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

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?

• Maybe he meant an LC filter, though the terms "minimum" and "cutoff" do confuse things. – SteveSh Dec 5 '19 at 15:00

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

$$Z=\sqrt2 R$$

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.

$$\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,