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    Voltage Drop

    Single-phase, three-phase and DC cable voltage drop with explained equations.

    EstimatePreliminary result, not a design verification.
    Before you calculate

    Understanding cable voltage drop

    As current flows through a cable, resistance and AC reactance cause a voltage difference between the supply and load. Calculate an estimated drop for one feeder, then explore how conductor temperature, cable length and power factor affect the result.

    Learn more — illustrated guide →
    Step 1

    Input parameters

    Enter one feeder’s operating conditions. Calculations update instantly.

    Enter distance to the load, not the return distance.
    Example: 3 × 240 mm² per phase. Use the TOTAL feeder current above.
    70 °C is an example, not an estimated actual temperature.
    Changes the sign of the reactance contribution.
    X for ONE cable/run. Layout and spacing affect AC reactance.
    Editable comparison only; not an automatic IEC compliance limit.
    Scope: Preliminary steady-state voltage-drop estimate for a single feeder. The automatic resistance is a simple material/area estimate, not a certified cable AC resistance or current-carrying-capacity check. Parallel runs assume identical cable material, area, length and impedance, equal current sharing, and equal return paths for two-wire circuits. For final design, verify cable arrangement, grouping/ampacity derating, manufacturer impedance, protection, upstream feeder drops and applicable rules.
    Step 2

    Calculated results

    Estimated voltage drop5.24 V
    Voltage drop percentage1.31%
    Estimated receiving voltage394.76 V
    Resistance R per run2.0629 Ω/km
    1 × 10 mm² Cu per phaseFeeder total: 32 A · Each run: 32 A · Equivalent R: 2.06289 Ω/km · Equivalent X: 0.08 Ω/km
    Compared with your target (5%)Within chosen target
    Maximum one-way length at your 5% limit (this cable, voltage drop only)191 mYour 50 m is 141 m inside this limit. Ampacity and short-circuit withstand are separate checks: see LV Cable Sizing.

    Reactance per run: 0.08 Ω/km · One-way length: 0.05 km · Balanced three-phase estimate

    Step 3

    Mathematical explanation

    Displayed values are rounded only after calculation. Units and input assumptions are shown explicitly.

    1. Determine conductor resistance

    Estimate resistance at 20 °C, then adjust it to the chosen conductor operating temperature.

    R20=1000×ρ20AR_{20}=\frac{1000\times\rho_{20}}{A}
    R20=1000×0.01724110≈1.7241 Ω/kmR_{20}=\frac{1000\times0.017241}{10}\approx1.7241\ \Omega/\mathrm{km}
    Rθ=R20×[1+α20×(θ−20)]R_{\theta}=R_{20}\times[1+\alpha_{20}\times(\theta-20)]
    Rθ=1.7241×[1+0.00393×(70−20)]≈2.0629 Ω/kmR_{\theta}=1.7241\times[1+0.00393\times(70-20)]\approx2.0629\ \Omega/\mathrm{km}

    2. Parallel cable current sharing

    A single conductor carries the total feeder current.

    Irun=ItotalnI_{\mathrm{run}}=\frac{I_{\mathrm{total}}}{n}
    Irun=321≈32 AI_{\mathrm{run}}=\frac{32}{1}\approx32\ \mathrm{A}
    Req=RrunnR_{\mathrm{eq}}=\frac{R_{\mathrm{run}}}{n}
    Req=2.062891≈2.06289 Ω/kmR_{\mathrm{eq}}=\frac{2.06289}{1}\approx2.06289\ \Omega/\mathrm{km}
    Xeq≈XrunnX_{\mathrm{eq}}\approx\frac{X_{\mathrm{run}}}{n}
    Xeq≈0.081≈0.08 Ω/kmX_{\mathrm{eq}}\approx\frac{0.08}{1}\approx0.08\ \Omega/\mathrm{km}

    3. Convert length and determine the equivalent impedance term

    Lkm=Lm1000L_{\mathrm{km}}=\frac{L_{\mathrm{m}}}{1000}
    Lkm=501000=0.05 kmL_{\mathrm{km}}=\frac{50}{1000}=0.05\ \mathrm{km}
    sin⁡φ=1−cos⁡2φ\sin\varphi=\sqrt{1-\cos^{2}\varphi}
    sin⁡φ=1−0.92≈0.4359\sin\varphi=\sqrt{1-0.9^{2}}\approx0.4359
    Zprojected=Reqcos⁡φ±Xeqsin⁡φZ_{\mathrm{projected}}=R_{\mathrm{eq}}\cos\varphi\pm X_{\mathrm{eq}}\sin\varphi
    Zprojected=2.06289×0.9+0.08×0.4359Z_{\mathrm{projected}}=2.06289\times0.9+0.08\times0.4359
    Zprojected≈1.89147 Ω/kmZ_{\mathrm{projected}}\approx1.89147\ \Omega/\mathrm{km}

    4. Calculate the voltage drop

    The √3 factor converts the per-phase voltage-drop projection to the line-to-line value.

    ΔU≈3×Itotal×L×(Reqcos⁡φ±Xeqsin⁡φ)\Delta U\approx\sqrt{3}\times I_{\mathrm{total}}\times L\times(R_{\mathrm{eq}}\cos\varphi\pm X_{\mathrm{eq}}\sin\varphi)
    ΔU≈3×32×0.05×1.89147\Delta U\approx\sqrt{3}\times32\times0.05\times1.89147
    ΔU≈5.242 V\boxed{\Delta U\approx5.242\ \mathrm{V}}

    5. Voltage drop as a percentage

    ΔU%=ΔUUsupply×100\Delta U_{\%}=\frac{\Delta U}{U_{\mathrm{supply}}}\times100
    ΔU%=5.2418400×100≈1.31%\Delta U_{\%}=\frac{5.2418}{400}\times100\approx1.31\%
    ULL −ΔU≈394.758 VU_{LL}\ -\Delta U\approx 394.758\ \mathrm{V}