A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances (R₁=R₂=R₃=R₄ ) . When R₃ resistance is heated to some temperature, its resistance value has gone up by 10 % . The potential difference (V_ a -V_ b ) (after R₃ is heated) is _ _ _ _ V.

Step-by-step solution

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: V_a - V_b = 0 V. When R₃ is heated and its resistance increases by 10%: R₃' = 1.1R The voltage at node a (between R₁ and R₂ ): V_a = 40 R₂ R₁ + R₂ = 40 R R + R = 40 1 2 = 20 V The voltage at node b (between R₃' and R₄ ): V_b = 40 R₄ R₃' + R₄ = 40 R 1.1R + R = 40 1 2.1 19.05 V Therefore: V_a - V_b = 20 - 19.05 = 0.95 V