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 step by step, we will analyze the circuit involving the capacitor, the resistor, and the battery.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- A capacitor of capacitance \( C \) is charged to a potential difference \( V_0 \).
- The charge \( Q \) on the capacitor is given by:
\[ Q = C \cdot V_0 \]

2. **Connecting the Circuit**:
- The capacitor is connected to a resistor \( R \) and a battery with EMF \( 3V_0 \).
- The positively charged plate of the capacitor is connected to the positive terminal of the battery.

3. **Applying Kirchhoff's Voltage Law**:
- We apply Kirchhoff's voltage law around the loop:
\[ -3V_0 + I \cdot R + \left( \frac{Q}{C} - V_0 \right) = 0 \]
- Rearranging gives:
\[ I \cdot R = 3V_0 - V_0 - \frac{Q}{C} \]
\[ I \cdot R = 2V_0 - \frac{Q}{C} \]

4. **Expressing Current in Terms of Charge**:
- The current \( I \) can be expressed as:
\[ I = -\frac{dQ}{dt} \]
- Substituting this into the equation gives:
\[ -\frac{dQ}{dt} \cdot R = 2V_0 - \frac{Q}{C} \]
- Rearranging leads to:
\[ \frac{dQ}{dt} = \frac{2CV_0 - Q}{R} \]

5. **Separating Variables and Integrating**:
- We separate variables:
\[ \frac{dQ}{2CV_0 - Q} = \frac{dt}{R} \]
- Integrating both sides:
\[ \int \frac{dQ}{2CV_0 - Q} = \int \frac{dt}{R} \]
- The left side integrates to:
\[ -\ln|2CV_0 - Q| = \frac{t}{R} + k \]
- Solving for \( Q \):
\[ 2CV_0 - Q = e^{-\frac{t}{R} + k} \]
\[ Q = 2CV_0 - e^{-\frac{t}{R} + k} \]

6. **Finding the Heat Generated**:
- The power \( P \) dissipated in the resistor is given by:
\[ P = I^2 R \]
- Substituting \( I = -\frac{dQ}{dt} \):
\[ P = \left(-\frac{dQ}{dt}\right)^2 R \]
- The total heat \( H \) generated until the current becomes zero is:
\[ H = \int_0^{\infty} P \, dt \]

7. **Calculating the Integral**:
- The integration leads to:
\[ H = \frac{4C V_0^2}{R} \cdot \left(\frac{R}{2}\right) = 2C V_0^2 \]

8. **Finding the Value of \( n \)**:
- From the expression for heat generated:
\[ H = n \cdot C V_0^2 \]
- Comparing both expressions gives:
\[ n = 2 \]

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