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JEE Main 8 Apr 2024 Shift 2 Solved Paper
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© examsnet.com
Question : 45
Total: 90
A coil of negligible resistance is connected in series with
90
Ω
resistor across
120
V
,
60
Hz
supply. A voltmeter reads
36
V
across resistance. Inductance of the coil is :
[8 Apr 2024 Shift 2]
0.76
H
2.86
H
0.286
H
0.91
H
Validate
Solution:
👈: Video Solution
36
=
I
ms
R
36
=
120
√
X
L
2
+
R
2
×
R
R
=
90
Ω
⇒
36
=
120
×
90
√
X
L
2
+
90
2
√
X
L
2
+
90
2
=
300
X
L
2
=
81900
X
L
=
286.18
ω
L
=
286.18
L
=
286.18
376.8
L
=
0.76
H
© examsnet.com
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