Initially, the Wheatstone bridge is balanced with R1=R2=R3=R4=R.
Since the bridge is balanced, the potential difference across the bridge is zero: Va−Vb=0 V.
When R3 is heated and its resistance increases by 10%: R3′=1.1R
The voltage at node a (between R1 and R2): Va=40×R1+R2R2=40×R+RR=40×21=20 V
The voltage at node b (between R3′ and R4): Vb=40×R3′+R4R4=40×1.1R+RR=40×2.11≈19.05 V
Therefore: Va−Vb=20−19.05=0.95 V
