Three-Phase Circuits

The advantages, for the same delivered power:

• Constant instantaneous power. The total power delivered to a balanced three-phase load does not fluctuate at all, whereas single-phase power pulsates at \(2\omega\) between zero and twice its average. A three-phase motor therefore produces steady torque; a single-phase motor vibrates. Section 12.7 proves this.

• Less conductor material. Three-phase transmission requires about 25 % less copper or aluminium than single-phase for the same power and the same losses — equivalently, single-phase needs 33 % more. Section 12.8 derives the figure.

• Better generation. A three-phase alternator has a higher power-to-weight ratio, so it is smaller, lighter and cheaper for a given output, and easier to transport and install.

• Self-starting motors. Three-phase currents in three windings produce a rotating magnetic field, so a three-phase induction motor starts by itself. A single-phase motor produces only a pulsating field and needs auxiliary starting arrangements — a capacitor, a shaded pole, or a starting winding — which add cost and reduce reliability.

• Better power factor in three-phase machines, and greater reliability under fault conditions, since a fault on one phase need not interrupt the others.

• Compatibility. A three-phase supply can feed a single-phase load simply by taking one phase and neutral. The converse is not possible.

Three-phase voltages are produced by an alternator whose rotor carries a magnet and whose stator carries three coils \(a\text{-}a'\), \(b\text{-}b'\) and \(c\text{-}c'\) placed physically 120° apart. As the rotor turns, its field cuts each coil in succession, inducing three voltages equal in magnitude but displaced in time by one third of a cycle.

Balanced phase voltages are equal in magnitude and 120° out of phase with one another. The two conditions together are what "balanced" means:

The second condition follows from the first plus the 120° spacing, and it is the fact that makes everything in this chapter work.

A balanced load is one whose phase impedances are equal in magnitude and in phase angle. For a Y-connected load and a Δ-connected load respectively:

The wye–delta transformation of Section 2.9, applied to three equal impedances, gives the conversion that will be used constantly:

Both the source and the load may be connected in either Y or Δ, giving four combinations: Y–Y, Y–Δ, Δ–Δ and Δ–Y. All four are analysed in the sections that follow, and all four reduce to the same per-phase calculation.

One practical note: a Δ-connected source is rare, because any small imbalance in the three generated voltages produces a large circulating current around the closed delta loop, with nothing to limit it but the winding impedance. Sources are almost always Y-connected. Loads are commonly either.

Take a Y-connected source feeding a Y-connected load, with the neutrals joined. This is the fundamental case; the other three reduce to it.

Line voltages. The phase voltages \(\mathbf{V}_{an}\), \(\mathbf{V}_{bn}\), \(\mathbf{V}_{cn}\) are measured from each line to the neutral. The line voltages are measured between lines, and follow by subtraction:

and similarly

Two facts, both essential: the line voltage magnitude is \(\sqrt{3}\) times the phase voltage, and the line voltages lead their corresponding phase voltages by 30°. The factor \(\sqrt{3} = 1.732\) is why a 400 V three-phase supply gives 230 V between any line and neutral, and both figures appear on the same distribution board.

Line currents. Each phase of the load sees its own phase voltage, so

In a Y connection the line current is the phase current, since the same conductor carries both:

The neutral carries nothing. Summing the three line currents:

The two neutral points are at the same potential regardless of the neutral wire's impedance, so the neutral wire can be removed entirely without changing anything. This is why three-phase transmission uses three conductors rather than four, and it is the single largest source of the material saving computed in Section 12.8. Note carefully that this holds only for a balanced load; Section 12.9 shows what happens otherwise.

Now a Y-connected source feeds a Δ-connected load. The load has no neutral point, so each phase of the load sees a line voltage:

The phase currents in the load follow directly:

Line currents. In a delta the line current is not the phase current; each line carries the difference of two phase currents. Applying KCL at node \(A\):

So the roles are exactly reversed from the Y connection. There, \(V_{L} = \sqrt{3}V_{p}\) and \(I_{L} = I_{p}\); here, \(V_{L} = V_{p}\) and \(I_{L} = \sqrt{3}I_{p}\). And where the line voltage led by 30° in the Y case, the line current lags by 30° in the Δ case.

This table is the single most useful thing to memorise in the chapter. A reliable mnemonic: the \(\sqrt{3}\) always goes with the quantity that is not shared. In a Y the three phases share the line conductors, so the currents match and the voltages differ; in a Δ the phases share the line terminals, so the voltages match and the currents differ.

With both source and load in delta, the source phase voltages are the line voltages, and they are applied directly across the load phases:

and the line currents follow by KCL at each corner:

giving \(I_{L} = \sqrt{3}I_{p}\) as before, with each line current lagging its corresponding phase current by 30°.

Here is the result that justifies the whole three-phase enterprise.

