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

We need to find the ratio of the charges on the capacitors $$C_1$$, $$C_2$$, and $$C_3$$ arranged in the given circuit configuration.

1. Analyze the Circuit Diagram Connections

Let's look closely at the nodes and connections across the network :

• Let the potential at terminal $$A$$ be $$V_A$$ and the potential at terminal $$B$$ be $$V_B$$. The total potential difference supplied by the battery source is $$V = V_A - V_B$$.
• Capacitor $$C_1$$: One plate is connected directly to node $$A$$ and the other plate is connected directly to node $$B$$. Therefore, $$C_1$$ is in a standalone parallel branch connected directly across the full potential difference $$V$$.
• Capacitors $$C_2$$ and $$C_3$$: They are connected in series with each other along the top branch. The left plate of $$C_2$$ connects to node $$A$$, their shared junction forms a common node $$D$$, and the right plate of $$C_3$$ connects to node $$B$$. Therefore, this entire combined series branch ($$C_2$$ and $$C_3$$) is in parallel with $$C_1$$ across the potential difference $$V$$.

2. Calculate the Charges on Each Capacitor

The values of the three capacitors are given as:

$$C_1 = 2\ \mu\text{F},\quad C_2 = 6\ \mu\text{F},\quad C_3 = 12\ \mu\text{F}$$

Charge on $$C_1$$ ($$Q_1$$):

Since the voltage across $$C_1$$ is the full voltage $$V$$:

$$Q_1 = C_1 \times V = 2V$$

Charges on $$C_2$$ and $$C_3$$ ($$Q_2$$ and $$Q_3$$):

Because $$C_2$$ and $$C_3$$ are connected together in series within their branch, the same amount of electric charge flows through both of them ($$Q_2 = Q_3$$). This shared charge is determined by the equivalent capacitance ($$C_{\text{eq, top}}$$) of that branch multiplied by the potential difference $$V$$:

$$\frac{1}{C_{\text{eq, top}}} = \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{6} + \frac{1}{12}$$

$$\frac{1}{C_{\text{eq, top}}} = \frac{2 + 1}{12} = \frac{3}{12} = \frac{1}{4} \implies C_{\text{eq, top}} = 4\ \mu\text{F}$$

Now, calculate the charge on this series branch:

$$Q_2 = Q_3 = C_{\text{eq, top}} \times V = 4V$$

3. Determine the Ratio of the Charges

We can now construct the required ratio of the charges on $$C_1$$, $$C_2$$, and $$C_3$$ respectively:

$$Q_1 : Q_2 : Q_3 = 2V : 4V : 4V$$

Dividing throughout by the common factor $$2V$$ gives:

$$Q_1 : Q_2 : Q_3 = 1 : 2 : 2$$

Final Answer: 1 : 2 : 2 (Option D)