Four resistances 40 Ω, 60 Ω, 90 Ω and 110 Ω make the arms of a quadrilateral ABCD. Across AC is a battery of emf 40 V and internal resistance negligible.The potential difference across BD in V is _______.

Solution

Physics sir · 4 steps

The two branches from A to C are independent, so find the drop from A to B and from A to D and compare.

1. Two parallel branches. There is no resistor between B and D, so current from A reaches C only through A→B→C or A→D→C. Each branch has the full 40 V across it.
2. Branch ABC. Total resistance is 40 + 60 = 100 Ω, so the current is 40/100 = 0.4 A. The drop across AB is 0.4 × 40 = 16 V, so V_B = V_A - 16.
3. Branch ADC. Total resistance is 90 + 110 = 200 Ω, so the current is 40/200 = 0.2 A. The drop across AD is 0.2 × 90 = 18 V, so V_D = V_A - 18.

   V_B - V_D = (V_A - 16) - (V_A - 18) = 2 V

4. So. The bridge is not balanced, because 40/60 ≠ 90/110. The potential difference across BD is 2 V.

The common slip: 6. Comparing the drop across AB (16 V) with the drop across DC (22 V) instead of across AD gives 6, not 2. Both drops must be measured from the same node A.

The idea to keep. To find the voltage between the two middle points of a bridge, measure both from the same end. Find the drop from A to each point, then subtract.

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