ECE 211 · Lecture 15: Balanced Three-Phase Circuits

§1 · The three-phase system

Three sources, four voltage names

A balanced three-phase source consists of three sinusoidal sources of equal magnitude whose phases are displaced by 120°. In the positive sequence, which is also called the abc sequence and is assumed throughout this lecture, the three phase voltages are

\tilde{V}_{an} = V_p \angle 0^\circ, \qquad \tilde{V}_{bn} = V_p \angle -120^\circ, \qquad \tilde{V}_{cn} = V_p \angle +120^\circ

Their sum vanishes, which is the algebraic fact behind everything that follows: $1\angle 0^\circ + 1\angle{-120^\circ} + 1\angle 120^\circ = 0$ .

Prerequisites carried over

Every quantity on this page is a phasor, written with a tilde, and every impedance is complex. Magnitudes are RMS values unless stated otherwise, so that the power formulas need no factor of one half. A time-domain source $v(t) = V_m\cos(\omega t + \phi)$ therefore becomes $\tilde{V} = (V_m/\sqrt{2}) \angle \phi$ .

The Y-Y system below carries the vocabulary of the lecture. Four different voltages are named, and confusing them is the most common source of error in three-phase work:

phase V @ source: V_an line V @ source: V_ab phase V @ load: V_AN line V @ load: V_AB

Lower-case letters name the source terminals and upper-case letters name the load terminals, exactly as in the board notes. A phase voltage is measured from a line terminal to the neutral point, and a line voltage is measured between two line terminals. The three line impedances model the conductors that connect the source to the load. Figure labels omit the phasor tilde for legibility; the equations retain it.

The neutral conductor carries nothing

Because the three source voltages sum to zero and the three load branches are identical, the three line currents also sum to zero, so the neutral conductor carries no current at all. It may therefore be removed, or replaced by any impedance whatsoever, without altering a single voltage or current in the balanced circuit. This observation is what makes the single-phase equivalent of Section 2 legitimate.

§2 · The single-phase equivalent

One loop replaces the whole system

Because the neutral point of the source and the neutral point of the load are held at the same potential, each phase may be extracted and solved on its own. The result is an ordinary single-loop AC circuit, and the remaining two phases follow by rotation rather than by further analysis.

The extracted phase

Phase a is cut out of the system: the source, the line impedance, the load impedance, and the neutral return that joins n to N. The animated dots represent the line current $\tilde{I}_a$ .

KVL around the single loop

-\tm{ph.van}{\tilde{V}_{an}} + \tm{ph.zl ph.ia}{Z_L \tilde{I}_a} + \tm{ph.zy ph.ia}{Z_Y \tilde{I}_a} = 0

The two impedances carry the same current, because the loop admits only one.

Solve for the line current

\tilde{I}_a = \frac{\tm{ph.van}{\tilde{V}_{an}}}{\tm{ph.zl}{Z_L} + \tm{ph.zy}{Z_Y}}

The load phase voltage follows at once as $\tilde{V}_{AN} = Z_Y \tilde{I}_a$ , which is smaller than $\tilde{V}_{an}$ by the drop across the line.

The other two phases are rotations

\tilde{I}_b = \frac{\tilde{V}_{bn}}{Z_L + Z_Y} = \tilde{I}_a \cdot 1\angle{-120^\circ}, \qquad \tilde{I}_c = \frac{\tilde{V}_{cn}}{Z_L + Z_Y} = \tilde{I}_a \cdot 1\angle 120^\circ

Only one calculation is ever performed. The three line currents have equal magnitudes and are separated by 120°, and they sum to zero.

For a Y connection, the line current is the phase current

The current that leaves terminal a on the line is the very same current that passes through the source phase and through the load branch $Z_Y$ , because those elements lie in series. This identity is the first half of the Y-Δ comparison assembled in Section 5.

§3 · The line voltage

Line to line: a factor of √3 and a 30° lead

The line voltage is defined between two line terminals, so it is obtained from two phase voltages by KVL rather than read off a single source. The derivation below is performed once for $\tilde{V}_{ab}$ ; the other two line voltages follow by rotation.

