Electrical homework RLC,SINGLE PHASE,MOTOR

Discussion in 'Homework Help' started by twoswordali, Jul 13, 2009.

  1. twoswordali

    Thread Starter New Member

    Aug 13, 2008
    4
    0
    Hello everybody I got these questions for homework and have no idea on how to approach them i would appreciate any help, a hint or an answer followed by an explanation. I dont wish to sound like an idiot but my brain has frozen in my electrical engineering classes urgh!! :) I would very much appreciate any help for any of these questions thank you
     
  2. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    I'll give the third one a try:

    The amp at the input is a comparator. It combines feedback from the tachometer and the voltage of the pot. The pot sets the voltage level of the comparator so that deviations reflected by the tachometer output are compensated for. If the motor is running too fast the output voltage of the comparator will be damped, thus slowing the motor. This provides a stable speed at the output since speed of a dc motor is proportional to the voltage it sees at its terminals.

    The driver is a power amplifier that is able to supply the current required by the motor without burning itself (the power amp) up. (The comparator stage cannot do this because it is a low current device.)

    The bevel drive turns the tachometer which is a transducer that creates a voltage proportional to speed. This voltage is sent to the comparator to automatically decrease the voltage output of the comparator if the motor is running too fast.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    In the first question proceed as follows

    1. Draw the circuit - at least to give an understanding of the layout.
    2. Convert the capcitor and inductor values to their equivalent reactance values Xc & Xl at 1KHz
    3. Convert the two series branches to an equivalent lumped impedance
    4. The capacitive branch will have the form Zc=Rc-jXc and the inductive branch Zl=Rl + jXl
    5. Calculate the current in each parallel branch
    6. Use Ic = (50@angle 0)/Zc and Il=(50@angle 0)/Zl
    7. Add Ic+Il (using complex notation) to find the source current Is.
    8. Is will have required phase angle, Theta
    9. Power Factor = cos(Theta)
     
  4. The Flavored Coffee Guy

    Member

    Nov 7, 2006
    13
    0
    1. Draw the Schematic.
    [​IMG]

    θ=arctan(XC/R)*-1 for the first branch.
    θ=arctan(XL/R) is the same for the second branch.

    And that's where it all drops off for me. Which means, I need to study again! It does wear off. I hope I die before I seriously forget.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Hi Coffe Guy,

    Vanilla, Caramel or Hazelnut?

    You missed the 15Ω loss resistance in the inductor.
     
  6. twoswordali

    Thread Starter New Member

    Aug 13, 2008
    4
    0
    thank you guys very much i thought thats how u do it;) man thanks again, what about the other questions how do i approach those??

    Again i cant express how grateful i am that you guys are helping me thank u very much.
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Q2 - Transformer problem

    Some assumptions are required ....

    1. Assuming 1000kVA rating is the maximum input rating at 6600V applied primary voltage - see note below.
    2. Given unity pf then rated primary current = 1,000,000VA/6600V = 151.5A
    3. The no-load iron losses = 3000W @ 6600V giving an equivalent loss resistance of (6600^2)/3000 = 14.52kΩ
    4. At rated input the primary series voltage drop = 151.5 x 0.4 = 60.6V
    5. Voltage (full load) at ideal primary transformer = 6600-60.6=6539.4V
    6. Iron loss current at full load = 6539.4/14.52k = 0.45A
    7. "Ideal" primary current at full load = 151.5-0.45=151.05A
    8. Turns ratio = 6600:400 = 16.5 : 1
    9. "Ideal" secondary voltage at full load = 6539.4/16.5 = 396.33V
    10. "Ideal" secondary current at full load = 151.05 x 16.5 = 2492.3A
    11. Secondary voltage drop at full load = 2492.3 x .005 = 12.46V
    12. Secondary output volatge (at full load) = 396.33-12.46 = 383.87V
    Note - If the rating is assumed for the secondary output = 1000kVA then the solution would be different.
     
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