It is required to construct a `10 mu` F capacitor which can be connected across a 200V battery . Capacitance `10 mu F` are available but they can withstand only 50V ,Design a combination which can yield the desired result .

Explanation

As each capacitor of `10 muF` can withstand only 50V, therefore to be connected across a 200V battery, four capacitors must be connected in series in a row. Capacity `C_(1)` of each row of four capacitors is
`(1)/(C_(1)) = (1)/(10) + (1)/(10) +(1)/(10) +(1)/(10) = (4)/(10)`
`C_(1) = (10)/(4) = 2.5 mu F`
For a total capacity of `10 muF`, four such rows of capacitors must be connected in parallel so that
`C_(p) = 4 C_(1) = 4xx2.5 = 10 mu F`.
Hence, we need 16 capacitors in series in each row and 4 such rows in parallel.