Three-Phase Power

The three phase powers. Each phase behaves as the single-phase load of Set 23, Problem 1, with its own voltage displaced by 120°:

Note the doubling: a 120° displacement in \(\omega t\) becomes 240° in \(2\omega t\). That is the crux.

Sum the oscillating terms. Reducing \(480°\) modulo 360° gives 120°, so the three arguments differ by 120° from each other:

Three unit phasors 120° apart sum to zero — the same identity that makes the three source voltages balance.

Hence the total:

The instantaneous power equals the average power at every instant. Each phase still pulsates violently; what one phase gives up, the others take on.

Why not two phases? Two sources 180° apart give \(2\times180° = 360° \equiv 0\) in the doubled angle — the two pulsations are in phase and add rather than cancel:

Any \(n \ge 3\) works. Three is chosen as the smallest that does — more phases would need more conductors for no further benefit.

The engineering consequences:

Phase sequence is the order in which the three voltages reach their positive maxima. Two orders are possible:

Note \(V\angle{+120°} = V\angle{-240°}\) — the same phasor, and legacy notation often uses the latter. Both describe \(V_c\) in positive sequence.

The balance conditions. A balanced set satisfies both:

Verifying the second for positive sequence:

This is the identity that made Problem 1 work, and it is why a balanced four-wire system carries no neutral current.

Why it matters:

Swapping any two of the three conductors reverses the sequence — which is how a motor's direction is changed, and why the connection must be checked after any rewiring.

The \(a\)-operator is a convenient shorthand:

Positive sequence is then \(\mathbf{V}_a,\ a^2\mathbf{V}_a,\ a\mathbf{V}_a\). Multiplying by \(a\) rotates a phasor 120° anticlockwise without changing its magnitude.

Determining sequence experimentally. A simple indicator uses two lamps and a capacitor in a star: the unbalanced star point shifts one way or the other depending on sequence, making one lamp brighter. Problem 15's neutral-displacement calculation is exactly the analysis of such a device.

The definitions. In a star, phase voltages are measured from each line to the neutral point; line voltages are measured between two lines:

Line voltage by KVL:

Convert to polar:

So the line voltage leads its phase voltage by 30° and is \(\sqrt3\) times larger.

The other two follow by symmetry:

Themselves a balanced set, 120° apart and summing to zero — as KVL round the three lines requires.

Where the \(\sqrt3\) comes from geometrically. Two phasors of equal length \(V\) separated by 120° have a difference of

The chord of a 120° arc. It is not an approximation and has nothing to do with the number of phases beyond the 120° displacement.

Currents are trivial in star:

because each line conductor is the phase conductor — there is nowhere else for the current to go. Numerically, a 400 V star system has \(V_{ph} = 400/\sqrt3 = 231\ \text{V}\), which is why domestic single-phase outlets in such systems read 230 V.

Voltages are immediate. Each delta branch is connected directly between two lines, so

There is no neutral and no other voltage available. A 400 V delta load has 400 V across every phase.

Currents require KCL at each corner of the delta. With phase currents \(\mathbf{I}_{ab}\), \(\mathbf{I}_{bc}\), \(\mathbf{I}_{ca}\) circulating round the triangle, the line current at corner \(a\) is

Current arrives along the line and splits into two branches — one leaving through \(ab\), one arriving from \(ca\).

With a balanced set \(\mathbf{I}_{ab} = I_{ph}\angle0°\), \(\mathbf{I}_{bc} = I_{ph}\angle{-120°}\), \(\mathbf{I}_{ca} = I_{ph}\angle{+120°}\):

The line current lags its phase current by 30° and is \(\sqrt3\) times larger — the exact mirror of the star voltage result.

Why the roles exchange. The two connections are duals:

Set 17's duality again: series and parallel, KVL and KCL, voltage and current. A star and a delta are the three-phase versions of the same pair.

The power formula is identical for both, which is worth checking. For delta:

and for star:

The \(\sqrt3\) lands on different quantities but the product is the same — which is why one power formula serves both connections.

Per-phase power. A balanced load shares the total equally:

For delta, the phase voltage is the line voltage:

Solve for the phase current:

The line current is \(\sqrt3\) larger in delta:

The phase impedance, with the angle from the power factor:

Check against the total-power formula:

and independently from the resistance:

A note on the impedance angle. It equals the power-factor angle exactly — the current lags the phase voltage by 36.87°. It does not lag the line current by that amount, nor the line voltage: those differ by the 30° of Problem 4. Only the per-phase relationship carries the impedance angle.

