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NCEES - PE Electrical and Computer 2025
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Question: 1
In a balanced three-phase system, the phase voltage is 277V and the total active power is 150 kW. If the power factor is 0.95 lagging, what is the line current?
1. 184 A
2. 209 A
61 A
wer: C
anation: VL = v3 * VP = v3 * 277 = 480 V 3 * VL * IL * cos f
000 = v3 * 480 * IL * 0.95
50,000 / (v3 * 480 * 0.95) = 319 A
stion: 2
is the primary function of the surge suppressor circuit found in some ble frequency drives (VFDs)?
limit the flow of harmonic currents. maintain power factor correction.
protect the VFD from high-voltage transients.
prevent the transmission of electromagnetic interference.
319 A
3. 3
Ans Expl
P = v
150,
IL = 1
Que
What varia
1. To
2. To
3. To
4. To
Answer: C
Explanation: The primary function of the surge suppressor circuit found in some variable frequency drives (VFDs) is to protect the VFD from high- voltage transients. These transients can be caused by switching events or other disturbances in the power system, and the surge suppressor helps to prevent
damage to the sensitive electronic components within the VFD.
Question: 3
In a balanced three-phase system, the phase voltage is 277V and the total reactive power is 75 kVAR. If the power factor is 0.8 lagging, what is the line current?
13 A
54 A
04 A
67 A
wer: C
anation: VL = v3 * VP = v3 * 277 = 480 V er factor = 0.8, so f = arccos(0.8) = 36.87� Q / sin f = 75,000 / sin(36.87�) = 125 kVA
S / (v3 * VL) = 125,000 / (v3 * 480) = 204 A
stion: 4
ch of the following is the most effective method to mitigate the impact dard 6-pulse VFD on the life expectancy of the power factor correctio citors in the facility?
1
1
2
2
Ans Expl
Pow S = IL =
Que
Whi of a
stan n
capa
1. Installing a line reactor
2. Upgrading to a 12-pulse VFD
3. Oversizing the capacitor bank
4. Implementing active harmonic filtering
Explanation: Upgrading from a 6-pulse to a 12-pulse VFD is the most effective method to mitigate the impact of a standard 6-pulse VFD on the life expectancy of the power factor correction capacitors in the facility. The 12-pulse configuration significantly reduces the harmonic content in the input current, which helps to minimize the additional stress and heating on the capacitors.
ree-phase, 460 V, 60 Hz, 10 hp induction motor has a full-load efficie 0% and a full-load power factor of 0.85. The full-load phase current of r is approximately:
9 A
0.5 A
2.1 A
3.7 A
wer: B
anation: To calculate the full-load phase current, we can use the form current (A) = (Full-load line current) / v3
re:
load line current = 15.5 A (from Question 15) ging in the value, we get:
current = 15.5 / v3 = 10.5 A
stion: 6
Question: 5
A th ncy
of 9 the
moto
1. 8.
2. 1
3. 1
4. 1
Ans
Expl ula:
Phase Whe Full- Plug Phase
Que
A 3-phase, 480 V, 60 Hz, 100 HP induction motor has a full-load efficiency of 94% and a full-load power factor of 0.90 lagging. What is the approximate full- load current of the motor?
1. 120 A
2. 130 A
3. 140 A
4. 150 A
Answer: B Explanation:
re:
horsepower = 100 HP ine voltage = 480 V
full-load efficiency = 0.94
full-load power factor = 0.90 lagging
ging in the values:
100 � 746) / (v3 � 480 � 0.94 � 0.90)
29.61 A
nding to the nearest 10 A, the answer is 130 A.
stion: 7
phase, 60 Hz, 480 V, 100 HP induction motor is operating at a power of 0.85 lagging and an efficiency of 92%. What is the line current dr e motor?
To calculate the full-load current of the motor, we can use the formula: I = (HP � 746) / (v3 � V � Eff � PF)
Whe HP =
V = l Eff = PF =
Plug I = ( I = 1
Rou Que A 3-
factor awn
by th
1. 132 A
2. 138 A
3. 144 A
4. 150 A
Explanation:
To find the line current drawn by the induction motor, we can use the formula: I = (P_in / (v3 � V � cos(?) � ?)) � 1000 / 746
Where:
I = Line current (in A)
P_in = Input power (in kW) V = Line voltage (480 V)
fficiency (0.92)
nput power can be calculated from the mechanical power and the iency:
= P_mech / ?
