ON OHM S LAW. 



43 



Formula of Reduction. 



Let the right-hand middle coil (No. 1) be taken to be 30 ohms, the bridge- 

 wire being -075 of the same units. Let t denote the resistance of this coil, 

 the unit being the resistance of a tenth of a millimetre of the bridge-wire, 

 therefore 



30x10000 



r=- 



•075 



: = 4000000. 



Let the resistances of 1, 2, 3, 4, 5 of the quintuple coil, measured in the 

 same units, be r-f-a, r + /3, r + y, r + o, 7--|-e. 



Hence, comparing middle coil 1 with 2, 1 being on the right, 



r + g _ r + 5 4-D + a? ,.. 



r + fl 7 + rOOOO + a-a?' ^ '' 



where r + D = resistance of middle coil 2, x the bridge-reading, a and h the 

 resistances of the connexions at its two ends. This gives 



«-/3={D-a~6 + 2(.r-5000)}{l-^'^^^^'}, • • (2) 



all other terms being negligible. 



Now the greatest possible value of 10000 — x is 6000, since the readings 

 never went below 4000, and D -|- 2ix — 5000) was never greater than 400. 



Hence the terra involving 



1000- 



--- is less than 



400x6000 



To' 



4000000 



and is therefore negligible, since we do not read beyond tenths of a milli- 

 metre. Hence we may use the formula 



o-/3 = D-^?^-|-2(^'-5000) (3) 



Similarly, in comparing one coil against four, we get the formula 



«-?(|3 + r+S+e) = D-^^-l-2(^-5000). . . (4) 

 To find a — h, the "bridge correction," a reading is taken with tlie coils 



Fig. 4. 



arranged as usual either for a single experiment or for a multiple-arc ex- 

 periment : let this reading be x. Then the connexions are crossed, as in the 



