2.2 AC Fundamentals, Reactance & Power Factor

AC Fundamentals, Reactance & Power Factor

Key Concept: Unlike direct current, alternating current (AC) voltages and currents continuously change in magnitude and periodically reverse direction. Because inductors and capacitors react to these continuous changes by storing and releasing energy in magnetic and electrostatic fields, AC circuits must be analyzed using vector relationships, reactance, impedance, and the power triangle.

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AC Sine Wave Architecture & Waveform Metrics

Commercial alternating current in North America is generated as a smooth sinusoidal wave by rotating synchronous alternators. The conductor loops rotate through stationary magnetic fields, producing an electromotive force proportional to the sine of the angle of rotation:

$v(t) = V_{\text{peak}} \sin(2\pi f t)$

Key Waveform Properties

1. Frequency ($f$): The number of complete cycles executed per second, measured in Hertz (Hz). The standard utility frequency throughout the United States is 60 Hz.
2. Period ($T$): The time required to complete one full electrical cycle ($360^\circ$ or $2\pi$ radians): $T = \frac{1}{f} = \frac{1}{60\text{ Hz}} \approx 0.01667\text{ seconds} = 16.67\text{ milliseconds}$
3. Angular Velocity ($\omega$): The rate of rotation expressed in radians per second: $\omega = 2\pi f = 2 \times \pi \times 60 \approx 377\text{ rad/s}$

Voltage
  ^
+Vpk|         ***
    |       *     *
    |      *       *
  0 +-----+---------+---------+-----> Time (ms)
    |    0 ms      8.33 ms   16.67 ms
    |                *       *
-Vpk|                 *     *
    |                   ***

The Four Voltage Values in AC Circuits

Electricians must clearly differentiate between the four ways an AC voltage is expressed:

• Peak Voltage ($V_{\text{peak}}$ or $V_{\text{max}}$): The maximum instantaneous voltage reached at the crest of the sine wave ($90^\circ$ and $270^\circ$).
• Peak-to-Peak Voltage ($V_{p-p}$): The total vertical voltage span from the positive peak to the negative trough: $V_{p-p} = 2 \times V_{\text{peak}}$
• Effective or RMS Voltage ($V_{\text{rms}}$): Root-Mean-Square voltage is the true practical measure of AC electrical work. One ampere of AC RMS current produces the exact same heating effect in a pure resistor as one ampere of steady DC current. All standard AC voltmeters, panel labels, and NEC ratings refer to RMS values unless explicitly designated otherwise: $V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} \approx 0.7071 \times V_{\text{peak}}$ $V_{\text{peak}} = \sqrt{2} \times V_{\text{rms}} \approx 1.4142 \times V_{\text{rms}}$
• Average Voltage ($V_{\text{avg}}$): The mathematical average of all instantaneous values over one half-cycle ($180^\circ$): $V_{\text{avg}} = \frac{2}{\pi} \times V_{\text{peak}} \approx 0.637 \times V_{\text{peak}} \approx 0.900 \times V_{\text{rms}}$

Nominal System Voltages and Peak Values

Exam Application: Conductor insulation ratings (such as 600V THHN/THWN-2) must withstand peak dielectric stresses of the waveform, not just the RMS value.

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Pure Elements and Phase Relationships: "ELI the ICE man"

In pure direct current, resistance is the sole factor limiting current. In AC systems, inductors and capacitors continuously interact with alternating waveforms, shifting the phase relationship between current and voltage:

1. Purely Resistive Circuit: Current and voltage reach their positive peaks, zero crossings, and negative peaks at the exact same instant. The phase angle ($\theta$) is $0^\circ$.
2. Purely Inductive Circuit: An inductor (coil) consists of turns of wire that establish a magnetic field. When alternating current flows, the expanding and collapsing magnetic field induces a Counter-Electromotive Force (CEMF) per Lenz's Law that opposes the change in current. As a result, current is delayed: Current lags voltage by $90^\circ$.
3. Purely Capacitive Circuit: A capacitor consists of two conductive plates separated by a dielectric insulator. When connected to AC, current flows at its maximum rate when the plates are uncharged and voltage across the plates is zero. As charge builds, voltage rises and current slows to zero. Therefore: Current leads voltage by $90^\circ$.

