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Resistive vs Reactive

Resistive vs reactive load comparison: power factor impact, current draw, energy billing, and correction methods with IEEE/IEC guidelines for load analysis.

Enginist Team
Published: October 17, 2025
Updated: November 27, 2025

Resistive vs Reactive Loads: Complete Engineering Comparison

Quick AnswerWhat is the difference between resistive and reactive loads?
Resistive loads (heaters, incandescent bulbs) have unity power factor (PF = 1.0) where current and voltage are in phase. Reactive loads (motors, transformers) have PF = 0.7-0.95 due to phase shift between current and voltage, requiring more current for the same real power. At PF = 0.8, a load draws 25% more current than an equivalent resistive load, increasing conductor sizing, losses, and utility bills.

Quick Verdict

Understanding the difference between resistive and reactive loads is fundamental to electrical system design, energy efficiency, and cost management. The distinction affects everything from conductor sizing to utility bills.

Bottom Line: Resistive loads are simpler—what you see is what you get. A 1000W heater draws ~4.2A at 240V, period. Reactive loads require more nuanced analysis. A 1000W motor at PF = 0.8 draws ~5.2A at 240V—25% more current for the same useful work. This difference compounds across facilities, affecting transformer sizing, cable costs, and monthly utility bills.

Most commercial and industrial facilities have predominantly reactive loads (motors account for 60-70% of industrial electricity use), making power factor management essential for cost control.

At-a-Glance Comparison Table

CharacteristicResistive LoadReactive LoadImpact
Power Factor1.0 (unity)0.7-0.95 typicalReactive needs correction
Current vs PowerI = P/VI = P/(V×PF)Reactive draws more current
Phase RelationshipV and I in phaseV and I out of phaseAffects metering
Power CalculationP=V×IP = V \times IP=V×I×PFP = V \times I \times PFReactive more complex
Heat GenerationAll power → heatReal power → work + heatDifferent efficiency
Utility BillingkW onlykW + kVA demand + PF penaltyReactive costs more
ExamplesHeaters, incandescentMotors, transformersKnow your load mix

Power Factor: The Fundamental Difference

Power factor is the ratio of real power (kW) doing useful work to apparent power (kVA) flowing in the circuit. It's the key metric distinguishing resistive from reactive loads.

Resistive Load Power Factor

Resistive loads convert electrical energy directly to heat (or light in incandescent bulbs). Current and voltage rise and fall together—they're "in phase."

Power factor = 1.0 means:

  • All current does useful work
  • Apparent power equals real power (kVA = kW)
  • Simple calculations: P=V×IP = V \times I
  • No reactive current component

Common resistive loads and their characteristics:

Load TypeTypical PFPower Relationship
Electric heaters1.0P=V×IP = V \times I directly
Incandescent bulbs1.0P=V×IP = V \times I directly
Electric ovens0.98-1.0Nearly pure resistive
Resistance welders0.95-1.0Slight inductance in cables
Water heaters1.0P=V×IP = V \times I directly

Reactive Load Power Factor

Reactive loads contain inductance (motors, transformers) or capacitance that stores and returns energy. Current and voltage are "out of phase"—current lags voltage in inductive loads, leads in capacitive.

Power factor < 1.0 means:

  • Some current doesn't do useful work (reactive current)
  • Apparent power exceeds real power (kVA > kW)
  • Current calculation requires PF: I=P/(V×PF)I = P / (V \times PF)
  • Reactive power (kVAR) oscillates between source and load

Common reactive loads and their characteristics:

Load TypeTypical PFNotes
Induction motors (full load)0.85-0.92PF drops at partial load
Induction motors (half load)0.70-0.80Significant PF drop
Transformers (loaded)0.80-0.95Depends on secondary load
Fluorescent lighting (magnetic)0.50-0.60Very low without correction
Fluorescent lighting (electronic)0.90-0.99Built-in PF correction
VFDs/Inverters0.95-0.99Input side typically high PF

Verdict: Power Factor

Winner: Resistive for simplicity — Resistive loads don't require power factor analysis or correction. However, reactive loads (especially motors) are essential for mechanical work. The goal isn't to avoid reactive loads but to manage their power factor impact through proper system design and correction.

