An electrical technician requires a capacitance of `2muF` in a circuit across a potential difference of `1kV`. A large number of `1muF`capacitors are available to him each of which can withstand a potential difference of not more than 400V. Suggest a possible arrangement that requires the minimum number of capacitors.

Options

A. 24
B. 32
C. 8
D. 16

Answer

B

Explanation

Here, the required capacitance `=2muF` and potential difference, `V=1.0kV=1000V`
According to required the capacitance, we draw the following circuit,

Given, capacitance of each capacitor, `C_(1)=1muF`, potential difference, `V_(1)=300V`
Now, from above the circuit difference across each capacitor,
`" "V_(1)=1000/n=300 rArr n=3.33`
So, for maximum bearable capacitance, we take n = 4. Hence, capacitance of each row in which all 4 capacitors are in series,
`" "1/C=1/1+1/1+1/1+1/1=4/1 rArr C^(.)=1/4muF`
The total capacitance of m rows,
`" "m times 1/4=m/4 muF`
Since, it is given that the total capacitance of circuit is `2muF`.
So, `" "2muF=m/4muF rArr m=8`
Thus, the minimum number of capacitor,
`" "=m times n =8 times 4=32`
Hence, the correct option is (b).