The quantity to be found

The line voltage $\tilde{V}_{ab}$ is the potential of terminal a relative to terminal b. No single element spans those two terminals, so a loop equation is required.

KVL around the a-n-b loop

-\tm{sv.va}{\tilde{V}_{an}} + \tm{sv.vab}{\tilde{V}_{ab}} + \tm{sv.vb}{\tilde{V}_{bn}} = 0 \;\Longrightarrow\; \tilde{V}_{ab} = \tilde{V}_{an} - \tilde{V}_{bn}

The loop is traversed from the neutral point up through the a phase, across the terminals, and back down through the b phase.

Positive sequence removes the second unknown

\tilde{V}_{bn} = \tilde{V}_{an} \cdot 1\angle{-120^\circ} \;\Longrightarrow\; \tilde{V}_{ab} = \tilde{V}_{an}\left[1 - 1\angle{-120^\circ}\right]

The bracket is a pure number, identical for every balanced positive-sequence system, so it is worth evaluating once and remembering.

Evaluate the bracket

1 - 1\angle{-120^\circ} = 1 - \left[\cos(-120^\circ) + j\sin(-120^\circ)\right] = 1 - \left[-\tfrac{1}{2} - j\tfrac{\sqrt{3}}{2}\right] = \tfrac{3}{2} + j\tfrac{\sqrt{3}}{2}

\left|\tfrac{3}{2} + j\tfrac{\sqrt{3}}{2}\right| = \sqrt{\tfrac{9}{4} + \tfrac{3}{4}} = \sqrt{3}, \qquad \arg = \tan^{-1}\!\frac{\sqrt{3}/2}{3/2} = 30^\circ \;\Longrightarrow\; \boxed{1 - 1\angle{-120^\circ} = \sqrt{3}\angle 30^\circ}

The relation, at the source and at the load

\boxed{\tilde{V}_{ab} = \tilde{V}_{an}\left[\sqrt{3}\angle 30^\circ\right]} \qquad\qquad \boxed{\tilde{V}_{AB} = \tilde{V}_{AN}\left[\sqrt{3}\angle 30^\circ\right]}

Nothing in the derivation referred to the source, so the same relation holds at the load terminals. The remaining line voltages are rotations of the first: $\tilde{V}_{bc}$ lags $\tilde{V}_{ab}$ by 120°, and $\tilde{V}_{ca}$ leads it by 120°.

The three phase voltages $\tilde{V}_{an}$ , $\tilde{V}_{bn}$ , and $\tilde{V}_{cn}$ form a symmetric star. The line voltage $\tilde{V}_{ab}$ is constructed as $\tilde{V}_{an} + (-\tilde{V}_{bn})$ , drawn dashed, and the completed parallelogram makes the two conclusions visible at once: the line voltage is longer by $\sqrt{3}$ and it leads by 30°.

|V_line| = √3 · |V_phase| √3 ≈ 1.732 the line voltage leads by 30°

A worked instance of the rule

A 120 V rms phase voltage produces a line voltage of $\sqrt{3} \cdot 120 = 207.8$ V rms, which is the familiar 120/208 V service. The 277/480 V service is the same rule applied again: $\sqrt{3} \cdot 277 = 479.7$ V.

§4 · The Δ circuit

The dual relation, carried by the currents

In a Δ connection the three elements are joined end to end in a triangle, and the three line terminals are the corners. The consequence is the exact dual of the Y result: the line voltage is now the phase voltage itself, while the line current differs from the phase current by $\sqrt{3}\angle{-30^\circ}$ .

The Δ load

Each impedance $Z_\Delta$ is connected directly between two line terminals, so the voltage across it is a line voltage: $\tilde{V}_{AB}$ appears across the branch from A to B. The currents $\tilde{I}_{AB}$ , $\tilde{I}_{BC}$ , and $\tilde{I}_{CA}$ circulate inside the triangle and are the phase currents; the currents $\tilde{I}_a$ , $\tilde{I}_b$ , and $\tilde{I}_c$ arrive on the lines.

