Electrical Circuit Analysis

Chapter 10
AC Power and Three-Phase Systems

Bachelors Program in Electrical Engineering
Electrical Circuit Analysis   •   Bachelor

AC Power and Three-Phase Systems

Introduction

Alternating-current power analysis connects circuit theory with generators, motors, transformers, industrial installations, renewable-energy converters, and utility systems. Unlike a direct-current circuit, an AC circuit may exchange stored energy with inductors and capacitors during each cycle. Consequently, voltage and current magnitude alone do not determine useful energy conversion; their phase relationship must also be considered.

Three-phase systems dominate electric-power generation and industrial distribution because they transmit nearly constant total power under balanced operation, use conductors efficiently, and naturally produce rotating magnetic fields. Engineers must distinguish line and phase quantities, select appropriate wye or delta relationships, and recognize when balanced-system shortcuts are invalid.

Unless stated otherwise, this chapter assumes sinusoidal steady state, RMS phasors, passive sign convention, and consistent reference directions.

Learning Objectives

After completing this chapter, students should be able to:

  1. Derive and calculate instantaneous and average AC power.
  2. Determine reactive, complex, and apparent power using RMS phasors.
  3. Interpret power factor and size basic compensation capacitors.
  4. Apply practical sign and metering conventions.
  5. Analyze balanced wye- and delta-connected three-phase systems.
  6. Analyze basic unbalanced four-wire wye, three-wire wye, and delta loads.
  7. Evaluate practical limitations involving harmonics, equipment ratings, safety, and compensation.

1. Instantaneous Power

For sinusoidal voltage and current, select voltage as the reference:

v(t) = √2V cos(ωt)
i(t) = √2I cos(ωt - φ)

Here, V and I are RMS values in volts and amperes, ω = 2πf is angular frequency in radians per second, f is frequency in hertz, and φ is the angle by which current lags voltage. A negative φ therefore represents leading current.

Instantaneous power is the product of instantaneous voltage and current:

p(t) = v(t)i(t)
= VI[cos φ + cos(2ωt - φ)]

This result follows from 2 cos A cos B = cos(A-B)+cos(A+B). It contains:

  • a constant component VI cos φ, and
  • a double-frequency component VI cos(2ωt-φ).

Instantaneous power is measured in watts. It can become negative during part of a cycle when stored energy is returned from a reactive element to the source.

Average or real power

The average of the double-frequency term over an integer number of periods is zero. Thus,

P = (1/T)∫0T p(t)dt = VI cos φ

where T = 1/f. Average power P, measured in watts (W), represents net energy converted per unit time into mechanical work, heat, light, or another nonrecoverable form.

For a pure resistor, φ=0°, so P=VI. For an ideal inductor or capacitor, |φ|=90° and average power is zero, although instantaneous power still oscillates.

2. Reactive, Complex, and Apparent Power

2.1 Reactive power

For sinusoidal steady-state operation, reactive power is

Q = VI sin φ

Reactive power is measured in volt-amperes reactive (var). Under the passive sign convention:

  • Q>0 for a net inductive load with lagging current;
  • Q<0 for a net capacitive load with leading current.

Reactive power does not represent net energy consumed per cycle. It quantifies cyclic energy exchange that nevertheless increases current and equipment loading.

2.2 Complex power

Let RMS phasors be V̅ and I̅. Complex power absorbed by a load is

S̅ = V̅I̅* = P + jQ

The asterisk denotes complex conjugation. The formula assumes passive sign convention: current enters the terminal assigned positive voltage polarity.

For an impedance Z = R+jX,

S̅ = I²Z = V²/Z*

In these expressions, I and V are RMS magnitudes. Therefore,

P = I²R
Q = I²X

Positive reactance gives positive reactive power; negative reactance gives negative reactive power.

2.3 Apparent power and the power triangle

The magnitude of complex power is apparent power:

|S̅| = S = VI = √(P²+Q²)

Apparent power is measured in volt-amperes (VA). Transformers, generators, cables, and converters are often rated in VA because heating and insulation stress depend strongly on voltage and current, irrespective of how much real power is delivered.

