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ELEC 273 LAB REPORT 3 COMPREHENSIVE EXAM PREP: AC
CIRCUIT ANALYSIS & FILTER DESIGN – PRACTICE
QUESTIONS WITH ANSWERS AND DETAILED RATIONALES
– COMPLETE STUDY GUIDE, LATEST 2026 EDITION
This complete practice examination is designed for students preparing for ELEC 273 Lab Report
3 on AC Circuit Analysis and Filter Design. It provides 150 advanced, exam-style multiple-
choice questions covering sinusoidal steady-state analysis, phasor representation, impedance
and admittance, series and parallel RLC circuits, resonance phenomena, passive filter design
(low-pass, high-pass, band-pass, band-stop), cutoff frequencies, Bode plots, quality factor,
bandwidth, transfer functions, and practical measurement techniques using oscilloscopes and
function generators. Each question includes a detailed rationale explaining the correct answer
and why alternatives are less suitable. The 2026 edition aligns with standard introductory
electrical engineering laboratory curricula and emphasizes both theoretical understanding and
hands-on experimental skills. Use this comprehensive review package to assess mastery, identify
knowledge gaps, and strengthen competencies for your lab report and practical assessments.
Perfect for students seeking excellence in AC circuit analysis and filter design.
Table of Contents
1. Sinusoidal Signals and Phasor Representation
2. Impedance and Admittance in AC Circuits
3. Series RC and RL Circuit Analysis
4. Series RLC Circuits and Resonance
5. Parallel RLC Circuits and Resonance
6. Passive Filter Fundamentals and Transfer Functions
7. Low-Pass and High-Pass Filter Design
8. Band-Pass and Band-Stop Filter Design
9. Bode Plots and Frequency Response Analysis
10. Experimental Measurements and Lab Techniques
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1: In sinusoidal steady-state analysis, a sinusoidal voltage v(t) = Vm cos(ωt + φ) is represented
by the phasor:
A) V = Vm ∠φ
B) V = Vm/√2 ∠φ
C) V = Vm ∠(ωt + φ)
D) V = Vm cos(φ)
Correct Answer: A
The phasor representation uses the peak amplitude Vm and phase angle φ, written as V =
Vm∠φ. Option B is the RMS phasor, which is sometimes used but the standard phasor uses
peak amplitude. Option C includes time dependence, which is removed in phasor form. Option
D is a scalar, not a phasor.
2: The impedance of a resistor in phasor form is:
A) R ∠90°
B) R ∠0°
C) R ∠-90°
D) jR
Correct Answer: B
A resistor's impedance is purely real with zero phase shift, so Z_R = R ∠0°. Option A is
inductive, C is capacitive, D is imaginary.
3: The impedance of an inductor with inductance L at angular frequency ω is:
A) jωL
B) 1/(jωL)
C) ωL ∠-90°
D) -jωL
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Correct Answer: A
Inductive impedance is Z_L = jωL, with phase +90°. Option B is capacitive reactance form, C
has wrong sign for phase, D has negative sign which would be capacitive.
4: The impedance of a capacitor with capacitance C at angular frequency ω is:
A) jωC
B) 1/(jωC) = -j/(ωC)
C) ωC ∠90°
D) j/(ωC)
Correct Answer: B
Capacitive impedance is Z_C = 1/(jωC) = -j/(ωC), with phase -90°. Option A is admittance, C
has positive phase, D has positive j which is incorrect.
5: The angular frequency ω for a 60 Hz source is approximately:
A) 377 rad/s
B) 60 rad/s
C) 314 rad/s
D) 120 rad/s
Correct Answer: A
ω = 2πf = 2π × 60 ≈ 377 rad/s. Option B is f in Hz, C is for 50 Hz (314), D is 2f without π.
6: In phasor analysis, the derivative of a sinusoidal voltage in the time domain corresponds to:
A) Dividing the phasor by jω
B) Multiplying the phasor by jω
C) Multiplying the phasor by ω
D) Dividing the phasor by ω
Correct Answer: B
Differentiation in time domain becomes multiplication by jω in phasor domain: d/dt ↔ jω.
Integration divides by jω. Options A, C, D are incorrect.
7: The phase angle between voltage and current in a purely inductive circuit is:
A) 0°
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B) +90° (current lags voltage)
C) -90° (current leads voltage)
D) 180°
Correct Answer: B
In a purely inductive circuit, voltage leads current by 90°, meaning current lags voltage.
Option A is resistive, C is capacitive, D is opposite phase.
8: The phase angle between voltage and current in a purely capacitive circuit is:
A) 0°
B) +90° (current lags voltage)
C) -90° (current leads voltage)
D) 180°
Correct Answer: C
In a purely capacitive circuit, current leads voltage by 90°, meaning the phase angle is -90°
when expressed as voltage relative to current. Option A resistive, B inductive, D opposite.
9: A sinusoidal voltage has an RMS value of 10 V. Its peak value is:
A) 14.14 V
B) 10 V
C) 7.07 V
D) 20 V
Correct Answer: A
V_peak = V_rms × √2 = 10 × 1.414 = 14.14 V. Option B is RMS, C is RMS/√2, D is double
RMS.
