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

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.

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1. Sinusoidal Waveforms, Frequency & RMS Values

A time-varying sinusoidal voltage is defined mathematically as:

$v(t) = V_{peak} \cos(\omega t + \theta_v)$

Where:

• $V_{peak}$ (or $V_m$) = Peak amplitude of the sinusoidal voltage (Volts).
• $\omega = 2\pi f$ = Angular frequency in radians per second ($\text{rad/s}$).
• $f$ = Cyclic frequency in Hertz ($\text{Hz} = \text{s}^{-1}$); in North America, $f = 60\text{ Hz} \implies \omega = 2\pi(60) \approx 376.991\text{ rad/s} \approx 377\text{ rad/s}$.
• $T = 1/f$ = Period of one full cycle; at 60 Hz, $T = 1/60\text{ s} = 16.67\text{ ms}$.
• $\theta_v$ = 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 $170\cos(377t)$"), all AC voltages and currents given in the PE Power exam (e.g., "a $480\text{ V}$ feeder" or "a $120\text{ V}$ branch") are RMS values.

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2. Phasor Representation & Complex Algebra

Phasor analysis transforms time-domain differential equations into algebraic equations involving complex numbers. Using Euler's identity ($e^{j\theta} = \cos\theta + j\sin\theta$):

$v(t) = \text{Re}\left\{ \sqrt{2} V_{rms} e^{j(\omega t + \theta_v)} \right\} \longleftrightarrow \mathbf{V} = V_{rms} e^{j\theta_v} = V_{rms}\angle\theta_v$

+-----------------------------------------------------------------------------------------+
|                           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:

$\mathbf{Z}_1 + \mathbf{Z}_2 = (R_1 + R_2) + j(X_1 + X_2)$

• Multiplication & Division: Performed most easily in polar form:

$\mathbf{Z}_1 \times \mathbf{Z}_2 = (|Z_1| \times |Z_2|)\angle(\theta_1 + \theta_2)$

$\frac{\mathbf{Z}_1}{\mathbf{Z}_2} = \left(\frac{|Z_1|}{|Z_2|}\right)\angle(\theta_1 - \theta_2)$

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3. Passive Element Impedance & Admittance

In the phasor domain, the ratio of phasor voltage to phasor current is the complex impedance $\mathbf{Z}$ (measured in Ohms, $\Omega$):

$\mathbf{Z} = \frac{\mathbf{V}}{\mathbf{I}} = R + jX$

+-----------------------------------------------------------------------------------------+
|                        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:

• $ ext{E L I}$: Voltage ($ ext{E}$) leads Current ($ ext{I}$) in an Inductor ($ ext{L}$).
• $ ext{I C E}$: Current ($ ext{I}$) leads Voltage ($ ext{E}$) in a Capacitor ($ ext{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 $\mathbf{Y}$ is the reciprocal of impedance, measured in Siemens ($ ext{S}$ or $\Omega^{-1}$):

$\mathbf{Y} = \frac{1}{\mathbf{Z}} = G + jB$

Where:

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

Converting rectangular impedance $\mathbf{Z} = R + jX$ to rectangular admittance $\mathbf{Y} = G + jB$:

$\mathbf{Y} = \frac{1}{R + jX} = \frac{R - jX}{(R + jX)(R - jX)} = \frac{R - jX}{R^2 + X^2} = \left(\frac{R}{R^2 + X^2}\right) + j\left(\frac{-X}{R^2 + X^2}\right)$

$\text{Conductance: } G = \frac{R}{R^2 + X^2}, \qquad \text{Susceptance: } B = \frac{-X}{R^2 + X^2}$

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

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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 $\mathbf{Z}$ and DC quantities with phasors $\mathbf{V}$ and $\mathbf{I}$:

• Phasor KCL: $\sum_{k=1}^K \mathbf{I}_k = 0$ at any node.
• Phasor KVL: $\sum_{k=1}^K \mathbf{V}_k = 0$ around any closed loop.
• Series Impedances: $\mathbf{Z}_{eq} = \mathbf{Z}_1 + \mathbf{Z}_2 + \dots + \mathbf{Z}_n$.
• Parallel Impedances: $\frac{1}{\mathbf{Z}{eq}} = \mathbf{Y}{eq} = \mathbf{Y}_1 + \mathbf{Y}_2 + \dots + \mathbf{Y}n$, or for two branches: $\mathbf{Z}{eq} = \frac{\mathbf{Z}_1 \mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2}$.
• Current Divider (Two Parallel Branches):

$\mathbf{I}_1 = \mathbf{I}_{total} \left( \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \right) = \mathbf{I}_{total} \left( \frac{\mathbf{Y}_1}{\mathbf{Y}_1 + \mathbf{Y}_2} \right)$

• Voltage Divider (Series Branches):

$\mathbf{V}_1 = \mathbf{V}_{total} \left( \frac{\mathbf{Z}_1}{\mathbf{Z}_1 + \mathbf{Z}_2} \right)$

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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 $\mathbf{V}_1$ at the non-reference node and the total current $\mathbf{I}_s$ 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:

• $\mathbf{V}s = 240\angle 0^\circ\text{ V}{rms}$ at $f = 60\text{ Hz}$.
• Series branch impedance: $\mathbf{Z}_1 = 4.0 + j3.0\ \Omega = 5.0\angle 36.87^\circ\ \Omega$.
• Load branch connected from node 1 to ground: $\mathbf{Z}_2 = 10.0 - j15.0\ \Omega = 18.028\angle -56.31^\circ\ \Omega$.

