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when two capacitors are connected in series, the effective capacitance is 2.4muF and when connected in parallel, the effective capacitance is 10muF. calculate the individual capacitances.​

Sagot :

Answer:

let one capacitor be x and other be y

[tex]in \: parallel \: connection : \\ \frac{x + y}{xy} = 10 \times {10}^{ - 6} - - - (a) \\ in \: series \: connection : \\ x + y = 2.4 \times {10}^{ - 6} - - - (b) \\ in \: (a) : \\ \frac{2.4 \times {10}^{ - 6} }{xy} = 10 \times {10}^{ - 6} \\ xy = 0.24 - - - (c) \\ from \: (b) : \\ y = (2.4 \times {10}^{ - 6} ) - x \\ \therefore \: in \: (c) : \\ x(2.4 \times {10}^{ - 6} - x) = 0.24 \\ 2.4 \times {10}^{ - 6} x - {x}^{2} = 0.24 \\ {x}^{2} - 2.4 \times {10}^{ - 6} x - 0.24 = 0 \\ x = 0.5 \\ 0.5y = 0.24 \\ y = 0.48 \\ { \boxed{ C _{1} = 0.5 \: farads}} \\ { \boxed{C _{2} = 0.48 \: farads}}[/tex]

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