Four capacitors of equal capacitance have an equivalent capacitance `C_(1)` when connected in series and an equivalent capacitance `C_(2)` when connected in parallel. The ratio `(C_(1))/(C_(2))` is

Options

A. `(1)/(4)`
B. `(1)/(16)`
C. `(1)/(8)`
D. `(1)/(12)`

Answer

To find the ratio of the equivalent capacitance \( \frac{C_1}{C_2} \) when four capacitors of equal capacitance \( C \) are connected in series and in parallel, we can follow these steps:

### Step 1: Calculate the equivalent capacitance in series

When capacitors are connected in series, the formula for the equivalent capacitance \( C_1 \) is given by:

\[ \frac{1}{C_1} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} \]

This simplifies to:

\[ \frac{1}{C_1} = \frac{4}{C} \]

Taking the reciprocal gives:

\[ C_1 = \frac{C}{4} \]

### Step 2: Calculate the equivalent capacitance in parallel

When capacitors are connected in parallel, the formula for the equivalent capacitance \( C_2 \) is:

\[ C_2 = C + C + C + C \]

This simplifies to:

\[ C_2 = 4C \]

### Step 3: Find the ratio \( \frac{C_1}{C_2} \)

Now we can find the ratio of the equivalent capacitances:

\[ \frac{C_1}{C_2} = \frac{\frac{C}{4}}{4C} \]

This simplifies to:

\[ \frac{C_1}{C_2} = \frac{C}{4} \cdot \frac{1}{4C} = \frac{1}{16} \]

### Conclusion

Thus, the ratio \( \frac{C_1}{C_2} \) is:

\[ \frac{C_1}{C_2} = \frac{1}{16} \]