A capacitor of capacitance `12(mu)F`is connected to a battery of emf 6.00V and internal resistance `1.00(Omega)`through resistanceless leads. `12.0(mu)s after the connections are made, what will be (a) the current in the circuit, (b)the power delivered by the battery ,(c)the power dissipated in heat and (d)the rate at which the energy stored in the capacitor is increasing.

Explanation

`c = 12.0 mu f = 12 xx 10^-6 F` .
` E_mf = 6.00 V `
` i= i_0 e^(-t/rc)`
` = Cv/t xx e^(-t/rc)`
` = (12 xx (10^-6)xx 6 / 12 xx (10^-6)) xx e^-1 `
` = 2.207 = 2.21A `
(b) Power delivered by battery
We know,
` V = V_0 e^ (-t/rc) `
` (where V & V_0 are potential V_1)`
` V_1 = V_0 e^(-t/rc)`
` rArr V= V_0 i xx e^-1 = 6 xx 6 xx e^-1`
` = 13.24 Omega `
` (c) U= (Cv^2/t) e^(-t/rC)`
` = (12xx (10^-6) xx 6/ 12 xx (10^-6)) xx (e^(-1))^2 `
` = 4.872` .