2 ms·
Need to multiply by the (average) voltage, and incorporate a fudge factor for the other components of the cell, I think. (P = UI, where P is power in Watt, U v
by FabHK 5y ago
Need to multiply by the (average) voltage, and incorporate a fudge factor for the other components of the cell, I think.
(P = UI, where P is power in Watt, U voltage in Joule per Coulomb, I current in Coulomb per second. And what we seek is specific energy, ie energy per mass, in J/kg or Wh/kg.)
- BenjiWiebe 5y agoWatts = Volts * Amps Any reason you've defined it as: (Amps * Seconds * Volts) / (Amps * Seconds) * (Amps * Seconds * Seconds) ? Fairly sure but not positive that I've reduced the units to base SI units correctly...
- FabHK 5y agoha, you're right, the Ampere is the SI base unit for purposes of definition, but the Coulomb is arguably the conceptually simpler one... (though Volt is definitely derived). So, in terms of "intuition" (left) and SI base units (right) you have: Watt = J s^-1 = N m s^-1 = kg m^2 s^-3 Volt = J C^-1 = N m A^-1 s^-1 = kg m^2 s^-3 A^-1 Ampere = C s^-1 = A