First I've finished my 3rd year of electric(al) engineering and I'm currently in Japan on an intership.
Second you are completely wrong. Say Vp is peak voltage of a sinusoidal voltage, Ip the peak sinusoidal current, Cos(X) the power factor, X is the phase difference betwen current and voltage.
P= (1/2)VpIpCos(X)
or
P=VICos(X) where V is the RMS voltage, I the RMS current.
The current and voltage given on the power brick are RMS (peak voltage for North America is around 170V). Now for the power factor, power factor is directly related to efficiency so when I say the entire power supply has 70% efficiency I don't have to take it into account.
So for 213W DC, if the efficiency is 70% then input AC power at the wall is 300W. And since I suspect they used the North American standard for voltage 110V the 2.7A number fits perfectly with the 300W number. Even at a lower efficiency like 66% the input power is 324W which would point to an equally valid voltage of 120V.
Second you are completely wrong. Say Vp is peak voltage of a sinusoidal voltage, Ip the peak sinusoidal current, Cos(X) the power factor, X is the phase difference betwen current and voltage.
P= (1/2)VpIpCos(X)
or
P=VICos(X) where V is the RMS voltage, I the RMS current.
The current and voltage given on the power brick are RMS (peak voltage for North America is around 170V). Now for the power factor, power factor is directly related to efficiency so when I say the entire power supply has 70% efficiency I don't have to take it into account.
So for 213W DC, if the efficiency is 70% then input AC power at the wall is 300W. And since I suspect they used the North American standard for voltage 110V the 2.7A number fits perfectly with the 300W number. Even at a lower efficiency like 66% the input power is 324W which would point to an equally valid voltage of 120V.