# 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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0
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?

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2. ### t_n_k AAC Fanatic!

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

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
84
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
84
0
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
84
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

Mar 6, 2009
5,448
790