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 (Vₐ-V_b) (after R₃ is heated) is _________ V.

Solution

Physics sir · 4 steps

Treat each side of the bridge as its own potential divider, then subtract the two midpoint potentials.

1. Two separate dividers. Take the bottom node as 0 V, so the top node is at 40 V. The left arm (R₁ then R₂) and the right arm (R₃ then R₄) each sit across the full 40 V, so each one works as an independent potential divider.
2. Point a. R₁ = R₂ = R, so the 40 V splits equally and Vₐ = 40 × R/2R = 20 V.
3. Point b. After heating, R₃ = 1.1R and R₄ = R. The potential of b is the drop across R₄.

   V_b = 40 × R₄/(R₃ + R₄) = 40 × R/2.1R = 19.05 V

4. So. Vₐ - V_b = 20 - 19.05 = 0.952 V ≈ 0.95 V, which is option (b).

The idea to keep. When a bridge goes out of balance, find each midpoint potential with the divider rule, V = V₀ R_(lower)/Rₜₒₜₐₗ, and subtract. A changed resistor changes the current too, not only its own drop.

Answer from the NTA final answer key. Solution by PrepWiser.