EEE 334 Circuits II Midterm Exam | 19 Questions
with Correct Answers & Worked Solutions | 2026
Updated | ASU
Table of Contents
Section Topic Area Questions
I Review of Circuit Fundamentals and Phasors 1–30
II Sinusoidal Steady-State Analysis 31–70
III AC Power Analysis 71–110
IV Three-Phase Circuits 111–145
V Frequency Response and Bode Plots 146–185
VI Resonance and Filter Circuits 186–225
VII Magnetically Coupled Circuits and Transformers 226–265
VIII Laplace Transform in Circuit Analysis 266–305
IX Transfer Functions and System Response 306–345
X Two-Port Networks 346–375
XI Comprehensive Review and Mixed Problems 376–400
,SECTION I: REVIEW OF CIRCUIT FUNDAMENTALS AND PHASORS
(Questions 1–30)
1. Convert the phasor V = 120∠30° V to rectangular form.
A. 60 + j103.92 V
B. 103.92 + j60 V
C. 120 + j30 V
D. 30 + j120 V
B. 103.92 + j60 V.
To convert from polar to rectangular form: V = 120∠30° = 120(cos 30° + j sin 30°) =
120(0.866 + j0.5) = 103.92 + j60 V.
2. Convert the rectangular form V = 6 + j8 V to polar form.
A. 10∠53.13° V
B. 14∠53.13° V
C. 10∠36.87° V
D. 14∠36.87° V
A. 10∠53.13° V.
Magnitude = √(6² + 8²) = √(36 + 64) = √100 = 10 V. Phase angle = tan⁻¹(8/6) =
tan⁻¹(1.333) = 53.13°. Therefore, V = 10∠53.13° V.
,3. The angular frequency ω of a sinusoidal signal with frequency f = 60 Hz is:
A. 60 rad/s
B. 120 rad/s
C. 377 rad/s
D. 60π rad/s
C. 377 rad/s.
Angular frequency ω = 2πf = 2π(60) = 377 rad/s. This is the standard angular frequency
for 60 Hz power systems.
4. A sinusoidal voltage v(t) = 50 sin(200πt + 30°) V. What is the frequency of the
signal?
A. 50 Hz
B. 100 Hz
C. 200 Hz
D. 200π Hz
B. 100 Hz.
The angular frequency ω = 200π rad/s. Since ω = 2πf, f = ω/2π = 200π/2π = 100 Hz.
5. The impedance of an ideal inductor L at angular frequency ω is:
A. jωL
, B. 1/(jωL)
C. ωL
D. 1/(ωL)
A. jωL.
The impedance of an inductor is Z_L = jωL, where ω is the angular frequency and L is the
inductance. This indicates that the inductor's impedance increases with frequency and has
a phase angle of +90°.
6. The impedance of an ideal capacitor C at angular frequency ω is:
A. jωC
B. 1/(jωC)
C. ωC
D. 1/(ωC)
B. 1/(jωC).
The impedance of a capacitor is Z_C = 1/(jωC) = -j/(ωC), where ω is the angular
frequency and C is the capacitance. This indicates that the capacitor's impedance
decreases with frequency and has a phase angle of -90°.
7. A resistor R = 100 Ω, inductor L = 50 mH, and capacitor C = 10 μF are connected
in series. The impedance at ω = 1000 rad/s is:
A. 100 + j50 Ω
B. 100 - j50 Ω
with Correct Answers & Worked Solutions | 2026
Updated | ASU
Table of Contents
Section Topic Area Questions
I Review of Circuit Fundamentals and Phasors 1–30
II Sinusoidal Steady-State Analysis 31–70
III AC Power Analysis 71–110
IV Three-Phase Circuits 111–145
V Frequency Response and Bode Plots 146–185
VI Resonance and Filter Circuits 186–225
VII Magnetically Coupled Circuits and Transformers 226–265
VIII Laplace Transform in Circuit Analysis 266–305
IX Transfer Functions and System Response 306–345
X Two-Port Networks 346–375
XI Comprehensive Review and Mixed Problems 376–400
,SECTION I: REVIEW OF CIRCUIT FUNDAMENTALS AND PHASORS
(Questions 1–30)
1. Convert the phasor V = 120∠30° V to rectangular form.
A. 60 + j103.92 V
B. 103.92 + j60 V
C. 120 + j30 V
D. 30 + j120 V
B. 103.92 + j60 V.
To convert from polar to rectangular form: V = 120∠30° = 120(cos 30° + j sin 30°) =
120(0.866 + j0.5) = 103.92 + j60 V.
2. Convert the rectangular form V = 6 + j8 V to polar form.
A. 10∠53.13° V
B. 14∠53.13° V
C. 10∠36.87° V
D. 14∠36.87° V
A. 10∠53.13° V.
Magnitude = √(6² + 8²) = √(36 + 64) = √100 = 10 V. Phase angle = tan⁻¹(8/6) =
tan⁻¹(1.333) = 53.13°. Therefore, V = 10∠53.13° V.
,3. The angular frequency ω of a sinusoidal signal with frequency f = 60 Hz is:
A. 60 rad/s
B. 120 rad/s
C. 377 rad/s
D. 60π rad/s
C. 377 rad/s.
Angular frequency ω = 2πf = 2π(60) = 377 rad/s. This is the standard angular frequency
for 60 Hz power systems.
4. A sinusoidal voltage v(t) = 50 sin(200πt + 30°) V. What is the frequency of the
signal?
A. 50 Hz
B. 100 Hz
C. 200 Hz
D. 200π Hz
B. 100 Hz.
The angular frequency ω = 200π rad/s. Since ω = 2πf, f = ω/2π = 200π/2π = 100 Hz.
5. The impedance of an ideal inductor L at angular frequency ω is:
A. jωL
, B. 1/(jωL)
C. ωL
D. 1/(ωL)
A. jωL.
The impedance of an inductor is Z_L = jωL, where ω is the angular frequency and L is the
inductance. This indicates that the inductor's impedance increases with frequency and has
a phase angle of +90°.
6. The impedance of an ideal capacitor C at angular frequency ω is:
A. jωC
B. 1/(jωC)
C. ωC
D. 1/(ωC)
B. 1/(jωC).
The impedance of a capacitor is Z_C = 1/(jωC) = -j/(ωC), where ω is the angular
frequency and C is the capacitance. This indicates that the capacitor's impedance
decreases with frequency and has a phase angle of -90°.
7. A resistor R = 100 Ω, inductor L = 50 mH, and capacitor C = 10 μF are connected
in series. The impedance at ω = 1000 rad/s is:
A. 100 + j50 Ω
B. 100 - j50 Ω
