Is there a way to find out what is the nominal current of this fan and what type of single phase induction motor is used?

  • \$\begingroup\$ The nominal current is around 82mA, if it works on 230V 50Hz, and the power rating given is 19W. Since it is given 10% tolerance, worst case would be about 91mA. Not sure about the motor \$\endgroup\$ – Atizs Mar 13 at 10:49
  • \$\begingroup\$ You have power consumption at nominal voltage, and you can calculate current from I = P/U. I = 19/230 = 0.083 A@50 Hz and I = 17/230 = 0.074 A@60 Hz. Inductor type isn't specified, as I can see. Why do you need the inductor type to know? \$\endgroup\$ – cyclone125 Mar 13 at 10:49

If you look at the RPM, you can see that the PU slip of the motor is (3000 - 2650)/3000 = 0.12 at 50 Hz and (3600 - 3100)/3600 = 0.14 at 60 Hz. There is also a graph of current vs. time that indicates the maximum starting current is about 3 three times the running current. That says that the motor is highly resistive and inefficient. It must be a shaded-pole motor.

Since the motor is highly resistive, you can assume that the leakage inductance does not influence the current very much and simply dividing the power by the voltage will provide a reasonable estimate for the operating current.

  • \$\begingroup\$ Why would being resistive matter here? Isnt current always P/V? I didn’t get why you mentioned motor being resisive and the slip to calculate the current. Isnt current always the ratio of given P to given V? Hope you might have some feedback. Thanks \$\endgroup\$ – panic attack Mar 27 at 17:45
  • \$\begingroup\$ For single-phase alternating current, P = V x I x pf or I = P/( V x pf). The power factor (pf) is a number between 0 and 1 that accounts for the voltage not being in phase with the current. I am assuming that the full-load pf for this motor is close enough to 1 to make I = P/V a reasonable estimate. \$\endgroup\$ – Charles Cowie Mar 27 at 19:44
  • \$\begingroup\$ Amazing thanks for the input. \$\endgroup\$ – panic attack Mar 30 at 15:11

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