# Three phase total power consumed

• Butterfly41398

#### Butterfly41398

Summary:: I don't seem to see any problem with my work. But my final answer does not match the correct answer. (In number 3)

Here's my work

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@Butterfly41398 , it's very difficult to make out the contents of your images (the handwritten workings), so I can't really help spot any errors. You'll have to clean up the images to make them useful, or better yet, type out the work so that others can quote it in replies.

PhDeezNutz
I don't seem to see any problem with my work.
As mentioned, it's best if you type your work into the forum to make it more legible. Please see the LaTeX Guide link below the Edit window for more information on how to format math equations at PF. Thanks.

PhDeezNutz
I agree with the berkeman, of course. However, I manage to read your calculation. I still have to recalculate, but at first glance, I have a general observation: to convert the arguments from negative to positive you have to add 360 degrees and not 180.

In my calculation Iy is correct, Ib is correct except the argument has to be 27.525 instead of 29.525. However Ir is quite different. Ir=28.406<-69.1127.

I agree with the berkeman, of course. However, I manage to read your calculation. I still have to recalculate, but at first glance, I have a general observation: to convert the arguments from negative to positive you have to add 360 degrees and not 180.
Following your entire calculation this it is not an issue. The only mistake it is wrong calculation of IR.
The power it is the real part of the sum of products: VR*conjugate(IR)+VY*conjugate(IY)+VB*conjugate(IB) where
VR=IR*ZR ;VY=IY*ZY;VB=IB*ZB