# From a supply of identical capacitors rated 8 mu F, 250 V, the minimum number of capacitors required to form a composite 16 mu F, 1000 V is? (A) 2 (B) 4 (C) 16 (D) 32

Jan 4, 2018

32

#### Explanation:

Recall

$\frac{1}{C} _ \text{eq} = \frac{1}{C} _ 1 + \frac{1}{C} _ 2 + \frac{1}{C} _ 3. . .$ in series

${C}_{\text{eq}} = {C}_{1} + {C}_{2} + {C}_{3} + \ldots$ in parallel

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

$\frac{1}{C} _ \text{eq} = \frac{1}{8 \mu F} + \frac{1}{8 \mu F} + \frac{1}{8 \mu F} + \frac{1}{8 \mu F} = \frac{1}{2 \mu F}$

$\Rightarrow {C}_{e q} = 2 \mu F$

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

$2 \mu F + 2 \mu F + 2 \mu F + \ldots = 16 \mu F$

$\Rightarrow$ total number of capacitors = 8 x 4 = 32

Can you prove that this is the minimum number of them needed?