# Prime Power Test Technician Level 2 Test 3 Of 3 (Testing Knowledge)

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Quizzes Created: 3 | Total Attempts: 4,161
Questions: 50 | Attempts: 921  Settings  Test Technician Level 2 Test 3. Test consists of 50 Questions. Test has 90 minute time limit. You may use a calculator and look up any formulas not given. Allow yourself at least an hour to complete as you cannot save and resume the test later. Good Luck!

• 1.

### Which of the following is not a standard voltage used for trip power on a medium voltage breaker?

• A.

48

• B.

120

• C.

60

• D.

240

• E.

None of the above

C. 60
Explanation
The question asks for a voltage that is not a standard voltage used for trip power on a medium voltage breaker. The options provided are 48, 120, 60, 240, and none of the above. The correct answer is 60 because it is not a standard voltage used for trip power on a medium voltage breaker.

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• 2.

### Which of the following is the ANSI device number for a phase sequence or phase balance voltage relay?

• A.

47

• B.

59

• C.

86

• D.

27

• E.

67

A. 47
Explanation
The ANSI device number 47 is assigned to a phase sequence or phase balance voltage relay. This relay is used to monitor the sequence and balance of voltage in a three-phase system. It detects any abnormalities in the phase sequence or imbalances in the voltage levels, providing protection against potential damage to equipment and ensuring proper functioning of the system.

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• 3.

### Measuring the resistivity of earth with a typical ground resistance test set is accomplished by:

• A.

The 2 point method

• B.

The fall of potential method

• C.

The 3 point method

• D.

The use of 4 electrodes

D. The use of 4 electrodes
Explanation
The use of 4 electrodes is the correct answer because it allows for a more accurate measurement of the resistivity of the earth. By using 4 electrodes, the test set can measure the voltage drop between two electrodes while also measuring the current flow between the other two electrodes. This method provides a more comprehensive understanding of the resistivity of the earth, as it takes into account both the voltage and current measurements. The 4 electrode method is considered more reliable and accurate compared to the other methods mentioned.

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• 4.

### Assuming each of the folliwing is the same length, which has the lowest resistance?

• A.

500 MCM copper cable

• B.

350 MCM aluminum cable

• C.

1/4 x 6 aluminum bus

• D.

1/4 x 6 copper bus

• E.

1/2 x 2 copper bus

D. 1/4 x 6 copper bus
Explanation
The 1/4 x 6 copper bus has the lowest resistance because copper has a lower resistivity compared to aluminum. Additionally, the larger cross-sectional area of the copper bus compared to the other options also contributes to its lower resistance.

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• 5.

### Metal enclosed medium voltage interrupter switches should be

• A.

Stored energy operated

• B.

Motor operated

• C.

Closing coil operated

• D.

• E.

Hook stick operated

A. Stored energy operated
Explanation
Metal enclosed medium voltage interrupter switches should be stored energy operated because this type of operation ensures that the switch can be opened and closed even when there is no external power source available. The stored energy, typically in the form of springs, provides the necessary force to operate the switch. This is important for safety and reliability, as it allows the switch to function independently and reliably in various situations, such as during power outages or when the switch needs to be operated remotely.

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• 6.

### Protection against extremely high fault current is typically achieved by use of

• A.

An instantaneous overcurrent relay

• B.

Phase balance relaying

• C.

Arrestors

• D.

Current limiting fuses

• E.

Equipment maximum voltage ratings

D. Current limiting fuses
Explanation
Current limiting fuses are typically used to protect against extremely high fault currents. When a fault occurs, the current through the fuse increases rapidly. The current limiting fuse is designed to have a low impedance, which causes it to quickly melt and open the circuit, limiting the amount of current that can flow. This helps to prevent damage to the equipment and reduces the risk of fire or other hazards. Instantaneous overcurrent relays, phase balance relaying, and arrestors may also provide some protection, but current limiting fuses are specifically designed for this purpose. Equipment maximum voltage ratings are not directly related to protection against fault currents.

