A capacitor of capacitance `C` is initially charged to a potential difference of `V` volt. Now it is connected to a battery of `2V` with oppoiste polarity. The ratio of heat generated to the final enegry stored in the capacitor will be

Options

A. 1.75
B. 2.25
C. 2.5
D. 0.5

Answer

To solve the problem, we need to determine the ratio of heat generated to the final energy stored in the capacitor after it is connected to a battery of `2V` with opposite polarity. Let's break down the solution step by step.

### Step 1: Initial Conditions
The capacitor has a capacitance `C` and is initially charged to a potential difference of `V`. The initial charge on the capacitor can be calculated as:
\[ Q_{\text{initial}} = C \times V \]

### Step 2: Final Conditions
When the capacitor is connected to a battery of `2V` with opposite polarity, the final charge on the capacitor will be:
\[ Q_{\text{final}} = C \times (-2V) \]
This means the capacitor will have a charge of `-2CV` on one plate and `+2CV` on the other plate.

### Step 3: Energy Stored in the Capacitor
The final energy stored in the capacitor can be calculated using the formula for the energy stored in a capacitor:
\[ U_{\text{final}} = \frac{1}{2} C V^2 \]
Substituting `2V` for `V`, we get:
\[ U_{\text{final}} = \frac{1}{2} C (2V)^2 = \frac{1}{2} C \cdot 4V^2 = 2CV^2 \]

### Step 4: Initial Energy Stored
The initial energy stored in the capacitor is:
\[ U_{\text{initial}} = \frac{1}{2} C V^2 \]

### Step 5: Change in Energy
The change in energy (which is the work done by the battery) can be calculated as:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values we found:
\[ \Delta U = 2CV^2 - \frac{1}{2} CV^2 = 2CV^2 - 0.5CV^2 = 1.5CV^2 \]

### Step 6: Charge Flowed from the Battery
The total charge that flows from the battery can be calculated as:
\[ Q = Q_{\text{final}} - Q_{\text{initial}} = (-2CV) - (CV) = -3CV \]
The negative sign indicates that the charge is flowing in the opposite direction.

### Step 7: Work Done by the Battery
The work done by the battery can be calculated as:
\[ W = Q \times V = 3CV \times 2V = 6CV^2 \]

### Step 8: Heat Generated
Using the energy conservation principle, we can find the heat generated:
\[ W = \Delta U + \text{Heat} \]
Rearranging gives us:
\[ \text{Heat} = W - \Delta U = 6CV^2 - 1.5CV^2 = 4.5CV^2 \]

### Step 9: Ratio of Heat Generated to Final Energy Stored
Now we can find the ratio of heat generated to the final energy stored:
\[ \text{Ratio} = \frac{\text{Heat}}{U_{\text{final}}} = \frac{4.5CV^2}{2CV^2} = \frac{4.5}{2} = 2.25 \]

### Final Answer
The ratio of heat generated to the final energy stored in the capacitor is:
\[ \text{Ratio} = 2.25 \]