# How to find rated current of a three phase induction motor

How to find the value of full load rated current and rated current for a three phase motor having its ratings as:

Stator resistance and inductance = [0.2147 ohm, 0.000991 H]
Rotor resistance and inductance = [0.2205 ohm, 0.000991 H]
Mutual inductance = 0.06419 H
Nominal power, voltage (line-line), and frequency = [1.5e+04 VA, 400 V, 50 Hz]
Pole pairs = 2


I am currently doing a project on simulating Blocked rotor test, DC test, No load test in MATLAB on three phase induction motor and in order to do that I need to find the rated current and machine ratings doesn't has its rated power factor or efficiency mentioned on it...

• Maybe this will work. Look online at a variety of 15kW motors. See what is the range of rated currents for 50 Hz motors. Try doing multiple simulations over a range of currents. Apr 17, 2021 at 21:43
• The range will not be that large. Apr 17, 2021 at 21:44
• For induction motors you often can use a 'conventional' power factor depending on the power, it's not a precision measurement anyway; I'd try starting with 0.65 or 0.7 Apr 19, 2021 at 5:51
• 5N·m torque seems low for a 15kVA motor. Apr 19, 2021 at 20:04
• @StainlessSteelRat yeah it is a 4 pole motor at 50 Hz so a bit under 1500 rpm. 1500 rpm is 157 rad/s. 5 Nm * 157 rad/s = 785 watts. I don't see how this question can be answered properly with so much uncertainty. Apr 19, 2021 at 21:55

You can not determine the rated current from the information given. The best you can do is to make an estimate based on typical motors available in the market. However a motor rated 15000 kW at 440V is not a motor that anyone could build. The required current would be something like 23,000 amps. A motor with that power rating would be rated something like 10,000 volts and 1000 amps.

By international standards, the power rating for a motor is the mechanical power that the motor can deliver under specified operating conditions. For a 3-phase motor, W = V X I X 1.73 X efficiency X power factor. Typical values of efficiency and power factor vary with influence due to power level, number of motor poles, type of enclosure and design choices made by the manufacturer. I found data for a 15 kW 400 V, 4-pole, 3-phase motor, at rated load, efficiency = 91.4%, pf = 0.83 current = 28.5 A.

If you solve the equivalent circuit, you can get a reasonable approximation of the rated current. Without the resistance in parallel with the magnetizing branch that represents the iron losses, you will be missing that contribution to current. There is also no way to determine mechanical losses from the equivalent circuit parameters.

• I just realized that I entered KW instead of W now can you suggest me the rated current Apr 17, 2021 at 20:27
• if there is no way to find the rated current using the above information then can you suggest me how should I perform this experiment, i.e.-DC test in matlab? Apr 17, 2021 at 20:29
• @TusharBharti, 'KW' would be kelvin-watts. You mean 'kW'. Capitals matter! Apr 17, 2021 at 20:39
• I revised my answer to include data for a motor that I found. You should revise the question to show 15 kW.
– user80875
Apr 17, 2021 at 21:04
• @CharlesCowie Now I have put on the machine's parameters too, please now can you give the answer accordingly Apr 19, 2021 at 17:44