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• 7.

### Switchgear interiors are compartmentalized for the purpose of

• A.

Worker safety

• B.

Dust and dirt control

• C.

Moisture control

• D.

Fault damage control

• E.

All of the above

E. All of the above
Explanation
Switchgear interiors are compartmentalized for the purpose of worker safety, dust and dirt control, moisture control, and fault damage control. Compartmentalization helps in ensuring the safety of workers by minimizing the risk of accidental contact with live parts. It also helps in keeping dust and dirt away, preventing any potential damage or malfunction. Additionally, compartmentalization helps in controlling moisture, which can cause corrosion and other electrical issues. Finally, it aids in containing faults and limiting their damage, ensuring the overall reliability and longevity of the switchgear.

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• 8.

### Directional relays are typically used

• A.

• B.

To prevent the motoring of a generator

• C.

To protect buses

• D.

On double ended substations with closed bus ties

• E.

B and C

E. B and C
Explanation
Directional relays are typically used on radial feeders to prevent the motoring of a generator and to protect buses on double-ended substations with closed bus ties. This means that directional relays are used in both scenarios to ensure the proper functioning and protection of the electrical system.

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• 9.

### A 3 phase transformer has a name plate voltage rating  of 13200-480Y/277. The actual incoming voltage is 13,750V. A secondary voltage of 500v is desired. Which of the follwing taps available on the no-load tap changer should be used to obtain the closest possible voltage to the desired voltage.

• A.

13,200

• B.

13,860

• C.

13,530

• D.

12,870

• E.

12,540

B. 13,860
Explanation
The correct answer is 13,860. The question asks for the tap on the no-load tap changer that will provide the closest possible voltage to the desired secondary voltage of 500V. Since the actual incoming voltage is 13,750V, the tap that is closest to this value is 13,860V.

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• 10.

### Which one of the follwing is damaging to many insulating materials used in high voltage rotating machines?

• A.

Hysteresis

• B.

Saturation

• C.

Corona

• D.

Eddy currents

• E.

None of the above

C. Corona
Explanation
Corona is damaging to many insulating materials used in high voltage rotating machines. Corona is a phenomenon that occurs when the electric field surrounding a conductor becomes ionized, creating a corona discharge. This discharge can cause degradation and breakdown of insulating materials, leading to insulation failure and damage to the machine. Therefore, corona is considered damaging to insulating materials in high voltage rotating machines.

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• 11.

### Which device may continue to be a source of harmful electrical energy after all switches and circuit breakers have been opened for a significant time?

• A.

Transformer

• B.

Capacitor

• C.

Resistor

• D.

Generator

• E.

Surge arrestor

B. Capacitor
Explanation
A capacitor is a device that stores electrical energy in an electric field. Even after all switches and circuit breakers have been opened for a significant time, a charged capacitor can still contain stored electrical energy. This is because capacitors can hold their charge for extended periods of time. Therefore, a capacitor may continue to be a source of harmful electrical energy even when all other devices have been disconnected or turned off.

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• 12.

### When measuring the ground resistance of an industrial grounding network, the ground resistance would normally be expected to be no greater than:

• A.

.01 ohms

• B.

.5 ohms

• C.

5 ohms

• D.

10 ohms

• E.

25 ohms

C. 5 ohms
Explanation
When measuring the ground resistance of an industrial grounding network, it is expected that the ground resistance should be as low as possible. A ground resistance of 5 ohms is considered acceptable in most cases because it indicates a good connection between the grounding system and the earth. A lower resistance would be even better, but anything above 5 ohms may indicate poor grounding and can lead to safety hazards and equipment damage. Therefore, a ground resistance of 5 ohms is the most appropriate and safe option.

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• 13.

### A watt-hour meter has a watt-hour constant of 1.2, it is used with 1200/5 ratio current transformers and 14,400/120 potential transfromers. What is the primary watt-hour constant Kh?

