2.2 Single-Phase AC Steady-State Analysis, Phasors & Impedance

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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)

2.2 Single-Phase AC Steady-State Analysis, Phasors & Impedance

Key Takeaways Explain This Topic AI

Sinusoidal AC power quantities in North America operate at frequency f = 60 Hz f = 60\text{ Hz} f = 60 Hz ( ω = 2 π f = 376.99 rad/s ≈ 377 rad/s \omega = 2\pi f = 376.99\text{ rad/s} \approx 377\text{ rad/s} ω = 2 π f = 376.99 rad/s ≈ 377 rad/s ), evaluated in Root-Mean-Square (RMS) values ( V r m s = V p e a k / 2 ≈ 0.707 V p e a k V_{rms} = V_{peak}/\sqrt{2} \approx 0.707 V_{peak} V r m s ​ = V p e ak ​ / 2 ​ ≈ 0.707 V p e ak ​ ).

Phasor transformation maps sinusoidal time-domain signals v ( t ) = V m cos ⁡ ( ω t + ϕ ) v(t) = V_m \cos(\omega t + \phi) v ( t ) = V m ​ cos ( ω t + ϕ ) into complex frequency-domain vectors V = V r m s ∠ ϕ = V r m s e j ϕ \mathbf{V} = V_{rms}\angle\phi = V_{rms}e^{j\phi} V = V r m s ​ ∠ ϕ = V r m s ​ e j ϕ , transforming differential equations into algebraic equations.

Impedance is defined as Z = R + j X \mathbf{Z} = R + jX Z = R + j X , where inductive reactance is X L = ω L X_L = \omega L X L ​ = ω L (producing + j ω L +j\omega L + jω L ) and capacitive reactance is X C = 1 / ( ω C ) X_C = 1/(\omega C) X C ​ = 1/ ( ω C ) (producing − j / ( ω C ) -j/(\omega C) − j / ( ω C ) ).

Phase angle relationships follow the mnemonic 'ELI the ICE man': voltage leads current by 90 degrees in pure inductors ( + 90 ∘ +90^\circ + 9 0 ∘ ), while current leads voltage by 90 degrees in pure capacitors ( − 90 ∘ -90^\circ − 9 0 ∘ ).

Admittance Y = 1 / Z = G + j B \mathbf{Y} = 1/\mathbf{Z} = G + jB Y = 1/ Z = G + j B simplifies parallel circuit calculations, where conductance G = R / ( R 2 + X 2 ) G = R/(R^2+X^2) G = R / ( R 2 + X 2 ) and susceptance B = − X / ( R 2 + X 2 ) B = -X/(R^2+X^2) B = − X / ( R 2 + X 2 ) (note inductive susceptance is negative, capacitive is positive).

Last updated: August 2026

Single-Phase AC Steady-State Analysis, Phasors & Impedance

Alternating Current (AC) steady-state analysis represents the core mathematical language of electric power systems. Generation, transmission, distribution, and utilization systems operate almost universally in AC at a synchronized power frequency (60 Hz in North America, 50 Hz in Europe and parts of Asia/South America).

The NCEES PE Power examination requires complete fluency in converting time-domain sinusoidal voltages and currents into the frequency-domain (phasor domain), manipulating complex numbers in rectangular and polar forms, calculating complex impedances and admittances, and applying network reduction laws to AC circuits.

1. Sinusoidal Waveforms, Frequency & RMS Values

A time-varying sinusoidal voltage is defined mathematically as: v ( t ) = V p e a k cos ⁡ ( ω t + θ v ) v(t) = V_{peak} \cos(\omega t + \theta_v) v ( t ) = V p e ak ​ cos ( ω t + θ v ​ )

Where:

V p e a k V_{peak} V p e ak ​ (or V m V_m V m ​ ) = Peak amplitude of the sinusoidal voltage (Volts).

