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# New NETA 4 Exam Questions

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Page 4 of 5 First ... 2345 Last 1. Originally Posted by ElectricalTestTech Just keep in mind, the new structure of the test recognizes which questions continuously get correct answers. The new structure throws out those questions and inputs different ones so the test can continuously change. While it is good to know the questions put up on this forum, do not expect them to be on your test.
This is an excellent point. It cant be emphasized enough that when taking the practice exam its important to focus on the subject matter rather than the actual question and answer. We are looking into implementing a similar feature that stops showing you questions after answering them correctly a certain number of times. This way users who frequently take exams will continue to see fresh questions in subsequent exams.  Reply With Quote

2. Originally Posted by ddd675 16. Which SCADA communication is only appropriate for short distances?
a. RS232
b. RS485
c. CAT6
d. Fibre

Here are the max distances for each:

RS232 (50 feet)
RS485 (1220 meters)
CAT6 (100 meters)
Fiber (2000 meters)  Reply With Quote

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Reputation  Originally Posted by Inglefles897 11. Two 1mH inductors are connected in parallel and in series with two 45uF capacitors connected in parallel. What is the total impedance of the circuit?

How did you come up with 0.41 milliohms? I don't see how to solve this without frequency to calculate reactance, assuming 60hz this is what i got

1mH = 2x 377 ohms in parallel= 188.50 ohms
45uF = 2x 58.9 ohms in parallel = 29.45 ohms
series 188.50 + 29.45 = 217.95 ohms
help?
It is 188.50 - 29.45 because they are 180 degrees apart in a 60 HZ system and when connected in series they subtract from each other. When they are in series the formula is Z = XL - XC. In parallel the formula is more complicated and is similar to R1xR2/(R1+R2). In this case it is (XL*-XC)/(XL-XC). This means that as XL-XC approach zero (resonance frequency) that the impedance of the parallel tank circuit will approach infinity at that resonant frequency, but since the reactor and capacitor have watt losses you will never get infinity and real world examples of a wave trap have been 2000 to 10000 ohms at resonant frequency. A wave trap blocks the resonant frequency from getting onto the operating bus and passes the 60 HZ, so a wave trap has to be rated to carry normal line loads. Your series resonant circuits are similar and we call those line tuners in the power line carrier world. As XL-XC approach resonant frequency the two cancel themselves out and it allows your resonant frequency to pass and it blocks your carrier frequency.

Resonant Frequency equals 1/(2 PI (LC)^.5). I know that this formula is not needed for the above mentioned problem, but it helps you understand why a problem like this may be important and helps you figure out what size capacitor and reactor is needed for a wave trap and line tuner.

So, the answer to this problem is actually 159.05.

My old Navy instructors told me that the Smithsonian had a tank circuit ran for decades and only seemed to lose power when they took measurements, but I but that was just an electronic legend, lol.  Reply With Quote

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Reputation ## Wattmeter question

Three coils with a resistance of 4 ohms and impedance of 13 ohms each are connected in wye to a 120/208 Volt panel. What will each wattmeter indicate if three single-phase wattmeters are used to measure the power?

Help: What will each wattmeter indicate?  Reply With Quote

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Reputation   Originally Posted by Vernon Didn't read all the answers but this one jumped out and just slapped me as very high. I think that if no three phase connection is given one assumes a single phase transformer. If I tell you I am standing here with a 480V to 120V transformer in my hand you would assume one transformer with a turns ratio of 4 to 1 and a primary voltage of 480 volts.

Ok - so the full load current at 138,000V and 2,500,000 VA is 18.116 A. At 8%Z with a shorted secondary you would multiply that by 12.5 so at rated voltage you have 226.45 amps Isc. But you are only applying 480V so you would get, proportionately, 0.787 AMPS. Generally, if I specify that a transformer is (for example) 10%Z that means that with the secondary shorted it will take 10% of the rated primary voltage to get full load current. www.cttestset.com
I can follow all the math except where the 12.5 number came from. Can anyone fill me in on what I missed?  Reply With Quote

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Reputation   Originally Posted by Vernon Didn't read all the answers but this one jumped out and just slapped me as very high. I think that if no three phase connection is given one assumes a single phase transformer. If I tell you I am standing here with a 480V to 120V transformer in my hand you would assume one transformer with a turns ratio of 4 to 1 and a primary voltage of 480 volts.

