# Electrical Engineering - Energy and Power - Discussion

Discussion Forum : Energy and Power - General Questions (Q.No. 5)
5.
A power supply produces a 0.6 W output with an input of 0.7 W. Its percentage of efficiency is
8.57%
42.85%
4.28%
85.7%
Explanation:
No answer description is available. Let's discuss.
Discussion:
30 comments Page 1 of 3.

AliBaBa said:   5 years ago
Why no one has manually solved this 0.6/0.7 to get 0.857.
06/10 Ã· 07/10.
06/10 * 10/07 cross multiplication will cancel both 10.
06/07 = 0.857,
For efficiency, %age multiply it with 100.
= 85.7%.
(1)

Efficiency = Output / Input
= 0.6 / 0.7
= 0.857.

Than in percentage efficiency = 0.857 * 100
= 85.7 %.

Rani said:   10 years ago
Option D was correct because the efficiency will be calculated by using formula (output/input).

So,
0.6/0.7 = 0.857.

Under % = 85.7%

E = output/input =output/output+losses.

So in the question they have to give that losses are neglected.

Efficiency(n) % = (Output Power / Input Power)*100
Efficiency(n) % = (0.6W / 0.7W)*100
Efficiency(n) % = 85.71

Dayu said:   5 years ago
eff=output power/input power * 100.
So output power is 0.6w and input power is 0.7 watt.
0.6/0.7 * 100 = 85.7%.
(4)

Efficiency(n)=output/input
Efficiency(n)=0.6W/0.7W (BOTH "w" CANCEL)
n=0.8571
n%=0.8571x100
n=85.71%

Adding to sandyachethu: ratio is always without units so thats why both 'w' would cancel eachother.

Ravi Purohit said:   1 decade ago
Because efficiency = output / input.

= 0.6 / 0.7 = 0.857 = 0.857*100 % = 85.7 %