Saved Bookmarks
| 1. |
Find the quality factorof a circuit with capacitance C=2.0 mu F and inductance L=5.0 mH if the maintenance of undamped oscillations in the circuit with the voltage amplitude across the capacitor being equal to V_(m)=1.0 Vrequires a power ( :P: )=0.10m W.The damping of oscillations is sufficiently low |
|
Answer» <P> Solution :Given `V=V_(m) E^(-BETAT)sin omegat, omega~= omega_(0) betaTlt lt 1`Power loss `=(" Energy loss per cylcle")/(T)` `~=(1)/(2) CV_(m)^(2)xx2 beta` `(` energy decreasesas `W_(0) e^(-2betat) ` so loss per cycle is `W_(0)xx2 betaT)` Thus `lt P gt =(1)/(2)CV_(m)^(2)xx(R)/(L)` or `R=(2lt P gt)/( V_(m)^(2))(L)/(C)` HENCE`Q=(1)/(R) sqrt((L)/(C))=sqrt((C)/(L))(V_(m)^(2))/(2lt P gt)=100`on putting the vales. |
|