Ok - so the full load current at 138,000V and 2,500,000 VA is 18.116 A. At 8%Z with a shorted secondary you would multiply that by 12.5 so at rated voltage you have 226.45 amps Isc. But you are only applying 480V so you would get, proportionately, 0.787 AMPS. Generally, if I specify that a transformer is (for example) 10%Z that means that with the secondary shorted it will take 10% of the rated primary voltage to get full load current. www.cttestset.com
I see it now...

(kVA x 100)/(kV x %Z)

Therefore,

(2,500,000VA x 100)/(138,000V x 8) = 226.449A

But at 480VAC vice 138,000VAC so,

138,000/480 = 287.5

So dividing the short-circuit primary amps by the voltage ratio difference:

226.449A/287.5 = 0.787A  Reply With Quote

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Reputation  Originally Posted by ddd675 7. A primary injection is to be performed on a transformer 138kV/25kV, 2500kva, 8%Z using a 480V generator connected to the high side with the low side grounded. The expected current in the high side is?
Assuming that the answer from the practice tests is correct, I believe I have finally satisfied myself with a reasonable solution to this problem.

- Assume the xfmr is single-phase since no other information is given.
- Using SC calculation, we find that the secondary fault current in a SC would be 1,250A.
- Knowing the secondary voltage and current, we can solve for the reactance in the secondary coil, giving 20 ohms @ 60Hz (another assumption).
- Changing gears; we now have a known secondary impedance since the secondary is grounded (20 ohms).
- With the fixed turns ratio of 5.52, we find that applying 480 to the primary gives us 86.9V on the secondary.
- Apply Ohm's Law to find secondary current over our 20 ohm impedance (4.35A).
- Again, use the TR of 5.52 to find what level of primary current will cause 4.35A in the secondary.

With all these assumptions, I come up with 0.787A of primary current.

This reference pointed me in the right direction. Until now, all the solutions I have seen on here were as clear as mud. https://www.eaton.com/ecm/groups/pub...wp009001en.pdf

Last edited by young_dad; 2 Weeks Ago at 07:13 PM.  Reply With Quote

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Reputation  Originally Posted by Inglefles897 11. Two 1mH inductors are connected in parallel and in series with two 45uF capacitors connected in parallel. What is the total impedance of the circuit?

How did you come up with 0.41 milliohms? I don't see how to solve this without frequency to calculate reactance, assuming 60hz this is what i got

1mH = 2x 377 ohms in parallel= 188.50 ohms
45uF = 2x 58.9 ohms in parallel = 29.45 ohms
series 188.50 + 29.45 = 217.95 ohms
help?
I came up with a completely different answer, but I see some issues in your work.

Xl = 2pifL = 2*pi*f*(.001H) = 0.377 ohms
Xc = 1/(2pifC) = 1/[2*pi*f*(.000045F)] = 58.9ohm

Inductors function like normal resistors, so parallel Xl of the same magnitude are halved ie. Xl total = 0.1885 ohm

Caps in parallel are additive, so Xc total = 117.8 ohm.

Since they act along the same axis in opposite directions, you can just subtract them now (I think) or go through whatever steps you need to resolve R and X.

I come up with about 117.6 ohm. <-90degree

Last edited by young_dad; 2 Weeks Ago at 08:32 AM.  Reply With Quote

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Reputation  Originally Posted by ddd675 31. While testing a 100KVAR 480V 3 phase wye connected capacitor. What would the expected uF be across A-B terminals.

Using E^2/R = P, you can manipulate to get Xc = E^2/Q (or P).

Plug that in to C=1/(2*pi*f*Xc) gives C = Q/(2*pi*f*E^2) to get phase capacitance of 1151uF. If you used 1/3 of bank KVAR (33.3kVar) and phase voltage (277), you would get a similar result. Both of these will give you a per phase uF.

To me, the straight forward solution is to look at the WYE and solve E^2/R = P. and plug R into Xc formula to get 1152uF.

Caps in series act like resistors in parallel, so take half of that to get the capacitance measure between A-B = 576

I also found this reference which helped me decided which method I will use on the test.

https://www.se.com/us/en/faqs/FA323640/

Last edited by young_dad; 2 Weeks Ago at 08:37 AM.  Reply With Quote

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Reputation  Originally Posted by ddd675 12. There is 60 amps measured on the line of a delta transformer secondary. How many amps are in the delta?
IL = 1.732IP in a delta, so solving for IP (in the delta) I get 34.6A.  Reply With Quote

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