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    Interactive calculator

    Power Factor

    Find power factor from kW, kvar or kVA, estimate reactive compensation and compare current before and after correction.

    Before you calculate

    Understanding power factor and reactive compensation

    Power factor compares useful active power (kW) with total apparent power (kVA). Inductive or capacitive reactive power (kvar) changes current loading. Estimate the correction needed to reach a selected target.

    Learn more — illustrated guide →
    Step 1

    Input parameters

    Enter the existing electrical power in kW or kVA, the current cos φ and your target cos φ. KiloVolt calculates the complete power triangle and supply current before and after correction.

    Active electrical input power.
    Greater than 0 and up to 1.
    Example: 0.95; select an appropriate site target.
    Advanced · Leading / lagging behavior

    Default: lagging → lagging. A leading load can require an inductor rather than capacitors to achieve the selected target.

    Step 2

    Calculated results

    Capacitive correction42.13 kvarTheoretical capacitive compensation at this operating point.

    Power and current comparison

    At 400 V (L-L)
    QuantityBeforeAfter
    Active power (P)100 kW100 kW
    Reactive power (Q)75 kvar32.87 kvar
    Apparent power (S)125 kVA105.26 kVA
    Line current (I)180.42 A151.93 A
    Power factor (cos φ)0.8 0.95
    Supply voltage (U)400 V400 V
    Original phase angle magnitude36.87°
    Supply current reduction28.49 A15.79% change
    Engineering scope: Balanced sinusoidal AC with constant active kW. The compensation kvar is theoretical—not a final capacitor-bank rating. Check changing load, harmonics, possible resonance, voltage rating, protection and utility requirements.

    1. Find the starting apparent power

    The entered power is active electrical kW.

    S1=Pcos⁡φ1S_1=\frac{P}{\cos\varphi_1}
    S1=1000.8≈125 kVAS_1=\frac{100}{0.8}\approx 125\,\mathrm{kVA}

    2. Calculate existing reactive power

    The sign is positive for lagging (inductive) and negative for leading (capacitive) current.

    Q1=±P×tan⁡ ⁣(arccos⁡(cos⁡φ1))Q_1=\pm P\times\tan\!\left(\arccos(\cos\varphi_1)\right)
    Q1=100 × 0.75≈75 kvarQ_1=100\,\times\,0.75\approx 75\,\mathrm{kvar}

    3. Calculate target reactive power

    The target cos φ sets the desired reactive component at the same active kW.

    Q2=±P×tan⁡ ⁣(arccos⁡(cos⁡φ2))Q_2=\pm P\times\tan\!\left(\arccos(\cos\varphi_2)\right)
    Q2=100 × 0.3287≈32.87 kvarQ_2=100\,\times\,0.3287\approx 32.87\,\mathrm{kvar}

    4. Determine the compensation required

    Positive difference: capacitive kvar injection is needed.

    Qcomp=Q1−Q2Q_{\mathrm{comp}}=Q_1-Q_2
    Qcomp=75 − (32.87)≈42.13 kvarQ_{\mathrm{comp}}=75\,-\,(32.87)\approx 42.13\,\mathrm{kvar}

    5. Compare apparent power and current

    The supply voltage and active load are held constant. The current is calculated from the apparent power.

    S2=Pcos⁡φ2I=S×10003 ULLS_2=\frac{P}{\cos\varphi_2}\qquad I=\frac{S\times1000}{\sqrt{3}\,U_{LL}}
    S2=1000.95≈105.26 kVAI1=180.42 AI2=151.93 AS_2=\frac{100}{0.95}\approx 105.26\,\mathrm{kVA}\qquad I_1=180.42\,\mathrm{A}\quad I_2=151.93\,\mathrm{A}

    Calculations use full precision internally; the displayed numbers are rounded for readability.