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Test Index
JEE Main Physics Class 12 Alternating Currents Part 1 Questions
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© examsnet.com
Question : 86
Total: 100
An inductor (L = 100 mH), a resistor (R = 100 Ω) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the short circuit is
[2006]
1/eA
eA
0.1 A
1 A
Validate
Solution:
Initially, when steady state is achieved,
i
=
E
R
Let
E
is short circuited at
t
=
0
. Then
At
t
=
0
Maximum current,
i
0
=
E
R
=
100
100
=
1
A
Let during decay of current at any time the current flowing is
−
L
d
i
d
t
−
i
R
=
0
⇒
d
i
i
=
−
R
L
d
t
⇒
i
∫
i
0
d
i
i
=
i
∫
i
0
−
R
L
d
t
⇒
log
e
i
i
0
=
−
R
L
t
⇒
i
=
i
0
e
−
R
L
t
⇒
i
=
E
R
e
−
R
L
t
=
1
×
e
−
100
×
10
−
3
100
×
10
−
3
=
1
e
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