Current Draw: The Practical Impact

The most immediate effect of power factor is on current draw. Lower power factor means higher current for the same real power output.

Current Calculation Comparison

For the same real power output, reactive loads draw more current:

Resistive Load (PF = 1.0): I=PV (single-phase)I = \frac{P}{V} \text{ (single-phase)} I=P3×V (three-phase)I = \frac{P}{\sqrt{3} \times V} \text{ (three-phase)}

Reactive Load (PF < 1.0): I=PV×PF (single-phase)I = \frac{P}{V \times PF} \text{ (single-phase)} I=P3×V×PF (three-phase)I = \frac{P}{\sqrt{3} \times V \times PF} \text{ (three-phase)}

Current Increase at Various Power Factors

Power FactorCurrent MultiplierExample: 10 kW at 400V 3φ
1.0 (resistive)1.00×14.4A
0.951.05×15.2A
0.901.11×16.0A
0.851.18×17.0A
0.801.25×18.0A
0.751.33×19.2A
0.701.43×20.6A

At PF = 0.70, current is 43% higher than resistive equivalent. This affects:

  • Conductor sizing: Larger cables needed
  • I²R losses: 2× current = 4× losses in cables
  • Transformer capacity: kVA rating, not kW
  • Circuit breaker sizing: Based on full current

Verdict: Current Draw

Winner: Resistive — Lower current for the same useful work means smaller conductors, lower losses, and simpler protection. However, proper power factor correction can bring reactive load current close to resistive equivalent.

Cost Impact: The Business Case

Power factor directly affects electricity costs through multiple mechanisms. Understanding these is essential for facility managers and engineers.

Utility Billing Components

Most commercial/industrial rates include:

  1. Energy Charge (kWh): Same for resistive and reactive—measures real work done
  2. Demand Charge (kW or kVA): If kVA-based, penalizes low PF
  3. Power Factor Penalty: Direct surcharge below threshold (typically 0.85-0.90)

Cost Comparison Example

Manufacturing Facility: 500 kW Load

Scenario A: Resistive-Dominated (PF = 0.95)

  • Real Power: 500 kW
  • Apparent Power: 500 / 0.95 = 526 kVA
  • Demand Charge (at $15/kVA): $7,890/month
  • PF Penalty: None (above 0.90 threshold)
  • Total Demand Cost: $7,890/month

Scenario B: Motor-Heavy (PF = 0.75)

  • Real Power: 500 kW
  • Apparent Power: 500 / 0.75 = 667 kVA
  • Demand Charge (at $15/kVA): $10,005/month
  • PF Penalty (20% of demand below 0.90): $2,000/month
  • Total Demand Cost: $12,005/month

Monthly Savings with PF Correction: $4,115 Annual Savings: $49,380

Power Factor Correction ROI

Correcting PF from 0.75 to 0.95 for the 500 kW facility:

Required capacitors: kVAR=500×(tan(cos1(0.75))tan(cos1(0.95)))kVAR = 500 \times (\tan(\cos^{-1}(0.75)) - \tan(\cos^{-1}(0.95))) kVAR=500×(0.8820.329)=277 kVARkVAR = 500 \times (0.882 - 0.329) = 277 \text{ kVAR}

  • Capacitor cost: ~$30-50/kVAR installed = $8,300-13,850
  • Annual savings: $49,380
  • Payback: 2-4 months

Verdict: Cost

Winner: Corrected reactive loads — While resistive loads inherently have lower costs per kW, most facilities need motors. Proper power factor correction brings reactive load costs close to resistive equivalent while maintaining necessary equipment.

The Power Triangle: Visualizing the Difference

The power triangle illustrates the relationship between real, reactive, and apparent power—essential for understanding why reactive loads behave differently.

