Consider the situation shown in the figure. The switch S is open for a long time and then closed and again steady state reached then

Text Solution
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C ε /2, – C ε , C ε 2 /2, C ε 2 /4, (e) C ε 2 /4
Sol. Situation when S is open

C eq. = C/2
Charge supplied Q = C eq. ε = 
Situation when S is closed

C eq. = C
Charge on capacitor C = C ε
After S is closed, voltage across 2C capacitor become zero, so charge on it also become zero.
So charge flown through the battery when the switch S is closed = C ε –
= 
the charge flown through the switch S is
= C ε (A to B) = – C ε (B to A)
Work done by the battery = Q supplied V =
ε = 
change in energy stored in the capacitors
energy stored in 2 nd case – energy stored in 1 st case =
C eq2 V 2 –
C eq1 V 2 =
.
ε 2 = 
(e) Heat developed in the system = Work done by battery – change in energy in the capacitors
=
–
= 
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