Question

A lightbulb has a resistance of 195 Ω and carries a current of 0.62 A.

The power rating of the lightbulb, to the nearest whole number, is
______ W.

Answer 1

75

Step-by-step explanation:

Power is current times voltage:

P = IV

Voltage is current times resistance:

V = IR

Therefore:

P = I²R

Given I = 0.62 A and R = 195 Ω:

P = (0.62 A)² (195 Ω)

P ≈ 75 W