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NETA 4 Exam Questions 7/2018

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  1. test11's Avatar
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    Quote Originally Posted by rdemarce View Post
    I agree that power is dependent on ohmís 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 ohmís law, power equals I(squared)*Resistance.

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

    In a transformer, power is your constant. Power is dependent on more than just ďohmís lawĒ 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.
    Ohm's law:Click image for larger version. 

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    Impedance, as you know, stays basically the same for the test as its simulated shorted to ground on the secondary side. Then, if you reduce the voltage, what has to happen to current? That is the current you plug into that power equation. So KVA is not constant.

    The KVA max OUTPUT staying constant at a large primary voltage range above percent impedance voltage is due to saturation and is outside the original scope of the question. It requires higher math to look at as primary impedance actually does change in this scenario (See SCPT or magnetic amplifiers at your peril).

    Again, at a tech level with one dependent, one independent, and one fixed variable, KVA will not be constant and power in will equal power out(unlike in saturation).

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    Quote Originally Posted by test11 View Post
    Ohm's law:Click image for larger version. 

Name:	Ohms-Law.png 
Views:	0 
Size:	10.7 KB 
ID:	451

    Impedance, as you know, stays basically the same for the test as its simulated shorted to ground on the secondary side. Then, if you reduce the voltage, what has to happen to current? That is the current you plug into that power equation. So KVA is not constant.

    The KVA max OUTPUT staying constant at a large primary voltage range above percent impedance voltage is due to saturation and is outside the original scope of the question. It requires higher math to look at as primary impedance actually does change in this scenario (See SCPT or magnetic amplifiers at your peril).

    Again, at a tech level with one dependent, one independent, and one fixed variable, KVA will not be constant and power in will equal power out(unlike in saturation).
    Just gonna put this here:

    https://www.electricaleasy.com/2014/...ansformer.html

    "The ammeter reading gives primary equivalent of full load current (Isc). The voltage applied for full load current is very small as compared to rated voltage. Hence, core loss due to small applied voltage can be neglected. Thus, the wattmeter reading can be taken as copper loss in the transformer."

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    Quote Originally Posted by 3Ashockers View Post
    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
    This is not how you do this. The percent impedance is a percent of the base impedance, which is calculated by using the base VA and V ratings. So you cannot use the "divide the FLA by the percent impedance" shortcut if you aren't using the base voltage to calculate the FLA since the percent impedance is a percent of the base impedance. Plus, if you were to short out the low side and apply 34.5kV to the high side, your high side current would be 334.69 A, so how does it make any sense that you would have 40x that when you are applying 1% of that voltage across the same amount of impedance?

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    Quote Originally Posted by rdemarce View Post
    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.
    The KVA rating of a transformer is based on how much KVA it can handle at the voltages listed on the nameplate and is based on it's construction (winding conductor ampacity, cooling capabilities, etc.). And transformers don't dictate how much flows through them, that is all dependent on how much the load demands and the source can supply. Certain machines, like motors, will draw more current at lower voltages to maintain the same horsepower output, but you can't treat transformers this way. If you were to apply half as much voltage to the transformer, it would cut down the VA rating because the windings won't magically increase the conductor size and ampacity and be able to handle twice as much current without generating more heat.

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    Quote Originally Posted by EricGoetz92 View Post
    This is not how you do this. The percent impedance is a percent of the base impedance, which is calculated by using the base VA and V ratings. So you cannot use the "divide the FLA by the percent impedance" shortcut if you aren't using the base voltage to calculate the FLA since the percent impedance is a percent of the base impedance. Plus, if you were to short out the low side and apply 34.5kV to the high side, your high side current would be 334.69 A, so how does it make any sense that you would have 40x that when you are applying 1% of that voltage across the same amount of impedance?
    If I recall correctly, you are correct and that was made clear in another post. I think maybe all the responses to this questions were collected somewhere else?

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    Quote Originally Posted by young_dad View Post
    If I recall correctly, you are correct and that was made clear in another post. I think maybe all the responses to this questions were collected somewhere else?
    If they were collected somewhere else, the TestGuy admins didn't see it because they still have a practice test question that doesn't have anywhere close to the right answer as a choice. And the "Help" for this question links to this thread, which shows a lot of incorrect information about how to do this problem.

    None of this is your fault of course, this is just one of my huge gripes about this site. They create practice test questions without even knowing what they are talking about. I'll end up submitting this question to the thread that is about correcting inaccurate test questions.

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    Show Correct Calculation

    Quote Originally Posted by EricGoetz92 View Post
    This is not how you do this. The percent impedance is a percent of the base impedance, which is calculated by using the base VA and V ratings. So you cannot use the "divide the FLA by the percent impedance" shortcut if you aren't using the base voltage to calculate the FLA since the percent impedance is a percent of the base impedance. Plus, if you were to short out the low side and apply 34.5kV to the high side, your high side current would be 334.69 A, so how does it make any sense that you would have 40x that when you are applying 1% of that voltage across the same amount of impedance?
    EricGoetz92,

    If that calculation is not right, can you respond with the correct calculation? It seems to make sense to me, please show the correct answer, in a line by line formula format. I got a similar answer for the primary current short circuit at 34.5 KV to 277 V / 1MVA 5% impedance, 334.69 amps SC primary, and I multiplied it by the ratio of the difference between 34,500 and 480, 71.875, and I got 24,054.4 amps for a 480 volt primary, or would you divide the ratio to get 4.656 amps, this always confuses me.
    Last edited by Primepower1; February 25, 2021 at 12:12 PM. Reason: My answer

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    Quote Originally Posted by Primepower1 View Post
    EricGoetz92,

    If that calculation is not right, can you respond with the correct calculation? It seems to make sense to me, please show the correct answer, in a line by line formula format.
    https://testguy.net/threads/6409-Tra...ng-Calculation

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    Quote Originally Posted by Primepower1 View Post
    EricGoetz92,

    If that calculation is not right, can you respond with the correct calculation? It seems to make sense to me, please show the correct answer, in a line by line formula format.
    Here is the problem and solution:

    1000kva D-Y 34.5kv to 480/277v secondary transformer 5% imp. Apply 480v to primary side and ground secondary. What is primary fault current?

    Find primary side Full Load Amps (FLA) at nameplate voltage:
    FLA = (VA nameplate)/(V nameplate*sqrt(3)) = 1000kVA/(34.5kV * sqrt(3)) = 16.734 A

    (FYI, when calculating the FLA on a three phase transformer, the formula is always the same whether it is the delta or wye winding.)

    Find the maximum short circuit current at rated voltage:
    I short circuit at rated voltage = FLA/(Percent Impedance) = 16.734 A/.05 = 334.69 A

    Find the short circuit current at the applied voltage:
    I short circuit at applied voltage = (V applied/V nameplate)*(I short circuit at rated voltage) = (480/34,500)*334.69 = 4.66 A

    In conclusion, the impedance of the transformer doesn't change based on the applied voltage, so the short circuit current reduces proportionally the applied voltage. So you need to first find the short circuit current at rated voltage since that is the only voltage level you can use the "I short circuit = FLA/Percent Impedance" formula at. Then you multiply that by the ratio of applied voltage to name plate voltage and you get your answer.

    If they asked you to find the amount of current flowing out of the secondary winding, you would do every step the exact same except during the first step where you calculate FLA, you would use 480V as the nameplate voltage instead of 34.5kV. You then use that FLA value moving forward. I won't write out the steps to that, but the answer would be 334.69 A.

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