ECE 211 · Lecture 14: Complex Power & Three-Phase Systems

§1 · Complex power in review

One complex number carries both halves of the story

In the sinusoidal steady state, part of the energy delivered to a load is converted irreversibly into heat or work, and part is merely exchanged back and forth with the inductances and capacitances. Complex power carries both parts at once. With RMS phasors,

S = \tilde{V}\tilde{I}^{*} = P + jQ

in which $P$ is the real (average) power actually consumed and $Q$ is the reactive power circulating between the source and the storage elements. The three quantities carry three different units, which is the traditional way of keeping them apart:

|S| apparent power · VA P real power · W Q reactive power · VAR

The RMS convention

Every phasor on this page is an RMS phasor, exactly as in the board notes: the magnitude of $\tilde{V}$ is the RMS value of the voltage, not the amplitude. The factor of one half in $S = \tfrac{1}{2} V I^{*}$ belongs to amplitude phasors and is absent throughout this lecture.

Three equivalent forms

S = \tilde{V}\tilde{I}^{*} = |\tilde{V}||\tilde{I}| \angle (\theta_v - \theta_I) = |\tilde{V}||\tilde{I}|\cos\theta_s + j|\tilde{V}||\tilde{I}|\sin\theta_s

The conjugate on the current is what makes the angle a difference. Without it the angle of $S$ would be $\theta_v + \theta_I$ , which carries no physical meaning.

Ohm's law $\tilde{I} = \tilde{V}/Z$ is substituted into the definition:

S = \tilde{V}\left(\frac{\tilde{V}}{Z}\right)^{*} = \frac{|\tilde{V}|^{2}}{Z^{*}}

This form is the fastest route from a known terminal voltage to the impedance that draws a prescribed complex power, and it is used twice on this page: once to size a correction capacitor, and once to collapse three parallel loads into one impedance.

Ohm's law is substituted the other way, $\tilde{V} = Z\tilde{I}$ :

S = Z\tilde{I}\tilde{I}^{*} = |\tilde{I}|^{2} Z = |\tilde{I}|^{2} R + j|\tilde{I}|^{2} X

Written out, this form shows that the sign of $Q$ follows the sign of the reactance directly: an inductive branch ( $X > 0$ ) absorbs positive reactive power, and a capacitive branch ( $X < 0$ ) absorbs negative reactive power.

Complex power adds

Conservation applies to the complex quantity as a whole, which is the single most useful fact in this lecture:

\sum S = 0 \qquad\Longleftrightarrow\qquad \sum P = 0 \;\;\text{and}\;\; \sum Q = 0

Summed over every element under the passive sign convention, the total is zero. Stated the other way, the complex power delivered by the sources equals the sum of the complex powers absorbed by the loads, and the real and reactive parts balance separately.

Apparent power does not add

Only $S$ adds, never $|S|$ . Two loads of 1000 VA each, one at $+36.87^\circ$ and one at $-36.87^\circ$ , present $S = 1600 + j0$ VA to the source, so the apparent power is 1600 VA and not 2000 VA. Magnitudes may be added only when the angles agree.

§2 · The power triangle

P, Q, and |S| are the sides of a right triangle

Because $S = P + jQ$ is a complex number, it may be drawn in the complex plane, and the drawing is called the power triangle: $P$ along the horizontal leg, $Q$ along the vertical leg, and $|S|$ as the hypotenuse. One angle appears in all three roles at once, which is what makes the picture worth memorizing:

\theta_s = \theta_v - \theta_I = \theta_Z = \cos^{-1}(\text{pf}) \qquad\qquad \cos\theta_s = \cos(\theta_v - \theta_I) = \cos\theta_Z = \text{pf}

The power factor is the cosine of that angle, so it is also the ratio $P/|S|$ : the fraction of the apparent power that performs useful work. The sliders drive the triangle, and the tabs select which way it leans.

The horizontal leg is the real power, the vertical leg is the reactive power, and the hypotenuse is the apparent power. The drawing is rescaled to fill the panel, so the shape of the triangle, not its size, carries the meaning.

