The Wheatstone Bridge is a very useful circuit. When the bridge is fully balanced, the right resistor is the same, the left resistor（R1=R3,and R2=Rx），The voltage across the bridge is zero. However, due to a small change in resistance, the bridge becomes unbalanced and a voltage difference occurs. Wheatstone bridge applications, such as strain gauges, pressure gauges, sensors and other equipment.

A differential amplifier can be used to extract the common mode signal while rejecting all common mode noise. As a very small signal change can be extracted from the bridge because common mode noise is easily rejected.

 The bridge voltage is calculated as follows：

V_B=  V_in*[Rx/(R3+Rx)- R2/(R1+R2)]

If R3=R1, 和 Rx= R2+delta, then

V_B=  V_in *[ (R2+delta)/(R1+R2+delta)-R2/(R1+R2)]

Now, if we assume delta is better than R1 + R2 is small, then

V_B = ~ V _in*[delta/(R1+R2)]

Therefore, we can see that the bridge voltage is about proportional errordelta，Divided by the sum of the resistors.

Due to the bridge voltage, we can calculate an unknown resistance value.

(R1+R2)*(R3+Rx)*V_B/V_in= Rx*(R1+R2)+ R2*(R3+Rx)

Rx*(R1+R2)*V_B/ V_in +  R3* (R1+R2)V_B/V_in= Rx*R1+Rx*R2 - R2*R3- Rx*R2

Rx*R1 - Rx*(R1+R2)*V_B/ V_in  = R2*R3 + R3* (R1+R2)V_B/V_in

Rx = (R2*R3 + R3* (R1+R2)V_B/V_in )/ (R1- (R1+R2)*V_B/ V_in)