2.4 Impedance, Power Triangles & Power Factor

2.4 Impedance, Power Triangles & Power Factor

Practical commercial and industrial AC systems rarely contain pure resistance, pure inductance, or pure capacitance in isolation. Electric discharge luminaires, commercial refrigeration compressors, three-phase induction motors, and electronic switch-mode power supplies present complex combinations of resistive and reactive elements. To properly size service conductors, transformers, and switchboards, an electrician must master vector impedance calculations and power factor correction principles.

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AC Impedance ($Z$) and Phasor Analysis

Impedance ($Z$, measured in Ohms, $\Omega$) is the total opposition an AC circuit presents to the flow of alternating current. Impedance combines pure resistance ($R$) and net reactance ($X$) into a single vector quantity.

Because resistance causes current and voltage to be in phase ($0^\circ$), inductive reactance causes voltage to lead current by $+90^\circ$ ($+j$), and capacitive reactance causes voltage to lag current by $-90^\circ$ ($-j$), these quantities cannot be summed with basic arithmetic. They must be added as vectors (phasors) in a complex plane:

           +j (Inductive Reactance, +XL)
               |
               |      * Phasor Z = R + j(XL - XC)
               |    * |
               |  *   | Net Reactance
               |* θ   | X = (XL - XC)
   ------------+------+------------ Real Axis (Resistance, R)
               |
               |
               |
           -j (Capacitive Reactance, -XC)

Series RLC Circuit Relationships

In a series RLC circuit:

1. Net Reactance ($X$): Because inductive and capacitive reactances are $180^\circ$ out of phase, they cancel each other out directly: $X = X_L - X_C$
   • If $X_L > X_C$, the circuit is net inductive (current lags voltage).
   • If $X_C > X_L$, the circuit is net capacitive (current leads voltage).
   • If $X_L = X_C$, the circuit is in resonance.
2. Total Impedance ($Z$): Derived from the Pythagorean theorem on the impedance triangle: $Z = \sqrt{R^2 + X^2} = \sqrt{R^2 + (X_L - X_C)^2}$
3. Phase Angle ($\theta$): The angular displacement between total applied voltage and circuit current: $\theta = \arctan\left(\frac{X_L - X_C}{R}\right)$
4. Ohm's Law for AC Circuits: $I = \frac{V}{Z} \qquad V = I \times Z \qquad Z = \frac{V}{I}$

Worked Example: Series RLC Circuit Calculation

Problem: A $240\text{V}, 60\text{ Hz}$ commercial branch circuit powers a series network consisting of a resistance $R = 30\ \Omega$, an inductive reactance $X_L = 70\ \Omega$, and a capacitive reactance $X_C = 30\ \Omega$. Determine net reactance, total impedance, circuit current, phase angle, and the voltage drop across each component.

1. Calculate net reactance ($X$): $X = X_L - X_C = 70\ \Omega - 30\ \Omega = 40\ \Omega\text{ (Inductive)}$
2. Calculate total impedance ($Z$): $Z = \sqrt{R^2 + X^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1,600} = \sqrt{2,500} = 50\ \Omega$
3. Calculate circuit current ($I$): $I = \frac{V}{Z} = \frac{240\text{ V}}{50\ \Omega} = 4.8\text{ Amps}$
4. Calculate phase angle ($\theta$): $\theta = \arctan\left(\frac{40}{30}\right) = \arctan(1.3333) \approx 53.13^\circ\text{ (lagging)}$
5. Calculate individual component voltage drops:
   • Across resistance: $V_R = I \times R = 4.8\text{ A} \times 30\ \Omega = 144\text{ V}$
   • Across inductor: $V_L = I \times X_L = 4.8\text{ A} \times 70\ \Omega = 336\text{ V}$
   • Across capacitor: $V_C = I \times X_C = 4.8\text{ A} \times 30\ \Omega = 144\text{ V}$
6. Verify total source voltage using vector addition: $V_S = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{144^2 + (336 - 144)^2} = \sqrt{144^2 + 192^2} = \sqrt{20,736 + 36,864} = \sqrt{57,600} = 240\text{ V}$