P (kW) 10

power factor 0.8

θ_s = 36.9° lagging |S| = P/pf = 12.50 kVA Q = P tan θ_s = 7.50 kVAR

The current lags the voltage, $\theta_s > 0$ , and $Q$ is positive, so the triangle stands above the horizontal. Motors, transformers, welders, and ballasts are inductive, which is why nearly every industrial load is lagging and why correction almost always means adding capacitance.

The current leads the voltage, $\theta_s < 0$ , and $Q$ is negative, so the triangle hangs below the horizontal. Capacitor banks and lightly loaded cables behave this way. A leading load supplies reactive power to the rest of the system rather than drawing it.

The power factor alone is ambiguous

The cosine is an even function, so a power factor of 0.8 describes both $\theta_s = +36.87^\circ$ and $\theta_s = -36.87^\circ$ . The word lagging or leading is part of the number and must always be carried with it; without it, the sign of $Q$ is unknown and the load cannot be reconstructed.

§3 · What a poor power factor costs

The same watts, delivered by a larger current

A load absorbing $P$ at a terminal voltage $V$ draws the current $P = V|\tilde{I}|\cos\theta$ , so

|\tilde{I}| = \frac{P}{V \cdot \text{pf}}

The line between the source and the load does not care about the angle: it dissipates $|\tilde{I}|^{2}R_{\text{line}}$ regardless. A poor power factor therefore buys nothing but extra current, and the extra current is paid for twice, once in the loss and once in the conductor size required to carry it.

The board example fixes $R_{\text{line}} = 0.2\ \Omega$ , a load of 11 kW, and a load voltage of 220 V rms, then asks for the power that must be generated.

|\tilde{I}| = \frac{11\,000}{220 \cdot 1} = 50\ \text{A} \qquad P_{\text{line}} = |\tilde{I}|^{2} R_{\text{line}} = 50^{2}(0.2) = 500\ \text{W}

P_{\text{gen}} = 11\,000 + 500 = 11.5\ \text{kW} \qquad \eta = \frac{11}{11.5} = \boxed{95.7\ \%}

|\tilde{I}| = \frac{11\,000}{220 \cdot 0.5} = 100\ \text{A} \qquad P_{\text{line}} = 100^{2}(0.2) = 2000\ \text{W}

P_{\text{gen}} = 11\,000 + 2000 = 13\ \text{kW} \qquad \eta = \frac{11}{13} = \boxed{84.6\ \%}

Halving the power factor doubles the current and quadruples the loss. The same 11 kW reaches the customer, and four times as much is burned in the line.

The moving dots represent the line current.

Efficiency against the load power factor.

load power factor 0.8

|I| = 62.5 A P_line = |I|² · (0.2 Ω) = 781 W P_gen = 11.78 kW η = 93.4 %

What the generator also supplies

Only the real power balance is written above, which is why the line resistance alone appears. The generator must additionally supply the reactive power of the load and of the line reactance; that part circulates rather than being consumed, but it occupies the same conductors and the same generator rating.

§4 · Power factor correction

Adding capacitance moves the tip of the triangle down

The utility charges more per kilowatt-hour for a load whose power factor is below 0.95, so the customer has an economic reason to correct it. Correction is performed by placing a capacitor in parallel with the load, where it changes the reactive power without disturbing the load itself: the load keeps its own voltage, its own current, and its own 1000 W.

The problem

NCSU has a load drawing 1 kW at a power factor of 0.8 lagging from a 200 V rms, 60 Hz supply. The power company charges more per kilowatt-hour for a load with a power factor below 0.95. The power factor is to be raised to 0.95.

Because the elements sit in parallel, their complex powers add: $S = \tilde{V}\tilde{I}^{*} = \tilde{V}[\tilde{I}_1^{*} + \tilde{I}_2^{*}] = S_1 + S_2$ .

