Skip to content

Chapter 8 Quiz — AC Signals and Sinusoidal Waveforms

Test your understanding of AC signals, sinusoidal parameters, RMS values, phase relationships, and complex number representations. Each question includes a detailed explanation.

Question 1

A sinusoidal voltage is given by \(v(t) = 120\sin(377t + 30°) \text{ V}\). Identify the peak amplitude \(V_m\), the angular frequency \(\omega\), the frequency \(f\), and the phase angle \(\phi\).

  1. \(V_m = 120 \text{ V}\), \(\omega = 377 \text{ rad/s}\), \(f = 60 \text{ Hz}\), \(\phi = 30°\)
  2. \(V_m = 84.9 \text{ V}\), \(\omega = 377 \text{ rad/s}\), \(f = 50 \text{ Hz}\), \(\phi = 30°\)
  3. \(V_m = 120 \text{ V}\), \(\omega = 60 \text{ rad/s}\), \(f = 377 \text{ Hz}\), \(\phi = 30°\)
  4. \(V_m = 169.7 \text{ V}\), \(\omega = 377 \text{ rad/s}\), \(f = 60 \text{ Hz}\), \(\phi = 0°\)
Show Answer

Correct Answer: A

From \(v(t) = V_m\sin(\omega t + \phi)\): \(V_m = 120 \text{ V}\) (the coefficient), \(\omega = 377 \text{ rad/s}\) (the coefficient of t), \(\phi = 30°\) (the phase offset). Frequency: \(f = \omega/(2\pi) = 377/(2\pi) \approx 60 \text{ Hz}\). Note: \(\omega = 377 \approx 2\pi \times 60\) is the standard US power frequency. The RMS value would be \(120/\sqrt{2} \approx 84.9 \text{ V}\), but \(V_m = 120 \text{ V}\) is the peak, not the RMS.

Concept Tested: Sinusoidal Parameters \(V_m\), \(\omega\), \(f\), \(\phi\)

Question 2

The US standard wall outlet delivers 120 V RMS at 60 Hz. What is the peak voltage \(V_m\) and the period \(T\)?

  1. \(V_m = 120 \text{ V}\), \(T = 16.67 \text{ ms}\)
  2. \(V_m = 170 \text{ V}\), \(T = 16.67 \text{ ms}\)
  3. \(V_m = 170 \text{ V}\), \(T = 60 \text{ ms}\)
  4. \(V_m = 84.9 \text{ V}\), \(T = 16.67 \text{ ms}\)
Show Answer

Correct Answer: B

Peak voltage: \(V_m = V_{RMS} \times \sqrt{2} = 120 \times 1.414 \approx 170 \text{ V}\). Period: \(T = 1/f = 1/60 = 16.67 \text{ ms}\). The wall outlet is described by its RMS value (120 V) because that is what matters for power delivery — it is the equivalent DC value. However, the actual voltage swings between +170 V and −170 V. High-voltage warning: 170 V peaks are lethal.

Concept Tested: \(V_m = V_{RMS}\sqrt{2}\) and Period \(T = 1/f\)

Question 3

Two signals are \(v_1(t) = 10\sin(\omega t)\) and \(v_2(t) = 10\sin(\omega t - 45°)\). How would you describe the phase relationship between them?

  1. \(v_1\) lags \(v_2\) by 45°
  2. \(v_2\) leads \(v_1\) by 45°
  3. \(v_1\) leads \(v_2\) by 45°
  4. The signals are in phase
Show Answer

Correct Answer: C

\(v_1\) has zero phase and \(v_2\) has a phase of \(-45°\). A negative phase means \(v_2\) is delayed (shifted right in time), so \(v_1\) reaches its peaks before \(v_2\). Therefore, \(v_1\) leads \(v_2\) by 45° (equivalently, \(v_2\) lags \(v_1\) by 45°). A positive phase = leading (earlier peaks); a negative phase = lagging (later peaks).

Concept Tested: Phase Relationships — Leading and Lagging

Question 4

Convert the complex number \(z = 5\angle 53.13°\) from polar form to rectangular form.

  1. \(z = 3 + j4\)
  2. \(z = 4 + j3\)
  3. \(z = 5 + j5\)
  4. \(z = 3 - j4\)
Show Answer

Correct Answer: A

Converting polar to rectangular: \(a = r\cos\theta = 5\cos(53.13°) = 5 \times 0.6 = 3\) and \(b = r\sin\theta = 5\sin(53.13°) = 5 \times 0.8 = 4\). So \(z = 3 + j4\). This is the classic 3-4-5 right triangle: \(\sqrt{3^2 + 4^2} = 5\) and \(\tan^{-1}(4/3) = 53.13°\). Memorizing this triangle saves time in many phasor calculations.

