8.2 AC Circuit Theory: Reactance, Impedance & Power Factor

AC Circuit Theory: Reactance, Impedance & Power Factor

When direct current flows through a closed circuit, the total opposition to current is governed entirely by electrical resistance ($R = V / I$). However, in alternating current systems, current continually alternates in direction and magnitude. This dynamic variation activates additional electromagnetic and electrostatic properties whenever conductors are coiled into inductors or separated across insulating dielectrics as capacitors. In AC circuits, total opposition to current flow—termed impedance ($Z$)—is a composite vector quantity incorporating both pure resistance and frequency-dependent reactance.

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Component Behaviour in AC Circuits

To understand composite AC circuits, an electrician must first isolate how pure resistance, pure inductance, and pure capacitance behave when energized by a sinusoidal alternating supply.

1. Pure Resistance ($R$)

In a purely resistive AC circuit (such as an incandescent filament or immersion heating element):

• The current waveform rises, peaks, and crosses zero at the exact same instant as the applied voltage waveform. Current and voltage are in phase; the phase angle is zero ($\phi = 0^\circ$).
• Ohm's law applies directly using RMS values: $V = I \times R$.
• Electrical energy is converted irreversibly into thermal energy or light; no energy is stored or returned to the supply.

2. Pure Inductance ($L$)

An inductor consists of a coil of wire (such as motor windings, chokes, or transformer coils). When alternating current flows through the coil, it establishes an alternating magnetic flux. By Faraday's Law and Lenz's Law, this changing flux induces a continuous counter-electromotive force (back-EMF) that opposes the change in current ($e = -L \frac{di}{dt}$).

• Phase Relationship: Because the back-EMF is greatest when the current is changing most rapidly (at zero crossings), the current is delayed relative to the applied voltage. In a purely inductive circuit, current lags applied voltage by exactly $90^\circ$ ($\frac{\pi}{2}$ radians, or one quarter of a cycle).
• Inductive Reactance ($X_L$): The opposition presented by an inductor to alternating current is called inductive reactance, measured in Ohms ($\Omega$): $X_L = 2\pi f L = \omega L$ Where:
  • $X_L$ = Inductive reactance in Ohms ($\Omega$)
  • $f$ = Supply frequency in Hertz ($\text{Hz}$)
  • $L$ = Inductance in Henrys ($\text{H}$)
• Frequency Dependence: Inductive reactance is directly proportional to frequency. At DC ($f = 0\text{ Hz}$), $X_L = 0\ \Omega$ (the inductor acts as a short circuit, offering only winding resistance). As frequency rises, the rate of flux change increases, and $X_L$ increases linearly.

3. Pure Capacitance ($C$)

A capacitor consists of two conductive plates separated by an insulating dielectric medium (such as power factor correction banks or motor run capacitors). When connected to an AC supply, the plates alternately charge, discharge, charge with reverse polarity, and discharge.

• Phase Relationship: Electric current flows onto the plates most rapidly when the applied voltage is crossing zero and changing at its highest rate ($i = C \frac{dv}{dt}$). As the capacitor voltage reaches peak, charging current ceases. In a purely capacitive circuit, current leads applied voltage by exactly $90^\circ$ ($\frac{\pi}{2}$ radians).
• Capacitive Reactance ($X_C$): The opposition presented by a capacitor to alternating current is called capacitive reactance, measured in Ohms ($\Omega$): $X_C = \frac{1}{2\pi f C} = \frac{1}{\omega C}$ Where:
  • $X_C$ = Capacitive reactance in Ohms ($\Omega$)
  • $f$ = Supply frequency in Hertz ($\text{Hz}$)
  • $C$ = Capacitance in Farads ($\text{F}$)
• Frequency Dependence: Capacitive reactance is inversely proportional to frequency. At DC ($f = 0\text{ Hz}$), $X_C = \infty\ \Omega$ (a capacitor completely blocks direct current). As frequency increases, the capacitor charges and discharges more frequently, allowing higher current to pass and reducing $X_C$.

The "CIVIL" Phase Mnemonic

Electricians use the standard industry mnemonic CIVIL to instantly recall phase relationships:

• In a Capacitor, current I leads voltage V (C - I - V).
• Voltage V leads current I in an inductive coil L (V - I - L).

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Series AC Circuits and Impedance ($Z$)

Practical circuits are rarely purely resistive or purely reactive. Real electrical equipment—such as motors, fluorescent luminaires, and transformers—possesses both resistance and inductance. When resistance and reactance occur in series, the total opposition to alternating current is called impedance (symbol $Z$, measured in Ohms, $\Omega$).

Why Arithmetic Addition Fails in AC Circuits

In a series DC circuit, component voltages add arithmetically ($V_T = V_1 + V_2$). In an AC circuit containing resistance and reactance, arithmetic addition cannot be used because the voltages are out of phase with each other:

• The voltage drop across the resistor ($V_R = I R$) is in phase with current.
• The voltage drop across the inductor ($V_L = I X_L$) leads current by $90^\circ$.
• The voltage drop across the capacitor ($V_C = I X_C$) lags current by $90^\circ$.

