# 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)

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)

How did you calculate it?

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

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

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

Nice explation sir.

Good solution.

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

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