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VITEEE 2016 Solved Paper
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© examsnet.com
Question : 4
Total: 125
In given circuit
C
1
=
C
2
=
C
3
= C initially Now, a dielectric slab of dielectric constant K =
3
2
is inserted in
C
2
The equivalent capacitance become
5
C
7
7
C
5
2
C
3
C
2
Validate
Solution:
When a dielectric slab of dielectric constant
K =
3
2
is inserted between the plates of
C
2
, its new capacitance
(
C
2
′
)
become
C
2
′
=
3
2
C
Equivalent capacitance of
C
2
′
and
C
3
is
C
e
q
=
C
2
′
+
C
3
=
3
2
C
+ C =
5
C
2
Now,
C
e
q
and
C
1
are in series. Therefore, their equivalent capacitance is
C
e
q
=
C
e
q
×
C
1
C
e
q
+
C
1
=
2
5
C
2
×
C
5
C
2
+
C
=
5
C
2
7
C
=
5
C
7
© examsnet.com
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