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.
A$0.10$
B$0.15$
C$0.20$
D$0.27$
IIT 2020, Diffcult
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D$0.27$
d $R _3^{\prime}=300(1+\alpha \Delta T )$
$=312 \Omega$
Now
$I _1=\frac{50}{372} \text { and } I _2=\frac{50}{600}$
$V _{ S }- V _{ T }=312 I _1-500 I _2$
$=41.94-41.67$
$=0.27 V$
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