CapacitanceHard
Question
A capacitor having initial charge q0 = CE/2 is connected to a cell of e.m.f. E through a resistor R as shown. Find the total heat generated in the circuit after the switch S is closed :-
Options
A.(1/12) CE2
B.(1/8)CE2
C.(1/4) CE2
D.None of these
Solution
Final charge on capacitor qf = CE
Charge flown = CE – CE/2 = CE/2
Work done by battery: Wb = E (CE/2) =
CE2
Change in energy of capacitor is
ᐃU =
[qf2 - q02]
=
[(CE)2 - (CE/2)2] = 
Heat = Wb - ᐃU =
CE2
Charge flown = CE – CE/2 = CE/2
Work done by battery: Wb = E (CE/2) =
CE2Change in energy of capacitor is
ᐃU =
[qf2 - q02]=
[(CE)2 - (CE/2)2] = Heat = Wb - ᐃU =
CE2Create a free account to view solution
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