Engineering guide

    Motor Current & Starting Methods

    Why a motor's kW is not its current, how the line and winding currents differ, and why the four starting-method numbers are four different things.

    KiloVolt Engineering Toolkit

    The kW on the nameplate is mechanical output

    An AC motor converts electrical input power into mechanical shaft power. The kW on a typical nameplate is the mechanical rated output, not the electrical power the motor draws. Because motors have losses and a power factor below one, the supply current must account for both efficiency and power factor. This guide covers three-phase squirrel-cage induction motors and the main ways to start them. The normal full-load current is a different quantity from the transient or reduced-voltage starting current.

    An induction motor turns because three-phase currents in the stator create a rotating magnetic field that induces rotor currents. In normal motoring the rotor runs a little below the synchronous speed; the difference is the slip. At standstill the electromagnetic conditions are very different, which is why a motor switched directly onto the line draws several times its rated current.

    1. Rated current from kW, voltage, efficiency and power factor

    For a balanced three-phase motor at full mechanical output on a sinusoidal supply:

    In=1000 Pshaft3 ULL η cos⁡φI_n=\frac{1000\,P_{\text{shaft}}}{\sqrt{3}\,U_{LL}\,\eta\,\cos\varphi}

    Pshaft is the rated mechanical output in kW, ULL the line-to-line voltage in V, η the efficiency as a decimal (92.2 % = 0.922) and cos φ the power factor at the rated point. A single-phase motor uses In = 1000 P/(U η cos φ) and must not reuse the three-phase reference values or the star-delta equations. The electrical quantities follow from the same data:

    Pin=Pη,S=Pincos⁡φ,Q=Pintan⁡(arccos⁡φ)P_{in}=\frac{P}{\eta},\qquad S=\frac{P_{in}}{\cos\varphi},\qquad Q=P_{in}\tan(\arccos\varphi)

    If you enter the nameplate current, it replaces the estimate everywhere. The calculator then shows the formula result only as a cross-check and warns when the two differ by more than 5 %, which usually means the data describe different motors.

    2. Worked example: 15 kW, 400 V

    Published ABB and WEG four-pole 50 Hz IE3 catalog rows for 15 kW:

    FieldABB 10-2021WEG W22KiloVolt midpoint
    Rated efficiency92.1 %92.3 %92.2 %
    Power factor0.820.830.825
    Manufacturer rated current28.5 A28.3 Aformula: 28.4633 A
    Locked-rotor ratio Is/In8.27.07.6
    In=150003×400×0.922×0.825=28.4633 AI_n=\frac{15000}{\sqrt{3}\times 400\times 0.922\times 0.825}=28.4633\ \text{A}

    That is 28.5 A on the screen, with Pin = 16.269 kW, S = 19.720 kVA and Q = 11.144 kvar. This is not the nameplate of a specific motor: real models differ.

    3. Rated line current versus delta winding current

    A three-phase motor has three supply leads, but the current in each stator winding depends on how the windings are connected. In a balanced delta connection the winding current is smaller than the line current by √3:

    Iwinding,Δ=Iline3I_{\text{winding},\Delta}=\frac{I_{\text{line}}}{\sqrt{3}}

    For the 15 kW example, 28.4633 A / √3 = 16.4333 A (16.4 A). In a star connection the winding current equals the line current. A 400 V supply does not say how the motor is wound: a 230Δ/400Y motor runs in star on 400 V, while a 400Δ/690Y motor can run in delta on 400 V. Follow the nameplate. The calculator therefore shows Delta winding current (if Δ-connected) unless you confirm the connection, and shows NOT APPLICABLE when the motor runs in star. If the actual connection is star, Iline/√3 must not be presented as the real winding current.

    This is the rated running winding current, not a starting current, and it does not change with the starting method. It matters for equipment that senses winding current, such as some inside-delta arrangements. For ordinary in-line motor equipment and for VFD output ratings, use the motor line current, not the lower winding current.

    4. The four starting methods

    Direct on line (DOL). Full line voltage is switched onto the terminals. The usual preliminary estimate is Ilocked-rotor ≈ In × Is/In. For the example, 28.4633 × 7.6 = 216.32 A. It is a standstill RMS line-current estimate: it excludes the first-cycle asymmetrical peak, the duration, the voltage dip and whether the load accelerates.

