Question

From a supply of identical capacitors rated $ 8{\text{ }}\mu F,250{\text{ }}V $ , the minimum number of capacitors required to form a composite $ {\text{16 }}\mu F,1000{\text{ }}V $ capacitor is:
A) 2
B) 4
C) 16
D) 32

Answer 1

Hint:
First, determine the number of capacitors required in series to handle a potential drop of $ 1000\,\,V $ . Then to have a net capacitance $ 16\,\mu F $ , we will have to connect multiple sets of capacitors in parallel to have the same potential drop but increase the capacitance.
Formula used:
For capacitors in series: $ \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}.... $
For capacitors in parallel: $ {C_{eq}} = {C_1} + {C_2} + {C_3}..... $

Complete step by step answer:
Let $ n $ be the number of capacitors required to form our composite system.
We’ve been given capacitors that are rated $ 8{\text{ }}\mu F,250{\text{ }}V $ . To generate a composite $ {\text{16 }}\mu F,1000{\text{ }}V $ capacitor, we need to first generate a system of capacitors that can handle a potential drop $ 1000\,V $ .
Since the potential drop across capacitors add up in series,
Capacitors required to handle $ 1000\,V $
$ = \dfrac{{{\text{Required}}\,{\text{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_{eq}}}} = \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} $
 $ \Rightarrow {C_{eq}} = 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
$ {\text{n = 4}} \times {\text{8 = 32}} $
Hence to form a composite capacitor of $ {\text{16 }}\mu F,1000{\text{ }}V $ using capacitors rated $ 8{\text{ }}\mu F,250{\text{ }}V $ , we need 32 capacitors which correspond to option (D).

Note:
When capacitors are connected in series, the net potential drop across them increases but the net capacitance decreases and when they are connected in parallel, the potential drop remains constant but the net capacitance increases. Option (A) seems like a lucrative option since 2 capacitors connected in parallel will have a net capacitance of $ 16\mu F $ however they won’t be able to handle $ 1000V $ without damage.