# 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 2 of 3.

Alpesh kachhadiya said:   8 years ago
Efficiency = Output/Input.
So, E =.6/.7*100.
= 85.7%.

MUBIANA said:   9 years ago
Efficiency = (output/input)*100.

Efficiency = (0.6/0.7)*100.

= 85.7%.

Jomel voltboy said:   9 years ago
E = Output/Input*100.
E = 0.6/0.7*100.
E = 85.7%

Baidya Nath Mohanta said:   10 years ago
Efficiency = OP/IP *100.
= (0.6/0.7)*100.
= 85.7.

Hemalatha said:   10 years ago
@Hemalatha absolutely right the answer is 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.

E = O/I.
= 0.6/0.7.
= 0.857.

E OF 100 % = 0.857*100.
= 85.7%.

Bhatruhari swain said:   1 decade ago
Efficiency = output/input.

= .6\.7.
= .857.

% efficiency = .857*100.
= 85.7%.