JEE Main 2020 · 9 January · morning · Single correct · +4 / −1
In the given circuit diagram, a wire is joining points B and D. The current in this wire is :
- (a) 4 A
- (b) 2 A
- (c) zero
- (d) 0.4 A
Answer and solution
Answer (b) 2 A
Solution
Physics sir · 4 steps
The wire makes B and D one point, so each pair of resistors on either side is in parallel.
- Redraw the circuit. The wire has no resistance, so B and D are at the same potential. Between A and B/D, the 1 Ω and 4 Ω are in parallel. Between B/D and C, the 2 Ω and 3 Ω are in parallel.
- Total current. Left pair: (1 × 4)/(1 + 4) = 0.8 Ω. Right pair: (2 × 3)/(2 + 3) = 1.2 Ω. In series they give 2 Ω, so the battery drives I = 20/2 = 10 A.
Resistor Voltage across it Current 1 Ω (A to B) 10 × 0.8 = 8 V 8 A into B 4 Ω (A to D) 8 V 2 A into D 2 Ω (B to C) 10 × 1.2 = 12 V 6 A out of B 3 Ω (D to C) 12 V 4 A out of D - Junction rule at B. 8 A arrives at B but only 6 A leaves through the 2 Ω. The extra 8 - 6 = 2 A must go down the wire to D. Check at D: 2 + 2 = 4 A in, 4 A out through the 3 Ω.
- So. The current in wire BD is 2 A, flowing from B to D. Option (b).
The idea to keep. A plain wire across two points fuses them into one node. Find the branch currents on each side, then the junction rule tells you what the wire must carry.
Answer from the NTA final answer key. Solution by PrepWiser.