Question

A lightbulb uses 98 J of electrical energy every second to produce 10 J of light energy in the same time frame. What is the percent efficiency of the lightbulb in converting electricity into light?

Answer 1

10.2 % efficiency.

Step-by-step explanation:

Percent efficiency formula: (output / input) * 100%

The output energy (in Joules) is how much was used eventually. - It is 10 J

The input energy (in Joules) is how much energy was put into the lightbulb initially. It is 98 J

(10 J / 98 J) * 100% = 10.2 % efficiency.