IIT JEE Advanced 2011 Paper 1

Let C1 = 2 μF and C2 = 8 μF. The charge stored in the capacitor before discharging is
Q = C1V = 2µC
E =

Q2

2C1

= 1 µJ
After closing switch 2, at study state, we have

Q−q

2µF

=

q

8µF

However, Q = 2 μC. Therefore,
1−

q

2µF

=

q

8µF

or q =

8

5

µC
Hence, the energy stored in capacitors C1 and C2 is
E' =

(Q−q)2

2C1

+

q2

2C2

=

[2−( 8 5 )]2

2×2

+

( 8 5 )2

2×8

= (

1

5

)µJ
Therefore, the percent of the energy dissipated is

E′−E

E

×100 =

( 1 5 )−1

1

× 100 = 80%