AC Theory - Stator Winding Motor Question

Discussion in 'Homework Help' started by Courtney Thomas, Apr 9, 2015.

  1. Courtney Thomas

    Thread Starter New Member

    Apr 9, 2015

    I've got this question on RL and RL-C circuits that doesn't state if they're in series or in parallel assuming that's where I'm going wrong with the question.

    The first half of the question is the RL: " The stator winding of an A.C motor has a resistance of _ and an inductance of _ and is connected to an 11kV 50Hz supply. A snippet of the answered questions answered for that are below where I've assumed from that question they're in series?
    And then the second half of the question is the RL-C where the capacitor is connected across the AC Motor. And I then had to calculate again the power drawn, power factor, and the weekly cost running the motor. But where the 2nd half of the question want you to comment on the saving. A snippet of the answers again is below.
    I think where I've gone wrong is the power factor calculation, where I'm using the same equation for both and not getting the right calculation for the weekly cost and instead of saving when adding the capacitor, it's more. Can anyone help me out with saying if that's the case?

    -college noob, help much appreciated
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Your maths presentation is somewhat lacking - both in rigour and clarity.
    For instance, you have in task (5a) that PF=12/79.45=0.2
    By my reckoning that should be PF=0.151. In this case it didn't matter, since you simply needed to find the effective load volt-amperes.

    Frankly, I can't really follow your task (5b) working.

    The process one follows might typically be:

    1. Without the capacitor, find the motor load real and reactive (lagging) current components.

    \small{\text{I_{motor_ {\(real\)}}=I_{rms} \cos \(\phi\) \ A}}

    \small{\text{I_{motor _{\(reactive\)}}=I_{rms} \sin \(\phi \) \ A}}


    \small{\text{\cos\(\phi \) =0.151}}
    \small{\tex{I_{rms}=138.45 \ A}}

    2. Calculate the (leading) current in the parallel capacitor (~ 76.026 A as you found).
    3. Calculate total load (motor impedance + parallel capacitor) current

     \small{\text{I_{load_{\(total\)}}=\sqrt{I_{motor_{\(real\)}}^2+\(I_{motor_{\(reactive\)}}-76.026 \)^2} \ A}}

    Proceed from there to find the new total load kVA, etc ....
    Last edited: Apr 10, 2015
    Courtney Thomas likes this.
  3. Courtney Thomas

    Thread Starter New Member

    Apr 9, 2015
    Cheers friend you pointed out my error and I think I've come out with the correct answers now for both. (Y)

    I understand it a bit more now in regards to what equations to use for next time!