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

Explanation

Total required capacitance , C=`2muF`
potential difference, V=1kv =1000 v
Capacitance of each capacitor , `C_(1)=1 muF`
each capacitor can withstand a potential difference `V_(1)=400 V`
Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. the potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. hence,number of capacitors in each row is given as
`1000/400=2.5`
Hence, there are three capacitors in each row.
Capacitance of each row `=1/(1+1+1)=1/3 muF`
Let there are n rows each hacing three capacitors, which are connected in parallel.
Hence, equivalent capacitance of the circuit is given as
`1/3+1/3+1/3+....n ` terms
`=n/3`
However, capacitance of the circuit is given as 2`muF`
`:. n/3=2`
`n=6`
Hence, 6 rows of three capacitors are present in the circuit. a minimum of 6 x 3 i.e., 18 capacitors are required for the given arragements.