HELP with Transformer Calculation

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Shon

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I have Mike Holt Exam Prep book and i am confused on transformer calculations. Unit 4 Question 16


A transformer primary winding operates at 240v,the secondary at 120v, and the load is 1,500w. If the transformer is rated at 92% efficient, whats the current flowing through the primary winding of this transformer?
 
Only if the load is purely resistive. What if that 1500W load is (or includes) an induction motor?

I hate poorly constructed test questions. For the sake of expediency they can leave people with the impression that things like this are simpler than they really are. All they had to do was to say the load was 1500VA instead of Watts if they didn’t want to require another question to get the right answer. But then again, was that the intent? Probably not if it’s a written test.
 
The question contains the tacit assumption that the transformer is operating at its rated load. The efficiency will not be constant over the load curve.
 
GoldDigger said:
The question contains the tacit assumption that the transformer is operating at its rated load.
Click to expand...
Two points.
A transformer, as J has pointed out,is rated in VA or kVA.
The question should not require the student to make assumptions.

Bad question.
 
These tests are designed for electricians not engineers. It was a simple question with the 4 choices below. I knew someone was going to make a mountain out of a molehill. :lol:
 
Besoeker said:
Two points.
A transformer, as J has pointed out,is rated in VA or kVA.
The question should not require the student to make assumptions.

Bad question.
Click to expand...

But it did require an assumption. KVA does not equal KW unless you ASSUME it is a resistive load. Nothing in the question indicates it is or isn't. Thus, you should assume it is not, such that you then must assume also a PF, which by industry standards 0.85 is typical. So, I would have to say the answer would be approximately 8A; but I don't have the luxury of knowing the answer choices so from an purely engineering perspective that is conservative abut a better answer than 6.8A.

I love moles!:D
 
they give you 4 numbers and seek the fifth
if they don't state it (or give the numbers to derive it, ie, isec) you can't use it
no assumptions allowed
vp = 240
vs = 120
sec load = 1500 w
eff = 0.92
ip = ?

as shown there are numerous ways to crunch the GIVEN numbers
losses = (1-0.92) x 1500 = 120
total input power = 1500 + 120 = 1620
vp x ip = 240 x ip = 1620
ip = 1620/240 = 6.75 rnd to 6.8

not a bad question but bad unwarranted assumptions
do they say 'assume a pf for a typ motor'? no
the load is 1500 w, do the give a ph ang or value for Q? no
do not ASS U ME, use only what is give

for all we know it could be DC
LOLWAITING FOR COMMENTS LOL
 
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kingpb said:
:thumbsup:
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no assumption required

S (va) = P (w) + jQ (var)
since they give P in watts you know pf = 1
they do not mention S or Q

if they gave S va then an assumption may be required: does eff include reactive 'losses'?
or is it all R losses?
you can still calc current but losses won't be accurate

if they only give Q var same as S
 
Ingenieur said:
nope
total power S is va, hence xfmr POWER capacity rating in va
reactive in var
but they are equivilent

what assumption must be made?

watt = J/sec = C V / sec = VA
Click to expand...
Please don't confuse real power (W) with apparent power (VA)
They are not equivalent.
Now, can we get back to the OP's question?
 
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