Question

From a supply of identical capacitors rated $8\mu \text{F},250\text{V}$, the minimum number of capacitors required to form a composite $16\mu \text{F}$, $1000\text{V}$ is

• A. 32
• B. 32.0
• C. 32.00

Answer 1

Answer: (A), (B), (C)

Let $n$ be the number of capacitors required to form our composite system. We've been given capacitors that are rated $8\mu \text{F}$, $250\text{V}$ . To generate a composite $16\mu \text{F},1000\text{V}$ capacitor, we need first to generate a system of capacitors that can handle a potential drop $1000\text{V}$.

Since the potential drop across capacitors add up in series, Capacitors required to handle $1000\text{V}$.

$=\frac{\text{Required potential drop}}{\text{Potential drop across one capacitor}}$

$=\frac{1000}{250}=4$

The net capacitance of these capacitors in series can be calculated as:

$\frac{1}{C_{\text{eq}}}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}$

$\Rightarrow C_{\text{eq}}=2\mu \text{F}$

To increase the net capacitance without changing the potential drop across the system, we must use the combination of $4$ such capacitors in parallel. Hence, to have a net capacitance $16\mu \text{F}$ , we require $8$ such sets in parallel so that the net capacitance $=2\times 8=16$ .
Hence, $n=\text{Capacitors in one set}\times \text{number of sets}$ $n=4\times 8=32$
Hence to form a composite capacitor of $16\mu \text{F},1000\text{V}$ using capacitors rated $8\mu \text{F},250\text{V}$ , we need 32 capacitors.