2.3 AC Power Relationships (Real, Reactive, Apparent Power & Power Factor Correction)

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1 Chapter 1: Exam Overview & Power Engineering Fundamentals

1.1 PE Power Exam Structure, Official Specification & NCEES Reference Strategy 1.2 Units, Conversions, Engineering Constants & Fundamental Power Equations 1.3 Measurement & Metering Principles (Power, Energy, Power Factor) 1.4 Instrument Testing & Diagnostics (Insulation Megger & Ground Resistance)

2 Chapter 2: Circuit Analysis Fundamentals

2.1 DC Circuit Laws, Network Theorems (Thevenin, Norton, Superposition) & Transients 2.2 Single-Phase AC Steady-State Analysis, Phasors & Impedance 2.3 AC Power Relationships (Real, Reactive, Apparent Power & Power Factor Correction) 2.4 Resonance, Frequency Response & Harmonic Distortion

3 Chapter 3: Three-Phase Systems & Symmetrical Components

3.1 Three-Phase Circuit Analysis (Wye and Delta Configurations & Transformations) 3.2 Balanced Three-Phase Power Calculations & Phasor Relationships 3.3 The Per-Unit (pu) System: Bases, Conversions & System Modeling 3.4 Symmetrical Components (Fortescue Transformation, Sequence Networks & Unbalance)

4 Chapter 4: Electric Power Transformers

4.1 Transformer Theory, Equivalent Circuits & Open/Short-Circuit Testing 4.2 Transformer Voltage Regulation, Efficiency & Loss Calculations 4.3 Three-Phase Transformer Connections, Phase Shifts & Paralleling 4.4 Autotransformers, Instrument Transformers (CTs & VTs) & Special Transformers

5 Chapter 5: Rotating Machines: Induction & Synchronous

5.1 Three-Phase Induction Motors: Principles, Equivalent Circuits & Slip-Torque Characteristics 5.2 Induction Motor Starting, Speed Control & Braking Methods 5.3 Synchronous Generators: Operation, Excitation, Power-Angle Equations & V-Curves 5.4 Synchronous Motors & Condensers for Power Factor Correction

6 Chapter 6: DC Machines & Power Devices

6.1 DC Generators & Motors: Types (Shunt, Series, Compound), Characteristics & Speed Control 6.2 Power Capacitors, Shunt Reactors & Harmonic Filter Design 6.3 Battery Energy Storage Systems (BESS) & Ultra-Capacitors 6.4 Renewable Generation Integration: Solar Photovoltaics (PV) & Wind Power Systems

7 Chapter 7: Power Electronics & Industrial Controls

7.1 Power Semiconductor Switching Devices (Diodes, Thyristors, MOSFETs, IGBTs) 7.2 AC-to-DC Converters (Uncontrolled & Controlled Rectifiers) 7.3 DC-to-DC Switch-Mode Converters (Buck, Boost, Buck-Boost) & DC-AC Inverters 7.4 Variable Frequency Drives (VFDs), Motor Starters & Ladder Logic / PLC Controls

8 Chapter 8: Transmission & Distribution Systems

8.1 Transmission Line Parameters (Resistance, Inductance, Capacitance & GMD/GMR) 8.2 Line Models (Short, Medium Nominal-π, Long Lines) & ABCD Two-Port Parameters 8.3 Surge Impedance, Surge Impedance Loading (SIL) & Voltage Regulation 8.4 Distribution System Architectures, Voltage Drops & Feeder Loading 8.5 Power Flow Studies: Bus Classification, Y-Bus & Newton-Raphson Solutions 8.6 Power System Stability: Swing Equation, Equal-Area Criterion & Critical Clearing Time

9 Chapter 9: Power System Fault Analysis

9.1 Fault Analysis Fundamentals & Subtransient/Transient Machine Reactances 9.2 Three-Phase Symmetrical Fault Calculations 9.3 Single Line-to-Ground (SLG) & Line-to-Line (LL) Fault Analysis 9.4 Double Line-to-Ground (DLG) Faults & Bus Impedance Matrix (Zbus) Methods

10 Chapter 10: Power System Protection & Coordination

10.1 Protective Relaying Principles, IEEE Device Designations & Instrument Transformers 10.2 Overcurrent Protection (50/51), Time-Current Curves (TCC) & Relay Coordination 10.3 Differential Protection (87) for Transformers, Busbars & Generators 10.4 Distance / Impedance Protection (21) for Transmission Lines & Directional Relays (67)

