Three-phase four-wire circuits

Flip on a single lamp in your house and, unnoticed, you are drawing current from just one of the three phases a power utility runs into your street. Millions of such single-phase loads — lights, sockets, appliances — are scattered unevenly across the three lines, switched on and off at random by millions of independent decisions. A pure three-phase source, wired with only its three line conductors, has no way to tolerate that imbalance gracefully: the load's electrical centre point would be dragged off-centre, and some appliances would see dangerously high voltage while others starved. The fix is a single extra wire — the neutral — turning a three-wire system into a three-phase four-wire system. This article asks what current that fourth wire actually carries, why it is exactly zero for a balanced load and generally nonzero otherwise, and why real distribution networks could not function without it.

The core insight

Adding a neutral conductor to a star-connected three-phase source lets each line independently supply single-phase loads at a fixed line-to-neutral voltage, because the neutral carries the phasor-sum imbalance current I⃗N=I⃗R+I⃗Y+I⃗B\vec I_N=\vec I_R+\vec I_Y+\vec I_B, which is exactly zero for a balanced load and nonzero — up to the size of a full phase current — for an unbalanced one.

The core idea

A three-phase three-wire system (only lines R, Y, B) suits balanced loads like motors, where no return conductor is needed. A three-phase four-wire system adds the neutral (N), tied to the source's star point and normally earthed, so each line can feed independent single-phase loads at the stable line-to-neutral voltage Vph=VL/3V_{ph}=V_L/\sqrt3, alongside any three-phase load across the lines.

Balanced case. When the three phase currents are equal in magnitude and exactly 120∘120^\circ apart, their phasor sum vanishes:

I⃗N=I⃗R+I⃗Y+I⃗B=0\vec I_N = \vec I_R+\vec I_Y+\vec I_B = 0

where: I⃗R,I⃗Y,I⃗B\vec I_R,\vec I_Y,\vec I_B are the phasor currents in the three lines. So a perfectly balanced load makes the neutral carry nothing — the four-wire and three-wire systems behave identically for it.

Unbalanced case. Real single-phase loads are rarely switched on symmetrically across the three phases, so the currents are generally unequal in magnitude and not exactly 120∘120^\circ apart:

I⃗N=I⃗R+I⃗Y+I⃗B≠0\vec I_N = \vec I_R+\vec I_Y+\vec I_B \neq 0

This nonzero resultant is precisely the imbalance current, and it must return through the neutral. Without that return path, an unbalanced star load's own star point would shift away from the source's, distorting the individual phase voltages seen by the loads — some overvoltaged, some undervoltaged. The neutral holds the load star point pinned at the source's neutral potential regardless of how the load is distributed.

Figure
Star connected source and unbalanced load joined by three lines and a neutral wire carrying the phasor sum of the three line currents
Three-phase four-wire distribution to unbalanced single-phase loads source star point R IR=10A Y IY=10A B IB=0A lamps R lamps Y B unloaded neutral N carries IN = IR+IY+IB = 10 A here

Why it matters practically. The neutral does two jobs: it fixes the line-to-neutral voltage regardless of load imbalance, and it carries the return imbalance current. That is why the neutral is never independently fused or switched, and is earthed in TN-type schemes, so a broken neutral cannot leave line-to-neutral loads exposed to arbitrary overvoltage.

Derivation ANeutral current as the phasor sum of the three line currents

The goal of this derivation is to show, starting only from Kirchhoff's current law at the load's star point and the definition of a balanced three-phase supply, why the neutral current is exactly zero for a balanced load and generally nonzero for an unbalanced one, so that the single phasor-sum formula for I⃗N\vec I_N can be trusted and applied directly to any combination of three line currents without re-deriving it each time.

Step 1. Apply Kirchhoff's current law at the load's star (neutral) node. Whatever current flows in along the three lines R, Y, B must flow out along the neutral wire back to the source, so the neutral current is simply the phasor sum: I⃗N=I⃗R+I⃗Y+I⃗B\vec I_N = \vec I_R+\vec I_Y+\vec I_B. This is a bookkeeping statement, true for any load, balanced or not.

Step 2. Specialise to a balanced load: equal magnitude IphI_{ph} and mutual displacement of exactly 120∘120^\circ, sequence R-Y-B, so I⃗R=Iph∠θ\vec I_R=I_{ph}\angle\theta, I⃗Y=Iph∠(θ−120∘)\vec I_Y=I_{ph}\angle(\theta-120^\circ), I⃗B=Iph∠(θ−240∘)\vec I_B=I_{ph}\angle(\theta-240^\circ).

Step 3. Resolve each into rectangular components and add. The three real parts are Iph[cos⁡θ+cos⁡(θ−120∘)+cos⁡(θ−240∘)]I_{ph}[\cos\theta+\cos(\theta-120^\circ)+\cos(\theta-240^\circ)], and the three imaginary parts follow the same pattern with sine. A standard trigonometric identity shows that three equal-magnitude phasors spaced exactly 120∘120^\circ apart always sum to zero, independent of θ\theta, so I⃗N=0\vec I_N=0 for any balanced load.