ω = 2 π f \omega = 2\pi f ω = 2 π f = Angular frequency in radians per second ( rad/s \text{rad/s} rad/s ).

f f f = Cyclic frequency in Hertz ( Hz = s − 1 \text{Hz} = \text{s}^{-1} Hz = s − 1 ); in North America, f = 60 Hz ⟹ ω = 2 π ( 60 ) ≈ 376.991 rad/s ≈ 377 rad/s f = 60\text{ Hz} \implies \omega = 2\pi(60) \approx 376.991\text{ rad/s} \approx 377\text{ rad/s} f = 60 Hz ⟹ ω = 2 π ( 60 ) ≈ 376.991 rad/s ≈ 377 rad/s .

T = 1 / f T = 1/f T = 1/ f = Period of one full cycle; at 60 Hz, T = 1 / 60 s = 16.67 ms T = 1/60\text{ s} = 16.67\text{ ms} T = 1/60 s = 16.67 ms .

θ v \theta_v θ v ​ = Phase angle in radians or degrees. +-----------------------------------------------------------------------------------------+

• | AC VOLTAGE METRICS & CONVERSIONS |
• | |
• | PEAK VALUE (V_peak): |
• | - Maximum instantaneous crest value above the zero baseline. |
• | |
• | ROOT-MEAN-SQUARE (RMS / EFFECTIVE VALUE) (V_rms): |
• | - Produces identical average thermal dissipation in a resistor as equivalent DC: |
• | V_rms = sqrt( (1/T) * integral_0^T [v(t)]^2 dt ) = V_peak / sqrt(2) |
• | V_rms = 0.7071 * V_peak |
• | V_peak = sqrt(2) * V_rms = 1.4142 * V_rms |
• | |
• | AVERAGE VALUE OF RECTIFIED SINE (V_avg): |
• | - Average over a positive half-cycle: V_avg = (2 / pi) * V_peak = 0.6366 * V_peak |
• | - Pure full-period sine wave average is ZERO: V_avg,full = 0 |
• +-----------------------------------------------------------------------------------------+

Important

NCEES Reference Handbook Standard: Unless explicitly stated otherwise (e.g., "an instantaneous voltage of 170 cos ⁡ ( 377 t ) 170\cos(377t) 170 cos ( 377 t ) "), all AC voltages and currents given in the PE Power exam (e.g., "a 480 V 480\text{ V} 480 V feeder" or "a 120 V 120\text{ V} 120 V branch") are RMS values .

2. Phasor Representation & Complex Algebra

Phasor analysis transforms time-domain differential equations into algebraic equations involving complex numbers. Using Euler's identity ( e j θ = cos ⁡ θ + j sin ⁡ θ e^{j\theta} = \cos\theta + j\sin\theta e j θ = cos θ + j sin θ ): v ( t ) = Re { 2 V r m s e j ( ω t + θ v ) } ⟷ V = V r m s e j θ v = V r m s ∠ θ v v(t) = \text{Re}\left\{ \sqrt{2} V_{rms} e^{j(\omega t + \theta_v)} \right\} \longleftrightarrow \mathbf{V} = V_{rms} e^{j\theta_v} = V_{rms}\angle\theta_v v ( t ) = Re { 2 ​ V r m s ​ e j ( ω t + θ v ​ ) } ⟷ V = V r m s ​ e j θ v ​ = V r m s ​ ∠ θ v ​ +-----------------------------------------------------------------------------------------+

• | COMPLEX NUMBER REPRESENTATIONS |
• | |
• | RECTANGULAR FORM: POLAR FORM: |
• | Z = R + jX Z = |Z| /_ theta_z = |Z| * e^(j*theta_z) |
• | - R = Real Part (Resistance) - |Z| = Magnitude = sqrt(R^2 + X^2) |
• | - X = Imaginary Part (Reactance) - theta_z = Phase Angle = arctan(X / R) |
• | |
• | CONVERSION IDENTITIES: |
• | - R = |Z| * cos(theta_z) |
• | - X = |Z| * sin(theta_z) |
• | - Complex Conjugate: Z* = R - jX = |Z| /_ -theta_z |
• +-----------------------------------------------------------------------------------------+
• Arithmetic Operations in AC Analysis

