﻿with 
  Alternate 
  Current 
  Measuring 
  Instruments. 
  141 
  

  

  which 
  is 
  impossible. 
  If 
  the 
  shunt 
  is 
  non-inductive 
  T 
  2 
  = 
  

  

  M— 
  1 
  

  

  and 
  tan 
  i/r 
  = 
  — 
  =-=— 
  T 
  lP 
  ; 
  while 
  if 
  the 
  instrument 
  is 
  non- 
  

   inductive 
  and 
  the 
  shunt 
  inductive, 
  

  

  * 
  r 
  M 
  ~ 
  1 
  T 
  

  

  tm 
  + 
  == 
  -M+T 
  f 
  V 
  Ta 
  ^ 
  

  

  Again, 
  if 
  both 
  T 
  : 
  p 
  and 
  T 
  p 
  are 
  small 
  compared 
  with 
  unitv, 
  

  

  M 
  — 
  1 
  

   the 
  expression 
  for 
  tan-^r 
  reduces 
  to 
  M 
  (T 
  1 
  — 
  T 
  2 
  )p. 
  This 
  

  

  of 
  course 
  means 
  that 
  the 
  tangent 
  of 
  the 
  phase-displacement 
  

  

  M 
  — 
  1 
  

  

  between 
  the 
  instrument 
  and 
  mains 
  is 
  the 
  fraction 
  — 
  ^- 
  of 
  

  

  that 
  between 
  the 
  instrument 
  and 
  shunt, 
  as 
  is 
  obvious 
  

   geometrically. 
  Finally, 
  if 
  T 
  1 
  and 
  T 
  3 
  are 
  nearly 
  equal, 
  

  

  M 
  — 
  1 
  

  

  tan 
  * 
  = 
  <r 
  = 
  M(1 
  + 
  T 
  y 
  ) 
  (T 
  1 
  -T 
  2 
  )p 
  . 
  . 
  . 
  (9) 
  

  

  where 
  T 
  is 
  the 
  mean 
  time 
  constant 
  as 
  before. 
  When 
  T 
  is 
  

   small 
  this 
  reduces 
  to 
  the 
  last 
  expression. 
  

  

  As 
  an 
  illustration 
  of 
  the 
  error 
  produced 
  by 
  shunting 
  a 
  

   wattmeter, 
  we 
  may 
  assume 
  a 
  case 
  where 
  the 
  time 
  constant 
  of 
  

   the 
  main 
  circuit 
  has 
  a 
  value 
  of 
  "0025 
  as 
  in 
  the 
  Kelvin 
  balance 
  

   above 
  cited, 
  and 
  suppose 
  that 
  it 
  is 
  shunted 
  with 
  a 
  non- 
  

   inductive 
  shunt 
  of 
  a 
  nominal 
  multiplying 
  power 
  of 
  10. 
  We 
  

   then 
  have 
  M 
  = 
  10, 
  T 
  2 
  = 
  0, 
  and 
  T 
  lP 
  at 
  50— 
  

  

  = 
  314 
  x 
  -0025 
  = 
  -785. 
  

   Hence 
  

  

  M 
  

  

  / 
  /M— 
  1\ 
  2 
  

  

  ' 
  = 
  M^/ 
  1 
  + 
  y-^-) 
  %y 
  = 
  10 
  VI 
  + 
  -81 
  x 
  -785* 
  = 
  12-25. 
  

  

  The 
  phase-displacement 
  is 
  

  

  M 
  — 
  1 
  9 
  

  

  ten* 
  i 
  = 
  l 
  - 
  w 
  - 
  T 
  lP 
  = 
  ^ 
  x 
  -785 
  = 
  -705 
  ; 
  

  

  so 
  that 
  the 
  current 
  in 
  the 
  wattmeter 
  lags 
  about 
  35° 
  behind 
  

   that 
  in 
  the 
  main 
  circuit. 
  Hence 
  the 
  instrument 
  would 
  read 
  

   22*5 
  per 
  cent, 
  too 
  low 
  owing 
  to 
  the 
  error 
  in 
  the 
  ratio, 
  while 
  

   the 
  phase-displacement 
  would 
  cause 
  the 
  power 
  factor 
  to 
  be 
  

   apparently 
  about 
  *5 
  when 
  actually 
  zero. 
  

  

  This 
  strikingly 
  illustrates 
  the 
  enormous 
  errors 
  which 
  may 
  

   be 
  produced 
  by 
  shunts 
  in 
  wattmeters. 
  For 
  any 
  given 
  small 
  

   displacement 
  ^r 
  we 
  have 
  from 
  (9) 
  

  

  T 
  x 
  -T 
  2 
  AT 
  M 
  f 
  l-TY 
  \ 
  , 
  nrv 
  

  

  