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Byju's Answer
Standard XII
Physics
EMF and EMF Devices
Two capacitor...
Question
Two capacitors,
C
1
=
2
μ
F
and
C
2
=
8
μ
F
are connected in series across a
300
V
source. Then
A
the charge on each capacitor is
4.8
×
10
−
4
C
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B
the potential difference across
C
1
is
60
V
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C
the potential difference across
C
2
is
240
V
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D
the energy stored in the system is
5.2
×
10
−
2
J
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Solution
The correct option is
A
the charge on each capacitor is
4.8
×
10
−
4
C
The equivalent capacitance is
C
e
q
=
C
1
C
2
C
1
+
C
2
=
2
×
8
2
+
8
=
1.6
μ
F
.
Equivalent charge is
Q
e
q
=
C
e
q
V
=
1.6
×
10
−
6
×
300
=
4.8
×
10
−
4
C
As the capacitors are in series so charge on each capacitor is
Q
e
q
=
4.8
×
10
−
4
C
.
The potential difference across
C
1
is
V
1
=
Q
e
q
C
1
=
4.8
×
10
−
4
2
×
10
−
6
=
240
V
The potential difference across
C
2
is
V
2
=
Q
e
q
C
2
=
4.8
×
10
−
4
8
×
10
−
6
=
60
V
Energy stored in the system is
U
=
1
2
C
e
q
V
2
=
1
2
×
1.6
×
10
−
6
×
(
300
)
2
=
7.2
×
10
−
2
J
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