltage rises by √3, so its current rises by √3, and the delta relation adds another √3. Line current and power both triple.
• Subscript order. Reversing the subtraction gives V_ba, 180° from V_ab.
• The 57.7% figure on the wrong base. It belongs to the three-unit bank. Applied to the two remaining units it counts the penalty twice.
• Assuming every conductor is the same voltage from neutral. On a four-wire delta with one winding center-tapped, the third corner sits at √3/2 of the line-to-line voltage from neutral.

How to study it

Take one set of three equal impedances. Solve it in wye and in delta on the same supply, then open one element in each and solve again. Write every voltage as a phasor from V_an at 0°, then repeat on acb sequence. Then draw a delta-wye bank from three single-phase units and derive its shift from the polarity marks. The same phasor sets feed symmetrical components and the per-unit system.

Part of the Circuit Analysis area of the PE Power exam. → Practice it on CastorPrep