13.  ALTERNATING-CURRENT  TRANSFORMER 

60.  The  alternating-current  transformer  consists  of  one  mag- 
netic circuit  interlinked  with  two  electric  circuits,  the  primary 
circuit  which  receives  energy,  and  the  secondary  circuit  which 
delivers  energy. 

Let  TI  =  resistance,    x\  =  2TrfSz  =  self-inductive    or  leakage 

reactance  of  secondary  circuit, 

r0  =  resistance,    XQ  =  2irfSi  =  self -inductive  or  leakage 
reactance  of  primary  circuit, 

where  S2  and  Si  refer  to  that  magnetic  flux  which  is  interlinked 
with  the  one  but  not  with  the  other  circuit. 


Let    a 


ratio  of  — — •  — -  turns  (ratio  of  transformation), 
primary 


An  alternating  e.m.f.  E0  impressed  upon  the  primary  electric 
circuit  causes  a  current,  which  produces  a  magnetic  flux  $  inter- 
linked with  primary  and  secondary  circuits.  This  flux  <l>  gener- 
ates e.m.fs.  EI  and  E{  in  secondary  and  in  primary  circuit, which 

Tjl 

are  to  each  other  as  the  ratio  of  turns,  thus  Ei  =  —  - 

Let  E  =  secondary  terminal  voltage,  I\  =  secondary  current, 
0i  =  lag  of  current  /i  behind  terminal  voltage  E  (where  B\  <  0 
denotes  leading  current). 

Denoting  then  in  Fig.  34  by  a  vector  OE  =  E  the  secondary 


68       ELEMENTS  OF  ELECTRICAL  ENGINEERING 


terminal  voltage,  01 1  =  l\  is  the  secondary  current  lagging  by 
the  angle  EOI  =  61. 

The  e.m.f.  consumed  by  the  secondary  resistance  7*1  is  OE'i  = 
E'i  =  Iiri  in  phase  with  /i. 

The  e.m.f.  consumed  by  the  secondary  reactance  Xi  is  OE"\  = 
E'\  =  I&i,  90  degrees  ahead  of  /i.  Thus  the  e.m.f.  con- 
sumed by  the  secondary  impedance  z\  =  Vn2  +  Xi2  is  the 
resultant  of  OE'i  and  OE"i,  or  OE"\  =  E"\  =JiZi. 

OE'"\  combined  with  the  terminal  voltage  OE  =  E  gives  the 
secondary  e.m.f.  OEi  =  E\. 

Proportional  thereto  by  the  ratio  of  turns  and  in  phase  there- 


FIG. 34.  —  Vector  diagram  of  e.m.fs.  and  currents  in  a  transformer. 
with  is  the  e.m.f.  generated  in  the  primary  OEi  =  Ef   where 


To  generate  e.m.f.  EI  and  Ei}  the  magnetic  flux  0$  =  <i>  is 
required,  90  time  degrees  ahead  of  OE\  and  OEi.  To  produce 
flux  $  the  m.m.f.  of  F  ampere-turns  is  required,  as  determined 
from  the  dimensions  of  the  magnetic  circuit,  and  thus  the 
primary  current  /oo,  represented  by  vector  O/oo,  leading  0$  by 
the  angle  a. 

Since  the  total  m.m.f.  of  the  transformer  is  given  by  the 
primary  exciting  current  70o,  there  must  be  a  component  of 
primary  current  /',  corresponding  to  the  secondary  current  /i, 
which  may  be  called  the  primary  load  current,  and  which  is 


ALTERNATING-CURRENT  TRANSFORMER 


69 


opposite  thereto  and  of  the  same  m.m.f.;  that  is,  of  the  intensity 
/'  =  a/i,  thus  represented  by  vector  01'  =  I'  =  a/i. 

O/oo,  the  primary  exciting  current,  and  the  primary  load 
current  O/',  or  component  of  primary  current  corresponding 
to  the  secondary  current,  combined,  give  the  total  primary 
current  O/o  =  /o- 

The  e.m.f.  consumed  by  resistance  in  the  primary  is  OE'Q  = 
E'Q  =  /Or0  in  phase  with  /0. 