11 Chapter 11: National Electrical Code (NEC) & Installation Design

11.1 NEC Fundamentals, Branch Circuit & Feeder Conductor Sizing (Ampacity & Derating) 11.2 Overcurrent Protection Sizing, Raceway Fill & Box Fill Calculations 11.3 Motor Branch Circuits, Controllers & Feeder Protection (NEC Article 430) 11.4 Grounding and Bonding Systems per NEC Article 250

12 Chapter 12: Electrical Safety, Standards & Facility Applications

12.1 Hazardous (Classified) Locations: Classes, Divisions, Zones, T-Codes & Sealing (NEC Articles 500-506) 12.2 Electrical Safety in the Workplace: NFPA 70E & Arc Flash Hazard Calculations 12.3 National Electrical Safety Code (NESC) Clearance & Utility Safety Requirements 12.4 Facility Applications: Illumination Design, Demand/Connected Loads & Lightning/Surge Protection 12.5 Special Occupancies & Systems: Health Care, Fire Pumps, Pools, Emergency & Standby (NEC Chapters 5-7)

2.3 AC Power Relationships (Real, Reactive, Apparent Power & Power Factor Correction)

Key Takeaways Explain This Topic AI

The Power Triangle relates Real Power ( P = V I cos ⁡ θ P = V I \cos\theta P = V I cos θ , Watts), Reactive Power ( Q = V I sin ⁡ θ Q = V I \sin\theta Q = V I sin θ , VARs), and Apparent Power ( S = V I S = V I S = V I , VA) through S = P 2 + Q 2 S = \sqrt{P^2 + Q^2} S = P 2 + Q 2 ​ and S = P + j Q \mathbf{S} = P + jQ S = P + j Q .

Complex power is defined rigorously as S = V I ∗ = ∣ I ∣ 2 Z = ∣ V ∣ 2 / Z ∗ \mathbf{S} = \mathbf{V} \mathbf{I}^* = |I|^2 \mathbf{Z} = |V|^2 / \mathbf{Z}^* S = V I ∗ = ∣ I ∣ 2 Z = ∣ V ∣ 2 / Z ∗ , where the complex conjugate of current ( I ∗ \mathbf{I}^* I ∗ ) ensures that inductive loads with lagging current consume positive reactive power ( + Q +Q + Q ).

Power Factor ( PF = cos ⁡ θ = P / S \text{PF} = \cos\theta = P/S PF = cos θ = P / S ) is lagging for inductive loads (current lags voltage, absorbing VARs) and leading for capacitive loads (current leads voltage, generating VARs).

Capacitor bank sizing for power factor correction uses Q c = P ( tan ⁡ θ 1 − tan ⁡ θ 2 ) Q_c = P(\tan\theta_1 - \tan\theta_2) Q c ​ = P ( tan θ 1 ​ − tan θ 2 ​ ) , requiring capacitance C = Q c / ( ω V 2 ) C = Q_c / (\omega V^2) C = Q c ​ / ( ω V 2 ) per phase, which alters reactive power without changing active power P P P .

Power factor correction reduces total line current ( I = S / V I = S/V I = S / V ), minimizes I 2 R I^2R I 2 R copper losses in upstream conductors and transformers, mitigates feeder voltage drop, releases electrical capacity headroom, and avoids utility low-PF economic penalties.

Last updated: August 2026

AC Power Relationships & Power Factor Correction

In alternating current (AC) power systems, electrical energy is delivered through the interaction of time-varying voltages and currents. Because practical industrial and commercial loads contain both resistive and reactive elements (primarily inductive windings in motors, transformers, and induction furnaces), power engineers must analyze active power, reactive power, apparent power, and power factor.

Power factor correction is one of the most frequently tested topics on the NCEES PE Electrical: Power examination, requiring candidates to design capacitor banks, calculate released capacity, quantify conductor loss reductions, and analyze voltage regulation improvements.

1. Instantaneous Power & The Derivation of AC Power Components

Consider an AC circuit excited by sinusoidal voltage v ( t ) = 2 V cos ⁡ ( ω t ) v(t) = \sqrt{2} V \cos(\omega t) v ( t ) = 2 ​ V cos ( ω t ) delivering current i ( t ) = 2 I cos ⁡ ( ω t − θ ) i(t) = \sqrt{2} I \cos(\omega t - \theta) i ( t ) = 2 ​ I cos ( ω t − θ ) , where V V V and I I I are RMS magnitudes and θ = θ v − θ i \theta = \theta_v - \theta_i θ = θ v ​ − θ i ​ is the impedance phase angle.

