Two capacitances of capacity `C_(1)`and `C_(2)` are connected in series and potential difference `V` is applied across it. Then the potential difference across `C_(1)` will be

Options

A. `V(C_(2))/(C_(1))`
B. `V(C_(1) + C_(2))/(C_(1))`
C. `V (C_(2))/(C_(1) + C_(2))`
D. `V(C_(1))/(C_(1) + C_(2))`

Explanation

To find the potential difference across capacitor \( C_1 \) when two capacitors \( C_1 \) and \( C_2 \) are connected in series with a total potential difference \( V \) applied across them, we can follow these steps:

### Step 1: Understand the series connection of capacitors
In a series connection, the same charge \( Q \) flows through both capacitors. The total potential difference \( V \) across the series combination is the sum of the potential differences across each capacitor.

### Step 2: Write the relationship for potential differences
Let \( V_1 \) be the potential difference across capacitor \( C_1 \) and \( V_2 \) be the potential difference across capacitor \( C_2 \). According to the series connection, we have:
\[ V = V_1 + V_2 \]

### Step 3: Relate charge and capacitance
For capacitors, the relationship between charge \( Q \), capacitance \( C \), and potential difference \( V \) is given by:
\[ Q = C \cdot V \]
Thus, for capacitor \( C_1 \):
\[ Q = C_1 \cdot V_1 \quad \text{(1)} \]
And for capacitor \( C_2 \):
\[ Q = C_2 \cdot V_2 \quad \text{(2)} \]

### Step 4: Set the charges equal
Since the same charge \( Q \) flows through both capacitors, we can equate the two expressions for \( Q \):
\[ C_1 \cdot V_1 = C_2 \cdot V_2 \]

### Step 5: Solve for \( V_2 \)
From the equation \( C_1 \cdot V_1 = C_2 \cdot V_2 \), we can express \( V_2 \) in terms of \( V_1 \):
\[ V_2 = \frac{C_1}{C_2} \cdot V_1 \quad \text{(3)} \]

### Step 6: Substitute \( V_2 \) back into the total voltage equation
Substituting equation (3) into the total voltage equation \( V = V_1 + V_2 \):
\[ V = V_1 + \frac{C_1}{C_2} \cdot V_1 \]
Factoring out \( V_1 \):
\[ V = V_1 \left(1 + \frac{C_1}{C_2}\right) \]

### Step 7: Solve for \( V_1 \)
Now, we can solve for \( V_1 \):
\[ V_1 = \frac{V}{1 + \frac{C_1}{C_2}} = \frac{V}{\frac{C_1 + C_2}{C_2}} = \frac{C_2 \cdot V}{C_1 + C_2} \]

### Final Result
Thus, the potential difference across capacitor \( C_1 \) is given by:
\[ V_1 = \frac{C_2 \cdot V}{C_1 + C_2} \]