Step 1: the complex power of the load as it stands

\theta_1 = \cos^{-1} 0.8 = 36.87^\circ \qquad |S_1| = \frac{P}{\text{pf}} = \frac{1000}{0.8} = \tm{pf.s1}{1250\ \text{VA}}

Q_1 = |S_1|\sin\theta_1 = 1250 (0.6) = \tm{pf.q1}{750\ \text{VAR}} \;\Longrightarrow\; S_1 = \tm{pf.p}{1000} + j\,\tm{pf.q1}{750}\ \text{VA}

The angle is positive because the load is lagging.

Step 2: the complex power that is wanted

The capacitor is ideal, so it absorbs no real power and the 1000 W stays exactly where it is. Only the reactive leg changes:

\cos\theta = 0.95 \;\Longrightarrow\; \theta = 18.19^\circ \qquad |S| = \frac{1000}{0.95} = 1052.6\ \text{VA}

\tan(18.19^\circ) = \frac{Q}{1000} \;\Longrightarrow\; Q = \tm{pf.stot}{329\ \text{VAR}} \;\Longrightarrow\; S_{\text{TOT}} = 1000 + j329\ \text{VA}

Step 3: the difference is the capacitor's assignment

S_{\text{TOT}} = S_1 + S_2 \;\Longrightarrow\; 1000 + j750 + S_2 = 1000 + j329 \;\Longrightarrow\; \boxed{S_2 = -j421\ \text{VAR}}

The capacitor must absorb 421 VAR of negative reactive power, which is the same statement as supplying 421 VAR to the load. Its complex power is purely imaginary, as it must be for a lossless element.

Step 4: from reactive power to capacitance

The form $S = |\tilde{V}|^{2}/Z^{*}$ converts the assignment into an impedance:

-j421 = \frac{200^{2}}{Z^{*}} \;\Longrightarrow\; Z^{*} = \frac{40\,000}{-j421} = +j94.9 \;\Longrightarrow\; Z = -j94.9\ \Omega

Z = \frac{1}{j\omega C} = \frac{-j}{\omega C} \;\Longrightarrow\; C = \frac{1}{\omega (94.9)} = \frac{1}{2\pi(60)(94.9)} \;\Longrightarrow\; \boxed{C = 27.9\ \mu\text{F}}

Step 5: what was bought

|\tilde{I}| = \frac{|S|}{|\tilde{V}|}: \qquad \frac{1250}{200} = 6.25\ \text{A} \;\longrightarrow\; \frac{1052.6}{200} = 5.26\ \text{A}

The source current falls by 16 %, and any loss upstream, proportional to the square of that current, falls by 29 %. The load still receives its 1000 W. The 421 VAR now shuttles between the capacitor and the load instead of travelling the whole way back to the generator.

Why the capacitor is placed in parallel and not in series

A series capacitor would carry the load current and would change the voltage across the load, altering the very quantity the customer needs held constant. A shunt capacitor sees the fixed terminal voltage, draws its own current $\tilde{I}_2$ independently, and leaves the load branch untouched. The parallel connection is also what makes the complex powers add so conveniently.

§5 · Several loads on one bus

Add the complex powers, then extract everything else

Loads are specified in the field by their power ratings, not by their impedances: so many kilowatts at such a power factor. The rule $\sum S = 0$ turns those ratings directly into a single equivalent load, from which the source current, the total impedance, and the power factor at the source all follow.

The problem

Three loads sit in parallel on a 110 V rms bus: 12 kW at 0.866 leading, 16 kW at 0.85 lagging, and 20 kVAR at 0.6 lagging. Find the source current $|\tilde{I}_s|$ , the total impedance, and the power factor at the source.

Load 1: leading, so the angle is negative

\cos\theta = 0.866 \;\Longrightarrow\; \theta = -30^\circ \qquad \frac{Q_1}{12\,000} = \tan(-30^\circ) \;\Longrightarrow\; Q_1 = -6930\ \text{VAR}

S_1 = 12\,000 - j6930\ \text{VA}

The negative sign is the entire content of the word leading, and it is the reason this load partially cancels the two that follow.