Addition & Subtraction: Must be performed in rectangular form : Z 1 + Z 2 = ( R 1 + R 2 ) + j ( X 1 + X 2 ) \mathbf{Z}_1 + \mathbf{Z}_2 = (R_1 + R_2) + j(X_1 + X_2) Z 1 ​ + Z 2 ​ = ( R 1 ​ + R 2 ​ ) + j ( X 1 ​ + X 2 ​ )

Multiplication & Division: Performed most easily in polar form : Z 1 × Z 2 = ( ∣ Z 1 ∣ × ∣ Z 2 ∣ ) ∠ ( θ 1 + θ 2 ) \mathbf{Z}_1 \times \mathbf{Z}_2 = (|Z_1| \times |Z_2|)\angle(\theta_1 + \theta_2) Z 1 ​ × Z 2 ​ = ( ∣ Z 1 ​ ∣ × ∣ Z 2 ​ ∣ ) ∠ ( θ 1 ​ + θ 2 ​ ) Z 1 Z 2 = ( ∣ Z 1 ∣ ∣ Z 2 ∣ ) ∠ ( θ 1 − θ 2 ) \frac{\mathbf{Z}_1}{\mathbf{Z}_2} = \left(\frac{|Z_1|}{|Z_2|}\right)\angle(\theta_1 - \theta_2) Z 2 ​ Z 1 ​ ​ = ( ∣ Z 2 ​ ∣ ∣ Z 1 ​ ∣ ​ ) ∠ ( θ 1 ​ − θ 2 ​ )

3. Passive Element Impedance & Admittance

In the phasor domain, the ratio of phasor voltage to phasor current is the complex impedance Z \mathbf{Z} Z (measured in Ohms, Ω \Omega Ω ): Z = V I = R + j X \mathbf{Z} = \frac{\mathbf{V}}{\mathbf{I}} = R + jX Z = I V ​ = R + j X +-----------------------------------------------------------------------------------------+

• | PASSIVE ELEMENT IMPEDANCE SUMMARY |
• | |
• | RESISTOR (R): |
• | - Time domain: v_R(t) = R * i_R(t) |
• | - Phasor impedance: Z_R = R = R /_ 0 deg |
• | - Voltage and current are IN PHASE (theta = 0 deg). |
• | |
• | INDUCTOR (L): |
• | - Time domain: v_L(t) = L * (di_L / dt) |
• | - Phasor impedance: Z_L = j*omega*L = j*X_L = X_L /_ +90 deg |
• | - Inductive Reactance: X_L = omega * L = 2 * pi * f * L (Ohms) |
• | - VOLTAGE LEADS CURRENT BY 90 DEGREES (ELI). |
• | |
• | CAPACITOR (C): |
• | - Time domain: i_C(t) = C * (dv_C / dt) |
• | - Phasor impedance: Z_C = 1 / (j*omega*C) = -j / (omega*C) = -j*X_C = X_C /_ -90 deg |
• | - Capacitive Reactance: X_C = 1 / (omega * C) = 1 / (2 * pi * f * C) (Ohms) |
• | - CURRENT LEADS VOLTAGE BY 90 DEGREES (ICE). |
• +-----------------------------------------------------------------------------------------+
• The "ELI the ICE man" Phase Rule

To quickly determine leading and lagging phase relationships:

E L I \text{E L I} E L I : Voltage ( E \text{E} E ) leads Current ( I \text{I} I ) in an Inductor ( L \text{L} L ).

