EEE 360 Midterm 2 Exam |
Questions and Answers | 2026
Update | 100% Correct - Arizona
State University.
Course
EEE 360
1. In a balanced three-phase system, the phase voltages are separated by:
A. 60°
B. 90°
C. 120°
D. 180°
Answer: C. 120°
Rationale: In a balanced three-phase system, the three sinusoidal phase quantities have equal
magnitude and are displaced by 120 electrical degrees.
2. A balanced three-phase Y-connected load has a line-to-line voltage of 480 V. What is the
approximate phase voltage?
A. 160 V
B. 277 V
C. 415 V
D. 480 V
Answer: B. 277 V
Rationale: For a balanced wye connection, V_phase = V_line/√3. Therefore, 480/√3 ≈ 277 V.
3. A balanced delta-connected load has a phase current of 10 A. What is the approximate
line-current magnitude?
A. 5.77 A
B. 10 A
C. 17.32 A
D. 30 A
Answer: C. 17.32 A
Rationale: For a balanced delta load, I_line = √3 I_phase. Thus, I_line = 1.732 × 10 ≈ 17.32 A.
4. A balanced wye-connected load has a line current of 12 A. What is the phase current?
,A. 6.93 A
B. 12 A
C. 20.78 A
D. 36 A
Answer: B. 12 A
Rationale: In a wye-connected load, line current equals phase current in magnitude.
5. A three-phase load consumes 20 kW at a power factor of 0.8 lagging. What is its apparent
power?
A. 16 kVA
B. 20 kVA
C. 25 kVA
D. 32 kVA
Answer: C. 25 kVA
Rationale: Apparent power S = P/pf = 20/0.8 = 25 kVA.
6. A three-phase load operates at 480 V line-to-line, 30 A line current, and 0.9 power factor.
What is its approximate real power?
A. 11.2 kW
B. 22.4 kW
C. 25.9 kW
D. 43.2 kW
Answer: C. 22.4 kW
Rationale: P = √3 V_L I_L pf = 1.732 × 480 × 30 × 0.9 ≈ 22.4 kW.
7. A balanced three-phase load has an impedance of 12 + j16 Ω per phase and is connected
in wye to a 480-V line-to-line source. What is the magnitude of the phase impedance?
A. 12 Ω
B. 16 Ω
C. 20 Ω
D. 28 Ω
Answer: C. 20 Ω
Rationale: |Z| = √(R² + X²) = √(12² + 16²) = √400 = 20 Ω.
8. For the load in Question 7, what is the approximate line-current magnitude?
, A. 6.0 A
B. 13.9 A
C. 20.0 A
D. 24.0 A
Answer: B. 13.9 A
Rationale: The phase voltage is 480/√3 ≈ 277 V. Since the load is wye, line current equals phase
current: I ≈ 277/20 = 13.9 A.
9. A three-phase load has a power factor of 0.6 lagging. The corresponding impedance angle
is approximately:
A. 36.9° lagging
B. 53.1° lagging
C. 60° leading
D. 90° leading
Answer: B. 53.1° lagging
Rationale: pf = cos θ. Therefore θ = cos⁻¹(0.6) ≈ 53.1°. An inductive load has a lagging power
factor.
10. What is the primary purpose of improving a power factor from 0.7 to 0.95 for a given real
power and voltage?
A. Increase the required line current
B. Reduce the required line current
C. Increase transmission losses
D. Eliminate all reactive power in every system
Answer: B. Reduce the required line current
Rationale: For fixed real power and voltage, higher power factor means lower current. Lower
current reduces I²R losses and voltage drop.
11. A transformer has 1000 primary turns and 200 secondary turns. If the primary voltage is
500 V, what is the ideal secondary voltage?
A. 50 V
B. 100 V
C. 250 V
D. 2500 V
Answer: B. 100 V
Rationale: For an ideal transformer, V₂/V₁ = N₂/N₁. Thus V₂ = 500(200/1000) = 100 V.
