EEE-360 Final Exam |Questions and Answers Latest
Updated 2026/2027 (Graded A+)- ASU
Question1(100pts): The f Use step by step method to find i0(t) for t>0 in the following first order circuit if switch
opens at t=0. 25 points
Following balanced three-phase wye-wye system is energized by the line-to-neutral voltage, which is at a
reference angle of zero for phase a. The system has a 4.5 kW load with a power factor of 0.7 lagging and a line
voltage of 480 V. Calculate the line current in phase a (in amps).
Enter only the current magnitude for automated grading.
Correct Answer: 7.73
Answer range +/- 0.01 (7.72 - 7.74)
Response The given information includes:
Feedback:
P3ph_load = 4500 W; pfload = 0.7 (lagging); Van = (480 V) / sqrt(3) = 277.1 V / 0°
A balanced three-phase Y-Y system can be represented by a single-phase equivalent circuit.
This circuit carries one-third of the total three-phase power.
P1ph = P3ph_load / 3 = (4500 W) / 3 = 1500 W
The load current is computed using Eq. (3.38) while referring to Table 3.4 for the proper sign of the
exponential for a lagging power factor. Note: Make sure your calculator is in radians mode for the following
step.
Ia = [P1ph / (conjg(Van) pfload)] exp[–j acos(pfload)]
= [1500 W / [(277.1 V / 0°)* (0.7)] exp[–j acos(0.7)] = (5.41 – 5.52 j) A
|Ia| = sqrt[ (5.41)2 + (–5.52)2 ] = 7.73 A
Note: switch calculator back to degree mode for following step.
/ Ia = arctan( –5..41) = –45.6°
The magnitude of the line current is the same in all three phases.
Question 2 (120 pts)
A balanced three-phase wye-wye system has the transformer secondary line-to-neutral voltage set as the
reference voltage. The transformer reduces the line voltage of 250 kV to supply a plant operating at a line
voltage of 50 kV. The plant draws 60 MW with a lagging power factor of 70 percent.
Calculate the line current (in amps) delivered by the generator.
Enter only the current magnitude for automated grading.
Correct Answer: 198
Answer range +/- 1 (197.0 - 199.0)
Response Feedback: Vload_an = 50 kV / sqrt(3) = 28.86 kV / 0°
Turns ratio: T = VP / VS = (250 kV) / (50 kV) = 5
On a per phase basis
Updated 2026/2027 (Graded A+)- ASU
Question1(100pts): The f Use step by step method to find i0(t) for t>0 in the following first order circuit if switch
opens at t=0. 25 points
Following balanced three-phase wye-wye system is energized by the line-to-neutral voltage, which is at a
reference angle of zero for phase a. The system has a 4.5 kW load with a power factor of 0.7 lagging and a line
voltage of 480 V. Calculate the line current in phase a (in amps).
Enter only the current magnitude for automated grading.
Correct Answer: 7.73
Answer range +/- 0.01 (7.72 - 7.74)
Response The given information includes:
Feedback:
P3ph_load = 4500 W; pfload = 0.7 (lagging); Van = (480 V) / sqrt(3) = 277.1 V / 0°
A balanced three-phase Y-Y system can be represented by a single-phase equivalent circuit.
This circuit carries one-third of the total three-phase power.
P1ph = P3ph_load / 3 = (4500 W) / 3 = 1500 W
The load current is computed using Eq. (3.38) while referring to Table 3.4 for the proper sign of the
exponential for a lagging power factor. Note: Make sure your calculator is in radians mode for the following
step.
Ia = [P1ph / (conjg(Van) pfload)] exp[–j acos(pfload)]
= [1500 W / [(277.1 V / 0°)* (0.7)] exp[–j acos(0.7)] = (5.41 – 5.52 j) A
|Ia| = sqrt[ (5.41)2 + (–5.52)2 ] = 7.73 A
Note: switch calculator back to degree mode for following step.
/ Ia = arctan( –5..41) = –45.6°
The magnitude of the line current is the same in all three phases.
Question 2 (120 pts)
A balanced three-phase wye-wye system has the transformer secondary line-to-neutral voltage set as the
reference voltage. The transformer reduces the line voltage of 250 kV to supply a plant operating at a line
voltage of 50 kV. The plant draws 60 MW with a lagging power factor of 70 percent.
Calculate the line current (in amps) delivered by the generator.
Enter only the current magnitude for automated grading.
Correct Answer: 198
Answer range +/- 1 (197.0 - 199.0)
Response Feedback: Vload_an = 50 kV / sqrt(3) = 28.86 kV / 0°
Turns ratio: T = VP / VS = (250 kV) / (50 kV) = 5
On a per phase basis
