Roger N. answered 08/16/21
. BE in Civil Engineering . Senior Structural/Civil Engineer
Solution:
The power in watt P = I2R = (IR)I = VI where V is the voltage in Volts (V) and I is the intensity of the current in Amps(A)
Note That 1W = 1V . 1A , then the intensity of the current is I = P/V = 40 W / 110 V = 0.36 A
for a 220 Volts source outlet , the power in the bulb is P = VI = ( 220 V)(0.36 A) = 79.2 W