The Mnemonic: "ELI the ICE man"

• E - L - I: In an inductive circuit (L), Voltage (E) leads Current (I).
• I - C - E: In a capacitive circuit (C), Current (I) leads Voltage (E).

Pure Inductive Circuit (ELI):          Pure Capacitive Circuit (ICE):
Voltage leads Current by 90°           Current leads Voltage by 90°

       +E (90°)                               +I (90°)
          ^                                      ^
          |                                      |
          |                                      |
  0° -----+-----> +I                     0° -----+-----> +E

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Reactance and Impedance Calculations

Inductive Reactance ($X_L$)

Inductive reactance is the opposition offered to alternating current flow by an inductor due to its magnetic counter-electromotive force. It is measured in Ohms ($\Omega$):

$X_L = 2\pi f L$

Where:

• $f = \text{frequency in Hertz (Hz)}$
• $L = \text{inductance in Henrys (H)}$

Observation: Inductive reactance is directly proportional to frequency. If frequency doubles, $X_L$ doubles. At DC ($f = 0$), $X_L = 0\ \Omega$ (an ideal inductor acts as a dead short to steady DC).

Capacitive Reactance ($X_C$)

Capacitive reactance is the opposition offered to alternating current flow by a capacitor due to the electrostatic counter-voltage developed across its dielectric plates. It is measured in Ohms ($\Omega$):

$X_C = \frac{1}{2\pi f C}$

Where:

• $f = \text{frequency in Hertz (Hz)}$
• $C = \text{capacitance in Farads (F)}$

Observation: Capacitive reactance is inversely proportional to frequency. As frequency increases, $X_C$ decreases. At DC ($f = 0$), $X_C = \infty$ (a capacitor completely blocks steady DC).

Total Circuit Impedance ($Z$)

Impedance is the total opposition to alternating current flow presented by a circuit containing resistance, inductance, and capacitance. Because resistance and net reactance act at right angles ($90^\circ$) in the complex plane, they cannot be added algebraically; they must be combined using vector geometry (Pythagorean theorem):

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

Where net reactance is $X = X_L - X_C$. The phase angle ($\theta$) between total voltage and total current is:

$\theta = \arctan\left(\frac{X_L - X_C}{R}\right)$

AC Ohm's Law

Once total impedance ($Z$) in ohms is determined, Ohm's Law applies directly in AC circuits:

$E = I \times Z \qquad I = \frac{E}{Z} \qquad Z = \frac{E}{I}$

Step-by-Step Impedance Worked Example

A series circuit connected across a 120V, 60 Hz AC source consists of a resistor $R = 12\ \Omega$, an inductor with $X_L = 25\ \Omega$, and a capacitor with $X_C = 9\ \Omega$.

• Step 1: Calculate Net Reactance ($X$): $X = X_L - X_C = 25\ \Omega - 9\ \Omega = 16\ \Omega\text{ (net inductive)}$
• Step 2: Calculate Total Impedance ($Z$): $Z = \sqrt{R^2 + X^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\ \Omega$
• Step 3: Calculate Circuit Current ($I$): $I = \frac{E}{Z} = \frac{120\text{ V}}{20\ \Omega} = 6.0\text{ A}$
• Step 4: Calculate Component Voltage Drops:
  • Resistor drop: $V_R = I \times R = 6.0\text{ A} \times 12\ \Omega = 72.0\text{ V}$
  • Inductor drop: $V_L = I \times X_L = 6.0\text{ A} \times 25\ \Omega = 150.0\text{ V}$
  • Capacitor drop: $V_C = I \times X_C = 6.0\text{ A} \times 9\ \Omega = 54.0\text{ V}$
  • Verify total voltage via vector addition (KVL for AC): $E = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{72^2 + (150 - 54)^2} = \sqrt{72^2 + 96^2} = \sqrt{5184 + 9216} = \sqrt{14400} = 120\text{ V}$ Note: Inductor voltage ($150\text{ V}$) actually exceeds source voltage ($120\text{ V}$). This is normal in reactive series circuits due to resonant energy exchange!

