Two condensers of capacities C and 3C are connected in parallel and then connected in series with a third condenser of capacity 3C. The combination is charged with a battery of 'V'volt. The charge on condenser of capacity C is (in coulomb)

Options

A. `1//2 (CV )`
B. `3CV //7`
C. `2CV `
D. `3//2 (CV)`

Answer

To solve the problem step-by-step, we will follow these steps:

### Step 1: Determine the Equivalent Capacitance of the Parallel Combination
We have two capacitors, one with capacitance \( C \) and the other with capacitance \( 3C \), connected in parallel. The equivalent capacitance \( C' \) for capacitors in parallel is given by:

\[ C' = C + 3C = 4C \]

### Step 2: Determine the Equivalent Capacitance of the Series Combination
Next, this equivalent capacitor \( C' = 4C \) is connected in series with a third capacitor of capacitance \( 3C \). The equivalent capacitance \( C'' \) for capacitors in series is given by:

\[ \frac{1}{C''} = \frac{1}{C'} + \frac{1}{3C} \]

Substituting \( C' \):

\[ \frac{1}{C''} = \frac{1}{4C} + \frac{1}{3C} \]

To combine these fractions, we find a common denominator, which is \( 12C \):

\[ \frac{1}{C''} = \frac{3}{12C} + \frac{4}{12C} = \frac{7}{12C} \]

Now, taking the reciprocal gives us:

\[ C'' = \frac{12C}{7} \]

### Step 3: Calculate the Total Charge in the Circuit
The total charge \( Q \) stored in the equivalent capacitor when connected to a battery of voltage \( V \) is given by:

\[ Q = C'' \cdot V = \frac{12C}{7} \cdot V = \frac{12CV}{7} \]

### Step 4: Determine the Charge on Each Capacitor in Series
In a series circuit, the charge \( Q \) on each capacitor is the same. Therefore, the charge on the capacitor with capacitance \( C \) (let's denote it as \( Q_1 \)) is given by:

\[ Q_1 = Q = \frac{12CV}{7} \]

### Step 5: Use the Charge Division Rule
Now, we need to find the charge on the capacitor \( C \) specifically. The charge on each capacitor in series is proportional to its capacitance. The charge on capacitor \( C \) can be expressed as:

\[ Q_1 = \frac{C}{C + 3C} \cdot Q = \frac{C}{4C} \cdot \frac{12CV}{7} = \frac{1}{4} \cdot \frac{12CV}{7} = \frac{12CV}{28} = \frac{3CV}{7} \]

### Final Answer
Thus, the charge on the capacitor of capacity \( C \) is:

\[ Q_1 = \frac{3CV}{7} \]