2.2 AC Fundamentals: Reactance, Impedance & Power Factor

2.2 AC Fundamentals: Reactance, Impedance & Power Factor

While direct current flows steadily in a single direction, commercial and residential utility power is generated and distributed as sinusoidal alternating current (AC). In alternating current systems, voltage and current continuously reverse polarity and direction. The presence of coils (magnetic fields) and dielectric surfaces (electrostatic fields) introduces reactive phenomena that fundamentally alter circuit calculations compared to basic DC resistive networks.

---

Nature of Alternating Current Waveforms

In North America, utility electrical systems operate at a standard frequency ($f$) of $60\text{ hertz}$ ($\text{Hz}$), completing 60 complete cycles of positive and negative alternation per second. The time required to complete one full cycle is the period ($T$):

$T = \frac{1}{f} = \frac{1}{60\text{ Hz}} \approx 0.01667\text{ s} = 16.67\text{ ms}$

Sinusoidal Voltage Measurements

Because an AC sine wave continuously changes value from zero to positive peak, back through zero to negative peak, and returns to zero, electricians use three distinct voltage metrics:

1. Peak Voltage ($V_{\text{peak}}$): The maximum instantaneous potential attained during the cycle, measured from the zero baseline to the crest.
2. Peak-to-Peak Voltage ($V_{\text{pk-pk}}$): The total voltage excursion from the positive crest to the negative trough: $V_{\text{pk-pk}} = 2 \times V_{\text{peak}}$.
3. Root-Mean-Square Voltage ($V_{\text{RMS}}$): Also termed the effective voltage, $V_{\text{RMS}}$ represents the DC equivalent voltage that would produce identical thermal heating in a pure resistance:

$V_{\text{RMS}} = \frac{V_{\text{peak}}}{\sqrt{2}} \approx 0.7071 \times V_{\text{peak}}$ $V_{\text{peak}} = \sqrt{2} \times V_{\text{RMS}} \approx 1.4142 \times V_{\text{RMS}}$

All standard multimeters, utility service ratings, and National Electrical Code voltage classifications ($120\text{ V}$, $208\text{ V}$, $240\text{ V}$, $277\text{ V}$, $480\text{ V}$) denote RMS values. For example, a nominal $120\text{ V}$ RMS branch circuit actually reaches a peak instantaneous potential of:

$V_{\text{peak}} = 120\text{ V} \times 1.4142 \approx 169.7\text{ V}$

---

Inductive and Capacitive Reactance

When pure resistance is energized by AC, current and voltage remain precisely in phase—both pass through zero and reach peak values simultaneously. However, electromagnetic coils and capacitors introduce opposition to alternating current known as reactance ($X$), measured in ohms ($\Omega$).

Inductive Reactance ($X_L$)

An inductor (such as a motor winding, transformer coil, or magnetic ballast) stores energy within a concentrated magnetic field. According to Faraday's and Lenz's Laws, any change in current induces a counter-electromotive force (CEMF) that directly opposes that change. This opposition is inductive reactance ($X_L$):

$X_L = 2\pi f L$

Where:

• $f$ = frequency in hertz ($\text{Hz}$)
• $L$ = inductance in henrys ($\text{H}$)
• $X_L$ = inductive reactance in ohms ($\Omega$)

Because an inductor opposes current changes, current is delayed behind voltage. In an ideal, purely inductive circuit, current lags voltage by exactly $90^\circ$.

Capacitive Reactance ($X_C$)

A capacitor stores energy electrostatically across conductive plates separated by an insulating dielectric. As alternating potential rises, charge rushes into the plates; when potential falls, charge discharges back into the line. The opposition offered to alternating current flow is capacitive reactance ($X_C$):

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

Where:

• $f$ = frequency in hertz ($\text{Hz}$)
• $C$ = capacitance in farads ($\text{F}$)
• $X_C$ = capacitive reactance in ohms ($\Omega$)

In a capacitor, maximum current flows when voltage is zero but changing at its fastest rate. Therefore, in an ideal purely capacitive circuit, current leads voltage by exactly $90^\circ$.

The "ELI the ICE man" Mnemonic

Electricians use a standard trade mnemonic to remember phase angle relationships:

• ELI: Voltage ($E$) Leads Current ($I$) in an Inductor ($L$).
• ICE: Current ($I$) Leads Voltage ($E$) in a Capacitor ($C$).

---

Circuit Impedance ($Z$)

In practical AC circuits containing resistance ($R$), inductance ($L$), and capacitance ($C$), total opposition to current cannot be found through simple arithmetic addition. Because resistance and net reactance act $90^\circ$ apart in time, they must be combined using vector addition (the Pythagorean theorem). The total opposition is impedance ($Z$), measured in ohms ($\Omega$).

The Series Impedance Formula

Because inductive reactance and capacitive reactance act $180^\circ$ opposite each other, they cancel directly:

$X_{\text{net}} = X_L - X_C$

Total circuit impedance is given by:

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

Applying Ohm's Law to AC circuits utilizes impedance in place of pure resistance:

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

Worked Example: Series RLC Circuit

A $120\text{ V}$, $60\text{ Hz}$ single-phase AC circuit contains a resistor with $R = 8\ \Omega$, an inductor with $X_L = 15\ \Omega$, and a capacitor with $X_C = 9\ \Omega$ connected in series.

