A capacitor of capacitance `10muF` is connected to a resistance `2Omega` and a battery of emf 5V of negligible interanl resistance. After `20mus` of comleting the circuit find.
a. power delivered by the battery
b. power dissipated as heat
c. rate of energy stored in the capacitor

Explanation

Here `C = 10muF`, `R = 2Omega, E = 5v, tau =RC = 2xx10xx10^(-6)`
`s =20 mus`
a. Current at` t = 20mu` s is
`I = (E )/(e )e^(9-t//tau) = (5)/(2) e^(-20//20) = (5)/(2e)`
Power delivered by battery is
`P_1 = EI =5xxx(5)/(2e) = (25)/(2e) =4.6W`
b. Power dissipated as heat is
`P_2 = I^2R = ((5)/(2e))^2 2 = (25)/(2e^2) = 1.7 W`
c. Energy on the capacitor is
`U = (q^(2))/(2C)`
Rate of energy stored in the capacitor is
`P_3 = (dU)/(dt) = (2q)/(2C)(dq)/(dt) = (q)/(C) (dp)/(dt) = (Q(1-e^(-t//tau)))/(C) I`
`= (EC)/(C)(1-e^(-t//tau)) I = EI(1 -e^(-t//tau))`
`=(5xx5)/(2e)(1-e^(-20//20))W =(25)/(2e) -(25)/(2e^2) = 2.9`