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First an examination of the resistancs show
that the bridge is nulled when I = 0 and Vx = 0
from the equation (Vx)+((R3*V)/(R1+R3))-(V*(R4+R5)/(R2+R4+R5))-(I*R5)=0(R3*V)/(R1+R3))=1.667 V
(V*(R4+R5)/(R2+R4+R5))=1.667 V
Thus,we can use equation Vx-IR5=0:
12mV-10I=0
Thus I = 1.2 mA
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