Question

1) A light bulb takes in 30 of energy per second. It transfers 3j as use

energy. Calculate the efficiency.
second. It transfers 3j as useful light energy and 27J as heat energy. Calculate the efficiency

Answer 1

`\boxed{\text{10 \%}}`

Step-by-step explanation:

The formula for efficiency is

`\begin{array}{rcl}\text{Efficiency} & = & \frac{\text{useful energy out}}{\text{energy in}} * 100 \,\% \\\\\eta & = & \frac{w_{\text{out}}}{w_{\text{in}} } * 100 \,\%\\\end{array}`

Data:

Useful energy = 3 J

Energy input = 30 J

Calculation:

`\begin{array}{rcl}\eta & = & \frac{\text{3 J}}{\text{30 J}} * 100 \,\%\\\\\eta & = & 10 \, \%\\\end{array}\\\text{ The efficiency is }\boxed{\textbf{10 \%}}`