Question

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.​

Answer 1

let one capacitor be x and other be y

`in \: parallel \: connection : \\ (x + y)/(xy) = 10 * {10}^( - 6) - - - (a) \\ in \: series \: connection : \\ x + y = 2.4 * {10}^( - 6) - - - (b) \\ in \: (a) : \\ \frac{2.4 * {10}^( - 6) }{xy} = 10 * {10}^( - 6) \\ xy = 0.24 - - - (c) \\ from \: (b) : \\ y = (2.4 * {10}^( - 6) ) - x \\ \therefore \: in \: (c) : \\ x(2.4 * {10}^( - 6) - x) = 0.24 \\ 2.4 * {10}^( - 6) x - {x}^(2) = 0.24 \\ {x}^(2) - 2.4 * {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}}`