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Series Resonance

Series resonance occurs at the specific operating frequency where inductive reactance exactly equals capacitive reactance ($X_L = X_C$). Under this condition:

$2\pi f_r L = \frac{1}{2\pi f_r C} \implies f_r = \frac{1}{2\pi \sqrt{LC}}$

Key Electrical Conditions at Series Resonance

1. Zero Net Reactance: $X_{net} = X_L - X_C = 0\ \Omega$.
2. Minimum Impedance: Total circuit impedance collapses to pure resistance: $Z = R$.
3. Maximum Current: Circuit current reaches its absolute upper limit: $I_{max} = \frac{V_S}{R}$.
4. Unity Power Factor: Current and voltage are perfectly in phase ($\theta = 0^\circ, \cos\theta = 1.0$).
5. Voltage Magnification ($Q$-factor): The Quality Factor ($Q = \frac{X_L}{R}$) can produce massive resonant overvoltages across $L$ and $C$ ($V_L = V_C = Q \times V_S$), capable of rupturing capacitor dielectric layers or flashing over insulation in commercial variable frequency drive (VFD) filter networks.

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The AC Power Triangle

In direct current circuits, power is simply $P = V \times I$. In alternating current circuits containing reactance, voltage and current are displaced by phase angle $\theta$, producing three distinct forms of electrical power that form the AC Power Triangle:

                     /| 
                    / | 
     Apparent Power/  | Reactive Power 
           (S, VA)/   | (Q, VAR)
                 /    | 
                /θ    | 
               +------+ 
              True Power (P, Watts)

1. True Power ($P$ / Real Power / Active Power)

• The actual rate at which electrical energy is transformed into mechanical work, heat, or light.
• Measured in Watts (W) or Kilowatts (kW).
• Formula: $P = V \times I \times \cos\theta = I^2 R$

2. Reactive Power ($Q$ / Quadrature Power)

• Power oscillating bidirectionally between the source and reactive storage fields (magnetic fields in inductors, electric fields in capacitors). Reactive power performs no useful mechanical work, but is required to sustain the alternating magnetic flux in motors, transformers, ballasts, and solenoids.
• Measured in Volt-Amperes Reactive (VAR) or kilovar (kVAR).
• Formula: $Q = V \times I \times \sin\theta = I^2 X$

3. Apparent Power ($S$ / Total Power)

• The total vector power supplied to the circuit by the source. All utility service equipment, transformers, switchgear, and generators are sized and rated in terms of Apparent Power.
• Measured in Volt-Amperes (VA) or kilovolt-amperes (kVA).
• Formula: $S = V \times I = I^2 Z = \sqrt{P^2 + Q^2}$

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Power Factor (PF) & Commercial Impacts

Power Factor represents the efficiency with which electrical power is delivered and utilized. It is defined as the mathematical ratio of True Power to Apparent Power:

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

Power factor ranges between $0.0$ and $1.0$ (or $0\%$ to $100\%$):

• Unity Power Factor ($\text{PF} = 1.0$): True power equals apparent power ($P = S$). All current performs useful work; phase angle $\theta = 0^\circ$.
• Lagging Power Factor: Current lags voltage (inductive loads such as induction motors, welders, and magnetic ballasts). This is the dominant state of commercial and industrial facilities.
• Leading Power Factor: Current leads voltage (capacitive loads, unloaded long underground cables, or over-excited synchronous motors).

Adverse Consequences of Poor Power Factor

Operating a commercial facility at poor power factor (e.g., $\text{PF} < 0.85$) imposes severe operational and economic burdens:

1. Excessive Line Current: To deliver a fixed amount of real work ($P$), lower power factor forces total line current to rise: $I = \frac{P}{V \times \text{PF}}$
2. Elevated Conductor Thermal Losses ($I^2R$): Higher line current causes heat dissipation in facility feeders and distribution transformers to surge with the square of current.
3. Premature Transformer & Switchgear Capacity Saturation: Distribution equipment is rated in kVA. Non-working reactive current consumes transformer ampacity, preventing the addition of new revenue-producing loads.
4. Increased System Voltage Drop: High line currents increase $I \times Z$ impedance drops across feeder conductors, causing brownouts and motor under-voltage.
5. Utility Demand Penalties: Commercial electric utility tariffs impose direct financial surcharges or bill based on peak kVA demand if average facility power factor falls below a contractual threshold (typically $0.90$ or $0.95$).

