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 capacitor 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.
Let a number of capacitors connected in series and these series circuits are connected in parallel row to each other. The potential difference across each row must be 1kV = 1000 V Potential difference across each capacitor= 300 V Number of capacitors in each row is thus = 1000/300 = 3.33 So, there are capacitors in each row. Capacitance of each row = 1+1+1+11=41μF Let there are n rows each having four capacitors connected in parallel, having equivalent capacitance = 4n Capacitance of the circuit = 2 μF ∴4n=2⇒n=8 Hence, 8 rows, each having 4 capacitors is the arrangement. So, total number of capacitors = 8 * 4 = 32
