1|Page
NEWEST 2026-2027 ELEVATOR MECHANIC Final
Exam Newest Exam Preparation With Complete
Questions And Correct Answers With Rationales |
Already Graded A+||Brand New Version!!
Question 1
A technician measures 480 VAC RMS on a waveform. What is the
approximate peak voltage of this waveform?
A) 480 V
B) 339 V
C) 679 V
D) 960 V
Answer: C – 679 V
Rationale: Peak voltage is calculated by multiplying the RMS value by
the square root of 2 (approximately 1.414). Therefore, 480 V × 1.414 =
678.7 V, which rounds to 679 V. This fundamental relationship in AC
theory is essential for elevator mechanics when measuring control
circuits and motor supplies, as it ensures proper voltage ratings for
components and safe troubleshooting procedures.
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Question 2
You are working on a brake coil with an inductance of 4 henries in a 60
Hz circuit. What is the inductive reactance?
A) 240 Ω
B) 1507 Ω
C) 15 Ω
D) 377 Ω
Answer: B – 1507 Ω
Rationale: Inductive reactance is calculated using the formula XL = 2πfL.
Substituting the values: XL = 2 × 3.1416 × 60 × 4 = 1507.96 Ω. This
calculation is essential when troubleshooting brake coils and solenoid
circuits, as it helps determine the impedance characteristics that affect
current flow and circuit performance.
Question 3
In a purely inductive AC circuit, by how many degrees does the current
lag the voltage?
A) 0°
B) 45°
C) 90°
D) 180°
Answer: C – 90°
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Rationale: Inductors oppose changes in current, causing the current to
reach its peak one-quarter cycle after the voltage. This 90° phase shift is
critical for understanding motor starting characteristics and power
factor correction in elevator systems. The phase relationship affects
how motors draw current and how power factor correction capacitors
must be sized.
Question 4
What is the voltage drop across a forward‑biased silicon diode?
A) 0.3 V
B) 0.7 V
C) 1.2 V
D) 2.0 V
Answer: B – 0.7 V
Rationale: A silicon diode has a forward voltage drop of approximately
0.7 volts due to the built‑in potential barrier of the PN junction. This
value is a constant reference for rectifier circuits, power supplies, and
logic inputs in elevator control boards.
Question 5
A three‑phase induction motor draws 40 amperes at 480 volts with a
power factor of 0.85. What is the true power (in kilowatts)?
A) 32.6 kW
B) 28.3 kW
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C) 19.2 kW
D) 45.7 kW
Answer: B – 28.3 kW
Rationale: The true power for a three‑phase system is P = √3 × V × I ×
PF. Substituting the values: P = 1.732 × 480 × 40 × 0.85 = 28,277 W, or
28.3 kW. This calculation is used to size feeders and determine the
actual load placed on the building supply.
Question 6
When testing a capacitor start motor, you observe that the auxiliary
winding does not drop out after the motor reaches 75% of rated speed.
What is the most likely cause?
A) The start capacitor is shorted
B) The centrifugal switch is welded closed
C) The run capacitor is open
D) The thermal overload is tripped
Answer: B – The centrifugal switch is welded closed
Rationale: The centrifugal switch is designed to open at approximately
70‑75% of synchronous speed, disconnecting the start winding. If it
remains closed, the start winding and capacitor stay in the circuit,
leading to overheating and possible motor failure. The mechanical
contacts may weld due to arcing or excessive current.
NEWEST 2026-2027 ELEVATOR MECHANIC Final
Exam Newest Exam Preparation With Complete
Questions And Correct Answers With Rationales |
Already Graded A+||Brand New Version!!
Question 1
A technician measures 480 VAC RMS on a waveform. What is the
approximate peak voltage of this waveform?
A) 480 V
B) 339 V
C) 679 V
D) 960 V
Answer: C – 679 V
Rationale: Peak voltage is calculated by multiplying the RMS value by
the square root of 2 (approximately 1.414). Therefore, 480 V × 1.414 =
678.7 V, which rounds to 679 V. This fundamental relationship in AC
theory is essential for elevator mechanics when measuring control
circuits and motor supplies, as it ensures proper voltage ratings for
components and safe troubleshooting procedures.
,2|Page
Question 2
You are working on a brake coil with an inductance of 4 henries in a 60
Hz circuit. What is the inductive reactance?
A) 240 Ω
B) 1507 Ω
C) 15 Ω
D) 377 Ω
Answer: B – 1507 Ω
Rationale: Inductive reactance is calculated using the formula XL = 2πfL.
Substituting the values: XL = 2 × 3.1416 × 60 × 4 = 1507.96 Ω. This
calculation is essential when troubleshooting brake coils and solenoid
circuits, as it helps determine the impedance characteristics that affect
current flow and circuit performance.
Question 3
In a purely inductive AC circuit, by how many degrees does the current
lag the voltage?
A) 0°
B) 45°
C) 90°
D) 180°
Answer: C – 90°
,3|Page
Rationale: Inductors oppose changes in current, causing the current to
reach its peak one-quarter cycle after the voltage. This 90° phase shift is
critical for understanding motor starting characteristics and power
factor correction in elevator systems. The phase relationship affects
how motors draw current and how power factor correction capacitors
must be sized.
Question 4
What is the voltage drop across a forward‑biased silicon diode?
A) 0.3 V
B) 0.7 V
C) 1.2 V
D) 2.0 V
Answer: B – 0.7 V
Rationale: A silicon diode has a forward voltage drop of approximately
0.7 volts due to the built‑in potential barrier of the PN junction. This
value is a constant reference for rectifier circuits, power supplies, and
logic inputs in elevator control boards.
Question 5
A three‑phase induction motor draws 40 amperes at 480 volts with a
power factor of 0.85. What is the true power (in kilowatts)?
A) 32.6 kW
B) 28.3 kW
, 4|Page
C) 19.2 kW
D) 45.7 kW
Answer: B – 28.3 kW
Rationale: The true power for a three‑phase system is P = √3 × V × I ×
PF. Substituting the values: P = 1.732 × 480 × 40 × 0.85 = 28,277 W, or
28.3 kW. This calculation is used to size feeders and determine the
actual load placed on the building supply.
Question 6
When testing a capacitor start motor, you observe that the auxiliary
winding does not drop out after the motor reaches 75% of rated speed.
What is the most likely cause?
A) The start capacitor is shorted
B) The centrifugal switch is welded closed
C) The run capacitor is open
D) The thermal overload is tripped
Answer: B – The centrifugal switch is welded closed
Rationale: The centrifugal switch is designed to open at approximately
70‑75% of synchronous speed, disconnecting the start winding. If it
remains closed, the start winding and capacitor stay in the circuit,
leading to overheating and possible motor failure. The mechanical
contacts may weld due to arcing or excessive current.
