Electronics - Transformers - Discussion

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?

[A]. 88%
[B]. 90%
[C]. 92%
[D]. 95%

Answer: Option B

Explanation:

No answer description available for this question.

Ricky said: (Oct 23, 2010)  
How did you calculate it?

Ramkumar said: (Oct 28, 2010)  
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: (Jan 26, 2011)  
Nice explation sir.

Kalai said: (Apr 27, 2011)  
Good solution.

Anuj Kumar said: (Aug 24, 2011)  
I don't know this formula whish is used to calculate P out?

So please help me.

Karpagamdevu said: (Sep 20, 2011)  
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: (Nov 9, 2011)  
Nice explanation.

Ece said: (Jul 8, 2012)  
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?

Ramesh said: (Aug 1, 2013)  
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.

Sarika said: (Jan 22, 2014)  
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.

Viyoo said: (Jul 13, 2014)  
Pin = 120*0.3=36W.
Pout = 18*18/10=32.4.

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

NAINA said: (Oct 18, 2014)  
Pin = v.i = 120*0.3 = 36.

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

n% = op/inp = 32.4/36 = 90%.

V Seenivasan said: (Jun 30, 2016)  
Fantastic explanation, Thank you all.

Ammu said: (Jul 12, 2016)  
Superb and helpful explanations.

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