OK, I understand now. But I'm not seeing a problem. The electrical measurement of voltage drop across the 1 ohm resistor leads one to calculate the current in the circuit, which of course will be the same everywhere, including the load. If you're seeing a problem, please be explicit.
Right. The current is correct.
The problem is that Rossi also uses the the voltage from the 1 ohm resistor as the entire circuit voltage, as well as for calculating the total circuit current.
What he is actually doing is calculating the power dissipated in the 1 ohm resistor. Then he uses that as the entire circuit power.