Discussion Forum

A capacitance of 2µF is required in an electrical circuit across a potential difference of 1kV. A large number of 1 µ F  capacitors are available which can withstand a potential difference of not more than 300 V. The  minimum number of capacitors required to achieve this is

Answer:

Explanation:

As a each capacitor cannot withstand more than 300V, so there should be four capacitors in each row became in the condition 1kV   i.e, 1000V will be divided by 4 (i.e. 250 not more than 300V)

    Now,equivalent capacitance of one row

                                         = $\frac{1}{4}\times 1\mu F=0.25\mu F$

                         [ $\therefore $ in series combination , $C_{eq}=\frac{c}{n}$ ]

Now, we need equivalent of 2μ F, so let

                we need n such rows

                                      $\therefore $    n × 0.25=2μF

                           [ $\therefore $  in parallel combination C_eq =nc]

                       $n=\frac{2}{0.25}$

                              =8

$\therefore $    Total number of capacitors = number of rows × number of capacitors in each row= 8 × 4=32