﻿Device 
  J 
  or 
  evaluating 
  Formidoi 
  and 
  solving 
  Equations. 
  297 
  

  

  deflexion 
  on 
  the 
  galvanometer. 
  In 
  this 
  case 
  let 
  the 
  reading 
  

   on 
  the 
  slide 
  AiBj 
  be 
  a,. 
  If 
  E 
  be 
  the 
  potential 
  at 
  R, 
  and 
  V 
  be 
  

   the 
  potential 
  at 
  P 
  or 
  Q, 
  then, 
  since 
  the 
  currents 
  in 
  PO 
  and 
  QO 
  

   are 
  equal, 
  we 
  have 
  

  

  (E-V)/(R/ai) 
  = 
  (E-V)/(E/a3) 
  + 
  (E-V)/(R/a3), 
  

  

  and 
  thus 
  aj 
  = 
  a^ 
  + 
  a^. 
  

  

  To 
  find 
  ^2— 
  «3, 
  all 
  we 
  have 
  to 
  do 
  is 
  to 
  change 
  the 
  connexion 
  

   of 
  the 
  finger 
  of 
  A3B3 
  from 
  Q 
  to 
  P. 
  Proceeding 
  as 
  ^before, 
  

   ai 
  now 
  gives 
  us 
  the 
  value 
  of 
  a2—a^. 
  

  

  When 
  subtracting 
  approximate 
  values, 
  it 
  has 
  to 
  be 
  re- 
  

   membered 
  that 
  the 
  percentage 
  error 
  in 
  the 
  result 
  may 
  be 
  

   large, 
  especially 
  when 
  a^ 
  — 
  a^ 
  is 
  small 
  compared 
  with 
  a^. 
  This 
  

   difficulty, 
  which 
  arises 
  when 
  subtracting 
  the 
  approximate 
  

   values 
  of 
  nearly 
  equal 
  quantities, 
  affects 
  all 
  approximate 
  

   methods. 
  In 
  our 
  case 
  it 
  can 
  only 
  be 
  overcome 
  by 
  increasing 
  

   the 
  accuracy 
  of 
  the 
  device. 
  

  

  VII. 
  The 
  Use 
  of 
  the 
  Arms 
  of 
  the 
  Bridge. 
  

   Terminals 
  are 
  connected 
  at 
  points 
  Pi, 
  Pg, 
  P3, 
  Q3, 
  Q2> 
  

   and 
  Qi 
  of 
  the 
  bridge 
  wire 
  (fig. 
  6). 
  These 
  points 
  are 
  chosen 
  

   so 
  thai 
  if 
  the 
  resistance 
  PiQi 
  be 
  2r, 
  the 
  resistances 
  of 
  OPi, 
  

   OP2, 
  and 
  OP3 
  are 
  equal 
  to 
  r, 
  r/10, 
  and 
  r/100, 
  and 
  also 
  to 
  the 
  

   resistances 
  of 
  OQj, 
  OQ2, 
  and 
  OQ3 
  respectively. 
  

  

  Fig. 
  6. 
  

  

  V 
  vV\A^— 
  =^^ 
  

  

  Let 
  the 
  magnitudes 
  of 
  the 
  currents 
  in 
  RPi, 
  PPg, 
  RP3, 
  

   RQ3; 
  and 
  RQ2 
  be 
  C„ 
  C2, 
  C3, 
  C^', 
  and 
  C2' 
  respectively. 
  

  

  Let 
  us 
  also 
  suppose 
  that 
  the 
  currents 
  in 
  the 
  two 
  branches 
  

   joining 
  R 
  and 
  Qi 
  are 
  C 
  and 
  C/ 
  respectively. 
  

  

  The 
  resistances 
  in 
  these 
  circuits 
  are 
  the 
  slide 
  resistances 
  

   described 
  above. 
  When 
  there 
  is 
  no 
  deflexion 
  on 
  the 
  

   galvanometer 
  joining 
  P^ 
  and 
  Qi, 
  the 
  potentials 
  of 
  these 
  two 
  

  

  