How is output power related to input power?
If a 100W cut takes 4x as long as 500W cut then a 100W cutter saves power but loses surgical time, but if it takes 5x as long then its the same amount of energy E=P*t.  Efficiency =Pout/Pin and is usually better for higher power equipment unless used at minimum levels. Talk to supplier for efficiency vs settings.

When not in use compare standby power when not cutting or measure it.


  I would go for the highest power model to save surgical time and use it sparingly and according to the surface area of the cut.


Although the cutter uses pulsed 380kHz or 480kHz at a rep rate of 33 kHz and possibly a much higher voltage to start ionization, I computed the rms voltage for each setting.

For a V source and rated load R [Ohms] and Power (Watts);
Vrms= √(P*R) from P=V²rms/R     

Battery capacity = Ah*V = Wh rated for 20h load.
Reduce capacity if discharge time is less.     

If R is open circuit, then Pd=0 except for internal idle power.
This can be tested with a DMM current meter.     

          Pd   R    Vrms    Mode


PURE  ....  400W 500Ω  447V    MONOPOLAR CUT
BLEND 1 320W 300Ω  310V
BLEND 2 240W 300Ω  268V
BLEND 3 160W 300Ω  219V      

FORCED  120W 500Ω  245V    MONOPOLAR COAG
SOFT  ...  120W 500Ω  245V
SPRAY .  100W 300Ω  173V           

STD  ....   100W 100Ω  100V    BIPOLAR
SOFT ....    80W 100Ω ,  89V
CUT ....    100W 200Ω  141V    

Of course these std load R values are dynamic and never constant.
tony.sunnysky@gmale   ;) EE since 1975