Question

Minimum number of $8\mu \text{F}$ and $250\text{V}$ capacitors used to make a combination of $16\mu \text{F}$ and $1000\text{V}$ are:

• A. $4$
• B. $8$
• C. $16$
• D. $32$

Answer 1

The correct option is D $32$

For this type of problems, remember that potential difference increases in series combination and equivalent capacitance increases in parallel combination.

So, if voltage across each $8\mu \text{F}$ is $250\text{V},$ then four such capacitors in series combination make up $1000\text{V}.$


From above combination or branch we get

$C_{eq}=(\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}{)}^{-1}$

$\Rightarrow C_{eq}=\frac{C}{4}=\frac{8}{4}=2\mu \text{F}$

And, potential difference,

$\Delta V=4V=4\times 250=1000\text{V}$
Since, required capacitance is $16\mu \text{F}$ and we know that parallel combination increases the capacitance.

So, $8$ branches of above combination need to be attached in parallel to get equivalent capacitance $16\mu \text{F}$. i.e. effective capacitance will become,

$C_{eq}^{\prime }=8\times C_{eq}=8\times 2=16\mu \text{F}$

And, $\Delta V=1000\text{V}$

So, total number of capacitors will be,

$\text{=number of capacitors in each branch}\times \text{number of branches}$

$=4\times 8$

$=32$

So, the total $32$ capacitors are required.

Hence, option (a) is the correct answer.