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JEE Advanced Model Paper 9 with solutions for online practice
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© examsnet.com
Question : 47
Total: 57
Parallel plate capacitor A of capacitance C is charged to a potential difference V.Another parallel plate capacitor B of capacitance 2C is charged to a potentialdifference. Charging battery is disconnected and two capacitors are connected inparallel such that positive terminal of A is connected to negative terminal of B. Theelectrostatic energy of the combination is now
6
C
V
2
. The potential difference towhich capacitor B is charged initially is x × 0.7 V . The value of ' x ' is _____.
Your Answer:
Validate
Solution:
E =
1
2
C
′
V
′
2
=
1
2
(3C)
V
′
2
= 6
C
V
2
(given)
So V' = 2V , But V' =
Q
′
C
′
=
Q
′
3
C
= 2V
Q' = 6CV
But Q' =
Q
2
−
Q
1
=
Q
2
- CV = 6CV
So,
Q
2
= 7CV
V
2
=
Q
2
C
2
=
7
C
V
2
C
= 3.5 V
⇒ 0.7 x = 3.5 ⇒ x = 5
© examsnet.com
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