PIERCE. THE COOPER HEWITT MERCURY INTERRUPTER. 399 



TABLE II. 

 23 Plates in Glass Condenser. 



Coil I and coil II are accurately wound, and their inductances, L x and L 2 

 have been calculated from their geometrical dimensions. 



L x = 1.06 X 10 5 magnetic units, 

 Z 2 = 14.1 x 10 5 " " . 



Let L = the unknown inductance of the leads, then by Thomson's 

 formula, we have from Table II, 



2tt V(A + L)G= 2.295 x 10" 5 

 2tt\/LC = -585 x 10- 5 



2ttV(L 2 + L) G= 8.03 x 10- 5 



(1) 

 (2) 

 (3) 



Eliminating C from (1) and (2), we have L = .073 X 10~ 5 , which, 

 substituted in (2), gives 



O = .1175 X 10~ 15 magnetic units, 

 = 1.05 X 10 5 cm. 



Likewise, from equation (3) 



C= 1.04 X 10 5 cm. 



