The value of current i₁ flowing from A to C in the circuit diagram is :

Solution

Physics sir · 4 steps

Use symmetry first: B, D and the centre all sit at the same potential, so the 5 Ω wires carry nothing.

1. Find equal potentials. Call the centre junction O. Take V_A = 8 V and V_C = 0. The circuit is mirror-symmetric left to right, so B, O and D each sit exactly halfway, at 4 V.
2. No bridge current. B, O and D are at the same potential, so no current flows in either 5 Ω resistor. You can remove them. That leaves three separate paths from A to C.

   Path	Resistance	Current = 8 V / R
   A → B → C	2 + 2 = 4 Ω	2 A
   A → O → C (carries i₁)	4 + 4 = 8 Ω	1 A
   A → D → C	2 + 2 = 4 Ω	2 A

3. Check. The total is 2 + 1 + 2 = 5 A. That matches 8 V across the equivalent resistance 1.6 Ω of 4, 8 and 4 Ω in parallel.
4. So. The middle path has 4 V across each 4 Ω resistor, so i₁ = (8 - 4)/4 = 1 A.

The idea to keep. Before you write Kirchhoff loops, look for symmetry. Points at equal potential carry no current between them, so those resistors can be removed and the network falls apart into simple paths.

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