1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The capacity of a condenser A is 10μF and it is charged using a battery of 100 V. The battery is disconnected and the condenser A is connected to a condenser B with common potential as 40 V. The capacity of B is:

A
8μF
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
15μF
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2μF
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1μF
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 15μFCA=10μFVA=100VSo, charge qA=CAVA=10×10−6×100 =10−3When the battery is disconnected, total charge must be constant.So, qA+qB=10−3Given common potential ΔV=40 VSo, qA=CAΔV=10×10−6×40 V=400×10−6 V And, qB=CBΔV⇒10−3=400×10−6+CB×40or, CB=0.6×10−340=15×10−6 F=15 μF

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
PHYSICS
Watch in App
Join BYJU'S Learning Program