Question

A certain oscillating circuit consits of a capacitor with capacitance `C`, a coil with inductance `L` and active resistance `R`, and a switch . When the swith was disconnected, the capacitor was charged, then the switch was closed and oscillations set in. Find the ratio of the voltage across the capacitor to its peak value at the moment immediatel after closing the switch.

Answer 1

The equation of the circuit is
`L(d^(2)Q)/(dt^(2))+R(d Q)/( dt)+(Q)/(C)=0`
where `Q=` charge on the capacitor,
This has the solution `Q=Q_(m)e^(-betat) sin ( omegat+ alpha)`
where` beta=(R)/(2L), omega=sqrt(omega_(0)^(2)-beta^(2)), omega_(0)^(2)=(1)/(LC)`
Now`I=(dQ)/(DT)=0` at` t=0`
so , `Q_(m) e^(-betat)(- beta sin ( omegat+ alpha) + omegacos ( omegat+ alpha))=0 ` at `t=0`
Thus `omega cos alpha= beta sin alpha `or `alpha= tan^(-1) ( (omega)/( beta))`
Now `V_(m) =(Q_(m))/(C)` and ` V_(0)=P.D.`at `=0=(Q_(m))/( C) sin alpha`
`:. (V_(0))/(V_(m))= sin alpha=(omega)/(sqrt(omega^(2)+ beta^(2)))=(omega)/( omega_(0))=sqrt(1-beta^(2)//omega_(0)^(2))=sqrt(1-(R^(2)C)/( 4 L^(2)))`