Three capacitors
Question:

Three capacitors $\mathrm{C}_{1}=2 \mu \mathrm{F}, \mathrm{C}_{2}=6 \mu \mathrm{F}$ and $\mathrm{C}_{3}=12 \mu \mathrm{F}$ are connected as shown in figure. Find the ratio of the charges on capacitors $C_{1}, C_{2}$ and $C_{3}$ respectively :

1. $2: 1: 1$

2. $2: 3: 3$

3. $1: 2: 2$

4. $3: 4: 4$

Correct Option: , 3

Solution:

$\left(V_{D}-V\right) C_{2}+\left(V_{D}-0\right) C_{3}=0$

$\left(V_{D}-V\right) 6+\left(V_{D}-0\right) 12=0$

$V_{D}-V+2 V_{D}=0$

$V_{D}=\frac{V}{3}$

$\mathrm{q}_{2}=\left(\mathrm{V}-\mathrm{V}_{\mathrm{D}}\right) \mathrm{C}_{2}=\left(\mathrm{V}-\frac{\mathrm{V}}{3}\right)(6 \mu \mathrm{F})$

$\mathrm{q}_{2}=(4 \mathrm{~V}) \mu \mathrm{F}$

$\mathrm{q}_{3}=\left(\mathrm{V}_{\mathrm{D}}-0\right) \mathrm{C}_{3}=\frac{\mathrm{V}}{3} \times 12 \mu \mathrm{F}=4 \mathrm{~V} \mu \mathrm{F}$

$\mathrm{q}_{1}=(\mathrm{V}-0) \mathrm{C}_{1}=\mathrm{V}(2 \mu \mathrm{F})$

$\mathrm{q}_{1}: \mathrm{q}_{2}: \mathrm{q}_{3}=2: 4: 4$

$\mathrm{q}_{1}: \mathrm{q}_{2}: \mathrm{q}_{3}=1: 2: 2$