Question

Three capacitors $C_1=2\mu F,C_2=6\mu F$ and $C_3=12\mu F$ are connected as shown in figure. Find the ratio of the charges on capacitors $C_1,C_2$ and $C_3$ respectively:

• A. 2 : 1 : 1
• B. 2 : 3 : 3
• C. 1 : 2 : 2
• D. 3 : 4 : 4

Answer 1

Answer: (C)

$\left(V_D-V\right)C_2+\left(V_D-0\right)C_3=0$
$\left(V_D-V\right)6+\left(V_D-0\right)12=0$
$V_D-V+2V_D=0$
$V_D=\frac{V}{3}$
$q_2=\left(V-V_D\right)C_2=\left(V-\frac{V}{3}\right)(6\mu F)$
$q_2=(4V)\mu F$
$q_3=\left(V_D-0\right)C_3=\frac{V}{3}\times 12\mu F=4V\mu F$
$q_1=(V-0)C_1=V(2\mu F)$
$q_1:q_2:q_3=2:4:4$
$q_1:q_2:q_3=1:2:2$