When the bulb is always marked with "220 V 40 W", the current passing through the filament is - A, one degree of electric energy for the lamp to work normally for - hours There must be a way to solve the problem!

When the bulb is always marked with "220 V 40 W", the current passing through the filament is - A, one degree of electric energy for the lamp to work normally for - hours There must be a way to solve the problem!

I = P / u, current = 40 / 220 = 0.1818, approximately equal to 0.182a
One degree of electricity is 1000 W. then 1000 / 40 = 25 hours