8.  POWER  IN  ALTERNATING-CURRENT  CIRCUITS 


of  effective  value  I  =  —7=-,  in  a  circuit  of  resistance  r  and  reac- 

V2 


39.  The  power  consumed  by  alternating  current  i  =  I0  sin  0, 
effective  value  I 

tance  x  =  2  nfL,  is 

p  =  ei, 

where  e  =  z!Q  sin  (0  +  00)  is  the  impressed  e.m.f.,  consisting  of 
the  components 

ei  =  r/0  sin  0,  the  e.m.f.  consumed  by  resistance 
and         62  =  x!Q  cos  0,  the  e.m.f.  consumed  by  reactance. 

z  =  \/r2  +  x2  is  the  impedance  and  tan  00  =  —  the  phase  angle 
of  the  circuit;  thus  the  power  is 

p  =  z/o2  sin  0  sin  (0  +  00) 

=  ^-  (€OS  00   -   COS  (20+   00)) 

=  zP  (cos  00  -  cos  (20  +  00)). 
Since  the  average  cos  (20  +  00)  =  zero,  the  average  power  is 

P  =  zP  cos  00 
=  rP  =  EJ-, 

that  is,  the  power  in  the  circuit  is  that  consumed  by  the  resistance, 
and  independent  of  the  reactance. 

Reactance  or  self-inductance  consumes  no  power,  and  the 
e.m.f.  of  self-inductance  is  a  wattless  or  reactive  e.m.f.,  while  the 
e.m.f.  of  resistance  is  a  power  or  active  e.m.f. 

The  wattless  e.m.f.  is  in  quadrature,  the  power  e.m.f.  in  phase 
with  the  current. 

In  general,  if  0  =  angle  of  time-phase  displacement  between 
the  resultant  e.m.f.  and  the  resultant  current  of  the  circuit, 
/  =  current,  E  =  impressed  e.m.f.,  consisting  of  two  com- 
ponents, one,  EI  =  E  cos  0,  in  phase  with  the  current,  the  other, 
1£2  =  E  sin  0,  in  quadrature  with  the  current,  the  power  in  the 
circuit  is  IEi  =  IE  cos  0,  and  the  e.m.f.  in  phase  with  the  current 
Ei  =  E  cos  0  is  a  power  e.m.f.,  the  e.m.f.  in  quadrature  with 
the  current  E2  =  E  sin  0  a  wattless  or  reactive  e.m.f. 


40       ELEMENTS  OF  ELECTRICAL  ENGINEERING 

40.  Thus  we  have  to  distinguish  power  e.m.f.  and  wattless  or 
reactive  e.m.f.,  or  power  component  of  e.m.f.,  in  phase  with  the 
current  and  wattless  or  reactive  component  of  e.m.f.,  in  quadra- 
ture with  the  current. 

Any  e.m.f.  can  be  considered  as  consisting  of  two  components, 
one,  the  power  component,  e\,  in  phase  with  the  current,  and 
the  other,  the  reactive  component,  ez,  in  quadrature  with  the 
current.  The  sum  of  instantaneous  values  of  the  two  compo- 
nents is  the  total  e.m.f. 

e  =  ei  +  e* 
If  E}  EI,  Ez  are  the  respective  effective  values,  we  have 


E    =       Ei*  +  E22,  since 
EI  =  E  cos  &, 
E2  =  E  sin  6, 

where  0  =  phase  angle  between  current  and  e.m.f. 

Analogously,  a  current  I  due  to  an  impressed  e.m.f.  E  with 
a  time-phase  angle  0  can  be  considered  as  consisting  of  two 
component  currents, 

1 1  =  I  cos  8,  the  active  or  power  component  of  the  current,  and 
J2  =  /  sin  0,  the  wattless  or  reactive  component  of  the  current. 

The  sum  of  instantaneous  values  of  the  power  and  reactive 
components  of  the  current  equals  the  instantaneous  value  of  the 
total  current, 

ii  +  iz  =  i, 

while  their  effective  values  have  the  relation 

i  =  V77+772. 

Thus  an  alternating  current  can  be  resolved  in  two  com- 
ponents, the  power  component,  in  phase  with  the  e.m.f.,  and  the 
wattless  or  reactive  component,  in  quadrature  with  the  e.m.f. 

An  alternating  e.m.f.  can  be  resolved  in  two  components: 
the  power  component,  in  phase  with  the  current,  and  the  watt- 
less or  reactive  component,  in  quadrature  with  the  current. 

The  power  in  the  circuit  is  the  current  times  the  e.m.f.  times 
the  cosine  of  the  time-phase  angle,  or  is  the  power  component 
of  the  current  times  the  total  e.m.f.,  or  the  power  component  of 
the  e.m.f.  times  the  total  current. 


VECTOR  DIAGRAMS  41 

EXAMPLES 

41.  (1)  What  is  the  power  received  over  the  transmission  line 
in  Section  7,  Example  2,  the  power  lost  in  the  line,  the  power 
put  into  the  line,  and  the  efficiency  of  transmission  with  non- 
inductive  load,  with  45-time-degree  lagging  load  and  45-degree 
leading  load? 

The  power  received  per  line  with  non-inductive  load  is  P  =  El 
=  3170  X  44  =  139  kw. 

With  a  load  of  45  degrees  phase  displacement,  P  =  El  cos 
45°  =  98  kw. 

The  power  lost  per  line  PI  =  PR  =  442  X  7.6  =  14.7  kw. 

Thus  the  input  into  the  line  P0  =  P  +  PI  =  151.7  kw.  at 
non-inductive  load, 

and       =  111.7  kw.  at  load  of  45  degrees  phase  displacement. 
The  efficiency  with  non-inductive  load  is 

P  14  7 

Po  =  l  -  15T7  =    )0-3  p 

and  with  a  load  of  45  degrees  phase  displacement  is 

P  14.7 

^-  =  1  —  111  -,  =  86.8  per  cent. 

L  Q  111./ 

The  total  output  is  3  P  =  411  kw.  and  291  kw.,  respectively. 
The  total  input  3  P0  =  451.1  kw.  and  335.1  kw.,  respectively.