13.2 AC Circuits, Phasors, Impedance, and Three-Phase Power

13.2 AC Circuits, Phasors, Impedance, and Three-Phase Power

Core FE Exam Principle: Steady-state sinusoidal AC analysis replaces time-domain differential equations with complex algebraic equations in the frequency domain using phasors and complex impedance ((\mathbf{Z} = R + j X)).

Sinusoidal Signals, RMS Values, and Phasor Notation

A sinusoidal voltage signal in the time domain is expressed as:

$v(t) = V_m \cos(\omega t + \phi_v) \quad [\text{Volts, V}]$

where:

• (V_m) = Peak voltage amplitude
• (\omega = 2\pi f = \frac{2\pi}{T}) = Angular frequency (rad/s), with cyclic frequency (f) (Hz) and period (T) (s)
• (\phi_v) = Phase angle (radians or degrees)

Root-Mean-Square (RMS) Values

The RMS (effective) value of a periodic waveform represents the DC equivalent voltage or current that delivers the exact same average power to a resistive load:

$V_{\text{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} v^2(t) dt} = \frac{V_m}{\sqrt{2}} \approx 0.7071 V_m$ $I_{\text{rms}} = \frac{I_m}{\sqrt{2}} \approx 0.7071 I_m$

Exam Convention: All standard AC voltage ratings (such as 120 V, 240 V, 480 V) and phasor magnitudes on the FE exam are RMS values unless explicitly specified as peak amplitudes.

Phasor Representation

Using Euler's identity ((e^{j\theta} = \cos\theta + j\sin\theta)), a sinusoidal quantity is represented in phasor form as a stationary complex vector in the frequency domain:

$\mathbf{V} = V_{\text{rms}} \angle \phi_v = V_{\text{rms}} e^{j \phi_v} = V_{\text{rms}} \left( \cos\phi_v + j \sin\phi_v \right)$

Complex Impedance and AC Frequency-Domain Analysis

In the frequency domain, Ohm's Law extends to AC circuits as:

$\mathbf{V} = \mathbf{I} \mathbf{Z} \implies \mathbf{Z} = \frac{\mathbf{V}}{\mathbf{I}} = R + j X$

where (\mathbf{Z}) is complex impedance ((\Omega)), (R) is resistance, and (X) is reactance.

Impedance of Fundamental Elements

Series RLC Resonance

For a series RLC circuit, total impedance is (\mathbf{Z} = R + j\left(\omega L - \frac{1}{\omega C}\right)). At the resonant frequency (\omega_0), the inductive and capacitive reactances cancel exactly ((X_L = X_C)):

$\omega_0 = \frac{1}{\sqrt{L C}} \quad [\text{rad/s}] \implies f_0 = \frac{1}{2\pi \sqrt{L C}} \quad [\text{Hz}]$

At resonance, impedance is purely resistive ((\mathbf{Z} = R)), minimizing impedance magnitude and maximizing current flow.

Complex Power and Power Factor Correction

AC power analysis involves three distinct components combined in the Complex Power Triangle:

                   Complex Power Triangle
                   
                   /|  S (Apparent Power, VA)
                  / |  
                 /  |  Q (Reactive Power, VAR)
                /   |
               /theta|
              +-----+ 
                 P (Real Power, Watts)