Power triangle P (W) Q (var) S (VA) φ Shown for an inductive load: Q > 0
Power triangle showing real, reactive, and apparent power
Figure: Power triangle showing real, reactive, and apparent power

Figure: Power triangle for a sinusoidal inductive load.

For a capacitive load, the reactive-power component points downward because Q<0.

Worked Example 1: Single-phase complex power

A load operates at 230 V RMS and draws 12 A at a lagging angle of 36.87°. Determine its complex, real, reactive, and apparent power.

Since cos 36.87°≈0.8 and sin 36.87°≈0.6,

S = VI = 230×12 = 2760 VA
P = 2760×0.8 = 2208 W
Q = 2760×0.6 = 1656 var

Thus,

S̅ = 2208 + j1656 VA

The positive reactive power confirms an inductive load.

3. Power Factor

For sinusoidal voltage and current, displacement power factor is

PF = cos φ = P/S

Power factor is described as lagging for inductive loads and leading for capacitive loads. Merely reporting “0.8 power factor” is incomplete unless the leading or lagging character is known.

At fixed real power and voltage,

I = P/(V PF)

for a single-phase sinusoidal load. A low power factor therefore increases current, conductor losses I²R, voltage drop, and required equipment rating.

3.1 Power-factor compensation

A shunt capacitor can supply negative reactive power locally, reducing reactive power drawn from the source. If a load with real power P is corrected from lagging angle φ1 to φ2, the required capacitor-bank rating magnitude is

QC = P(tan φ1 - tan φ2)

Here, QC is stated as a positive capacitor-bank rating in var. In signed complex-power calculations, the capacitor contributes -jQC. The formula assumes real load power remains unchanged and both initial and target conditions are sinusoidal.

For a capacitor,

|QC| = ωCV²

where V is the RMS voltage across that capacitor.

For a balanced three-phase bank at line voltage VL:

QC,Y = ωCYVL²
QC,Δ = 3ωC_ΔVL²

In these equations, CY or C_Δ is the capacitance of each identical phase unit. The wye formula follows because each capacitor sees VL/√3, whereas each delta capacitor sees VL.

Worked Example 2: Capacitor-bank sizing

A 40 kW, 400 V, 50 Hz balanced load operates at 0.75 lagging power factor. Find the delta-connected capacitance per phase required to improve the power factor to 0.95 lagging.

φ1 = cos-1(0.75), tan φ1 ≈ 0.8819
φ2 = cos-1(0.95), tan φ2 ≈ 0.3287

Therefore,

QC = 40000(0.8819-0.3287) ≈ 22130 var

For a delta bank,

C_Δ = QC/(3ωVL²)
C_Δ = 22130/[3(2π×50)(400²)] ≈ 146.8 μF

Each delta branch therefore requires approximately 147 μF. A commercial bank must also satisfy voltage, current, thermal, switching, and harmonic-duty requirements.

3.2 Harmonics and true power factor

The relation PF=cos φ applies only when voltage and current are sinusoidal. With distorted waveforms,

PFtrue = P/(VrmsIrms)

True power factor includes both phase displacement and waveform distortion. For periodic waveforms, average power may be evaluated by summing same-frequency harmonic contributions:

P = ∑h VhIh cos φh

This assumes compatible RMS harmonic phasors and no cross-frequency average terms. Reactive power under nonsinusoidal conditions requires a clearly stated standard or definition; the simple sinusoidal power triangle may no longer fully characterize the load.

4. Balanced Three-Phase Sources

A balanced positive-sequence source has equal phase-voltage magnitudes separated by 120°:

V̅AN = Vph∠0°
V̅BN = Vph∠-120°
V̅CN = Vph∠120°

The order A-B-C is called positive phase sequence. Interchanging any two lines reverses the sequence, an important consideration for motor rotation.

A system is balanced only when the source phase quantities are balanced and the three load impedances are identical. Under this condition, one phase may be analyzed and the remaining quantities obtained by 120° shifts.

