A 12 V battery is connected across a network with a 4 Ω resistor in series with two 6 Ω resistors in parallel. What is the current through the 4 Ω resistor?
A2.0 A
B1.71 A
C3.0 A
D0.86 A
Explanation
📌 R_parallel = 3 Ω. R_total = 4 + 3 = 7 Ω. I = V/R = 12/7 = 1.71 A. Distractor A: ignored parallel; C: only used 4 Ω; D: half.