- #1

camino

- 42

- 0

## Homework Statement

http://img42.imageshack.us/img42/3537/powersyst.jpg

## Homework Equations

P = I^2 R

## The Attempt at a Solution

Here's what I have so far:

I1 = 49.68A<-68.6 degrees (top loop)

I2 = 49.68A<-128.6 degrees (bottom loop)

I3 = 28.68<-98.6 degrees (triangle loop)

a)

PA12 = (49.68A)^2(12) = 29,617.23 W

PB12 = (49.68A-49.68A)^2(12) = 0 W

PC12 = (49.68A)^2(12) = 29,617.23 W

Ptotal12 = 29,617.23W + 0W + 29,617.23W = [59,234 W] answer

b)

Pab400 = (49.68A-28.68A)^2(400) = 176,400 W

Pbc400 = (49.68A-28.68A)^2(400) = 176,400 W

Pca400 = (28.68A)^2(400) = 329,017 W

Ptotal400 = 176,400W + 176,400W + 329,017W = [681,817 W] answer

Now c) and d) I am unsure of. I was thinking power factor would be 59,234 W/681,817W which would give me a p.f. of 0.087 but that doesn't seem right, and efficiency I have no clue how to calculate.

I'm not 100% sure I even did parts a) and b) right. Any help/explanations would be greatly appreciated!

Last edited by a moderator: