Two cells of emfs 1 V and 2 V and internal resistances 2 Ω and 1 Ω respectively are connected in series with an external resistance of 6 Ω, giving current I₁. The same cells are then connected in parallel with the same external resistance, giving current I₂. If I₁/I₂ = x/3, find x.
Physics · Current electricity · combination of cells and maximum power theorem
JEE Mains 2025 3 April Evening Shift

Question

Two cells of emfs 1 V and 2 V and internal resistances 2 Ω and 1 Ω respectively are connected in series with an external resistance of 6 Ω, giving current I₁. The same cells are then connected in parallel with the same external resistance, giving current I₂. If I₁/I₂ = x/3, find x.

Physics · Current electricity · combination of cells and maximum power theorem

JEE Mains 2025 3 April Evening Shift

G1 Physics · Accepted Answer

Series combination: Total emf = 1+2 = 3 V. Total internal resistance = 2+1 = 3 Ω. With external resistance 6 Ω, I₁ = 3/(3+6) = 1/3 A. For the parallel combination of unequal cells, the Thevenin equivalent emf is E e q E_eq E e ​ q = (E₁/r₁ + E₂/r₂)/(1/r₁ + 1/r₂) = (1/2 + 2/1)/(1/2 + 1) = (2.5)/(1.5) = 5/3 V. Equivalent internal resistance: r e q r_eq r e ​ q = (2×1)/(2+1) = 2/3 Ω. Therefore, I₂ = (5/3)/(6+2/3) = (5/3)/(20/3) = 1/4 A. Thus, I₁/I₂ = (1/3)/(1/4) = 4/3. Given I₁/I₂ = x/3, x = 4.