Question

Four strain gauges $R_A, R_B, R_C, R_D$ (nominal $R$) are in a bridge. Under force: $R_A$ and $R_D$ increase by $\Delta R$, $R_B$ and $R_C$ decrease by $\Delta R$. A potentiometer of total resistance $R_v$ is connected as shown to rebalance the bridge. If $R=100~\Omega$ and $\Delta R=1~\Omega$, the minimum $R_v$ required to balance the bridge is _____ $\Omega$ (rounded off to two decimal places).

Hint:

In a strain-gauge bridge with opposite changes ($\pm\Delta R$), rebalancing by adding a small series resistance to the {decreasing} arm restores the ratio $R_A/R_B=R_C/R_D$; a handy result is $R_v=\dfrac{4R\Delta R}{R-\Delta R}$.

Answer 1

Correct Answer: 4

For balance in a Wheatstone bridge, the ratio condition must hold: \[ \frac{R_A}{R_B^{(\text{eff})}}=\frac{R_C}{R_D}. \] With the compensating potentiometer effectively {in series} with the $R_B$ arm, \[ R_A=R+\Delta R, R_C=R-\Delta R, R_D=R+\Delta R, R_B^{(\text{eff})}=R_B+R_v=(R-\Delta R)+R_v. \] Hence \[ \frac{R+\Delta R}{(R-\Delta R)+R_v}=\frac{R-\Delta R}{R+\Delta R} \;\Rightarrow\; (R-\Delta R)+R_v=\frac{(R+\Delta R)^2}{R-\Delta R}. \] Therefore \[ R_v=\frac{(R+\Delta R)^2-(R-\Delta R)^2}{R-\Delta R} =\frac{4R\Delta R}{R-\Delta R}. \] Substitute $R=100~\Omega,\ \Delta R=1~\Omega$: \[ R_v=\frac{4(100)(1)}{100-1}=\frac{400}{99}=4.04~\Omega. \]