The instantaneous power p ( t ) = v ( t ) ⋅ i ( t ) p(t) = v(t) \cdot i(t) p ( t ) = v ( t ) ⋅ i ( t ) is: p ( t ) = 2 V I cos ⁡ ( ω t ) cos ⁡ ( ω t − θ ) p(t) = 2 V I \cos(\omega t) \cos(\omega t - \theta) p ( t ) = 2 V I cos ( ω t ) cos ( ω t − θ )

Applying the trigonometric product-to-sum identity cos ⁡ ( A ) cos ⁡ ( B ) = 1 2 [ cos ⁡ ( A − B ) + cos ⁡ ( A + B ) ] \cos(A)\cos(B) = \frac{1}{2}[\cos(A-B) + \cos(A+B)] cos ( A ) cos ( B ) = 2 1 ​ [ cos ( A − B ) + cos ( A + B )] : p ( t ) = V I cos ⁡ θ + V I cos ⁡ ( 2 ω t − θ ) p(t) = V I \cos\theta + V I \cos(2\omega t - \theta) p ( t ) = V I cos θ + V I cos ( 2 ω t − θ )

Expanding cos ⁡ ( 2 ω t − θ ) = cos ⁡ ( 2 ω t ) cos ⁡ θ + sin ⁡ ( 2 ω t ) sin ⁡ θ \cos(2\omega t - \theta) = \cos(2\omega t)\cos\theta + \sin(2\omega t)\sin\theta cos ( 2 ω t − θ ) = cos ( 2 ω t ) cos θ + sin ( 2 ω t ) sin θ : p ( t ) = V I cos ⁡ θ [ 1 + cos ⁡ ( 2 ω t ) ] ⏟ Real Power Component (Always ≥ 0 ) − V I sin ⁡ θ sin ⁡ ( 2 ω t ) ⏟ Reactive Power Component (Alternating Swing) p(t) = \underbrace{V I \cos\theta \left[1 + \cos(2\omega t)\right]}_{\text{Real Power Component (Always } \ge 0\text{)}} - \underbrace{V I \sin\theta \sin(2\omega t)}_{\text{Reactive Power Component (Alternating Swing)}} p ( t ) = Real Power Component (Always ≥ 0 ) V I cos θ [ 1 + cos ( 2 ω t ) ] ​ ​ − Reactive Power Component (Alternating Swing) V I sin θ sin ( 2 ω t ) ​ ​ +-----------------------------------------------------------------------------------------+

• | THE THREE AC POWER COMPONENTS |
• | |
• | 1. REAL / ACTIVE POWER (P): |
• | - P = V * I * cos(theta) (Watts, kW, MW) |
• | - Net unidirectional energy transferred to the load per unit time. |
• | - Converted into mechanical work, heat, light, or chemical energy. |
• | |
• | 2. REACTIVE POWER (Q): |
• | - Q = V * I * sin(theta) (VAR, kVAR, MVAR - Volt-Amperes Reactive) |
• | - Energy that continuously oscillates between source and load magnetic/electric |
• | fields at twice the line frequency (2*omega). Net average real work = 0. |
• | - Sustains electromagnetic flux in motors and transformers. |
• | |
• | 3. APPARENT POWER (S): |
• | - S = V * I (Volt-Amperes, kVA, MVA) |
• | - Total rating capacity required for electrical equipment (generators, cables, |
• | transformers) to deliver both P and Q without thermal overload. |
• +-----------------------------------------------------------------------------------------+
• 2. The Power Triangle & Complex Power Equations

The geometrical relationship between P P P , Q Q Q , and S S S forms the Power Triangle in the complex plane: THE POWER TRIANGLE (INDUCTIVE / LAGGING LOAD)

• + ^ +j (Reactive Power Q, +kVAR)
• /| |
• / | | * S = P + jQ = |S| /_ theta
• / | | /|
• Apparent Power / | Reactive Power | / |
• S (kVA) / | Q (kVAR) | S / | Q (+kVAR, Inductive)
• / | | / |
• / theta| | /theta|
• +-------+ +---+-----+-----> +Re (Real Power P, kW)
• Real Power P (kW) | P (kW)

Mathematical Formulation of the Power Triangle S = P 2 + Q 2 S = \sqrt{P^2 + Q^2} S = P 2 + Q 2 ​ P = S cos ⁡ θ = ∣ S ∣ cos ⁡ θ P = S \cos\theta = |\mathbf{S}| \cos\theta P = S cos θ = ∣ S ∣ cos θ Q = S sin ⁡ θ = ∣ S ∣ sin ⁡ θ Q = S \sin\theta = |\mathbf{S}| \sin\theta Q = S sin θ = ∣ S ∣ sin θ tan ⁡ θ = Q P ⟹ θ = arctan ⁡ ( Q P ) \tan\theta = \frac{Q}{P} \implies \theta = \arctan\left(\frac{Q}{P}\right) tan θ = P Q ​ ⟹ θ = arctan ( P Q ​ )

Complex Power ( S \mathbf{S} S ) and the Conjugate Current Rule

Complex power S \mathbf{S} S combines real and reactive power into a single complex variable: S = V I ∗ = P + j Q = ∣ S ∣ ∠ θ \mathbf{S} = \mathbf{V} \mathbf{I}^* = P + jQ = |S|\angle\theta S = V I ∗ = P + j Q = ∣ S ∣∠ θ

Where:

V = V ∠ θ v \mathbf{V} = V\angle\theta_v V = V ∠ θ v ​ = RMS voltage phasor.