5. Balanced Wye-Connected Loads

In a wye or star load, one end of each phase impedance is connected to a common neutral point. The line-to-line voltage is the phasor difference between two phase voltages. For positive sequence,

V̅AB = V̅AN-V̅BN = √3Vph∠30°

Therefore,

VL = √3Vph

For a wye load,

IL = Iph

Thus, for each impedance Zph,

I̅A = V̅AN/Zph

The neutral current is zero in an ideal balanced system because the three phase currents sum to zero.

Worked Example 3: Balanced wye load

A balanced wye load with Zph=8+j6 Ω is connected to a 400 V line-to-line, positive-sequence supply. Find the line current and total power.

The impedance magnitude and angle are

Zph = 10∠36.87° Ω

The phase voltage is

Vph = 400/√3 = 230.94 V

Taking V̅AN as the reference,

I̅A = 230.94∠0°/10∠36.87°
= 23.09∠-36.87° A

Because the load is wye connected, the line-current magnitude is 23.09 A. The total complex power is

S̅3φ = 3V̅phI̅ph* = 3Iph²Zph

Hence,

P3φ = √3VLIL cos φ = 12.8 kW
Q3φ = √3VLIL sin φ = 9.6 kvar
S3φ = √3VLIL = 16.0 kVA

6. Balanced Delta-Connected Loads

A delta load has three impedances connected between pairs of line conductors. Every branch therefore receives line-to-line voltage:

Vph = VL

The line current is the phasor difference between adjacent branch currents. With branch-current directions A→B, B→C, and C→A,

I̅A = I̅AB-I̅CA

For a balanced positive-sequence delta load,

IL = √3Iph

and the A-line current lags the A–B branch current by 30°. This angular statement depends on the stated phase sequence and current-reference directions; the magnitude relation does not.

Delta-connected three-phase load A B C Z_AB Z_CA Z_BC Each line terminates at a vertex; every side contains a load impedance.
Correct delta-connected load with each supply line connected to a delta vertex and an impedance in every branch
Figure: Correct delta-connected load with each supply line connected to a delta vertex and an impedance in every branch

Figure: Correct topology of a delta-connected load. The B supply lead connects directly to vertex B, and each branch contains an explicit impedance rather than a short circuit.

For either a balanced wye or balanced delta load,

S̅3φ = 3V̅phI̅ph*

and, using line magnitudes,

P3φ = √3VLIL cos φ
Q3φ = √3VLIL sin φ
S3φ = √3VLIL

These line-quantity formulas require a balanced sinusoidal load. They must not be applied indiscriminately to unbalanced systems.

7. Measuring Three-Phase Power

7.1 Three-wattmeter method

With an accessible neutral, one wattmeter may be connected per phase. Total real power is the algebraic sum:

Ptotal = WA+WB+WC

This method is valid for balanced or unbalanced four-wire systems when each wattmeter is correctly referenced to neutral.

7.2 Two-wattmeter method

Two wattmeters can measure total real power in any three-wire system, balanced or unbalanced:

Ptotal = W1+W2

For one common connection convention and a balanced load,

W1 = VLIL cos(30°-φ)
W2 = VLIL cos(30°+φ)

Consequently,

Qtotal = √3(W1-W2)
tan φ = √3(W1-W2)/(W1+W2)

Changing wattmeter coil assignments may reverse the difference, so the connection convention must be documented. The reactive-power and power-factor formulas above require a balanced sinusoidal load, whereas the sum still gives total real power for a three-wire load. At lagging power factor below 0.5, one wattmeter reading is negative under the stated convention.

8. Basic Unbalanced Three-Phase Analysis

Unbalance may arise from unequal single-phase loads, unequal impedances, conductor faults, or asymmetric supply voltages. Under unbalance, each phase must generally be treated separately.

8.1 Four-wire unbalanced wye load

If the source and load neutral points are connected by an ideal conductor, each phase load receives its corresponding source phase voltage:

I̅A = V̅AN/ZA,
I̅B = V̅BN/ZB,
I̅C = V̅CN/ZC

For line currents directed from source to load and neutral current directed from load to source,

I̅N = I̅A+I̅B+I̅C

The neutral current is therefore a phasor sum, not an arithmetic sum of magnitudes.