The  e.m.f.  consumed  by  the  primary  reactance  is  OE"o  =  E"Q 
=  /0£o,  90  degrees  ahead  of  O/o. 

OE'Q  and  OE"o  combined  gives  OE'"Q,  the  e.m.f.  consumed  by 
the  primary  impedance. 


FIG.  35. — Vector  diagram  of  transformer  with  lagging  load  current. 

Equal  and  opposite  to  the  primary  counter-generated  e.m.f. 
OEi  is  the  component  of  primary  e.m.f.,  OEf,  consumed  thereby. 

OE'  combined  with  OE"'Q  gives  OEQ  =  EQj  the  primary  im- 
pressed e.m.f.,  and  angle  0o  =  -#o#/o,  the  phase  angle  of  the 
primary  circuit. 

Figs.  35,  36,  and  37  give  the  polar  diagrams  of  0i  =  45°  or 
lagging  current,  0i  =  zero  or  non-inductive  circuit,  and  6  = 
—  45°  or  leading  current. 

61.  As  seen,  the  primary  impressed  e.m.f.  E0  required  to  pro- 
duce the  same  secondary  terminal  voltage  E  at  the  same  current 
1 1  is  larger  with  lagging  or  inductive  and  smaller  with  leading 


70       ELEMENTS  OF  ELECTRICAL  ENGINEERING 


current  than  on  a  non-inductive  secondary  circuit;  or,  inversely, 
at  the  same  secondary  current  I\  the  secondary  terminal  voltage 
E  with  lagging  current  is  less  and  with  leading  current  more 
than  with  non-inductive  secondary  circuit,  at  the  same  primary 
impressed  e.m.f.  EQ. 

The  calculation  of  numerical  values  is  not  practicable  by 
measurement  from  the  diagram,  since  the  magnitudes  of  the 
different  quantities  are  too  different,  E'i:E"iiEi:Eo  being 
frequently  in  the  proportion  1  :  10  :  100  :2000. 

Trigonometrically,  the  calculation  is  thus: 


FIG.  36. — Vector  diagram  of  transformer  with  non-inductive  loading. 
In  triangle  OEEi,  Fig.  34,  writing 

tan  0'  =  ^ 
we  have, 

also, 


OEi2  =  OE2  +  EEi2  -  2  OE  EEi  cos 


EEi  = 

i  =  180  -  tf  +  0i, 


hence, 

E£  =  E2  +  7i2zi2  +  2  #7iZi  cos  (0'  -  00 . 

This  gives  the  secondary  e.m.f.,  EI,  and  therefrom  the  primary 
counter-generated  e.m.f. 

Ei  =  — • 
In  triangle  EOEi  we  have 

sin  E\OE  -T-  sin  E\EO  =  .E^n  -f-  EiO 


ALTERNATING-CURRENT  TRANSFORMER         71 

thus,  writing 

£  E&E  =  0", 
we  have 

sin  0"  -4-  sin  (0'  -  0i)  =  hz  +  Ei, 

wherefrom  we  get 

%  6",  and  £  E1OIl  =  6  =  0,  +  0", 

the  phase  displacement  between  secondary  current  and  secondary 
e.m.f. 


FIG.   37. — Vector  diagram  of  transformer  with  leading  load  current. 

In  triangle  O/oo/o  we  have 


since 


and 


0/02  =  O/oo2  +  /oo/o2  -  2  O/oo/oo/o  COS  O/oo/o, 

£  #i00  =  90°, 
$  O/oo/o  =  90  +  0  +  a, 


J00/0  =  r  =  al 


O/oo  =  IQQ  =  exciting  current, 

calculated  from  the  dimensions  of  the  magnetic  circuit.     Thus 
the  primary  current  is 

J02  =  J002  +  a2/l2  _{_  2  a/ i/oo  sin  (0  +  a). 
In  triangle  O/oo/o  we  have 

sin  /ooO/o  -T-  sin  O/oo/o  =  /oo/o  -^  0/0; 
writing 

this  becomes 

sin  0"0  -^  sin  (0  +  a)  =  a/i  -?-  /0; 

therefrom  we  get  0"0,  and  thus 

£  #'0/0  =  02  =  90°  -  a  -  0"0. 