I C E \text{I C E} I C E : Current ( I \text{I} I ) leads Voltage ( E \text{E} E ) in a Capacitor ( C \text{C} C ). INDUCTIVE (ELI) CAPACITIVE (ICE)

Voltage Leads Current Current Leads Voltage

• +j (Im) +j (Im)
• ^ ^
• | V phasor | I phasor
• | |
• +-------> +Re +-------> +Re
• / /
• / /
• v I phasor v V phasor
• (I lags V by theta) (I leads V by theta)
• Admittance, Conductance & Susceptance

Admittance Y \mathbf{Y} Y is the reciprocal of impedance, measured in Siemens ( S \text{S} S or Ω − 1 \Omega^{-1} Ω − 1 ): Y = 1 Z = G + j B \mathbf{Y} = \frac{1}{\mathbf{Z}} = G + jB Y = Z 1 ​ = G + j B

Where:

• G G G = Conductance (Siemens, S)
• B B B = Susceptance (Siemens, S)

Converting rectangular impedance Z = R + j X \mathbf{Z} = R + jX Z = R + j X to rectangular admittance Y = G + j B \mathbf{Y} = G + jB Y = G + j B : Y = 1 R + j X = R − j X ( R + j X ) ( R − j X ) = R − j X R 2 + X 2 = ( R R 2 + X 2 ) + j ( − X R 2 + X 2 ) \mathbf{Y} = \frac{1}{R + jX} = \frac{R - jX}{(R + jX)(R - jX)} = \frac{R - jX}{R^2 + X^2} = \left(\frac{R}{R^2 + X^2}\right) + j\left(\frac{-X}{R^2 + X^2}\right) Y = R + j X 1 ​ = ( R + j X ) ( R − j X ) R − j X ​ = R 2 + X 2 R − j X ​ = ( R 2 + X 2 R ​ ) + j ( R 2 + X 2 − X ​ ) Conductance: G = R R 2 + X 2 , Susceptance: B = − X R 2 + X 2 \text{Conductance: } G = \frac{R}{R^2 + X^2}, \qquad \text{Susceptance: } B = \frac{-X}{R^2 + X^2} Conductance: G = R 2 + X 2 R ​ , Susceptance: B = R 2 + X 2 − X ​

Warning

Sign Inversion in Susceptance: Notice that for an inductive impedance ( X = + ω L > 0 X = +\omega L > 0 X = + ω L > 0 ), the resulting inductive susceptance is negative ( B L = − 1 / ( ω L ) < 0 B_L = -1/(\omega L) < 0 B L ​ = − 1/ ( ω L ) < 0 ). For a capacitive impedance ( X = − 1 / ( ω C ) < 0 X = -1/(\omega C) < 0 X = − 1/ ( ω C ) < 0 ), the capacitive susceptance is positive ( B C = + ω C > 0 B_C = +\omega C > 0 B C ​ = + ω C > 0 ). This sign inversion is a frequent source of errors on the PE exam.

4. AC Circuit Analysis Laws & Impedance Combinations

All DC circuit laws generalize directly to the AC phasor domain by replacing real resistances with complex impedances Z \mathbf{Z} Z and DC quantities with phasors V \mathbf{V} V and I \mathbf{I} I :

Phasor KCL: ∑ k = 1 K I k = 0 \sum_{k=1}^K \mathbf{I}_k = 0 ∑ k = 1 K ​ I k ​ = 0 at any node.

Phasor KVL: ∑ k = 1 K V k = 0 \sum_{k=1}^K \mathbf{V}_k = 0 ∑ k = 1 K ​ V k ​ = 0 around any closed loop.

Series Impedances: Z e q = Z 1 + Z 2 + ⋯ + Z n \mathbf{Z}_{eq} = \mathbf{Z}_1 + \mathbf{Z}_2 + \dots + \mathbf{Z}_n Z e q ​ = Z 1 ​ + Z 2 ​ + ⋯ + Z n ​ .

