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First-principles AC power-triangle relations6 min read

Worked Example: Converting a Known 100 kW Load into kVA, kVAR and Line Current

Starting from real power and power factor alone, the complete AC power triangle for a 415 V three-phase load — apparent power, reactive power and current, all in one pass.

Scenario

Known quantity100 kW real power
System415 V three-phase
Power factor0.85 lagging

Step-by-step calculation

Step 1: Find apparent power from real power and power factor

S (kVA) = P / PF
100 / 0.85
S = 117.65 kVA

Step 2: Find reactive power from the power triangle

Q = √(S² - P²)
√(117.65² - 100²)
Q = 61.97 kVAr

Step 3: Find line current

I = (S x 1000) / (√3 x V)
117,650 / (1.732 x 415)
I = 163.67 A

Result summary

CheckRequirementActualStatus
Apparent powern/a (this is the conversion result)117.65 kVAāœ“ PASS
Reactive powern/a (this is the conversion result)61.97 kVArāœ“ PASS
Line currentn/a (this is the conversion result)163.67 Aāœ“ PASS
A 100 kW load at 0.85 lagging power factor on a 415 V three-phase supply draws 117.65 kVA of apparent power, 61.97 kVAr of reactive power, and 163.67 A of line current.

Key insight: Any two of {P, Q, S, PF} (plus voltage, for current) fully determine the rest of the power triangle — this calculator can start from whichever one quantity is actually known (kW, kVA, kVAr, or measured amps) and derive everything else, rather than requiring the input to always be real power specifically.

Try it with your own numbers

Every input in this example is editable in the live calculator — free, no signup.

Open kW / kVA / kVAR / Amps Converter calculator →

Frequently asked questions

What if power factor were exactly 1.0 (unity)?

At unity power factor, apparent power equals real power exactly (S = P, since PF = 1), and reactive power drops to zero — physically, this represents a purely resistive load with no magnetizing or capacitive reactive component, which is why the calculator flags kVAr as the known quantity as mathematically undefined at PF = 1 (any apparent power value would give zero reactive power, so there's nothing to solve for from that direction).

Does 'leading' vs 'lagging' power factor change these numbers?

No — the magnitude relationships between P, Q, S and I are identical whether the load is inductive (lagging, drawing reactive power) or capacitive (leading, supplying reactive power back). The lagging/leading distinction matters for how that reactive power interacts with the rest of the system (e.g. whether it helps or hurts overall power factor when combined with other loads), not for this load's own P/Q/S/I magnitudes in isolation.

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