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# Level III Questions, seen recently (from memory bear with me), help?

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1. Member
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Originally Posted by SecondGen
Step 1: Calculate the needed Voltage Drop by subtracting the 2 voltages

125V - 24V = 101V

Step 2: Calculate the needed Resistance using ohms law: R = V / I

101V / 0.500 = 202 Ohms

Step 3: Calculate Resistor Wattage from the Current and Resistance using P = I2R

0.500 x 0.500 x 202 = 50.5W
is that means that two resistors connected in series?

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Originally Posted by jwdrowe
An ATS does not transfer from utility to generator. The generator is running and producing voltage. Which of the following is the cause of it not transferring?
A) The breaker feeding the ATS did not open
B) battery does not have enough charge to operate the ATS
C) Frequency of the generator is too low
D) Service only lost one phase
I just completed the Exam today and this question was on it. However, it had more to the scenario. Something to the effect of the utility power was lost at the start. This to me helped me to rule out D which implies the power was not completely lost. I chose C in the end.

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Originally Posted by jwdrowe
According to the NFPA 70, what is the max allowable time delay results for a ground fault relay to operate?
A) 1 second at 1200 amps
B) 3 sec at 3000 amps
C) 5 sec at 1200 amps
D) At the actual available fault current
i found that the book said 1 sec at 3000+ amps
This one was also on mine and one of the choices was 1 sec at 3000 amps which is exactly what you found. I would like to know what book it was found in??

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## SF6 Gage measures what?

Originally Posted by dvansick2345
I would of said density or level but I'm not sure
The answer is Density. I remember this bening a topic of discussion from one of the pop quizzes in the NETA World publications in the last couple years. Its also somewhere in the study guide that NETA has. The one with a combined set of pop quizzes from all past issues.

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Originally Posted by NoWorldOrder
The correct answer is D. All I can think of is that the question is asking about an ATS that is only monitoring one phase. Also to consider, in MTM transfer schemes (which I've written the logic for many times) the end-users do not want a transfer on a single phase going to ground. They want the HRG to handle it and then to let the motors run on two phases while they track it down.

Anyway, the correct answer is D. Had the same question.
I humbly disagree with NoWorldOrder in that the question states that the generator is running and is at voltage. If the system is monitoring a different phase than a 'single phase event', there would be no start command to the genset. I believe the answer is that frequency is not in the transfer band to give the ATS a permissive. IMHO anyways, for what that might be worth.

6. Originally Posted by SecondGen
Step 1: Calculate the needed Voltage Drop by subtracting the 2 voltages

125V - 24V = 101V

Step 2: Calculate the needed Resistance using ohms law: R = V / I

101V / 0.500 = 202 Ohms

Step 3: Calculate Resistor Wattage from the Current and Resistance using P = I2R

0.500 x 0.500 x 202 = 50.5W
Late to the game response, I know, but you don't even need to calculate the resistance. Do the KVL to figure out voltage drop of 101 volts, then multiply by 500mA since total current flows through both components, and power is as easy as PIE (P=I x E).

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Originally Posted by ElectricalTestTech
Well here is my attempt at helping you. Someone please feel free to correct me if I'm wrong.

1. What is an application for a circuit switcher with a pre-insertion inductor?
A) Transient control w/ high fault currents
B) Transient control for capacitor bank switching
C) Current damping at 12.4kV
D) Current damping above 34.5kV

A pre-insertion inductor arrangement is provided for a circuit switcher to reduce audible and electrical noise and to limit transient inrush current and/or voltages upon closing of the circuit by the circuit switcher. Pre-insertion inductors have been introduced as a cost-effective alternative to pre-insertion resistors for controlling capacitor bank energization overvoltages.

I have to go with B

2. How is lagging power factor improved using a synchronous motor?
A) Raise excitation of field windings
B) Lower excitation of field windings
C) Raise excitation of amortisseur winding
D) Lower excitation of amortisseur winding
A Synchronous Motor can be made to operate at unity and leading power factor by just increasing its excitation voltage by increasing the field current. I believe the answer is A. But someone please chime in on this one
In the old days the utility used synchronous motors to control reactive power. They would not drive a load. By control of the excitation current they could off-set and bring the power factor close to unity.

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Originally Posted by cjones09
Looking for help on this question. These are my thoughts, what do you think?

A. This could be possible, the question doesn't state if the utility available light is still on or not
B. Not sure what type of ATS we are dealing with
C. Question doesn't specify the application but I would think any sync issues would be resolved at the gen switchboard
D. This would be a cause for transfer, so this answer is probably incorrect

Thanks in advance.
My though is that a battery should not be involved with an ATS, loss of phase should cause a transfer. I would think it would be a sync check issue. If it could not sync it would not transfer.

9. Originally Posted by ray12taylor
This one was also on mine and one of the choices was 1 sec at 3000 amps which is exactly what you found. I would like to know what book it was found in??
Wouldnt it be 1 second at 1200amps? The max allowable ground fault setting is 1200amps.

10. Originally Posted by Warrengarber
Wouldnt it be 1 second at 1200amps? The max allowable ground fault setting is 1200amps.
See NEC 230.95 (A) Max pick up setting is 1200A. Maximum time delay is 1 sec for fault current of 3000A or more. What the answers seem to imply is that 1 second it the maximum delay from pick up setting to 3000A. Most GFR's I can think of on bolted pressure switch setups have a definitely time delay after pickup with max of 1 sec, so the time delay would be 1 sec at 1200A pickup. Electronic breakers with GF can have I^2T slope functions, and the knee point at pickup can be above 1 sec, but is always below 1 sec by 3000A. But this would allow for a timing at a 1200A pickup of 2 secs in many cases. Code only specifies that the time delay has to be 1 second or less at 3000A.

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