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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 4RLet ω0 be the resonance frequencyminimum impedance is at ω0Zmin=XR=RAt ω=2ω0, Z=√10Zmin√R2+(ωL−1ωC)2=√10×RR2+(2ω0L−12ω0×C)2=10×R2because at ω0 , ω0×L=1ω0C , R2+(2ω0L−ω0L2)2=10×R2(32ω0L)2=9×R232×ω0L=3×RXL=ωL=2ω0L=4R (at ω=2ω0)

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