Load 2: lagging

\cos\theta = 0.85 \;\Longrightarrow\; \theta = 31.8^\circ \qquad \frac{Q_2}{16\,000} = \tan(31.8^\circ) \;\Longrightarrow\; Q_2 = 9916\ \text{VAR}

S_2 = 16\,000 + j9916\ \text{VA}

Load 3: specified by its reactive power instead

\cos\theta = 0.6 \;\Longrightarrow\; \theta = 53.13^\circ \qquad \tan(53.13^\circ) = \frac{20\,000}{P_3} \;\Longrightarrow\; P_3 = 15\,000\ \text{W}

S_3 = 15\,000 + j20\,000\ \text{VA}

Any two of $P$ , $Q$ , $|S|$ , and the power factor fix the triangle; here the vertical leg and the angle were given, so the horizontal leg was recovered.

Add the three, tip to tail

S_T = \tm{ld.s1}{S_1} + \tm{ld.s2}{S_2} + \tm{ld.s3}{S_3} = (12 + 16 + 15) + j(-6.93 + 9.92 + 20)

\Longrightarrow\; \boxed{S_T = 43 + j23\ \text{kVA}} \qquad \text{pf} = \cos\left[\tan^{-1}\frac{23}{43}\right] = \boxed{0.88\ \text{lagging}}

The sum leans less steeply than loads 2 and 3 alone, because the leading load contributed negative reactive power. That is precisely the mechanism of Section 4, supplied here by a neighbour instead of by a capacitor bank.

The source current follows from the magnitude

|S_T| = \sqrt{43\,000^{2} + 23\,000^{2}} = 48.8\ \text{kVA} \qquad |S_T| = |\tilde{V}||\tilde{I}_s|

|\tilde{I}_s| = \frac{48\,760}{110} \;\Longrightarrow\; \boxed{|\tilde{I}_s| = 443\ \text{A rms}}

The equivalent impedance closes the problem

S_T = \frac{|\tilde{V}|^{2}}{Z_T^{*}} \;\Longrightarrow\; Z_T^{*} = \frac{110^{2}}{43\,000 + j23\,000} = \frac{12\,100\,(43\,000 - j23\,000)}{43\,000^{2} + 23\,000^{2}}

Z_T^{*} = 0.219 - j0.117 \;\Longrightarrow\; \boxed{Z_T = 0.219 + j0.117\ \Omega}

The result is checked against the current already found: $|\tilde{I}_s| = 110/\sqrt{0.219^{2} + 0.117^{2}} = 443$ A rms ✓. The positive reactance confirms the lagging power factor, since $\theta_Z = \theta_s$ .

§6 · Three-phase power

Three sources, one system

A single-phase source delivers power that pulses at twice the line frequency, and it requires a full return conductor for the current it sends out. A three-phase system uses three sources of equal amplitude, displaced by 120 degrees, so that the total instantaneous power is constant and the three return currents cancel one another. That cancellation is the reason essentially all generation and transmission is three-phase.

A balanced system (the assumption throughout)

The three line impedances $Z_{\text{line}}$ are all equal.
The three load impedances $Z_{\text{load}}$ are all equal.
The generator impedance is negligible, $Z_g \approx 0$ .
The generated voltages are equal in amplitude and 120 degrees out of phase.

Phase sequence

The three generated voltages are distinguished only by their angles. What matters is the order in which the phasors pass a fixed point as the whole set rotates counterclockwise at $\omega$ , and that order is called the phase sequence. It fixes the direction in which every three-phase motor on the system turns, so it is not a bookkeeping detail.

The phasor set rotates counterclockwise at $\omega$ . Hover over a phasor to link it to its name: $V_{an}$ , $V_{bn}$ , $V_{cn}$ .

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

The order in which the phasors pass a fixed point is a, b, c. This is the positive (or abc) sequence, and it is the default assumption unless a problem states otherwise.

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

The order is now a, c, b. Interchanging any two of the three connections converts one sequence into the other, which is exactly how the direction of rotation of an induction motor is reversed in practice.

V_an + V_bn + V_cn = 0 1∠0° + 1∠−120° + 1∠120° = 0 true in either sequence

That the three phasors sum to zero is the algebraic root of everything that follows: in a balanced system the three line currents also sum to zero, so the neutral conductor carries nothing at all.

