CHAPTER XXXVI 
THREE-PHASE SYSTEM 

308. With equal load of the same phase displacement in all 
three branches, the symmetrical three-phase system offers no 
special featm-es over those of three equally loaded single-phase 
systems, and can be treated as such; since the mutual reactions 
between the three phases balance at equal distribution of load, 
that is, since each phase is acted upon by the preceding phase in 
an equal but opposite manner as by the following phase. 

With unequal distribution of load between the different 
branches, the voltages and phase differences become more or less 
unequal. These unbalancing effects are obviously maximum if 
some of the phases are fully loaded, others unloaded. 

Let E = e.m.f. between branches 1 and 2 of a three-phaser. 
Then 

e E = e.m.f. between 2 and 3, 
€^E = e.m.f. between 3 and 1; 

where e — v^ = —-^ . 

Let 

Zi, Z2, Z3 = impedances of the lines issuing from generator 
terminals 1, 2, 3, 
and 1^1, Yo, Y3 = admittances of the consumer circuits con- 
nected between lines 2 and 3, 3 and 1, 1 and 2. 
If then, 

Ii, 1 2, 1 3, are the currents issuing from the generator termi- 
nals into the lines, it is, 

Ix + h + h = 0. (1) 

If, I'l, I'l, I's = currents through the admittances, Fi, ¥2, Y3, 
' from 2 to 3, 3 to 1, 1 to 2, it is, 

h = // - /'2, or, h + /'2 - /'a = 0 

'h = I'l - 'I'z, or, 72 + I'z - /'i = 0 [ (2) 

73=>2-i'x, or, /3 + h-r2 = 0 

457 


458 ALTERNATING-CURRENT PHENOMENA 

These three equations (2) added, give (1) as dependent 
equation. 

At the ends of the hnes 1, 2, 3, it is: 


E'l = El — Zili + ^3/3 
E 1 — El — Zils + Zili 
E 3 = E3 — Zili + Z2/2 

the differences of potential, and : 

I\ = E\Yi 

I i = E 2^2 

I 3 — E 3Y3 


(3) 


(4) 


the currents in the receiver circuits. 

These nine equations (2), (3), (4), determine the nine quan- 
tities: /i, h, 1 3, //, h', I/, El', EoJ, E/. 

Equations (4) substituted in (2) give: 

Ix = E'3Y3 - E'^Y^ 

'h = E\Yi - E'^Yz \ (5) 

h = E\Y-, - E'lYx 


These equations (5) substituted in (3), and transposed, give: 
as e.m.fs. at the generator terminals. 


since Ei = e E 
E2 = €^'e 

E3 = e' 


€ E - E\(l + YiZ2 + F1Z3) + E'^Y^Zs + E'3Y3Z, = 0 
e^E - ^'2(1 + Y2Z3 + Y2ZO + E'3Y3Zi + E\YiZ3 = 0 [ (6) 
E - 'e'3(1 + F3Z1 + F3Z2) + E\YiZ, + ^'2^2^! = 0 

as three linear equations with the three quantities, E'l, E'l, E'3. 


THREE-PHASE SYSTEM 

Substituting the abbreviations: 


459 


K = 


Ky = 


Ko = 


Kz^ 


- (1 + F1Z2 + Y^Za), Y^Za, YzZ^ 

Y,Zs, - (I + F2Z3+ F2Z1), F3Z1 

F1Z2, F2Z1, - (1 + FsZi + F3Z2) 


6, F2Z3 


F3Z2 


e\ - (1 + F2Z3 + F2Z1), F3ZX 

1, FaZi, - (1 + F3Z1 + F3Z2) 

- (I + F1Z2+ FiZ3),6, F3Z2 


F1Z3, 

F1Z2, 


6^ F3Z1 

1, - (1 + F3Z1 + F3Z2). 


(7) 


- (l + FiZ2+F,Z3), F2Z3, 


F1Z3, 
F1Z2, 


- (1 + F2Z3 + F2ZO, e' 
Y,Z,, 1 


we have: 


EKi 
?'- K 

EK2 
?'- K 

EK3 

^'= K 

YiEK 
T 

1 

f'- K 

Y2EK 
T 

2 

^.'- K 

Y3EK3 

^'- k 1 

Y3K3 - Y2K2E 

T 

^'- K 

YiKi - YJCzE 

^- K 

J. Y2K2 — YiK\i!j 

(8) 


(9) 


(10) 


hence, 


E\ + E'2 + ^'3 = 0 I 


(11) 


460 ALTERNATING-CURRENT PHENOMENA 
309. Special Cases. 

A. Balanced System 
Y,= Y,= Y,= Y 

Substituting this in (6), and transposing: 


1 + 3FZ 


E\ = 

E 2 = 

E 3 = 

h = 

h = 
h = 


iE 


1 +3 FZ 


2E 


1 + 3 rz 

E 
1 +3 FZ 

e^e - 1)EY 

1 + 3 FZ 
(6- 1)EY 

1 + 3 FZ 
e (e - 1) EY 

1 + 3FZ 


(12) 


The equations of the symmetrical balanced three-phase system. 

B. One Circuit Loaded, Two Unloaded 

Fi = F2 = 0, Ys = Y, 
Zi = Z2 = Z3 = Z. 


Substituted in equations (6) : 

eE - E'i-\- Ez'YZ = 0 


unloaded branches. 


t'E - E\ -h E\YZ = 0, 
E - E'sil +2 YZ) = 0, loaded branch. 


hence: 
E', = 
E 2 = 
E 3 = 


E\e-{-{l+2e)YZ} 

1 + 2 FZ 
E{e^-\-(l +2e2) YZ} 


1+2 FZ 


E 


1 +2FZ 


unloaded ; 
loaded; 


all three 
e.m.fs. 
unequal, and (13) 

of unequal 
phase angles. 


THREE-PHASE SYSTEM 


461 


I\ = /'2 = 0 

EY 
I'z = 


Ii = 


1 +2YZ 

EY 
1 +2FZ 
EY 


f' 1 + 2 FZ 

/a = 0 


C. Two Circuits Loaded, One Unloaded 

Fi = F2 = F, F3 = 0, 
Zi = Z2 ^= Z3 = z. 

Substituting this in equations (6), it is 

e E - E\{1 + 2 FZ) + E'2YZ = 0 
e^E - E'^il + 2 FZ) + i;'iFZ = 0 

E - E's -\- {E\ + ^'2)FZ = 0 unloaded branch. 


loaded branches. 


or, since 


{E\ + E'2) = - E',; 
E- E's'- E'sYZ = 0, 

E' 3 = 


1 + YZ' 


thus, 


je;6{i + (2 + 6)fzi 

E\ = - 


E 2 = 


E 3 = 


1 + 4 FZ + 3 Y'~Z' 
Ee^l 4-(2 + ei) YZ\ 
1 + 4 FZ + 3 Y^Z^ 
E 


1 + YZ 


loaded branches. 


unloaded branch. 


(13) 


(13) 


(14) 


As seen, with unsymmetrical distribution of load, all three 
branches become more or less unequal, and the phase displace- 
ment between them unequal also.