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Question -

An electrical technician requires acapacitance of 2 µF in a circuit across a potential difference of 1 kV. A largenumber of 1 µF capacitors are available to him each of which can withstand apotential difference of not more than 400 V. Suggest a possible arrangementthat requires the minimum number of capacitors.



Answer -

Total required capacitance, C =2 µF

Potential difference, V =1 kV = 1000 V

Capacitance of each capacitor, C1 =1µF

Each capacitor can withstand a potentialdifference, V1 = 400 V

Suppose a number of capacitors are connected in seriesand these series circuits are connected in parallel (row) to each other. Thepotential difference across each row must be 1000 V and potential differenceacross each capacitor must be 400 V. Hence, number of capacitors in each row isgiven as

Hence, there are three capacitors in each row.

Capacitance of each row

Let there are n rows, eachhaving three capacitors, which are connected in parallel. Hence, equivalentcapacitance of the circuit is given as

Hence, 6 rows of three capacitors arepresent in the circuit. A minimum of 6 × 3 i.e., 18 capacitors are required forthe given arrangement.

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