Balanced wye-wye system

1. Jun 21, 2006

I would like to know what I'm doing wrong here. I am not getting what the book has.

Q: In a balanced three-phase wye-wye system, the load impedance is $8+j4\Omega$. The source has phase sequence abc and $\bar V_{an} = 120<0\,\,V_{rms}$. If the load voltage $\bar V_{AN} = 116.62<-1.33\,\,V_{rms}$ determine the line impedence.

Please excuse me being lazy and not looking up how to properly represent polar numbers in LaTeX. Thus $X < 90$ would mean a magnitude of $X$ with a phase angle of $90$ (in degrees).

A:
This is how I'm going about it:
$$\bar Z_{load} = 8+j4 \Omega$$
$$\bar V_{an} = 120 < 0 \,\,V_{rms}$$
$$\bar V_{AN} = 111.62 < -1.33 \,\,V_{rms}$$
$$\bar Z_{line} = ?$$

So I simply setup a voltage divider:
$$\bar V_{AN} = \bar V_{an}\left( \frac{\bar Z_{load}}{\bar Z_{line} + \bar Z_{load}}\right)$$

Solving for $\bar Z_{line}$ yields:

$$\bar Z_{line} = \frac{\bar V_{an}\bar Z_{load}}{\bar V_{AN}}-\bar Z_{load} = \frac{(120<0)(8+j4)}{(116.62<-1.33)}-8+j4 =0.134+0.306j \Omega$$

The book gets $0.5 + 0.5j \Omega$

Last edited: Jun 21, 2006
2. Jul 17, 2006

interested_learner

I am not a power expert, but it looks right to me. You math is right too. Tell me what your prof says.

3. Jul 20, 2006

Staff: Mentor

I'm guessing the difference has to do with the Y shape of the load and line impedances. When they say that the load impedance is 8+j4, is that each of the three Y impedances, or the parallel combination of them, or some other variation? I haven't worked with Y-delta stuff much, so I don't know what the convention is. But maybe that's why the book has a different answer.

EDIT -- Oops, I see now that this question was from last month. Sorry for the slow response, FrogPad. What turned out to be the error?

4. Jul 28, 2006