Question

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 400v

Answer 1

Answer:

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+1

1

=

4

1

μF

Let there are n rows each having four capacitors connected in parallel, having equivalent capacitance =

4

n

Capacitance of the circuit = 2 μF

∴

4

n

=2⇒n=8

Hence, 8 rows, each having 4 capacitors is the arrangement.

So, total number of capacitors = 8 * 4 = 32