Each individual phase delivers a pulsating power, exactly as in Chapter 11: a constant part plus a term at \(2\omega\). But the three \(2\omega\) terms are themselves displaced by 240° (twice 120°), so they sum to zero. Adding the three instantaneous powers leaves only the constant parts:

The total instantaneous power in a balanced three-phase system does not vary with time at all, even though each phase's contribution pulsates. The result holds whether the load is Y- or Δ-connected.

The practical consequence is enormous. A three-phase motor receives energy at a perfectly steady rate and therefore produces steady torque; a single-phase motor of the same rating receives power that swings between zero and twice its average a hundred times a second, and vibrates accordingly. This is the principal reason three-phase is used for generation and distribution.

The economic case for three-phase transmission can be made precisely. Compare two systems delivering the same power \(P_{L}\) at the same line voltage \(V_{L}\) over the same distance, and require them to suffer the same losses.

Single-phase, two conductors:

Three-phase, three conductors: the line current is smaller by \(\sqrt{3}\) since \(P_{L} = \sqrt{3}V_{L}I'_{L}\), so

Requiring the losses to be equal gives

Since \(R = \rho\ell/A\), a resistance twice as large means a cross-section half as large: \(A' = A/2\). Now count the material, which is cross-section times length times number of conductors:

Single-phase transmission requires 33 % more conductor material than three-phase for the same power, voltage, distance and losses — equivalently, three-phase saves 25 % of the material. On a national transmission network that difference is measured in hundreds of thousands of tonnes of aluminium.

Two points about the derivation. It assumes the neutral can be dispensed with, which Section 12.4 established for balanced loads — a four-wire system would not show the same saving. And it holds the line voltage constant, which is the fair comparison because that is what insulation and clearances are designed for.

An unbalanced system is one in which the load impedances are not all equal. Real distribution systems are always somewhat unbalanced, because single-phase loads are connected between individual lines and neutral and never divide perfectly among the three.

Everything convenient about the balanced case now fails:

• The three line currents differ in both magnitude and angle, so they no longer sum to zero.

• The neutral carries current, and its value must be computed:

  \[\mathbf{I}_{n} = -\left(\mathbf{I}_{a}+\mathbf{I}_{b}+\mathbf{I}_{c}\right)\]

• Per-phase analysis is invalid — there is no rotation that maps one phase onto another.

• The formula \(P = \sqrt{3}V_{L}I_{L}\cos\theta\) does not apply. Each phase must be computed separately and the complex powers added.

The remedy is to fall back on the general methods of Chapter 10: mesh or nodal analysis on the complete three-phase circuit. There is nothing special about it — it is simply a larger AC circuit, and the phasor machinery handles it.

Why this matters in practice. The neutral conductor of a four-wire distribution system must be sized for the worst expected imbalance. Worse, a broken neutral on an unbalanced system allows the load's star point to shift away from earth potential, so that lightly loaded phases see voltages far above their rating — a well-known cause of appliance destruction across a whole building.

Measuring three-phase power might seem to need three wattmeters, one per phase. It does not. Two wattmeters suffice for any three-wire system, balanced or unbalanced, Y- or Δ-connected — and they yield the power factor as a bonus.

Connection. Each wattmeter's current coil carries one line current. Its voltage coil is connected between that line and the third line — the one carrying neither current coil. The \(\pm\) terminal of each voltage coil goes to the line whose current the corresponding current coil carries.

The word algebraic is important: one reading can be negative, and it must then be subtracted. Note also that the individual readings no longer correspond to the power in any particular phase — only their sum has meaning.

The chain from generator to socket makes the theory concrete. A power station generates at perhaps 11 kV three-phase, which is transformed up to 220 kV or 400 kV for transmission — high voltage means low current for the same power, and hence low \(I^{2}R\) losses. It is stepped down in stages to a local distribution voltage, and finally to the four-wire system that reaches consumers.

In India that final system is 415 V between lines and 240 V between any line and neutral, the two figures related by \(\sqrt{3}\): \(415/\sqrt{3} = 239.6~\mathrm{V}\). In Europe the corresponding pair is 400 V and 230 V, and in North America 208 V and 120 V. Every one of these pairs is a \(\sqrt{3}\) ratio, and recognising it is a quick way to confirm that a system is three-phase four-wire.

A domestic consumer receives one phase and the neutral — hence "single-phase supply" — while an industrial consumer takes all three. Domestic loads are distributed among the three phases of the street feeder so that the system remains approximately balanced overall, which is why your neighbours may be on a different phase from you.

Loads that must be balanced. A three-phase motor is inherently balanced. Large single-phase loads — electric arc furnaces, railway traction, induction heaters — are not, and connecting one to a three-phase network causes voltage imbalance that other consumers experience as flicker. Utilities therefore impose limits on single-phase load size and require large ones to be connected through balancing arrangements.

