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Tardigrade
Question
Physics
An inductor of inductance 2 H and a resistance of 10 Ω are connected in series to an ac source of 110 V, 60 Hz. The current in the circuit will be
Q. An inductor of inductance
2
H
and a resistance of
10
Ω
are connected in series to an ac source of
110
V
,
60
Hz
. The current in the circuit will be
2745
209
Alternating Current
Report Error
A
0.32 A
0%
B
0.15 A
67%
C
0.48 A
0%
D
0.80 A
33%
Solution:
X
L
=
ω
L
=
2
π
∪
L
=
2
π
×
60
×
2
=
240
π
Ω
,
R
=
10Ω
Z
=
R
2
+
X
L
2
=
1
0
2
+
(
240
π
)
2
≈
240
π
I
r
m
s
=
Z
V
r
m
s
=
240
π
110
=
0.15
A