FLORIDA BOARD OF PROFESSIONAL
ENGINEERS FUNDAMENTALS OF
ENGINEERING 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 sinusoidal voltage is given by v(t)=170sin(377t+30∘) V. What is
the RMS voltage?
A. 85.0 V
B. 120.2 V
C. 170 V
D. 240.4 V
Answer: B. 120.2 V
Rationale: For a sinusoidal waveform, Vrms=Vpeak/2. Therefore,
170/2=120.2 V. The phase angle does not affect the RMS magnitude.
2. An impedance of Z=6+j8 Ω is connected to a 120-V RMS source.
What is the magnitude of the resulting current?
A. 8 A
B. 10 A
C. 12 A
D. 20 A
Answer: B. 12 A
1
,Rationale: The impedance magnitude is ∣Z∣=62+82=10 Ω. Ohm's law
for AC circuits gives I=V/∣Z∣=120/10=12 A.
3. A 24-V ideal voltage source is connected in series with a 6-Ω
resistor. What power is dissipated by the resistor?
A. 24 W
B. 48 W
C. 96 W
D. 144 W
Answer: C. 96 W
Rationale: The current is I=V/R=24/6=4 A. Resistor power is
P=I2R=(4)2(6)=96 W.
4. A 10-Ω resistor and a 20-Ω resistor are connected in parallel across
a 60-V source. What is the total source current?
A. 3 A
B. 6 A
C. 9 A
D. 12 A
Answer: C. 9 A
Rationale: The equivalent resistance is Req=(10×20)/(10+20)=6.667 Ω.
Therefore, IT=60/6.667=9 A.
5. A capacitor of 50 μF is connected to a 400-Hz sinusoidal source.
What is its capacitive reactance?
A. 3.98 Ω
B. 7.96 Ω
2
,C. 15.9 Ω
D. 31.8 Ω
Answer: B. 7.96 Ω
Rationale: Capacitive reactance is XC=1/(2πfC). Substitution gives XC
=1/[2π(400)(50×10−6)]≈7.96 Ω.
6. A 0.20-H inductor is connected to a 60-Hz source. What is its
inductive reactance?
A. 12.6 Ω
B. 37.7 Ω
C. 75.4 Ω
D. 120.0 Ω
Answer: B. 75.4 Ω
Rationale: Inductive reactance is XL=2πfL. Thus XL
=2π(60)(0.20)=75.4 Ω.
7. A first-order RC circuit has R=10kΩ and C=20μF. What is its time
constant?
A. 0.002 s
B. 0.02 s
C. 0.20 s
D. 2.0 s
Answer: C. 0.20 s
Rationale: For an RC circuit, τ=RC. Therefore,
τ=(10,000)(20×10−6)=0.20 s. After one time constant during charging,
the capacitor voltage reaches approximately 63.2% of its final value.
3
, 8. A Thevenin equivalent circuit consists of a 12-V source in series
with a 4-Ω resistance. What current flows when a 2-Ω load is
connected?
A. 1 A
B. 2 A
C. 3 A
D. 6 A
Answer: B. 2 A
Rationale: The total resistance is 4+2=6 Ω. Thus I=12/6=2 A. The load
voltage would be 2(2)=4 V.
9. A Norton equivalent circuit has a Norton current of 8 A and
Norton resistance of 5 Ω. What is the corresponding Thevenin
voltage?
A. 1.6 V
B. 13 V
C. 40 V
D. 64 V
Answer: C. 40 V
Rationale: The Norton-to-Thevenin relationship is VTH=INRN. Thus
VTH=8(5)=40 V, with RTH=RN=5 Ω.
10. A load resistance receives maximum power from a Thevenin
source when the load resistance is:
A. Zero
B. Half the Thevenin resistance
C. Equal to the Thevenin resistance
D. Twice the Thevenin resistance
4
ENGINEERS FUNDAMENTALS OF
ENGINEERING 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 sinusoidal voltage is given by v(t)=170sin(377t+30∘) V. What is
the RMS voltage?
A. 85.0 V
B. 120.2 V
C. 170 V
D. 240.4 V
Answer: B. 120.2 V
Rationale: For a sinusoidal waveform, Vrms=Vpeak/2. Therefore,
170/2=120.2 V. The phase angle does not affect the RMS magnitude.
2. An impedance of Z=6+j8 Ω is connected to a 120-V RMS source.
What is the magnitude of the resulting current?
A. 8 A
B. 10 A
C. 12 A
D. 20 A
Answer: B. 12 A
1
,Rationale: The impedance magnitude is ∣Z∣=62+82=10 Ω. Ohm's law
for AC circuits gives I=V/∣Z∣=120/10=12 A.
3. A 24-V ideal voltage source is connected in series with a 6-Ω
resistor. What power is dissipated by the resistor?
A. 24 W
B. 48 W
C. 96 W
D. 144 W
Answer: C. 96 W
Rationale: The current is I=V/R=24/6=4 A. Resistor power is
P=I2R=(4)2(6)=96 W.
4. A 10-Ω resistor and a 20-Ω resistor are connected in parallel across
a 60-V source. What is the total source current?
A. 3 A
B. 6 A
C. 9 A
D. 12 A
Answer: C. 9 A
Rationale: The equivalent resistance is Req=(10×20)/(10+20)=6.667 Ω.
Therefore, IT=60/6.667=9 A.
5. A capacitor of 50 μF is connected to a 400-Hz sinusoidal source.
What is its capacitive reactance?
A. 3.98 Ω
B. 7.96 Ω
2
,C. 15.9 Ω
D. 31.8 Ω
Answer: B. 7.96 Ω
Rationale: Capacitive reactance is XC=1/(2πfC). Substitution gives XC
=1/[2π(400)(50×10−6)]≈7.96 Ω.
6. A 0.20-H inductor is connected to a 60-Hz source. What is its
inductive reactance?
A. 12.6 Ω
B. 37.7 Ω
C. 75.4 Ω
D. 120.0 Ω
Answer: B. 75.4 Ω
Rationale: Inductive reactance is XL=2πfL. Thus XL
=2π(60)(0.20)=75.4 Ω.
7. A first-order RC circuit has R=10kΩ and C=20μF. What is its time
constant?
A. 0.002 s
B. 0.02 s
C. 0.20 s
D. 2.0 s
Answer: C. 0.20 s
Rationale: For an RC circuit, τ=RC. Therefore,
τ=(10,000)(20×10−6)=0.20 s. After one time constant during charging,
the capacitor voltage reaches approximately 63.2% of its final value.
3
, 8. A Thevenin equivalent circuit consists of a 12-V source in series
with a 4-Ω resistance. What current flows when a 2-Ω load is
connected?
A. 1 A
B. 2 A
C. 3 A
D. 6 A
Answer: B. 2 A
Rationale: The total resistance is 4+2=6 Ω. Thus I=12/6=2 A. The load
voltage would be 2(2)=4 V.
9. A Norton equivalent circuit has a Norton current of 8 A and
Norton resistance of 5 Ω. What is the corresponding Thevenin
voltage?
A. 1.6 V
B. 13 V
C. 40 V
D. 64 V
Answer: C. 40 V
Rationale: The Norton-to-Thevenin relationship is VTH=INRN. Thus
VTH=8(5)=40 V, with RTH=RN=5 Ω.
10. A load resistance receives maximum power from a Thevenin
source when the load resistance is:
A. Zero
B. Half the Thevenin resistance
C. Equal to the Thevenin resistance
D. Twice the Thevenin resistance
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