• A.

720

• B.

2.4

• C.

1.2

• D.

34,560

• E.

Nobody knows!

D. 34,560
Explanation
The primary watt-hour constant Kh can be calculated by multiplying the watt-hour constant of the meter (1.2) with the ratio of the current transformers (1200/5) and the ratio of the potential transformers (14,400/120). Therefore, the calculation would be 1.2 * (1200/5) * (14,400/120) = 34,560.

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• 14.

### NETA acceptance testing specifications require that analog field instruments be tested for accuracy:

• A.

Every month

• B.

Every 3 months

• C.

Every 6 months

• D.

Every 12 months

• E.

Every 24 months

C. Every 6 months
Explanation
According to NETA acceptance testing specifications, analog field instruments need to be tested for accuracy every 6 months. This means that these instruments should undergo a calibration process to ensure their readings are accurate and reliable twice a year. Regular testing is necessary to identify any deviations or errors in the instrument's measurements and to maintain the overall quality and performance of the instrument. By conducting these tests every 6 months, any potential issues can be detected and addressed in a timely manner, ensuring the accuracy of the instrument's readings.

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• 15.

### The most important reason secondary circuits of current transformers should never be left open is:

• A.

Fuse could blow

• B.

Ammeter may not work

• C.

Voltmeter could be affected

• D.

Dangerous voltages could be produced

• E.

None of the above

D. Dangerous voltages could be produced
Explanation
Leaving the secondary circuits of current transformers open can lead to the production of dangerous voltages. Current transformers are designed to step down high currents to a safe level for measurement or protection purposes. When the secondary circuits are left open, the transformer can become disconnected from the load, causing the voltage to rise significantly. This can create a hazardous situation as the high voltage can pose a risk to equipment and personnel. It is therefore crucial to ensure that the secondary circuits of current transformers are always properly connected and not left open.

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• 16.

### Polarization index is defined as the ratio of the measured insulation resistance at what 2 specific times?

• A.

10 mins to 1 min

• B.

60 secs to 30 secs

• C.

10 mins to 5 mins

• D.

5 mins to 1 min

A. 10 mins to 1 min
Explanation
The polarization index is defined as the ratio of the insulation resistance measured at two specific times, namely 10 minutes and 1 minute. This ratio helps to assess the quality and condition of the insulation in electrical equipment. By comparing the resistance values at these two time intervals, any significant decrease in resistance can indicate potential insulation issues such as moisture, dirt, or degradation. Therefore, monitoring the polarization index can provide valuable insights into the health of the insulation system and help prevent electrical failures.

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• 17.

### Which of the following will have the greatest effect on an  insulation resistance reading?

• A.

The size of the device being tested

• B.

The temperature of the device being tested

• C.

The ambient temperature

• D.

The relative humidity

B. The temperature of the device being tested
Explanation
The temperature of the device being tested will have the greatest effect on an insulation resistance reading. Insulation resistance is a measure of how well a material can resist the flow of electric current. As temperature increases, the resistance of the insulation material decreases, leading to a lower insulation resistance reading. This is because higher temperatures can cause the insulation material to degrade or become more conductive, reducing its ability to resist the flow of current. Therefore, the temperature of the device being tested is the most significant factor in determining the insulation resistance reading.

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• 18.

### A synchronous motor used without load in an electric system serves to:

• A.

Balance the high voltage drop

• B.

Improve the power factor of the system

• C.

Provide inductive reactive power

• D.

Regulate the voltage for variable speed machines

B. Improve the power factor of the system
Explanation
A synchronous motor used without load in an electric system improves the power factor of the system. The power factor is a measure of how effectively the electrical power is being used in a system. A low power factor can result in inefficient energy usage and increased electricity costs. By using a synchronous motor without load, the power factor can be improved as the motor helps to compensate for reactive power, thereby reducing the amount of reactive power flowing through the system. This leads to a more efficient use of electrical power and an improved power factor.

