Three cells of emf 2 V each, internal resistance 0.5 Ω each, are connected in series. Total emf and internal resistance:
AE = 6V, r = 0.5 Ω
BE = 6V, r = 1.5 Ω
CE = 2V, r = 1.5 Ω
DE = 6V, r = 1/6 Ω
Answer & Solution
Correct answer: B. E = 6V, r = 1.5 Ω
In series: E_eq = E₁ + E₂ + E₃ = 6V. r_eq = r₁ + r₂ + r₃ = 1.5 Ω. (Series adds both emf and internal resistance.)
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