KCL at the corner

\tm{dl.ia}{\tilde{I}_a} = \tm{dl.iab}{\tilde{I}_{AB}} - \tm{dl.ica}{\tilde{I}_{CA}}

The dashed surface encloses corner A. One current arrives on the line, one leaves into the A-B branch, and one arrives from the C-A branch.

Positive sequence removes the second unknown

\tilde{I}_{CA} = \tilde{I}_{AB} \cdot 1\angle 120^\circ \;\Longrightarrow\; \tilde{I}_a = \tilde{I}_{AB}\left[1 - 1\angle 120^\circ\right]

The phase currents inherit the sequence of the voltages that drive them, because all three branches carry the same impedance.

Evaluate the bracket

1 - 1\angle 120^\circ = 1 - \left[-\tfrac{1}{2} + j\tfrac{\sqrt{3}}{2}\right] = \tfrac{3}{2} - j\tfrac{\sqrt{3}}{2} = \sqrt{3}\angle{-30^\circ}

The bracket is the complex conjugate of the one obtained in Section 3, so the magnitude is again $\sqrt{3}$ and the shift is again 30°, but of opposite sign.

The relation

\boxed{\underbrace{\tilde{I}_a}_{\text{line } I} = \underbrace{\tilde{I}_{AB}}_{\text{phase } I} \left[\sqrt{3}\angle{-30^\circ}\right]}

The line current is larger by $\sqrt{3}$ and lags the phase current by 30°.

The current diagram of the Δ load mirrors the voltage diagram of the Y source. The phase currents $\tilde{I}_{AB}$ , $\tilde{I}_{BC}$ , and $\tilde{I}_{CA}$ form the star, and the line current $\tilde{I}_a$ is completed by adding the dashed $-\tilde{I}_{CA}$ to $\tilde{I}_{AB}$ .

§5 · Δ against Y

Which quantity is shared, and which is transformed

One rule per connection is sufficient, because the other quantity is then the one that carries the $\sqrt{3}$ factor. The two rules are mirror images:

Δ: line V = phase V, line I ≠ phase I

Y: line I = phase I, line V ≠ phase V

\tilde{I}_{\text{line}} = \tilde{I}_{\text{phase}} \qquad\qquad \tilde{V}_{\text{line}} = \tilde{V}_{\text{phase}}\left[\sqrt{3}\angle 30^\circ\right]

The three branches meet at a neutral point, so each branch carries its own line current. The line voltage spans two branches and therefore acquires the $\sqrt{3}\angle 30^\circ$ factor.

\tilde{V}_{\text{line}} = \tilde{V}_{\text{phase}} \qquad\qquad \tilde{I}_{\text{line}} = \tilde{I}_{\text{phase}}\left[\sqrt{3}\angle{-30^\circ}\right]

Each branch is connected directly between two line terminals, so each branch sees a line voltage. The line current splits between two branches and therefore acquires the $\sqrt{3}\angle{-30^\circ}$ factor.

Converting a load between the two connections

A balanced Δ load and a balanced Y load are interchangeable, and the transformation is the balanced Y-Δ rule of Lecture 2 with resistances replaced by impedances:

\boxed{Z_\Delta = 3 Z_Y} \qquad\qquad \boxed{Z_Y = \frac{Z_\Delta}{3}}

Converting a Δ load to its Y equivalent is the standard first move, because the single-phase equivalent of Section 2 requires a Y load and a neutral point to work with.

The Δ connected source

The three sources are joined into a closed triangle, and the terminals a, b, and c are the corners. The source phase voltage $\tilde{V}_{ab}$ is the line voltage itself, which is the defining property of the Δ source.

A Δ source may also be replaced by an equivalent Y source, whose phase voltage is obtained by inverting the relation of Section 3:

\tilde{V}_{an} = \frac{\tilde{V}_{ab}}{\sqrt{3}\angle 30^\circ}

The Y equivalent voltage is a fiction of the conversion

The quantity $\tilde{V}_{an}$ computed above is a Δ-Y conversion equivalent voltage only. The actual phase voltage of the Δ source remains the line voltage $\tilde{V}_{ab}$ , and any power calculation performed on the source itself must use that value. The equivalent is legitimate for finding line currents and load quantities, which is exactly what it is built for.

