AC power problems.

Discussion in 'Homework Help' started by eleceng, Aug 3, 2009.

  1. eleceng

    Thread Starter New Member

    Aug 3, 2009
    4
    0
    Studying for a big final exam and have no solutions for this questions and unable to work out the last few questions. The circuit diagram is attached and i need to calculate the following:
    -total impeadance of the circuit (i calculate 5.8 ohm)
    -the load current (i calculate 68.65A)
    -rms load voltage (Vl) (i caculate 242.7v)
    -the phase angle of load voltage relative to the supply voltage
    -load power
    -load reactive power
    -power drawn from supply
    -reactive power drawn from supply
    -power factor of supply
    -power lost in connecting cable
     
    • q6.jpg
      q6.jpg
      File size:
      43.1 KB
      Views:
      29
  2. yourownfree

    Active Member

    Jul 16, 2008
    89
    0
    When I calculate separate I get .741401 for the 2.23 mh and resistor and 5 for the 9.55 and 4 ohm total of 5.741401 Ohms. So I would have to say 5.74 Ohms. or 5.7 Ohms.
     
    Last edited: Aug 3, 2009
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The total complex impedance is 5.83 (angle:39.43°) ohms - so you are correct eleceng.

    You have the RMS current correct but not the load RMS voltage.

    I have V_load = 343.3 volts

    Since ...

    Vload will be (I_load * Z_load) where Z_load is 5.0 (angle: 36.87°) ohms and I_load = 68.66 amps. I think you forgot to include the load inductance in the total load calculation.
     
  4. eleceng

    Thread Starter New Member

    Aug 3, 2009
    4
    0
    I intially had a the voltage of 343.3 Volts but assumed that to get the RMS i had to multiply the value by 1/√2 - clearly i was wrong. Therefore the load power is 23.57KW ? (P = I^2 * R) and the power drawn from supply is 27.342KW. The load reactive power is (P = I(68.66)^2 * (2.93)XL) therefore 13.817Kvar and reactive power drawn from supply is (P = I(68.66)^2 * (3.7)XL) therefore 17.446KVAr?
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782

    I get .....

    Load power = I^2*R_load = (68.66)^2*4 = 18.85kW
    Source power = I^2*R_total = (68.66)^2*4.5 = 21.21kW
    Load Reactive Power = I^2*X_load = (68.66)^2*3.0 = 14.14kVar
    Source Reactive Power = I^2*X_total = (68.66)^2*3.7 = 17.44kVar
    Source apparent power = V_source*I=400*68.66=27.46kVA
    Source pf = Real Power / Apparent Power = 21.21/27.46 = 0.772

    Verify pf using total impedance angle = 39.43°

    pf = cos(39.43°)= 0.772
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    pf is lagging BTW
     
  7. eleceng

    Thread Starter New Member

    Aug 3, 2009
    4
    0
    Thank you very much for your help, can now see where I have been going wrong! Thanks.
     
Loading...