Tim said:

True, but the next harmonic is the 3rd, easily blocked by inductance above

resonance.

Okay, if you are only interested in the fundamental, lets see what we

know.

The voltage across the capacitor is identical to the voltage across

the resistor: Vr=Vc

The current through the capacitor is at right angles to the current

through the resistor, and the vector sum of those two must equal the

inductor current, since it is in series.

By Pythagorous: Il^2=Ic^2+Ir^2

The vector sum of either the resistor voltage or the identical

capacitor voltage and the inductor voltage is equal to the source

voltage. Vs=Vr+Vl or Vs=Vc+Vl

The inductor current and capacitor currents are at right angles to

their voltages (in opposite directions).

Since you are assuming resonance, the magnitude of the ratio of

inductor voltage divided by inductor current equals the magnitude of

the capacitor voltage divided by the capacitor current:

|Vl/Il|=|Vc/Ic|

or |j*(2*pi*f*L)|=|-j/(2*pi*f*C)|

or 2*pi*f*L=1/(2*pi*f*C

Unfortunately, I am too out of practice to combine these facts in a

phasor diagram so all I can suggest is a pair of loop equations solved

for the two complex currents, Ia and Ib.

Vs=Ia*[(j*2*pi*L)+R] -Ib*R

0=Ib*[R-j/(2*pi*f*C)] -Ia*R

and at resonance, I assume 2*pi*f*L=1/(2*pi*f*C)

Ztotal = Vs/Ia

But then, Ztotal = j*2*pi*f*L + 1/[(1/R) + (2*pi*f*C)/(-j)]

However, I would just plug the component values into LTspice and see

what current the simulation passes per volt of source.