• Three-phase systems deliver constant instantaneous power, save 25 % of conductor material, and permit self-starting motors.

• Balanced voltages are equal in magnitude, 120° apart, and sum to zero. The sequence is abc (positive) or acb (negative).

• For a Y connection: \(V_{L} = \sqrt{3}V_{p}\) with the line voltage leading by 30°, and \(I_{L} = I_{p}\).

• For a Δ connection: \(V_{L} = V_{p}\), and \(I_{L} = \sqrt{3}I_{p}\) with the line current lagging by 30°.

• For balanced loads \(\mathbf{Z}_{\Delta} = 3\mathbf{Z}_{Y}\). Convert everything to Y and analyse per-phase.

• In a balanced Y–Y system the neutral current is zero and the neutral wire may be omitted.

• Total power is \(P = \sqrt{3}V_{L}I_{L}\cos\theta\) for either connection, with \(\theta\) the load impedance angle.

• The same impedances draw three times the power in Δ as in Y — the basis of the star–delta starter.

• Unbalanced systems require mesh or nodal analysis; the neutral carries current and the \(\sqrt{3}\) power formula fails.

• Two wattmeters measure any three-wire system, giving \(P_{T} = P_{1}+P_{2}\), \(Q_{T} = \sqrt{3}\left(P_{2}-P_{1}\right)\) and the power factor.

• Putting the \(\sqrt{3}\) in the wrong place. In Y it belongs to the voltage; in Δ to the current. Getting this backwards is the commonest error in the chapter.

• Forgetting the 30° angle. Line and phase quantities differ in angle as well as magnitude. A magnitude-only answer is incomplete.

• Using the line-to-line angle as \(\theta\) in the power formula. The \(\theta\) in \(P = \sqrt{3}V_{L}I_{L}\cos\theta\) is the load impedance angle, not the angle between \(\mathbf{V}_{L}\) and \(\mathbf{I}_{L}\).

• Writing \(\mathbf{Z}_{Y} = 3\mathbf{Z}_{\Delta}\). It is the other way round: the delta impedance is the larger, \(\mathbf{Z}_{\Delta} = 3\mathbf{Z}_{Y}\).

• Assuming the neutral current is zero in an unbalanced system. That result holds only for balanced loads, and the neutral conductor must be sized for the imbalance.

• Applying \(P = \sqrt{3}V_{L}I_{L}\cos\theta\) to an unbalanced load. There is no single \(I_{L}\) or \(\theta\). Compute each phase separately and add the complex powers.

• Adding two wattmeter readings when one is negative. The sum is algebraic; a negative reading must be subtracted.

• Believing each wattmeter reads one phase's power. Neither does. Example 12.7 shows readings of 653 W and 1651 W where every phase carries 768 W.

• Using the two-wattmeter method on a four-wire system with a non-zero neutral current. It is valid for three-wire systems only.

• Confusing 240 V with 415 V. One is line-to-neutral, the other line-to-line, and they differ by \(\sqrt{3}\). Check which the problem means before computing anything.

Work these before opening the answers. Assume the abc sequence throughout and state whether each power factor is leading or lagging.

1. P12.1 A balanced Y-connected source has \(V_{L} = 415~\mathrm{V}\). Find the phase voltage, and write all three phase voltages taking \(\mathbf{V}_{an}\) as reference.

   Show answer

   \[V_{p} = \frac{415}{\sqrt{3}} = 239.6~\mathrm{V}\]

   \[\mathbf{V}_{an} = 239.6\angle 0^{\circ}, \quad \mathbf{V}_{bn} = 239.6\angle-120^{\circ}, \quad \mathbf{V}_{cn} = 239.6\angle 120^{\circ}~\mathrm{V}\]

   This is the standard Indian domestic and industrial supply.

2. P12.2 A balanced Y load of \(\left(9+j12\right)~\Omega\) per phase is fed from \(\mathbf{V}_{an} = 220\angle 0^{\circ}~\mathrm{V}\). Find the three line currents.

   Show answer

   \[\mathbf{Z}_{Y} = 9+j12 = 15\angle 53.13^{\circ}~\Omega\]

   \[\mathbf{I}_{a} = \frac{220\angle 0^{\circ}}{15\angle 53.13^{\circ}} = 14.67\angle-53.13^{\circ}~\mathrm{A}\]

   \[\mathbf{I}_{b} = 14.67\angle-173.1^{\circ}~\mathrm{A}, \qquad \mathbf{I}_{c} = 14.67\angle 66.87^{\circ}~\mathrm{A}\]

3. P12.3 Find the total \(P\), \(Q\) and \(S\) for the load of P12.2.

   Show answer

   \[P = 3\left(220\right)\left(14.67\right)\left(0.6\right) = 5808~\mathrm{W}\]

   \[Q = 3\left(220\right)\left(14.67\right)\left(0.8\right) = 7744~\mathrm{VAR}, \qquad S = 3\left(220\right)\left(14.67\right) = 9680~\mathrm{VA}\]

   Check: \(\sqrt{5808^{2}+7744^{2}} = 9680\) VA \(\checkmark\), and the pf is 0.6 lagging.