I ∗ = I ∠ ( − θ i ) \mathbf{I}^* = I\angle(-\theta_i) I ∗ = I ∠ ( − θ i ​ ) = Complex conjugate of the RMS current phasor.

θ = θ v − θ i \theta = \theta_v - \theta_i θ = θ v ​ − θ i ​ = Impedance angle / power factor angle. +-----------------------------------------------------------------------------------------+

• | WHY THE COMPLEX CONJUGATE (I*) IS MANDATORY |
• | |
• | If S = V * I (without conjugate): |
• | - Angle would be (theta_v + theta_i), which depends on arbitrary reference angle! |
• | |
• | With S = V * I*: |
• | - Angle is (theta_v - theta_i) = theta_impedance (an absolute physical property). |
• | - Inductive load (current lags voltage, theta_i < theta_v): |
• | theta = theta_v - theta_i > 0 -> S = P + jQ (POSITIVE Q absorbed). |
• | - Capacitive load (current leads voltage, theta_i > theta_v): |
• | theta = theta_v - theta_i < 0 -> S = P - jQ (NEGATIVE Q absorbed / generated). |
• +-----------------------------------------------------------------------------------------+

Alternative formulas for complex power using impedance Z = R + j X \mathbf{Z} = R + jX Z = R + j X : S = ∣ I ∣ 2 Z = I r m s 2 R + j I r m s 2 X \mathbf{S} = |\mathbf{I}|^2 \mathbf{Z} = I_{rms}^2 R + j I_{rms}^2 X S = ∣ I ∣ 2 Z = I r m s 2 ​ R + j I r m s 2 ​ X S = ∣ V ∣ 2 Z ∗ = V r m s 2 R − j X \mathbf{S} = \frac{|\mathbf{V}|^2}{\mathbf{Z}^*} = \frac{V_{rms}^2}{R - jX} S = Z ∗ ∣ V ∣ 2 ​ = R − j X V r m s 2 ​ ​

3. Power Factor: Lagging vs. Leading Conventions

Power Factor (PF) is the ratio of real active power to total apparent power: PF = P S = P P 2 + Q 2 = cos ⁡ θ \text{PF} = \frac{P}{S} = \frac{P}{\sqrt{P^2 + Q^2}} = \cos\theta PF = S P ​ = P 2 + Q 2 ​ P ​ = cos θ

Power Factor Type Current vs. Voltage Phase Load Nature Reactive Power ( Q Q Q ) Typical Equipment

Lagging PF Current lags Voltage ( θ i < θ v \theta_i < \theta_v θ i ​ < θ v ​ ) Inductive Absorbs + Q +Q + Q ( + j VAR +j\text{VAR} + j VAR ) Induction motors, transformers, ballasts

Leading PF Current leads Voltage ( θ i > θ v \theta_i > \theta_v θ i ​ > θ v ​ ) Capacitive Generates + Q +Q + Q (absorbs − Q -Q − Q ) Capacitor banks, overexcited synchronous motors

Unity PF ( 1.0 1.0 1.0 ) Current in phase with Voltage ( θ i = θ v \theta_i = \theta_v θ i ​ = θ v ​ ) Purely Resistive Q = 0 Q = 0 Q = 0 Electric heaters, incandescent lighting

4. Power Factor Correction Engineering & Capacitor Sizing

Industrial facilities with large inductive motor loads typically operate at low lagging power factors ( 0.70 − 0.85 0.70 - 0.85 0.70 − 0.85 lagging). Connecting shunt capacitor banks in parallel with the load supplies the required magnetizing reactive power ( Q c Q_c Q c ​ ) locally, reducing the total reactive power drawn from the utility. POWER FACTOR CORRECTION MECHANISM

Uncorrected Load (PF_1) Corrected System (PF_2)

• + +
• /| /|
• / | / |
• S_1/ | Q_1 S_2/ | Q_2 = Q_1 - Q_c
• / | / |
• /th1 | /th2 |
• +-----+ +-----+
• P (Unchanged) P (Unchanged)
• |
• +---[ Q_c = Q_1 - Q_2 ]
• Step-by-Step Capacitor Sizing Derivation