    Star-delta (Y/Δ). The windings start in star, so each sees the phase voltage. Under the idealised assumption the supply line current in the initial star stage is one third of the DOL line current at the same line voltage:

    Istart,Y≈Istart,DOL3I_{start,Y}\approx \frac{I_{start,DOL}}{3}

    Example: 216.32 / 3 = 72.11 A. The starting torque is also about one third, which can make star-delta unsuitable for loads that need high starting torque; an open-transition changeover adds transients that are not modelled. Delta running on 400 V needs a compatible winding (for example 400Δ/690Y with six accessible terminals); a 230Δ/400Y motor must not be switched to delta on a 400 V line.

    Soft starter. A soft starter reduces the applied voltage electronically while starting, and many have an adjustable current limit. KiloVolt uses an illustrative 350 % of In as an assumed setting: 3.5 × 28.4633 = 99.62 A. It is not a prediction that the current will reach 99.62 A or that the motor will accelerate. See the Soft Starter Sizing guide.

    Variable-frequency drive (VFD). A drive controls the frequency and voltage. KiloVolt shows 1.0 × In (28.46 A) as a nominal motor-side reference only, never as a start-current prediction and never as the drive's mains input current. See the VFD Sizing guide.

    5. Comparison (not performance guarantees)

    Method15 kW exampleWhat it is
    DOL216.32 AEstimated locked-rotor RMS motor line current
    Y/Δ72.11 AEstimated initial star-stage supply current; excludes the changeover
    Soft starter99.62 AAssumed current-limit setting at 350 % of In
    VFD28.46 ANominal motor-side reference, not an actual starting current

    6. Speed, slip and rated torque

    ns=120 fp,s=ns−nrns×100 %,TN=9550 Pshaftnrn_s=\frac{120\,f}{p},\qquad s=\frac{n_s-n_r}{n_s}\times 100\,\%,\qquad T_{N}=\frac{9550\,P_{\text{shaft}}}{n_r}

    At 50 Hz and 4 poles ns = 1500 rpm; at a rated speed of 1465 rpm the slip is 2.333 % and the 15 kW shaft torque is 97.8 N·m (9550 is a rounded constant). These are rated-output quantities, not the locked-rotor torque and not an acceleration-time prediction. A rated speed at or above the synchronous speed is not normal induction-motor operation and is reported as NOT ASSESSED. With a VFD use the actual output frequency, not 50 Hz.

    7. How the built-in reference model works

    Without a nameplate the calculator uses published four-pole IE3 400 V 50 Hz catalog rows from two manufacturer series (ABB General Performance, edition 10-2021, and WEG W22, Rev. 34 of 04/2024) for 22 ratings from 0.75 to 200 kW. At a listed rating it uses the plain arithmetic mean of the two series for efficiency, power factor and Is/In; between two ratings it interpolates linearly; it never extrapolates. Other frequencies, pole numbers, voltages and motor types are NOT ASSESSED unless you enter the datasheet values.

    The figure is a two-series reference estimate. It is not a statistical mean of the market, not an IEC guaranteed value (IEC 60034-30-1 defines efficiency classes, not power factor or locked-rotor current) and not your motor. Locked-rotor ratios vary widely: at 22 kW ABB lists 10.1 and WEG 7.7. Later ABB editions print different power factors and Is/In for some small ratings, so the edition matters. All 22 rows of both series were compared value by value with the original catalog pages (ABB 10-2021; WEG catalogue Rev. 34 of 04/2024). WEG, too, revises its data between editions. The nameplate or datasheet always overrides the model.

    8. Questions and common mistakes

    Why is my motor's current not exactly the calculator's? Winding, rotor and frame designs differ; the reference is not your manufacturer's model.

    Will a star-delta starter always start the motor? No. The torque is about a third, and the changeover can cause transients.

    Does the 350 % soft-starter setting mean the current will reach 350 %? No. It is a limit setting; the actual current depends on the device and the load.

    Can I size a breaker or generator from these four numbers? Not safely. That needs the protective curves, the starting time, the cable impedance, the voltage dip and the manufacturer's data.

    Can I use the four-pole 50 Hz defaults for a six-pole 60 Hz motor? No. Enter the datasheet values.

    Why 28.5 A and 16.4 A for the same motor? The first is the rated current in each supply line. The second is the conditional current in each delta-connected winding, 28.46/√3. They describe different conductors, not two different loads. And 16.4 A is not the current while starting in star.

    Use the numbers carefully: for a fast preliminary load check use the rated current; use the DOL and star-delta values as order-of-magnitude comparisons and the soft-starter and VFD figures as control reference points only. Final protection, feeder, generator, torque and procurement decisions need the nameplate and the manufacturers' data. Related tools: AC Power, Voltage Drop, LV Cable Sizing. Sources: IEC 60034-30-1:2025, IEC 60034-12:2024, ABB 9AKK108196 EN 10-2021 and the WEG W22 technical catalogue.

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