﻿of a Balanced Alternating-Current Bridge. 1029 

 Item No. 2 (see Note at the end) : 



(r 3 + ;'a)L 3 — ;o)M 3 ) j(o^/L l = (?•, + /a>L l ~/&)M 1 ) < y'toM 3 , 



from which 



L 3 M 1 =L 1 M 3 (12) 



and ^M^^Mg (13) 



These equations, combined, give 



M 1 /M 3 =L 1 /L 3 =r 1 / n (14) 



The bridge cannot be balanced on direct current, but the 

 ratio of the resistances must be equal to that of the self- 

 inductances, before two mutual inductances can be compared. 



Item No. 3 : 



(r 3 + j&)L 3 ->M 3 ) r 2 = i\{i\ +;&)L 4 +;©M 3 ) . 



Consequently r 3 r 2 = T\r A (15) 



r 2 (L 3 -M 3 ) = n (L 4 + M 3 ) . , . . (16) 

 so that 



n/'- 2 = »-3K=(L 3 -M 3 )/L 4 + M 3 ). • • (17) 

 Item No. 4 : 



->r 2 /ft)G 3 = '—jriltoGte 



?<A=nC 3 (18) 



Item No. 5 : 



(r 3 +j(oJj 3 ) r 2 = ?v* 4 (l +jr 2 coCa), 

 from which 



r 3 f 2 =*W . (19) 



L 3 = r in C G (20) 



Item No. 6. 



r 2( r 3+> L a) =n[>4(-l +>Ca(r 2 -[-?' 6 ) +y©0«r s r6]. 

 Equating the real and the imaginary parts, we get 



?V'3 = V 4 ■ (21) 



r 2 L 3 = nG a [n(r 2 + n) + r 2 r 5 ] .... (22) 



or, combining the two, 



L 3 =C a [?v', + nK + r 3 )] (23) 



"When n = 0, this expression becomes identical with eq. (20). 

 Item No. 7 : 



(r 3 - //a)0 3 ) ywMi = (r, + t /ft)L! ->M,)?V 



