A capacitor of capacitance C charged by battery at V volt and then disconnected. At t = 0, it is connected to an uncharged capacitor of capacitance 2C through a resistance R. The charge on the second capacitor as a function of time is given by `q=(alpha CV)/(3) (1-e^(-(3t)/(beta RC)))` then fing the value of `(alpha )/(beta)`.

Answer

To solve the problem, we need to analyze the situation involving the two capacitors and the resistor. Let's break it down step by step.

### Step 1: Understand the Initial Conditions
We have a capacitor of capacitance \( C \) charged to a voltage \( V \). The charge on this capacitor when fully charged is given by:
\[ Q_0 = C \cdot V \]
At \( t = 0 \), this charged capacitor is disconnected from the battery and connected to an uncharged capacitor of capacitance \( 2C \) through a resistor \( R \).

**Hint:** Remember that the initial charge on the first capacitor is \( Q_0 = C \cdot V \).

### Step 2: Set Up the Charge Conservation Equation
When the charged capacitor \( C \) is connected to the uncharged capacitor \( 2C \), charge will redistribute between them. Let \( q \) be the charge on the \( 2C \) capacitor at time \( t \). The charge on the \( C \) capacitor will then be \( Q_0 - q \).

Using Kirchhoff's law, we can write the equation for the charge:
\[ \frac{Q_0 - q}{C} + \frac{q}{2C} = 0 \]
This simplifies to:
\[ \frac{Q_0 - q}{C} = -\frac{q}{2C} \]

**Hint:** Use Kirchhoff's voltage law to set up the equation based on charge conservation.

### Step 3: Rearranging the Equation
From the equation above, we can rearrange it to find a relationship between \( q \) and \( Q_0 \):
\[ Q_0 - q = -\frac{q}{2} \]
Multiplying through by \( 2C \) gives:
\[ 2C(Q_0 - q) = -q \]
Expanding this, we have:
\[ 2CQ_0 - 2Cq = -q \]
Rearranging leads to:
\[ 2CQ_0 = 2Cq - q \implies 2CQ_0 = q(2C + 1) \]
Thus,
\[ q = \frac{2CQ_0}{2C + 1} \]

**Hint:** Rearranging the charge conservation equation helps in isolating \( q \).

### Step 4: Deriving the Differential Equation
The current \( I \) flowing through the resistor can be expressed as:
\[ I = -\frac{dq}{dt} \]
Using Ohm's law, we have:
\[ I = \frac{Q_0 - q}{R} \]
Setting these equal gives:
\[ -\frac{dq}{dt} = \frac{Q_0 - q}{R} \]

**Hint:** Relate the current to the change in charge over time.

### Step 5: Solve the Differential Equation
Rearranging gives:
\[ \frac{dq}{Q_0 - q} = -\frac{dt}{R} \]
Integrating both sides:
\[ -\ln|Q_0 - q| = \frac{t}{R} + C \]
Exponentiating leads to:
\[ Q_0 - q = K e^{-\frac{t}{R}} \]
where \( K \) is a constant determined by initial conditions.

**Hint:** Use integration to solve the differential equation.

### Step 6: Apply Initial Conditions
At \( t = 0 \), \( q = 0 \):
\[ Q_0 - 0 = K e^{0} \implies K = Q_0 \]
Thus, we have:
\[ Q_0 - q = Q_0 e^{-\frac{t}{R}} \implies q = Q_0(1 - e^{-\frac{t}{R}}) \]

**Hint:** Use the initial condition to find the constant \( K \).

### Step 7: Relate to Given Function
The problem states that:
\[ q = \frac{\alpha CV}{3} (1 - e^{-\frac{3t}{\beta RC}}) \]
Comparing this with our derived equation:
\[ q = Q_0(1 - e^{-\frac{t}{R}}) \]
We see that:
\[ Q_0 = CV \implies \frac{\alpha CV}{3} = CV \implies \alpha = 3 \]
And from the exponent:
\[ -\frac{t}{R} = -\frac{3t}{\beta RC} \implies \beta = 3 \]

### Step 8: Find \( \frac{\alpha}{\beta} \)
Now we can find:
\[ \frac{\alpha}{\beta} = \frac{3}{3} = 1 \]

### Final Answer
Thus, the value of \( \frac{\alpha}{\beta} \) is:
\[ \boxed{1} \]