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## Main Question or Discussion Point

Most people in electronics know that the resonant frequency for a parallel LC network is given by w=1/sqrt(LC). But, few realize that when there is a resistor in series with the inductor, the resonant frequency is shifted downward.

When I solve the equation for the resulting impedance, I get:

Z=(R+jwL)/[(1-w

Z=[R - j(wR

Since the phase at resonance is 0, I can equate the imaginary part to 0 and solve for w|

[w

=> w

So, here's my question....according to this equation, if R=0, we get the standard, w

So, what would it mean physically, if R > than this limit allows? What should the expected behavior of the circuit be?

I suspect if I plug this negative frequency into the original equation for the total impedance I'd just get a different

When I solve the equation for the resulting impedance, I get:

Z=(R+jwL)/[(1-w

^{2}LC)+jwRC. I multiply the top and bottom by the complex conjugate and get:Z=[R - j(wR

^{2}C-wL+w^{3}LC)]/[(1-w^{2}LC)+(wRC)^{2}]Since the phase at resonance is 0, I can equate the imaginary part to 0 and solve for w|

_{w0}:[w

_{0}R^{2}C-w_{0}L+w_{0}^{3}LC/]/[(1-w_{0}^{2}LC)+(w_{0}RC)^{2}] = 0=> w

_{0}=sqrt(1/LC -R^{2}/L^{2}).So, here's my question....according to this equation, if R=0, we get the standard, w

_{0}=sqrt(1/LC). √. For any R, the resonant frequency will be shifted to a lower frequency. However, this is only true as long as the a 1/LC - R^{2}/L^{2}is ≥0, which only occurs for R ≤ sqrt(L/C), since frequency must be real.So, what would it mean physically, if R > than this limit allows? What should the expected behavior of the circuit be?

I suspect if I plug this negative frequency into the original equation for the total impedance I'd just get a different

**complex impedance**at that frequency, instead of the**resistance**I'd get if I plugged in a**real**w that satisfies the equation for w_{0}, but I"m not sure what that means.....no resonance at any value of R that results in w_{0}being imaginary?