The minimum number of 8 and 250 V capacitors which are used to make...

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



Also effective voltage is

$V^{′′} = 1000 = n \times 250$

$⇒ n = \dfrac{1000}{250} = 4$

Also these four capacitor are connected in series then effective capacitance is

$\dfrac{1}{C′′} = \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} = \dfrac{4}{8}$

$⇒ C^{′′} = 2\muF$

$∴ C^{′′′′} = 16 = 2 \times m$

$⇒ m = \dfrac{16}{2} = 8$

Hence $N = m \times n = 8 \times 4 = 32$