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CalcTree
Welcome! This AC current calculation will allow you to calculate the full load current of an AC induction motor. It will utilise supply voltage, power rating, power factor, efficiency and the type of phase to calculate a current value in amps. Let's begin!

Calculation

Inputs



P
:10kW



V_LN
:230volts



V_LL
:415



Cos θ (PF)
:0.85



n
:0.85


Output



I (S-Ph)
:60.1775237ampere



I (3-Ph), A
:19.2560039ampere



Explanation

The full load motor current is the maximum amperage the induction motor windings are designed to draw given ideal conditions (determined by the manufacturer).
Single Phase Induction Motors

I=1000PkWVLNcosθnI = \frac {1000*P_{kW}} {V_{LN} * cos \: \theta * n |}
Three-Phase Induction Motors

I=1000PkWVLLcosθnI = \frac {1000*P_{kW}} {V_{LL} * cos \: \theta * n |}
Where:

I=MotorCurrent(A)PkW=PowerRating(kW)VLN=LinetoNeutralVoltage(V)VLL=LinetoLineVoltage(V)cosθ=PowerFactorn=EfficiencyI = Motor \: Current \: (A) \\ P_{kW} = Power \: Rating \: (kW) \\ V_{LN} = Line \: to \: Neutral \: Voltage \: (V) \\ V_{LL} = Line \: to \: Line \: Voltage \: (V) \\ cos \: \theta \: = Power \: Factor \\ n = Efficiency

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