# The capacitors A and B are connected in series with a battery. When S is closed two capacitor gets fully charged , then :

(A) Ratio of charges on A and B is 3:2

(B)Ratio of electrical energies stored in A and B is 2:3

(C)The potential difference across A is 6 V and across B is 4V

(D)The potential difference across A is 4 V and across B is 6V

Total capacitance in series $=\large\frac{20 \times 30}{20+30}$
$\qquad= 12 \mu F$
$Q= CV$
$\quad= 12 \times 10= 120 \mu C$
$V_A= \large\frac{Q}{C_A} =\frac{120}{20}$$=6V V_B= \large\frac{Q}{C_B} =\frac{120}{30}$$=4V$
Hence C is the correct answer.