Step 1: Compute Total Equivalent Impedance ($\mathbf{Z}_{total}$)

$\mathbf{Z}_{total} = \mathbf{Z}_1 + \mathbf{Z}_2 = (4.0 + j3.0) + (10.0 - j15.0) = 14.0 - j12.0\ \Omega$

Convert $\mathbf{Z}_{total}$ to polar form:

$|Z_{total}| = \sqrt{14.0^2 + (-12.0)^2} = \sqrt{196.0 + 144.0} = \sqrt{340.0} = 18.439\ \Omega$

$\theta_{total} = \arctan\left(\frac{-12.0}{14.0}\right) = -40.601^\circ$

$\mathbf{Z}_{total} = 18.439\angle -40.60^\circ\ \Omega$

Step 2: Calculate Total Source Current ($\mathbf{I}_s$)

$\mathbf{I}_s = \frac{\mathbf{V}_s}{\mathbf{Z}_{total}} = \frac{240\angle 0^\circ\text{ V}}{18.439\angle -40.60^\circ\ \Omega} = 13.016\angle +40.60^\circ\text{ A}_{rms}$

Step 3: Calculate Node Voltage ($\mathbf{V}_1$) Using Voltage Division

$\mathbf{V}_1 = \mathbf{V}_s \left(\frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2}\right) = (240\angle 0^\circ) \left(\frac{18.028\angle -56.31^\circ}{18.439\angle -40.60^\circ}\right)$

$\mathbf{V}_1 = 240 \times \left(\frac{18.028}{18.439}\right) \angle (0^\circ - 56.31^\circ - (-40.60^\circ))$

$\mathbf{V}_1 = 234.65\angle -15.71^\circ\text{ V}_{rms}$

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Example 2: Equivalent Admittance of Parallel RLC Branches

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

• Branch A: An inductive branch with $R_A = 12\ \Omega$ in series with $L_A = 42.44\text{ mH}$.
• Branch B: A capacitive branch with $R_B = 20\ \Omega$ in series with $C_B = 132.63\ \mu\text{F}$. Calculate the total admittance $\mathbf{Y}{total}$, equivalent parallel impedance $\mathbf{Z}{eq}$, and total current $\mathbf{I}_{total}$.

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

• Inductive reactance: $X_{LA} = \omega L_A = 377.0 \times 0.04244 = 16.0\ \Omega$.
  • $\mathbf{Z}_A = 12.0 + j16.0\ \Omega = 20.0\angle 53.13^\circ\ \Omega$.
• Capacitive reactance: $X_{CB} = \frac{1}{\omega C_B} = \frac{1}{377.0 \times 132.63 \times 10^{-6}} = 20.0\ \Omega$.
  • $\mathbf{Z}_B = 20.0 - j20.0\ \Omega = 28.284\angle -45.0^\circ\ \Omega$.

Step 2: Calculate Individual Branch Admittances

$\mathbf{Y}_A = \frac{1}{\mathbf{Z}_A} = \frac{1}{20.0\angle 53.13^\circ} = 0.050\angle -53.13^\circ\text{ S} = (0.030 - j0.040)\text{ S}$

$\mathbf{Y}_B = \frac{1}{\mathbf{Z}_B} = \frac{1}{28.284\angle -45.0^\circ} = 0.03536\angle +45.0^\circ\text{ S} = (0.025 + j0.025)\text{ S}$

Step 3: Combine Parallel Admittances

$\mathbf{Y}_{total} = \mathbf{Y}_A + \mathbf{Y}_B = (0.030 + 0.025) + j(-0.040 + 0.025) = 0.055 - j0.015\text{ S}$

Convert $\mathbf{Y}_{total}$ to polar form:

$|Y_{total}| = \sqrt{0.055^2 + (-0.015)^2} = \sqrt{0.003025 + 0.000225} = \sqrt{0.003250} = 0.05701\text{ S}$

$\theta_Y = \arctan\left(\frac{-0.015}{0.055}\right) = -15.255^\circ$

$\mathbf{Y}_{total} = 0.05701\angle -15.26^\circ\text{ S}$

Step 4: Calculate Total Current with $\mathbf{V} = 208\angle 0^\circ\text{ V}$

$\mathbf{I}_{total} = \mathbf{V} \times \mathbf{Y}_{total} = (208\angle 0^\circ) \times (0.05701\angle -15.26^\circ) = 11.86\angle -15.26^\circ\text{ A}_{rms}$

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

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6. Common PE Exam Traps & Pitfalls

1. Degree vs. Radian Calculator Mode: Ensure your calculator is set to Degrees when computing polar angles (e.g., $\angle 36.87^\circ$), but remember to use Radians if evaluating $\omega t$ directly in trigonometric arguments.
2. Peak vs. RMS Confusion: Peak voltage $V_{peak} = \sqrt{2} V_{rms}$. If a problem specifies $v(t) = 170\cos(377t)$, the RMS phasor magnitude is $170/\sqrt{2} = 120\text{ V}$. Using 170 directly in power formulas will overestimate power by a factor of 2 ($170^2 / 120^2 = 2.0$).
3. Capacitive Susceptance Sign: Remember $Z_C = -jX_C \implies Y_C = +jB_C = +j\omega C$. Do not place a negative sign in front of capacitive susceptance.