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JEE Main Physics Class 12 Alternating Currents Part 1 Questions
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© examsnet.com
Question : 22
Total: 100
In a series
L
−
C
−
R
resonant circuit, the quality factor is measured as 100 . If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be ......... .
[26 Feb 2021 Shift 1]
Your Answer:
Validate
Solution:
Given, initial quality factor
(
Q
i
)
=
100
Let initial inductance
(
x
L
i
)
=
x
Final inductance
(
x
L
f
)
=
2
x
and initial resistance
(
R
i
)
=
R
Final resistance
(
R
f
)
=
R
2
Final quality factor
=
Q
f
Since,
Q
i
=
X
L
R
and
Q
f
=
2
X
L
R
∕
2
⇒
Q
f
=
4
X
L
R
=
4
Q
i
=
4
×
100
Hence, final quality factor will be 400 .
© examsnet.com
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