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JEE Main Physics Class 12 Electrostatic Potential and Capacitance Part 3 Questions
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Section:
Physics
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© examsnet.com
Question : 2
Total: 36
A capacitor of capacitance
100
µ
F
is charged to a potential of
12
V
and connected to a
6.4
mH
inductor to produce oscillations. The maximum current in the circuit would be :
[29-Jan-2024 Shift 1]
3.2
A
1.5
A
2.0
A
1.2
A
Validate
Solution:
👈: Video Solution
By energy conservation
1
2
CV
2
=
1
2
LI
max
2
I
max
=
√
C
L
V
=
√
100
×
10
−
6
6.4
×
10
−
3
×
12
=
12
8
=
3
2
=
1.5
A
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