Power Components

  • Real Power (P, kW): Does useful work—runs motors, heats elements, produces light
  • Reactive Power (Q, kVAR): Stored/returned in magnetic or electric fields—does no useful work but is necessary for motor operation
  • Apparent Power (S, kVA): Total power flowing in circuit = √(P² + Q²)

Triangle Relationships

S=P2+Q2S = \sqrt{P^2 + Q^2}

PF=cos(θ)=PS=kWkVAPF = \cos(\theta) = \frac{P}{S} = \frac{kW}{kVA}

Q=P×tan(θ)Q = P \times \tan(\theta)

For resistive loads: Q = 0, so S = P and PF = 1.0 For reactive loads: Q > 0, so S > P and PF < 1.0

Practical Interpretation

Load ConditionP (kW)Q (kVAR)S (kVA)PF
Pure resistive10001001.00
Motor at 0.85 PF100621180.85
Motor at 0.75 PF100881330.75
Uncorrected fluorescent1001732000.50

The motor at PF = 0.75 requires 33% more apparent power (and current) than a resistive load doing the same real work.

Application-Specific Recommendations

When Resistive Loads Dominate

Facilities with primarily resistive loads have simpler electrical systems:

Typical Applications:

  • Electric heating plants
  • Smelting operations with resistance furnaces
  • Incandescent lighting (increasingly rare)
  • Electric hot water systems
  • Food processing with heating elements

Design Considerations:

  • Power factor correction rarely needed
  • Simple load calculations (kW = kVA)
  • Transformer and cable sizing based on kW
  • No capacitor banks required
  • Harmonics from electronic controls may be the main concern

When Reactive Loads Dominate

Most industrial and commercial facilities fall into this category:

Typical Applications:

  • Manufacturing with motor-driven equipment
  • HVAC systems (compressors, fans, pumps)
  • Water/wastewater treatment plants
  • Commercial buildings with elevators and HVAC
  • Data centers (UPS systems, cooling)

Design Considerations:

  • Size transformers and cables for kVA, not kW
  • Plan for power factor correction
  • Consider automatic capacitor switching for variable loads
  • Account for motor starting current (6-8× full load)
  • Evaluate harmonic impact from VFDs

Power Factor Correction Methods

Capacitor Banks

The most common correction method—capacitors supply reactive current locally:

Advantages:

  • Simple, proven technology
  • Low maintenance (no moving parts)
  • Can be sized precisely for the load
  • Fast response with automatic switching

Sizing: kVARrequired=kW×(tan(θ1)tan(θ2))kVAR_{required} = kW \times (\tan(\theta_1) - \tan(\theta_2))

Where θ₁ = original PF angle, θ₂ = target PF angle

Synchronous Condensers

Over-excited synchronous motors that generate reactive power:

Advantages:

  • Can both absorb and supply reactive power
  • Provides inertia for grid stability
  • No harmonic amplification concerns

Disadvantages:

  • Higher maintenance than capacitors
  • Larger footprint
  • Higher cost for small applications

Active Power Factor Correction

Electronic circuits that shape current waveform:

Advantages:

  • Addresses harmonics and power factor together
  • Fast response to load changes
  • Common in modern electronic equipment

Applications:

  • Built into VFDs, UPS systems, LED drivers
  • Standalone active filters for large facilities

Common Mistakes to Avoid

MistakeImpactPrevention
Ignoring power factor in designUndersized cables and transformersSize for kVA, not kW
Over-correcting power factorLeading PF, voltage rise, capacitor damageTarget 0.95, not 1.0
Fixed capacitors on variable loadsOver-correction at light loadUse automatic switching
Capacitors on VFD outputHarmonic resonance, equipment damageInstall on line side only
Ignoring utility rate structureMissing savings opportunityAnalyze actual rate schedule

Use these calculators to analyze and correct power factor:

Key Takeaways

  • Power factor: Resistive PF = 1.0; reactive typically 0.7-0.95
  • Current impact: At PF = 0.8, current is 25% higher than resistive equivalent
  • Cost impact: Low PF increases demand charges and triggers utility penalties
  • Correction: Capacitors typically pay back in 1-2 years for industrial facilities
  • Design rule: Size cables and transformers for kVA (apparent), not kW (real)

Further Reading

References & Standards

  • IEEE 1459: Standard Definitions for the Measurement of Electric Power Quantities
  • IEC 61000-3-2: Limits for harmonic current emissions
  • IEEE 141 (Red Book): Recommended Practice for Electric Power Distribution for Industrial Plants
  • NEMA MG 1: Motors and Generators standard

Disclaimer: This comparison provides general technical guidance based on international standards. Actual performance and costs depend on specific installation conditions and utility rate structures. Always consult with licensed engineers for system design decisions.

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