Concept Tested: Polar to Rectangular Conversion \(a = r\cos\theta\), \(b = r\sin\theta\)

Question 5

Convert the complex number \(z = -5 + j12\) to polar form.

  1. \(13\angle 67.4°\)
  2. \(13\angle -67.4°\)
  3. \(13\angle 112.6°\)
  4. \(17\angle 112.6°\)
Show Answer

Correct Answer: C

Magnitude: \(r = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\). Angle: the point \((-5, 12)\) is in the second quadrant. \(\theta = 180° - \tan^{-1}(12/5) = 180° - 67.4° = 112.6°\). So \(z = 13\angle 112.6°\). When using \(\tan^{-1}\), always check which quadrant the point is in — the calculator will only give you angles in the range −90° to +90°, and you must adjust for quadrants II, III, and IV.

Concept Tested: Rectangular to Polar Conversion with Quadrant Awareness

Question 6

Multiply \(z_1 = 4\angle 30°\) by \(z_2 = 3\angle 60°\). Express the result in both polar and rectangular form.

  1. \(12\angle 90° = j12\)
  2. \(12\angle 30° = 6\sqrt{3} + j6\)
  3. \(7\angle 90° = j7\)
  4. \(12\angle 1800°\)
Show Answer

Correct Answer: A

For polar multiplication: multiply magnitudes and add angles. \(z_1 \times z_2 = (4 \times 3)\angle(30° + 60°) = 12\angle 90°\). In rectangular: \(12\cos(90°) + j12\sin(90°) = 0 + j12 = j12\). This result is purely imaginary, which makes physical sense: rotating a vector by 90° moves it from the real axis to the imaginary axis. Polar form is always the easiest for multiplication and division.

Concept Tested: Complex Number Multiplication in Polar Form

Question 7

A signal has a DC component of 4 V and an AC (sinusoidal) component with a peak of 3 V. What is the total RMS voltage of the combined signal?

  1. 7 V
  2. 5 V
  3. 4.53 V
  4. 6.12 V
Show Answer

Correct Answer: C

For a signal with both DC and AC components, the total RMS is: \(V_{rms,total} = \sqrt{V_{DC}^2 + V_{AC,rms}^2}\). The AC RMS is \(V_{AC,rms} = 3/\sqrt{2} = 2.121 \text{ V}\). Therefore: \(V_{rms,total} = \sqrt{4^2 + (2.121)^2} = \sqrt{16 + 4.5} = \sqrt{20.5} \approx 4.53 \text{ V}\). You cannot simply add DC and AC RMS values — they combine as the square root of the sum of squares, just like perpendicular vectors.

Concept Tested: Total RMS with DC and AC Components

Question 8

According to Euler's formula, \(e^{j\theta} = \cos\theta + j\sin\theta\). What is the significance of this relationship for AC circuit analysis?

  1. It allows us to convert DC circuits to AC by substituting \(j\omega\)
  2. It shows that complex exponentials are the same as DC sources
  3. It enables differentiation and integration of sinusoids to become simple multiplication by \(j\omega\) and division by \(j\omega\), making phasor analysis possible
  4. It proves that all AC signals are periodic
Show Answer

Correct Answer: C

Euler's formula is the mathematical bridge that makes phasor analysis work. Since \(\frac{d}{dt}e^{j\omega t} = j\omega e^{j\omega t}\), differentiation of a complex exponential (which represents a sinusoid) is just multiplication by \(j\omega\). This converts differential equations — which would require solving ODEs — into simple algebraic equations with complex numbers. Kirchhoff's laws in AC then look just like DC laws but with impedances (\(Z = R + jX\)) instead of pure resistances.

Concept Tested: Euler's Formula and its Role in Phasor Analysis

Answers Summary

Question Answer Concept Tested
1 A Sinusoidal Parameters
2 B Peak Voltage and Period
3 C Phase Relationships
4 A Polar to Rectangular Conversion
5 C Rectangular to Polar Conversion
6 A Complex Multiplication in Polar Form
7 C Total RMS with DC and AC Components \(\sqrt{V_{DC}^2 + V_{AC,rms}^2}\)
8 C Euler's Formula and Phasor Analysis