Worked Example 4: Neutral current

A 230 V phase-voltage source supplies three resistive wye loads of 10 Ω, 20 Ω, and 30 Ω through an ideal neutral. With positive sequence,

I̅A = 23∠0° A
I̅B = 11.5∠-120° A
I̅C = 7.667∠120° A

Rectangular addition gives

I̅N ≈ 13.417-j3.319 A

Thus,

I̅N ≈ 13.82∠-13.9° A

Although every load is resistive, unequal phase currents produce substantial neutral current.

8.2 Three-wire unbalanced wye load

If the load neutral is isolated, its voltage is not generally equal to the source neutral voltage. Let source neutral be N and floating load neutral be n. With phase admittances YA, YB, and YC, Kirchhoff’s current law gives

V̅nN = (YAV̅AN+YBV̅BN+YCV̅CN)/(YA+YB+YC)

provided the denominator is nonzero. The load phase voltages are then

V̅An = V̅AN-V̅nN

with analogous expressions for phases B and C. Neutral displacement can cause some phase loads to experience overvoltage even when line-to-line voltages remain nominal.

8.3 Unbalanced delta load

Each delta branch remains directly across a line-to-line voltage:

I̅AB=V̅AB/ZAB,
I̅BC=V̅BC/ZBC,
I̅CA=V̅CA/ZCA

The line currents follow from Kirchhoff’s current law:

I̅A=I̅AB-I̅CA
I̅B=I̅BC-I̅AB
I̅C=I̅CA-I̅BC

For any unbalanced load, total complex power should be found by summing branch or phase complex powers:

S̅total = ∑k V̅kI̅k*

9. Applied Case: Correcting an Industrial Motor Installation

Background

This is an illustrative case. A manufacturing facility operates a 75 kW mechanical-output induction motor from a 400 V, 50 Hz three-phase supply. Motor efficiency is 92%, and operating power factor is 0.78 lagging. The facility seeks to improve power factor to 0.95 lagging.

Problem

Determine the existing line current, required capacitor-bank rating, corrected line current, and approximate delta-connected capacitance per phase.

Analysis

The electrical input real power is

Pin = 75/0.92 = 81.52 kW

The initial line current is

I1 = Pin/(√3VLPF1)
I1 = 81520/(√3×400×0.78) ≈ 150.9 A

The relevant tangent values are approximately

tan(cos-10.78) = 0.8023
tan(cos-10.95) = 0.3287

Therefore,

QC = 81.52(0.8023-0.3287) ≈ 38.61 kvar

The corrected current is

I2 = 81520/(√3×400×0.95) ≈ 123.9 A

For a delta bank,

C_Δ = 38610/[3(2π×50)(400²)] ≈ 256 μF

Approach and outcome

A nominal 38.6 kvar bank would theoretically meet the target at this operating point. The line current falls from approximately 150.9 A to 123.9 A, a reduction of about 17.9%. This lowers conductor losses and releases upstream capacity.

In practice, engineers would select standard switched steps rather than blindly installing a fixed bank. Motor loading varies, and excessive capacitance can produce leading power factor, switching transients, self-excitation after disconnection, or harmonic resonance.

Learning points

  • Compensation is based on electrical input power, not mechanical output power.
  • Capacitors reduce source current but do not reduce the motor’s real-power requirement.
  • A calculated capacitance is only the beginning of equipment selection.
  • Harmonic surveys and protection coordination may justify detuned capacitor banks or active compensation.

10. Engineering Practice, Limitations, and Contemporary Developments

Best-practice workflow

  1. Confirm whether values are RMS, peak, phase, or line quantities.
  2. Draw the actual wye or delta topology and assign current directions.
  3. State phase sequence and phasor reference.
  4. Use complex conjugation when calculating complex power.
  5. Check whether balanced and sinusoidal assumptions are valid.
  6. Sum phase powers for unbalanced systems.
  7. Verify results through conservation of power and plausible equipment ratings.
  8. Document meter polarity and two-wattmeter connection conventions.