Mathematical Definitions

1. Complex Power ((\mathbf{S})): $\mathbf{S} = \mathbf{V}_{\text{rms}} \mathbf{I}_{\text{rms}}^* = P + j Q = |\mathbf{S}| \angle \theta \quad [\text{VA}]$ where (\mathbf{I}^*) denotes the complex conjugate of the current phasor.
2. Real / Active Power ((P)): Work-producing power dissipated as heat or work: $P = Re\\{\mathbf{S}\\} = V_{\text{rms}} I_{\text{rms}} \cos\theta = I_{\text{rms}}^2 R \quad [\text{Watts, W}]$
3. Reactive Power ((Q)): Energy continuously exchanged between source and reactive fields: $Q = Im\\{\mathbf{S}\\} = V_{\text{rms}} I_{\text{rms}} \sin\theta = I_{\text{rms}}^2 X \quad [\text{VAR}]$ (Positive (Q) indicates inductive load; negative (Q) indicates capacitive load).
4. Apparent Power ((S = |\mathbf{S}|)): Magnitude of complex power: $S = |\mathbf{S}| = \sqrt{P^2 + Q^2} = V_{\text{rms}} I_{\text{rms}} \quad [\text{Volt-Amperes, VA}]$
5. Power Factor ((pf)): Ratio of real power to apparent power: $pf = \cos\theta = \frac{P}{S}$
   • Lagging pf: Inductive loads (current lags voltage, (\theta > 0)).
   • Leading pf: Capacitive loads (current leads voltage, (\theta < 0)).

Power Factor Correction

Low power factor increases utility line currents and transmission losses. Parallel capacitor banks are installed across inductive loads to supply local reactive power (Q_C), raising power factor from (\cos\theta_1) to (\cos\theta_2):

$Q_C = P \left( \tan\theta_1 - \tan\theta_2 \right) \quad [\text{VAR}]$ $C = \frac{Q_C}{\omega V_{\text{rms}}^2} = \frac{P \left( \tan\theta_1 - \tan\theta_2 \right)}{\omega V_{\text{rms}}^2} \quad [\text{Farads, F}]$

Balanced Three-Phase (3-Phase) Power Systems

Three-phase electrical systems utilize three sinusoidal voltages of equal magnitude separated in phase by (120^\circ). Balanced three-phase loads are configured in either Wye (Y) or Delta ((\Delta)) topologies.

Wye (Y) and Delta ((\Delta)) Relationships

Total Three-Phase Power Formulas

Regardless of whether the balanced load is connected in Wye or Delta, total three-phase power equations in terms of line-to-line quantities are identical:

• Total Real Power: $P_{\text{total}} = \sqrt{3} V_{\text{line}} I_{\text{line}} \cos\theta = 3 P_{\text{phase}} = 3 V_{\text{phase}} I_{\text{phase}} \cos\theta \quad [\text{W}]$
• Total Reactive Power: $Q_{\text{total}} = \sqrt{3} V_{\text{line}} I_{\text{line}} \sin\theta = 3 Q_{\text{phase}} = 3 V_{\text{phase}} I_{\text{phase}} \sin\theta \quad [\text{VAR}]$
• Total Apparent Power: $S_{\text{total}} = \sqrt{3} V_{\text{line}} I_{\text{line}} = 3 S_{\text{phase}} = 3 V_{\text{phase}} I_{\text{phase}} \quad [\text{VA}]$

Comprehensive Worked Engineering Example

Problem Statement

A balanced three-phase industrial plant load is connected in Wye (Y) configuration to a 3-phase 480 V (line-to-line RMS), 60 Hz utility line. Each phase branch of the load consists of a resistor (R = 30.0 \ \Omega) in series with an inductor reactance (X_L = 40.0 \ \Omega).

Calculate:

1. The phase voltage (V_{\text{phase}}), phase impedance magnitude (|\mathbf{Z}{\phi}|), line current (I{\text{line}}), and load power factor (pf).
2. The total 3-phase active power (P_{\text{total}}), reactive power (Q_{\text{total}}), and apparent power (S_{\text{total}}).
3. The total required capacitive reactive power (Q_{C,\text{total}}) and per-phase capacitance (C_{\phi}) of a Wye-connected capacitor bank needed to correct the plant power factor to 0.95 lagging.

