Question

In the balanced condition, the values of the resistances of the four arms of a Wheatstone bridge are shown in the figure below. The resistance $R_3$ has temperature coefficient ${0.0004}^{\circ }{\text{C}}^{-1}$. If the temperature of $R_3$ is increased by ${100}^{\circ }{\text{C}}^{-1}$, the voltage developed between $S$ and $T$ will be ______ $\text{volt}$.

Answer 1

After the temperature rise the new resistance is,
$R_3^{\prime }=R_3(1+\alpha \Delta T)$
$R_3^{\prime }=300(1+0.004\times 100$)
$R_3^{\prime }=312\Omega$
Now redrawing the circuit,
The current $I_1$ goes to the series connection of $60\Omega$ and $312\Omega$
$I_1=\frac{V}{R_{eq}}=\frac{50}{372}$
and $I_2=\frac{50}{600}$
Writing the equation of potential along path $S\to T$,
$V_S+60I_1-100I_2=V_T$
$V_S-V_T=100\times I_2–60\times I_1$
$\text{Or,}{V}_S-V_T=\frac{50}{6}-\frac{3000}{372}$
$\therefore V_S-V_T=0.268\simeq 0.270\text{V}$