Saved Bookmarks
| 1. |
Obtain an expression for quality factor of series RLC circuit connected to ac source. |
|
Answer» Solution :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)` |
|