Step 4. Drop the balance assumption. If any phase current differs in magnitude or angle from this pattern, the cancellation in Step 3 no longer occurs exactly, and I⃗N=I⃗R+I⃗Y+I⃗B\vec I_N=\vec I_R+\vec I_Y+\vec I_B is a nonzero phasor — the imbalance current that must physically return through the neutral wire.

Worked example

A four-wire system supplies three single-phase resistive loads connected line-to-neutral: 10 A10\ \text{A} in phase R, 10 A10\ \text{A} in phase Y, and phase B unloaded (current zero). Voltages are balanced, 120∘120^\circ apart, sequence R-Y-B, with V⃗R\vec V_R as reference; currents are in phase with their respective voltages since the loads are resistive. Find the neutral current.

Step 1. Write the phase currents in rectangular form. I⃗R=10∠0∘=10+j0\vec I_R = 10\angle0^\circ = 10+j0. I⃗Y=10∠(−120)∘=−5−j8.66\vec I_Y = 10\angle(-120)^\circ = -5-j8.66. I⃗B=0\vec I_B = 0.

Step 2. Sum to find the neutral current: I⃗N=I⃗R+I⃗Y+I⃗B=(10−5+0)+j(0−8.66+0)=5−j8.66\vec I_N = \vec I_R+\vec I_Y+\vec I_B = (10-5+0)+j(0-8.66+0) = 5-j8.66.

Step 3. Convert to magnitude and angle. ∣I⃗N∣=52+8.662=25+75=100=10 A|\vec I_N| = \sqrt{5^2+8.66^2}=\sqrt{25+75}=\sqrt{100}=10\ \text{A}, and ∠I⃗N=tan⁡−1(−8.66/5)=−60∘\angle\vec I_N = \tan^{-1}(-8.66/5) = -60^\circ. The neutral carries a full 10 A10\ \text{A} — this single unloaded phase produces a neutral current equal in magnitude to a fully loaded phase, confirming the neutral must be sized for substantial current, never treated as negligible.

Competitive stretch

Suppose phase B is instead loaded with 5 A5\ \text{A} resistive (rather than left open) while R and Y stay as above. Then I⃗B=5∠(−240)∘=5∠120∘=−2.5+j4.33\vec I_B = 5\angle(-240)^\circ = 5\angle120^\circ = -2.5+j4.33, so I⃗N=(10−5−2.5)+j(0−8.66+4.33)=2.5−j4.33\vec I_N = (10-5-2.5)+j(0-8.66+4.33) = 2.5-j4.33, giving ∣I⃗N∣=2.52+4.332=6.25+18.75=25=5 A|\vec I_N|=\sqrt{2.5^2+4.33^2}=\sqrt{6.25+18.75}=\sqrt{25}=5\ \text{A}. Partially loading the idle phase halved the neutral current — a useful check that adding load symmetrically reduces imbalance.

Common traps

  • Assuming the neutral is purely a safety/earthing wire that never carries current — under any real unbalanced loading it routinely does, and must be sized accordingly.
  • Believing a three-wire system can safely feed unbalanced single-phase loads — without a return path the load's star point shifts and phase voltages become unequal.
  • Confusing "unbalanced" with "faulty" — everyday unequal switching of single-phase loads across three phases is normal operation the four-wire system is designed for.
  • Treating neutral current as a scalar sum of the phase-current magnitudes instead of a phasor sum — depending on the angles, INI_N can be much smaller than, up to equal to, or in extreme cases larger than the biggest individual phase current.

Ask the tutor: want to see numerically what happens to each load's voltage if the neutral connection breaks under this same unbalanced example, or try another neutral-current phasor-sum problem?

Why does a balanced three-phase load not need a neutral wire at all?

Because with equal-magnitude currents spaced exactly 120 degrees apart, Kirchhoff's current law at the star point shows their phasor sum is always zero, whatever the common phase angle happens to be. Physically, whatever current returns from one line is exactly balanced by the combined return from the other two at every instant, so no net current ever needs the neutral path — that is why motors and other balanced three-phase equipment are typically wired three-wire.

Does a bigger imbalance always mean a bigger neutral current?

Not necessarily in a simple proportional sense, because INI_N is a phasor sum, not a scalar difference. Two phases could each be heavily loaded but nearly cancel if their angular relationship works out, while a small change in just one phase's current can shift INI_N substantially, as seen when adding a modest 5 A to an unloaded phase B halved the neutral current in the stretch example. Always resolve into components and add before judging the size of the imbalance current.

Why is the neutral wire never fused or switched independently like the line conductors?

If the neutral opened while an unbalanced load was still connected, the load's star point would no longer be held at the source's neutral potential and would drift depending on the relative impedances of the loads on each phase, potentially exposing some line-to-neutral loads to voltages well above their rated value. Because that failure mode is dangerous rather than merely inconvenient, distribution practice keeps the neutral solidly connected (and earthed in TN-type schemes) rather than treating it as just another switchable conductor.

Before this

Where this leads