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 $$^\circ C^{−1}$$. If the temperature of $$𝑅_3$$ is increased by 100 $$^\circ C$$, the voltage developed between 𝑆 and 𝑇 will be __________ volt.

Initially the bridge is balanced, that is

$$\frac{R_1}{R_2}=\frac{R_3}{R_4}\qquad -(1)$$

From the resistance values printed in the figure (all four arms carry the same resistance) we have

$$R_1 = R_2 = R_3 = R_4 = R$$

Let the d.c. source connected across the points $$P$$ and $$Q$$ have an emf $$E$$. While the bridge is balanced the potentials of the junctions $$S$$ and $$T$$ are equal, so no current flows through the galvanometer.

The temperature of the resistor $$R_3$$ is raised by $$\Delta T = 100^{\circ}\!C$$. Its temperature coefficient is $$\alpha = 0.0004\ ^{\circ}C^{-1}$$, therefore its resistance becomes

$$R_3' = R\bigl(1+\alpha \Delta T\bigr) = R\bigl(1+0.0004\times 100\bigr) = R\,(1+0.04) = 1.04\,R$$

The other three resistances remain equal to $$R$$. After the change, the currents in the two arms are different, so the potentials of $$S$$ and $$T$$ are no longer identical.

Potential of junction $$S$$
The two resistors $$R_1$$ and $$R_3'$$ are in series in the left branch. With point $$Q$$ taken as reference (its potential is 0), the potential of $$S$$ is

$$V_S = E\,\frac{R_3'}{R_1 + R_3'} = E\,\frac{1.04\,R}{R + 1.04\,R} = E\,\frac{1.04}{1 + 1.04} = E\,\frac{1.04}{2.04} = 0.5098\,E$$

Potential of junction $$T$$
The resistors $$R_2$$ and $$R_4$$ form the right branch, still in balance with each other. Hence

$$V_T = E\,\frac{R_4}{R_2 + R_4} = E\,\frac{R}{R + R} = E\,\frac{1}{2} = 0.5000\,E$$

Voltage developed between $$S$$ and $$T$$

$$V_{ST} = V_S - V_T = 0.5098\,E - 0.5000\,E = 0.0098\,E$$

Thus the temperature rise of $$100^{\circ}C$$ produces a potential difference of about $$1\%$$ of the supply voltage between the junctions:

$$\boxed{V_{ST} \approx 9.8\times10^{-3}\,E\ \text{volt}}$$

If the e.m.f. of the battery is, for example, $$E = 10\ \text{V}$$ the galvanometer terminals will be at a difference of $$\approx 0.098\ \text{V}$$.