Question

The minimum number of $8\mu F$ and 250 V capacitors which are used to make a combination of capacitance 16 $\mu F$ and voltage 1000 V is

• A. 4
• B. 32
• C. 8
• D. 3

Answer 1

Answer: (B)

Let $m$ rows of $n$ series capacitor be taken then minimum number of capacitors required is

Also effective voltage is
$V^{\prime \prime }=1000=n\times 250$
$\Rightarrow n=\frac{1000}{250}=4$
Also these four capacitor are connected in series then effective capacitance is
$\frac{1}{C^{\prime \prime }}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{4}{8}$
$\Rightarrow C^{\prime \prime }=2\mu F$
$\therefore C^{\prime \prime \prime \prime }=16=2\times m$
$\Rightarrow m=\frac{16}{2}=8$
Hence $N=m\times n=8\times 4=32$