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-w2LC)+jwRC. I multiply the top and bottom by the complex conjugate and get: Z=[R - j(wR2C-wL+w3LC)]/[(1-w2LC)+(wRC)2] Since the phase at resonance is 0, I can equate the imaginary part to 0 and solve for w|w0: [w0R2C-w0L+w03LC/]/[(1-w02LC)+(w0RC)2] = 0 => w0=sqrt(1/LC -R2/L2). So, here's my question....according to this equation, if R=0, we get the standard, w0=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 - R2/L2 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 w0, but I"m not sure what that means.....no resonance at any value of R that results in w0 being imaginary?