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 $100^{\circ} C^{- 1}$ , the voltage developed between $S$ and $T$ will be ______ $volt$ .

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Solution

After the temperature rise the new resistance is,
$R_{3}^{'} = R_{3} (1 + α Δ T)$
$R_{3}^{'} = 300 (1 + 0.004 \times 100$ )
$R_{3}^{'} = 312 Ω$
Now redrawing the circuit,
The current $I_{1}$ goes to the series connection of $60 Ω$ and $312 Ω$
$I_{1} = \frac{V}{R_{e q}} = \frac{50}{372}$
and $I_{2} = \frac{50}{600}$
Writing the equation of potential along path $S \to T$ ,
$V_{S} + 60 I_{1} - 100 I_{2} = V_{T}$
$V_{S} - V_{T} = 100 \times I_{2} - 60 \times I_{1}$
$Or, V_{S} - V_{T} = \frac{50}{6} - \frac{3000}{372}$
$∴ V_{S} - V_{T} = 0.268 ≃ 0.270 V$

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