Question

Question#7:a) A television uses 1500J of electric energy to produce 1100J 1 point

that's converted to sound and light. The TV has an efficiency of
- 73.3%
- 26.6%
- 136.3%
Question#7:b)In this TV,
% is wasted energy
- 73.3%
- 26.6%
- 36.3%​

Answer 1

a.73.3%

b.26.6%

Step-by-step explanation:

a. The efficiency is the energy output, divided by the energy input, and expressed as a percentage.

#Given the energy input as 1500J and the subsequent output as 1100J, the TV's efficiency is calculated as:

`\eta=(E_o)/(E_i)\\\\=(1100)/(1500)* 100\%\\\\=73.33\%`

Hence, the TV has an efficiency of 33.3%

b. From a above, we know that the TV has an output of 1100J and input of 1500J.

-We therefore calculate it's wastage as the difference between output and input:

Wastage=Energy Input-Energy Output=1500J-1400J=400J

`E_w=(E_(wasted))/(E_i)* 100\%\\\\=(400/1500)*100\%\\\\=26.67\%`

Hence, the TV has wasted of 26.6%