﻿140 
  Dr. 
  Drysdale 
  on 
  the 
  Use 
  of 
  Shunts 
  and 
  Transformers 
  

  

  In 
  the 
  case 
  of 
  the 
  Kelvin 
  centiampere 
  balance 
  before 
  

   mentioned, 
  T 
  = 
  '00258 
  at 
  a 
  frequency 
  of 
  50, 
  therefore 
  

   Tp 
  = 
  314 
  x 
  -00258 
  = 
  -810 
  and 
  T 
  2 
  p 
  2 
  = 
  '655. 
  Consequently 
  

   the 
  time 
  constant 
  or 
  the 
  self-induction 
  of 
  the 
  shunt 
  must 
  be 
  

  

  adjusted 
  to 
  _ 
  = 
  1*53 
  per 
  cent, 
  for 
  an 
  accuracy 
  of 
  1 
  per 
  

  

  cent, 
  in 
  the 
  multiplying 
  power. 
  The 
  simplest 
  method 
  of 
  

   adjusting 
  or 
  checking 
  the 
  shunt 
  is 
  of 
  course 
  to 
  test 
  the 
  ratio 
  

   with 
  both 
  D.C. 
  and 
  A.C. 
  or 
  with 
  A.O. 
  of 
  two 
  different 
  

   frequencies. 
  

  

  Phase 
  Displacement. 
  — 
  Returning 
  to 
  formula 
  (1) 
  and 
  

   rationalizing 
  the 
  denominator, 
  we 
  get 
  

  

  V 
  

  

  J-l 
  J- 
  2 
  

  

  v 
  

  

  = 
  T2T2 
  { 
  ( 
  r 
  i 
  + 
  r 
  2) 
  (*Vs— 
  #1%) 
  + 
  Oi 
  + 
  # 
  2 
  )Oi#2 
  + 
  r 
  8«l) 
  

   ll 
  -L2 
  

  

  +j[ 
  (rj 
  + 
  r 
  2 
  )(i\x 
  2 
  + 
  r 
  2 
  #i) 
  — 
  (#1 
  + 
  #2) 
  (*Vs 
  — 
  a?^)] 
  } 
  , 
  

  

  and 
  consequently 
  

  

  tan 
  ^ 
  = 
  Q'i 
  + 
  r 
  2 
  ) 
  (r^ 
  2 
  + 
  r 
  2 
  Xi) 
  - 
  ( 
  A'i 
  + 
  ii? 
  2 
  ) 
  Qv* 
  2 
  ~ 
  ^1^2) 
  

  

  ( 
  **] 
  + 
  ^2) 
  (*v 
  8 
  — 
  #] 
  #2) 
  + 
  ( 
  «i 
  + 
  #2) 
  (n^2 
  + 
  Vi) 
  

  

  where 
  (/> 
  is 
  the 
  angle 
  of 
  lag 
  of 
  the 
  main 
  current 
  behind 
  V. 
  

   Inserting 
  the 
  values 
  of 
  M, 
  T 
  1? 
  and 
  T 
  2 
  as 
  before, 
  we 
  get 
  

  

  tan 
  A 
  _ 
  T, 
  + 
  (Jt-l)T 
  t 
  + 
  {(M-l)Ti+T,}T 
  1 
  1Vp 
  » 
  

   tan 
  * 
  — 
  M+{(M-1)T^+T 
  3 
  ^ 
  p 
  ' 
  

  

  Similarly 
  tanc/) 
  1 
  = 
  T 
  1 
  p 
  where 
  fa 
  is 
  the 
  lag 
  in 
  the 
  instrument. 
  

  

  Hence 
  

  

  * 
  /jl 
  ^ 
  tan 
  c^!— 
  tan 
  ^ 
  

   tan 
  (6 
  1 
  -(b) 
  = 
  — 
  / 
  x 
  , 
  ^ 
  — 
  *-z 
  

   ir 
  r/ 
  1 
  + 
  tan 
  </>! 
  tunc/) 
  

  

  is 
  the 
  tangent 
  of 
  the 
  angle 
  of 
  lag 
  of 
  the 
  current 
  in 
  the 
  instru- 
  

   ment 
  behind 
  the 
  main 
  current. 
  Putting 
  in 
  the 
  values 
  of 
  

   tan 
  (/>! 
  and 
  tan 
  6 
  and 
  simplifying, 
  we 
  have 
  as 
  the 
  result 
  

  

  Uiwf 
  = 
  tan(0! 
  — 
  fa) 
  

  

  1 
  -f 
  T 
  2 
  p- 
  

  

  M 
  - 
  1 
  <T 
  * 
  * 
  J 
  l 
  (l+T 
  1 
  y)^(M-l)T 
  1 
  +T.}T,^ 
  \ 
  S) 
  

  

  = 
  ~M~ 
  (il 
  - 
  }/ 
  1 
  M.+ 
  {Miy+C* 
  - 
  l/rxTj+T,"} 
  *-+ 
  { 
  (M 
  - 
  l^+T.tfiyT,/*' 
  P 
  ' 
  

  

  The 
  phase-displacemeni 
  is 
  consequently 
  zero 
  Eor 
  either 
  

  

  M 
  = 
  l 
  or 
  T 
  1 
  = 
  T.., 
  as 
  is 
  obvious, 
  and 
  also 
  for 
  l 
  + 
  T 
  l 
  2 
  j>- 
  = 
  0, 
  

  

  