Step-by-Step Solution

Step 1: Compute Phase Voltage, Impedance, Line Current, and Power Factor

• For a Wye connection, phase voltage is line voltage divided by (\sqrt{3}): $V_{\text{phase}} = \frac{V_{\text{line}}}{\sqrt{3}} = \frac{480 \text{ V}}{\sqrt{3}} = 277.13 \text{ V}$
• Phase impedance magnitude (|\mathbf{Z}_{\phi}|): $|\mathbf{Z}_{\phi}| = \sqrt{R^2 + X_L^2} = \sqrt{30.0^2 + 40.0^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50.0 \\ \Omega$
• Phase angle (\theta): $\theta = \arctan\left(\frac{X_L}{R}\right) = \arctan\left(\frac{40.0}{30.0}\right) = 53.13^\circ$
• Line current (equal to phase current for Wye): $I_{\text{line}} = I_{\text{phase}} = \frac{V_{\text{phase}}}{|\mathbf{Z}_{\phi}|} = \frac{277.13 \text{ V}}{50.0 \\ \Omega} = 5.5426 \text{ A}$
• Uncorrected Power Factor: $pf_1 = \cos(53.13^\circ) = \frac{R}{|\mathbf{Z}_{\phi}|} = \frac{30.0}{50.0} = 0.600 \text{ lagging}$

Step 2: Compute Total 3-Phase Powers

• Total Real Power (P_{\text{total}}): $P_{\text{total}} = 3 I_{\text{phase}}^2 R = 3 \times (5.5426 \text{ A})^2 \times 30.0 \\ \Omega = 3 \times 30.720 \times 30.0 = 2764.8 \text{ W} = 2.765 \text{ kW}$
• Total Reactive Power (Q_{\text{total}}): $Q_{\text{total}} = 3 I_{\text{phase}}^2 X_L = 3 \times (5.5426 \text{ A})^2 \times 40.0 \\ \Omega = 3 \times 30.720 \times 40.0 = 3686.4 \text{ VAR} = 3.686 \text{ kVAR}$
• Total Apparent Power (S_{\text{total}}): $S_{\text{total}} = \sqrt{P_{\text{total}}^2 + Q_{\text{total}}^2} = \sqrt{2.7648^2 + 3.6864^2} = 4.608 \text{ kVA}$

Step 3: Compute Power Factor Correction Capacitor Bank

• Target power factor angle (\theta_2 = \arccos(0.95) = 18.195^\circ).
• Trigonometric tangents: (\tan\theta_1 = \tan(53.13^\circ) = 1.3333), (\tan\theta_2 = \tan(18.195^\circ) = 0.3287).
• Total required capacitive reactive power (Q_{C,\text{total}}): $Q_{C,\text{total}} = P_{\text{total}} \left( \tan\theta_1 - \tan\theta_2 \right) = 2764.8 \text{ W} \times (1.3333 - 0.3287) = 2764.8 \times 1.0046 = 2777.5 \text{ VAR}$
• Capacitive reactive power per phase (Wye bank): $Q_{C,\phi} = \frac{Q_{C,\text{total}}}{3} = \frac{2777.5}{3} = 925.83 \text{ VAR}$
• Per-phase capacitance (C_{\phi}) at (f = 60 \text{ Hz}) ((\omega = 376.99 \text{ rad/s})): $C_{\phi} = \frac{Q_{C,\phi}}{\omega V_{\text{phase}}^2} = \frac{925.83 \text{ VAR}}{376.99 \text{ rad/s} \times (277.13 \text{ V})^2} = \frac{925.83}{376.99 \times 76801} = \frac{925.83}{2.8954 \times 10^7} = 3.20 \times 10^{-5} \text{ F} = 32.0 \\ \mu\text{F}$

Final Answer: (I_{\text{line}} = 5.54 \text{ A}), (pf_1 = 0.60 \text{ lagging}), (P_{\text{total}} = 2.765 \text{ kW}), (Q_{\text{total}} = 3.686 \text{ kVAR}), (S_{\text{total}} = 4.608 \text{ kVA}), total capacitor rating (Q_{C,\text{total}} = 2.778 \text{ kVAR}), and Wye phase capacitance (C_{\phi} = 32.0 \ \mu\text{F}).