Consider the situation shown in figure. The switch S is open for a long time and then closed.
a) Find the charge flown through the battery when the switch S is closed.
b) Find the work done by the battery.
c) Find the change in energy stored in the capacitors.
d) Find the heat developed in the system.
Given circuit as shown below -
a) Charge flown through the battery when the switch S is closed. Since the switch was open for a long time, hence the charge flown must be due to the both.
When the switch is closed, the capacitor is in series, the equivalent capacitance is given by
Now, we know the relation between capacitance, charge q and voltage v given by ,
⇒ 
b) Work done by the battery
We know, work done is given by
Where q is charge stored and v is the applied voltage
We have, 
and 
Substituting the values, we get,
= 
c) Change in energy stored in the capacitors
Energy stored in capacitor is given by –
Initially, the energy stored in the capacitor is given by
1)
After closing the switch, the capacitance changes to
Energy stored after closing the switch is given by -
2)
From 1) and 2)
Change in energy stored in the capacitors
= 
= 
3)
d) Heat developed in the system
The net change in the stored energy is wasted as heat developed in the system,
Hence, heat developed in the systems is given as-
H=∆E
Where, H is the heat developed and ∆E is the change in the stored energy in the capacitor
From 3)
⇒ 
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