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An LCR circuit has L = 10 mL, R = 3 Omeg...

An `LCR` circuit has `L = 10 mL, R = 3 Omega`, and `C = 1 mu F` connected in series to a source of `15 cos omega t` volt. The current amplitude at a frequency that is 10% lower then the resonant frequency is

Text Solution

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The resonance frequency
`omega_(r)=sqrt(1/(LC))=sqrt(1/(10xx10^(-3)xx1xx10^(6)))`
`=10^(4)` rad/s
Thus `omega=0.90omega_(r)=9xx10^(3)` rad/s
`X_(L)=omegaL=9xx10^(3)xx(10xx10^(-3))`
`-90Omega`
and `X_(C)=1/(omegaC)=1/(9xx10^(3)xx10^(-6))`
`=111.11Omega`
The impenance `Z=sqrt(R^(2)+(X_(L)~X_(C))^(2))`
`=sqrt(3^(2)+(111.11-90^(2)))`
`=21.32Omega`
The current amplitude
`i_(0)=(V_(0))/Z=15/21.32=0.704A`
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