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

Given, total capacitance required for the combination, `C=2muF`
Capacitance of each cpacitor, `C.=1muF`
Potential difference across the circuit, `V=1kV`
Voltage rating of each `1muF` capacitor, `V.=400V`
Let the combination of the capacitors consists consists of m capacitors in series and n such rows of capacitors in series are connected in parallel.
Thus, total number of capacitors, `N=mn`.
As it is given that the maximum voltage that each capacitor can withstand is 400V, if m capacitors are connected in series, the series combination of the capacitors can withstand voltage equal to `400 xx m` volts.
Now, each row of capacitors will be connected across a potential difference of 1,000 V.
`400m=1,000`
`m=2.5`
As the number of capacitor cannot be fractional, as well as it cannot be equal to 2 as if it were equal to 2, then the potential difference across each capacitor would become equal to 500 V that exceeds the maximum value of the voltage that each capacitor can withstand.
Thus, `m=3`
If `C_(s)` is total capacitance in a row, then:
`(1)/(C )=(1)/(1)+(1)/(1)+(1)/(1)=3`
`C_(s)=(1)/(3)muF`
Thus, the total capacitance of the n rows of the capacitors is given by:
`C=nC_(s)`
`2=n xx(1)/(3)`
Thus, `n=6`
Therefore, the total number of `1muF` capacitors required by the technician is 18, where 6 rows of 3 capacitors connected are to be in parallel.