Because $V_R$ and $V_L$ act at right angles ($90^\circ$) to each other on a phasor diagram, they must be combined using Pythagoras' Theorem.

Series R-L Circuit Formulas

• Total Circuit Impedance: $Z = \sqrt{R^2 + X_L^2}$
• Total Supply Voltage: $V_T = \sqrt{V_R^2 + V_L^2}$
• Circuit Current: $I = \frac{V_T}{Z}$
• Phase Angle: $\cos\phi = \frac{R}{Z} = \frac{V_R}{V_T}, \qquad \tan\phi = \frac{X_L}{R}$

Series R-C Circuit Formulas

• Total Circuit Impedance: $Z = \sqrt{R^2 + X_C^2}$
• Total Supply Voltage: $V_T = \sqrt{V_R^2 + V_C^2}$
• Phase Angle: Current leads voltage by $\phi$, where $\cos\phi = \frac{R}{Z}$.

Series R-L-C Circuit and Electrical Resonance

When resistance, inductance, and capacitance are connected in series, the inductive and capacitive reactances point in diametrically opposite directions on the phasor diagram ($180^\circ$ apart). They directly oppose each other. The net reactance ($X$) of the circuit is their arithmetic difference:

$X = X_L - X_C$

The total impedance of a series R-L-C circuit is:

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

The Condition of Electrical Series Resonance

A unique operating condition occurs when the supply frequency is adjusted until inductive reactance equals capacitive reactance:

$X_L = X_C$

When this occurs:

1. The net reactance cancels to zero: $X = X_L - X_C = 0\ \Omega$.
2. Circuit impedance collapses to its absolute minimum value, which is equal to pure resistance alone: $Z = R$.
3. Circuit current reaches its absolute maximum possible value: $I = \frac{V_T}{R}$.
4. Current and voltage are completely in phase ($\phi = 0^\circ$, unity power factor, $\cos\phi = 1.0$).

The frequency at which resonance occurs is the resonant frequency ($f_0$):

$2\pi f_0 L = \frac{1}{2\pi f_0 C} \implies f_0^2 = \frac{1}{4\pi^2 L C} \implies f_0 = \frac{1}{2\pi\sqrt{LC}}$

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Power in AC Circuits: The Power Triangle

In DC circuits, electrical power is calculated simply as $P = V \times I$. In AC circuits containing inductive or capacitive reactances, current and voltage are displaced by phase angle $\phi$. Consequently, power divides into three distinct components forming the Power Triangle:

1. True Power ($P$)

Also known as active power or real power, this is the actual rate at which electrical energy is converted into useful work (mechanical rotation, light) or heat. It is dissipated exclusively in the resistive elements of the circuit.

$P = V \times I \times \cos\phi = I^2 R$

True power is measured in Watts ($\text{W}$) or kilowatts ($\text{kW}$, where $1\text{ kW} = 1,000\text{ W}$).

2. Reactive Power ($Q$)

Reactive power is the power absorbed and returned by reactive components (inductors and capacitors) as their magnetic and electrostatic fields build up and collapse each half-cycle. It performs no net useful work, but is physically necessary to sustain the alternating magnetic flux in motors and transformers.

$Q = V \times I \times \sin\phi = I^2 X$

Reactive power is measured in Volt-Amperes Reactive ($\text{VAR}$) or kilovar ($\text{kVAR}$). By convention, inductive loads consume lagging VARs, while capacitors generate leading VARs.

3. Apparent Power ($S$)

Apparent power is the total product of the RMS supply voltage and RMS supply current. It represents the total generating and cable capacity that electricity suppliers must provide to energize the installation.

$S = V \times I = \sqrt{P^2 + Q^2} = I^2 Z$

Apparent power is measured in Volt-Amperes ($\text{VA}$) or kilovolt-amperes ($\text{kVA}$). Transformers and backup generators are rated in kVA because their heating limits depend on total voltage and current, regardless of phase angle.

4. Power Factor ($\text{PF} = \cos\phi$)

The power factor is defined as the ratio of true active power to total apparent power:

$\text{Power Factor (PF)} = \frac{\text{True Power } (P)}{\text{Apparent Power } (S)} = \frac{R}{Z} = \cos\phi$

Power factor is a dimensionless number between $0.0$ and $1.0$:

• Unity Power Factor ($1.0$): Purely resistive load; all power is converted into useful work ($P = S, Q = 0$).
• Lagging Power Factor: Occurs in inductive circuits (motors, transformers, ballasts) where current lags voltage.
• Leading Power Factor: Occurs in capacitive circuits where current leads voltage.

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Practical Consequences of Low Power Factor & Power Factor Correction

Most industrial and commercial loads consist of AC induction motors, welding plant, and fluorescent lighting ballasts. These loads are heavily inductive, operating with uncorrected lagging power factors typically between $0.60$ and $0.80$.

