ECE 211 · Lecture 16: Balanced Three-Phase Power

§1 · The per-phase circuit

One phase carries the whole story

A balanced three-phase system is solved one phase at a time. The three sources are equal in magnitude and separated by 120°, the three load branches are identical, and the neutral therefore carries no current. Phase a alone may be drawn as an ordinary single-loop circuit, in which the neutral connection serves as the return path:

The single-phase Y-Y equivalent circuit. Lowercase subscripts denote the source terminals and uppercase subscripts denote the load terminals, so $\tilde V_{an}$ is a source phase voltage and $\tilde V_{AN}$ is a load phase voltage. The line impedance carries the difference between them.

Two conversions are performed before the circuit is drawn. A Δ-connected load is replaced by its Y equivalent, and a line voltage is replaced by the phase voltage that produces it:

\tm{pp.zy}{Z_Y} = \frac{Z_\Delta}{3} \qquad \tm{pp.van}{\tilde V_{an}} = \frac{\tilde V_{ab}}{\sqrt{3}\,\angle 30^\circ} \qquad \tm{pp.vload}{\tilde V_{AN}} = \frac{\tilde V_{AB}}{\sqrt{3}\,\angle 30^\circ}

The second and third relations are read off the phasor diagram. The line voltage is the difference of two phase voltages, $\tilde V_{AB} = \tilde V_{AN} - \tilde V_{BN}$ , and for a balanced positive sequence that difference is larger by $\sqrt{3}$ and rotated forward by 30°:

Adding $-\tilde V_{BN}$ to $\tilde V_{AN}$ closes the triangle on $\tilde V_{AB}$ . The three phase voltages are equal in magnitude and 120° apart, so the resulting line voltage leads its phase voltage by 30° and exceeds it by the factor $\sqrt{3}$ . Dividing by $\sqrt{3}\angle 30^\circ$ reverses the construction.

Sequence

A positive (abc) sequence is assumed throughout, meaning that $\tilde V_{BN}$ lags $\tilde V_{AN}$ by 120° and $\tilde V_{CN}$ leads it by 120°. For a negative sequence the magnitude factor $\sqrt{3}$ is unchanged and the 30° rotation reverses sign.

§2 · Line and phase quantities

The connection changes the split, not the product

The line quantities are the same for both connections, because the two loads are attached to the same three wires. What differs is how each connection divides the line voltage and the line current among its three branches:

Y: each branch spans a phase voltage

Δ: each branch spans a line voltage

Each phase impedance is placed between a line and the neutral, so it stands across a phase voltage and carries the full line current:

|\tilde V_{1\phi}| = |\tilde V_{AN}| = \frac{|\tilde V_{AB}|}{\sqrt{3}} \qquad\qquad |\tilde I_{1\phi}| = |\tilde I_a|

Each phase impedance is placed between two lines, so it stands across a line voltage, and the line current divides between the two branches that meet at each terminal:

|\tilde V_{1\phi}| = |\tilde V_{AB}| \qquad\qquad |\tilde I_{1\phi}| = |\tilde I_{AB}| = \frac{|\tilde I_a|}{\sqrt{3}}

The product of the two per-phase magnitudes is identical in the two rows, which is the reason a load may be analyzed without knowing how it is connected:

Per-phase quantity	Y connection	Δ connection
Phase voltage \|V_1φ\|	\|V_AB\| / √3	\|V_AB\|
Phase current \|I_1φ\|	\|I_a\|	\|I_a\| / √3
Apparent power \|S_1φ\|	\|V_AB\| \|I_a\| / √3	\|V_AB\| \|I_a\| / √3
Three-phase total	√3 \|V_AB\| \|I_a\|	√3 \|V_AB\| \|I_a\|

|S_1φ| = |Ṽ_1φ| · |Ĩ_1φ| |S_3φ| = √3 · |Ṽ_AB| · |Ĩ_a|

§3 · Power in a balanced load

Three routes to the same number

The total power of a balanced load is three times the power of one phase. Which of the three expressions is used depends only on which data are available:

