transformer losses

Discussion in 'Homework Help' started by yoamocuy, Oct 26, 2009.

  1. yoamocuy

    Thread Starter Active Member

    Oct 7, 2009
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    I need to find the following for the transformer shown
    a)core loss
    b)copper loss
    c)active power supplied
    d)reactive power supplied

    I know that the copper loss is equal to Re*I1 and I have no problems finding the ative or reactive powe, but I am not sure about the core loss.

    Would the core loss just be Ic*Ro?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The core loss is "modeled" by the resistor Ro. Calculate the power dissipation in Ro.

    What were your other values?

    It's not clear whether parts (c) & (d) refer to those quantities supplied to the secondary side or those drawn from the 150V supply.
     
  3. yoamocuy

    Thread Starter Active Member

    Oct 7, 2009
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    0
    My turns ratio was 2:1 so moving my secondary impedance over to the primary sides gives me (8-j12)+(1+j24)=(9+j12)

    Vs/(9+j12)=I2
    after converting the impedance to polar and dividing I get a current I2 of 10A at an angle of -53.13

    Io=Vs/total impedance of open circuit
    1/(total impedance)=1/250+1/j100
    total impedance=34.48+j86.21
    total impedance in polar=92.85Ω at an angle of 68.20°
    Io=150/92.85=1.62A at an angle of -68.20°

    I1=Io+I2
    after converting both Io and I2 back to rectangular and adding together I get I1=9.26 at an angle of 44.54°

    b) copper loss=Re*I1
    copper loss=1*9.26=9.26 W

    (for both parts c and d I'm finding power supplied)
    c)active power supplied=VIcosθ
    P=150*9.26*cos(44.54)
    P=990.2 W

    d)Q=-VIsinθ
    Q=-150*9.26*sin(44.54)
    Q=-1145 var

    a) Ok you said that core loss is the power dissipated in Ro, so would that just be P=Vs^2/Ro?
    P=150^2/250=90 W?
     
  4. yoamocuy

    Thread Starter Active Member

    Oct 7, 2009
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    OK, I solved for my primary current and got 11.57A at an angle of
    -55.21°.

    a) so if core loss is the power dissipated across Ro, would that just make the power equal to Vs^2/ Ro? which would be 150^2/250=90 W

    b) for the copper loss I was thinking that it was Re*I1^2 however now Im thinking it would be I2^2 because if I used I1 I would be neglecting the current lost across the open circuit, which normally wouldn't make a huge difference but for this problem the open circuit curent is 1.6 so it does make a difference.

    Therefore, copper loss= 1*10^2=100 W

    I'm not sure about this answer because it seems like Vs^2/Re and Vs*I2 should all equal the same thing but I get different answers for each, so is there a problem with my method of solving?

    For both c and d I am finding the power supplied
    c) P=VIcosθ
    P=(150)*(11.57)*cos(-55.21)
    P=990.2 W
    d)Q=-VIsuinθ
    Q=-(150)*(11.57)*sin(-55.12)
    Q=-1.43 kvars
     
  5. yoamocuy

    Thread Starter Active Member

    Oct 7, 2009
    80
    0
    I got 10A at an angle of -53.1 for I'2
    Io=1.62 at an angle of -68.20
    I1=11.57 at an angle of -55.21

    a) if core loss is equal to power dissipated in Ro then it is equal to Vs^2/Ro? so core loss=90 W?

    b) Originally I though copper loss was equal to Re*I1^2 but now im thinking it would be I'2^2 because using I1 neglects the current that goes through the open circuit. When I do this, however, I get 100 W which seems ok but if I calculate the power disippated across Re using either Vs^2/Re or Vs*I'2 then I get two completely different values.

    for parts c and d I'm supposed to find the power supplied
    c) P=VIcos(phi)
    P=(150)*(11.57)*cos(-55.21)
    P=990.2 W
    d) Q=-VIsin(phi)
    Q=-(150)*(11.57)*sin(-55.21)
    Q=1.425 kvars

    So I'm pretty sure about parts c and d, and if core loss is = to power dissipated in Ro then part a shoudl be ok but I don't know what happened on part b
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I agree with all your answers - copper loss=100W:)
     
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