Balanced 380 VLL, 30 an-bn-cn Source a Three-Line, Impedance Diagram of the Sample System 0.5 +1.50 WWW Taa 0.5 +1.50 WWW~C Ibb 0.5 +1.50 WWW Icc 30 Load 1: Y - Connected 30 + j12 Ω per Phase c' Į Ic'n a' Ja'n b' ↓ Ib'n 0.8 + j2 Ω WWW 0.8 + j2 Ω WWW-c 0.8 + j2 Ω WWW~C la'a" a" Ib'b" c" 30 Load 2: A-Connected 24 + 190 02 per Phase c'c" la"b" Ib"c" b" OD M

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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FIND THE UNKNOWN CURRENTS AND VOLTAGES. ATTATCHED ARE THE FINAL ANSWERS. SHOW THE SOLUTION

 

Balanced
380 VLL, 30
an-bn-cn
Source
b
Three-Line,
Impedance
Diagram of the
Sample System
0.5 + j1.5 Ω
WWW~C
ww
Taa
0.5 + j1.5 Ω
-WWW-
ww
0.5+j1.5 0
Ibb.
Icc
30 Load 1:
Y - Connected
30 + j12 Ω
per Phase
c'
↓
Ic'n
a'
la'n
b'
↓
Ib'n
0.8 + j2 Ω
WWW
0.8 + j2 Ω
WWW
0.8 + j2 Ω
WW~C
la'a" a"
Ib'b"
Ic'c"
c"
30 Load 2:
A - Connected
24 + j90 Ω
per Phase
b"
la"b"
Ip"c"
Transcribed Image Text:Balanced 380 VLL, 30 an-bn-cn Source b Three-Line, Impedance Diagram of the Sample System 0.5 + j1.5 Ω WWW~C ww Taa 0.5 + j1.5 Ω -WWW- ww 0.5+j1.5 0 Ibb. Icc 30 Load 1: Y - Connected 30 + j12 Ω per Phase c' ↓ Ic'n a' la'n b' ↓ Ib'n 0.8 + j2 Ω WWW 0.8 + j2 Ω WWW 0.8 + j2 Ω WW~C la'a" a" Ib'b" Ic'c" c" 30 Load 2: A - Connected 24 + j90 Ω per Phase b" la"b" Ip"c"
Sample Problem with Final Answers
Source:
Van = 219.39320° v
Vab= 380230° v
Line 1:
Jaa' = 11.1162 - 49.68° A
Vaa' = 17.576221.89⁰ v
Line 2:
Ia'a" = 6.1222 - 76.47° A
Va'a": = 13.1884 - 8.27° v
Load 1:
la'n = 6.2892 23.65° A
Va'n 203.1912 - 1.85° v
Va'b' = 351.935228.15° v
Vbn
Vbc
=
= 219.3934 - 120° v
= 380-90° v
Ibb' = 11.1162 - 169.68° A
Vbb': = 17.5762-98.11° v
Ib'b" = 6.1222163.53° A
Vb'b" = 13.1882 - 128.27° v
Sample Problem with Final Answers
Ib'n = 6.2892 - 143.65° A
Vb'n 203.1912 - 121.85° v
Vb'c' = 351.9352 - 91.85° v
Ven 219.3932120° v
Vca
3802150° v
=
Ib"c" = 3.5352 - 46.49° A
V"c" = 329.2462 - 91.4⁰ v
Icc':
= 11.116270.32° A
Vcc= 17.5762141.89° v
Ic'c" = 6.122243.53° A
Vec" = 13.188/111.73⁰ v
Vbn = 219.3934 - 120° v
Vbc = 380-90° v
Load 2:
*** Line currents of Load 2 are the same as the line currents of Line 2.
la"b" = 3.5352 - 46.49° A
Va"b" = 329.246228.6° v
Ic'n = 6.289296.35° A
Ve'n 203.1912118.15° v
Vc'a' = 351.9352148.15° v
=
Ic"a" = 3.5352 - 46.49° A
Ve" a": = 329.2462148.6° v
For comparison of voltages, the source voltages are again placed here:
Source:
Van = 219.39320° v
Vab= 380230° v
Ven
Vca = 3802150° v
= 219.3932120° v
Transcribed Image Text:Sample Problem with Final Answers Source: Van = 219.39320° v Vab= 380230° v Line 1: Jaa' = 11.1162 - 49.68° A Vaa' = 17.576221.89⁰ v Line 2: Ia'a" = 6.1222 - 76.47° A Va'a": = 13.1884 - 8.27° v Load 1: la'n = 6.2892 23.65° A Va'n 203.1912 - 1.85° v Va'b' = 351.935228.15° v Vbn Vbc = = 219.3934 - 120° v = 380-90° v Ibb' = 11.1162 - 169.68° A Vbb': = 17.5762-98.11° v Ib'b" = 6.1222163.53° A Vb'b" = 13.1882 - 128.27° v Sample Problem with Final Answers Ib'n = 6.2892 - 143.65° A Vb'n 203.1912 - 121.85° v Vb'c' = 351.9352 - 91.85° v Ven 219.3932120° v Vca 3802150° v = Ib"c" = 3.5352 - 46.49° A V"c" = 329.2462 - 91.4⁰ v Icc': = 11.116270.32° A Vcc= 17.5762141.89° v Ic'c" = 6.122243.53° A Vec" = 13.188/111.73⁰ v Vbn = 219.3934 - 120° v Vbc = 380-90° v Load 2: *** Line currents of Load 2 are the same as the line currents of Line 2. la"b" = 3.5352 - 46.49° A Va"b" = 329.246228.6° v Ic'n = 6.289296.35° A Ve'n 203.1912118.15° v Vc'a' = 351.9352148.15° v = Ic"a" = 3.5352 - 46.49° A Ve" a": = 329.2462148.6° v For comparison of voltages, the source voltages are again placed here: Source: Van = 219.39320° v Vab= 380230° v Ven Vca = 3802150° v = 219.3932120° v
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FIND THE UNKNOWN CURRENTS AND VOLTAGES. ATTATCHED ARE THE FINAL ANSWERS. SHOW THE SOLUTION. SOLVE FOR SUB PARTS LOAD 1 - LOAD 2 PART 

