An infinite number of capacitors, having capacitances `1muF, 2muF, 4muF and 8muF…..` are connected in series. The equivalent capacitance of the system is

Options

A. infinite
B. `0.25muF`
C. `0.5muF`
D. `2muF`

Explanation

To find the equivalent capacitance of an infinite number of capacitors connected in series with capacitances of \(1 \mu F, 2 \mu F, 4 \mu F, 8 \mu F, \ldots\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Series Capacitance Formula**:
For capacitors in series, the reciprocal of the equivalent capacitance \(C_s\) is given by:
\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \ldots \]

2. **List the Capacitors**:
The given capacitors are:
\[ C_1 = 1 \mu F, \quad C_2 = 2 \mu F, \quad C_3 = 4 \mu F, \quad C_4 = 8 \mu F, \ldots \]
This sequence can be expressed as:
\[ C_n = 2^{n-1} \mu F \quad \text{for } n = 1, 2, 3, \ldots \]

3. **Write the Series Capacitance Equation**:
Substitute the values into the series capacitance formula:
\[ \frac{1}{C_s} = \frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \]

4. **Recognize the Series as a Geometric Series**:
The series can be recognized as a geometric series where:
- First term \(a = 1\)
- Common ratio \(r = \frac{1}{2}\)

5. **Sum the Infinite Geometric Series**:
The sum \(S\) of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Here, substituting the values:
\[ S = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \]

6. **Calculate the Equivalent Capacitance**:
Thus, we have:
\[ \frac{1}{C_s} = 2 \implies C_s = \frac{1}{2} \mu F = 0.5 \mu F \]

### Final Answer:
The equivalent capacitance of the system is:
\[ C_s = 0.5 \mu F \]