Gauge resistance change and quarter-bridge output

Given: $R_g = 120\ \Omega$, $G = 2.1$, $\varepsilon = 1500\ \mu\varepsilon = 1.5\times10^{-3}$.

Change in resistance

$$\frac{\Delta R}{R} = G\varepsilon = 2.1 \times 1.5\times10^{-3} = 3.15\times10^{-3} = 0.315\%$$ $$\Delta R = 120 \times 3.15\times10^{-3} = 0.378\ \Omega$$

Quarter-bridge output

All arms are 120 $\Omega$ and $V_{ex} = 5$ V. With one active arm of resistance $R + \Delta R$:

$$V_o = V_{ex}\left[\frac{R + \Delta R}{2R + \Delta R} - \frac{1}{2}\right] = V_{ex}\frac{\Delta R}{2(2R + \Delta R)}$$ $$V_o = 5\times\frac{0.378}{2(240.378)} = 3.931\times10^{-3}\ \text{V}$$

The usual small-change approximation gives

$$V_o \approx \frac{V_{ex}}{4}\frac{\Delta R}{R} = \frac{5}{4}(3.15\times10^{-3}) = 3.94\ \text{mV}$$

          +---- R1 = 120 ----B---- R_g = 120 + dR ----+
          |                  |                        |
  5 V (+) A                 V_o (between B and D)     C (-)
          |                  |                        |
          +---- R3 = 120 ----D------ R2 = 120 --------+
   (5 V excitation applied across A and C)

Answer: $\Delta R = 0.378\ \Omega$ ($0.315\%$); bridge output $\approx 3.93$ mV (3.94 mV by the approximate formula).