Questions and Answers | 2026
Update | 100% Correct - Arizona
State University.
Course
EEE 360
1. In a balanced three-phase system, the phase voltages are separated by:
A. 60°
B. 90°
C. 120°
D. 180°
Answer: C. 120°
Rationale: In a balanced three-phase system, the three sinusoidal phase quantities have equal
magnitude and are displaced by 120 electrical degrees.
2. A balanced three-phase Y-connected load has a line-to-line voltage of 480 V. What is the
approximate phase voltage?
A. 160 V
B. 277 V
C. 415 V
D. 480 V
Answer: B. 277 V
Rationale: For a balanced wye connection, V_phase = V_line/√3. Therefore, 480/√3 ≈ 277 V.
3. A balanced delta-connected load has a phase current of 10 A. What is the approximate
line-current magnitude?
A. 5.77 A
B. 10 A
C. 17.32 A
D. 30 A
Answer: C. 17.32 A
Rationale: For a balanced delta load, I_line = √3 I_phase. Thus, I_line = 1.732 × 10 ≈ 17.32 A.
4. A balanced wye-connected load has a line current of 12 A. What is the phase current?
,A. 6.93 A
B. 12 A
C. 20.78 A
D. 36 A
Answer: B. 12 A
Rationale: In a wye-connected load, line current equals phase current in magnitude.
5. A three-phase load consumes 20 kW at a power factor of 0.8 lagging. What is its apparent
power?
A. 16 kVA
B. 20 kVA
C. 25 kVA
D. 32 kVA
Answer: C. 25 kVA
Rationale: Apparent power S = P/pf = 20/0.8 = 25 kVA.
6. A three-phase load operates at 480 V line-to-line, 30 A line current, and 0.9 power factor.
What is its approximate real power?
A. 11.2 kW
B. 22.4 kW
C. 25.9 kW
D. 43.2 kW
Answer: C. 22.4 kW
Rationale: P = √3 V_L I_L pf = 1.732 × 480 × 30 × 0.9 ≈ 22.4 kW.
7. A balanced three-phase load has an impedance of 12 + j16 Ω per phase and is connected
in wye to a 480-V line-to-line source. What is the magnitude of the phase impedance?
A. 12 Ω
B. 16 Ω
C. 20 Ω
D. 28 Ω
Answer: C. 20 Ω
Rationale: |Z| = √(R² + X²) = √(12² + 16²) = √400 = 20 Ω.
8. For the load in Question 7, what is the approximate line-current magnitude?
, A. 6.0 A
B. 13.9 A
C. 20.0 A
D. 24.0 A
Answer: B. 13.9 A
Rationale: The phase voltage is 480/√3 ≈ 277 V. Since the load is wye, line current equals phase
current: I ≈ 277/20 = 13.9 A.
9. A three-phase load has a power factor of 0.6 lagging. The corresponding impedance angle
is approximately:
A. 36.9° lagging
B. 53.1° lagging
C. 60° leading
D. 90° leading
Answer: B. 53.1° lagging
Rationale: pf = cos θ. Therefore θ = cos⁻¹(0.6) ≈ 53.1°. An inductive load has a lagging power
factor.
10. What is the primary purpose of improving a power factor from 0.7 to 0.95 for a given real
power and voltage?
A. Increase the required line current
B. Reduce the required line current
C. Increase transmission losses
D. Eliminate all reactive power in every system
Answer: B. Reduce the required line current
Rationale: For fixed real power and voltage, higher power factor means lower current. Lower
current reduces I²R losses and voltage drop.
11. A transformer has 1000 primary turns and 200 secondary turns. If the primary voltage is
500 V, what is the ideal secondary voltage?
A. 50 V
B. 100 V
C. 250 V
D. 2500 V
Answer: B. 100 V
Rationale: For an ideal transformer, V₂/V₁ = N₂/N₁. Thus V₂ = 500(200/1000) = 100 V.
