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)

$\left( V _{ D }- V \right) C _{2}+\left( V _{ D }-0\right) C _{3}=0$
$\left( V _{ D }- V \right) 6+\left( V _{ D }-0\right) 12=0$
$V _{ D }- V +2 V _{ D }=0$
$V _{ D }=\frac{ V }{3}$
$q _{2}=\left( V - V _{ D }\right) C _{2}=\left( V -\frac{ V }{3}\right)(6 \,\mu F )$
$q _{2}=(4 V )\, \mu F$
$q _{3}=\left( V _{ D }-0\right) C _{3}=\frac{ V }{3} \times 12 \,\mu F =4 \,V\, \mu F$
$q _{1}=( V -0) C _{1}= V (2 \,\mu F )$
$q _{1}: q _{2}: q _{3}=2: 4: 4$
$q _{1}: q _{2}: q _{3}=1: 2: 2$