10: The period of a 1 kHz sinusoidal signal is:
A) 1 ms
B) 1 s
C) 0.1 ms
D) 10 ms
ELEC 273 LAB REPORT 3 COMPREHENSIVE EXAM PREP: AC
CIRCUIT ANALYSIS & FILTER DESIGN – PRACTICE
QUESTIONS WITH ANSWERS AND DETAILED RATIONALES
– COMPLETE STUDY GUIDE, LATEST 2026 EDITION
This complete practice examination is designed for students preparing for ELEC 273 Lab Report
3 on AC Circuit Analysis and Filter Design. It provides 150 advanced, exam-style multiple-
choice questions covering sinusoidal steady-state analysis, phasor representation, impedance
and admittance, series and parallel RLC circuits, resonance phenomena, passive filter design
(low-pass, high-pass, band-pass, band-stop), cutoff frequencies, Bode plots, quality factor,
bandwidth, transfer functions, and practical measurement techniques using oscilloscopes and
function generators. Each question includes a detailed rationale explaining the correct answer
and why alternatives are less suitable. The 2026 edition aligns with standard introductory
electrical engineering laboratory curricula and emphasizes both theoretical understanding and
hands-on experimental skills. Use this comprehensive review package to assess mastery, identify
knowledge gaps, and strengthen competencies for your lab report and practical assessments.
Perfect for students seeking excellence in AC circuit analysis and filter design.
Table of Contents
1. Sinusoidal Signals and Phasor Representation
2. Impedance and Admittance in AC Circuits
3. Series RC and RL Circuit Analysis
4. Series RLC Circuits and Resonance
5. Parallel RLC Circuits and Resonance
6. Passive Filter Fundamentals and Transfer Functions
7. Low-Pass and High-Pass Filter Design
8. Band-Pass and Band-Stop Filter Design
9. Bode Plots and Frequency Response Analysis
10. Experimental Measurements and Lab Techniques
,2|Page
1: In sinusoidal steady-state analysis, a sinusoidal voltage v(t) = Vm cos(ωt + φ) is represented
by the phasor:
A) V = Vm ∠φ
B) V = Vm/√2 ∠φ
C) V = Vm ∠(ωt + φ)
D) V = Vm cos(φ)
Correct Answer: A
The phasor representation uses the peak amplitude Vm and phase angle φ, written as V =
Vm∠φ. Option B is the RMS phasor, which is sometimes used but the standard phasor uses
peak amplitude. Option C includes time dependence, which is removed in phasor form. Option
D is a scalar, not a phasor.
2: The impedance of a resistor in phasor form is:
A) R ∠90°
B) R ∠0°
C) R ∠-90°
D) jR
Correct Answer: B
A resistor's impedance is purely real with zero phase shift, so Z_R = R ∠0°. Option A is
inductive, C is capacitive, D is imaginary.
3: The impedance of an inductor with inductance L at angular frequency ω is:
A) jωL
B) 1/(jωL)
C) ωL ∠-90°
D) -jωL
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Correct Answer: A
Inductive impedance is Z_L = jωL, with phase +90°. Option B is capacitive reactance form, C
has wrong sign for phase, D has negative sign which would be capacitive.
4: The impedance of a capacitor with capacitance C at angular frequency ω is:
A) jωC
B) 1/(jωC) = -j/(ωC)
C) ωC ∠90°
D) j/(ωC)
Correct Answer: B
Capacitive impedance is Z_C = 1/(jωC) = -j/(ωC), with phase -90°. Option A is admittance, C
has positive phase, D has positive j which is incorrect.
5: The angular frequency ω for a 60 Hz source is approximately:
A) 377 rad/s
B) 60 rad/s
C) 314 rad/s
D) 120 rad/s
Correct Answer: A
ω = 2πf = 2π × 60 ≈ 377 rad/s. Option B is f in Hz, C is for 50 Hz (314), D is 2f without π.
6: In phasor analysis, the derivative of a sinusoidal voltage in the time domain corresponds to:
A) Dividing the phasor by jω
B) Multiplying the phasor by jω
C) Multiplying the phasor by ω
D) Dividing the phasor by ω
Correct Answer: B
Differentiation in time domain becomes multiplication by jω in phasor domain: d/dt ↔ jω.
Integration divides by jω. Options A, C, D are incorrect.
7: The phase angle between voltage and current in a purely inductive circuit is:
A) 0°
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B) +90° (current lags voltage)
C) -90° (current leads voltage)
D) 180°
Correct Answer: B
In a purely inductive circuit, voltage leads current by 90°, meaning current lags voltage.
Option A is resistive, C is capacitive, D is opposite phase.
8: The phase angle between voltage and current in a purely capacitive circuit is:
A) 0°
B) +90° (current lags voltage)
C) -90° (current leads voltage)
D) 180°
Correct Answer: C
In a purely capacitive circuit, current leads voltage by 90°, meaning the phase angle is -90°
when expressed as voltage relative to current. Option A resistive, B inductive, D opposite.
9: A sinusoidal voltage has an RMS value of 10 V. Its peak value is:
A) 14.14 V
B) 10 V
C) 7.07 V
D) 20 V
Correct Answer: A
V_peak = V_rms × √2 = 10 × 1.414 = 14.14 V. Option B is RMS, C is RMS/√2, D is double
RMS.
10: The period of a 1 kHz sinusoidal signal is:
A) 1 ms
B) 1 s
C) 0.1 ms
D) 10 ms