P_{3\phi} = 3 P_{1\phi} = 3\,\mathrm{Re}\{\tm{pp.vload}{\tilde V_{AN}}\,\tm{pp.ia}{\tilde I_a^{*}}\}

P_{3\phi} = 3\,\mathrm{Re}\{|\tm{pp.ia}{\tilde I_a}|^2\,\tm{pp.zy}{Z_Y}\} \qquad\qquad P_{3\phi} = 3\,\mathrm{Re}\left\{\frac{|\tilde V_{AB}|^2}{Z_\Delta^{*}}\right\}

The first form uses the per-phase voltage and current of a Y load. The second uses the line current, which is also the phase current of a Y load. The third is written entirely in Δ quantities: the branch voltage of a Δ load is the line voltage, so no conversion of the given data is required at all.

The power factor is the cosine of the impedance angle, and the impedance whose angle is taken depends on where the power factor is measured:

\text{load pf} = \cos\theta_{\tm{pp.zy}{Z_Y}} = \cos\theta_{Z_\Delta} \qquad\qquad \text{source pf} = \cos\theta_{\tm{pp.zy}{Z_Y} + \tm{pp.zl}{Z_L}}

The Y-to-Δ transformation scales an impedance by a real factor of three and therefore leaves its angle unchanged, which is why the load power factor may be evaluated in either connection. The source sees the line impedance in series with the load, so its power factor differs from that of the load.

Lagging and leading

An inductive load draws a lagging current, so $\theta$ is positive and the reactive power $Q$ is positive. A capacitive load draws a leading current, so $\theta$ is negative and $Q$ is negative. The power factor alone is ambiguous, since $\cos(+\theta) = \cos(-\theta)$ , and the word lagging or leading must therefore accompany every quoted value.

The power triangle

The three-phase apparent power is $|S_{3\phi}| = \sqrt{3}\,|\tilde V_{AB}|\,|\tilde I_a|$ regardless of the connection, and the load angle resolves it into real and reactive parts. The sliders below drive the triangle:

The horizontal leg is the real power, the vertical leg is the reactive power, and the hypotenuse is the apparent power. The default values are those of Example A.

Line voltage |Ṽ_AB| (V rms) 208

Line current |Ĩ_a| (A rms) 73.8

Load angle θ (degrees) 31.5

|S_3φ| = √3 · 208 V · 73.8 A = 26.59 kVA

P_3φ = 22.67 kW Q_3φ = 13.89 kvar pf = cos θ = 0.853

§4 · When the per-phase circuit is required

Draw it only when the two ends must be connected

The Y-Y circuit is required when load information is given and source information is requested, or the reverse. The two ends differ by the drop across the line impedance, and only the circuit relates them.
If a load voltage or a load current is given together with the load impedance, the load power follows without the circuit. Everything needed already refers to the same terminals, so one of the three power expressions of §3 applies directly.
Only the magnitudes of the line voltage or the line current are usually given, and no generality is lost. Every magnitude and every power in a balanced system depends on impedance angles, never on the arbitrary reference angle of the source. When a phasor answer is required, a reference is chosen, conventionally $\tilde V_{AN} = |\tilde V_{AN}|\angle 0^\circ$ .

A consequence worth remembering

In a per-phase circuit the source, the line, and the load form a single series path, so $\tilde V_{an} = \tilde I_a (Z_L + Z_Y)$ and therefore $|\tilde V_{an}| = |\tilde I_a|\,|Z_L + Z_Y|$ . A source voltage magnitude can be obtained from a load voltage magnitude by multiplying magnitudes; no phasor addition is needed. Phasor addition becomes necessary only when two branches with different angles are combined, as in Example B.