Parallel Impedances: 1 Z e q = Y e q = Y 1 + Y 2 + ⋯ + Y n \frac{1}{\mathbf{Z}_{eq}} = \mathbf{Y}_{eq} = \mathbf{Y}_1 + \mathbf{Y}_2 + \dots + \mathbf{Y}_n Z e q ​ 1 ​ = Y e q ​ = Y 1 ​ + Y 2 ​ + ⋯ + Y n ​ , or for two branches: Z e q = Z 1 Z 2 Z 1 + Z 2 \mathbf{Z}_{eq} = \frac{\mathbf{Z}_1 \mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} Z e q ​ = Z 1 ​ + Z 2 ​ Z 1 ​ Z 2 ​ ​ .

Current Divider (Two Parallel Branches): I 1 = I t o t a l ( Z 2 Z 1 + Z 2 ) = I t o t a l ( Y 1 Y 1 + Y 2 ) \mathbf{I}_1 = \mathbf{I}_{total} \left( \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \right) = \mathbf{I}_{total} \left( \frac{\mathbf{Y}_1}{\mathbf{Y}_1 + \mathbf{Y}_2} \right) I 1 ​ = I t o t a l ​ ( Z 1 ​ + Z 2 ​ Z 2 ​ ​ ) = I t o t a l ​ ( Y 1 ​ + Y 2 ​ Y 1 ​ ​ )

Voltage Divider (Series Branches): V 1 = V t o t a l ( Z 1 Z 1 + Z 2 ) \mathbf{V}_1 = \mathbf{V}_{total} \left( \frac{\mathbf{Z}_1}{\mathbf{Z}_1 + \mathbf{Z}_2} \right) V 1 ​ = V t o t a l ​ ( Z 1 ​ + Z 2 ​ Z 1 ​ ​ )

• 5. Comprehensive Worked Mathematical Examples
• Example 1: AC Nodal Analysis with Reactive Elements at 60 Hz

Problem: In the 60 Hz single-phase AC circuit shown below, calculate the steady-state phasor voltage V 1 \mathbf{V}_1 V 1 ​ at the non-reference node and the total current I s \mathbf{I}_s I s ​ supplied by the source. I_s Z_1 = 4 + j3 ohms

• +-------->----------[ R_1 + jX_L1 ]----+ V_1
• | |
• ( + ) +---+
• V_s = 240 /_ 0 deg V | | Z_2 = 10 - j15 ohms
• ( - ) | | (Parallel branch 2)
• | +---+
• | |
• +--------------------------------------+ (Reference Ground 0 V)

Given Parameters:

V s = 240 ∠ 0 ∘ V r m s \mathbf{V}_s = 240\angle 0^\circ\text{ V}_{rms} V s ​ = 240∠ 0 ∘ V r m s ​ at f = 60 Hz f = 60\text{ Hz} f = 60 Hz .

Series branch impedance: Z 1 = 4.0 + j 3.0 Ω = 5.0 ∠ 36.87 ∘ Ω \mathbf{Z}_1 = 4.0 + j3.0\ \Omega = 5.0\angle 36.87^\circ\ \Omega Z 1 ​ = 4.0 + j 3.0 Ω = 5.0∠36.8 7 ∘ Ω .

Load branch connected from node 1 to ground: Z 2 = 10.0 − j 15.0 Ω = 18.028 ∠ − 56.31 ∘ Ω \mathbf{Z}_2 = 10.0 - j15.0\ \Omega = 18.028\angle -56.31^\circ\ \Omega Z 2 ​ = 10.0 − j 15.0 Ω = 18.028∠ − 56.3 1 ∘ Ω .

