Saved Bookmarks
| 1. |
Give analytical solution to a series RLC circuit connected to ac. |
Answer» Solution : The voltage equation for the CIRCUIT is `L(di)/(dt)+Ri+(q)/(c)=v=v_(m)SINOMEGAT` The solution to the equation will be `q=q_(m)SIN(omegat+theta)` `therefore q_(m)OMEGA[Rcos(omegat+theta)+(X_(C)-X_(L))sin(omegat+theta)]=v_(m)sinomegat` where, `X_(C)=(1)/(Comega)andX_(L)=Lomega` Also, `q_(m)omegaZ[(R)/(Z)cos(omegat+theta)+(X_(C)-X_(L))/(Z)sin(omegat+theta)]=v_(m)sinomegat`  Put, `(R)/(Z)=cosphiand((X_(C)-X_(L))/(Z))=sinphi""...(1)` Hence, `q_(m)omegaZ(cos(omegat+theta-phi))=v_(m)sinomegat` i.e., `[q_(m)omega(cos(omegat+theta-phi)]Z=v_(m)sinomegat` If `theta-phi=-pi//2`, then `cos(omegat+theta-phi)=cos(pi//2-omegat)=sin(omegat)` `therefore v_(m)=q_(m)omegaZ,i_(m)=q_(m)omegaandZ=(v_(m))/(i_(m))` Since, `i=(dq)/(dt)=(d)/(dt)(q_(m)sin(omegat+theta)=q_(m)omegacos(omegat+theta)` i.e., `i=i_(m)cos(omegat-(pi)/(2)+phi)=i_(m)cos((pi)/(2)-(omegat+phi))` i.e., `i=i_(m)sin(omegat+phi)` By using (1), `tanphi=(X_(C)-X_(L))/(Z)` graphically we represent it as in fig (i) and fig (ii) Hence, `v^(2)=v_(R)^(2)+(V_(C)-V_(L))^(2)` using fig (ii) or `Z=((v_(m))/(i_(m)))=sqrt(R^(2)+(X_(C)-X_(L))^(2))` 
|
|