72       ELEMENTS  OF  ELECTRICAL  ENGINEERING 

In  triangle  OE'EQ  we  have 

OEQ2  =  OE~'2  +  WWo*  -20E'  WW0  cos  OE'  #0; 
writing 

tan  0'0  =  — ' 

7*0 

we  have 

£  OE'E0  =  180°  -  0'  +  02, 

OE'  =  E<  =      > 


thus  the  impressed  e.m.f.  is 

ET    9  ^l2       I        T    9        9       I       *^UO*0  /„    /  „    s 

£V    =    ^2"   +  ^02^02   H COS  (00     —    02). 

In  triangle  OE'EQ 

sin  E'OEo  -T-  sin 
thus,  writing 


we  have 

sin  Q'\  +  sin  (0'0  -  02)  =  /o£o  ^  ^0; 
herefrom  we  get  ^  0"i,  and 

^    00    =    02    +    0"l, 

the  phase  displacement  between  primary  current  and  impressed 
e.m.f. 

As  seen,  the  trigonometric  method  of  transformer  calculation 
is  rather  complicated. 

62.  Somewhat  simpler  is  the  algebraic  method  of  resolving 
into  rectangular  components. 

Considering  first  the  secondary  circuit,  of  current  7i  lagging 
behind  the  terminal  voltage  E  by  angle  0i. 

The  terminal  voltage  E  has  the  components  E  cos  0i  in  phase, 
E  sin  0i  in  quadrature  with  and  ahead  of  the  current  /i. 

The  e.m.f.  consumed  by  resistance  r1}  I-pi,  is  in  phase. 

The  e.m.f.  consumed  by  reactance  x\,  I&i,  is  in  quadrature 
ahead  of  /i. 

Thus  the  secondary  e.m.f.  has  the  components 

E  cos  0i  +  /ifi  in  phase, 

E  sin  0i  +  IiXi  in  quadrature  ahead  of  the  current  /i,  and 
the  total  value, 

Ei  =  V(E  cos  0i  +  /in)2  +  (E  sin  0i  +  /iZi)2, 


ALTERNATING-CURRENT  TRANSFORMER         73 

and  the  tangent  of  the  phase  angle  of  the  secondary  circuit  is 

E  sin  0i  + 


tan  e  = 


E  COS  61  + 


Resolving  all  quantities  into  components  in  phase  and  in 
quadrature  with  the  secondary  e.m.f.  EI,  or  in  horizontal  and 
in  vertical  components,  choosing  the  magnetism  or  mutual  flux 
as  vertical  axis,  and  denoting  the  direction  to  the  right  and 
upward  as  positive,  to  the  left  and  downward  as  negative,  we 
have 

Horizontal  Vertical 

component         component 

Secondary  current,  /i,  —  Ii  cos  0  +  /i  sin  0 

Secondary  e.m.f.,  EI,  —  EI  0 

Primary  counter-generated  e.m.f., 

El  =  ^  -  -1  0 

a  a 

Primary    e.m.f.    consumed    thereby, 

E'  =  -  E,,  +  ^  0 

Primary   load  current,  /'  =  —  a/i,+  al\  cos  6  —  al\  sin  0 
Magnetic  flux,  $>,  0  —  <£ 

Primary  exciting  current,   /Oo,   con- 
sisting of  core  loss  current,  /oo  sin  a 
magnetizing  current,  —  /oo  cos  a. 
hence,  total  primary  current,  J0, 

Horizontal  component  Vertical  component 

all  cos  0i  +  /oo  sin  a  —  (all  sin  0i  +  /oo  cos  a) 