Initial system state at uncorrected power factor PF 1 = cos ⁡ θ 1 \text{PF}_1 = \cos\theta_1 PF 1 ​ = cos θ 1 ​ : θ 1 = arccos ⁡ ( PF 1 ) , Q 1 = P tan ⁡ θ 1 \theta_1 = \arccos(\text{PF}_1), \qquad Q_1 = P \tan\theta_1 θ 1 ​ = arccos ( PF 1 ​ ) , Q 1 ​ = P tan θ 1 ​

Target system state at improved power factor PF 2 = cos ⁡ θ 2 \text{PF}_2 = \cos\theta_2 PF 2 ​ = cos θ 2 ​ : θ 2 = arccos ⁡ ( PF 2 ) , Q 2 = P tan ⁡ θ 2 \theta_2 = \arccos(\text{PF}_2), \qquad Q_2 = P \tan\theta_2 θ 2 ​ = arccos ( PF 2 ​ ) , Q 2 ​ = P tan θ 2 ​

Required capacitor bank reactive power ( Q c Q_c Q c ​ in VAR \text{VAR} VAR or kVAR \text{kVAR} kVAR ): Q c = Q 1 − Q 2 = P ( tan ⁡ θ 1 − tan ⁡ θ 2 ) Q_c = Q_1 - Q_2 = P\left(\tan\theta_1 - \tan\theta_2\right) Q c ​ = Q 1 ​ − Q 2 ​ = P ( tan θ 1 ​ − tan θ 2 ​ )

Required capacitance ( C C C in Farads) for a single-phase system with operating voltage V r m s V_{rms} V r m s ​ : Q c = V r m s 2 X C = V r m s 2 ( ω C ) = 2 π f V r m s 2 C Q_c = \frac{V_{rms}^2}{X_C} = V_{rms}^2 (\omega C) = 2\pi f V_{rms}^2 C Q c ​ = X C ​ V r m s 2 ​ ​ = V r m s 2 ​ ( ω C ) = 2 π f V r m s 2 ​ C C = Q c ω V r m s 2 = Q c 2 π f V r m s 2 C = \frac{Q_c}{\omega V_{rms}^2} = \frac{Q_c}{2\pi f V_{rms}^2} C = ω V r m s 2 ​ Q c ​ ​ = 2 π f V r m s 2 ​ Q c ​ ​

Tip

Active Power Invariance: Adding ideal shunt capacitors does NOT change the real active power P P P demanded by the facility's mechanical processes ( P 1 = P 2 = P P_1 = P_2 = P P 1 ​ = P 2 ​ = P ). Only the reactive power ( Q Q Q ) and apparent power ( S S S ) are reduced.

5. Operational & Economic Benefits of PF Correction

Improving power factor delivers substantial engineering and economic advantages: +-----------------------------------------------------------------------------------------+

• | BENEFITS OF POWER FACTOR CORRECTION |
• | |
• | 1. LINE CURRENT REDUCTION: |
• | - Current is inversely proportional to PF: I = P / (V * PF) |
• | - Current reduction ratio: I_2 / I_1 = PF_1 / PF_2 = S_2 / S_1 |
• | |
• | 2. COPPER LOSS REDUCTION (I^2 * R): |
• | - Conductor & transformer thermal power losses decrease by: |
• | P_loss,2 / P_loss,1 = (I_2 / I_1)^2 = (PF_1 / PF_2)^2 |
• | - Example: Correcting PF from 0.70 to 0.95 reduces conductor losses by 45.7%! |
• | |
• | 3. VOLTAGE REGULATION IMPROVEMENT: |
• | - Feeder voltage drop approximation: Delta V approx I * (R*cos(theta) + X*sin(theta))|
• | - Reducing reactive current dramatically cuts feeder voltage drop. |
• | |
• | 4. RELEASE OF SYSTEM CAPACITY (HEADROOM): |
• | - Released kVA capacity = S_1 - S_2 = P * (1/PF_1 - 1/PF_2) |
• | - Allows existing transformers/switchgear to serve additional plant expansion. |
• | |
• | 5. ELIMINATION OF UTILITY TARIFF PENALTIES: |
• | - Electric utilities impose billing surcharges or peak kVA demand charges when |
• | customer power factor drops below a contractual threshold (typically 0.90-0.95). |
• +-----------------------------------------------------------------------------------------+
• 6. Comprehensive Worked Mathematical Examples
• Example 1: Multi-Load Facility Power Aggregation

Problem: A 480 V 480\text{ V} 480 V , 60 Hz 60\text{ Hz} 60 Hz single-phase industrial feeder supplies three parallel loads:

Load 1: 120 kW 120\text{ kW} 120 kW heating load at unity power factor ( PF = 1.0 \text{PF} = 1.0 PF = 1.0 ).

