JEE Main 2021 · 27 August · morning · Single correct · +4 / −1
Five identical cells each of internal resistance 1Ω and emf 5V are connected in series and in parallel with an external resistance 'R'. For what value of 'R', current in series and parallel combination will remain the same?
- (a) 1 Ω
- (b) 25 Ω
- (c) 5 Ω
- (d) 10 Ω
Answer and solution
Answer (a) 1 Ω
Solution
Physics sir · 4 steps
Write the current for each arrangement, set the two equal, and R drops out in one line.
- Series combination. In series the emfs add and the internal resistances add: total emf 5 × 5 = 25 V, total internal resistance 5 × 1 = 5 Ω. So Iₛ = 25/(R + 5).
- Parallel combination. In parallel the emf stays 5 V, and the five 1 Ω internal resistances combine to 1/5 = 0.2 Ω. So Iₚ = 5/(R + 0.2).
25/(R + 5) = 5/(R + 0.2)
- Solve. Cross-multiply: 25R + 5 = 5R + 25, so 20R = 20 and R = 1 Ω. Check: both currents come to 25/6 A.
- So. The currents match when R = 1 Ω, which is option (a). This equals the internal resistance of one cell.
The idea to keep. For n identical cells, series and parallel give the same current exactly when the external resistance equals the internal resistance of one cell, R = r. Remember it, but derive it once to trust it.
Answer from the NTA final answer key. Solution by PrepWiser.