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4Ω series with (3Ω ∥ 6Ω), Battery = 12V current through 3Ω
R_parallel = (3×6)/(3+6) = 18/9 = 2Ω R_total = 4 + 2 = 6Ω I_total = 12/6 = 2A V across parallel = 2 × 2 = 4V I through 3Ω = 4/3 = 1.33 A