The Attempt at a Solution
1) Ummm ... well ...no idea :tongue2:
2) I guess no. Because if a capacitor is connected with battery, work done by batery is QE (EMF) but thermal energy is just QE/2. Rest QE/2energy is stored in capacitor.
3) Same as (2)
4) i = E/(R+r)
--- r is internal resistance of battery
--- R is external resistance.
--- E is EMF
Work done by battery = QV = Q(E - ir) where r is internal resistance of battery
Thermal energy across in resistor = i2Rt = E2RT/(R+r)2
Now i dont know what to do.
But i still guess that for non-ideal its no and for ideal its yes ... don't know why
5) Well i thought of some explanation before i started typing but now i cant remember it
6) Yes, only for non ideal battery. At time of charging, V = E + ir