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• 19.

### A delta connected 3 phase 3 wire system with balanced phase currents of 542 amps has a line currents of approximately:

• A.

313 amps

• B.

767 amps

• C.

939 amps

• D.

542 amps

• E.

271 amps

C. 939 amps
Explanation
In a delta connected 3 phase 3 wire system, the line current is equal to the phase current. Since the phase currents are balanced and each phase has a current of 542 amps, the line current will also be 542 amps. Therefore, the correct answer is 542 amps, not 939 amps.

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• 20.

### Capacitors are used in AC circuits to:

• A.

provide power factor correction

• B.

Decrease system kVA demand

• C.

Provide more efficient utilization of feeders

• D.

All of the above

D. All of the above
Explanation
Capacitors are used in AC circuits to provide power factor correction, decrease system kVA demand, and provide more efficient utilization of feeders. Power factor correction is important in AC circuits to minimize reactive power and improve the efficiency of power transmission. By adding capacitors, the system's power factor can be improved, reducing the kVA demand and allowing for more efficient utilization of feeders. Therefore, all of the given options are correct.

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• 21.

### What is normally considered the maximum acceptable power factor for new insulating oil?

• A.

.05 %

• B.

.5 %

• C.

1.0%

• D.

1.5%

A. .05 %
Explanation
The correct answer is .5%. Power factor is a measure of how effectively electrical power is being used in a system. A power factor of 1 means that all the power being supplied is being used effectively, while a power factor less than 1 indicates that there is some power being wasted. In the case of insulating oil, a power factor of .5% is considered the maximum acceptable level, as it indicates that the oil is effectively insulating and not wasting electrical power. A power factor of .05% would be extremely low and would suggest a highly inefficient or faulty insulating oil.

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• 22.

### What is the ANSI device number is for an under voltage relay?

• A.

47

• B.

59

• C.

86

• D.

87

• E.

27

E. 27
Explanation
The ANSI device number 27 is assigned to an under voltage relay. This relay is used to detect when the voltage of a system falls below a certain threshold and initiates protective actions to prevent damage to equipment or interruption of power supply.

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• 23.

### In ccordance with NETA standards, what is the correct acceptance test voltage for 5kV rated EPR 133% cable?

• A.

65kV

• B.

45kV

• C.

55kV

• D.

25kV

• E.

15kV

D. 25kV
Explanation
According to NETA standards, the correct acceptance test voltage for a 5kV rated EPR 133% cable is 25kV. This means that during the acceptance test, the cable should be subjected to a voltage of 25kV to ensure its proper functioning and adherence to standards.

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• 24.

### If a watt-hour meter has a disk constant of 3.6 and is used with 600:5 ampere current transformers and 20:1 potential transformers, the primary Kh is?

• A.

133.3 whr per revolution

• B.

8640 whr per revolution

• C.

1728 whr per revolution

• D.

43200 whr per revolution

B. 8640 whr per revolution
Explanation
The primary Kh, or primary constant, is a measure of the energy consumed per revolution of the watt-hour meter's disk. In this case, the disk constant is given as 3.6, which means that for every revolution of the disk, 3.6 watt-hours of energy are consumed. The current transformers have a ratio of 600:5, meaning that for every 600 amperes of primary current, there is 5 amperes of secondary current. Similarly, the potential transformers have a ratio of 20:1. By multiplying the ratios and the disk constant, we can calculate the primary Kh as 3.6 * (600/5) * (20/1) = 8640 whr per revolution.

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• 25.

### Interfacial tension

• A.

Is measured to detect metals in oil

• B.

Is measured in MgKOH/gm

• C.

Is measured in dynes/cm

• D.

Determines the resistivity of oil

• E.

