﻿1026 Prof. V. Karapetoff on General Equations 



Assuming the currents and the impedances to be expressed 

 as complex quantities, we have the following three funda- 

 mental equations of the voltage drop in the parts of the 

 bridge: — 



I Z 1 -(I„ + I,)X„ 1 =IZ« (A) 



IZ 6 + I„Z 4 =I(Z 8 -(I« + WX,„ 3 . . . (B) 

 (I„-I)Z 2 =I(Z„ + Z 6 ) (C) 



Eliminating the currents from these equations, we obtain 

 the following general relationship among the impedances of a 

 balanced bridge : — 



(Z 3 — X,„ 3 ) [(Z 2 4- X ml ) 4- Y X ml (Z 2 + Zfe)] = (Z, — X ml ) 



[(Z 4 + X w3 )(l + Y a (Z 2 + Z 6 ))+Y a Z 2 Z 6 ] . . (D) 



In this equation the admittance Y a is used in place of the 

 impedance Z a ; the relationship between the two is Z a Y a =l. 

 While in any special case eq. (D) may be applied directly, 

 there are some typical special cases for which it is more 

 convenient to write simplified forms of eq. (D). Eight such 

 cases are considered below. 



(1) No mutual inductances and a single-path branch 2. 

 This means t! at X TOl = X„ l3 = 0, and Y a = ; we then have a 

 simple bridge consisting of four impedances. Eq. (D) 

 becomes 



z 3 z 2 =z,z 4 (1) 



(2) No mutual inductance and a single-shunt branch 2. 

 Here again X wl = X m3 = and Z& = 0. The points A 2 and 

 A 2 ' coincide, and Z a is in parallel vvith Z 2 . Eq. (D) becomes 



ZsZ^Z^l + YA) (2) 



(3) No mutual inductance and a double shunt in branch 2, 

 as shown in tig. 1. In this case the only simplification is 

 that X TOl = X 7Jl3 = 0, and eq. (D) gives 



Z 5 Z 2 = Z 1 [Z 4 (l + Y a (Z 2 + Z 6 ))+Y a Z 2 Z 6 ]. . . (3) 



(4) Mutual inductance in branch 3 only and no shunt in 

 branch 2. We have X ml = and Y a .= ; eq. (D) becomes 



(Z3-X wl3 )Z 2 = Z 1 (Z 4 + X m3 ) .... (4) 



(5) Mutual inductance in branch 3 only and a single shunt 

 in branch 2. In this case X nil = Z 6 = and Z« is in parallel 

 with Z 2 . Eq. (D) gives 



(Z 3 -X Hl3 )Z 2 = Z 1 [(Z 4 + X OT3 )(l + Y a Z 2 )] . . (5) 