§5 · Example A: a power factor from meter data

Find the power factor of the load

The total power of a three-phase load is measured as 22,659 W, the line voltage at the load is 208 V rms, and the line current is 73.8 A rms. The power factor of the load is requested. All three data refer to the load terminals, so the per-phase circuit is not needed.

The data

P_{3\phi} = 22{,}659\ \text{W} \qquad |\tilde V_{AB}| = 208\ \text{V rms} \qquad |\tilde I_a| = 73.8\ \text{A rms}

The connection of the load is not stated, and it will turn out not to matter.

The real power of one phase

P_{1\phi} = \frac{P_{3\phi}}{3} = \frac{22{,}659}{3} = \tm{xa.p1}{7553\ \text{W}}

This is the horizontal leg of the per-phase power triangle.

The apparent power of one phase

|S_{1\phi}| = |\tilde V_{1\phi}|\,|\tilde I_{1\phi}^{*}| = \frac{|\tilde V_{AB}|\,|\tilde I_a|}{\sqrt{3}} = \frac{(208)(73.8)}{\sqrt{3}} = \tm{xa.s1}{8863\ \text{VA}}

The connection may be assumed freely, as the tabs below confirm. The hypotenuse of the triangle is now fixed, and the reactive leg follows as $\tm{xa.q1}{Q_{1\phi} = 4636\ \text{var}}$ .

The angle between them

\theta_s = \arccos\!\left(\frac{P_{1\phi}}{|S_{1\phi}|}\right) = 31.5^\circ

The angle of the complex power equals the angle of the load impedance, which is what the power factor reports.

The power factor

\text{pf} = \cos\theta_s = \frac{P_{1\phi}}{|S_{1\phi}|} = \frac{7553}{8863} = \boxed{0.852}

The sign of the angle is not determined by these data. A load of this kind is almost always inductive, so the value is quoted as 0.852 lagging.

Check with the three-phase form

\sqrt{3}\,|\tilde V_{AB}|\,|\tilde I_a|\,\text{pf} = \sqrt{3}\,(208)(73.8)(0.852) = 22{,}652\ \text{W} \approx P_{3\phi}\ \checkmark

The residual difference is rounding in the quoted power factor. Working the same problem in one line, $\text{pf} = P_{3\phi} / (\sqrt{3}\,|\tilde V_{AB}|\,|\tilde I_a|)$ , is the fastest route once the identity of §2 is trusted.

The phase voltage is the line voltage divided by $\sqrt{3}$ , and the phase current is the line current:

|S_{1\phi}| = \frac{208}{\sqrt{3}} \cdot 73.8 = (120.1)(73.8) = 8863\ \text{VA}

The phase voltage is the line voltage, and the phase current is the line current divided by $\sqrt{3}$ :

|S_{1\phi}| = 208 \cdot \frac{73.8}{\sqrt{3}} = (208)(42.61) = 8863\ \text{VA}

The two assumptions place the factor $\sqrt{3}$ on different quantities and return the same apparent power, exactly as the third row of the table in §2 predicts.

§6 · Example B: two loads in parallel

Find the total power and the power factor

Two balanced three-phase loads are supplied by the same line. Load 1 is Δ-connected with an impedance of $24 + j18\ \Omega$ per phase, and load 2 is Y-connected with an impedance of $6 + j4\ \Omega$ per phase. The line voltage at the load is 208 V rms.

The two loads share a line voltage

Z_{\Delta 1} = 24 + j18\ \Omega \qquad Z_Y = 6 + j4\ \Omega \qquad |\tilde V_{AB}| = 208\ \text{V rms}

Both loads are attached to the same three wires, so they are in parallel. Loads may be combined only after they are expressed in the same connection. Each Δ is drawn flat in the figure, as one impedance between each pair of lines, with a jump marking the place where the a-c branch crosses line b without touching it.

