EAMCET Engineering 2005 Solved Paper

Charge stored at the capacitor
q=C1V1=4×200=800µC
When this capacitor is connected with a uncharged capacitor, then common potential on both capacitors
V=‌

C1V1+C2V2

C1+C2

=‌

800+0

4+2

=‌

800

6

V
Loss in energy = Initial energy - Final energy
=‌‌

1

2

C1V12−‌

1

2

(C1+C2)V2
=‌‌

1

2

×4×10−6×(200)2
‌‌‌−‌

1

2

(4+2)×10−6×(‌

800

6

)2
=‌2×10−6×4×104−‌

3×10−6×64×104

36

=‌8×10−2−‌

64

12

×10−2
=‌8×10−2−5.33×10−2
=‌2.67×10−2J