Hello everybody.The exam question was why before the inverter(between power lines and inverter)we measure 390 volts and 0.6 A but after the inverter(between inverter and motor)we measure 390 volts and 2.6 A.How can we explain the difference between these 2 currents...?

 

Hello,

Can you post a schematic on how you did measure the currents and where?

Bertus

 

actually the question has no schematic.the measurments were taken in the lab with voltometer and amperometer...imagine 3 boxes power line network-inverter-motor..

 

something like this

 

Hello,

It could be the power factor that makes the difference.
The motor is an inductive load.
The inverter can improve the power factor.
Read this page chapter on power factor of the eBook:
http://www.allaboutcircuits.com/vol_2/chpt_11/index.html

Bertus

 

Waveshape can also have a tremendous effect on measuring both voltage and current. I would bet that the output waveform of the inverter is not a sine wave. Rather it will be more like variable width pulses of DC trying to approximate the power curve of a sine wave.

 

I was thinking along the same lines, billb3857.

Zeke13, is this a "True Sine-Wave inverter" a "Modified Sine-Wave inverter" or just a "AC to DC inverter" ?

The types of equipment that can use and the accuracy of the inversion is related directly to the type of output created by the inversion.

 

the inverter is dc to ac and the output voltage waveform is not true sine wave, but i think that the problem is that the ac motor which we use like a load has a very poor power factor 0.65 or 0.75 .

 

I just realized I posted "AC to DC inverter".. I obviously meant to say "DC to AC"

As for the power factor, if your inverters sine wave "efficency" and the poor power factor of the motor may be the problem.

The motor will try to draw many times its rated-run amperage when starting. If the sine wave isnt exactly where the motor expects, it wont have the umph to get moving from a stall. LRA or Locked Rotor Amps is the rating you will have to contend with when choosing an inverter. Also, the more finicky the motor, the more true of a sine wave you will need.

 

I just want to remind that the whole problem was a lab experiment,and the question was why there is such current difference...also the A.C motor has no mechanical load during the experiment..My conclusion is that the difference comes from the power factor from the formula p=^3VIcosa