PROBLEM OF WHEATSTONE 7 S BRIDGE. 347 



Breaking the circuit produces the same effect, but in the op- 

 posite direction. In both cases the needle will be stationary if 

 we have 



(25) ^.k'^.Iv. 



^ 5) b V a a' 



Induced currents are not, however, generally of such short 

 duration that the needle has not time to move a little in the 

 direction of the first effect. If we want to balance the bridge 

 so that the current in the galvanometer is always zero, the nume- 

 rator of the equation must be always zero that is to say, that 

 we have identically 



di 



From this follows first = , which gives, for any given period, 

 at 



a = p and a' = /3', 

 or 



da d$ da! dp 



- and -~~' 



Equation (26) then becomes 



- __ . 



a dt b a dt 



As the currents a and a' are independent, it follows, from the 

 ratio ab' = bd) that 



( 27 ) ** = and i^. 



a b a b 



Besides the ordinary condition of conjugate diagonals, the 

 complete equilibrium of the bridge requires then, both for variable 

 and for constant currents, that the coefficients of self-induction 

 of the four branches of the bridge are respectively proportional to 

 their resistances. 



