A tank circuit uses a 0.09 uH inductor and a 0.4-uF capacitor. The resistance of the inductor is 0.3 ohms. Would the quality of the inductor be 158 and the bandwidth be 5.3?

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The quality factor Q of a component is by DEFINITION the rapport between the power stored and the power loss in that particular component at resonance. Higher the quality factor is, higher the amplitude of oscillations are.

Q = Energy Stored/Energy Loss = Power Stored/Power Loss =Ps/Pl

at the resonant frequency of

`F_r = 1/(2*pi*sqrt(L*C)) =1/(2*pi*sqrt(9*10^(-8)*4*10^(-7)))=`

`=838.82 KHz`

If we write the quality factor as

`Q = F_r/(Delta(F))`

we can find the bandwidth of the circuit as

`Delta(F) = F_r/Q`

For a SERIES circuit (which is the case here because the  inductor resistance is always considered in series with the inductance) the current I through all components is the same and the stored an loss power are

`Ps = I^2*X(L) = I^2*omega*L`

`Pl = I^2*R`

which gives

`Q = (omega_r*L)/R = (2*pi*Fr*L)/R = (2*pi*838820*9*10^(-8))/0.3= 1.5811`

The bandwidth of the circuit is

`Delta(F) = F_r/Q =838820/1.5811=530516 Hz = 530.5 KHz`

For the circuit to have a quality factor Q =150 the values of the components need to be

L =0.09 miliH (not microH)

C =0.4 microF

but in this case the bandwidth will be `Delta(F) = 53 Hz`

Approved by eNotes Editorial Team

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