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.

Explanation

Here total capacitance, `C = 2 mu F`
Potential difference V = 1Kv = 1000 volt
Capacity of each capacitor `C_(1) = 1 mu F`
Maximum potential difference acorss each V = 400 volt

Let n capacitors of `1 mu F` each be connected in series in a row and m such rows be connected in parallel as shown in the figure. As potential difference in each row = 1000 volt
`:.` Potential difference across each capacitor `= (1000)/(n) = 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 condensors of `1 mu F`
Each is series = 1/3
Total capacitance of m such rows in parallel `= (m)/(3)`
`:. (m)/(3) = 2 (mu F)` or m = 6
`n xx m = 3 xx 6 = 18`.
Hence `1 mu F` capacitors should be connected in six parallel rows, each row containing three capactiros in series.