§6 · Balanced three-phase power

One phase, then multiply by three

The complex power of a single phase follows the same three forms as in single-phase AC analysis, with $\tilde{V}_\phi$ and $\tilde{I}_\phi$ taken across and through one load branch:

S_{1\phi} = \tilde{V}_\phi \tilde{I}_\phi^{\,*} = \left|\tilde{I}_\phi\right|^2 Z_\phi = \frac{\left|\tilde{V}_\phi\right|^2}{Z_\phi^{\,*}}

For a balanced system the three phases are identical, so the total is three times the single-phase result:

\boxed{S_{3\phi} = 3 S_{1\phi} = P_{3\phi} + j Q_{3\phi}}

The two practical specializations are worth writing out, because each avoids a conversion step. For a Y load the line current is the phase current, and for a Δ load the line voltage is the phase voltage:

Y load: S_3φ = 3 |I_a|² · Z_Y Δ load: S_3φ = 3 |V_AB|² / Z_Δ* Z_Δ = 3 Z_Y connects the two

The line-quantity form

Substituting either rule of Section 5 into $3|\tilde{V}_\phi||\tilde{I}_\phi|$ gives one expression that holds for both connections: $\left|S_{3\phi}\right| = \sqrt{3}\,\left|\tilde{V}_{\text{line}}\right| \left|\tilde{I}_{\text{line}}\right|$ , with $P_{3\phi} = \left|S_{3\phi}\right|\cos\theta$ and $Q_{3\phi} = \left|S_{3\phi}\right|\sin\theta$ , where $\theta$ is the angle of the load impedance. The $\sqrt{3}$ appears exactly once, whichever connection is in use.

The power triangle

The line voltage is held at 208 V rms below. The current slider scales the triangle, and the impedance angle $\theta$ divides the apparent power between the real power and the reactive power. A positive angle denotes an inductive load, which absorbs reactive power.

The horizontal leg is the real power $P_{3\phi}$ , the vertical leg is the reactive power $Q_{3\phi}$ , and the hypotenuse is the apparent power $\left|S_{3\phi}\right|$ . The power factor is the cosine of the angle at the origin.

|I_line| (A rms) 8

θ = ∠Z (degrees) 30

|S_3φ| = 2.882 kVA P_3φ = 2.496 kW Q_3φ = 1.441 kvar pf = 0.866 lagging, inductive

Reactive power is not wasted power

A load at $\theta = 60^\circ$ draws the same line current as a load at $\theta = 0^\circ$ yet converts only half of the apparent power into real power. The line loss, which depends on the current alone, is unchanged. This is the whole motivation for power factor correction, and it is why utilities meter reactive power separately.

§7 · Worked example A: three line currents and the load power

A Y-Y system specified in the time domain

The problem

A Y-Y, three-phase, positive-sequence system has the a-phase source voltage $v_{an}(t) = 100\sqrt{2}\cos(2\pi 60 t)$ V, a line impedance of 10 Ω, and a load consisting of a 40 Ω resistor in series with a 40 mH inductor. The three line currents and the real power delivered to the load are required.

Phasors first

\tilde{V}_{an} = \frac{100\sqrt{2}}{\sqrt{2}}\angle 0^\circ = 100\angle 0^\circ \ \text{V rms}, \qquad \omega = 2\pi(60) = 377\ \text{rad/s}

\tm{ea.zy}{Z_Y} = 40 + j\omega L = 40 + j(377)(0.04) = 40 + j15.08\ \Omega

The RMS convention is adopted because the source amplitude was written as $100\sqrt{2}$ , which is exactly the form that produces a round 100 V rms.

