Final answer:
The hot resistance of a 25-W light bulb that runs on 120-V AC is 576 Ω.
Step-by-step explanation:
The hot resistance of a light bulb can be determined using Ohm's Law, which states that the resistance is equal to the voltage divided by the current. In this case, we are given the power (25 W) and voltage (120 V) of the light bulb. Since power is equal to voltage multiplied by current, we can rearrange the equation to solve for current, and then substitute the values to find resistance.
Using P = IV, we can find the current: I = P / V = 25 W / 120 V = 0.208 A.
Now, using Ohm's Law, we can find the resistance: R = V / I = 120 V / 0.208 A = 576 Ω.
Therefore, the hot resistance of the 25-W light bulb is 576 Ω.