How to measure input power with an AC source and tank circuit?

[PhysicsProf]

   I've been playing around lately with AC sources, including the grid and my own AC generators. 

   But I'm having some trouble convincing myself that I'm measuring the input power correctly.  Let me pose a few questions:

  An AC source (say the grid) starts an LC tank circuit going and stays connected. The LC circuit resonant frequency is the same as the frequency of the AC source.

  A.  Assuming small resistances throughout (and low transmitted power), there will be very little input power required.  Sure, there will be currents sloshing back and forth, but not much power consumed.  So we put in a small CSR to measure the current-flow from the AC source (call it Iin), and measure the voltage (Vin) from the AC source.  Does Vin * Iin accurately tell us the input power?  If not, how can one accurately measure Pin for this AC circuit?

  B.  Now keep resistances in the circuit low, but let the circuit transmit power to a near-by matched tank-circuit (with significant load R in this pick-up circuit).  Now does Vin * Iin (as defined in A) accurately tell us the input power?  If not, how can one accurately measure Pin for this AC circuit?

   C.  For the set-up in B, can we determine Pout from the voltage drop across R, that is,
Pout = V^2 / R ??

[Slider2732]

It may be an idea to have a commercial solution do the measuring too, as a verification ?
Am thinking Wattmeter, that is highly accurate.
You may already have a good one, but this thing looks to be a featured device for $20. It has the bonus ability of being swiftly kicked back if it doesn't work, via Amazon Prime !
https://www.amazon.com/dp/B078JJBW3P/ref=sspa_dk_detail_3?psc=1&pd_rd_i=B078JJBW3P&pd_rd_wg=YLwY5&pd_rd_r=Y9Z2ZXB6P3DB6EH6HE3A&pd_rd_w=1Ekw2

What stands out is this section of the description:

Current measurement range: 0.001 A to 15 A
Power measurement range: 0.23 W–1800 W
Frequency display: 46–65 Hz
Resolving power: 0.001 W

If your circuit does draw 1/4W, it should show the measurements that would tally with the CSR and other testing methods. That 1/1000W resolving power figure looks to be ideal. 
Also, if the circuit might not draw 1/4W, a lamp with constant draw could also plug into it and then any change from the DUT would be noted on the meter. If your circuit flipped into OU then the numbers would go down (just thinking aloud).

[Smudge]

The average value of Vin*Iin taken over an exact full cycle (or a number of full cycles) should give you the input power.  Use a scope where you can set cursors to extract exact cycles.  Don't use meters.
Smudge

[PhysicsProf]

  Thanks for responses - so far in my experiments, the currents are too small for that meter, Mark.  I do have "watt-meters", different makers - the kind you plug in, then plug your DUT into that.  But my currents are too small, at present anyway.

   Smudge - let me come back to this tank circuit and see whether we agree.  With DC, its easy to visualize Vin * Iin, but with AC, the current sloshes back and forth, not so easy to see.
  An AC source (say the grid) starts an LC tank circuit going and stays connected. The LC circuit resonant frequency is the same as the frequency of the AC source.

  A.  Assuming small resistances throughout (and low transmitted power), there will be very little input power required.  Sure, there will be currents sloshing back and forth, but not much power consumed.  So we put in a small CSR to measure the current-flow from the AC source (call it Iin), and measure the voltage (Vin) from the AC source.  Does Vin * Iin accurately tell us the input power?

I'm not interested in the power consumed by the CSResistor - but rather in the power consumed IN the tank circuit - that's what I want to measure.  But how?

[Smudge]

Current sloshing back and forth is not power sloshing back and forth, so forget the sloshing.  Just think positive and negative, same thing with voltage.  And the right combination of positive or negative voltage with positive or negative current tells you the direction and value of the instantaneous power flow, and that can flow in either direction.  And I stand by my simple explanation of using the average value of V and I taken over a full cycle which should tell you the input power under any sort of load, inductive, capacitive, resonant, whatever.  Luckily modern scopes do the sampling and do the math for you.
Smudge