Basic Electrical & Electronics Engineering — Alternating Current Fundamentals, NEC licence examination syllabus (Nepal Engineering Council).
Three Phase Systems: why industry doesn't use single phase
Three wires carrying three staggered sine waves — and suddenly motors self-start.
Start here: why a fan needs a push and a factory motor does not
Some old single-phase motors will not start on their own — they hum, and need a nudge before they run. Once turning, they are fine. It is only starting that defeats them.
The reason: a single AC supply pushes back and forth along one line. Back and forth is not the same as round and round, so there is nothing to tell the motor which way to turn.
Three-phase fixes this completely. Three supplies, each peaking a third of a cycle after the last, arranged around the motor. As one pulls, the next takes over, then the next — so the pull genuinely rotates. A three-phase motor starts by itself because the field is already going round.
💡 The everyday picture: one person shoving a roundabout back and forth achieves nothing. Three people spaced around it, each pushing in turn as it comes past, spin it up easily. Three-phase is those three people.
Why 120° apart
A full cycle is 360°. Share it equally among three, and each gets 360 ÷ 3 = 120°.
Equal spacing matters for a reason worth knowing: with the three phases evenly spread and the load balanced, the three currents cancel exactly at any instant. One is positive while the other two are negative, and they always sum to zero.
💡 That cancellation is why a balanced three-phase system needs almost no neutral current — the return path carries nothing, because the three phases return through each other. It is also why an unbalanced load is a problem: the cancellation fails and the neutral suddenly has to carry real current.
The practical payoff
⚡ What three-phase buys
Motors start themselvesThe rotating field needs no starting mechanism, which removes the part most likely to fail.
Smoother powerSingle-phase power drops to zero twice a cycle. With three phases staggered, total power is nearly constant — so motors do not vibrate.
Less copperThree phases deliver far more power per kilogram of wire than three separate single-phase circuits, because they share return paths.
Star and delta: two ways to wire the same three
Given three supplies, there are exactly two sensible ways to connect them, and the difference is easy to picture.
🔌 The two connections
Star (Y)All three joined at a common centre point, like three spokes from a hub. That centre is the neutral.
Delta (Δ)Joined end to end in a closed triangle. No centre, no neutral.
STAR line voltage = √3 × phase voltage
line current = phase current
DELTA line voltage = phase voltage
line current = √3 × phase current
The √3 appears in ONE place each time — and it is
the opposite place. That is the whole difference.
💡 Where does √3 come from? The two phases you measure between are 120° apart, not aligned, so they do not simply add — you are adding them at an angle. Working the geometry gives exactly √3 ≈ 1.732. It is the same reason walking 1 km east then 1 km north leaves you 1.41 km away rather than 2 km.
Your house runs on a single sine wave. A factory doesn't. Instead it gets three sine waves of equal size, each shifted 120° from the next. This arrangement is so much better for delivering power that essentially all generation and transmission on earth uses it. The reasons are worth knowing, because "state the advantages of three-phase" is a stock exam question.
The three voltages
v_R = V_m sin(ωt)
v_Y = V_m sin(ωt − 120°)
v_B = V_m sin(ωt − 240°)
At every instant: v_R + v_Y + v_B = 0
That last identity is the whole trick — it's why a balanced
system needs no return current in the neutral.
✅ Why three phase wins
Constant powerSingle-phase power pulses at twice the supply frequency (it hits zero 100 times a second). Three-phase total power is constant — smoother torque, less vibration in motors.
Self-starting motorsThree staggered phases create a rotating magnetic field naturally. A single-phase motor needs extra windings or a capacitor just to know which way to turn.
Less copperFor the same power delivered over the same distance, three-phase transmission uses roughly 75% of the conductor material of single phase.
Higher ratingA three-phase machine of the same physical size delivers about 1.5× the output of a single-phase one.
Star (Y) connection
Three windings joined at a common point (the NEUTRAL).
V_line = √3 × V_phase (line voltage is bigger)
I_line = I_phase (same current)
Line voltage LEADS phase voltage by 30°
Delta (Δ) connection
Three windings joined end-to-end in a closed loop.