Step 1: Compute Total Equivalent Impedance ( Z t o t a l \mathbf{Z}_{total} Z t o t a l ​ ) Z t o t a l = Z 1 + Z 2 = ( 4.0 + j 3.0 ) + ( 10.0 − j 15.0 ) = 14.0 − j 12.0 Ω \mathbf{Z}_{total} = \mathbf{Z}_1 + \mathbf{Z}_2 = (4.0 + j3.0) + (10.0 - j15.0) = 14.0 - j12.0\ \Omega Z t o t a l ​ = Z 1 ​ + Z 2 ​ = ( 4.0 + j 3.0 ) + ( 10.0 − j 15.0 ) = 14.0 − j 12.0 Ω

Convert Z t o t a l \mathbf{Z}_{total} Z t o t a l ​ to polar form: ∣ Z t o t a l ∣ = 14.0 2 + ( − 12.0 ) 2 = 196.0 + 144.0 = 340.0 = 18.439 Ω |Z_{total}| = \sqrt{14.0^2 + (-12.0)^2} = \sqrt{196.0 + 144.0} = \sqrt{340.0} = 18.439\ \Omega ∣ Z t o t a l ​ ∣ = 14. 0 2 + ( − 12.0 ) 2 ​ = 196.0 + 144.0 ​ = 340.0 ​ = 18.439 Ω θ t o t a l = arctan ⁡ ( − 12.0 14.0 ) = − 40.601 ∘ \theta_{total} = \arctan\left(\frac{-12.0}{14.0}\right) = -40.601^\circ θ t o t a l ​ = arctan ( 14.0 − 12.0 ​ ) = − 40.60 1 ∘ Z t o t a l = 18.439 ∠ − 40.60 ∘ Ω \mathbf{Z}_{total} = 18.439\angle -40.60^\circ\ \Omega Z t o t a l ​ = 18.439∠ − 40.6 0 ∘ Ω

Step 2: Calculate Total Source Current ( I s \mathbf{I}_s I s ​ ) I s = V s Z t o t a l = 240 ∠ 0 ∘ V 18.439 ∠ − 40.60 ∘ Ω = 13.016 ∠ + 40.60 ∘ A r m s \mathbf{I}_s = \frac{\mathbf{V}_s}{\mathbf{Z}_{total}} = \frac{240\angle 0^\circ\text{ V}}{18.439\angle -40.60^\circ\ \Omega} = 13.016\angle +40.60^\circ\text{ A}_{rms} I s ​ = Z t o t a l ​ V s ​ ​ = 18.439∠ − 40.6 0 ∘ Ω 240∠ 0 ∘ V ​ = 13.016∠ + 40.6 0 ∘ A r m s ​

Step 3: Calculate Node Voltage ( V 1 \mathbf{V}_1 V 1 ​ ) Using Voltage Division V 1 = V s ( Z 2 Z 1 + Z 2 ) = ( 240 ∠ 0 ∘ ) ( 18.028 ∠ − 56.31 ∘ 18.439 ∠ − 40.60 ∘ ) \mathbf{V}_1 = \mathbf{V}_s \left(\frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2}\right) = (240\angle 0^\circ) \left(\frac{18.028\angle -56.31^\circ}{18.439\angle -40.60^\circ}\right) V 1 ​ = V s ​ ( Z 1 ​ + Z 2 ​ Z 2 ​ ​ ) = ( 240∠ 0 ∘ ) ( 18.439∠ − 40.6 0 ∘ 18.028∠ − 56.3 1 ∘ ​ ) V 1 = 240 × ( 18.028 18.439 ) ∠ ( 0 ∘ − 56.31 ∘ − ( − 40.60 ∘ ) ) \mathbf{V}_1 = 240 \times \left(\frac{18.028}{18.439}\right) \angle (0^\circ - 56.31^\circ - (-40.60^\circ)) V 1 ​ = 240 × ( 18.439 18.028 ​ ) ∠ ( 0 ∘ − 56.3 1 ∘ − ( − 40.6 0 ∘ )) V 1 = 234.65 ∠ − 15.71 ∘ V r m s \mathbf{V}_1 = 234.65\angle -15.71^\circ\text{ V}_{rms} V 1 ​ = 234.65∠ − 15.7 1 ∘ V r m s ​

Example 2: Equivalent Admittance of Parallel RLC Branches

Problem: A 208 V 208\text{ V} 208 V , 60 Hz 60\text{ Hz} 60 Hz single-phase bus supplies two parallel branch loads:

Branch A: An inductive branch with R A = 12 Ω R_A = 12\ \Omega R A ​ = 12 Ω in series with L A = 42.44 mH L_A = 42.44\text{ mH} L A ​ = 42.44 mH .

