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TEXAS PROFESSIONAL ENGINEER ELECTRICAL ENGINEERING LICENSING EXAM ACTUAL 2026 VERIFIED QUESTIONS WITH ANSWERS COMPREHENSIVE & RATIONALE|INSTANT DOWNLOAD

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TEXAS PROFESSIONAL ENGINEER ELECTRICAL ENGINEERING LICENSING EXAM ACTUAL 2026 VERIFIED QUESTIONS WITH ANSWERS COMPREHENSIVE & RATIONALE|INSTANT DOWNLOAD TEXAS PROFESSIONAL ENGINEER ELECTRICAL ENGINEERING LICENSING EXAM ACTUAL 2026 VERIFIED QUESTIONS WITH ANSWERS COMPREHENSIVE & RATIONALE|INSTANT DOWNLOAD TEXAS PROFESSIONAL ENGINEER ELECTRICAL ENGINEERING LICENSING EXAM ACTUAL 2026 VERIFIED QUESTIONS WITH ANSWERS COMPREHENSIVE & RATIONALE|INSTANT DOWNLOAD

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TEXAS PROFESSIONAL ENGINEER
ELECTRICAL ENGINEERING LICENSING
EXAM ACTUAL 2026 VERIFIED QUESTIONS
WITH ANSWERS COMPREHENSIVE &
RATIONALE|INSTANT DOWNLOAD

1. A balanced three-phase power system has a line-to-line voltage of
480 V. What is the phase voltage in a wye-connected system?
A. 120 V
B. 277 V
C. 480 V
D. 831 V
Answer: B. 277 V
Rationale: In a balanced wye-connected three-phase system, phase
voltage equals line voltage divided by √3. Therefore, Vphase = 480/√3
= 277 V. This relationship is fundamental in power distribution
calculations.


2. A 60 Hz transformer has 500 primary turns and 100 secondary
turns. If the primary voltage is 2400 V, what is the secondary
voltage?
A. 120 V
B. 240 V
C. 480 V
D. 600 V
Answer: D. 480 V


1

,Rationale: Transformer voltage ratio follows the turns ratio: Vp/Vs =
Np/Ns. Therefore, 2400/Vs = 500/100, giving Vs = 480 V.


3. In an AC circuit, the real power consumed by a load is calculated
as:
A. VI
B. VI sinθ
C. VI cosθ
D. V/I
Answer: C. VI cosθ
Rationale: Real power is the portion of apparent power converted into
useful work. The formula is P = VI cosθ, where cosθ is the power
factor.


4. A motor draws 50 A from a 480 V three-phase supply with a
power factor of 0.85. What is the approximate input power?
A. 17.7 kW
B. 35.3 kW
C. 53.4 kW
D. 70 kW
Answer: B. 35.3 kW
Rationale: Three-phase power is calculated using P = √3VIcosθ.
Substituting values: P = 1.732 × 480 × 50 × 0.85 = 35,346 W or
approximately 35.3 kW.


5. The primary purpose of a circuit breaker in an electrical system is
to:

2

,A. Increase voltage
B. Improve power factor
C. Interrupt fault currents
D. Reduce harmonic distortion
Answer: C. Interrupt fault currents
Rationale: Circuit breakers protect electrical systems by opening the
circuit during abnormal conditions such as short circuits and
overloads.


6. According to Ohm’s Law, current through a resistor is:
A. Directly proportional to resistance
B. Voltage divided by resistance
C. Resistance divided by voltage
D. Independent of voltage
Answer: B. Voltage divided by resistance
Rationale: Ohm’s Law states I = V/R. Current increases with voltage
and decreases with resistance.


7. The impedance of an ideal inductor is:
A. R
B. 1/jωC
C. jωL
D. 1/L
Answer: C. jωL
Rationale: Inductive reactance increases with frequency and
inductance. The impedance is ZL = jωL.


3

, 8. A capacitor in an AC circuit causes current to:
A. Lag voltage by 90 degrees
B. Lead voltage by 90 degrees
C. Be in phase with voltage
D. Become zero
Answer: B. Lead voltage by 90 degrees
Rationale: In a purely capacitive circuit, current leads voltage by 90
degrees because capacitor current depends on the rate of voltage
change.


9. The main function of a transformer core is to:
A. Reduce resistance
B. Increase capacitance
C. Provide a magnetic path
D. Store electrical energy
Answer: C. Provide a magnetic path
Rationale: The iron core provides a low-reluctance path for magnetic
flux, improving transformer efficiency.


10. The short-circuit current of a power system is mainly limited by:
A. Frequency
B. System impedance
C. Voltage rating
D. Power factor
Answer: B. System impedance
Rationale: Fault current is determined by the available voltage divided
by the impedance between the source and fault point.

4

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