if capacitance for capacitor is 4
micro farad and potential is 1000 volt,
then find the energy stored in capacitor
8.0
4.0
6.0
2.0
Answers
Answered by
0
Explanation:
ANSWER
As Q=CV
∴ Initially charge on each capacitor is
Q
1
=C
1
V
1
=(3μF)(6V)=18μC and Q
2
=C
2
V
2
=(4μF)(6V)=24μC
When two capacitors are joined to each other such that negative plate of one is attached with the positive plate of the other. The charges Q
1
and Q
2
are redistributed till they attain the common potential which is given by
Common potential V=
Totalcapacitiance
Totalcharge
=
3μF+4μF
24μC−18μC
=
7
6
V
Final energy stored
U
f
=
2
1
(C
1
+C
2
)V
2
=
2
1
[3×10
−6
+4×10
−6
]×(
7
6
)
2
=
2
1
×7×10
−6
×
49
36
=2.57×10
−6
J
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