Branch B: A capacitive branch with R B = 20 Ω R_B = 20\ \Omega R B ​ = 20 Ω in series with C B = 132.63 μ F C_B = 132.63\ \mu\text{F} C B ​ = 132.63 μ F .

Calculate the total admittance Y t o t a l \mathbf{Y}_{total} Y t o t a l ​ , equivalent parallel impedance Z e q \mathbf{Z}_{eq} Z e q ​ , and total current I t o t a l \mathbf{I}_{total} I t o t a l ​ .

Step 1: Calculate Reactances at 60 Hz ( ω = 377.0 rad/s \omega = 377.0\text{ rad/s} ω = 377.0 rad/s )

Inductive reactance: X L A = ω L A = 377.0 × 0.04244 = 16.0 Ω X_{LA} = \omega L_A = 377.0 \times 0.04244 = 16.0\ \Omega X L A ​ = ω L A ​ = 377.0 × 0.04244 = 16.0 Ω .

Z A = 12.0 + j 16.0 Ω = 20.0 ∠ 53.13 ∘ Ω \mathbf{Z}_A = 12.0 + j16.0\ \Omega = 20.0\angle 53.13^\circ\ \Omega Z A ​ = 12.0 + j 16.0 Ω = 20.0∠53.1 3 ∘ Ω .

Capacitive reactance: X C B = 1 ω C B = 1 377.0 × 132.63 × 10 − 6 = 20.0 Ω X_{CB} = \frac{1}{\omega C_B} = \frac{1}{377.0 \times 132.63 \times 10^{-6}} = 20.0\ \Omega X CB ​ = ω C B ​ 1 ​ = 377.0 × 132.63 × 1 0 − 6 1 ​ = 20.0 Ω .

Z B = 20.0 − j 20.0 Ω = 28.284 ∠ − 45.0 ∘ Ω \mathbf{Z}_B = 20.0 - j20.0\ \Omega = 28.284\angle -45.0^\circ\ \Omega Z B ​ = 20.0 − j 20.0 Ω = 28.284∠ − 45. 0 ∘ Ω .

Step 2: Calculate Individual Branch Admittances Y A = 1 Z A = 1 20.0 ∠ 53.13 ∘ = 0.050 ∠ − 53.13 ∘ S = ( 0.030 − j 0.040 ) S \mathbf{Y}_A = \frac{1}{\mathbf{Z}_A} = \frac{1}{20.0\angle 53.13^\circ} = 0.050\angle -53.13^\circ\text{ S} = (0.030 - j0.040)\text{ S} Y A ​ = Z A ​ 1 ​ = 20.0∠53.1 3 ∘ 1 ​ = 0.050∠ − 53.1 3 ∘ S = ( 0.030 − j 0.040 ) S Y B = 1 Z B = 1 28.284 ∠ − 45.0 ∘ = 0.03536 ∠ + 45.0 ∘ S = ( 0.025 + j 0.025 ) S \mathbf{Y}_B = \frac{1}{\mathbf{Z}_B} = \frac{1}{28.284\angle -45.0^\circ} = 0.03536\angle +45.0^\circ\text{ S} = (0.025 + j0.025)\text{ S} Y B ​ = Z B ​ 1 ​ = 28.284∠ − 45. 0 ∘ 1 ​ = 0.03536∠ + 45. 0 ∘ S = ( 0.025 + j 0.025 ) S

Step 3: Combine Parallel Admittances Y t o t a l = Y A + Y B = ( 0.030 + 0.025 ) + j ( − 0.040 + 0.025 ) = 0.055 − j 0.015 S \mathbf{Y}_{total} = \mathbf{Y}_A + \mathbf{Y}_B = (0.030 + 0.025) + j(-0.040 + 0.025) = 0.055 - j0.015\text{ S} Y t o t a l ​ = Y A ​ + Y B ​ = ( 0.030 + 0.025 ) + j ( − 0.040 + 0.025 ) = 0.055 − j 0.015 S

