﻿of a Balanced Alternating- Current Bridge. 



L027 



(6) Mutual inductance in branch 1 only and no shun/ in 

 branch 2. We have X m3 =0 and Y rt = ; eq. (D) becomes 



Z;;(Z 2 + X wl ) ■— (Zi— X Iwl )Z 4 



(6) 



(7) Mutual inductance in branch 1 only and a single slum/ 

 in branch 2. In this case X TO8 = Z& = and Z a is in parallel 

 with Z 2 . Eq. (D) gives 



Z 3 [(Z 2 + X ml ) + Y«X ml Z 2 ] = (Z 1 -X ml )(l + Y a Z 2 )Z 4 . 



(7) 



(8) Mutual inductance in both branches, but the branch 2 is 

 not shunted. In this case Y tt = 0, and eq. (D) becomes 



(Z 3 -X„, 3 )(Z 2 + X,„ 1 ) = (Z 1 -X mi )(Z 4 + X„ 1 3). . (8) 



In the following table the special applications of the bridge 

 are those discussed by Dr. Poole in the article mentioned 

 above, and the references are to the figures in his article. All 

 these applications are covered by the foregoing eight special 

 cases of formula D, and the case number is stated for each 

 application. In some applications the four resistances of the 

 bridge have first to be balanced with direct current, in other 

 cases it is not feasible. This is indicated by " yes " or " no " 

 in one of the columns. 



It will now be shown how readily the familiar formulae for 

 the measurements shown in the table can be derived from 

 the formulae (1) to (8), all of which are specific cases of eq. 

 (D). The item numbers below refer to the items in the 

 table, and in each case an equation is used as indicated in 

 the table. All the results check with those given by 

 Dr. Poole. 



Item Jfo. 1 : 



(r 3 +JG>L fi )r 2 =r 1 (r i +ja>L±). 



Separating the real and the imaginary parts, gives 



?- 3 r 2 = ?V'4 ....... (9) 



UH^rJn (10) 



These equations combined give 



L 8 /L 4 =r 1 /r- 2 =r 3 /i 



(11) 



Eq. (9) shows that the bridge must be balanced on direct 

 current as well as on alternating current. 