No neutral point exists.
V_line = V_phase (same voltage)
I_line = √3 × I_phase (line current is bigger)
Line current LAGS phase current by 30°
The pattern is symmetric and easy to hold: star boosts voltage, delta boosts current, each by √3. In both cases the total power uses the same formula: P = √3 V_L I_L cos φ. So the √3 never disappears. It simply moves between the voltage term and the current term.
Power in three-phase — one formula for both
Per phase: P_ph = V_ph × I_ph × cos φ
Total: P = 3 V_ph I_ph cos φ
In LINE quantities (works for BOTH star and delta):
P = √3 × V_L × I_L × cos φ
Q = √3 × V_L × I_L × sin φ (VAR)
S = √3 × V_L × I_L (VA)
Worked numerical 1 — star-connected load
A balanced star load of 10 Ω resistance and 10 Ω inductive reactance per phase is fed from a 400 V, 50 Hz three-phase supply. Find phase voltage, phase current, line current, power factor and total power.
Star, so V_L = 400 V is the LINE voltage:
V_ph = V_L/√3 = 400/1.732 = 231 V
Impedance per phase:
Z_ph = √(R² + X_L²) = √(10² + 10²) = √200 = 14.14 Ω
Phase current:
I_ph = V_ph/Z_ph = 231/14.14 = 16.34 A
I_L = I_ph = 16.34 A (star)
Power factor:
cos φ = R/Z = 10/14.14 = 0.707 lagging (φ = 45°)
Total active power:
P = √3 V_L I_L cos φ
= 1.732 × 400 × 16.34 × 0.707
= 8003 W ≈ 8.0 kW
Check per-phase: P = 3 × I_ph² × R
= 3 × (16.34)² × 10 = 8010 W ✔
Reactive power:
Q = √3 × 400 × 16.34 × 0.707 = 8.0 kVAR
Apparent:
S = √3 × 400 × 16.34 = 11.32 kVA
Check: √(8.0² + 8.0²) = 11.3 kVA ✔
Worked numerical 2 — the same load in delta
Reconnect that identical load in delta on the same 400 V supply. Find the new currents and power.
Delta, so each phase sees the FULL line voltage:
V_ph = V_L = 400 V
I_ph = V_ph/Z_ph = 400/14.14 = 28.29 A
I_L = √3 × I_ph = 1.732 × 28.29 = 48.99 A
P = √3 V_L I_L cos φ
= 1.732 × 400 × 48.99 × 0.707
= 24 000 W = 24 kW
COMPARE: star 8 kW → delta 24 kW
Delta draws exactly THREE TIMES the power of star for the
same load impedance and supply. That is why large induction
motors start in star (low current) and switch to delta once
running — the classic "star-delta starter".
That 3× ratio is one of the most useful facts in the syllabus. Because V_ph is √3 larger in delta, and I_ph is therefore √3 larger too, power goes up by √3 × √3 = 3. A star-delta starter exploits this to limit starting current to a third of its delta value.
Worked numerical 3 — working backwards from power
A three-phase motor delivers 15 kW at 0.85 power factor from a 415 V supply, with 90% efficiency. Find the line current.
Output power = 15 kW, so INPUT power is larger:
P_in = 15/0.90 = 16.67 kW
From P = √3 V_L I_L cos φ:
I_L = P/(√3 × V_L × cos φ)
= 16 670/(1.732 × 415 × 0.85)
= 16 670/610.9
= 27.3 A
Apparent power drawn:
S = √3 × 415 × 27.3 = 19.6 kVA
Note the efficiency step: the supply must deliver more than the mechanical output, and forgetting to divide by η is a very common lost mark.
💡 Exam angle: worth 8–10 marks, and often asked as "solve for star, then repeat for delta". Write down which connection you're in and immediately fix V_ph and I_ph — that single line prevents most errors. Memorise P = √3 V_L I_L cos φ (identical for both), the √3 relations, and the delta = 3 × star power result. If the question mentions efficiency, remember input = output/η.
Syllabus points
Star & delta connections
Line vs phase voltage/current (numerical)
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