Load 2: 200 kW 200\text{ kW} 200 kW induction motor load operating at 0.80 lagging 0.80\text{ lagging} 0.80 lagging power factor.

Load 3: 80 kVA 80\text{ kVA} 80 kVA synchronous motor operating at 0.75 leading 0.75\text{ leading} 0.75 leading power factor.

Calculate the total real power P t o t a l P_{total} P t o t a l ​ , total reactive power Q t o t a l Q_{total} Q t o t a l ​ , total apparent power S t o t a l S_{total} S t o t a l ​ , overall facility power factor, and total line current I l i n e I_{line} I l in e ​ .

Step 1: Compute Power Components for Each Load

Load 1:

• P 1 = 120.0 kW P_1 = 120.0\text{ kW} P 1 ​ = 120.0 kW
• Q 1 = 0.0 kVAR Q_1 = 0.0\text{ kVAR} Q 1 ​ = 0.0 kVAR
• S 1 = 120.0 + j 0.0 kVA \mathbf{S}_1 = 120.0 + j0.0\text{ kVA} S 1 ​ = 120.0 + j 0.0 kVA

Load 2:

P 2 = 200.0 kW P_2 = 200.0\text{ kW} P 2 ​ = 200.0 kW

S 2 = P 2 PF 2 = 200.0 0.80 = 250.0 kVA S_2 = \frac{P_2}{\text{PF}_2} = \frac{200.0}{0.80} = 250.0\text{ kVA} S 2 ​ = PF 2 ​ P 2 ​ ​ = 0.80 200.0 ​ = 250.0 kVA

θ 2 = arccos ⁡ ( 0.80 ) = + 36.87 ∘ \theta_2 = \arccos(0.80) = +36.87^\circ θ 2 ​ = arccos ( 0.80 ) = + 36.8 7 ∘

Q 2 = P 2 tan ⁡ θ 2 = 200.0 × tan ⁡ ( 36.87 ∘ ) = + 150.0 kVAR Q_2 = P_2 \tan\theta_2 = 200.0 \times \tan(36.87^\circ) = +150.0\text{ kVAR} Q 2 ​ = P 2 ​ tan θ 2 ​ = 200.0 × tan ( 36.8 7 ∘ ) = + 150.0 kVAR (Inductive, + j +j + j )

S 2 = 200.0 + j 150.0 kVA \mathbf{S}_2 = 200.0 + j150.0\text{ kVA} S 2 ​ = 200.0 + j 150.0 kVA

Load 3:

S 3 = 80.0 kVA S_3 = 80.0\text{ kVA} S 3 ​ = 80.0 kVA

PF 3 = 0.75 leading ⟹ θ 3 = − arccos ⁡ ( 0.75 ) = − 41.41 ∘ \text{PF}_3 = 0.75\text{ leading} \implies \theta_3 = -\arccos(0.75) = -41.41^\circ PF 3 ​ = 0.75 leading ⟹ θ 3 ​ = − arccos ( 0.75 ) = − 41.4 1 ∘

P 3 = S 3 cos ⁡ θ 3 = 80.0 × 0.75 = 60.0 kW P_3 = S_3 \cos\theta_3 = 80.0 \times 0.75 = 60.0\text{ kW} P 3 ​ = S 3 ​ cos θ 3 ​ = 80.0 × 0.75 = 60.0 kW

Q 3 = − S 3 sin ⁡ θ 3 = − 80.0 × sin ⁡ ( 41.41 ∘ ) = − 52.915 kVAR Q_3 = -S_3 \sin\theta_3 = -80.0 \times \sin(41.41^\circ) = -52.915\text{ kVAR} Q 3 ​ = − S 3 ​ sin θ 3 ​ = − 80.0 × sin ( 41.4 1 ∘ ) = − 52.915 kVAR (Capacitive, − j -j − j )

S 3 = 60.0 − j 52.915 kVA \mathbf{S}_3 = 60.0 - j52.915\text{ kVA} S 3 ​ = 60.0 − j 52.915 kVA