= (100 HP � 746 W/HP) / 0.92
= 81.3 kW
tituting the values, we get:
81.3 kW / (v3 � 480 V � 0.85 � 0.92)) � 1000 / 746
38 A
efore, the line current drawn by the motor is 138 A.
stion: 8
phase, 480 V, 60 Hz, 4-pole induction motor has a full-load efficiency and a full-load power factor of 0.85. What is the approximate efficien otor at 75% of full load?
? = Power factor angle (cos^-1(0.85) = 31.79�)
? = E
The i effic P_in P_in P_in
Subs I = ( I = 1
Ther
Que
A 3- of
92% cy of
the m
1. 90%
2. 91%
3. 92%
4. 93%
Answer: B
Explanation: The efficiency of an induction motor typically increases as the load increases, reaching a maximum around the full-load point. For a motor with a full-load efficiency of 92%, the efficiency at 75% of full load is typically around 91%.
Question: 9
he output waveform is pulsating and unidirectional. he output voltage is always positive.
he average output voltage is equal to the peak value of the input volta he output waveform contains a large amount of ripple.
wer: C
anation: In a half-wave rectifier, the average output voltage is not equ eak value of the input voltage. The average output voltage of a half-w fier is approximately 0.318 times the peak value of the input voltage. T e to the fact that the rectifier only allows one half-cycle of the input A eform to pass through, resulting in a pulsating unidirectional output.
stion: 10
ree-phase, 480 V, 60 Hz, 100 hp induction motor has the following meters:
Which of the following is not a characteristic of a half-wave rectifier?
1. T
2. T
3. T ge.
4. T
Ans
Expl al to
the p ave
recti his
is du C
wav
Que A th para
Stator resistance (Rs) = 0.2 O/phase Rotor resistance (Rr) = 0.15 O/phase Stator reactance (Xs) = 1.2 O/phase Rotor reactance (Xr) = 1.0 O/phase
Magnetizing reactance (Xm) = 25 O/phase
The motor is operating at 0.85 power factor, lagging. What is the motor's full- load current?
1. 100 A
2. 120 A
3. 140 A
4. 160 A
anation:
alculate the motor's full-load current, we need to first determine the valent circuit parameters and then use them to find the current.
quivalent circuit parameters can be calculated as follows: valent resistance (R_eq) = Rs + Rr = 0.2 O + 0.15 O = 0.35 O
valent reactance (X_eq) = Xs + Xr = 1.2 O + 1.0 O = 2.2 O
valent impedance (Z_eq) = sqrt(R_eq^2 + X_eq^2) = sqrt(0.35^2 + 2. 23 O
er factor = cos(atan(X_eq/R_eq)) = cos(atan(2.2/0.35)) = 0.85 (laggin ull-load current can be calculated as:
load current = (100 hp � 746 W/hp) / (v3 � 480 V � 0.85) = 120 A efore, the motor's full-load current is 120 A.
stion: 11
Answer: B Expl
To c equi
The e Equi
Equi
Equi 2^2)
= 2.
Pow g)
The f Full-
Ther
Que
In a balanced three-phase system, the line voltage is 13.8 kV and the total apparent power is 20 MVA. If the power factor is 0.9 lagging, what is the line current?
1. 679 A
2. 836 A
3. 940 A
Answer: B
Explanation: S = v3 * VL * IL 20,000,000 = v3 * 13,800 * IL
IL = 20,000,000 / (v3 * 13,800) = 836 A
stion: 12
is the purpose of cold cranking amps (CCA) in a battery specificatio easure the battery's capacity in ampere-hours
dicate the maximum current the battery can provide
etermine the battery's ability to start an engine in cold weather epresent the battery's internal resistance
wer: C
anation: The cold cranking amps (CCA) specification for a battery ates its ability to start an engine in cold weather. CCA measures the nt a battery can deliver for 30 seconds at 0�F (-18�C) while maintaini mum voltage, which is important for reliable engine starting in low
eratures. The other options do not accurately describe the purpose of C oes not measure capacity, maximum current, or internal resistance.
