Question

For a Physics project, a student requires a $4\mu F$ capacitor in a circuit across a potential difference of 1 kV. Unfortunately, $4\mu F$ capacitors are out of stock in the lab. However, $2\mu F$ capacitors which can withstand a potential difference of 400 V are available in plenty. Find the minimum number of capacitors required.

Answer 1

$1000V=n_1\times 400V$
$\Rightarrow n_1=2.5$
$\therefore$ 3 capacitors are required in each row (Equivalent capacitance in each row = $\frac{2}{3}\mu F$)
Now, $C_{e}q=4$
$\Rightarrow n_2.\frac{2}{3}=4\Rightarrow n_2=6$
$\therefore$ Total capacitors $=6\times 3=18$