Practical limitations

Real systems contain feeder impedance, voltage unbalance, frequency variation, harmonics, and time-varying loads. Capacitor tolerance and aging alter compensation. Neutral conductors can carry high triplen-harmonic currents even when fundamental-frequency phase currents are balanced. Thermal ratings must therefore be based on measured RMS current and harmonic content, not only fundamental phasors.

Safety, governance, and regulation

Three-phase installations can deliver hazardous fault energy. Measurement and switching require correctly rated instruments, current transformers, protective equipment, isolation procedures, and compliance with applicable electrical codes and arc-flash practices. Capacitors retain dangerous charge after disconnection and require discharge provisions.

Utilities may impose power-factor tariffs, harmonic limits, and interconnection requirements. Engineers must report measurements honestly, preserve meter calibration records, and avoid compensation strategies that transfer unacceptable distortion or resonance risk to other users.

Contemporary technology

Digital power-quality meters simultaneously estimate RMS values, real power, harmonic spectra, unbalance, and true power factor. Smart capacitor controllers switch steps according to measured reactive demand. Active power filters and inverter-based resources can provide dynamic reactive-power control. Data analytics can identify deteriorating power factor or abnormal phase imbalance, but automated recommendations must be validated against network models, protection settings, and field measurements.

Chapter Summary

Instantaneous AC power is the product of voltage and current and contains average and double-frequency components for sinusoidal signals. Real power represents net energy conversion, reactive power describes cyclic energy exchange, and complex power combines both. Apparent power determines major equipment-loading requirements, while power factor indicates how effectively current capacity supports real-power transfer.

Shunt capacitors can reduce inductive reactive demand, but their ratings must be calculated from the desired power-factor change and checked for harmonics, resonance, switching, and operating variation. In balanced three-phase systems, wye loads satisfy VL=√3Vph and IL=Iph, whereas delta loads satisfy VL=Vph and IL=√3Iph. Balanced total power can be calculated from line quantities using √3VLIL.

Unbalanced systems require phase-by-phase phasor analysis. Four-wire wye systems may carry neutral current, isolated-neutral wye systems experience neutral displacement, and delta line currents must be obtained from branch-current differences. Reliable engineering practice depends on correct topology, explicit conventions, validated assumptions, suitable metering, and compliance with safety and power-quality requirements.

Key Terms

Instantaneous power: The time-dependent product of instantaneous voltage and current.

Average power: Net electrical energy converted per unit time, measured in watts.

Reactive power: Measure of cyclic energy exchange, expressed in var.

Complex power: Phasor-domain quantity S̅=P+jQ.

Apparent power: Magnitude of complex power, measured in VA.

Power factor: Ratio of real power to apparent power.

Displacement power factor: Cosine of the fundamental voltage-current phase angle.

True power factor: Ratio P/(VrmsIrms) including distortion effects.

Passive sign convention: Convention in which absorbed power is positive when current enters the positive-voltage terminal.

Power-factor compensation: Local provision of reactive power to improve source-side power factor.

Phase sequence: Order in which three-phase quantities reach corresponding phase positions.

Balanced system: System with equal phase magnitudes and 120° separation together with equal load impedances.

Line voltage: Voltage measured between two line conductors.

Phase voltage: Voltage across one phase of a source or load.

Wye connection: Three-phase connection with one terminal of each phase joined at a common point.

Delta connection: Closed three-branch connection between the three line conductors.

Neutral displacement: Voltage shift of a floating load neutral relative to the source neutral.

Two-wattmeter method: Technique using two wattmeters to measure total real power in a three-wire system.

Review Questions

Short Answer Questions

  1. Distinguish instantaneous, average, reactive, and apparent power.
  2. Why is complex conjugation applied to current in the complex-power formula?
  3. State the line-to-phase voltage and current relationships for balanced wye and delta loads.
  4. Why can an unbalanced four-wire wye load carry neutral current?
  5. Distinguish displacement power factor from true power factor.

