I would like to know what I'm doing wrong here. I am not getting what the book has.(adsbygoogle = window.adsbygoogle || []).push({});

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

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

A:

This is how I'm going about it:

[tex] \bar Z_{load} = 8+j4 \Omega [/tex]

[tex] \bar V_{an} = 120 < 0 \,\,V_{rms} [/tex]

[tex] \bar V_{AN} = 111.62 < -1.33 \,\,V_{rms} [/tex]

[tex] \bar Z_{line} = ? [/tex]

So I simply setup a voltage divider:

[tex] \bar V_{AN} = \bar V_{an}\left( \frac{\bar Z_{load}}{\bar Z_{line} + \bar Z_{load}}\right) [/tex]

Solving for [itex] \bar Z_{line} [/itex] yields:

[tex] \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[/tex]

The book gets [itex] 0.5 + 0.5j \Omega [/itex]

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# Homework Help: Balanced wye-wye system

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