Per-phase power:

For star, the phase voltage is reduced by \(\sqrt3\):

This is the single change from Problem 5, and it is the whole difference between the two connections at this stage.

Phase current, which equals the line current in star:

The phase impedance:

Compare with Problem 5's delta. The two problems have different total powers, so the comparison must be normalised. For the same load power on the same line voltage:

The last row is the important one: two loads drawing equal power from equal line voltage draw equal line current, whichever way they are connected. The supply cannot tell the difference.

Verify the \(Z_\Delta = 3Z_Y\) relation on these numbers, scaling Problem 5's load to 1200 W:

Impedance scales inversely with power at fixed voltage, so the scaling is legitimate. The factor of three is exact and is the basis of Problem 19.

Delta, so the phase voltage is the line voltage:

Per-phase impedance:

Reading the power factor straight off the impedance triangle avoids computing an angle at all.

aCurrents:

bThe three powers, from the line quantities:

Two independent checks. From the per-phase elements:

and from the power triangle:

The \(3I_{ph}^2R\) route is the most reliable — it needs no \(\sqrt3\), no power factor and no angle, only the phase current and the resistance.

Note the poor power factor. At 0.6 lagging the supply carries 48 kVA to deliver 28.8 kW — 67% more current than a unity-power-factor load of the same wattage would need. Problem 17 corrects it.

Phase voltage:

Per-phase impedance, the special case \(R = X\):

Equal resistance and reactance always give exactly 45°, hence \(\text{pf} = \cos45° = 0.7071\) and \(P = Q\).

Phase current, equal to line current:

Power per phase — using \(I^2R\), which needs no power factor:

A closed form worth noting. Substituting \(V_{ph} = V_L/\sqrt3\) and \(I_{ph} = V_{ph}/|\mathbf{Z}|\):

The three and the \(\sqrt3^2\) cancel exactly, leaving a formula in line voltage alone — convenient, and free of any \(\sqrt3\) to misplace.

The reactive power equals the real power here, since \(R = X\):

So the supply must carry 41% more apparent power than the load consumes — a direct consequence of the 0.707 power factor.

Phase voltage and impedance:

A 3-4-5 triangle scaled by two — the arithmetic is exact throughout.

Line current, equal to phase current in star:

Power factor from the impedance triangle:

Total power and apparent power:

Confirm by the closed form of Problem 8:

Both agreeing, and neither requiring a power factor or a \(\sqrt3\).

What if the same impedance were connected in delta? By Problem 6's equivalence, a delta of \((8+j6)\) behaves as a star of \((8+j6)/3\), so every current and power triples:

Which is Problem 12's result, previewed. Note that this is the same load impedance reconnected, not a different load.

Per-phase power and voltage:

Phase current:

Impedance magnitude:

Resolve into components using \(\cos\phi = 0.8\), \(\sin\phi = 0.6\):

Convert reactance to inductance at 50 Hz:

Check by an independent route. The resistance can be found directly from the power without ever computing the impedance:

and the reactive power gives the reactance:

A caution about the connection. The same 12 kW at 0.8 lagging in a delta would give \(R = 25.6\ \Omega\) and \(L = 61.1\ \text{mH}\) — exactly three times each, by the equivalence of Problem 6. The nameplate data alone cannot tell you which connection is inside the machine, so it must be stated.

Delta, so the phase voltage is the line voltage:

Per-phase impedance:

Another 3-4-5 triangle, this time scaled by three.

Currents:

Powers:

Check with the delta closed form. Adapting Problem 8's result — for delta the phase voltage is \(V_L\) rather than \(V_L/\sqrt3\), so a factor of three appears:

Agreeing to four figures, and computed without a power factor, a \(\sqrt3\) or a current.

The two closed forms side by side:

The factor of three between them is the whole content of Problem 12.

The power factor is the same in both cases, being a property of the impedance alone:

aStar connection:

bDelta connection:

The ratios:

Why exactly three, traced through the two \(\sqrt3\) factors:

Each phase sees \(\sqrt3\) times the voltage and therefore carries \(\sqrt3\) times the current, and power is their product.

The practical application — star–delta starting. An induction motor started in star draws only a third of the current it would draw in delta, and develops a third of the torque. Once running, it is switched to delta for full performance:

This limits the inrush that would otherwise dip the supply voltage for every other customer on the feeder — at the cost of reduced starting torque, so it suits only motors that start on light load.