Principally detects impurities in the oil

C. Is measured in dynes/cm
Explanation
Interfacial tension is a property that measures the force of attraction between two immiscible phases, such as oil and water. It is measured in dynes/cm, which is a unit of force per unit length. By measuring the interfacial tension, one can determine the level of attraction or repulsion between the oil and other substances present, such as impurities or metals. Therefore, measuring interfacial tension is a method used to detect impurities in the oil and assess its quality.

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• 26.

### Which of the follwing conditions on a system would you expect to find during a short circuit analysis?

• A.

• B.

Breakers with insufficient interrupting capacity

• C.

Protective relays with the wrong settings

• D.

Defective equipment grounds

B. Breakers with insufficient interrupting capacity
Explanation
During a short circuit analysis, one would expect to find breakers with insufficient interrupting capacity. This is because a short circuit can cause a sudden surge of current, and if the breaker's interrupting capacity is not sufficient, it may fail to interrupt the flow of current and protect the system from damage. This can lead to further issues such as equipment damage, fire hazards, or even system failure. Therefore, it is crucial to ensure that breakers have adequate interrupting capacity to handle short circuit conditions effectively.

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• 27.

### A gas likely to be used in an inert-gas system of a power transformer is?

• A.

SF6

• B.

Air

• C.

Nitrogen

• D.

Oxygen

• E.

Hydrogen

C. Nitrogen
Explanation
Nitrogen is likely to be used in an inert-gas system of a power transformer because it is an inert gas, meaning it does not react with other substances. This makes it suitable for creating an oxygen-free environment inside the transformer, which helps prevent the formation of flammable gases and reduces the risk of fire or explosion. Additionally, nitrogen has good cooling properties and can dissipate heat effectively, making it an ideal choice for maintaining optimal operating temperatures in the transformer.

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• 28.

### What is the value of the voltage in a 540 turn secondary winding of a transformer if 230 volts is applied to its 270 turn primary winding?

• A.

269v

• B.

115v

• C.

460v

• D.

622v

• E.

135v

C. 460v
Explanation
The voltage in the secondary winding of a transformer is determined by the turns ratio between the primary and secondary windings. In this case, the turns ratio is 540:270, which simplifies to 2:1. This means that for every 2 turns in the primary winding, there is 1 turn in the secondary winding. Since 230 volts is applied to the primary winding, the voltage in the secondary winding will be half of that, which is 115 volts. Therefore, the correct answer is 115v.

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• 29.

### The arcing contacts on a circuit breaker are used to:

• A.

Increase the magnetic field of the blowout coil

• B.

Prevent damage to the main contacts

• C.

Ensure that a closed circuit is made

• D.

Reduce the heating of the trip coil

• E.

Reduce the main contact pressure

B. Prevent damage to the main contacts
Explanation
The arcing contacts on a circuit breaker are used to prevent damage to the main contacts. When a circuit breaker interrupts the flow of current, an arc is formed between the contacts. This arc can cause damage to the main contacts over time due to the high temperatures and electrical stress. The arcing contacts are designed to handle and extinguish this arc, protecting the main contacts from damage. By absorbing and redirecting the arc energy, the arcing contacts ensure the longevity and proper functioning of the circuit breaker.

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• 30.

### If a current of 30 amps is measured with a 100 amp 3% ammeter, the maximum possible error in terms of actual error is:

• A.

3%

• B.

4.6%

• C.

10%

• D.

3.33%

• E.

65

C. 10%
Explanation
The maximum possible error in terms of actual error is 10%. This means that the ammeter can have a deviation of up to 10% from the actual value of the current being measured. For example, if the actual current is 30 amps, the ammeter could display a reading anywhere between 27 amps and 33 amps.

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• 31.

### What is the primary purpose of ground fault protectors?

• A.

Prevent ground faults

• B.

Detect low energy ground faults

• C.

Protect personnel from shocks

• D.

Provide sensitive short circuit protection

• E.

