Illinois Electrical Engineer Examination
Advanced 100-Question Practice Exam
2026 | Questions & Answers with
Detailed Rationales | Complete Exam
Prep & Study Guide
1. In a balanced three-phase, four-wire system supplying a linear balanced
load, the neutral current is:
A. Equal to the phase current
B. Equal to the sum of the phase-current magnitudes
C. Zero
D. Equal to √3 times the phase current
Answer: Zero
Rationale: In a balanced three-phase system, the three phase currents are equal
in magnitude and 120° apart, so their phasor sum is zero.
2. A 480-V, three-phase motor draws 72 A at 0.86 power factor. What is the
approximate real input power?
,A. 41.2 kW
B. 51.4 kW
C. 51.4 kW
D. 59.8 kW
Answer: 51.4 kW
Rationale: P = √3VIpf = 1.732(480)(72)(0.86) ≈ 51.4 kW.
3. A lagging power factor on an industrial electrical system is primarily
associated with:
A. Capacitive loads
B. Resistive loads
C. Inductive loads
D. Harmonic filters
Answer: Inductive loads
Rationale: Motors, transformers, and reactors consume reactive power and
typically produce lagging power factor.
4. A capacitor bank is installed across an inductive load primarily to:
A. Increase reactive power consumption
B. Increase system frequency
C. Supply reactive power locally and improve power factor
D. Increase conductor resistance
Answer: Supply reactive power locally and improve power factor
Rationale: Capacitors provide leading reactive current that offsets lagging
reactive current from inductive equipment.
5. For a sinusoidal voltage waveform, the RMS value is related to its peak
value by:
,A. Vrms = 2Vpk
B. Vrms = Vpk/2
C. Vrms = Vpk/√2
D. Vrms = √2Vpk
Answer: Vrms = Vpk/√2
Rationale: For a sinusoidal waveform, RMS magnitude equals approximately
0.707 times the peak magnitude.
6. An ideal transformer has 480 V on its primary and a turns ratio of 4:1,
primary-to-secondary. The secondary voltage is:
A. 60 V
B. 96 V
C. 120 V
D. 1,920 V
Answer: 120 V
Rationale: V₂/V₁ = N₂/N₁. Thus V₂ = 480(1/4) = 120 V.
7. A transformer has a 13.8-kV primary and a 480-V secondary. Its primary-to-
secondary turns ratio is approximately:
A. 2.88:1
B. 13.8:1
C. 28.75:1
D. 57.5:1
Answer: 28.75:1
Rationale: The ideal turns ratio equals the voltage ratio: 13,800/480 = 28.75.
8. In a three-phase system, the relationship between line voltage and phase
voltage for a balanced wye-connected source is:
, A. VL = VP
B. VL = VP/√3
C. VL = √3VP
D. VL = 3VP
Answer: VL = √3VP
Rationale: In a balanced wye system, line voltage magnitude is √3 times phase
voltage magnitude.
9. A balanced delta-connected load has a line current that is:
A. Equal to phase current
B. √3 times the phase current
C. Three times the phase current
D. Phase current divided by √3
Answer: √3 times the phase current
Rationale: For a balanced delta load, IL = √3 IP.
10.What is the primary purpose of an equipment grounding conductor?
A. Carry normal load current
B. Stabilize system frequency
C. Provide a low-impedance path for fault current
D. Increase transformer efficiency
Answer: Provide a low-impedance path for fault current
Rationale: The equipment grounding path facilitates operation of overcurrent
protective devices during ground faults.
11.A circuit breaker has an interrupting rating of 22 kA. If the available fault
current at its terminals is 18 kA, the breaker:
A. Is necessarily undersized
B. Has sufficient interrupting capacity
Advanced 100-Question Practice Exam
2026 | Questions & Answers with
Detailed Rationales | Complete Exam
Prep & Study Guide
1. In a balanced three-phase, four-wire system supplying a linear balanced
load, the neutral current is:
A. Equal to the phase current
B. Equal to the sum of the phase-current magnitudes
C. Zero
D. Equal to √3 times the phase current
Answer: Zero
Rationale: In a balanced three-phase system, the three phase currents are equal
in magnitude and 120° apart, so their phasor sum is zero.
2. A 480-V, three-phase motor draws 72 A at 0.86 power factor. What is the
approximate real input power?
,A. 41.2 kW
B. 51.4 kW
C. 51.4 kW
D. 59.8 kW
Answer: 51.4 kW
Rationale: P = √3VIpf = 1.732(480)(72)(0.86) ≈ 51.4 kW.
3. A lagging power factor on an industrial electrical system is primarily
associated with:
A. Capacitive loads
B. Resistive loads
C. Inductive loads
D. Harmonic filters
Answer: Inductive loads
Rationale: Motors, transformers, and reactors consume reactive power and
typically produce lagging power factor.
4. A capacitor bank is installed across an inductive load primarily to:
A. Increase reactive power consumption
B. Increase system frequency
C. Supply reactive power locally and improve power factor
D. Increase conductor resistance
Answer: Supply reactive power locally and improve power factor
Rationale: Capacitors provide leading reactive current that offsets lagging
reactive current from inductive equipment.
5. For a sinusoidal voltage waveform, the RMS value is related to its peak
value by:
,A. Vrms = 2Vpk
B. Vrms = Vpk/2
C. Vrms = Vpk/√2
D. Vrms = √2Vpk
Answer: Vrms = Vpk/√2
Rationale: For a sinusoidal waveform, RMS magnitude equals approximately
0.707 times the peak magnitude.
6. An ideal transformer has 480 V on its primary and a turns ratio of 4:1,
primary-to-secondary. The secondary voltage is:
A. 60 V
B. 96 V
C. 120 V
D. 1,920 V
Answer: 120 V
Rationale: V₂/V₁ = N₂/N₁. Thus V₂ = 480(1/4) = 120 V.
7. A transformer has a 13.8-kV primary and a 480-V secondary. Its primary-to-
secondary turns ratio is approximately:
A. 2.88:1
B. 13.8:1
C. 28.75:1
D. 57.5:1
Answer: 28.75:1
Rationale: The ideal turns ratio equals the voltage ratio: 13,800/480 = 28.75.
8. In a three-phase system, the relationship between line voltage and phase
voltage for a balanced wye-connected source is:
, A. VL = VP
B. VL = VP/√3
C. VL = √3VP
D. VL = 3VP
Answer: VL = √3VP
Rationale: In a balanced wye system, line voltage magnitude is √3 times phase
voltage magnitude.
9. A balanced delta-connected load has a line current that is:
A. Equal to phase current
B. √3 times the phase current
C. Three times the phase current
D. Phase current divided by √3
Answer: √3 times the phase current
Rationale: For a balanced delta load, IL = √3 IP.
10.What is the primary purpose of an equipment grounding conductor?
A. Carry normal load current
B. Stabilize system frequency
C. Provide a low-impedance path for fault current
D. Increase transformer efficiency
Answer: Provide a low-impedance path for fault current
Rationale: The equipment grounding path facilitates operation of overcurrent
protective devices during ground faults.
11.A circuit breaker has an interrupting rating of 22 kA. If the available fault
current at its terminals is 18 kA, the breaker:
A. Is necessarily undersized
B. Has sufficient interrupting capacity
