EEE 360 MIDTERM EXAM 1 COMPLETE QUESTIONS WITH 100%
VERIFIED ANSWERS
Arizona State University
Closed book. Only one sheet (both sides) of handwritten notes and a calculator
are permitted. No book or computer use is allowed. All questions are
compulsory. There is only one correct answer.
Time: 90 minutes + 10 minutes to scan and upload
Question 1
If the load impedance is 𝑍 = 60 + 𝑗60 Ω, the power factor of the load is:
A. ∠ − 45∘
B. 1.0
C. 0.7071
D. None of the above
Correct Answer: C
Detailed Explanation:
The power factor (PF) is the cosine of the angle between voltage and current,
which is also the cosine of the impedance angle.
Given 𝑍 = 60 + 𝑗60 Ω:
1. Calculate the impedance angle:
Imaginary 60
𝜃 = tan−1 ( ) = tan−1 ( ) = tan−1 (1) = 45∘
Real 60
2. Since both real and imaginary parts are positive, the impedance is
inductive, and the current lags voltage. The power factor angle is +45∘ .
, 3. Calculate power factor:
𝑃𝐹 = cos(𝜃) = cos(45∘ ) = 0.7071
Option A (∠ − 45∘ ) is incorrect because it shows the angle but not the power
factor value, and the angle is positive, not negative.
Option B (1.0) would require zero reactance (purely resistive load).
Option D is incorrect because 0.7071 is correct.
Thus, the correct answer is C. 0.7071.
Question 2
For a well-designed power system, which of the following is most desirable?
A. Large voltage regulation, and High Power Factor
B. Small voltage regulation, and High Power Factor
C. Large voltage regulation, and Low Power Factor
D. Small voltage regulation, and Low Power Factor
Correct Answer: B
Detailed Explanation:
This question tests understanding of desirable power system characteristics.
Voltage Regulation:
• Voltage regulation measures the change in load voltage from no-load to
full-load.
-Small voltage regulation means the load voltage remains nearly constant
regardless of load changes.
• Large voltage regulation means voltage drops significantly under load,
which is undesirable because equipment requires stable voltage.
Power Factor:
, • High power factor (close to 1.0) means most of the apparent power is real
power (useful work).
• Low power factor means more reactive power flows, causing higher line
currents, increased losses, and reduced system capacity.
• Utilities penalize low power factor and reward high power factor.
Conclusion: A well-designed power system should have small voltage
regulation (stable voltage) and high power factor (efficient power delivery).
Thus, the correct answer is B. Small voltage regulation, and High Power Factor.
Question 3
With regards to the open circuit test on real power transformers, which of the
following sentence is most appropriate?
A. In an open circuit test, the low voltage side is open circuited and we use the
test results to calculate the parameters of the magnetizing branch
B. In an open circuit test, the high voltage side is open circuited and we use the
test results to calculate the parameters of the equivalent series branch
C. In an open circuit test, the low voltage side is short circuited and we use the
test results to calculate the parameters of the equivalent series branch
D. In an open circuit test, the high voltage side is open circuited and we use the
test results to calculate the parameters of the magnetizing branch
Correct Answer: D
Detailed Explanation:
The open circuit (OC) test is performed to determine the magnetizing branch
parameters (core loss resistance 𝑅𝑐 and magnetizing reactance 𝑋𝑚 ).
Procedure:
1. The high voltage (HV) side is left open circuited (no load connected).
2. Rated voltage is applied to the low voltage (LV) side.
, 3. Measurements taken: input voltage (V_OC), input current (I_OC), and input
power (P_OC).
4. Since the HV side is open, the input current is very small (only magnetizing
current and core loss current).
5. The equivalent series branch (copper losses) is negligible because the
current is small.
Calculations:
• 𝑅𝑐 is calculated from core loss.
• 𝑋𝑚 is calculated from magnetizing current.
Why other options are wrong:
• A: Incorrect because the LV side is NOT open circuited; voltage is applied to
LV, and HV is open.
• B: Incorrect because the OC test does NOT calculate series branch
parameters (that's the short circuit test).
• C: Incorrect because the OC test does NOT involve short-circuiting any side.
Thus, the correct answer is D. In an open circuit test, the high voltage side is
open circuited and we use the test results to calculate the parameters of the
magnetizing branch.
Question 4
For a three-phase system, in which the generator is wye and the load is delta,
which of the following statement is true?
