Question

Minimum number of capacitors each of $8 \mu \mathrm{F}$ and 250 $\mathrm{V}$ used to make a composite capacitor of $16 \mu \mathrm{F}$ and $1000 \mathrm{~V}$ are,
(a) 8
(b) 32
(c) 16
(d) 24

Answer 1

Hint: First, determine the number of capacitors required in a series to handle a $1000 \mathrm{~V}$ potential drop. Then we will have to connect multiple sets of capacitors in parallel to have the same potential drop but increase the capacitance in order to have a net capacitance of $16 \mu F$.

Formula used:
For capacitors in series: $\dfrac{1}{C_{e q}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}} \ldots$
For capacitors in parallel: $C_{e q}=C_{1}+C_{2}+C_{3} \ldots \ldots$

Complete solution:
Let $n$ be the number of capacitors required to form our composite system.
We've been given capacitors that are rated $8 \mu F, 250 \mathrm{~V}$.

To generate a composite $16 \mu \mathrm{F}, 1000 \mathrm{~V}$ capacitor, we need to first generate a system of capacitors that can handle a potential drop of $1000 \mathrm{~V}$.

Since the potential drop across capacitors add up in series,Capacitors required to handle $1000 \mathrm{~V}$
$=\dfrac{\text { Required potential drop }}{\text { potential drop across one capacitor }}$ $=\dfrac{1000}{250}=4$

The net capacitance of these capacitors in series can be calculated as:
$\dfrac{1}{C_{e q}}=\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}$
$\Rightarrow C_{e q}=2 \mu F$

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 $16 \mu F$, we require 8 such sets in parallel so that the net capacitance $=2 \times 8=16$.
Hence
$n=$ Capacitors in one set $\times$ number of sets
$\mathbf{n}=4 \times 8=32$

Hence to form a composite capacitor of $16 \mu F, 1000 V$ using capacitors rated $8 \mu F, 250 V,$ we need 32 capacitors which correspond to option (b).

Note:
The net potential drop across them increases when capacitors are connected in series, but the net capacitance decreases and when they are connected in parallel, the potential drop remains constant, but the net capacitance increases. Option (A) appears to be a lucrative option as 2 condensers connected in parallel will have a net capacitance of $16 \mu F$, but without damage they will not be able to handle $1000 \mathrm{~V}$.