wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600pF capacitor. What is the common potential (in V) and energy lost (in J) after reconnection?

A
100,6×106
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
200,6×105
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
200,5×106
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
100,6×105
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 100,6×106
Total charge remains constant.
So, q1=q2
Common potential, VC=C1V1C1+C2=100V
Energy lost =UiUf=12C1V1212(C1+C2)Vc2 =6×106J

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Series RLC
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon