A capacitor of capacitance `8.0(mu)F` is connected to a bettery of emf 6.0V through a resistance of `24(Omega)`.Find the current in the circuit (a)just after the connections are made and (b)one time constant after the connections are made.

Explanation

`C = 8 mu f , E = 6V, R = 24 Omega `
` (a) i= V/R = 6/24 = 0.25 A `
` (b) q = Qe^-(t/RC)`
` = (8 xx (10^-6) xx 6) e^-1`
` = 48 xx (10^-6) xx 0.63 `
` 3.024 xx (10^-5) `
` V = Q/C = (3.024 xx (10^-5)/ 8 xx (10^-5)) = 3.78 `
` E = V + iR `
` rArr 6 = 3.7 + i24 `
` rArr i = 0.09A ` .