Ideal source and lines (no internal or line impedance); every number is an exact complex-number (phasor) calculation, RMS values. Colors: L1 red, L2 yellow, L3 blue, N gray; graphs also use solid / dashed / dotted lines. Wattmeters: current coils in L1 and L2, voltage coils to L3.
The diagram behind every three-phase exam question: a wye (star) generator with its neutral, three line conductors, a delta load and a single-phase load. Switch between the labeled diagram, a blank one to test your memory, and a quiz where you place each label yourself. You can also download both versions or print a label worksheet.
Tap a word, then tap the numbered box it belongs to (on the diagram or in the list below). Tap a filled box to take the word back. On a phone, swipe the diagram sideways to see every number.
New numbers every time. Solve on paper, type your answer for the quantity asked, then open the step-by-step solution.
Fifteen questions (from a bank of twenty), numeric and conceptual: wye and delta, √3, neutral current, power, power factor, capacitor banks and the two-wattmeter method.
A single AC voltage reaches zero twice every cycle, so the power it delivers pulses: a single-phase motor gets a jerky push. Put three coils round the generator, 120° apart, and the spinning magnet makes three voltages that peak one after another: L1, then L2, then L3. At every moment at least two of them are working, and the total power is perfectly smooth. Because the three currents cancel when the loads are equal, the “return” wire carries nothing and can be thin or left out. In 1891 a three-phase line of roughly 175 km carried power from a hydroelectric plant at Lauffen am Neckar to the International Electrotechnical Exhibition in Frankfurt, one of the first long-distance demonstrations of AC transmission; within a few decades three-phase AC became the standard for power grids.
The three coils (or loads) can be joined in two ways. In wye (Y, star) one end of each is joined at the star point, which can be connected to a neutral wire. Each coil gets the phase voltage Vph; between two lines you measure the line voltage VL = √3 Vph, because two phasors of the same size 120° apart differ by √3 times that size. That is why Brazil’s 220/380 V supply gives 220 V between a line and neutral and 380 V between two lines (220 × 1.732 = 381). In delta (Δ) the three are joined in a triangle: each one sits between two lines, so Vph = VL, but each line current is the difference of two branch currents, IL = √3 Iph. A delta has no star point, so it has no neutral.
Each phase has an impedance Z = R + jX. The current lags the phase voltage by the impedance angle φ (inductive load) or leads it (capacitive). Total power: P = √3 VL IL cos φ (watts), Q = √3 VL IL sin φ (var) and S = √3 VL IL (VA), the same formula for wye and delta. The power factor cos φ = P/S tells you how much of the current does useful work. Motors are inductive (PF about 0.8), so they draw extra current that only shuttles energy back and forth. A capacitor bank supplies that reactive power locally: the real power stays the same, but the line current falls, and so do the cable losses and the transformer load.
With unequal loads the currents no longer cancel. On a 4-wire supply the neutral carries the phasor sum IN = Ia + Ib + Ic and keeps every phase at Vph. Without the neutral, the load’s star point floats to the voltage given by Millman’s theorem, and the phase voltages become unequal: lightly loaded phases get over-voltage. Power in any 3-wire system is measured with just two wattmeters (Blondel): P = W1 + W2. For a balanced load W1 = VLIL cos(30° − φ) and W2 = VLIL cos(30° + φ), so tan φ = √3 (W1 − W2)/(W1 + W2), and W2 goes negative below PF 0.5. Reversing two lines reverses the phase sequence (abc → acb): a motor turns the other way, and the two wattmeter readings swap.
Press Balanced Y, tick Rotate with ωt and watch the phasors spin while the white power curve stays flat. Switch to Same load in Δ: three times the current. Press Unbalanced + neutral, then Broken neutral and watch the lamps. Try PF correction and drag the target to 1.00. Finally press Low PF: a wattmeter reads negative, and the sum is still right.
| Quantity | Wye (Y, star) | Delta (Δ) |
|---|---|---|
| Phase voltage (across one Z) | Vph = VL/√3 | Vph = VL |
| Phase current (in one Z) | Iph = IL | Iph = IL/√3 |
| Phase angle shift (sequence abc) | Vab leads Van by 30° | Ia lags Iab by 30° |
| Neutral | possible (star point); IN = Ia + Ib + Ic | none |
| Total power (balanced) | P = √3 VL IL cos φ = 3 Vph Iph cos φ · Q = √3 VL IL sin φ · S = √3 VL IL (φ = angle of Z) | |
| Equivalent impedance | ZY | ZY = ZΔ/3 |
| Capacitor bank for QC | C = QC/(3ωVph²) | C = QC/(3ωVL²) = CY/3 |
For the same power, the same line-to-line voltage and the same losses, a three-phase 3-wire line needs 75% of the conductor material of a single-phase 2-wire line: each of the three wires carries 1/√3 of the single-phase current, so it can have twice the resistance (half the cross-section). A two-phase system (two voltages 90° apart, used in some early networks) also gives constant power and a rotating field, but needs four wires, or three with an unequal shared wire. Three phases is the smallest symmetric system that does it with three identical wires, which is what makes the simple, rugged induction motor and economical lines possible. More phases add wires without much gain.
A motor whose windings are rated for the line voltage in delta can be started in wye: each winding then gets VL/√3, so the starting line current is one third of the direct delta start (and the starting torque is one third too). Compare the presets “Balanced Y” (21.94 A) and “Same load in Δ” (65.82 A).
Brazil uses 60 Hz with low-voltage networks of 127/220 V or 220/380 V, depending on the region and the distribution company. Europe and most IEC countries use 230/400 V, 50 Hz (IEC 60038). North America uses 60 Hz with 120/208 V wye in commercial buildings and 277/480 V in industry (homes get single-phase 120/240 V). The nominal pairs are rounded: 380/√3 = 219.4 V, 400/√3 = 230.9 V. Wire colors: IEC 60445 uses brown, black and gray for L1, L2, L3 and blue for the neutral; this lab draws L1 red, L2 yellow and L3 blue (an older convention still found in many textbooks) because those stay visible on a dark screen, and it names every wire.
