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A series circuit consisting of a capacitor and a coil with active resistance is connected to a source of harmonic voltage whose frequency can be aried, keeping the voltage amplitdue are `n` times less than the resonance amplitude. Find `:` `(a)` the resonance frequency, `(b)` the quality factor of the circuit. |
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Answer» At resonance `omega_(0)L=(omega_(0)C)^(-1)` or `omega_(0)=(1)/(sqrt(LC))` `(I_(m))_(res)=(V_(m))/(R)`. Now `(V_(m))/( nR)=(V_(m))/sqrt(R^(2)+(omega_(1)L-(1)/(omega_(1)C))^(2))=(V_(m))/( sqrt(R^(2)+(omega_(2)L-(1)/( omega_(2)C))^(2)))` Then `omega_(1)L-(1)/( omega_(1)C)=sqrt(n^(2)-1)R` `omega_(2)L-(1)/( omega_(2)C)=sqrt(n^(2)-1)R` ( assuming `omega_(2) gt omega_(2))` or `omega_(1)-(omega_(0)^(2))/( omega_(1))=- omega_(2)+(omega_(0)^(2))/( omega_(2))=- sqrt (n^(2)-1 )(R)/(L)` or` omega_(1)+ omega_(2)=( omega_(0)^(2))/( omega_(1)omega_(2))(omega_(1)+ omega_(2))implies omega_(0)=sqrt( omega_(1) omega_(2))` and ` omega_(2) - omega_(1)= sqrt( n^(2)-1) (R)/(L)` `beta=(R)/(2L) = ( omega_(2)-omega_(1))/( 2 sqrt( n^(2)-1))` and `Q=sqrt( (omega_(0)^(2))/(4 beta^(2))-(1)/(4))=sqrt(((n^(2)-1)omega_(1)omega_(2))/( (omega_(2)-omega_(1))^(2))=(1)/(4))` |
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