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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 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
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Solution
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 the potential difference across each capacitor must be 400 V. Hence, the number of capacitors in each row is given as
`1000/400 = 2.5`
Hence, there are three capacitors in each row.
Capacitance of each row
= `1/(1 + 1 + 1)`
= `1/3 mu "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"`
= `"n"/3`
However, the capacitance of the circuit is given as 2 `mu"F"`
∴ `"n"/3 = 2`
n = 6
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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