Why is the impedance zero at cutoff frequencies in a series RLC resonance circuit?

Question

I see that, in many derivations in the frequency analysis of series RLC circuits, \$X_L - X_c = R\$ or \$Xc - X_L = R\$ is considered at lower and higher cutoff frequencies. So does that mean that the impedance \$Z = \sqrt{(X_L-Xc)^2 + R^2}\$ is zero at cutoff frequencies?

Answer 1

So does that mean that the impedance (Z = sqrt((Xl-Xc)^2 + R^2)) is zero at cutoff frequencies?

No, your math is wrong.

If \$|X_L - X_C| = R\$ at either of the 3 dB points (cut-off frequencies) then : -

$$Z=\sqrt{R^2 + R^2}$$

or

$$Z=\sqrt2 R$$

Answer 2

The impedance is only zero at the cutoff (resonant) frequency for ideal capacitors and inductors and when R=0. For real caps and inductors, and when R is not zero, the impedance at resonance may be the minimum, but is not zero.

Answer 3

\$Z_{load}= X_L + X_c + R~~~~~~~~ X_L=jωL,~~ Xc= 1/(jωC)= -j/ωC \$

\$Z_{load} = j(ωL-1/ωC) + R \$

\$Z = R ~~~@~ ω=ω_o = 1/\sqrt{LC}\$

\$Q=ω_oL/R \$

Thus at series resonance the LC reactance nulls to 0 and load reduces to R and attenuation rises to 0 (for ideal source), while current maximizes to I=V/R as shown below.

Sim

If you move the output between L & C , you see a low pass filter with same peaking factor or Q with voltage gain of 10 due to the impedance ratio of Xc/R at resonance,