A capacitance of `2.0muF` is required in an electrical circuit across a potential differences of 1.0 kV. A large number of `1muF` capacitors are available which can withstand a potential differences of not more than 300V. The minimum number of capacitors required to achieve this is

Options

A. 16
B. 24
C. 32
D. 2

Answer

C

Explanation

As each capacitors cannot withstand more than 300 V, so there should be four capacitors in each row became in this condition 1 kV i.e., 1000 V will be divided by 4 (i.e., 250 not more than 300 V).
Now, equivalent capacitance of one row `=(1)/(4)xx1muF=0.25muF` [`because` In series combination `C_(eq)=c//n`]
Now, we need equivalent of `2muF`, so let we need n such rows
`therefore nxx0.25=2muF` [`because` In series combination `C_(eq)=nc`] , `n=(2)/(0.25)=8`
`therefore`Total number of capacitors=number of rows`xx`number of capacitors in each row
`=8xx4=32`