JEE Main 2024 · 29 January · evening · Single correct · +4 / −1
In the given circuit, the current in resistance R₃ is :
- (a) 2 A
- (b) 1.5 A
- (c) 1 A
- (d) 2.5 A
Answer and solution
Answer (c) 1 A
Solution
Physics sir · 4 steps
Collapse the two equal 4 Ω resistors into one, find the battery current, then split it in half.
- Parallel pair. R₂ and R₃ connect between the same two junctions, so they are in parallel. Two equal resistors in parallel give half of one: (4 × 4)/(4 + 4) = 2 Ω.
- Total resistance. This 2 Ω combination sits in series with R₁ and R₄, so Rₜₒₜₐₗ = 2 + 2 + 1 = 5 Ω.
I = V/Rₜₒₜₐₗ = (10 V)/(5 Ω) = 2 A
- Split the current. All 2 A reaches the parallel pair. R₂ and R₃ are equal and share the same voltage, so each carries half of it. You can also check by voltage: the pair drops 2 × 2 = 4 V, and 4 V across 4 Ω gives 1 A.
- So. The current in R₃ is 1 A, option (c).
The idea to keep. In a series–parallel circuit, first reduce everything to one resistance to get the main current. Then work backwards: equal branches in parallel share the current equally.
Answer from the NTA final answer key. Solution by PrepWiser.