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Question

An AC source of variable frequency is applied across a series L-C-R circuit. At a frequency double the resonance frequency, the impedance is 10 times the minimum impedance. The inductive reactance is

A
R
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B
2R
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C
3R
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D
4R
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Solution

The correct option is D 4R
Let ω0 be the resonance frequency
minimum impedance is at ω0

Zmin=XR=R

At ω=2ω0,

Z=10Zmin

R2+(ωL1ωC)2=10×R

R2+(2ω0L12ω0×C)2=10×R2

because at ω0 , ω0×L=1ω0C ,

R2+(2ω0Lω0L2)2=10×R2

(32ω0L)2=9×R2

32×ω0L=3×R

XL=ωL=2ω0L=4R (at ω=2ω0)

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