Ans)
A) Given
![VAB = -VBA 240Z - 60° V](//img.homeworklib.com/questions/f6499cd0-0401-11eb-b880-4f203da1ccd3.png?x-oss-process=image/resize,w_560)
the line to neutral voltage is given as
phase voltage lags line voltage by 30 degrees
poistive sequence balanced source other phase voltages are
![VBA 138.562 - 90° - 120° 120º = 138.562 - 2100 V](//img.homeworklib.com/questions/f7e140a0-0401-11eb-9c87-0b1f91b44609.png?x-oss-process=image/resize,w_560)
![V_{Cn'}=138.56\angle -90^{o}+120^{o}=138.56\angle 30^{o}\: \: V](//img.homeworklib.com/questions/f83e8de0-0401-11eb-b0db-99a0352bca47.png?x-oss-process=image/resize,w_560)
now Kirchofff's current law at node 'n' gives
![\frac{V_{nn'}-V_{An'}}{5+j15}+\frac{V_{nn'}-V_{Bn'}}{12-j12}+\frac{V_{nn'}-V_{Cn'}}{9+j7}=0](//img.homeworklib.com/questions/f89f0850-0401-11eb-b2ef-9b991d2e7417.png?x-oss-process=image/resize,w_560)
![(\frac{1}{5+j15}+\frac{1}{12-j12}+\frac{1}{9+j7})V_{nn'}=\frac{V_{An'}}{5+j15}+\frac{V_{Bn'}}{12-j12}+\frac{V_{Cn'}}{9+j7}](//img.homeworklib.com/questions/f9083ee0-0401-11eb-bd7a-dbfc661dd50d.png?x-oss-process=image/resize,w_560)
![0.1495\angle -28.87^{o}V_{nn'}=\frac{138.56\angle -90^{o}}{5+j15}+\frac{138.56\angle -210^{o}}{12-j12}+\frac{138.56\angle 30^{o}}{9+j7}](//img.homeworklib.com/questions/f97602f0-0401-11eb-bbfc-5fdd6aaacab7.png?x-oss-process=image/resize,w_560)
![0.1495\angle -28.87^{o}V_{nn'}=7.76\angle -122.44^{o}](//img.homeworklib.com/questions/f9cd8ca0-0401-11eb-b359-93932a93e099.png?x-oss-process=image/resize,w_560)
![V_{nn'}=\frac{7.76\angle -122.44^{o}}{0.1495\angle -28.87^{o}}](//img.homeworklib.com/questions/fa2f1d00-0401-11eb-8848-8b9843a80e85.png?x-oss-process=image/resize,w_560)
is the answer
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B)
first we will find current through parallel load connected given
S=10kVA fp=0.6 + leading
as load is connected between A and B the voltage is
the current through this load is given as
![I_{AB}=(\frac{S\angle -cos^{-1}f_{p}}{V_{AB}})^{*}=(\frac{10k\angle -cos^{-1}0.6}{240\angle -60^{o}})^{*}](//img.homeworklib.com/questions/fb38fb80-0401-11eb-9a05-55a06aba4314.png?x-oss-process=image/resize,w_560)
![I_{AB}=41.67\angle -6.87^{o}\: \: A](//img.homeworklib.com/questions/fb9b5cd0-0401-11eb-9d81-fdcb8f492650.png?x-oss-process=image/resize,w_560)
now currents in load impedances connected in star network is
![I_{An}=\frac{V_{An'}-V_{nn'}}{5+j15}=5.49\angle -159.44^{o}\: \: A](//img.homeworklib.com/questions/fbf2b900-0401-11eb-899a-33153620b2b6.png?x-oss-process=image/resize,w_560)
![I_{Bn}=\frac{V_{Bn'}-V_{nn'}}{12-j12}=9.91\angle 178.96^{o}\: \: A](//img.homeworklib.com/questions/fc550d20-0401-11eb-b30a-8b5580f9bad6.png?x-oss-process=image/resize,w_560)
![I_{Cn}=\frac{V_{Cn'}-V_{nn'}}{9+j7}=15.15\angle 6.62^{o}\: \: A](//img.homeworklib.com/questions/fcb7d730-0401-11eb-957d-0fd5bd514043.png?x-oss-process=image/resize,w_560)
now two loads in place using Kirchhoff's current law the
currents drawn from source is
![I_{B}=I_{Bn}-I_{AB}=51.54\angle 174.25^{o}\: \: A](//img.homeworklib.com/questions/fd1c80b0-0401-11eb-b67b-83400c1ce0f1.png?x-oss-process=image/resize,w_560)
as the load is not connected in this branch
given
![V_{AB}=240\angle -60^{o}\: \: V](//img.homeworklib.com/questions/fae4e610-0401-11eb-b9ae-9dc97b0fa865.png?x-oss-process=image/resize,w_560)
![V_{CA}=240\angle -60^{o}+120^{o}\: \: V](//img.homeworklib.com/questions/fe63cf20-0401-11eb-a9eb-bf3d5c6d0b52.png?x-oss-process=image/resize,w_560)
![V_{CA}=240\angle 60^{o}\: \: V](//img.homeworklib.com/questions/fec701e0-0401-11eb-b378-01084c8573c9.png?x-oss-process=image/resize,w_560)
Now wattmeter reading in B is
![W_{B}=Real(V_{BA}*I_{B}^{*})=7226.8\: \: W](//img.homeworklib.com/questions/ff17d7e0-0401-11eb-bd4f-556772bfc8e5.png?x-oss-process=image/resize,w_560)
![\mathbf{W_{B}=7226.8\: \: W}](//img.homeworklib.com/questions/ff757b60-0401-11eb-89ed-5d795c10f3df.png?x-oss-process=image/resize,w_560)
Now wattmeter reading in C is
![W_{C}=Real(V_{CA}*I_{C}^{*})=2169.6\: \: W](//img.homeworklib.com/questions/ffd81e20-0401-11eb-b852-cd00d709ea2a.png?x-oss-process=image/resize,w_560)
![\mathbf{W_{C}=2169.6\: \: W}](//img.homeworklib.com/questions/003977a0-0402-11eb-aa6a-a75b8a72600f.png?x-oss-process=image/resize,w_560)
VBA 2402120° V
VAB = -VBA 240Z - 60° V
VAB VAN -2 -30° V3 240 L-60° -30° V V3
Van' 138.562 - 90° V
VBA 138.562 - 90° - 120° 120º = 138.562 - 2100 V
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VAN VBn' + Ven' 9+17 Vn 12 - 312 9+17 5+j15 5+j15 12 - 312
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51.922 – 93.56° V nn'
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VBA 2402120° V
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