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Test Index
JEE Main Physics Class 12 Alternating Currents Part 1 Questions
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© examsnet.com
Question : 59
Total: 100
A sinusoidal voltage of peak value
283
V
and angular frequency
320
∕
s
is applied to a series
L
C
R
circuit. Given that
R
=
5
Ω
,
L
=
25
m
H
and
C
=
1000
µ
F
. The total impedance, and phase difference between the voltage across the source and the current will respectively be :
[Online April 9, 2017]
10
Ω
and
tan
−
1
(
5
3
)
7
Ω
and
45
°
10
Ω
and
tan
−
1
(
8
3
)
7
Ω
and
tan
−
1
(
5
3
)
Validate
Solution:
Given,
V
0
=
283
volt,
ω
=
320
,
R
=
5
Ω
,
L
=
25
m
H
,
C
=
1000
µ
F
x
L
=
ω
L
=
320
×
25
×
10
−
3
=
8
Ω
x
C
=
1
ω
C
=
1
320
×
1000
×
10
−
6
=
3.1
Ω
Total impedance of the circuit:
Z
=
√
R
2
+
(
X
L
−
X
C
)
2
=
√
25
+
(
4.9
)
2
=
7
Ω
Phase difference between the voltage and current
tan
ϕ
=
X
L
−
X
C
R
tan
ϕ
=
4.9
5
≈
1
⇒
ϕ
=
45
°
© examsnet.com
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