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The Difference Between kW, kVA, and kVAr — And Why Getting It Wrong Costs You 20% More Cable

Engineers who size cables using kW instead of kVA consistently undersize by 15-25%. A 100kW motor at 0.85 PF draws 118kVA — and that's the number your cable must carry. Here's why this matters.

KholisFebruary 24, 20268 min read

This article addresses a fundamental error that I encounter in at least one design review every month: sizing cables and protective devices using kilowatts (kW) instead of kilovolt-amperes (kVA).

Cables carry current. Current is determined by apparent power (kVA), not real power (kW). If a load consumes 100 kW at 0.85 power factor, the cable must carry 118 kVA worth of current — 18% more than the kW figure suggests. Engineers who receive a load schedule in kW and calculate current directly from the kW value will undersize every cable on a motor-heavy installation.

This is not an obscure technical point. It's the difference between a cable that's adequate and one that overheats. On an industrial plant with overall power factor of 0.80, the error is 25%.

The Power Triangle

Three types of power exist in AC circuits:

  • Real power (P, measured in kW) — the power that does useful work (turning motors, producing heat, generating light)
  • Reactive power (Q, measured in kVAr) — the power that sustains magnetic fields in motors and transformers (does no useful work but must be supplied)
  • Apparent power (S, measured in kVA) — the total power that the source must deliver and the cable must carry

These three are related by the power triangle:

Power Triangle Relationship

S² = P² + Q² → kVA = √(kW² + kVAr²)

The power factor (PF) is the ratio of real power to apparent power:

Power Factor

PF = P / S = kW / kVA = cosφ

Why Cables Carry kVA, Not kW

A cable doesn't know whether the current flowing through it is doing "useful" work or sustaining a magnetic field. It carries total current, and total current corresponds to apparent power.

The current drawn by a three-phase load is:

Current from Apparent Power

I = S / (√3 × V) = kVA × 1000 / (√3 × V)

NOT:

Incorrect Current Calculation

I ≠ P / (√3 × V) = kW × 1000 / (√3 × V) ← WRONG for PF ≠ 1

The correct formula that accounts for power factor:

Correct Current Calculation

I = P / (√3 × V × PF) = kW × 1000 / (√3 × V × PF)

Dividing by PF is equivalent to using kVA instead of kW. At PF = 0.85, dividing by 0.85 increases the current by 18%. At PF = 0.80, the increase is 25%.

The Practical Impact

Scenario: A load schedule shows 100 kW, 415 V, 3-phase, PF = 0.80.

Wrong approach (kW-based):

Incorrect Sizing

I = 100,000 / (√3 × 415) = 139 A

From BS 7671 Table 4D4A (multicore PVC on tray): 35 mm² = 143 A. Select 35 mm².

Correct approach (kVA-based):

Correct Sizing

I = 100,000 / (√3 × 415 × 0.80) = 174 A

From same table: 50 mm² = 181 A. Select 50 mm².

The difference: 50 mm² instead of 35 mm² — a full cable size larger. The 35 mm² cable, rated at 143 A, would carry 174 A — overloaded by 22%. It would overheat, degrade the insulation, and potentially cause a fire.

This Error is Invisible in Load Schedules

Many load schedules are provided in kW only. The power factor is either listed separately, assumed, or omitted entirely. If you receive a load schedule in kW and don't ask "what is the power factor?", you cannot calculate the correct cable size. Always request or assume a power factor for every motor, transformer, and inductive load.

Typical Power Factors by Load Type

Load TypeTypical Power FactorkVA/kW Ratio
Resistive heater1.001.00
Incandescent lamp1.001.00
LED lighting (with PFC)0.95–0.991.01–1.05
Induction motor (full load)0.80–0.901.11–1.25
Induction motor (50% load)0.65–0.751.33–1.54
Induction motor (25% load)0.40–0.551.82–2.50
Transformer (no load)0.10–0.205.0–10.0
Welding equipment0.40–0.601.67–2.50
Compressor0.80–0.851.18–1.25
VFD (at motor full load)0.95–0.981.02–1.05

Partially Loaded Motors Are Worse

A motor's power factor drops significantly at part load. A motor rated 0.85 PF at full load may operate at 0.65 PF at 50% load. If the motor typically runs at partial load (common for pumps and fans sized for maximum conditions), the actual current is higher than calculated at rated power factor. Always use the power factor at the expected operating load, not the nameplate value.

