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

Here. Total capacitance, `C = 2 mu F`,
Potential difference, `V = 1 kV = 1000 vol t`
Capacity of each capacitor, `C_(1) = 1 muF`,
Maximum potential difference across each.
`V' = 400 vol t`
Let n capacitors fo `1 muF` each be connected in series in a row and m such rows be connected in parallel as shown in Fig.
As potential difference acoross each row = 1000 volt
`:.` Potential difference across each capacitors `= (1000)/(n) = V' = 400`
`:. n = (1000)/(400) = 2.5`
As n has to be a whole number (not less than 2.5), therefore , `n = 3`.
Capacitance of each row of 3 conderes of `11 muF` each, in series `= 1//3`.
Total capacitance of m such rows in parallel `= (m)/(3) :. (m)/(3) = 2 (muF) or m = 6`
`:.` total number capacitors `= n xx m = 3xx6 = 18`
Hence `1 muF` capacitors should be connected in six parallel rows, each row contaning three capacitors in series.