As C1= 6µF
C2 = 12µF
V = 200V
Ceq = 6(12)/18 = 4µF
Q = CV = 4 ×10^-6 × 200 = 8 × 10^-4C
In series, charge remains same, so same charge will flow through both capacitors.
Hence, option A is correct.
As C1= 6µF
C2 = 12µF
V = 200V
Ceq = 6(12)/18 = 4µF
Q = CV = 4 ×10^-6 × 200 = 8 × 10^-4C
In series, charge remains same, so same charge will flow through both capacitors.
Hence, option A is correct.