NETA 4 Exam Questions 7/2018

[Primepower1]

I took the NETA IV exam this Monday. There was maybe one or two questions from this entire site on the exam!!!

This site is not for memorizing, it is to see questions that are similar to what would be on the test. It makes sense that the people that make the test questions would monitor this site and intentionally change numbers and wording, but still cover the same subjects.

[MartyCoyne]

This site is not for memorizing, it is to see questions that are similar to what would be on the test. It makes sense that the people that make the test questions would monitor this site and intentionally change numbers and wording, but still cover the same subjects.

[young_dad]

Can someone explain to me how to differentiate between the wording of the test question and the two results other than looking at the magnitude of the answers? I have re-read the questions many times and cannot tell (other than answer choice) which method you would choose. They both appear to ask for the short circuit primary current, and short of the answers, I would get the answer of 4.65A as my answer. I mean, let's face it, if I hook up a 480 source to the primary with the secondary grounded, I'm going to get one or the other, not both.

Check out this write up on calculating transformer SC currents. It is essentially an application of Ohm's law using the impedance of the transformer.

http://apps.geindustrial.com/publibr...ircuit|generic

[chils15]

Check out this write up on calculating transformer SC currents. It is essentially an application of Ohm's law using the impedance of the transformer.

http://apps.geindustrial.com/publibr...ircuit|generic

good read worth the time

[bnhocking]

1000kva 34.5kv to 480v secondary transformer 5% imp. apply 480v to primary side and short secondary to ground. What is primary fault current?

I have seen this question very similar in the level 4 test bank.

[JesseJames]

Can someone explain to me how to differentiate between the wording of the test question and the two results other than looking at the magnitude of the answers? I have re-read the questions many times and cannot tell (other than answer choice) which method you would choose. They both appear to ask for the short circuit primary current, and short of the answers, I would get the answer of 4.65A as my answer. I mean, let's face it, if I hook up a 480 source to the primary with the secondary grounded, I'm going to get one or the other, not both.

The question is what is the current in the primary winding when you put 480VAC across it....somehow the word "FAULT" got typed in there before "CURRENT"...
I think the 4.65A is correct.

[test11]

Assuming D-Y

Turns Ratio 34.5k to 277 = 124.5:1
Sec. phase voltage at 480 primary = 3.86V
Line Voltage from 480 primary = 6.68V
1,000,000VA / (1.732*6.7V) = 86.4kA
Secondary SC Current = 86.4/.05 = 1,729kA
Primary Current = 1,729,000 / 124.5 = 13.88kA SC

These answers are much higher than normal current in the transformer. This is impossible as the impedance of the transformer is unchanged and the voltage decreased significantly. Using ohms law, the current must be much lower than normal which is 16.73A. Therefore, the KVA of the transformer is different at this voltage and cannot be used directly as a basis in finding the current at the new lower voltage.

[rdemarce]

These answers are much higher than normal current in the transformer. This is impossible as the impedance of the transformer is unchanged and the voltage decreased significantly. Using ohms law, the current must be much lower than normal which is 16.73A. Therefore, the KVA of the transformer is different at this voltage and cannot be used directly as a basis in finding the current at the new lower voltage.

The KVA of the transformer is the same regardless of what voltage is on the transformer. As your voltage drops, your current goes up in order to achieve the same KVA rating. You all have to remember that transformers are rated in Power, not voltage or current by themselves. The transformer doesn't care what voltage is on the windings, it's going to provide the amount of power that it can supply - once you factor in the impedance of the transformer.

[test11]

The KVA of the transformer is the same regardless of what voltage is on the transformer. As your voltage drops, your current goes up in order to achieve the same KVA rating. You all have to remember that transformers are rated in Power, not voltage or current by themselves. The transformer doesn't care what voltage is on the windings, it's going to provide the amount of power that it can supply - once you factor in the impedance of the transformer.

No. Transformer impedance stays the same, but voltage is reduced. Therefore current goes down with voltage. As a matter of fact, you can look up how percent impedance is calculated. They just raise voltage on the primary with a shorted secondary winding and see what percentage of rated voltage it takes to produce the rated KVA.

In short, power draw is dependent on all of ohm's law. You have to look at it accordingly.

[rdemarce]

No. Transformer impedance stays the same, but voltage is reduced. Therefore current goes down with voltage. As a matter of fact, you can look up how percent impedance is calculated. They just raise voltage on the primary with a shorted secondary winding and see what percentage of rated voltage it takes to produce the rated KVA.

In short, power draw is dependent on all of ohm's law. You have to look at it accordingly.

I agree that power is dependent on ohm’s law and the transformer impedance stays the same. I 100% disagree with the rest of your post.

The available fault current is the same no matter what voltage is applied to the primary windings of the transformer. Assume an infinite bus supplying the transformer and a shirt to ground on the secondary. The short to ground will take all of the current the transformer can throw at it, therefore the current on the primary windings will be the same amount of current as there would be if the rated voltage was applied to the transformer.

Every time there is a ground fault, what happens? Using ohm’s law, power equals I(squared)*Resistance.

If the resistance approaches zero (we know it’s not actually zero) then current has to approach infinity.

In a transformer, power is your constant. Power is dependent on more than just “ohm’s law” because your statement implies power can only be calculated by volts and amps. But if you use current and resistance you see what actually happens.

To answer the question, you find the available fault current on the primary winding (at whatever voltage you want) and that is the answer no matter what voltage is used on the primary windings.