Question 2.23:

An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF 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.


 
Total required capacitance, C = 2 μF Potential difference, V = 1, kV = 1000 V
Capacitance of each capacitor, C1 = 1 μF
Each capacitor can withstand a potential difference, V1 = 400 V
Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and the potential difference across each capacitor must be 400 V.
Hence, the number of capacitors in each row is given as 1000400 = 2.5
Hence, there are three capacitors in each row.
The capacitance of each row 11 + 1 + 1 = 13 μF
Let there are n rows, each having three capacitors, = 2.5 which are connected in parallel.
Hence the capacitance of the circuit is given as 
13 + 13 + 13 + ...................n term
However. the capacitance of the circuit is given as 2 μ
 n3 = 2 
       n = 6
Hence, 6 rows of three capacitors are present in the circuit. A Minimum of 6 x 3 i.e., 18 capacitors are required for the given arrangement.