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JEE Main 26-Aug-2021 Shift 1 Solved Paper
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© examsnet.com
Question : 7
Total: 90
An inductor coil stores
64
J
of magnetic field energy and dissipates energy at the rate of
640
W
when a current of
8
A
is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds.
[26 Aug 2021 Shift 1]
0.4
0.8
0.125
0.2
Validate
Solution:
Magnetic energy is stored in an inductor, whereas energy isdissipated through resistor.
Given,
U
m
=
64
J
i = 8 A and
P = 640 W
As, magnetic energy,
U
m
=
1
2
Li
2
⇒
64
=
1
2
×
L
×
64
⇒ L = 2 H
Energy dissipated,
P
=
i
2
R
⇒ 640 64 = R
R = 10 Ω
Time constant of R-L circuit is given by
τ
=
L
R
⇒
τ
=
2
10
=
1
5
=
0.2
s
© examsnet.com
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