The difficulty and its resolution. The line impedance is in the line conductors while the load is in delta — they cannot be added directly. Convert the delta to its star equivalent, and everything becomes a single series path per phase.

Build the per-phase circuit. Take one phase from the source neutral through the line to the load neutral:

This is legitimate because in a balanced system no current flows between the two neutrals, so they may be joined by an imaginary wire of zero impedance.

The line current:

The voltage at the load:

A drop from 400 V to 375 V — 6.2%, which would be unacceptable in practice and indicates an undersized cable.

Powers and efficiency, counting all three phases:

Note that the load's delta resistance is 30 Ω but the star-equivalent 10 Ω must be used with the line current — mixing the two is the standard error here.

Why the per-phase method works, stated carefully:

If either source or load is unbalanced the method fails, and Problems 14 to 16 must be used instead.

The four-wire system decouples. Because the neutral holds each load's return at the star point, every phase behaves as an independent single-phase circuit at \(V_{ph}\):

This is what makes the problem easy — and it is exactly what fails in Problem 15.

The three phase voltages:

Each load is resistive, so its current is in phase with its own phase voltage, with magnitude \(P/V_{ph}\):

The angle matches the voltage because the load is purely resistive. For a reactive load each current would be displaced by its own impedance angle, and the arithmetic below would be unchanged in form.

The neutral carries the phasor sum by KCL at the star point:

Resolving into rectangular components:

Exactly 30 A — the numbers were chosen to make it so.

A check worth doing. Had the three loads been equal, the currents would form a balanced set and

by the identity of Problem 2. The 30 A here measures the departure from balance, not the size of the load — note that it is smaller than any of the three phase currents.

The common error is to compute each current as \(P/\mathbf{V}\) with the voltage as a phasor in the denominator, which conjugates the angle and flips the sign of every imaginary part. Dividing a scalar power by a phasor is not a valid operation; the correct statement is \(\mathbf{I} = (S/\mathbf{V})^{*}\), and for a resistive load this reduces to a magnitude at the voltage's own angle.

Why this is harder. Without a neutral, the load's star point is not tied to the source's. It floats to whatever potential makes the three currents sum to zero — and that potential is generally not zero.

Millman's theorem (Set 14) gives the displacement directly:

This is simply KCL at the floating node, solved for its potential — the supernode method of Set 7 applied to a three-branch star.

The admittances:

Evaluating with \(V_{ph} = 230.9\ \text{V}\):

The load's star point sits 29 V away from the source neutral — a substantial displacement, and the entire consequence of removing the neutral wire.

The voltages actually across each load:

One load is over-volted by 11% and another under-volted by 10.5% — both well outside the ±6% that equipment is normally designed to tolerate.

Verify by KCL, which is the condition that defined \(\mathbf{V}_N\):

With no neutral there is nowhere for a residual current to go, so this must hold exactly — and it is the check that confirms the displacement was computed correctly.

Contrast the two systems:

Its function is to hold every phase voltage at \(V_L/\sqrt3\) regardless of imbalance. Problems 14 and 15 are the two cases side by side: with a neutral, imbalance produces a modest neutral current; without one, it produces voltage errors of over 10%.

Why it is never fused or switched. Consider what happens if the neutral opens while the phases stay connected:

The system converts instantly from Problem 14's case to Problem 15's. With a badly unbalanced load — say one heavily loaded phase and one lightly loaded — the displacement can be severe:

In the limit of one phase open-circuit and the other two loaded, the lightly loaded phase approaches the full line voltage. A broken neutral is one of the most destructive faults in low-voltage distribution, and it damages customers' equipment rather than the network's.

The harmonic problem. The balanced-current cancellation relies on the three currents being 120° apart. For the third harmonic and its multiples, the displacement is

so the three third-harmonic currents are in phase. They do not cancel — they add arithmetically in the neutral.

The consequence:

So a load whose current is 33% third harmonic — typical of switched-mode power supplies and LED drivers — gives a neutral current comparable to the phase current even when the fundamental is perfectly balanced:

The neutral carries nearly the full phase current while being sized — under older practice — at half the phase conductor's cross-section. Overheated neutrals in office buildings full of computers were a recognised problem before standards changed.

Why the third harmonic behaves this way is worth stating generally. Harmonic \(h\) has phase displacement \(120h\) degrees:

Multiples of three are called triplen harmonics for this reason, and they are the ones that matter for neutral sizing.

