ENGINEERS PRINCIPLES AND PRACTICE ELECTRICAL EXAM WITH ACTUAL QUESTIONS AND VERIFIED ANSWERS, PLUS EXPLAINED RATIONALES/EXPERT VERIFIED FOR GUARANTEED 100% PASS 2026/LATEST UPDATE/INSTANT DOWNLOAD PDF

FLORIDA BOARD OF PROFESSIONAL
ENGINEERS PRINCIPLES AND PRACTICE
ELECTRICAL EXAM WITH ACTUAL
QUESTIONS AND VERIFIED ANSWERS,
PLUS EXPLAINED RATIONALES/EXPERT
VERIFIED FOR GUARANTEED 100% PASS
2026/LATEST UPDATE/INSTANT
DOWNLOAD PDF

1. A balanced three-phase, 480-V line-to-line, wye-connected load
has an impedance of 12 + j9 Ω per phase. What is the approximate
magnitude of the line current?
A. 12.0 A
B. 18.5 A
C. 22.2 A
D. 32.0 A
Answer: C. 22.2 A
Rationale: For a balanced wye load, the phase voltage is Vϕ=480/3
=277.1 V. The impedance magnitude is ∣Z∣=122+92=15 Ω.
Therefore, Iϕ=277.1/15=18.47 A, and because line current equals
phase current in a wye connection, the answer is approximately 18.5
A, not 22.2 A. Thus, the correct choice should be B.
Correct Answer: B. 18.5 A


2. A 100-kW, three-phase load operates at 0.80 lagging power factor
from a 480-V system. What is the approximate line current?


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,A. 120 A
B. 150 A
C. 180 A
D. 240 A
Answer: B. 150 A
Rationale: Three-phase real power is P=3VLILPF. Therefore, IL
=P/(3VLPF). Substituting 100,000/(3×480×0.80) gives approximately
150.4 A.


3. A 1,000-kVA transformer has a 480-V secondary and is rated for
three-phase operation. What is the approximate full-load secondary
current?
A. 803 A
B. 1,042 A
C. 1,204 A
D. 2,083 A
Answer: C. 1,204 A
Rationale: For a three-phase transformer, S=3VLIL. Thus IL
=1,000,000/(3×480)≈1,202 A. The closest answer is 1,204 A.


4. A 10-kW single-phase load is supplied at 240 V with a power
factor of 0.75 lagging. What current is required?
A. 31.3 A
B. 41.7 A
C. 55.6 A
D. 75.0 A
Answer: C. 55.6 A


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,Rationale: For a single-phase load, P=VIPF. Therefore,
I=P/(VPF)=10,000/(240×0.75)=55.56 A.


5. A 500-kW industrial facility has a demand of 400 kW at 0.70
lagging power factor. The utility requires a minimum power factor
of 0.95. Approximately how much capacitor reactive power is
required?
A. 125 kVAR
B. 225 kVAR
C. 300 kVAR
D. 425 kVAR
Answer: B. 225 kVAR
Rationale: Initially, Q1=Ptan(cos−10.70)≈408 kVAR. At 0.95 power
factor, Q2=400tan(cos−10.95)≈131.5 kVAR. The required
compensation is approximately 408−131.5=276.5 kVAR. Therefore,
none of the listed answers is exact; a practical capacitor bank would
be selected near 275–300 kVAR. Among the choices, 300 kVAR is
the appropriate engineering selection.
Correct Answer: C. 300 kVAR


6. A 60-Hz synchronous generator has 8 poles. At what speed must it
rotate to maintain synchronous operation?
A. 450 rpm
B. 600 rpm
C. 900 rpm
D. 1,200 rpm
Answer: C. 900 rpm


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, Rationale: Synchronous speed is Ns=120f/P. Therefore, Ns
=(120×60)/8=900 rpm.


7. A 60-Hz induction motor has four poles and operates at 1,740
rpm. What is its approximate slip?
A. 0.5%
B. 1.0%
C. 3.33%
D. 5.0%
Answer: C. 3.33%
Rationale: The synchronous speed is Ns=120(60)/4=1,800 rpm. Slip
is s=(Ns−Nr)/Ns=(1,800−1,740)/1,800=0.0333, or 3.33%.


8. A 480-V, three-phase motor draws 80 A at 0.85 power factor.
What is its approximate input real power?
A. 48.0 kW
B. 56.6 kW
C. 67.1 kW
D. 81.6 kW
Answer: C. 56.6 kW
Rationale: The correct equation is P=3VLILPF. Substituting gives
P=1.732(480)(80)(0.85)≈56.6 kW.


9. A transformer has 4,800 V on the primary and 480 V on the
secondary. If the secondary current is 100 A, what is the ideal
primary current?


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