JEE Main 2020 · 7 January · morning · Single correct · +4 / −1
The current I₁ (in A) flowing through 1 Ω resistor in the following circuit is :
- (a) 0.4
- (b) 0.25
- (c) 0.2
- (d) 0.5
Answer and solution
Answer (c) 0.2
Solution
Physics sir · 4 steps
The battery puts 1 V across the whole top branch, so find that branch's current first, then split it.
- Spot the branches. The 1 V battery connects the left and right wires directly. So the lower 2 Ω resistor and the whole upper branch each have the full 1 V across them, independently.
- Upper branch resistance. The two 1 Ω resistors are in parallel, giving (1 × 1)/(1 + 1) = 0.5 Ω. This is in series with the upper 2 Ω, so the branch has 2.5 Ω in total.
I_(branch) = (1 V)/(2.5 Ω) = 0.4 A
- Split the current. The two parallel 1 Ω resistors are equal, so they share the 0.4 A equally. Each carries 0.2 A.
- So. I₁ = 0.2 A, which is option (c).
The idea to keep. A branch connected straight across an ideal battery gets the full emf, whatever else is in parallel with it. Solve that branch on its own, then divide current among parallel resistors in inverse ratio of resistance.
Answer from the NTA final answer key. Solution by PrepWiser.