Two condensers of capacity `C` and `2C` are connected in parallel and these are charged upto `V` volt. If the battery is removed and dielectric medium of constant `K` is put between the plates of first condenser, then the potential at each condenser is.

Options

A. `(V)/(k +2)`
B. `2 + (k)/(3V)`
C. `(2V)/(k+2)`
D. `(3V)/(k+2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the initial charge on each capacitor
When the two capacitors \(C\) and \(2C\) are charged to a voltage \(V\), the charges on them can be calculated as follows:
- Charge on the first capacitor \(Q_1 = C \cdot V\)
- Charge on the second capacitor \(Q_2 = 2C \cdot V\)

### Step 2: Calculate the total charge in the system
Since the capacitors are connected in parallel, the total charge \(Q_{total}\) when the battery is removed is:
\[ Q_{total} = Q_1 + Q_2 = C \cdot V + 2C \cdot V = 3C \cdot V \]

### Step 3: Analyze the effect of the dielectric on the first capacitor
When a dielectric of constant \(K\) is introduced into the first capacitor, its capacitance changes from \(C\) to \(C' = K \cdot C\). The charge on the first capacitor remains the same because the battery is removed, so:
\[ Q_1 = C' \cdot V' \]
Where \(V'\) is the new potential across the first capacitor.

### Step 4: Set up the equation for the first capacitor
Substituting \(C' = K \cdot C\) into the equation gives:
\[ C \cdot V = K \cdot C \cdot V' \]
From this, we can solve for \(V'\):
\[ V' = \frac{V}{K} \]

### Step 5: Determine the potential across the second capacitor
Since the total charge in the system remains constant and is distributed between the two capacitors, we can express the charge on the second capacitor as:
\[ Q_2 = 2C \cdot V'' \]
Where \(V''\) is the potential across the second capacitor. The total charge must equal the initial total charge:
\[ Q_{total} = Q_1 + Q_2 \implies 3C \cdot V = K \cdot C \cdot V' + 2C \cdot V'' \]
Substituting \(V' = \frac{V}{K}\):
\[ 3C \cdot V = K \cdot C \cdot \frac{V}{K} + 2C \cdot V'' \]
This simplifies to:
\[ 3C \cdot V = C \cdot V + 2C \cdot V'' \]
\[ 2C \cdot V = 2C \cdot V'' \]
Dividing both sides by \(2C\):
\[ V'' = V \]

### Final Results
- The potential across the first capacitor after the dielectric is introduced is \(V' = \frac{V}{K}\).
- The potential across the second capacitor remains \(V'' = V\).