Step 2: Sum Real and Reactive Powers P t o t a l = P 1 + P 2 + P 3 = 120.0 + 200.0 + 60.0 = 380.0 kW P_{total} = P_1 + P_2 + P_3 = 120.0 + 200.0 + 60.0 = 380.0\text{ kW} P t o t a l ​ = P 1 ​ + P 2 ​ + P 3 ​ = 120.0 + 200.0 + 60.0 = 380.0 kW Q t o t a l = Q 1 + Q 2 + Q 3 = 0.0 + 150.0 + ( − 52.915 ) = + 97.085 kVAR Q_{total} = Q_1 + Q_2 + Q_3 = 0.0 + 150.0 + (-52.915) = +97.085\text{ kVAR} Q t o t a l ​ = Q 1 ​ + Q 2 ​ + Q 3 ​ = 0.0 + 150.0 + ( − 52.915 ) = + 97.085 kVAR S t o t a l = 380.0 + j 97.085 kVA \mathbf{S}_{total} = 380.0 + j97.085\text{ kVA} S t o t a l ​ = 380.0 + j 97.085 kVA

Step 3: Calculate Total Apparent Power & Power Factor S t o t a l = P t o t a l 2 + Q t o t a l 2 = 380.0 2 + 97.085 2 = 144400 + 9425.5 = 153825.5 = 392.21 kVA S_{total} = \sqrt{P_{total}^2 + Q_{total}^2} = \sqrt{380.0^2 + 97.085^2} = \sqrt{144400 + 9425.5} = \sqrt{153825.5} = 392.21\text{ kVA} S t o t a l ​ = P t o t a l 2 ​ + Q t o t a l 2 ​ ​ = 380. 0 2 + 97.08 5 2 ​ = 144400 + 9425.5 ​ = 153825.5 ​ = 392.21 kVA PF t o t a l = P t o t a l S t o t a l = 380.0 392.21 = 0.9689 ≈ 0.969 Lagging (since Q > 0 ) \text{PF}_{total} = \frac{P_{total}}{S_{total}} = \frac{380.0}{392.21} = 0.9689 \approx 0.969\text{ Lagging (since } Q > 0\text{)} PF t o t a l ​ = S t o t a l ​ P t o t a l ​ ​ = 392.21 380.0 ​ = 0.9689 ≈ 0.969 Lagging (since Q > 0 )

Step 4: Calculate Feeder Line Current I l i n e = S t o t a l V = 392,210 VA 480 V = 817.10 A r m s I_{line} = \frac{S_{total}}{V} = \frac{392{,}210\text{ VA}}{480\text{ V}} = 817.10\text{ A}_{rms} I l in e ​ = V S t o t a l ​ ​ = 480 V 392 , 210 VA ​ = 817.10 A r m s ​

Example 2: Industrial Capacitor Bank Sizing & Loss Reduction

Problem: An industrial manufacturing plant operates at 480 V 480\text{ V} 480 V , 60 Hz 60\text{ Hz} 60 Hz single-phase and consumes P = 600 kW P = 600\text{ kW} P = 600 kW at an uncorrected power factor of PF 1 = 0.72 lagging \text{PF}_1 = 0.72\text{ lagging} PF 1 ​ = 0.72 lagging .

Calculate the required capacitor bank rating ( Q c Q_c Q c ​ in kVAR \text{kVAR} kVAR ) and capacitance ( C C C in μ F \mu\text{F} μ F ) to improve the plant power factor to PF 2 = 0.95 lagging \text{PF}_2 = 0.95\text{ lagging} PF 2 ​ = 0.95 lagging .

Determine the percentage reduction in feeder line current and conductor I 2 R I^2R I 2 R power loss.

Step 1: Calculate Initial and Target Angles

Initial angle: θ 1 = arccos ⁡ ( 0.72 ) = 43.945 ∘ ⟹ tan ⁡ θ 1 = tan ⁡ ( 43.945 ∘ ) = 0.9638 \theta_1 = \arccos(0.72) = 43.945^\circ \implies \tan\theta_1 = \tan(43.945^\circ) = 0.9638 θ 1 ​ = arccos ( 0.72 ) = 43.94 5 ∘ ⟹ tan θ 1 ​ = tan ( 43.94 5 ∘ ) = 0.9638

Target angle: θ 2 = arccos ⁡ ( 0.95 ) = 18.195 ∘ ⟹ tan ⁡ θ 2 = tan ⁡ ( 18.195 ∘ ) = 0.3287 \theta_2 = \arccos(0.95) = 18.195^\circ \implies \tan\theta_2 = \tan(18.195^\circ) = 0.3287 θ 2 ​ = arccos ( 0.95 ) = 18.19 5 ∘ ⟹ tan θ 2 ​ = tan ( 18.19 5 ∘ ) = 0.3287