Long Answer Questions

  1. Derive the instantaneous and average power expressions for sinusoidal voltage and current.
  2. Explain how power-factor correction affects current, losses, voltage regulation, and equipment capacity.
  3. Derive the balanced three-phase power formula using both wye and delta connections.
  4. Explain the two-wattmeter method, including its applicability and sign limitations.
  5. Compare the analysis of unbalanced four-wire wye, isolated-neutral wye, and delta loads.

Analytical / Application Questions

  1. A 230 V single-phase load absorbs 5 kW at 0.70 lagging power factor. Calculate current, reactive power, apparent power, and the capacitor rating required to reach 0.95 lagging.
  2. A balanced delta load of 12+j9 Ω per phase is supplied at 415 V line voltage. Determine branch currents, line-current magnitude, and total complex power.
  3. Three unequal impedances are connected in wye to a balanced source without a neutral conductor. Develop a calculation procedure for neutral displacement, phase voltages, currents, and total complex power.

Multiple Choice Questions with Answers

Question 1

A sinusoidal voltage and current are v(t)=√2V cos(ωt) and i(t)=√2I cos(ωt-φ). Which expression gives instantaneous power?

A. p(t)=VI cos φ

B. p(t)=VI[cos φ+cos(2ωt-φ)]

C. p(t)=VI sin(2ωt-φ)

D. p(t)=2VI cos(ωt-φ)

Correct Answer: B. p(t)=VI[cos φ+cos(2ωt-φ)]

Question 2

A 230 V RMS load draws 10 A at 0.8 lagging power factor. What complex power does it absorb?

A. 1.84-j1.38 kVA

B. 2.30+j1.84 kVA

C. 1.84+j1.38 kVA

D. 2.30+j1.38 kVA

Correct Answer: C. 1.84+j1.38 kVA

Question 3

Which statement correctly describes shunt-capacitor compensation of an inductive load?

A. It increases the load's real-power requirement.

B. It supplies negative reactive power and can reduce source current.

C. It always eliminates harmonic distortion.

D. It changes the supply frequency.

Correct Answer: B. It supplies negative reactive power and can reduce source current.

Question 4

A balanced wye load is supplied from a 400 V line-to-line source. What is the RMS voltage across each phase impedance?

A. 133.3 V

B. 230.9 V

C. 400 V

D. 692.8 V

Correct Answer: B. 230.9 V

Question 5

A balanced delta load has a branch-current magnitude of 10 A. What is its line-current magnitude?

A. 5.77 A

B. 10.0 A

C. 14.14 A

D. 17.32 A

Correct Answer: D. 17.32 A

Question 6

A balanced three-phase load draws 10 A from a 400 V line-to-line supply at 0.8 lagging power factor. What real power does it absorb?

A. 3.20 kW

B. 5.54 kW

C. 6.93 kW

D. 9.60 kW

Correct Answer: B. 5.54 kW

Question 7

Which statement about the two-wattmeter method is correct?

A. The wattmeter sum gives total real power for a three-wire load, including an unbalanced load.

B. It requires an accessible neutral conductor.

C. Its wattmeter difference always gives reactive power for any unbalanced load.

D. Both readings must remain positive at every power factor.

Correct Answer: A. The wattmeter sum gives total real power for a three-wire load, including an unbalanced load.

Question 8

For an unbalanced four-wire wye load with ideal neutral and line currents directed toward the load, how is the returning neutral-current phasor found?

A. By adding the three current magnitudes arithmetically

B. By multiplying one phase current by three

C. By taking the phasor sum of the three line currents

D. By dividing the largest phase current by √3

Correct Answer: C. By taking the phasor sum of the three line currents

Question 9

Approximately what capacitance per delta branch supplies 10 kvar at 400 V and 50 Hz in a balanced capacitor bank?

A. 22.1 μF

B. 38.3 μF

C. 66.3 μF

D. 198.9 μF

Correct Answer: C. 66.3 μF

Question 10

For distorted periodic voltage and current, which expression defines true power factor?

A. cos φ1 only

B. Q/P

C. VrmsIrms/P

D. P/(VrmsIrms)

Correct Answer: D. P/(VrmsIrms)

Electrical Circuit Analysis
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