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JEE Main Physics Class 12 Alternating Currents Part 2 Questions
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© examsnet.com
Question : 57
Total: 75
An alternating emf
E
=
440
sin
100
π
t
is applied to a circuit containing an inductance of
√
2
π
H
. If an a.c. ammeter is connected in the circuit, its reading will be :
[29-Jul-2022-Shift-1]
4.4
A
1.55
A
2.2
A
3.11
A
Validate
Solution:
E
=
440
Sin
100
π
t
,
L
=
√
2
π
H
X
L
=
ω
L
=
100
π
√
2
π
=
100
√
2
Ω
Peak current
I
0
=
E
0
X
L
=
440
100
√
2
=
2.2
√
2
A
AC ammeter reads
RMS
value therefore reading will be
I
rms
I
rms
=
I
0
√
2
=
2.2
A
© examsnet.com
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