None of the above

C. Protect personnel from shocks
Explanation
The primary purpose of ground fault protectors is to protect personnel from shocks. Ground fault protectors are designed to detect any leakage of electrical current to the ground and quickly shut off power to prevent electric shocks. This helps ensure the safety of individuals working with or around electrical equipment and prevents potential injuries or fatalities.

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• 32.

### If a watt-hour meter has a disk constant of 3.6 and is used with 150:5 ampere current transformers and 4:1 potential transformers, the primary Kh is:

• A.

.03 whr per rev

• B.

.9 whr per rev

• C.

14.4 whr per rev

• D.

432 whr per rev

• E.

864 whr per rev

D. 432 whr per rev
Explanation
The primary Kh is calculated by multiplying the disk constant (3.6) with the ratio of the current transformers (150:5) and the ratio of the potential transformers (4:1).

So, primary Kh = 3.6 * (150/5) * (4/1) = 3.6 * 30 * 4 = 432 whr per rev.

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• 33.

### In accordance with NETA standards, the correct maintenance test voltage for 15kv rated poly 133% insulation level cable is:

• A.

65kV

• B.

45kV

• C.

55kV

• D.

49kV

• E.

25kV

C. 55kV
Explanation
According to NETA standards, the correct maintenance test voltage for a 15kV rated poly 133% insulation level cable is 55kV. This voltage is determined based on the insulation level of the cable and is necessary to ensure that the cable is functioning properly and can withstand the expected operating conditions. It is important to adhere to these standards to prevent any potential electrical failures or hazards.

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• 34.

### What is the maximum symmetrical fault current available on the secondary side of a 3 phase transformer with data plate: 1000kVA , 12,470-480/277Y, 6.25% Impedance?                   Fault Current = FLA x (100/impedance)  FLA = kVa / (1.732 x (L-LkV))

• A.

17,184

• B.

7,410

• C.

12,077

• D.

19,246

• E.

16,000

D. 19,246
Explanation
The fault current is calculated using the formula Fault Current = FLA x (100/impedance), where FLA is the full load amperage and impedance is given as 6.25%. FLA is calculated using the formula FLA = kVa / (1.732 x (L-LkV)), where kVa is given as 1000 and L-LkV is given as 12,470-480. By substituting the given values into the formulas, we can calculate the fault current to be 19,246.

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• 35.

### A 3 phase voltage is 600V, power factor is .75 and the balanced load consumption is 21,200 watts. The load current is:

• A.

26.4 A

• B.

27.2 A

• C.

54.4 A

• D.

35.3 A

• E.

29 A

B. 27.2 A
Explanation
The load current can be calculated using the formula: Load current = Power / (sqrt(3) x Voltage x Power Factor). Plugging in the given values, we get Load current = 21200 / (sqrt(3) x 600 x 0.75) = 27.2 A.

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• 36.

### What is the watts produced in one pole of a circuit breaker which has a resistance of 80 microhms and a load current of 700 amps?

• A.

39.2 milliwatts

• B.

39.2 watts

• C.

8.75 watts

• D.

56 watts

• E.

None of the above

B. 39.2 watts
Explanation
The power (in watts) produced in one pole of a circuit breaker can be calculated using the formula P = I^2 * R, where P is the power, I is the current, and R is the resistance. In this case, the resistance is given as 80 microhms (or 0.00008 ohms) and the load current is 700 amps. Plugging these values into the formula, we get P = (700^2) * 0.00008 = 39.2 watts. Therefore, the correct answer is 39.2 watts.

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• 37.

### A delta-delta bank of 3 single phase transformers rated at 7500kVA has one transformer shorted out. If reconnected in an Open-Delta configuration, what load will the remaining transfromers be able to carry safely?

• A.

2,887 kVA

• B.

4,328 kVA

• C.

6,495 kVA

• D.

7,500 kVA

• E.

