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ALBERTA ELECTRICIAN APPREnticeship CERTIFICATION EXAM with Questions and Answers/Plus a Rationale Updated 2026 A+/Instant Download PDF

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ALBERTA ELECTRICIAN APPREnticeship CERTIFICATION EXAM with Questions and Answers/Plus a Rationale Updated 2026 A+/Instant Download PDF

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ALBERTA ELECTRICIAN APPREnticeship
CERTIFICATION EXAM with Questions and Answers/Plus
a Rationale Updated 2026 A+/Instant Download PDF
EXAM COVERAGE - 1. Electrical Theory and Circuit
Calculations - 2. Canadian Electrical Code (CEC) Part I
Regulations - 3. Motor Controls and Protection Systems - 4.
Transformers and Distribution Systems - 5. Industrial
Electronics and Instrumentation
1. A 3-phase, 600V, 50 HP induction motor operating at full load draws a current of 45A with a
power factor of 0.85 lagging. What is the total active power consumed by the motor under these
operating conditions?

A. 32.1 kW

B. 39.7 kW

C. 46.8 kW

D. 52.3 kW

CORRECT ANSWER : B

Rationale: Active power for a 3-phase system is calculated using the formula $P = \sqrt{3}
\times V_L \times I_L \times \cos(\theta)$. Substituting the given values yields $P = \sqrt{3}
\times 600 \times 45 \times 0.85 = 39,675 \, \text{W}$, or approximately 39.7 kW. Options A, C,
and D reflect calculation errors from omitting the square root of 3, misapplying the power
factor, or confusing active and apparent power.

2. According to the Canadian Electrical Code (CEC) Part I, what is the minimum ampacity required
for conductors supplying a continuous-duty motor with a full-load current rating of 50A?

A. 50.0A

B. 57.5A

C. 62.5A

D. 75.0A

CORRECT ANSWER : C

, Rationale: Rule 28-104 of the CEC requires conductors supplying a continuous-duty motor to
have an ampacity of not less than 125% of the motor's full-load current rating. Calculating
$125\% \times 50\text{A} = 62.5\text{A}$. Option A fails to apply the required safety factor for
continuous loads, while options B and D incorrectly apply other percentage multipliers.

3. When sizing overcurrent protection for a capacitor bank connected across a motor terminal to
improve power factor, what is the maximum allowable setting of the branch-circuit fuses as a
percentage of the capacitor's rated current under the CEC?

A. 135%

B. 150%

C. 250%

D. 300%

CORRECT ANSWER : C

Rationale: CEC Rule 26-222 specifies that overcurrent protection for capacitors shall not exceed
250% of the rated capacitor current to handle transient inrush currents safely. Options A, B, and
D represent incorrect multiplier values specified elsewhere in the code for different equipment
types.

4. A 240V single-phase branch circuit feeds a resistive heating load of 4.8 kW in a commercial
building. What is the minimum standard overcurrent device rating required for this circuit if it is
designated as a continuous load?

A. 20A

B. 25A

C. 30A

D. 40A

CORRECT ANSWER : C

Rationale: For continuous loads, the circuit must be rated for 125% of the load current. The
load current is $4800\text{W} / 240\text{V} = 20\text{A}$. Multiplying by 1.25 gives 25A, but
standard overcurrent device ratings require selecting the next higher standard size, which is 30A
per CEC rules. Options A, B, and D either miscalculate the continuous load factor or fail to
select the correct standard fuse/breaker size.

,5. In a three-phase, four-wire wye distribution system supplying unbalanced single-phase line-to-
neutral loads, what is the primary consequence if the neutral conductor experiences an open
circuit?

A. Phase voltages remain balanced due to line impedance.

B. Phase voltages will shift, leading to overvoltage on lightly loaded phases and
undervoltage on heavily loaded phases.

C. Total system current increases to offset the open neutral.

D. Motor loads connected to the system will experience immediate thermal overload without
phase loss.

CORRECT ANSWER : B

Rationale: An open neutral in an unbalanced 3-phase 4-wire system causes a neutral shift,
forcing the phase-to-neutral voltages to divide unevenly based on load impedances. This creates
dangerous overvoltages on phases with lighter loads and drops voltage on heavier loads.
Options A, C, and D are technically incorrect because line voltages do not remain stable and
total current cannot increase magically without a return path.

6. Which of the following defines the maximum allowable voltage drop for a branch circuit and
feeder combined, as recommended by the Appendix B notes of the Canadian Electrical Code?

A. 2% for branch circuits and 1% for feeders

B. 3% for branch circuits or feeders, and a maximum of 5% total

C. 5% for branch circuits and 5% for feeders

D. 10% total system drop under peak motor starting conditions

CORRECT ANSWER : B

Rationale: CEC Appendix B guidelines recommend that voltage drop should not exceed 3% in a
branch circuit or feeder, with a maximum combined total of 5% to ensure proper operation of
electrical equipment. Options A, C, and D misstate these standard design recommendations.

7. What is the minimum burial depth required for direct-buried underground cables operating at
300V to ground, installed under a residential driveway subject to vehicular traffic, according to
the CEC?

A. 450 mm

B. 600 mm

, C. 900 mm

D. 1000 mm

CORRECT ANSWER : C

Rationale: CEC Table 53 specifies that direct-buried cables subject to heavy vehicular traffic
(like driveways) must be buried at a minimum depth of 900 mm. Options A and B apply to lesser-
traffic areas or raceway installations, while option D exceeds standard requirements.

8. When troubleshooting a 3-phase squirrel-cage induction motor that hums and fails to start while
drawing locked-rotor current on all three lines, what is the most likely root cause?

A. Single-phasing due to a blown fuse on one line

B. Mechanical seizure of the motor bearings or driven load

C. An open-circuit condition in the control circuit holding coil

D. Excessive voltage drop across the supply lines

CORRECT ANSWER : B

Rationale: If current is drawn on all three lines but the motor only hums and fails to rotate, the
stator is energized, pointing to a severe mechanical lock preventing rotor movement. A single-
phase fault would cause current draw on only two lines, and a control circuit issue would
prevent any contactor closure.

9. A transformer has a turns ratio of 5:1 (primary to secondary). If 120V is applied to the primary
winding and the secondary feeds an impedance of 4 ohms, what is the primary reflected
impedance?

A. 20 ohms

B. 48 ohms

C. 100 ohms

D. 400 ohms

CORRECT ANSWER : C

Rationale: Reflected impedance in a transformer is scaled by the square of the turns ratio ($Z_p
= a^2 \times Z_s$, where $a = 5/1$). Thus, $Z_p = 5^2 \times 4 = 25 \times 4 = 100\text{
ohms}$. Options A, B, and D miscalculate by failing to square the turns ratio or inverting the
formula.

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Subido en
23 de julio de 2026
Número de páginas
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2025/2026
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