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 emf`E= E_(o) e^(jomegat)=E_(o) sin omegat`
The 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 circuit
`I=(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))]=0`
or`C=(L)/(R^(2)+omega_(r)^(2)L^(2))`(writing `omega_(r)`for `omega`at resonance)
This gives resonant angular frequency
`omega_(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 resonance
QFind the impedance of AC circuit at resonance shown in the adjacent figure

A coil of inductance 100 muH and 60 Omega effective resistance is connected in parallel with a condenser of 100 pF. The resonant circuit so formed is excited from an alternating current source. The frequency of the source is varied until the maximum impendence of the circuit is obtained. The frequency of the generator at which this occurs is approximately

Options

A. `(10^(2)/(2pi) Hz`
B. `(10^(7)/(2pi)Hz`
C. `(10^(5)/(2pi) HZ`
D. `(10^(3)/(2pi)Hz`

Answer

B