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The AC Power Triangle: True, Apparent, and Reactive Power

In reactive AC circuits, voltage and current are displaced by phase angle $\theta$. As a result, not all current entering the circuit produces real mechanical or thermal work. We define three distinct types of power that form the Power Triangle:

                    /|  Apparent Power (S)
                   / |  in Volt-Amperes (VA or kVA)
                  /  |  S = E x I
                 /   |  
                /    |  Reactive Power (Q)
               /     |  in VAR or kVAR
              /      |  Q = E x I x sin(θ)
             / θ     |  
            +--------+
          True Power (P)
       in Watts (W or kW)
       P = E x I x cos(θ)

1. True Power / Real Power ($P$): Measured in Watts (W) or Kilowatts (kW). True power represents actual useful work performed—turning a motor shaft, heating an oven element, or producing illumination. It is consumed entirely by resistance: $P = E \times I \times \cos\theta$
2. Reactive Power ($Q$): Measured in Volt-Amperes Reactive (VAR) or Kilovolt-Amperes Reactive (kVAR). Reactive power represents the "bouncing" energy stored in magnetic fields (inductors) and electrostatic fields (capacitors) that alternates back and forth between source and load every half-cycle without performing useful work: $Q = E \times I \times \sin\theta$
3. Apparent Power ($S$): Measured in Volt-Amperes (VA) or Kilovolt-Amperes (kVA). Apparent power represents the total capacity that generators, transformers, switchgear, and feeder conductors must deliver to satisfy both the true and reactive power demands: $S = E \times I$ $S = \sqrt{P^2 + Q^2}$

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Power Factor (PF) & Phase Angle Relationships

Power Factor (PF) is the ratio of real working power to the total apparent power delivered by the system:

$PF = \frac{\text{True Power}}{\text{Apparent Power}} = \frac{P}{S} = \frac{\text{Watts}}{\text{Volt-Amperes}} = \frac{\text{kW}}{\text{kVA}} = \cos\theta$

Power factor is expressed either as a decimal between 0 and 1.0 (e.g., 0.85) or as a percentage (e.g., 85%).

Lagging versus Leading Power Factor

• Lagging Power Factor: Occurs when load current lags voltage. This is caused by inductive loads, which dominate commercial and industrial facilities: induction motors, transformers, welding equipment, and magnetic ballasts.
• Leading Power Factor: Occurs when load current leads voltage. This is caused by capacitive loads: power factor correction capacitor banks, lightly loaded underground transmission cables, and overexcited synchronous motors.

The Operational Penalty of Low Power Factor

Consider an industrial facility drawing $100\text{ kW}$ of true power at 480V single-phase:

• At 1.0 (Unity) Power Factor: $S = \frac{100\text{ kW}}{1.0} = 100\text{ kVA} \implies I = \frac{100,000\text{ VA}}{480\text{ V}} = 208.3\text{ A}$
• At 0.60 Lagging Power Factor: $S = \frac{100\text{ kW}}{0.60} = 166.7\text{ kVA} \implies I = \frac{166,667\text{ VA}}{480\text{ V}} = 347.2\text{ A}$

Notice that at 0.60 PF, the facility performs the exact same $100\text{ kW}$ of useful work, but the supply conductors must carry $347.2\text{ A}$ instead of $208.3\text{ A}$—an increase of $138.9\text{ A}$ (67% more current!).

