Section 4
4-6
(1) and nanovoltmeter offset (2)
Ic + (Ix) x (Nx) = (Is) x (N)
Is = (Ix) x (Nx / N)+Ic / N
nV = (Is) x (Rs) - (Ix) x (Rx) + Vos
Then we can state the generalized formula:
nV = Rs Ix.
Nx
N
+
Ic
N
-Ix.Rx+Vos
(4)
If at the approximate null of the bridge Nx = Tn and we switch ±(t) turns to winding Nx,
then assuming ampere-turn balance is maintained, the bridge voltage imbalance (nV) for
current (+) becomes nV(+t)(+) and nV(-t)(+). Where:
nV(-t)(+)= Rs.
Tn-t
N
-Ix.Rx+Vos(-t)
(5)
and:
nV(+t)(+)= Rs.
Tn+ t
N
-Ix.Rx+Vos(+t)
(6)
Subtracting (5) from (6):
nV(+t)(+)-nV(-t)(+)= Rs.
2.t
N
.Ix
(7)
If we now reverse the direction of the test current to Ixx (-), where Ixx is not necessarily
equal to Ix (+) then:
nV(-t)(-)= Rs.
-Tn+ t
N
+Ixx.Rx+Vos(-t)
(8)
and:
nV(+t)(-)= Rs.
-Tn - t
N
+Ixx.Rx+Vos(+t)
(9)
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