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. (a) 8
• A. (b) 32
• A. (c) 16
• A. (d) 24

Answer 1

(b): Minimum number of capacitors in each row

$= \frac{1000}{250} = 4$

Therefore, 4 capacitors connected in series. If there are m such rows, then total capacity

$= m \times 2 = 16\therefore m = 16/2 = 8$

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