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R₃=5Ω series with (R₁=3Ω ∥ R₂=2Ω), Battery = 2.5V
1/R_p = 1/3 + 1/2 = 5/6 → R_p = 1.2Ω R_total = 1.2 + 5 = 6.2Ω I = 2.5/6.2 = ≈ 0.40 A
(i) Total resistance = 6.2 Ω
(ii) Current = 25/62 ≈ 0.40 A