Transform the Y load into a Δ

Z_{\Delta 2} = 3 Z_Y = 3(6 + j4) = \tm{xb.zd2}{18 + j12\ \Omega}

The Δ form is chosen because the branch voltage of a Δ is the line voltage, which is the given datum.

Combine the two Δ loads

Z_\Delta = \frac{(24 + j18)(18 + j12)}{(24 + j18) + (18 + j12)} = \frac{216 + j612}{42 + j30} = \tm{xb.zdt}{10.30 + j7.22\ \Omega}

Corresponding branches of the two Δ loads join the same pair of lines, so they combine by the product-over-sum rule of any two parallel impedances.

The total power

P_{3\phi} = 3\,\mathrm{Re}\left\{\frac{|\tilde V_{AB}|^2}{Z_\Delta^{*}}\right\} = 3 \cdot \frac{(208)^2 (10.297)}{(10.297)^2 + (7.216)^2} = \boxed{8453\ \text{W}}

Multiplying numerator and denominator by $Z_\Delta$ turns the reciprocal into $Z_\Delta / |Z_\Delta|^2$ , whose real part is $R / |Z_\Delta|^2$ .

The power factor of the combination

\text{pf} = \cos\theta_{Z_\Delta} = \cos\left[\arctan\!\left(\frac{7.216}{10.297}\right)\right] = \cos 35.0^\circ = \boxed{0.819\ \text{lagging}}

Both loads are inductive, so the combination is inductive and the power factor lags.

Check by adding the two powers

Power is additive, so the two loads may also be evaluated separately, load 2 in its original Y form:

P_1 = 3\,\frac{(208)^2 (24)}{24^2 + 18^2} = 3461\ \text{W} \qquad P_2 = 3\,\frac{(208/\sqrt{3})^2 (6)}{6^2 + 4^2} = 4992\ \text{W}

P_1 + P_2 = 8453\ \text{W}\ \checkmark

The agreement confirms both the transformation and the parallel combination, and it uses the Y form of load 2, so the check is independent of the step that transformed it.

A rounding trap

Rounding the equivalent impedance to $10.32 + j7.23\ \Omega$ before computing the power, as the board notes do, yields 8438 W. The exact combination is $Z_\Delta = (381 + j267)/37 = 10.2973 + j7.2162\ \Omega$ , which gives 8453 W and agrees with the independent check above. Intermediate quantities should be carried at full precision and rounded once, at the end.

§7 · Example C: a Y-Δ system with a line impedance

Five questions about one circuit

A balanced three-phase Y-Δ system has a line voltage of 90 V rms at the source, a line impedance of $2\ \Omega$ , and a load impedance of $60 + j20\ \Omega$ per Δ branch. Here source data are given and load data are requested, so the per-phase circuit is required.

(a) Draw the single-phase Y-Y circuit

\tilde V_{an} = \frac{\tilde V_{ab}}{\sqrt{3}\angle 30^\circ} \;\Rightarrow\; |\tilde V_{an}| = \frac{90}{\sqrt{3}} = 51.96\ \text{V} \qquad Z_Y = \frac{Z_\Delta}{3} = \frac{60 + j20}{3} = 20 + j6.67\ \Omega

The load is converted to Y and the source to a phase voltage. The line impedance is already a per-phase quantity and is carried over unchanged.

(b) The line current

|\tilde I_a| = \left|\frac{90/\sqrt{3}}{2 + 20 + j6.67}\right| = \frac{51.96}{\sqrt{22^2 + 6.67^2}} = \frac{51.96}{22.99} = \boxed{2.26\ \text{A rms}}

One loop with one impedance in series with the source, exactly as in a single-phase circuit.

(c) The phase current at the load

|\tilde I_{AB}| = \frac{|\tilde I_a|}{\sqrt{3}} = \frac{2.26}{\sqrt{3}} = \boxed{1.30\ \text{A rms}}

The load is physically Δ-connected, so its branch current is smaller than the line current by $\sqrt{3}$ . The Y equivalent exists only on paper and carries the line current instead.

