The equivalent resistance between the points A and B in the following circuit is x/5Ω. The value of x is ____ .

Solution

Physics sir · 4 steps

This is an unbalanced Wheatstone bridge, so use its symmetry and a one-node KCL.

1. Spot the bridge. Call the top middle junction C and the bottom middle junction D. The arms are A–C 6 Ω, A–D 3 Ω, C–B 3 Ω, D–B 6 Ω, with 3 Ω across C–D. Since 6/3 ≠ 3/6, the bridge is not balanced and the middle resistor carries current.
2. Use symmetry. Put A at potential V = 1 and B at 0. Swapping A with B and C with D leaves every resistor unchanged, so V_D = 1 - V_C.

   (1 - V_C)/6 - V_C/3 + ((1 - V_C) - V_C)/3 = 0 ⇒ 7V_C = 3 ⇒ V_C = 3/7

3. Current from A. This gives V_D = 4/7. The current leaving A is (1 - 3/7)/6 + (1 - 4/7)/3 = 2/21 + 3/21 = 5/21 A.
4. So. R_(AB) = V/I = 1/(5/21) = 21/5 Ω, so x = 21.

The common slip: 22.5. Treating the bridge as balanced and dropping the middle 3 Ω gives (6+3) ∥ (3+6) = 4.5 Ω, so x = 22.5. The ratios are 6:3 and 3:6, so the bridge is not balanced.

The idea to keep. Check the bridge ratio before you delete the middle resistor. If it is unbalanced, set A = 1 V and B = 0, use symmetry to link the middle potentials, and apply KCL at one node.

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