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

Explanation

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

### Step-by-Step Solution:

1. **Understanding Series Connection**:
In a series connection of capacitors, the charge \( Q \) on each capacitor is the same. The total potential difference \( V \) across the series combination is the sum of the potential differences across each capacitor.

2. **Charge Relationship**:
Since both capacitors have the same charge \( Q \), we can express the charge on each capacitor:
\[ Q = C_1 V_1 = C_2 V_2 \]
where \( V_1 \) is the potential difference across \( C_1 \) and \( V_2 \) is the potential difference across \( C_2 \).

3. **Total Voltage**:
The total voltage across the series combination is given by:
\[ V = V_1 + V_2 \]

4. **Expressing \( V_2 \)**:
From the charge relationship, we can express \( V_2 \) in terms of \( Q \):
\[ V_2 = \frac{Q}{C_2} \]

5. **Substituting \( V_2 \)**:
Substitute \( V_2 \) into the total voltage equation:
\[ V = V_1 + \frac{Q}{C_2} \]

6. **Expressing \( V_1 \)**:
From the charge relationship, we can also express \( Q \) in terms of \( V_1 \):
\[ Q = C_1 V_1 \]
Substitute this into the equation for \( V \):
\[ V = V_1 + \frac{C_1 V_1}{C_2} \]

7. **Factoring Out \( V_1 \)**:
Factor \( V_1 \) out from the right-hand side:
\[ V = V_1 \left(1 + \frac{C_1}{C_2}\right) \]

8. **Solving for \( V_1 \)**:
Now, solve for \( V_1 \):
\[ V_1 = \frac{V}{1 + \frac{C_1}{C_2}} = \frac{V C_2}{C_1 + C_2} \]

9. **Final Expression**:
Therefore, the potential difference across capacitor \( C_1 \) is:
\[ V_1 = \frac{C_2}{C_1 + C_2} V \]

### Final Answer:
The potential difference across capacitor \( C_1 \) is given by:
\[ V_1 = \frac{C_2}{C_1 + C_2} V \]