§7 · The balanced Y-Y circuit

One phase is the whole problem

Both the source and the load below are connected in Y, that is, three branches meeting at a common node: n at the generator and N at the load. The two neutrals are joined by the neutral conductor. The three lines carry the phase currents to the load.

The circuit

Three sources in Y, three equal line impedances, and three equal load impedances in Y. The generator impedance is taken as zero, so the source terminals a, b, and c carry the generated voltages themselves.

KVL around the top loop

The highlighted chain is the loop: it leaves n through the a-phase source, crosses the line and the load, and returns along the neutral.

-\tm{yy.sa}{\tilde{V}_{an}} + \tm{yy.zla}{Z_L \tilde{I}_a} + \tm{yy.zya}{Z_Y \tilde{I}_a} = 0 \;\Longrightarrow\; \boxed{\tilde{I}_a = \frac{\tilde{V}_{an}}{Z_L + Z_Y}}

The neutral contributes nothing to this equation, because $Z_g \approx 0$ places n and N at the same potential in a balanced system.

The other two phases are copies

The b and c loops are identical except for the source angle, so no further work is required:

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

The three currents are equal in magnitude and 120 degrees apart, inheriting the symmetry of the sources.

The neutral carries nothing

\tilde{I}_N = \tilde{I}_a + \tilde{I}_b + \tilde{I}_c = \tilde{I}_a\left[1 + 1\angle{-120^\circ} + 1\angle{120^\circ}\right] = \boxed{0\ \text{A}}

Because no current flows in it, the neutral conductor may be removed, or replaced by an impedance of any value, without changing a single current or voltage in a balanced system. This is why transmission lines carry three conductors and not four, and why three-phase delivers power with 75 % of the copper a single-phase system would need for the same load.

The total power is three times the phase power, and it is steady

P_T = 3\,|\tilde{V}_p||\tilde{I}_p|\cos\theta \qquad Q_T = 3\,|\tilde{V}_p||\tilde{I}_p|\sin\theta \qquad S_T = 3\,\tilde{V}_p\tilde{I}_p^{*}

The three pulsating single-phase powers, displaced by 120 degrees, add to a constant. A three-phase motor therefore receives smooth power and produces smooth torque, which a single-phase motor cannot do.

The per-phase circuit

Since the analysis of one phase settles all three, the whole system is replaced by a single loop for the a phase, with n and N tied together directly. Every three-phase problem in this course reduces to this drawing plus two rotations by 120 degrees.

The per-phase equivalent: $V_{an}$ drives $Z_L$ in series with $Z_Y$ , and the current is $\tilde{I}_a$ . The b and c results follow by subtracting and adding 120 degrees.

§8 · Your turn

Practice A: correct a feeder

A 240 V rms, 60 Hz feeder with a line resistance of 0.1 Ω supplies a load of 24 kW at a power factor of 0.6 lagging, measured at the load terminals. Find the line loss and the delivery efficiency, then size the shunt capacitor that raises the power factor to 0.95 lagging, and recompute the loss. The problem should be attempted on paper before the hints are opened, in order.

The 0.1 Ω line feeds the 24 kW load, whose terminal voltage is 240 V rms. The capacitor is to be sized.

Hint 1: the current comes before everything else

The load voltage is fixed at 240 V, so $|\tilde{I}| = P/(V \cdot \text{pf})$ gives the line current directly. The line dissipates $|\tilde{I}|^{2}R_{\text{line}}$ , and the efficiency compares the 24 kW delivered with the sum of the two.

Hint 2: the capacitor changes only the vertical leg

The real power stays at 24 kW throughout. Compute $Q$ before correction from $Q = P\tan\theta$ , compute the $Q$ that a power factor of 0.95 requires, and take the difference. That difference is the reactive power the capacitor must absorb, and it is negative.

Hint 3: from VAR to farads

Use $S = |\tilde{V}|^{2}/Z^{*}$ with the load voltage, or equivalently the shortcut $|Q_C| = |\tilde{V}|^{2}/X_C$ with $X_C = 1/(\omega C)$ . The capacitor sees the full 240 V, not the source voltage.

