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 omega.gif load. What is the efficiency of the transformer?
88%
90%
92%
95%
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
14 comments Page 1 of 2.

Sarika said:   1 decade 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)

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)

Ricky said:   1 decade ago
How did you calculate it?

Ramkumar said:   1 decade ago
pin= v*i = 120*.3 = 36

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

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

Yogeshwaran said:   1 decade ago
Nice explation sir.

Kalai said:   1 decade ago
Good solution.

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

So please help me.

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

Shibli said:   1 decade ago
Nice explanation.

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?


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