Question

Minimum number of capacitors each of 8 $\mu F$ and 250 V used to make a composite capacitor of 16 $\mu F$ and 1000 V are:

• A. 8
• B. 32
• C. 16
• D. 24

Answer 1

The correct option is B 32

To have $1000V$ tolerance, you need to stack up four $8\mu F$ capacitor in series. The equivalent capacitance of 4 such capacitors is:

$⟹C_{eq}=\frac{8\mu F}{4}=2\mu F$

To make a total of $16\mu F$ that tolerates $1000V$, you need $8$ set of them arranged in parallel such that

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

$\therefore$ Total number of capacitors $=8\times 4=32$