One division delivers the a-phase current

\tilde{I}_a = \frac{\tm{ea.van}{\tilde{V}_{an}}}{\tm{ea.zl}{Z_L} + \tm{ea.zy}{Z_Y}} = \frac{100\angle 0^\circ}{50 + j15.08} = \frac{100\angle 0^\circ}{52.22\angle 16.79^\circ} \;\Longrightarrow\; \boxed{\tilde{I}_a = 1.915\angle{-16.79^\circ}\ \text{A rms}}

The current lags the source voltage, as an inductive load requires.

The other two currents are rotations

\tilde{I}_b = 1.915\angle{-136.79^\circ}\ \text{A rms}, \qquad \tilde{I}_c = 1.915\angle{103.21^\circ}\ \text{A rms}

Each is obtained from $\tilde{I}_a$ by subtracting and adding 120° respectively, which is the whole benefit of the balanced assumption.

The real power delivered to the load

S_{1\phi} = \left|\tilde{I}_a\right|^2 Z_Y = (1.915)^2 (40 + j15.08) \;\Longrightarrow\; P_{1\phi} = (1.915)^2 (40) = 146.7\ \text{W}

P_{3\phi} = 3\left|\tilde{I}_a\right|^2 (40) = 3(3.666)(40) \;\Longrightarrow\; \boxed{P_{3\phi} = 440\ \text{W}}

Only the resistive part of $Z_Y$ appears, because the inductor absorbs no real power. The reactive part gives $Q_{3\phi} = 3(3.666)(15.08) = 166$ var.

Check: the source supplies exactly what the load and the line consume

The three line resistances absorb $3\left|\tilde{I}_a\right|^2 (10) = 110$ W, so the load and the line together require 550 W. The source delivers

S_{3\phi,\text{src}} = 3 \tilde{V}_{an} \tilde{I}_a^{\,*} = 3(100)(1.915)\angle 16.79^\circ = 574.5\angle 16.79^\circ = 550 + j166\ \text{VA}

The real parts agree at 550 W, and the reactive parts agree at 166 var, since the line is purely resistive here and the inductor is the only reactive element. ✓

Rounding early is expensive

Truncating the current to 1.9 A before squaring gives $3(1.9)^2(40) = 433$ W, an error of 7 W on a 440 W answer. Magnitudes should be carried at full precision until the final multiplication, because the power formulas square them.

§8 · Worked example B: starting from the load voltage

When the given quantity sits at the far end

The problem

A three-phase, Y-Y, positive-sequence system has a line voltage at the load of 100 V rms, a line impedance of 10 Ω, and a Y-connected load impedance of $40 + j15.1\ \Omega$ . The real power delivered to the load is required.

The line impedance is not needed

The load voltage is given directly, so the source voltage never enters the calculation. The 10 Ω line impedance would be required only if the source voltage were the given quantity, or if the line loss were requested.

Convert the load to Δ, so that the line voltage is the phase voltage

\tm{eb.zd}{Z_\Delta} = 3 \tm{eb.zy}{Z_Y} = 120 + j45.3\ \Omega

S_{1\phi} = \frac{\left|\tilde{V}_{AB}\right|^2}{Z_\Delta^{\,*}} \;\Longrightarrow\; S_{3\phi} = \frac{3\left|\tilde{V}_{AB}\right|^2}{Z_\Delta^{\,*}}

The conversion is worth the one line it costs: the given line voltage becomes a phase voltage of the Δ equivalent, so no $\sqrt{3}$ factor is needed anywhere in the calculation.

Rationalize and take the real part

S_{3\phi} = \frac{3(100)^2}{120 - j45.3} \cdot \frac{120 + j45.3}{120 + j45.3} = \frac{3(100)^2 (120 + j45.3)}{120^2 + 45.3^2}

P_{3\phi} = \frac{3(100^2)(120)}{16452}

The answer

\boxed{P_{3\phi} = 218.8\ \text{W} \approx 219\ \text{W}}

The reactive part follows from the same fraction: $Q_{3\phi} = 3(100^2)(45.3)/16452 = 82.6$ var.

