Given,
Capacitance of capacitor, C= 12.0 μF = 12 × 10−6 F
Emf of battery, V0 = 6.00 V
Internal resistance of battery, R = 1 Ω
Time interval, t = 12 μs
(a)
Charging current in the circuit is given as,
i = i0e−t/RC
Current at, t = 12.0 μs
(b)
During charging, charge on the capacitor at any time ''t'' is given as
Work done by battery in in time delivering this charge is,
W = QV0
Power deliver by the battery in time ''t'' is,
Putting, t = 12 μs
(c)
Energy stroed in the capacitor at any instant of time is given as,
Rate at which the energy stored in the capacitor is,
So, the power dissipated in heat =
= 13.25 8.37 = 4.87 W
(d)
Rate at which the energy stored in the capacitor is increasing =