Question

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

• A. $32$
• B. $2$
• C. $16$
• D. $24$

Answer 1

Answer: (A)

$32$

$m\times 300>1000$

$m>\frac{10}{3}$

or $m=4$

For $m=4$, capacitance of 1 branch is $\frac{1\mu F}{4}=0.25\mu F$

For overall capacitance of $2\mu F$, $8$ such branches or equivalent

is $C_{eq}=8\times 0.25=2\mu F$

Total capacitors required= $8\times 4=32$