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

The potential difference can only be increased by connecting capacitors in series, while capacitance can only be increased by connecting capacitance in parallel.
To acquire the required arrangement let there be m rows, connected in parallel, each row containing n capacitors in series. Then total number of capacitors `N=mn`.
If V is the net potential difference and `V_(0)`, the potential difference across each capacitor, then
`V=nV_(0)`, i.e., `n=(V)/(V_(0))(1kV)/(400V)=(1000V)/(400V)=2.5`
An n cannot be a friction we must taken `n=3`. if `C_(0)` is capacitance of each capacitor, the capacitance of a row=`(C_(0))/(n)`.
As m rows are connected in parallel, net capacitance `C=(mC_(0))/(n)`
Given `C=2muF and C_(0)=1muF, n=3 " "therefore 2muF=(mxx(1muF))/(3)or " "m=(2xx3)/(1)=6`
Minimum number of capacitors, `N=mn=3x xx6=18`.