8 
Příklad řešení pro n = 4 (proud čtyrfásový). 
Budiž dáno: n = 4, tedy #■= 90°, E = 100, tedy e = 70-71 
r 0 = 0-8 r' = 0-1 r 1 = 3 r 2 = 4 r 3 = 10 r 4 = 6 vesměs 
^ 0 =1 q' — 0 ^ = 4 — 3 (>3 = 0 p 4 = 8 v ohmech. 
Najdeme 
2? 0 =l-3 #' = 0-1 #,= 5 # 2 =5 #3=10 # 4 =10 
G9 0 = 51°20' ó = 0 = 53°8' ra 2 = 36°52' co 3 = 0 co 4 = m 1 = 53°8'. 
Dosadíme do rovnic (14), obdržíme soustavu 8 lineárních rovnic, 
jichž řešení dává 
A 1 = 0-09884 
A 2 = ^0-09658 
A 3 = — 0-09015 
A 4 = 0-06984 
B ± = 0-12076 
B 2 = 0-12139 
# 3 = — 0-00921 
# 4 = — 0-05402 
P x = 0-23520 
P 2 = 0-23957 
# 3 = 0-11322 
# 4 = 0-13197 
Q x = 0-04488 
0 2 = — 0-00872 
Q 3 = — 0-08030 
<? 4 = 0-02656 
Sečtouce hodnoty pod sebou stojící, obdržíme 
A = Z A k = — 0-01805, B = 2J B k = 0-17892, P = 2 P k = 0-71996, 
Q = Z Q k = — 0-01758. 
Nyní najdeme z rovnic (15) 
a = — 0-52106 p = 5-45775 
a z rovnic (16), (17), (18), (19) a (20) 
«i = 9-517 
« 2 = hšjS 9-735 
a 3 = — 8-636 
« 4 = 6-770 
13-337 
13-450 
0-261 
4-696 
= 16-38 
16-60 
8-64 
8-24 
J 2 = 
J,= 
J,= 
4 = 11-58 
i 2 = 11-74 
i 3 = 6-11 
u = 5-83 
(Proudy pracovní vesměs v ampérech.) 
a k = a k —■ a, p k = p k p . 
= 54° 29' 
<p 2 = 125° 54' 
q> 3 = 181° 44' 
cp 4 = .325° 15' 
- 
10-038 
K = 
7-879 
Jqi — 
12-76 
i 01 = 9-02 
<Poi = 38° 8' 
í?2 — “ 
9-214 
K = 
7-992 
^02 ~ 
12-20 
» 02 = 8-62 
^>02 = 139° 4' 
a 3 = 
8-115 
b 3 = 
5-719 
^03 = 
9-93 
*03= 7-02 
<jP 03 =215° 10' 
čř 4 := ' 
7-291 
\ = 
—-10-154 
^04 = 
12-50 
í 04 = 8-84 
<Poi = 305" 41' 
(Proudy íásové.) 
CCk = Cí k -2 - 1, bk = Pk- 2 - Pk—1 • 
«i' = 
— 15-406 
4-435 
j i = 
16-03 
4' = 11-33 
tp^ = 163° 56' 
< = 
— 2-747 
A'« 
— 18-033 
j‘ = 
18-24 
4' = 12-90 
cp 2 ' = 261° 20 ' 
< = 
19-252 
/V = 
—- 0-113 
j 3 = 
19-25 
4' = 13-61 
(p 3 ' = 359° 40' 
< = 
— 1-100 
13-711 
13-75 
4' = 9-73 
<p 4 ' = 94° 35' 
(Proudy síťové.) 
XIV. 
