Consider the parallel resonant circuit shown in adjacent figure. One branch contains an inductor of inductance L and a small ohmic resistance R, whereas the other branch contains a capacitor of capacitance C. The circuit is fed by a source of alternating emfE= E_(o) e^(jomegat)=E_(o)sin omegatThe impedance of inductor branch, Z (1)= R + jomegaL The impendence of capacitor branch, Z(2) = (1//jomegaC) therefour Net impedance Z of the two parallel branches is given by(1)/(Z)=(1)/(Z_(1))+ (1)/)Z_(2)=(1)/(R+jomegaC=(R-jomegaL)/(jomegaL)(R-jomegaL)+jomegaC=(R)/(R^(2)+omega^(2)+L^(2))+jomega[C-(L)/(R^(2)+omegaL^(2))]The current flowing in the circuitI=(E)/(Z)=(E)[(R)/(R^(2)+omega^(2)L^(2))+jomega(C-(L)/(R^(2)+omega^(2)L^(2)))]For resonance to occur, the current must be in phase with the applied emf. For this, the reactive component of current should be zero, l.e.omega[C-(L)/(R^(2)+omega^(2)L^(2))]=0orC=(L)/(R^(2)+omega_(r)^(2)L^(2))(writing omega_(r)for omegaat resonance) This gives resonant angular frequencyomega_(r)=sqrt((1)/(LC)-R^(2)/(L^(2))) At parallel circuit resonance, theimpendence is maximum and current is minimum. Parallel resonant circuit is sometimes called the anti-resonance in order to distinguish from series resonanceQFind the impedance of AC circuit at resonance shown in the adjacent figure Find the impedance of AC circuit at resonance shown in the adjacent figure