4. P12.4 A Δ load of \(\left(30+j40\right)~\Omega\) per phase is connected to a 400 V three-phase supply. Find the phase current, the line current, and the total power.

   Show answer

   \[\left|\mathbf{Z}_{\Delta}\right| = 50~\Omega, \qquad \theta = 53.13^{\circ}, \qquad I_{p} = \frac{400}{50} = 8~\mathrm{A}\]

   \[I_{L} = \sqrt{3}\left(8\right) = 13.86~\mathrm{A}\]

   \[P = \sqrt{3}\left(400\right)\left(13.86\right)\left(0.6\right) = 5760~\mathrm{W}\]

5. P12.5 The same impedances of P12.4 are reconnected in star across the same supply. Find the new total power.

   Show answer

   \[P_{Y} = \frac{P_{\Delta}}{3} = \frac{5760}{3} = 1920~\mathrm{W}\]

   Confirming directly: \(V_{p} = 400/\sqrt{3} = 230.9~\mathrm{V}\), \(I_{p} = 230.9/50 = 4.619~\mathrm{A}\), so \(P = 3(230.9)(4.619)(0.6) = 1920~\mathrm{W}\) \(\checkmark\)

6. P12.6 Convert a balanced Δ load of \(\left(36+j45\right)~\Omega\) per phase to its Y equivalent, and state the line current if \(V_{p} = 240~\mathrm{V}\).

   Show answer

   \[\mathbf{Z}_{Y} = \frac{36+j45}{3} = \left(12+j15\right)~\Omega = 19.21\angle 51.34^{\circ}~\Omega\]

   \[I_{L} = \frac{240}{19.21} = 12.49~\mathrm{A}\]

7. P12.7 Two wattmeters read \(P_{1} = 1200~\mathrm{W}\) and \(P_{2} = 2400~\mathrm{W}\). Find the total power, the reactive power, and the power factor.

   Show answer

   \[P_{T} = 1200+2400 = 3600~\mathrm{W}, \qquad Q_{T} = \sqrt{3}\left(1200\right) = 2078~\mathrm{VAR}\]

   \[\tan\theta = \sqrt{3}\frac{1200}{3600} = 0.5774 \;\Longrightarrow\; \theta = 30.00^{\circ}, \qquad \mathrm{pf} = 0.8660\ \text{lagging}\]

   Lagging because \(P_{2} > P_{1}\).

8. P12.8 Two wattmeters read \(P_{1} = -500~\mathrm{W}\) and \(P_{2} = 2500~\mathrm{W}\). Find the total power and power factor, and comment.

   Show answer

   \[P_{T} = -500+2500 = 2000~\mathrm{W}, \qquad Q_{T} = \sqrt{3}\left(3000\right) = 5196~\mathrm{VAR}\]

   \[\tan\theta = \sqrt{3}\frac{3000}{2000} = 2.598 \;\Longrightarrow\; \theta = 68.95^{\circ}, \qquad \mathrm{pf} = 0.3592\ \text{lagging}\]

   The negative reading signals a power factor below 0.5. In the laboratory the meter would try to deflect backwards; reverse its current-coil connections, read it as positive, and subtract.

9. P12.9 An unbalanced Y load with a neutral has line currents \(\mathbf{I}_{a} = 10\angle 0^{\circ}\), \(\mathbf{I}_{b} = 12\angle-120^{\circ}\) and \(\mathbf{I}_{c} = 8\angle 120^{\circ}~\mathrm{A}\). Find the neutral current.

   Show answer

   Converting to rectangular: \(10\), \(-6-j10.39\), \(-4+j6.928\). Summing:

   \[\mathbf{I}_{a}+\mathbf{I}_{b}+\mathbf{I}_{c} = 0 - j3.464~\mathrm{A}\]

   \[\mathbf{I}_{n} = -\left(0-j3.464\right) = 3.464\angle 90^{\circ}~\mathrm{A}\]

   Non-zero, as expected for an unbalanced load.

10. P12.10 A 415 V three-phase supply feeds a balanced load drawing 15 kW at 0.85 pf lagging. Find the line current and the reactive power.

    Show answer

    \[I_{L} = \frac{P}{\sqrt{3}V_{L}\cos\theta} = \frac{15\,000}{\sqrt{3}\left(415\right)\left(0.85\right)} = 24.55~\mathrm{A}\]

    \[\theta = \cos^{-1}0.85 = 31.79^{\circ}, \qquad Q = P\tan\theta = 15\,000\left(0.6197\right) = 9296~\mathrm{VAR}\]