Question

A number of capacitors labelled as $8 \, \mu F, 250 \, V$ are given. Minimum number of capacitors needed to get an arrangement equivalent to $10 \, \mu F, 1000 \, V$ is

• A. 64
• B. 20
• C. 32
• D. 16

Answer 1

Answer: (B)

20

Answer 2

Let II capacitors of $8 \, \mu F, 250 \, V$ each be connected in series in a row and m such rows be connected in parallel.
As the voltage needed in the arrangement is 1000 V , so, potential difference across each capacitor can have 250 V across it.
Therefore, number of capacitors in each row, $n = \frac{1000}{250} = 4$
Equivalent capacitance of each row
$\frac{1}{C'} = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{4}{8}$ or $C' = 2 \mu F$
Equivalent capacitance of the circuit, $C" = m C'$
$m = \frac{C"}{C' } = \frac{10 \mu F}{2\mu F} = 5$
$\therefore$ Minimum number of capacitors required
$= n \times m = 4 \times5 = 20$