A capacitor of capacitance C is charged to a potential difference `V_(0)` and then connected with a resistance and a battery of EMF `3V_(0)` such that the positively charged plate of the capacitor is connected to the positive terminal of the battery. The total heat generated in the resistance until the current in the circuit becomes zero is `n(CV_(0)^(2))`. the value of n is ________.

Explanation

To solve the problem, we need to analyze the situation step by step.

### Step 1: Initial Energy in the Capacitor
The initial potential energy (U_initial) stored in the capacitor when charged to a potential difference \( V_0 \) is given by the formula:
\[ U_{\text{initial}} = \frac{1}{2} C V_0^2 \]

### Step 2: Final Configuration
When the capacitor is connected to a battery of EMF \( 3V_0 \), the positively charged plate of the capacitor is connected to the positive terminal of the battery. This means that the potential difference across the capacitor will change.

### Step 3: Final Energy in the Capacitor
The final potential energy (U_final) stored in the capacitor after connecting to the battery can be calculated using the final voltage \( 3V_0 \):
\[ U_{\text{final}} = \frac{1}{2} C (3V_0)^2 = \frac{1}{2} C \cdot 9V_0^2 = \frac{9}{2} C V_0^2 \]

### Step 4: Work Done by the Battery
The work done by the battery (W) can be calculated based on the charge that flows through the battery and the potential difference across it. The initial charge on the capacitor is \( Q_{\text{initial}} = C V_0 \) and the final charge on the capacitor after connecting to the battery is \( Q_{\text{final}} = C \cdot 3V_0 = 3C V_0 \).

The charge that flows through the battery is:
\[ Q_{\text{flow}} = Q_{\text{final}} - Q_{\text{initial}} = 3C V_0 - C V_0 = 2C V_0 \]

The work done by the battery is then:
\[ W = Q_{\text{flow}} \cdot \text{EMF} = (2C V_0) \cdot (3V_0) = 6C V_0^2 \]

### Step 5: Energy Conservation
The work done by the battery is used to increase the potential energy of the capacitor and also to heat the resistor. Thus, we can write:
\[ W = U_{\text{final}} - U_{\text{initial}} + \text{Heat generated in the resistor} \]
Substituting the values we have:
\[ 6C V_0^2 = \left(\frac{9}{2} C V_0^2\right) - \left(\frac{1}{2} C V_0^2\right) + \text{Heat} \]

### Step 6: Calculate the Heat Generated
Now, simplifying the equation:
\[ 6C V_0^2 = \left(\frac{9}{2} - \frac{1}{2}\right) C V_0^2 + \text{Heat} \]
\[ 6C V_0^2 = 4C V_0^2 + \text{Heat} \]
\[ \text{Heat} = 6C V_0^2 - 4C V_0^2 = 2C V_0^2 \]

### Step 7: Relate Heat to Given Expression
According to the problem, the total heat generated in the resistance is given as:
\[ \text{Heat} = n(C V_0^2) \]
From our calculation, we found that:
\[ \text{Heat} = 2C V_0^2 \]
Thus, equating both expressions, we have:
\[ n(C V_0^2) = 2C V_0^2 \]
Dividing both sides by \( C V_0^2 \) (assuming \( C \) and \( V_0 \) are not zero):
\[ n = 2 \]

### Final Answer
The value of \( n \) is:
\[ \boxed{2} \]