Two resistors 4 Ω and 12 Ω are in parallel; this combination is in series with 8 Ω. Total resistance:
A20 Ω
B11 Ω
C12 Ω
D8 Ω
Answer & Solution
Correct answer: B. 11 Ω
Parallel: (4 × 12)/(4 + 12) = 48/16 = 3 Ω. Then series with 8: 3 + 8 = 11 Ω.
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