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when two chapacitors are joined in series and resultance capacity is 4.8mueF and when the same twoare joined in parallel the resultant capacity is 30mueF their individual capacities are

Kota manasa 29th Jan, 2021

Answer (1)

ayanmukherjee045 30th Jan, 2021

Dear aspirant,

To solve this question you have to follow certain steps-

1-First form an equation when the capacitors are joined in series. For eg. (1/x)+(1/y)=(1/4.8) where x and y are capacitance of both capacitors.

2-Similarly form an equation when they are joined in parallel

3-Solve these equations simultaneously to get each capacitance.

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