Convert Y t o t a l \mathbf{Y}_{total} Y t o t a l ​ to polar form: ∣ Y t o t a l ∣ = 0.055 2 + ( − 0.015 ) 2 = 0.003025 + 0.000225 = 0.003250 = 0.05701 S |Y_{total}| = \sqrt{0.055^2 + (-0.015)^2} = \sqrt{0.003025 + 0.000225} = \sqrt{0.003250} = 0.05701\text{ S} ∣ Y t o t a l ​ ∣ = 0.05 5 2 + ( − 0.015 ) 2 ​ = 0.003025 + 0.000225 ​ = 0.003250 ​ = 0.05701 S θ Y = arctan ⁡ ( − 0.015 0.055 ) = − 15.255 ∘ \theta_Y = \arctan\left(\frac{-0.015}{0.055}\right) = -15.255^\circ θ Y ​ = arctan ( 0.055 − 0.015 ​ ) = − 15.25 5 ∘ Y t o t a l = 0.05701 ∠ − 15.26 ∘ S \mathbf{Y}_{total} = 0.05701\angle -15.26^\circ\text{ S} Y t o t a l ​ = 0.05701∠ − 15.2 6 ∘ S

Step 4: Calculate Total Current with V = 208 ∠ 0 ∘ V \mathbf{V} = 208\angle 0^\circ\text{ V} V = 208∠ 0 ∘ V I t o t a l = V × Y t o t a l = ( 208 ∠ 0 ∘ ) × ( 0.05701 ∠ − 15.26 ∘ ) = 11.86 ∠ − 15.26 ∘ A r m s \mathbf{I}_{total} = \mathbf{V} \times \mathbf{Y}_{total} = (208\angle 0^\circ) \times (0.05701\angle -15.26^\circ) = 11.86\angle -15.26^\circ\text{ A}_{rms} I t o t a l ​ = V × Y t o t a l ​ = ( 208∠ 0 ∘ ) × ( 0.05701∠ − 15.2 6 ∘ ) = 11.86∠ − 15.2 6 ∘ A r m s ​

Because current angle ( − 15.26 ∘ -15.26^\circ − 15.2 6 ∘ ) is negative relative to voltage ( 0 ∘ 0^\circ 0 ∘ ), the total network remains slightly inductive (current lags voltage).

6. Common PE Exam Traps & Pitfalls

Degree vs. Radian Calculator Mode: Ensure your calculator is set to Degrees when computing polar angles (e.g., ∠ 36.87 ∘ \angle 36.87^\circ ∠36.8 7 ∘ ), but remember to use Radians if evaluating ω t \omega t ω t directly in trigonometric arguments.

Peak vs. RMS Confusion: Peak voltage V p e a k = 2 V r m s V_{peak} = \sqrt{2} V_{rms} V p e ak ​ = 2 ​ V r m s ​ . If a problem specifies v ( t ) = 170 cos ⁡ ( 377 t ) v(t) = 170\cos(377t) v ( t ) = 170 cos ( 377 t ) , the RMS phasor magnitude is 170 / 2 = 120 V 170/\sqrt{2} = 120\text{ V} 170/ 2 ​ = 120 V . Using 170 directly in power formulas will overestimate power by a factor of 2 ( 170 2 / 120 2 = 2.0 170^2 / 120^2 = 2.0 17 0 2 /12 0 2 = 2.0 ).

Capacitive Susceptance Sign: Remember Z C = − j X C ⟹ Y C = + j B C = + j ω C Z_C = -jX_C \implies Y_C = +jB_C = +j\omega C Z C ​ = − j X C ​ ⟹ Y C ​ = + j B C ​ = + jω C . Do not place a negative sign in front of capacitive susceptance.