Que
What n?
1. M
2. In
3. D
4. R
Ans Expl
indic
curre ng a
mini
temp CA
- it d
Question: 13
What is the main advantage of a boost converter over a linear voltage regulator?
1. Higher efficiency
2. Ability to produce an output voltage higher than the input voltage
3. Smaller size and weight
Answer: B
Explanation: The primary advantage of a boost converter over a linear voltage regulator is the ability to produce an output voltage that is higher than the input voltage. This makes boost converters useful for applications where the load requires a higher voltage than the available power source can provide.
stion: 14
phase, 60 Hz, wye-connected, 4,160 V generator is supplying power t e, 60 Hz, delta-connected load. The generator line current is 100 A. W
load kVA?
18 kVA
43 kVA
80 kVA
98 kVA
wer: B anation:
nd the load kVA, we need to use the formula: 3 � V � I
re:
pparent power (in kVA)
Que
A 3- o a 3-
phas hat
is the
1. 7
2. 7
3. 7
4. 7
Ans Expl
To fi S = v Whe S = A
V = Line voltage (4,160 V) I = Line current (100 A)
Since the generator is wye-connected and the load is delta-connected, the line voltage on the load side is the same as the phase voltage on the generator side, which is 4,160 V.
Substituting the values, we get: S = v3 � 4,160 V � 100 A
S = 743 kVA
Therefore, the load kVA is 743 kVA.
Question: 15
ree-phase, 460V, 60 Hz, 4-pole induction motor has the following meters:
r resistance (Rs) = 0.8 O
resistance (Rr) = 0.6 O r reactance (Xs) = 2.0 O reactance (Xr) = 1.8 O power = 100 HP
speed = 1800 rpm
ming the motor is operating at full load, what is the phase current of t r?
A A A A
wer: A
A th para
Stato Rotor Stato Rotor Rated Rated
Assu he
moto
1. 72
2. 76
3. 80
4. 84
Ans Explanation:
To calculate the phase current of the induction motor at full load, we can use the following formula:
I_phase = I_line / v3
Where:
I_phase = Phase current (A) I_line = Line current (A)
ase = 124 A / v3 ase = 71.6 A
efore, the phase current of the motor at full load is approximately 72 A
stion: 16
arehouse has an area of 150 ft x 250 ft and requires an average illumin FC. The luminaires are mounted at a height of 30 ft and have a light of 0.75. What is the minimum total lumens output required from the ing fixtures?
80,000 lumens
10,000 lumens
40,000 lumens
70,000 lumens
wer: B
From the previous question, we found that the line current at full load is 124 A. Substituting the value, we get:
I_ph I_ph
Ther .
Que
A w ance
of 20 loss
factor light
1. 1
2. 2
3. 2
4. 2
Ans
Explanation: To calculate the minimum total lumens output required, we can use the Lumen method:
Total Lumens = (Desired Illuminance � Area) / (Luminaire Efficiency � Light Loss Factor)
Given information:
Warehouse area: 150 ft x 250 ft = 37,500 sq ft Desired average illuminance: 20 FC Luminaire mounting height: 30 ft
Light loss factor: 0.75
: Calculate the total lumens required.
Lumens = (20 FC � 37,500 sq ft) / (0.40 � 0.75) = 208,333 lumens nding up, the correct answer is B) 210,000 lumens.
stion: 17
ch of the following is a common application for a Type 3 SPD? ervice entrance protection
istribution panel protection eceptacle-level protection ll of the above
wer: C
anation: Type 3 SPDs are typically used for receptacle-level protectio iding surge protection for individual electronic devices or equipment.
Step 1: Calculate the Luminaire Efficiency using the mounting height. Luminaire Efficiency = 0.40 (for a mounting height of 30 ft)
Step 2 Total
Rou
Que
Whi
1. S
2. D
3. R
4. A
Ans
Expl n,
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