Question Detail
Question
6 mF and 3 mF capacitors in series across 18 V. Charge on each?
Correct Answer
Option A is the correct answer.
Detailed Explanation
C_eq = (6×3)/(6+3) = 18/9 = 2 mF = 2000 µF. Q = C_eq V = 2000e-6 × 18 = 0.036 C = 36000 µC. Options are 36,46,75,96 µC. If C in µF? 6µF and 3µF in series → C_eq=2µF, Q=2e-6×18=36e-6 C=36 µC. Yes A.
Hint
Series: same Q, C_eq = product/sum.