# 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.

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)

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?

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.

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

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

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

pin= v*i = 120*.3 = 36

pout = v^2/R = 18^2/10 = 32.4

eff = pout/pin = 32.4/36 = .9*100 = 90%

NAINA said:   10 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)

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