Step 2: Compute Required Capacitor Bank kVAR ( Q c Q_c Q c ​ ) Q c = P ( tan ⁡ θ 1 − tan ⁡ θ 2 ) = 600 kW × ( 0.9638 − 0.3287 ) = 600 × 0.6351 = 381.07 kVAR Q_c = P\left(\tan\theta_1 - \tan\theta_2\right) = 600\text{ kW} \times (0.9638 - 0.3287) = 600 \times 0.6351 = 381.07\text{ kVAR} Q c ​ = P ( tan θ 1 ​ − tan θ 2 ​ ) = 600 kW × ( 0.9638 − 0.3287 ) = 600 × 0.6351 = 381.07 kVAR

Step 3: Calculate Required Capacitance ( C C C ) C = Q c 2 π f V 2 = 381,070 VAR 2 π ( 60 ) ( 480 V ) 2 = 381,070 376.991 × 230,400 = 381,070 86,858,726 = 4.387 × 10 − 3 F = 4387 μ F C = \frac{Q_c}{2\pi f V^2} = \frac{381{,}070\text{ VAR}}{2\pi(60)(480\text{ V})^2} = \frac{381{,}070}{376.991 \times 230{,}400} = \frac{381{,}070}{86{,}858{,}726} = 4.387 \times 10^{-3}\text{ F} = 4387\ \mu\text{F} C = 2 π f V 2 Q c ​ ​ = 2 π ( 60 ) ( 480 V ) 2 381 , 070 VAR ​ = 376.991 × 230 , 400 381 , 070 ​ = 86 , 858 , 726 381 , 070 ​ = 4.387 × 1 0 − 3 F = 4387 μ F

Step 4: Calculate Current and Conductor Loss Reductions

Initial apparent power: S 1 = 600 kW 0.72 = 833.33 kVA ⟹ I 1 = 833,333 480 = 1736.1 A S_1 = \frac{600\text{ kW}}{0.72} = 833.33\text{ kVA} \implies I_1 = \frac{833{,}333}{480} = 1736.1\text{ A} S 1 ​ = 0.72 600 kW ​ = 833.33 kVA ⟹ I 1 ​ = 480 833 , 333 ​ = 1736.1 A

Corrected apparent power: S 2 = 600 kW 0.95 = 631.58 kVA ⟹ I 2 = 631,580 480 = 1315.8 A S_2 = \frac{600\text{ kW}}{0.95} = 631.58\text{ kVA} \implies I_2 = \frac{631{,}580}{480} = 1315.8\text{ A} S 2 ​ = 0.95 600 kW ​ = 631.58 kVA ⟹ I 2 ​ = 480 631 , 580 ​ = 1315.8 A Line Current Reduction = ( 1 − I 2 I 1 ) × 100 % = ( 1 − 0.72 0.95 ) × 100 % = ( 1 − 0.7579 ) × 100 % = 24.21 % \text{Line Current Reduction} = \left(1 - \frac{I_2}{I_1}\right) \times 100\% = \left(1 - \frac{0.72}{0.95}\right) \times 100\% = \left(1 - 0.7579\right) \times 100\% = 24.21\% Line Current Reduction = ( 1 − I 1 ​ I 2 ​ ​ ) × 100% = ( 1 − 0.95 0.72 ​ ) × 100% = ( 1 − 0.7579 ) × 100% = 24.21% Feeder I 2 R Loss Ratio = ( I 2 I 1 ) 2 = ( 0.7579 ) 2 = 0.5744 \text{Feeder } I^2 R \text{ Loss Ratio} = \left(\frac{I_2}{I_1}\right)^2 = (0.7579)^2 = 0.5744 Feeder I 2 R Loss Ratio = ( I 1 ​ I 2 ​ ​ ) 2 = ( 0.7579 ) 2 = 0.5744 Feeder Loss Reduction = ( 1 − 0.5744 ) × 100 % = 42.56 % \text{Feeder Loss Reduction} = (1 - 0.5744) \times 100\% = 42.56\% Feeder Loss Reduction = ( 1 − 0.5744 ) × 100% = 42.56%

7. Common PE Exam Traps & Pitfalls

Forgetting the Current Conjugate in Complex Power: Calculating S = V I \mathbf{S} = \mathbf{V}\mathbf{I} S = VI instead of S = V I ∗ \mathbf{S} = \mathbf{V}\mathbf{I}^* S = V I ∗ flips the sign of reactive power, erroneously making inductive loads appear capacitive.

Altering Real Power During Power Factor Correction: Adding capacitors supplies reactive power only. Do not change real power P P P unless the capacitor is explicitly specified with internal dielectric resistance.

Overcorrecting Past Unity: Adding excessive capacitance ( Q c > Q 1 Q_c > Q_1 Q c ​ > Q 1 ​ ) drives the circuit into a leading power factor ( Q < 0 Q < 0 Q < 0 ), which can cause system overvoltages (Ferranti effect on lightly loaded cables) and unwanted resonance with distribution transformers.