Severe Consequences of Low Power Factor

1. Conductor Sizing: Feeders, transformers, and switchboards must be substantially oversized to handle non-working reactive current.
2. Conductor Thermal Losses: Line heating increases with $I^2 R$. Operating at 0.60 PF increases feeder heat losses by $(\frac{347.2}{208.3})^2 = 2.78$ times (a 178% increase in wasted energy!).
3. Voltage Regulation: Higher line currents cause excessive voltage drop across distribution feeders.
4. Utility Penalties: Electric utilities install kVAR/kVA demand meters and levy heavy financial surcharges on commercial customers with power factors below 0.90 or 0.95.

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Step-by-Step Power Factor Correction Calculation

To correct poor power factor caused by inductive loads, electricians install shunt capacitor banks in parallel with the load. Because capacitive reactive power ($Q_C$) leads by $90^\circ$ while inductive reactive power ($Q_L$) lags by $90^\circ$, they directly cancel each other out ($Q_{\text{net}} = Q_L - Q_C$).

             Inductive Load
             (P = 120 kW, PF = 0.65)
             +----------------------+
[ 480 V ] ---+                      +--- [ Ground/Neutral ]
             +---[ Capacitor Bank ]-+
                   (Supplies kVAR)

Practical Worked Problem

A 480V, single-phase industrial plant operates at an active load of $120\text{ kW}$ with an existing lagging power factor of $0.65$. Plant management wants to install a capacitor bank to improve the power factor to $0.95$ lagging to eliminate utility billing penalties. Determine:

1. The existing Apparent Power ($S_1$) and line current ($I_1$).
2. The existing Reactive Power ($Q_1$).
3. The target Apparent Power ($S_2$) and target Reactive Power ($Q_2$).
4. The required capacitor rating in kVAR.
5. The line current reduction achieved.

Solution Steps

• Step 1: Calculate Initial Apparent Power ($S_1$) and Line Current ($I_1$): $S_1 = \frac{P}{PF_1} = \frac{120\text{ kW}}{0.65} \approx 184.62\text{ kVA}$ $I_1 = \frac{S_1}{V} = \frac{184,620\text{ VA}}{480\text{ V}} \approx 384.6\text{ A}$

• Step 2: Calculate Initial Reactive Power ($Q_1$):

  • Initial phase angle: $\theta_1 = \arccos(0.65) \approx 49.46^\circ$
  • Using trigonometry: $Q_1 = P \times \tan(\theta_1)$ $\tan(49.46^\circ) \approx 1.1691$ $Q_1 = 120\text{ kW} \times 1.1691 \approx 140.29\text{ kVAR}$

• Step 3: Calculate Target Apparent Power ($S_2$) and Target Reactive Power ($Q_2$):

  • At desired $PF_2 = 0.95$: $S_2 = \frac{P}{PF_2} = \frac{120\text{ kW}}{0.95} \approx 126.32\text{ kVA}$
  • Target phase angle: $\theta_2 = \arccos(0.95) \approx 18.19^\circ$ $\tan(18.19^\circ) \approx 0.3287$ $Q_2 = P \times \tan(\theta_2) = 120\text{ kW} \times 0.3287 \approx 39.44\text{ kVAR}$

• Step 4: Calculate Required Capacitor Rating ($Q_C$): The capacitor must supply the difference between existing and target reactive power: $Q_C = Q_1 - Q_2 = 140.29\text{ kVAR} - 39.44\text{ kVAR} = 100.85\text{ kVAR}$ Result: Install a standard $100\text{ kVAR}$ shunt capacitor bank.

• Step 5: Determine Line Current Reduction:

  • New line current at 0.95 PF: $I_2 = \frac{S_2}{V} = \frac{126,320\text{ VA}}{480\text{ V}} \approx 263.2\text{ A}$
  • Current reduction: $\Delta I = 384.6\text{ A} - 263.2\text{ A} = 121.4\text{ A}$ Summary: Adding the capacitor bank slashes feeder current by $121.4\text{ A}$ (a 31.6% reduction) without altering motor speeds, shaft output, or heating performance.