5,000 kVA

B. 4,328 kVA
Explanation
When one transformer in a delta-delta bank is shorted out, the remaining transformers can be reconnected in an open-delta configuration. In this configuration, the total kVA capacity of the bank is reduced to two-thirds of the original capacity. Since the original capacity of the bank is 7500kVA, the remaining transformers will be able to carry a load of 7500kVA * (2/3) = 5000kVA. Therefore, the correct answer is 5,000 kVA.

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• 38.

### A correctly performed turns to turns ratio test of a transformer can determine which of the following?

• A.

Polarity of the windings

• B.

Number of turns in each winding

• C.

The turns ration for each tap setting

• D.

A,B, and C

• E.

A and C only

E. A and C only
Explanation
A correctly performed turns ratio test of a transformer can determine the polarity of the windings and the turns ratio for each tap setting. This test helps identify the direction of the windings and ensures that the transformer is functioning properly. By measuring the turns ratio, it is possible to determine if the transformer is stepping up or stepping down the voltage as intended.

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• 39.

• A.

5 V

• B.

10 V

• C.

15 V

• D.

20 V

• E.

25 V

A. 5 V
• 40.

### The system voltage is 480/277 volts. Which potential transformer intallation can provide 120/69 volts on the secondary?

• A.

3 PTs 4-1 ratio, connected delta-delta

• B.

3 PTs 4-1 ratio, connected delta-wye

• C.

3 PTs 4-1 ratio, connected wye-wye

• D.

3 PTs 2.4-1 ratio, connected wye-wye

C. 3 PTs 4-1 ratio, connected wye-wye
Explanation
When the system voltage is 480/277 volts, the potential transformer installation that can provide 120/69 volts on the secondary is 3 PTs 4-1 ratio, connected wye-wye. This is because the wye-wye connection allows for a step-down ratio of 4-1, which means the secondary voltage is 1/4th of the primary voltage. Therefore, if the primary voltage is 480 volts, the secondary voltage will be 120 volts. Similarly, if the primary voltage is 277 volts, the secondary voltage will be 69 volts.

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• 41.

### A relay device number 51 is connected to an 800/5 CT. The relay is set to time dial 3 and tap 10. What primary current must flow to cause the relay to start to "time out"?

• A.

8000 A

• B.

4000 A

• C.

1600 A

• D.

2000 A

• E.

800 A

C. 1600 A
Explanation
The relay device number 51 is connected to an 800/5 CT, which means that the CT ratio is 800/5. The relay is set to time dial 3 and tap 10. The time dial and tap settings determine the operating time of the relay. In this case, the primary current required to cause the relay to start to "time out" can be calculated by multiplying the CT ratio (800/5) by the tap setting (10) and dividing it by the time dial setting (3). Therefore, the primary current required is (800/5) x 10 / 3 = 1600 A.

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• 42.

### Insulation resistance testing of control wiring should be performed at:

• A.

250 VDC

• B.

500 VDC

• C.

600 VDC

• D.

1000 VDC

• E.

5000 VDC

D. 1000 VDC
Explanation
Insulation resistance testing is performed to determine the integrity of the insulation material used in control wiring. Higher voltage levels are used during testing to ensure accurate results. The correct answer of 1000 VDC indicates that this is the appropriate voltage level at which the insulation resistance testing of control wiring should be conducted.

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• 43.

### During the perfomance of a thermographic scan of similar components within the switchboard that are under similar load, at what temperature difference does NETA recommend investigation?

• A.

1-2 deg F

• B.

1-3 deg C

• C.

2-5 deg F

• D.

2-5 deg C

• E.

3-5 deg C

B. 1-3 deg C
Explanation
NETA recommends investigation when there is a temperature difference of 1-3 degrees Celsius during the performance of a thermographic scan of similar components within the switchboard that are under similar load. This temperature difference may indicate potential issues or abnormalities in the switchboard that need further investigation to ensure its proper functioning and safety.

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• 44.

### On a typical 1000kVA delta-delta transformer rated 13,800/480 V, the voltage ratio is:

• A.

