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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<br /> =0.134+0.306j \Omega
The book gets 0.5 + 0.5j \Omega
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<br /> =0.134+0.306j \Omega
The book gets 0.5 + 0.5j \Omega
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