Where the Error Compounds

The kW/kVA error is most damaging in maximum demand calculations, where the error propagates through every level of the distribution system.

Consider a factory with:

  • 10 × 45 kW motors at PF = 0.82
  • 50 kW of lighting at PF = 0.98
  • 30 kW of heating at PF = 1.00
  • Diversity factor: 0.75

Wrong (kW-based): Total load = (10 × 45 + 50 + 30) × 0.75 = 397.5 kW Current = 397,500 / (√3 × 415) = 553 A

Correct (kVA-based): Motor load: 10 × 45/0.82 = 549 kVA Lighting: 50/0.98 = 51 kVA Heating: 30/1.0 = 30 kVA Total with diversity: (549 + 51 + 30) × 0.75 = 472.5 kVA Current = 472,500 / (√3 × 415) = 657 A

The difference: 657 A vs 553 A — the correct value is 19% higher. This affects the main cable, the transformer sizing, and the switchgear rating. A transformer sized at 553 A (400 kVA) would be overloaded by a 657 A demand.

Why Generator and Transformer Nameplates Show kVA

Have you ever wondered why generators are rated in kVA (or MVA) and not kW? Because generators are thermally limited by current, not by the power factor of the load. A 500 kVA generator can supply 500 kVA regardless of whether the power factor is 0.8 (400 kW useful) or 0.6 (300 kW useful). The current in the windings is the same — 500 kVA worth.

Transformers are the same. A 1000 kVA transformer can supply 1000 kVA. At PF = 0.85, that's 850 kW of useful power. At PF = 0.70, it's only 700 kW. The transformer doesn't care about the power factor — it cares about the current, which is determined by kVA.

This is why power factor correction can "free up" capacity on existing transformers and generators. If you improve the power factor from 0.80 to 0.95, the same 1000 kVA transformer can now deliver 950 kW instead of 800 kW — a 19% increase in useful capacity without any hardware change.

Power Factor Correction as a Cable Sizing Strategy

In some cases, it's cheaper to install power factor correction capacitors than to upsize cables. Consider:

Without PFC: 400 kW at PF 0.75 → 533 kVA → 742 A at 415 V → requires 300 mm² cable With PFC to 0.95: 400 kW at PF 0.95 → 421 kVA → 586 A at 415 V → requires 240 mm² cable

The cable cost saving for a 100 m run might be $5,000–$8,000. A 120 kVAr capacitor bank to achieve this improvement costs approximately $3,000–$5,000. The PFC pays for itself in cable savings alone, before considering the electricity tariff savings (most commercial tariffs penalise low power factor).

The One Rule to Remember

Cables carry current. Current = kVA, not kW. Always convert kW to kVA before calculating current.

The Complete Formula

I = kW × 1000 / (√3 × V × PF × η)

Where η is the efficiency (for motor loads). Never skip the PF term. Never assume PF = 1 unless the load is genuinely resistive.

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Frequently Asked Questions

What is the difference between kW, kVA, and kVAR?

kW is real power (does useful work), kVAR is reactive power (sustains magnetic/electric fields), and kVA is apparent power (vector sum of kW and kVAR). Cables must be sized for kVA, not kW.

Why does power factor affect cable sizing?

Lower power factor means higher current for the same real power output. A motor drawing 100 kW at PF 0.7 requires 43% more current (and a larger cable) than at PF 1.0.

How do I convert between kW and kVA?

kVA = kW / Power Factor. For example, 100 kW at PF 0.85 = 117.6 kVA. This kVA value determines the cable sizing current using I = kVA / (V × sqrt(3)) for three-phase.

Kholis

Kholis

Lead Electrical & Instrumentation Engineer

18+ years of experience in electrical engineering at large-scale mining operations. Specializing in power systems design, cable sizing, and protection coordination across BS 7671, IEC 60364, NEC, and AS/NZS standards.

18+ years electrical engineering experienceLead E&I Engineer at major mining operationECalPro founder & developer