Given that ;
Energy (work ) Produced by bulb =20 KJ = 20000 J
Input power = 15 KW
Output power given by the bulb = Rate of doing work
= Work ÷ time
= 20000 ÷ 1800 <em>(since 1 min = 60 seconds)</em>
= 11.11 W
Noe, efficiency is the ratio of output power to input power
η = output power ÷ input power
= 11.11 ÷ 15
= 0.74
= 74%
<em>Efficiency of light bulb is 74%</em>
<em>If you liked the procedure, please give me brainliest</em>