A. For the delta connected load, the magnitude of the line current is √3 times the
magnitude of the delta load (phase) current
B. For the delta connected load, the line current leads the delta load (phase)
current by 30∘
C. For the wye connected generator, the magnitude of the line-to-neutral (phase)
VERIFIED ANSWERS
Arizona State University
Closed book. Only one sheet (both sides) of handwritten notes and a calculator
are permitted. No book or computer use is allowed. All questions are
compulsory. There is only one correct answer.
Time: 90 minutes + 10 minutes to scan and upload
Question 1
If the load impedance is 𝑍 = 60 + 𝑗60 Ω, the power factor of the load is:
A. ∠ − 45∘
B. 1.0
C. 0.7071
D. None of the above
Correct Answer: C
Detailed Explanation:
The power factor (PF) is the cosine of the angle between voltage and current,
which is also the cosine of the impedance angle.
Given 𝑍 = 60 + 𝑗60 Ω:
1. Calculate the impedance angle:
Imaginary 60
𝜃 = tan−1 ( ) = tan−1 ( ) = tan−1 (1) = 45∘
Real 60
2. Since both real and imaginary parts are positive, the impedance is
inductive, and the current lags voltage. The power factor angle is +45∘ .
, 3. Calculate power factor:
𝑃𝐹 = cos(𝜃) = cos(45∘ ) = 0.7071
Option A (∠ − 45∘ ) is incorrect because it shows the angle but not the power
factor value, and the angle is positive, not negative.
Option B (1.0) would require zero reactance (purely resistive load).
Option D is incorrect because 0.7071 is correct.
Thus, the correct answer is C. 0.7071.
Question 2
For a well-designed power system, which of the following is most desirable?
A. Large voltage regulation, and High Power Factor
B. Small voltage regulation, and High Power Factor
C. Large voltage regulation, and Low Power Factor
D. Small voltage regulation, and Low Power Factor
Correct Answer: B
Detailed Explanation:
This question tests understanding of desirable power system characteristics.
Voltage Regulation:
• Voltage regulation measures the change in load voltage from no-load to
full-load.
-Small voltage regulation means the load voltage remains nearly constant
regardless of load changes.
• Large voltage regulation means voltage drops significantly under load,
which is undesirable because equipment requires stable voltage.
Power Factor:
, • High power factor (close to 1.0) means most of the apparent power is real
power (useful work).
• Low power factor means more reactive power flows, causing higher line
currents, increased losses, and reduced system capacity.
• Utilities penalize low power factor and reward high power factor.
Conclusion: A well-designed power system should have small voltage
regulation (stable voltage) and high power factor (efficient power delivery).
Thus, the correct answer is B. Small voltage regulation, and High Power Factor.
Question 3
With regards to the open circuit test on real power transformers, which of the
following sentence is most appropriate?
A. In an open circuit test, the low voltage side is open circuited and we use the
test results to calculate the parameters of the magnetizing branch
B. In an open circuit test, the high voltage side is open circuited and we use the
test results to calculate the parameters of the equivalent series branch
C. In an open circuit test, the low voltage side is short circuited and we use the
test results to calculate the parameters of the equivalent series branch
D. In an open circuit test, the high voltage side is open circuited and we use the
test results to calculate the parameters of the magnetizing branch
Correct Answer: D
Detailed Explanation:
The open circuit (OC) test is performed to determine the magnetizing branch
parameters (core loss resistance 𝑅𝑐 and magnetizing reactance 𝑋𝑚 ).
Procedure:
1. The high voltage (HV) side is left open circuited (no load connected).
2. Rated voltage is applied to the low voltage (LV) side.
, 3. Measurements taken: input voltage (V_OC), input current (I_OC), and input
power (P_OC).
4. Since the HV side is open, the input current is very small (only magnetizing
current and core loss current).
5. The equivalent series branch (copper losses) is negligible because the
current is small.
Calculations:
• 𝑅𝑐 is calculated from core loss.
• 𝑋𝑚 is calculated from magnetizing current.
Why other options are wrong:
• A: Incorrect because the LV side is NOT open circuited; voltage is applied to
LV, and HV is open.
• B: Incorrect because the OC test does NOT calculate series branch
parameters (that's the short circuit test).
• C: Incorrect because the OC test does NOT involve short-circuiting any side.
Thus, the correct answer is D. In an open circuit test, the high voltage side is
open circuited and we use the test results to calculate the parameters of the
magnetizing branch.
Question 4
For a three-phase system, in which the generator is wye and the load is delta,
which of the following statement is true?
A. For the delta connected load, the magnitude of the line current is √3 times the
magnitude of the delta load (phase) current
B. For the delta connected load, the line current leads the delta load (phase)
current by 30∘
C. For the wye connected generator, the magnitude of the line-to-neutral (phase)