Operational and Economic Penalties of Low Power Factor

To deliver a specified quantity of active power ($P$) to a factory, the line current drawn from the supply is:

$I = \frac{P}{V \times \cos\phi}$

If the power factor drops from $1.0$ to $0.5$, the line current drawn by the installation doubles to perform the exact same useful mechanical work:

1. Increased Cable Heating and Losses: Cable power loss increases with the square of current ($I^2R$). Doubling line current quadruples thermal distribution losses.
2. Excessive Voltage Drop: Higher current causes larger voltage drops along distribution feeders ($V_d = I Z$), impairing terminal voltage at machines and causing luminaire flicker.
3. Overloaded Infrastructure: Distribution transformers, switchgear, and circuit breakers must be significantly larger (higher kVA ratings) to accommodate the extra reactive current.
4. Commercial Surcharges: Electricity supply companies (DNOs) meter both active energy (kWh) and reactive energy (kVARh). If a commercial facility operates below a prescribed threshold (typically $0.90$ or $0.95$ lagging), heavy financial penalties are levied on monthly bills.

Power Factor Correction (PFC)

To eliminate these penalties, electrical contractors install Power Factor Correction (PFC) capacitor banks connected in parallel across the incoming factory distribution switchboard or directly at motor terminals.

• Because capacitors draw leading reactive current, they supply the magnetizing VARs required by the inductive motor locally.
• The leading reactive power of the capacitor cancels the lagging reactive power of the inductor ($Q_{\text{net}} = Q_L - Q_C$).
• The supply network is relieved of transmitting reactive power, causing total line current to fall and raising overall power factor close to unity (typically $0.95$ to $0.98$ lagging).

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Step-by-Step Worked Calculations

Worked Example 1: Inductive Reactance and Capacitive Reactance

Problem: A circuit fed from a $230\text{ V}$, $50\text{ Hz}$ supply contains a choke coil with an inductance of $0.15\text{ H}$ and a capacitor with a capacitance of $68\ \mu\text{F}$. Calculate the inductive reactance ($X_L$) of the choke and the capacitive reactance ($X_C$) of the capacitor.

Solution:

1. Inductive reactance: $X_L = 2\pi f L = 2 \times \pi \times 50 \times 0.15 = 314.16 \times 0.15 = 47.12\ \Omega$
2. Capacitive reactance (convert $\mu\text{F}$ to Farads: $68\ \mu\text{F} = 68 \times 10^{-6}\text{ F}$): $X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times \pi \times 50 \times (68 \times 10^{-6})} = \frac{1}{314.16 \times 6.8 \times 10^{-5}} = \frac{1}{0.02136} = 46.81\ \Omega$

Worked Example 2: Series R-L Circuit Analysis

Problem: A contactor coil has an internal resistance of $30\ \Omega$ and an inductive reactance of $40\ \Omega$. It is connected across a $230\text{ V}$, $50\text{ Hz}$ AC supply. Calculate:

1. Total circuit impedance ($Z$).
2. Circuit current ($I$).
3. Resistor voltage drop ($V_R$) and inductor voltage drop ($V_L$).
4. Circuit power factor ($\cos\phi$).
5. True power dissipated ($P$).

Solution:

1. Total impedance: $Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1,600} = \sqrt{2,500} = 50\ \Omega$
2. Circuit current: $I = \frac{V_T}{Z} = \frac{230\text{ V}}{50\ \Omega} = 4.60\text{ A}$
3. Component voltages: $V_R = I \times R = 4.60\text{ A} \times 30\ \Omega = 138\text{ V}$ $V_L = I \times X_L = 4.60\text{ A} \times 40\ \Omega = 184\text{ V}$ (Verification: $\sqrt{138^2 + 184^2} = \sqrt{19,044 + 33,856} = \sqrt{52,900} = 230\text{ V}$)
4. Power factor: $\cos\phi = \frac{R}{Z} = \frac{30}{50} = 0.60\text{ lagging}$
5. True power: $P = V_T \times I \times \cos\phi = 230 \times 4.60 \times 0.60 = 634.8\text{ W}$ (Alternatively: $P = I^2 R = 4.60^2 \times 30 = 21.16 \times 30 = 634.8\text{ W}$)

Worked Example 3: Power Triangle and PFC Benefits

Problem: A commercial sub-main supplies a $230\text{ V}$ load drawing an active power of $9.2\text{ kW}$ at an uncorrected power factor of $0.64$ lagging. Calculate the line current drawn. If parallel power factor correction capacitors improve the power factor to $0.96$ lagging, determine the new line current and the current reduction achieved.

Solution:

1. Uncorrected line current ($I_1$): $I_1 = \frac{P}{V \times \cos\phi_1} = \frac{9,200\text{ W}}{230\text{ V} \times 0.64} = \frac{9,200}{147.2} = 62.50\text{ A}$
2. Corrected line current ($I_2$): $I_2 = \frac{P}{V \times \cos\phi_2} = \frac{9,200\text{ W}}{230\text{ V} \times 0.96} = \frac{9,200}{220.8} = 41.67\text{ A}$
3. Reduction in line current: $\Delta I = 62.50\text{ A} - 41.67\text{ A} = 20.83\text{ A} \quad (33.3\%\text{ reduction})$ This significant current reduction allows existing cables to run cooler, reduces voltage drop, and avoids DNO reactive power penalty charges.