Ideal sources and ideal lines (no internal or line impedance), linear loads, sinusoidal steady state, RMS phasors. Lamps are modeled as fixed resistances (a real filament’s resistance rises as it heats up). The wattmeters are ideal and connected on the supply side, so they see the load and the capacitor bank together. A delta source is assumed to have identical windings, so no current circulates inside the delta; its winding currents are then Iab = (Ia − Ib)/3, which is IL/√3 when balanced. The capacitor bank is sized from the load’s own P and Q (for an unbalanced load the corrected PF is then approximate). The animation is slowed down: the real rotor of a 2-pole 60 Hz machine turns 3 600 times a minute. The meters show 3–4 significant figures because the model is exact; real instruments are typically accurate to about 0.5–2%.
tp_model.js by nodal analysis with complex numbers (Van = Vph∠0° for a wye source; Vab = VL∠0° for a delta source; sequence abc unless acb is chosen) and checked against the hand calculations above in an automated test.Three-phase power is AC delivered as three voltages of the same size and frequency, 120 degrees apart, made by three coils in one generator. It is used for almost all generation and transmission and for motors and large loads, because it delivers constant power, needs less conductor material than single-phase for the same power, and produces a rotating magnetic field that drives induction motors.
Key takeaway: three equal AC voltages 120° apart.In a wye connection the line voltage is the difference of two phase voltages that are 120 degrees apart. Two phasors of size V with 120 degrees between them differ by √3 × V (the base of an isosceles triangle with a 120 degree apex is 2V cos 30° = √3 V). So 220 V phase-to-neutral gives 220 × 1.732 = 380 V between two lines, and the line voltage leads the phase voltage by 30 degrees.
Key takeaway: V_L = √3 V_ph in wye, from 120° phasor geometry.In wye, one end of each winding or load is joined at a star point that can be connected to a neutral wire; each one gets the phase voltage V_L/√3 and carries the line current. In delta, the three are joined in a triangle, each across two lines: each gets the full line voltage, and the line current is √3 times the branch current. A delta has no neutral point.
Key takeaway: wye: V_ph = V_L/√3, I_L = I_ph; delta: V_ph = V_L, I_L = √3 I_ph.The neutral current is the phasor sum of the three line currents. In a balanced system they are equal in size and 120 degrees apart, and three such phasors add up to zero at every instant. That is why high-voltage transmission lines need only three conductors.
Key takeaway: balanced currents cancel: I_a + I_b + I_c = 0.The star point of the load is no longer held at the supply neutral. It moves to the voltage given by Millman’s theorem, and the phase voltages become unequal. In the lab, lamps of 10, 20 and 40 ohms on a 220/380 V supply get 143.6, 248.8 and 287.3 V instead of 219.4 V: the most lightly loaded phase gets an over-voltage that can destroy appliances. That is why the neutral must never be fused or switched on its own.
Key takeaway: a lost neutral makes the phase voltages unequal.For a balanced load P = √3 × V_L × I_L × cos φ, Q = √3 × V_L × I_L × sin φ and S = √3 × V_L × I_L, for both wye and delta. Here φ is the angle of the load impedance, the angle between the phase voltage and the phase current, not between line voltage and line current. Example: 380 V, 21.94 A, PF 0.8 gives P = 11.55 kW.
Key takeaway: P = √3 V_L I_L cos φ, with φ the impedance angle.Find the reactive power to remove, Q_C = P (tan φ1 − tan φ2), then share it between three capacitors: C = Q_C / (3 ω V²), with V the voltage across each capacitor (V_ph for a wye bank, V_L for a delta bank, so a delta bank needs one third of the capacitance). Raising an 11.55 kW load from PF 0.80 to 0.95 at 220/380 V, 60 Hz needs 4.87 kvar: 89.4 µF per phase in wye, and the line current falls from 21.94 A to 18.48 A.
Key takeaway: Q_C = P(tan φ1 − tan φ2), C = Q_C/(3ωV²).In any 3-wire system two wattmeters measure the total power: current coils in two lines, voltage coils to the third, and P = W1 + W2 (Blondel’s theorem). For a balanced load W1 = V_L I_L cos(30° − φ) and W2 = V_L I_L cos(30° + φ). For a lagging load with φ more than 60 degrees (power factor below 0.5), cos(30° + φ) is negative, so W2 reads negative (for a leading load it is W1). The meter is fine; add the readings with their signs. The power factor follows from tan φ = √3 (W1 − W2)/(W1 + W2).
Key takeaway: P = W1 + W2 with signs; one reading is negative when PF < 0.5.The power of each phase is p = VI cos φ plus a term at twice the supply frequency. The three double-frequency terms are shifted by 240 degrees from each other and cancel, so p_a + p_b + p_c = 3 V I cos φ at every instant. A three-phase motor therefore gets a smooth torque without the 100 or 120 Hz pulsation of a single-phase machine, which means less vibration and noise.
Key takeaway: balanced three-phase power does not pulsate.Brazil uses 60 Hz with low-voltage networks of 127/220 V or 220/380 V, depending on the region (the first number is phase-to-neutral, the second line-to-line). Europe and most IEC countries use 230/400 V at 50 Hz. North America uses 60 Hz with 120/208 V in commercial buildings and 277/480 V in industry. All of them are wye systems with V_L = √3 V_ph.
Key takeaway: Brazil 127/220 or 220/380 V, 60 Hz; Europe 230/400 V, 50 Hz.