(d) The line voltage at the load

|\tilde V_{AN}| = |\tilde I_a|\,|Z_Y| = 2.26\sqrt{20^2 + 6.67^2} = 47.65\ \text{V}

|\tilde V_{AB}| = \sqrt{3}\,|\tilde V_{AN}| = \boxed{82.5\ \text{V rms}}

The load receives 82.5 V of the 90 V available at the source; the remainder is dropped across the line.

(e) The power factor at the load

\text{pf} = \cos\theta_{Z_\Delta} = \cos\left[\arctan\!\left(\frac{20}{60}\right)\right] = \cos 18.4^\circ = \boxed{0.95\ \text{lagging}}

The angle of $Z_Y$ is the same as the angle of $Z_\Delta$ , so either impedance answers the question.

Check, and one observation

The load line voltage may be recovered from the Δ branch directly, without the Y equivalent:

|\tilde V_{AB}| = |\tilde I_{AB}|\,|Z_\Delta| = 1.30\sqrt{60^2 + 20^2} = (1.30)(63.25) = 82.5\ \text{V}\ \checkmark

The source power factor is $\cos[\arctan(6.67/22)] = 0.957$ , slightly higher than the 0.95 seen by the load: the line contributes resistance but no reactance, which pulls the total impedance angle toward zero.

§8 · Example D: line loss and the source voltage

Find the line loss and the line voltage at the source

The line voltage at the load is 100 V rms, the line impedance is $10\ \Omega$ , and the Y-connected load impedance is $40 + j15.1\ \Omega$ . Load data are given and source data are requested, so the per-phase circuit is again required.

The data

|\tilde V_{AB}| = 100\ \text{V rms at the load} \qquad Z_L = 10\ \Omega \qquad Z_Y = 40 + j15.1\ \Omega

The line loss and the source line voltage are requested, and both lie on the source side of the line impedance.

The load phase voltage

|\tilde V_{AN}| = \frac{|\tilde V_{AB}|}{\sqrt{3}} = \frac{100}{\sqrt{3}} = 57.74\ \text{V rms}

The load is already Y-connected, so only the voltage requires conversion.

The line current

|\tilde I_a| = \frac{|\tilde V_{AN}|}{|Z_Y|} = \frac{57.74}{\sqrt{40^2 + 15.1^2}} = \frac{57.74}{42.76} = 1.350\ \text{A rms}

The line impedance plays no part here, because the load voltage and the load impedance are both known at the same terminals.

The three-phase line loss

P_{1\phi,\text{line}} = |\tilde I_a|^2 R_L = (1.350)^2 (10) = 18.2\ \text{W}

P_{3\phi,\text{line}} = 3 (1.350)^2 (10) = \boxed{54.7\ \text{W}}

All three lines carry the same current magnitude, so the total loss is three times the loss of one line.

The line voltage at the source

|\tilde V_{an}| = |\tilde I_a|\,|Z_L + Z_Y| = 1.350\sqrt{50^2 + 15.1^2} = (1.350)(52.23) = 70.51\ \text{V}

|\tilde V_{ab}| = \sqrt{3}\,|\tilde V_{an}| = \boxed{122\ \text{V rms}}

The source and the load lie on one series path, so the magnitudes multiply and no phasor addition is required.

What the numbers mean

P_{3\phi,\text{load}} = 3 (1.350)^2 (40) = 218.7\ \text{W} \qquad \frac{P_{\text{line}}}{P_{\text{line}} + P_{\text{load}}} = \frac{54.7}{273.4} = 20\%

A fifth of the generated power is dissipated in the line, and the source must be raised 22% above the load line voltage to deliver it. This is precisely why transmission is performed at high voltage: for a given power, raising the line voltage lowers the line current, and the loss falls with the square of that current.

