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InterviewSolution
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Obtain on expression for impedance and current in series LCR circuit. Deduce an expression for the resonating frequency of an LCR series reasonating circuit. |
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Answer» Solution :Circuit consists of resistor of RESISTANCE R, inductor of inductance L and capacitor of Let i be the current and q be the charge at any instant t. Back e.m.f across inductor is `-L (di)/(dt)` and across capacitor is `(-q)/(C )` TOTAL e.m.f `= V_(m)sin omega t - L (di)/(dt)-(q)/(C )`  According to Ohms law, this must be equal to iR `V_(m)sin omega t-L(di)/(dt)-(q)/(C )=iR` `L(di)/(dt)+iR+(q)/(C )=V_(m)sin omega t`.....(1) The current i any instant in the circuit is `i=i_(m)sin(omega t - phi)` if `omega L ge (1)/(omega C)` which is POSSIBLE at HIGH frequencies. `i=i_(m)sin(omega t+phi)` if `omega L lt(1)/(omega C)` which is possible at low frequencies. The maximum current `(i_(m))` is given by `i_(m)=(V_(m))/(sqrt(R^(2)+((1)/(omega c)-omega L)^(2)))` ....(2) Its impedance `(Z)=(V_(m))/(i_(m))=sqrt(R^(2)+((1)/(omega C)-omega L)^(2))` ....(3) Let `phi` be the phase difference between current and e.m.f `phi = tan^(-1)(((1)/(omega C)-omega L)/(R ))` ....(4) RESONANT freuency `(f_(0))` : At this frequency, the impedance of LCR circuit is minimum and is equal to R. At this frequency current is maximum. This frequency is called resonant frequency `(f_(0))`. At resonance `omega L = (1)/(omega C)` `omega^(2)=(1)/(LC)` (or) `omega = (1)/(sqrt(LC))` `2pi f_(0)=(1)/(sqrt(LC))` `f_(0)=(1)/(2pi sqrt(LC))`  In frequency response curve, at resonant frequency `(f_(0))`, current is maximum. This series resonant circuit is called acceptor circuit. |
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