Independent check: the same answer through the Y load

The Y route uses the phase voltage at the load, which is smaller than the line voltage by $\sqrt{3}$ :

\left|\tilde{V}_{AN}\right| = \frac{100}{\sqrt{3}} = 57.74\ \text{V rms}, \qquad \left|\tilde{I}_a\right| = \frac{57.74}{\left|40 + j15.1\right|} = \frac{57.74}{42.76} = 1.350\ \text{A rms}

P_{3\phi} = 3\left|\tilde{I}_a\right|^2 (40) = 3(1.823)(40) = 218.8\ \text{W} \ \checkmark

Two routes that share no arithmetic reach the same number, which is the strongest check available. As a further remark, the source voltage may now be recovered if desired: $\left|\tilde{V}_{an}\right| = \left|\tilde{I}_a\right| \left|50 + j15.1\right| = 1.350(52.23) = 70.5$ V rms, a line voltage of 122.2 V rms at the source.

§9 · Your turn

Practice: a Δ load behind a line impedance

The problem below combines every result of the lecture: a Δ load must be converted, a single-phase equivalent must be solved, and a phase current must be recovered from a line current. It should be attempted on paper before the hints are opened, in order.

A balanced Δ load of $60 + j45\ \Omega$ per branch is fed through three line impedances of $1 + j2\ \Omega$ from a Y-connected, positive-sequence source with $\tilde{V}_{an} = 120\angle 0^\circ$ V rms. Find the line current $\tilde{I}_a$ , the load phase current $\tilde{I}_{AB}$ , and the total real power delivered to the load.

Hint 1: Make the load a Y

The single-phase equivalent needs a neutral point, and a Δ load has none. Convert first: $Z_Y = Z_\Delta / 3$ . The line impedance is not touched by the conversion, because it lies outside the load.

Hint 2: Solve one phase

The per-phase loop of Section 2 gives $\tilde{I}_a = \tilde{V}_{an} / (Z_L + Z_Y)$ . The line current is unchanged by the Δ-Y conversion, which is precisely why the conversion is legitimate.

Hint 3: Return to the Δ for the phase current

Section 4 gives $\tilde{I}_a = \tilde{I}_{AB}\left[\sqrt{3}\angle{-30^\circ}\right]$ , so the phase current is recovered by dividing. For the power, either $3\left|\tilde{I}_{AB}\right|^2 (60)$ or $3\left|\tilde{I}_a\right|^2 (20)$ may be used; they must agree.

Solution

The Y equivalent of the load is $Z_Y = (60 + j45)/3 = 20 + j15\ \Omega$ , so the per-phase loop carries $Z_L + Z_Y = 21 + j17 = 27.02\angle 38.99^\circ\ \Omega$ and

\tilde{I}_a = \frac{120\angle 0^\circ}{27.02\angle 38.99^\circ} \;\Longrightarrow\; \boxed{\tilde{I}_a = 4.441\angle{-38.99^\circ}\ \text{A rms}}

\tilde{I}_{AB} = \frac{\tilde{I}_a}{\sqrt{3}\angle{-30^\circ}} = \frac{4.441\angle{-38.99^\circ}}{1.732\angle{-30^\circ}} \;\Longrightarrow\; \boxed{\tilde{I}_{AB} = 2.564\angle{-8.99^\circ}\ \text{A rms}}

The real power delivered to the load, computed on the Δ branches and then independently on the Y equivalent:

P_{3\phi} = 3\left|\tilde{I}_{AB}\right|^2 (60) = 3(6.576)(60) = 1184\ \text{W} \qquad P_{3\phi} = 3\left|\tilde{I}_a\right|^2 (20) = 3(19.73)(20) = 1184\ \text{W} \ \checkmark

A third check closes the accounting. The three line resistances absorb $3\left|\tilde{I}_a\right|^2 (1) = 59.2$ W, while the source supplies $3(120)(4.441)\cos(38.99^\circ) = 1242.8$ W, and $1184 + 59.2 = 1243$ W ✓. The load line voltage, if wanted, is $\left|\tilde{V}_{AB}\right| = \left|\tilde{I}_{AB}\right| \left|Z_\Delta\right| = 2.564(75) = 192.3$ V rms.