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Power Factor Correction Engineering

Because induction motors and transformers require inductive magnetizing current (lagging by $90^\circ$), connecting shunt capacitor banks in parallel across facility distribution buses supplies this reactive current locally. The capacitor draws leading current that cancels out the lagging current of the inductive loads, freeing upstream utility lines and transformers to carry only true active power.

Sizing Power Factor Correction Capacitors

To correct facility power factor from an initial lagging value ($\text{PF}_1 = \cos\theta_1$) to an improved target value ($\text{PF}_2 = \cos\theta_2$):

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

Where:

• $Q_C = \text{required capacitive rating in kVAR}$
• $P = \text{facility true power in kW}$
• $\theta_1 = \arccos(\text{PF}_1) = \text{initial phase angle}$
• $\theta_2 = \arccos(\text{PF}_2) = \text{target phase angle}$

Worked Example: Commercial Facility Power Factor Correction

Problem: A commercial fabrication facility operates on a $480\text{V}$, 3-phase service with a steady active load $P = 300\text{ kW}$ at an uncorrected power factor of $0.72\text{ lagging}$ ($\text{PF}_1 = 0.72$). Management wishes to install a shunt capacitor bank to raise the facility power factor to $0.95\text{ lagging}$ ($\text{PF}_2 = 0.95$) to eliminate utility low-power-factor penalties.

1. Calculate initial apparent power ($S_1$) and line current ($I_1$): $S_1 = \frac{P}{\text{PF}_1} = \frac{300\text{ kW}}{0.72} \approx 416.67\text{ kVA}$ $I_1 = \frac{S_1}{\sqrt{3} \times V} = \frac{416,667\text{ VA}}{1.732 \times 480\text{ V}} = \frac{416,667}{831.38} \approx 501.2\text{ Amps}$
2. Determine initial and target phase angles: $\theta_1 = \arccos(0.72) \approx 43.95^\circ \implies \tan(43.95^\circ) \approx 0.9639$ $\theta_2 = \arccos(0.95) \approx 18.19^\circ \implies \tan(18.19^\circ) \approx 0.3287$
3. Calculate initial and target reactive power: $Q_1 = P \times \tan\theta_1 = 300\text{ kW} \times 0.9639 \approx 289.17\text{ kVAR}$ $Q_2 = P \times \tan\theta_2 = 300\text{ kW} \times 0.3287 \approx 98.61\text{ kVAR}$
4. Calculate required capacitor bank rating ($Q_C$): $Q_C = Q_1 - Q_2 = 289.17\text{ kVAR} - 98.61\text{ kVAR} = 190.56\text{ kVAR}$ (Specify a standard commercial $200\text{ kVAR}$ switched capacitor bank).
5. Calculate post-correction apparent power ($S_2$) and line current ($I_2$): $S_2 = \frac{P}{\text{PF}_2} = \frac{300\text{ kW}}{0.95} \approx 315.79\text{ kVA}$ $I_2 = \frac{S_2}{\sqrt{3} \times V} = \frac{315,789\text{ VA}}{831.38} \approx 379.8\text{ Amps}$
6. Evaluation of Results: Correcting the power factor drops feeder line current from $501.2\text{ A}$ to $379.8\text{ A}$—an immediate reduction of $121.4\text{ Amps}$ ($24.2\%$ current reduction). Conductor $I^2R$ thermal losses throughout the facility main service decrease by: $\text{Loss Reduction} = 1 - \left(\frac{379.8}{501.2}\right)^2 = 1 - (0.7578)^2 = 1 - 0.5743 = 42.57\%$