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JEE Main Physics Class 12 Electrostatic Potential and Capacitance Part 1 Questions
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Question : 1
Total: 100
Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :
[Sep. 05, 2020 (I)]
25
6
C
V
2
3
2
C
V
2
zero
9
2
C
V
2
Validate
Solution:
When capacitors C and 2C capacitance are charged to V and 2V respectively.
Q
1
=
C
V
Q
2
=
2
C
×
2
V
=
4
C
V
When connected in parallel
By conservation of charge
4
C
V
−
C
V
=
(
C
+
2
C
)
V
common
V
common
=
3
C
V
3
C
=
V
Therefore final energy of this configuration,
U
f
=
(
1
2
C
V
2
+
1
2
×
2
C
V
2
)
=
3
2
C
V
2
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