The reactive powers, exactly as in Set 23, Problem 11 but for the whole three-phase load:

Confirming \(Q_1 = 38.4\) kvar against Problem 7's direct calculation.

aDelta-connected capacitors. Each sees the full line voltage:

bStar-connected capacitors. Each sees only \(V_L/\sqrt3\):

The comparison:

Delta needs one third of the capacitance but at \(\sqrt3\) times the voltage. Since a capacitor's physical size and cost scale roughly with \(CV^2\) — the stored energy — the two arrangements are comparable in bulk, and the last row makes this precise.

Why delta is usually chosen in practice, despite the equal energy:

Star is preferred where the available capacitor voltage rating is limited, or in high-voltage banks where dividing the voltage is itself the objective.

The result:

A 37% reduction in current, and hence a 60% reduction in line losses — from a passive bank drawing no real power at all.

Fix what must be held equal: the power \(P\), the voltage \(V\) between conductors, the power factor, the distance \(\ell\), and the total \(I^2R\) loss. What is then compared is the total volume of copper.

Single-phase, two wires:

Three-phase, three wires:

The three-phase line current is smaller by \(\sqrt3\) — the single most important consequence of the whole arrangement.

Equate the losses:

Each three-phase conductor may have twice the resistance, because it carries less current and there are three of them rather than two.

Convert to cross-sectional area. Since \(R = \rho\ell/A\), doubling the resistance halves the area:

Compare the total volume of conductor:

Extending the comparison:

Which is why transmission — where every load is balanced three-phase machinery — uses three wires and no neutral, while distribution to individual premises accepts the fourth conductor's cost in exchange for the voltage stability of Problem 16.

The saving is larger than 25% in practice, because the smaller current also permits smaller insulators, lighter towers and lower-rated switchgear. The copper figure is a lower bound on the economic advantage, not the whole of it.

The general transformation from Set 2, for a delta of \(\mathbf{Z}_{ab}\), \(\mathbf{Z}_{bc}\), \(\mathbf{Z}_{ca}\) to an equivalent star:

The star arm at a node is the product of the two delta arms meeting there, divided by the sum of all three.

For a balanced delta, all three are \(\mathbf{Z}_\Delta\):

One line of algebra, and it holds for complex impedances exactly as for resistances — the transformation was derived from Kirchhoff's laws and linearity, so Set 22, Problem 1's argument applies.

Verify it independently by power. A delta on line voltage \(V_L\) draws

and a star of \(\mathbf{Z}_\Delta/3\) draws

Identical, confirming the equivalence is genuine and not merely a definition.

Where it is used:

A worked instance. A star load of \((6+j8)\ \Omega\) in parallel with a delta load of \((30+j15)\ \Omega\) per phase, on 400 V. Convert the delta:

Now combine per phase, as two impedances across the same \(V_{ph}\):

Using \(P = 3|\mathbf{V}|^2G\) — the conductance form of Set 23, Problem 15 — which avoids inverting the admittance at all.

When the general form is needed. Only when the load is unbalanced. Then the three star arms differ, each given by its own product-over-sum, and no shortcut applies. An unbalanced delta cannot be handled by the per-phase method in any case, so the transformation is usually a step towards a mesh or nodal analysis rather than a simplification in itself.

The gains, each established in this set:

The costs:

What is not new. Balanced three-phase analysis introduced no new circuit theory whatever. Problem 13's per-phase equivalent reduces every balanced problem to the single-phase case, so all of Sets 20 to 23 applies unchanged:

Only the unbalanced cases needed genuinely different treatment, and even those used Millman's theorem from Set 14.

The measurement problem. Set 23, Problem 17 established that a wattmeter reads the average of \(vi\) for one pair of terminals. To measure three-phase power the obvious approach uses three wattmeters, one per phase — but this needs access to the neutral, which a three-wire system does not have:

Why two suffice — the result Set 25 will prove. With no neutral, KCL forces

so one current is determined by the other two, and only two independent measurements are required. This is Blondel's theorem: an \(n\)-wire system needs \(n-1\) wattmeters. Problem 15's KCL check was the same constraint used for a different purpose.

The bonus that makes the method valuable. The two readings do not merely sum to the total power — their difference yields the reactive power and hence the power factor:

So two instruments give \(P\), \(Q\) and the power factor of a balanced load, with no phase-measuring equipment at all. One of them reads negative below 0.5 power factor, which is where the 30° shift of Problem 3 finally earns its keep.