I have the following problem:


Okay on this site, https://www.jcalc.net/motor-current-calculator, I found a formula for the motor current with full load and a 3-phase supply:

\$I=\frac{P\cdot 1000}{\sqrt{3}\cdot V\cdot pf \cdot \eta} \$, where \$ P\$ is the motor power rating, \$ V\$ is voltage, \$pf \$ is the power factor, \$\eta \$ is the efficiency. Plugging my values into the equation we get:

\$I=\frac{P\cdot 1000}{\sqrt{3}\cdot V\cdot pf \cdot \eta} =\frac{12\text{kW} \cdot 1000}{\sqrt{3} \cdot 220 \text{V} \cdot 0.89 \cdot 0.87}=40.67 \text{A}\$

This seems like a VERY large current and it made me doubt my answer.

Have I used the correct formula, and used the information correctly - at the moment I don't use the fact that there is a slip of \$5 \% \$. I hope someone can help me with this.


2 Answers 2


Your calculation is correct. Your use of line-to-line voltage and division by the square root of 3 takes care of the per-phase aspect of the problem. You can probably find motors that size that have a bit higher efficiency and a bit better power factor, but that current is about typical for a 3-phase motor that size.

The slip is not needed, but it is a hint that this motor is not as efficient as you might expect. Typical slip for a 3-phase motor is about 2% to 3%. Slip is directly proportional to power lost in the rotor resistance. Another hint about efficiency and power factor is the number of motor poles. 4-pole motors generally have the highest efficiency and power factor. Motors with more than 6 poles generally have a significantly lower power factor.

  • \$\begingroup\$ Great! Thanks for the confirmation, @Charles Cowie. \$\endgroup\$
    – Carl
    May 20, 2020 at 13:09
  • \$\begingroup\$ It's worth pointing out here that the question, AS ASKED, cannot be answered with any accuracy. Current drawn by an AC induction motor (or any motor for that matter) is a function of the LOAD on the motor. The question made no mention of the motor load.The only way to effectively answer that question would be to start by saying "Assuming the motor is fully loaded..." \$\endgroup\$
    – JRaef
    May 26, 2020 at 19:07
  • \$\begingroup\$ @JRaef I take nominal to mean "rated" or "nameplate." The question asks for nominal input current corresponding to nominal output power, efficiency, power factor and voltage. I believe the wording means "what is the current for full-load nameplate operation." The calculation should be as accurate as the nameplate data. \$\endgroup\$
    – user80875
    May 26, 2020 at 19:27
  • \$\begingroup\$ Fair point. I don't consider "rated" and "nominal" to be synonymous, but I know that's not a universal interpretation.To me, "rated" is what's on the nameplate, "nominal" is what it actually draws under load, and in this case the load was not mentioned. But no worries, you and lots of other people see it differently. \$\endgroup\$
    – JRaef
    May 26, 2020 at 19:36
  • \$\begingroup\$ @JRaef I think "nominal" is too context sensitive and subject to interpretation to be used in a lot of situations including the original question. However the consistency of use within the question saves it from disaster. \$\endgroup\$
    – user80875
    May 26, 2020 at 19:47

The power per phase is 4 kW. The phase voltage is \$220/\sqrt3\$ = 127 volts. In simple terms that means a phase current of: -

$$\dfrac{4000}{127} = 31.5 \text{ amps}$$

So, it's going to be bigger than that taking into account power factor and efficiency.

Both power factor and efficiency converts the real power of 4 kW into an apparent power of: -

$$\dfrac{4000}{0.87 \times 0.89} = 5166\text{ VA}$$

So plugging that back into the top equation gives a current of 40.67 amps.

  • \$\begingroup\$ Thanks @Andy aka. I have another question out of curiosity. In the problem it says a "star connected induction motor". Would the calculation change if it instead of a star connection were, for example, a delta connected induction motor? \$\endgroup\$
    – Carl
    May 20, 2020 at 13:13
  • 2
    \$\begingroup\$ The line current would be the same at 40.67 amps but the individual current into each of the delta connected windings would \$\sqrt3\$ lower at 23.5 amps. That then means a VA into each winding of 5166 VA (23.5 x 220) and of course that matches the "per phase" VA in my answer for the star connected motor. \$\endgroup\$
    – Andy aka
    May 20, 2020 at 13:20
  • \$\begingroup\$ @Carl; this analysis is like breaking the motor into three single-phase, 4 kW motors. Analysis of the motor's equivalent circuit is done that way. A single-phase equivalent circuit in analyzed using the line-to-neutral voltage. The resultant power and loss calculations are multiplied by three to get the results for the complete motor. \$\endgroup\$
    – user80875
    May 20, 2020 at 14:28

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