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Test Index
JEE Main Physics Class 12 Electrostatic Potential and Capacitance Part 2 Questions
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© examsnet.com
Question : 4
Total: 100
A capacitor of capacitance
900
µ
F
is charged by a
100
V
battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as
x
×
10
−
2
J
. The value of
x
is ______.
[30-Jan-2023 Shift 1]
Your Answer:
Validate
Solution:
C
=
900
µ
F
Q
=
CV
=
900
×
10
−
6
×
100
=
9
×
10
−
2
=
90
MC
Now
Common potential will be developed across both capacitors by
kVL
Total charge on left plates of capacitors should be conserved.
∴
90
mc
+
0
=
2
cv
0
cv
0
=
45
mc
Heat dissipated
=
U
i
−
U
f
[Change in energy stored in the capacitors]
=
1
2
(
90
mc
)
2
900
µ
F
−
2
×
1
2
(
45
mc
)
2
900
µ
F
[
U
=
Q
2
2
c
]
=
1
2
×
900
×
10
−
6
(
8100
−
4050
)
×
10
−
6
=
2.25
Joule
OR
Heat
=
1
2
C
1
C
2
C
1
+
C
2
(
V
1
−
V
2
)
2
=
1
2
C
2
2
C
(
100
−
0
)
2
=
1
2
900
×
10
−
6
2
×
10
4
=
9
4
Joule
=
2.25
Joule
© examsnet.com
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