# 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:
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Discussion:
30 comments Page 1 of 3.

Efficiency=output/input

=. 6\.7
=. 857

% efficiency =. 857*100
= 85.7%

@Ankita Very nice.

Bharat Prajapati said:   1 decade ago
Ankita is Right

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.

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

Efficiency =o/p/i/p=0.6*100/0.7=8.71

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

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

Efficiency = Output/Input.

=0.6/0.7 = 0.857.

In Percentage, 0.857*100 = 85.7%.