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 given circuit and the charge transfer between the capacitors. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
Initially, we have a capacitor of capacitance \( C \) charged to a voltage \( V \). The charge on this capacitor is given by:
\[ Q_0 = CV \]
This capacitor is then disconnected from the battery.

### Step 2: Connect the Capacitors
At \( t = 0 \), the charged capacitor \( C \) is connected to an uncharged capacitor \( 2C \) through a resistor \( R \). The charge \( Q \) will flow from capacitor \( C \) to capacitor \( 2C \).

### Step 3: Charge Conservation
The initial charge on capacitor \( C \) is \( CV \). When charge \( Q \) flows to capacitor \( 2C \), the remaining charge on capacitor \( C \) becomes:
\[ Q_{\text{remaining}} = CV - Q \]

### Step 4: Voltage Across the Capacitors
The voltage across capacitor \( C \) is:
\[ V_C = \frac{Q_{\text{remaining}}}{C} = \frac{CV - Q}{C} = V - \frac{Q}{C} \]
The voltage across capacitor \( 2C \) is:
\[ V_{2C} = \frac{Q}{2C} \]

### Step 5: Apply Kirchhoff's Voltage Law
According to Kirchhoff's law, the sum of the potential differences around the loop must equal zero:
\[ V_C = V_{2C} + IR \]
Substituting the expressions for \( V_C \) and \( V_{2C} \):
\[ V - \frac{Q}{C} = \frac{Q}{2C} + IR \]
Where \( I = -\frac{dQ}{dt} \).

### Step 6: Rearranging the Equation
Rearranging gives:
\[ V - \frac{Q}{C} - \frac{Q}{2C} = -R \frac{dQ}{dt} \]
Combining terms:
\[ V - \frac{3Q}{2C} = -R \frac{dQ}{dt} \]

### Step 7: Separate Variables
Rearranging for separation of variables:
\[ \frac{dQ}{\frac{2CV - 3Q}{2C}} = -\frac{dt}{R} \]

### Step 8: Integrate
Integrating both sides:
\[ \int \frac{2C}{2CV - 3Q} dQ = -\int \frac{dt}{R} \]
This leads to:
\[ \ln(2CV - 3Q) = -\frac{t}{R} + k \]
Where \( k \) is the constant of integration.

### Step 9: Solve for \( Q \)
Exponentiating both sides gives:
\[ 2CV - 3Q = e^{-\frac{t}{R} + k} \]
At \( t = 0 \), \( Q = 0 \):
\[ 2CV = e^k \implies k = \ln(2CV) \]
Thus:
\[ 2CV - 3Q = 2CV e^{-\frac{t}{R}} \]
Rearranging gives:
\[ Q = \frac{2CV}{3}(1 - e^{-\frac{3t}{2RC}}) \]

### Step 10: Identify \( \alpha \) and \( \beta \)
Comparing with the given function:
\[ Q = \frac{\alpha CV}{3}(1 - e^{-\frac{3t}{\beta RC}}) \]
We see that:
- \( \alpha = 2 \)
- \( \beta = 2 \)

### Step 11: Calculate \( \frac{\alpha}{\beta} \)
Thus:
\[ \frac{\alpha}{\beta} = \frac{2}{2} = 1 \]

### Final Answer
The value of \( \frac{\alpha}{\beta} \) is \( 1 \).

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