Solution

Before correction. $|\tilde{I}| = 24\,000/(240 \cdot 0.6) = 166.7$ A rms, so

P_{\text{line}} = 166.7^{2}(0.1) = 2778\ \text{W} \qquad \eta = \frac{24\,000}{26\,778} = \boxed{89.6\ \%}

Sizing the capacitor. With $\theta_1 = 53.13^\circ$ and $\theta_2 = 18.19^\circ$ ,

Q_1 = 24\,000\tan(53.13^\circ) = 32\,000\ \text{VAR} \qquad Q_2 = 24\,000\tan(18.19^\circ) = 7888\ \text{VAR}

Q_C = Q_2 - Q_1 = -24\,112\ \text{VAR} \qquad X_C = \frac{240^{2}}{24\,112} = 2.389\ \Omega \qquad \boxed{C = \frac{1}{2\pi(60)(2.389)} = 1110\ \mu\text{F}}

After correction. $|\tilde{I}| = 24\,000/(240 \cdot 0.95) = 105.3$ A rms, so $P_{\text{line}} = 105.3^{2}(0.1) = 1108$ W and

\eta = \frac{24\,000}{25\,108} = \boxed{95.6\ \%}

The loss falls to 40 % of its former value, exactly the square of the ratio of the currents, $(105.3/166.7)^{2} = 0.399$ ✓. A capacitor rated 24.1 kVAR at 240 V is required, which is a substantial bank; the size is the price of a 0.6 power factor.

Practice B: a balanced Y-Y system

A balanced, positive-sequence Y-connected source with $\tilde{V}_{an} = 120\angle 0^\circ$ V rms feeds a balanced Y-connected load $Z_Y = 11 + j10\ \Omega$ through three lines of $Z_L = 1 + j2\ \Omega$ each. Find the three line currents, the neutral current, the total power absorbed by the load, and the power lost in the lines.

The source drives the load through the lines. The neutral conductor joins n to N.

Hint 1: draw one phase

Only the a phase needs to be solved. Its loop contains $\tilde{V}_{an}$ , $Z_L$ , and $Z_Y$ in series, with n and N at the same potential.

Hint 2: the total impedance is convenient here

$Z_L + Z_Y = 12 + j12$ , whose polar form has an angle of exactly 45 degrees and a magnitude of $12\sqrt{2}$ . The arithmetic stays exact all the way to the end.

Hint 3: power, phase by phase

Each phase of the load absorbs $|\tilde{I}|^{2}Z_Y$ , so the three-phase total is three times that. The same statement holds for the line resistance. A check is available: the complex power delivered by the three sources must equal the sum of the two.

Solution

Z_L + Z_Y = 12 + j12 = 12\sqrt{2}\,\angle 45^\circ\ \Omega \qquad \tilde{I}_a = \frac{120\angle 0^\circ}{12\sqrt{2}\,\angle 45^\circ} = \boxed{7.07\angle -45^\circ\ \text{A rms}}

\tilde{I}_b = 7.07\angle -165^\circ\ \text{A rms} \qquad \tilde{I}_c = 7.07\angle 75^\circ\ \text{A rms} \qquad \tilde{I}_N = 0\ \text{A}

With $|\tilde{I}|^{2} = 50$ A² exactly, the load and the lines take

P_{\text{load}} = 3(50)(11) = \boxed{1650\ \text{W}} \qquad Q_{\text{load}} = 3(50)(10) = 1500\ \text{VAR} \qquad P_{\text{line}} = 3(50)(1) = \boxed{150\ \text{W}}

Check. The sources deliver $S_T = 3\tilde{V}_p\tilde{I}_p^{*} = 3(120)(7.07)\angle 45^\circ = 1800 + j1800$ VA. The real parts balance, $1650 + 150 = 1800$ W ✓, and so do the imaginary parts, $1500 + 3(50)(2) = 1800$ VAR ✓. The power factor at the source is $\cos 45^\circ = 0.707$ lagging, which this system would want corrected.