RICE: ALTERNATING CURRENTS IN WHEATSTONE’S BRIDGE. 33 
Suppose first that the arms of the bridge con- Pe 9. 
tain only resistances. Then 
2,1, 2313 
gt 5 Z,==!, 
aud equation [3] becomes rjr,—1,f,, [4] 
the same as for continuous currents. 
Suppose next that each arm of the bridge contains a resistance 
and an inductance, then z,=r,—jx,, Z,=f,—jxX,, Z,=1!,—jX3) 
Z,—=1,—jx,, and equation [3] becomes 
,t3—-X.X,—](f,X» hex, = tit —-*%, 4, J etx) [5] 
Transposing, and equating the reals to zero and the imaginaries 
to zero, gives the two equations of condition 
Pols —V¥,0,4—=XgX3-—_X{Xyq [6] 
Ee, hk i oe exe [7] 
Tier Wm I Wee 3%eTheX3 
Hence, in general, if any six of the constants of the bridge be 
given, the other two may be obtained from equations [6] and [7]. 
For example, if the resistances of three arms be given, the neces- 
sary resistance and inductance of the fourth arm my be calculated. 
Or, if the four resistances and two of the inductances are known, 
the other two inductances may be calculated. 
It is evident from equation [6] that the bridge need not be bal- 
anced for continuous currents in order to be balanced for alter- 
nating currents. But if the bridge is first balanced for continuous 
currents, equations [6] and [7] reduce to 
X1X4—X_Xe [8] 
ba Ces eee set ee [6] 
I, ee 1G ones 
In this case also, if two of the inductances, say x, and Kigg ake 
given, one and only one pair of values of x, and x, can be obtained 
2 ? Lote r, , 
that will balance the bridge. But if —'—-—!, equation [g] reduces 
Xo Ty 
xs ick Ve : d 3 ‘ 
(@ —=== 4 == a which is the same as [8]; consequently in this 
X4 4 2 
Palit: b P 
case the ratio a only can be determined from [8] and [9], and 
4 
