Question

A lightbulb's output is 60 W when it operates at 120 V.

If the voltage is cut in half and the power dropped to 15 W during a brownout, what is the ratio of the bulb's resistance at full power to its resistance during the brownout?

Express your answer using two significant figures.

Answer 1

Here ,

for the 120 V

let the resistance is R

as Power = V^2/R

60 = 120^2/R

R = 240 Ohm

Now, for the 60 V

Power = V^2/R1

15 = 60^2/R1

R1 = 240 Ohm

ratio of bulb's resistance = 240/240

ratio of bulb's resistance = 1

the ratio of the bulb's resistance at full power to its resistance during the brownout is 1