Electronics - Transformers - Discussion

Discussion Forum : Transformers - General Questions (Q.No. 2)
2.
A transformer is plugged into a 120 V rms source and has a primary current of 300 mA rms. The secondary is providing 18 V across a 10 load. What is the efficiency of the transformer?
88%
90%
92%
95%
Explanation:
No answer description is available. Let's discuss.
Discussion:
14 comments Page 1 of 2.

Ammu said:   7 years ago
Superb and helpful explanations.

V seenivasan said:   7 years ago
Fantastic explanation, Thank you all.

NAINA said:   9 years ago
Pin = v.i = 120*0.3 = 36.

Pout = v*(v%r) = 18*(18/10) = 32.4.

n% = op/inp = 32.4/36 = 90%.
(1)

Viyoo said:   9 years ago
Pin = 120*0.3=36W.
Pout = 18*18/10=32.4.

efficiency = Pin/Pout.
= (32.4/36)*100.
= 90%.

Sarika said:   10 years ago
Input power = 120*3*0.001 = 36.
Output current = v/r = 18/10 = 1.8.

Output power = 18*1.8 = 32.4.

n% = O/P power / I/P power = (32.4/36)*100.
= 0.90*100.
= 90%.

Thank you.
(1)

Ramesh said:   1 decade ago
Input watt 120*.3(V*I) = 36.

Output watt :
So I = V/R(18/10) = 1.8.

18*1.8 = 32.4 watt.

Eff = output/input.

32.4/36 = 0.90.

So the answer is B.

Ece said:   1 decade ago
Should it be that both voltage and current in the primary must be in rms? what if they were in average value? or what if the other one is rms and the other one is average?

Shibli said:   1 decade ago
Nice explanation.

Karpagamdevu said:   1 decade ago
Formula for P out:

We know that p=v.i --->1
and i=v/r --->2
sub 2 in 1

So p=v*(v/r)
Hence (v^2)/r

Anuj Kumar said:   1 decade ago
I don't know this formula whish is used to calculate P out?