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Standard XII
Physics
Energy Stored in an Inductor
Energy stored...
Question
Energy stored in a coil of self – inductance 40mH carrying a steady current of 2 A is
A
0.8 J
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B
8 J
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C
0.08 J
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D
80 J
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Solution
The correct option is
C
0.08 J
U
=
1
2
L
i
2
=
1
2
40
×
10
−
3
×
(
2
)
2
=
80
×
10
−
3
U
=
0.08
J
Suggest Corrections
0
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