The effective capacitance of two capacitors of capacitances `C_(1)` and `C_(2)(C_(2)gtC_(1))` connected in parallel is `(25)/(6)` times the effective capacitance when they are connected in series. The ratio `(C_(2))/(C_(1))` is

Options

A. `(3)/(2)`
B. `(4)/(3)`
C. `(5)/(3)`
D. `(25)/(6)`

Answer

To solve the problem, we need to find the ratio \( \frac{C_2}{C_1} \) given that the effective capacitance of two capacitors \( C_1 \) and \( C_2 \) (where \( C_2 > C_1 \)) connected in parallel is \( \frac{25}{6} \) times the effective capacitance when they are connected in series.

### Step-by-step Solution:

1. **Write the formulas for capacitance in series and parallel**:
- For capacitors in series:
\[ C_S = \frac{C_1 C_2}{C_1 + C_2} \]
- For capacitors in parallel:
\[ C_P = C_1 + C_2 \]

2. **Set up the equation based on the problem statement**:
According to the problem, the effective capacitance in parallel is \( \frac{25}{6} \) times the effective capacitance in series:
\[ C_P = \frac{25}{6} C_S \]

3. **Substitute the expressions for \( C_P \) and \( C_S \)**:
\[ C_1 + C_2 = \frac{25}{6} \left( \frac{C_1 C_2}{C_1 + C_2} \right) \]

4. **Cross-multiply to eliminate the fraction**:
\[ (C_1 + C_2)^2 = \frac{25}{6} C_1 C_2 \]

5. **Expand the left side**:
\[ C_1^2 + 2C_1C_2 + C_2^2 = \frac{25}{6} C_1 C_2 \]

6. **Rearrange the equation**:
\[ C_1^2 + C_2^2 + 2C_1C_2 - \frac{25}{6} C_1 C_2 = 0 \]
This simplifies to:
\[ C_1^2 + C_2^2 + \left(2 - \frac{25}{6}\right) C_1 C_2 = 0 \]
Which further simplifies to:
\[ C_1^2 + C_2^2 - \frac{13}{6} C_1 C_2 = 0 \]

7. **Divide the entire equation by \( C_1^2 \)**:
\[ 1 + \frac{C_2^2}{C_1^2} - \frac{13}{6} \frac{C_2}{C_1} = 0 \]

8. **Let \( x = \frac{C_2}{C_1} \)**:
The equation becomes:
\[ 1 + x^2 - \frac{13}{6} x = 0 \]

9. **Rearranging gives us a standard quadratic equation**:
\[ x^2 - \frac{13}{6} x + 1 = 0 \]

10. **Use the quadratic formula to solve for \( x \)**:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -\frac{13}{6}, c = 1 \):
\[ x = \frac{\frac{13}{6} \pm \sqrt{\left(-\frac{13}{6}\right)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ x = \frac{\frac{13}{6} \pm \sqrt{\frac{169}{36} - 4}}{2} \]
\[ x = \frac{\frac{13}{6} \pm \sqrt{\frac{169 - 144}{36}}}{2} \]
\[ x = \frac{\frac{13}{6} \pm \sqrt{\frac{25}{36}}}{2} \]
\[ x = \frac{\frac{13}{6} \pm \frac{5}{6}}{2} \]

11. **Calculating the two possible values for \( x \)**:
- First value:
\[ x_1 = \frac{18/6}{2} = \frac{3}{2} \]
- Second value:
\[ x_2 = \frac{8/6}{2} = \frac{2}{3} \]

12. **Determine the correct value based on the condition \( C_2 > C_1 \)**:
Since \( C_2 > C_1 \), we take \( x = \frac{3}{2} \).

### Final Answer:
\[ \frac{C_2}{C_1} = \frac{3}{2} \]