Balanced
380 VLL, 30
an-bn-cn
Source
b
Three-Line,
Impedance
Diagram of the
Sample System
0.5 + j1.5 Ω
WWW~C
ww
Taa
0.5 + j1.5 Ω
-WWW-
ww
0.5+j1.5 0
Ibb.
Icc
30 Load 1:
Y - Connected
30 + j12 Ω
per Phase
c'
↓
Ic'n
a'
la'n
b'
↓
Ib'n
0.8 + j2 Ω
WWW
0.8 + j2 Ω
WWW
0.8 + j2 Ω
WW~C
la'a" a"
Ib'b"
Ic'c"
c"
30 Load 2:
A - Connected
24 + j90 Ω
per Phase
b"
la"b"
Ip"c"
Transcribed Image Text:Balanced 380 VLL, 30 an-bn-cn Source b Three-Line, Impedance Diagram of the Sample System 0.5 + j1.5 Ω WWW~C ww Taa 0.5 + j1.5 Ω -WWW- ww 0.5+j1.5 0 Ibb. Icc 30 Load 1: Y - Connected 30 + j12 Ω per Phase c' ↓ Ic'n a' la'n b' ↓ Ib'n 0.8 + j2 Ω WWW 0.8 + j2 Ω WWW 0.8 + j2 Ω WW~C la'a" a" Ib'b" Ic'c" c" 30 Load 2: A - Connected 24 + j90 Ω per Phase b" la"b" Ip"c"
Sample Problem with Final Answers
Source:
Van = 219.39320° v
Vab= 380230° v
Line 1:
Jaa' = 11.1162 - 49.68° A
Vaa' = 17.576221.89⁰ v
Line 2:
Ia'a" = 6.1222 - 76.47° A
Va'a": = 13.1884 - 8.27° v
Load 1:
la'n = 6.2892 23.65° A
Va'n 203.1912 - 1.85° v
Va'b' = 351.935228.15° v
Vbn
Vbc
=
= 219.3934 - 120° v
= 380-90° v
Ibb' = 11.1162 - 169.68° A
Vbb': = 17.5762-98.11° v
Ib'b" = 6.1222163.53° A
Vb'b" = 13.1882 - 128.27° v
Sample Problem with Final Answers
Ib'n = 6.2892 - 143.65° A
Vb'n 203.1912 - 121.85° v
Vb'c' = 351.9352 - 91.85° v
Ven 219.3932120° v
Vca
3802150° v
=
Ib"c" = 3.5352 - 46.49° A
V"c" = 329.2462 - 91.4⁰ v
Icc':
= 11.116270.32° A
Vcc= 17.5762141.89° v
Ic'c" = 6.122243.53° A
Vec" = 13.188/111.73⁰ v
Vbn = 219.3934 - 120° v
Vbc = 380-90° v
Load 2:
*** Line currents of Load 2 are the same as the line currents of Line 2.
la"b" = 3.5352 - 46.49° A
Va"b" = 329.246228.6° v
Ic'n = 6.289296.35° A
Ve'n 203.1912118.15° v
Vc'a' = 351.9352148.15° v
=
Ic"a" = 3.5352 - 46.49° A
Ve" a": = 329.2462148.6° v
For comparison of voltages, the source voltages are again placed here:
Source:
Van = 219.39320° v
Vab= 380230° v
Ven
Vca = 3802150° v
= 219.3932120° v
Transcribed Image Text:Sample Problem with Final Answers Source: Van = 219.39320° v Vab= 380230° v Line 1: Jaa' = 11.1162 - 49.68° A Vaa' = 17.576221.89⁰ v Line 2: Ia'a" = 6.1222 - 76.47° A Va'a": = 13.1884 - 8.27° v Load 1: la'n = 6.2892 23.65° A Va'n 203.1912 - 1.85° v Va'b' = 351.935228.15° v Vbn Vbc = = 219.3934 - 120° v = 380-90° v Ibb' = 11.1162 - 169.68° A Vbb': = 17.5762-98.11° v Ib'b" = 6.1222163.53° A Vb'b" = 13.1882 - 128.27° v Sample Problem with Final Answers Ib'n = 6.2892 - 143.65° A Vb'n 203.1912 - 121.85° v Vb'c' = 351.9352 - 91.85° v Ven 219.3932120° v Vca 3802150° v = Ib"c" = 3.5352 - 46.49° A V"c" = 329.2462 - 91.4⁰ v Icc': = 11.116270.32° A Vcc= 17.5762141.89° v Ic'c" = 6.122243.53° A Vec" = 13.188/111.73⁰ v Vbn = 219.3934 - 120° v Vbc = 380-90° v Load 2: *** Line currents of Load 2 are the same as the line currents of Line 2. la"b" = 3.5352 - 46.49° A Va"b" = 329.246228.6° v Ic'n = 6.289296.35° A Ve'n 203.1912118.15° v Vc'a' = 351.9352148.15° v = Ic"a" = 3.5352 - 46.49° A Ve" a": = 329.2462148.6° v For comparison of voltages, the source voltages are again placed here: Source: Van = 219.39320° v Vab= 380230° v Ven Vca = 3802150° v = 219.3932120° v
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