E.m.f.  consumed  by  primary  resistance  r0,  E'Q  =  Ior0  in  phase 
with  /o, 

Horizontal  component  Vertical  component 

r0a/i  cos  0  +  r0/oo  sin  a  —  (r0a/i  sin  0  +  r0/oo  cos  a) 

E.m.f.   consumed  by  primary  reactance  x0,   E0  =  IoXQ,  90° 
ahead  of  /o, 

Horizontal  component  Vertical  component 

X0ali  sin  0  +  Zo/oo  cos  a          +  X0ali  cos  0  +  rr0/oo  sin  a 

Tjl 

E.m.f.    consumed    by    primary    generated    e.m.f.,    E'  =  -—• 
horizontal. 


74       ELEMENTS  OF  ELECTRICAL  ENGINEERING 
The  total  primary  impressed  e.m.f.,  EQ, 

Horizontal  component 
ET 

—  +  ol\  (r0  cos  0  •+•  #o  sin  6)  +  /oo  (TO  sin  a  +  XQ  cos  a). 

Vertical  component 

a/i  (r0  sin  0  —  £0  cos  0)  +  70o  (ro  cos  a  —  XQ  sin  a), 

or  writing  tan  0'o  =  —  > 

?*o 

since 

\A*o2  +  £02  =  ZQ,  sin  0'0  =  —  ,  and  cos  0'0  =  -°' 

20  ZQ 


Substituting  this  value,  the  horizontal  component  of  E0  is 
•pi 

—  +  a20/i  cos  (0  —  0'0)  +  2o/oo  sin  (a  +  0r0)  ; 

the  vertical  component  of  E0  is 

azo/i  sin  (0  —  0'0)  +  20/oo  cos  (a  +  0'0), 
and,  the  total  primary  impressed  e.m.f.  is 

£o=\/r—  +azo/icos(0—  0'o)+*oloosin(a+0'o)]>+ 


2azo/oo  .  .  ,  o*zo2/i2  ,  a22o2/oo2     .  2a"zo2/i/oo  . 

Mn(a+g/o)+—        -  +—      —  +  - 


Combining  the  two  components,  the  total  primary  current  is 
0  +  /oo  sin  a)2  +  (a/i  sin  0  +  /oo  cos  a) 


Since  the  tangent  of  the  phase  angle  is  the  ratio  of  vertical 
component  to  horizontal  component,  we  have,  primary  e.m.f. 
phase, 

_         azp/i  sin  (0  —  0'0)  -f  ZO/OQ  cos  («  +  0'o) 
tan  0    —  -j=, 

—  +  020/1  cos  (0  -  0'0)  +  20/oo  sin  (a  -  0'0) 


primary  current  phase, 
„ 

and  lag  of  primary  current  behind  impressed  e.m.f., 


„  _  ali  sin  0  -f  /oo  cos  a 
~  ali  cos  0  +  /oo  sin  a 


ALTERNATING-CURRENT  TRANSFORMER        75 


EXAMPLES 

63.  (1)  In  a  20-kw.  transformer  the  ratio  of  turns  is  20  -f-  1, 
and  100  volts  is  produced  at  the  secondary  terminals  at  full 
load.  What  is  the  primary  current  at  full  load,  and  the  regu- 
lation, that  is,  the  rise  of  secondary  voltage  from  full  load  to  no 
load,  at  constant  primary  voltage,  and  what  is  this  primary 
voltage? 

(a)  at  non-inductive  secondary  load, 

(6)  with  60  degrees  time  lag  in  the  external  secondary  circuit, 

(c)  with  60  degrees  time  lead  in  the  external  secondary  circuit. 

The  exciting  current  is  0.5  amp.,  the  core  loss  600  watts, 
the  primary  resistance  2  ohms,  the  primary  reactance  5  ohms, 
the  secondary  resistance  0.004  ohm,  the  secondary  reactance 
0.01  ohm. 

