Question

A capacitance of $2 \, \mu F$ is required in an electrical circuit across a potential difference of $1.0\, kV$. A large number of $1 \, \mu F$ capacitors are available which can withstand a potential difference of not more than $300\, V$. The minimum number of capacitors required to achieve this is :

• A. 2
• B. 16
• C. 24
• D. 32

Answer 1

The Correct Option is D

To solve this problem, we need to determine how to arrange multiple capacitors of 1 \, \mu F, each capable of withstanding a maximum of 300 \, V, to achieve an effective capacitance of 2 \, \mu F across a potential difference of 1.0 \, kV.

First, let's consider the constraints:

• Each capacitor can only handle up to 300 \, V.
• We need the combined capacitors to handle 1.0 \, kV = 1000 \, V.

We need to connect enough capacitors in series to withstand the full 1000 \, V. The formula for the total voltage across capacitors in series is:

V_{\text{total}} = V_1 + V_2 + \ldots + V_n

Given each capacitor can handle 300 \, V, we calculate the minimum number of capacitors in series:

n = \frac{1000 \, V}{300 \, V} \approx 3.33

Therefore, we need at least 4 capacitors in series to ensure that the voltage does not exceed the rating of any single capacitor.

When capacitors are connected in series, the equivalent capacitance C_{\text{series}} is given by:

\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}

Substituting C_1 = C_2 = \ldots = C_n = 1 \, \mu F, we have:

\frac{1}{C_{\text{series}}} = \frac{4}{1 \, \mu F} \Rightarrow C_{\text{series}} = 0.25 \, \mu F

To achieve an equivalent capacitance of 2 \, \mu F, capacitors in series must be grouped in parallel, as capacitors in parallel sum directly:

C_{\text{parallel}} = n \times C_{\text{series}}

Substituting the required total capacitance:

2 \, \mu F = n \times 0.25 \, \mu F \Rightarrow n = \frac{2}{0.25} = 8

The total number of capacitors needed is the product of capacitors in series and those sets connected in parallel:

4 \, (\text{series}) \times 8 \, (\text{parallel}) = 32

Thus, the minimum number of capacitors required is 32.