28.00

• B.

28.75

• C.

29.00

• D.

20.80

• E.

38.75

B. 28.75
Explanation
The voltage ratio of a transformer is determined by the ratio of the primary voltage to the secondary voltage. In this case, the primary voltage is 13,800 V and the secondary voltage is 480 V. To find the voltage ratio, we divide the primary voltage by the secondary voltage: 13,800/480 = 28.75. Therefore, the correct answer is 28.75.

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• 45.

### A wattmeter measures:

• A.

Current magnitude times voltage magnitude

• B.

Current magnitude times voltage magnitude times % power factor

• C.

System energy use

• D.

Current magnitude plus voltage magnitude

B. Current magnitude times voltage magnitude times % power factor
Explanation
A wattmeter measures the power consumed by an electrical device or system. Power is calculated by multiplying the current magnitude (in amperes) by the voltage magnitude (in volts). However, since power factor represents the efficiency of the system, it is also taken into account. Therefore, the correct answer is the product of current magnitude, voltage magnitude, and % power factor.

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• 46.

### If 230 volts is applied to a 160 turn primary winding, what is the voltage in the 320 secondary winding of the transformer?

• A.

230

• B.

460

• C.

120

• D.

480

• E.

600

B. 460
Explanation
The voltage in the secondary winding of the transformer can be determined using the turns ratio. In this case, the primary winding has 160 turns and the secondary winding has 320 turns. Since the turns ratio is 2:1, the voltage in the secondary winding will be twice the voltage in the primary winding. Therefore, if the voltage applied to the primary winding is 230 volts, the voltage in the secondary winding will be 460 volts.

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• 47.

### What is the primary and secondary phase displacement of a 3 phase transformer connected delta-wye?

• A.

120 deg

• B.

180 deg

• C.

0 deg

• D.

30 deg

• E.

90 deg

D. 30 deg
Explanation
In a 3 phase transformer connected delta-wye, the primary and secondary phase displacement is 30 degrees. This means that the voltage waveforms of the primary and secondary phases are out of phase with each other by 30 degrees. This phase displacement is important for ensuring proper synchronization and operation of the transformer in a three-phase power system.

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• 48.

### What is the primary and secondary phase displacement of a 3 phase transformer connected delta-delta?

• A.

120 deg

• B.

180 deg

• C.

30 deg

• D.

0 deg

• E.

90 deg

D. 0 deg
Explanation
The primary and secondary phase displacement of a delta-delta connected 3-phase transformer is 0 degrees. This means that the voltage waveforms in the primary and secondary windings are in phase with each other. In other words, the peaks and troughs of the voltage waveforms occur at the same time in both windings. This is because the delta connection does not introduce any phase shift between the primary and secondary sides of the transformer.

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• 49.

### What is the primary and secondary phase displacement of a 3 phase transformer connected wye-wye?

• A.

120 deg

• B.

180 deg

• C.

30 deg

• D.

0 deg

• E.

90 deg

D. 0 deg
Explanation
For a 3 phase transformer connected wye-wye, the primary and secondary phase displacement is 0 degrees. This means that the phase angles of the primary and secondary voltages are in perfect alignment with each other.

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• 50.

### The type of bridging impedance most commonly used on a load-tap changer is:

• A.

Series capacitor

• B.

Resistor

• C.

Contactor

• D.

Preventative auto-transformer

• E.

Step inductor

D. Preventative auto-transformer
Explanation
A load-tap changer is a device used to vary the ratio of a transformer's turns ratio and adjust the output voltage. The most commonly used type of bridging impedance on a load-tap changer is a preventative auto-transformer. This type of auto-transformer is designed to provide voltage regulation and prevent voltage fluctuations in the system. It achieves this by tapping into the system voltage and adjusting it to the desired level. By using a preventative auto-transformer, the load-tap changer can effectively regulate the voltage and ensure stable operation of the transformer.

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