Obtain an expression for quality factor of series RLC circuit connected to ac source.

Explanation

We know that for maximum current
`i_(m)=(v_(m))/(sqrt(R^(2)+(omega_(L)-(1)/(omega_(C)))^(2)))" "...(1)`
From the current frequency curve, the band width `omega_(1)-omega_(2)=2Deltaomega`. The quantity `((omega_(0))/(2Deltaomega))` is regarded as a measure of the sharpness of resonance.
Hence, `omega_(1)=omega_(o)+Deltaomegaandomega_(2)=omega_(0)-Deltaomega`
i.e., `i_(rms)=(i_(m))/(sqrt(2))=(v_(m))/(sqrt(2))=(v_(m))/(sqrt(R^(2)+(omega_(1)L-(1)/(omega_(1)C))^(2)))" "...(2)`
we note that `omega_(1)andomega_(2)` are obtained by drawing perpendicular to the frequency axis corresponding to the `(1)/(sqrt(2))i_(m)`
Equation (2) may be simplified as
`Rsqrt(2)=sqrt(R^(2)+(Comega_(1)L=(1)/(omega_(2)C))^(2))`. Squaring both side,
`2R^(2)=R^(2)+(omega_(1)L-(1)/(omega_(1)C))^(2)`
or `R=omega_(1)L-(1)/(omega_(1)C)`
i.e., `R=(omega_(o)+Deltaomega)L-(1)/((omega_(0)+Deltaomega)C)`
i.e., `R=omega_(0)L(1+(Deltaomega)/(omega_(0)))-(1)/(omega_(0)C(1+(Deltaomega)/(omega_(0))))`
but `omega_(0)^(2)=(1)/(LC)oromega_(0)L=(1)/(omega_(0)C)`
Hence, `R=omega_(0)L(1+(Deltaomega)/(omega_(0)))-omega_(0)L(1-(Deltaomega)/(omega_(0)))`
Simplifying we get,
`((omega_(0)L)/(R))=((1)/(2Deltaomega))omega_(0)`
The ratio `(omega_(0)L)/(R)` is called the quality factor 'Q' of the circuit.
`Q=(omega_(0)L)/(R)`
i.e., `Q=(1)/(R)sqrt((L)/(C))=(1)/(omega_(0)RC)`