alleryTwo capacitors 2 µF and 4 µF connected in SERIES across 12 V. Charge on each:
A6 µC
B16 µC
C8 µC
D2.67 µC × 12
Answer & Solution
Correct answer: B. 16 µC
**Series:** 1/C_eq = 1/2 + 1/4 = 3/4 → C_eq = 4/3 µF.
Charge Q = C_eq V = (4/3) × 12 = 16 µC.
Same Q on each cap in series.
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