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 = 1µF
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 potential difference across each capacitor must be 400 V. Hence, number of capacitors in each row is given as
Hence, there are three capacitors in each row.
Capacitance of each row `=1/(1+1+1)=1/3mu F`
Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as
`1/3+1/3+1/3+..................n "" terms`
However, capacitance of the circuit is given as 2 `muF`
Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 × 3 i.e., 18 capacitors are required for the given arrangement.
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