§9 · Your turn

Practice: a rated load behind a line impedance

A balanced three-phase Δ-connected load absorbs 15 kW at a power factor of 0.8 lagging. The line voltage at the load is 480 V rms, and each line has an impedance of $0.3 + j0.9\ \Omega$ . The problem should be attempted on paper before the hints are opened, in order.

Find (a) the magnitude of the line current, (b) the per-phase impedance of the Δ load, (c) the three-phase line loss, and (d) the line voltage at the source.

Hint 1: Start from the apparent power

The load power and the load power factor give the apparent power, and the identity of §2 converts it into a line current: $|S_{3\phi}| = P_{3\phi} / \text{pf}$ and $|\tilde I_a| = |S_{3\phi}| / (\sqrt{3}\,|\tilde V_{AB}|)$ .

Hint 2: The impedance follows from one branch

A Δ branch stands across the line voltage and carries $|\tilde I_a| / \sqrt{3}$ , so its magnitude is the ratio of the two. Its angle is $\arccos(0.8)$ , positive because the power factor lags.

Hint 3: The line

The line loss uses the line current in all three lines. The source voltage uses the series path of the per-phase circuit, with $Z_Y = Z_\Delta / 3$ in series with $Z_L = 0.3 + j0.9\ \Omega$ .

Solution

(a) The apparent power and the line current:

|S_{3\phi}| = \frac{15{,}000}{0.8} = 18{,}750\ \text{VA} \qquad |\tilde I_a| = \frac{18{,}750}{\sqrt{3}\,(480)} = \boxed{22.55\ \text{A rms}}

(b) One Δ branch carries $|\tilde I_{AB}| = 22.55/\sqrt{3} = 13.02$ A across 480 V, and its angle is $\arccos(0.8) = 36.87^\circ$ :

|Z_\Delta| = \frac{480}{13.02} = 36.86\ \Omega \qquad Z_\Delta = 36.86\,(0.8 + j0.6) = \boxed{29.49 + j22.12\ \Omega}

(c) The line loss, with $R_L = 0.3\ \Omega$ in each line:

P_{3\phi,\text{line}} = 3 (22.55)^2 (0.3) = \boxed{458\ \text{W}}

(d) With $Z_Y = Z_\Delta / 3 = 9.830 + j7.373\ \Omega$ , the per-phase series path is $Z_L + Z_Y = 10.130 + j8.273\ \Omega$ , whose magnitude is $13.08\ \Omega$ :

|\tilde V_{an}| = (22.55)(13.08) = 294.9\ \text{V} \qquad |\tilde V_{ab}| = \sqrt{3}\,(294.9) = \boxed{511\ \text{V rms}}

Check by phasors. Taking $\tilde V_{AN} = 277.1\angle 0^\circ$ V as the reference, the current lags by 36.87°, so $\tilde I_a = 22.55\angle{-36.87^\circ} = 18.04 - j13.53$ A and the line drop is $\tilde I_a (0.3 + j0.9) = 17.59 + j12.18$ V. Adding it,

\tilde V_{an} = 294.7 + j12.18\ \text{V} \qquad |\tilde V_{an}| = 294.9\ \text{V}\ \checkmark

Check by power. The three-phase power delivered by the source is $3(22.55)^2(10.130) = 15{,}458$ W, which equals the 15,000 W absorbed by the load plus the 458 W lost in the line. ✓ The source power factor is $\cos[\arctan(8.273/10.130)] = 0.774$ lagging, well below the 0.8 of the load, because the line is far more reactive than resistive.

Per-phase quantity	Y connection	Δ connection
Phase voltage \|V_1φ\|	\|V_AB\| / √3	\|V_AB\|
Phase current \|I_1φ\|	\|I_a\|	\|I_a\| / √3
Apparent power \|S_1φ\|	\|V_AB\| \|I_a\| / √3	\|V_AB\| \|I_a\| / √3
Three-phase total	√3 \|V_AB\| \|I_a\|	√3 \|V_AB\| \|I_a\|