1. Calculate Net Reactance ($X_{\text{net}}$): $X_{\text{net}} = X_L - X_C = 15\ \Omega - 9\ \Omega = 6\ \Omega\text{ (net inductive)}$
2. Calculate Total Impedance ($Z$): $Z = \sqrt{R^2 + X_{\text{net}}^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\ \Omega$
3. Calculate Circuit Current ($I$): $I = \frac{V}{Z} = \frac{120\text{ V}}{10\ \Omega} = 12\text{ A}$
4. Calculate Component Voltage Drops:
   • $V_R = I \times R = 12\text{ A} \times 8\ \Omega = 96\text{ V}$
   • $V_L = I \times X_L = 12\text{ A} \times 15\ \Omega = 180\text{ V}$
   • $V_C = I \times X_C = 12\text{ A} \times 9\ \Omega = 108\text{ V}$
   • Verification via vector sum: $V_T = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{96^2 + (180 - 108)^2} = \sqrt{9,216 + 5,184} = \sqrt{14,400} = 120\text{ V}$

---

The AC Power Triangle and Power Factor

In AC circuits containing reactance, voltage and current waveforms are out of phase. As a result, not all apparent energy delivered to the circuit performs useful mechanical or thermal work.

The Three Components of AC Power

1. True Power ($P$): Also known as real power or active power, measured in watts ($\text{W}$) or kilowatts ($\text{kW}$). True power represents the rate of actual energy consumed and transformed into mechanical shaft rotation, heat, or light: $P = V \times I \times \cos \theta = I^2 \times R$
2. Reactive Power ($Q$): Measured in volt-amperes reactive ($\text{VAR}$) or kilovolt-amperes reactive ($\text{kVAR}$). Reactive power represents the energy that oscillates back and forth between the power source and the magnetic/electrostatic fields without doing net work: $Q = V \times I \times \sin \theta = I^2 \times X$
3. Apparent Power ($S$): Measured in volt-amperes ($\text{VA}$) or kilovolt-amperes ($\text{kVA}$). Apparent power represents the total vector combination of true and reactive power, reflecting the total capacity that utility generators, transformers, and conductors must supply: $S = V \times I = I^2 \times Z = \sqrt{P^2 + Q^2}$

Understanding Power Factor ($\text{PF}$)

Power factor is the ratio of true work-performing power to apparent power supplied:

$\text{PF} = \frac{P}{S} = \cos \theta$

Where $\theta$ is the phase angle between voltage and current waveforms. Power factor is expressed either as a decimal between $0.0$ and $1.0$, or as a percentage from $0\%$ to $100\%$.

• Lagging Power Factor: Occurs in inductive circuits (e.g., induction motors, welders, HID ballasts) where current lags voltage.
• Leading Power Factor: Occurs in capacitive circuits where current leads voltage.
• Unity Power Factor ($1.0$ or $100\%$): Occurs in purely resistive circuits where voltage and current are in phase ($S = P$).

Practical Trade Impact: Conductor Sizing and Power Factor Correction

A low power factor forces conductors, circuit breakers, and distribution transformers to carry excessive current to deliver a given amount of true power. For instance, consider a $480\text{ V}$ industrial pump drawing $48\text{ kW}$ of true power:

• At 0.60 Power Factor: $S = \frac{48\text{ kW}}{0.60} = 80\text{ kVA} \implies I = \frac{80,000\text{ VA}}{480\text{ V}} \approx 166.7\text{ A}$
• At 1.00 Power Factor (Corrected): $S = \frac{48\text{ kW}}{1.00} = 48\text{ kVA} \implies I = \frac{48,000\text{ VA}}{480\text{ V}} = 100.0\text{ A}$

By installing power factor correction capacitors in parallel with inductive loads, the capacitive leading current ($I_C$) cancels out the inductive lagging current ($I_L$). Conductor current drops by $66.7\text{ A}$ ($40\%$), reducing $I^2 R$ heat loss, freeing up feeder capacity, and eliminating utility low-power-factor penalties.

Sizing Correction Capacitors (kVAR)

To raise a load's power factor, a capacitor bank supplies the reactive power that the load would otherwise draw from the source. The capacitor rating is the difference between the reactive power before and after correction:

$Q_C = P \times (\tan\theta_1 - \tan\theta_2)$

where $\theta_1 = \cos^{-1}(\text{PF}_1)$ and $\theta_2 = \cos^{-1}(\text{PF}_2)$.

Example: A 100 kW load runs at 0.70 PF and is to be corrected to 0.95 PF.

• $\theta_1 = \cos^{-1}(0.70) = 45.57^\circ$, so $\tan\theta_1 = 1.020$
• $\theta_2 = \cos^{-1}(0.95) = 18.19^\circ$, so $\tan\theta_2 = 0.329$
• $Q_C = 100 \times (1.020 - 0.329) = 69.1$ kVAR

Apparent power drops from $100 \div 0.70 = 142.9$ kVA to $100 \div 0.95 = 105.3$ kVA, so the source and feeder current fall by about 26%. The capacitor installation itself is covered by NEC Article 460, which requires capacitor circuit conductors to have an ampacity of at least 135% of the capacitor's rated current.

---

Summary of AC Power Quantities