Exciting  current  and  core  loss  may  be  assumed  as  constant. 

600  watts  at  2000  volts  gives  0.3  amp.  core  loss  current, 
hence  V0.52  —  32  =  0.4  amp.  magnetizing  current. 

We  have  thus 


r0  =  2 

XQ   —   5 


=  0.004 
=  0.01 


/oo  cos  a  =  0.3 
/oo  sin  a  =  0.4 
/oo  =  0.5 


0.05 


1.  Secondary  current  as  horizontal  axis: 


Non-inductive, 
0i  =0 

01  -*+W 

Le*d, 

0l  =   -  60° 

Hor. 

Vert. 

Hor. 

Vert. 

Hor. 

Vert. 

Secondary  current,  /i.  . 
Secondary  terminal 
voltage,  E. 

200 

100 

0.8 
0 
100.8 

0 

% 

0 
0 
+2.0 
+2.0 

200 

50 
0.8 

0 
50.8 

0 

+86.6 

° 

+  2.0 

+88.6 

200 

50 
0.8 
0 
50.8 

0 

-86.6 
0 
+  2.0 
-84.6 

Resistance  voltage,  I\T\. 
Reactance  voltage,  I\xi. 
Secondary  e.m.f  .,  E\... 

Secondary  e.m.f.,  total 
tan  0 

100.80 
+0.0198 

+  1.1° 

102.13 

+  1.745 
+60.2° 

98.68 
-   1.665 
-59.0° 

76       ELEMENTS  OF  ELECTRICAL  ENGINEERING 
2.  Magnetic  flux  as  vertical  axis: 


Non-inductive, 
61  =  0 

Lag, 

0!  =   +  60° 

Lead, 

0i  =-  60° 

Hor. 

Vert. 

Hor. 

Vert. 

Hor. 

Vert. 

Secondary  gen- 
erated    e.m.f., 
E  

-100.80 
-200 

+  10 

0.3 
+  10.3 

20.6 
3.0 

2016 
2039.6 

0 

+  4 

+  0.2 
0.4 
-  0.6 

1.2 
+51.3 

0 

+50.1 

-102.13 
-  99.4 

+     4.97 
0.3 
+     5.27 

10.54 
45.20 

2042.6 
2098.34 

0 

-172.8 

-     8.64 
-     0.4 
-     9.04 

-   18.08 
+  26.35 

0 

+     8.27 

-  98.68 
-103 

+     5.15 
0.3 
+     5.45 

10.90 
-  40.85 

1973.6 
1943.65 

0 
-171.4 

+  8.57 
-  0.4 
+  8.17 

+  16.34 

+27.25 

0 

+43.59 

Secondary   cur- 
rent, 1  1  
Primary       load 
current,     /'  = 
-a/i 

Primary    excit- 
ing current,  70o 
Total     primary 
current,  /o  .  .  .  . 
Primary   resist- 
ance,   voltage, 
/oPo  

Primary    react- 
ance,   voltage, 
Iox0  
E.m.f.  consum- 
ed by  primary 
counter  e.m.f., 
-El 

a 
Total     primary 
impressed  e.m.f., 
E°    

Hence, 


Non-inductive, 

0i  =  0 

Lag, 

0i  =   +  60° 

Lead, 

0!    =     -    60° 

Resultant  Eo                               .... 

2040  1 

2098  3 

1944  2 

Resultant  /o 

10  32 

10  47 

9  82 

Phase  of  E0       

-1.4° 

-  0  2° 

—   1  2° 

Phase  of  /o 

+3.3° 

+59  8° 

—56  3° 

Primary  lag   60 

+4.7° 

+60  0° 

—  55  1° 

W 

Rpsrulation  

1  02005 

1  04915 

0  9721 

2000'  ' 
Drop  of  voltage,  per  cent 

2  005 

4  915 

—  2  79 

Change  of  phase,  do  —  0i  

4.7° 

0 

4.9° 

RECTANGULAR  COORDINATES  77