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Chapter 8 Practice Problems

Practice Problems

Problem 1 — Sinusoidal Parameters

A sinusoidal voltage is given by \(v(t) = 325\cos(314.16t - 30°)\) volts.

(a) Identify the peak amplitude, angular frequency, frequency, period, and phase angle.

(b) Calculate the RMS voltage.

(c) What is the value of \(v\) at t = 0?

(d) At what time does the first positive peak occur?

Solution

(a) From the equation:

  • Peak amplitude: \(V_m = 325\text{ V}\)
  • Angular frequency: \(\omega = 314.16\text{ rad/s}\)
  • Frequency: \(f = \omega/(2\pi) = 314.16/6.283 = 50\text{ Hz}\)
  • Period: \(T = 1/f = 0.02\text{ s} = 20\text{ ms}\)
  • Phase angle: \(\phi = -30°\) (the signal lags a reference cosine by 30°)

(b) RMS voltage:

\[V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{325}{\sqrt{2}} = 229.8\text{ V} \approx 230\text{ V}\]

(This is the European grid voltage.)

(c) At t = 0:

\[v(0) = 325\cos(-30°) = 325 \times 0.866 = 281.6\text{ V}\]

(d) Peak occurs when \(314.16t - 30° = 0°\), so \(t = 30°/314.16 = (30\pi/180)/314.16 = 0.524/314.16 = 1.667\text{ ms}\).

Problem 2 — Complex Number Operations

Perform the following complex number calculations, expressing results in both rectangular and polar form.

(a) \((3 + j4) + (1 - j2)\)

(b) \((3 + j4) \times (1 - j2)\)

(c) \(\dfrac{10\angle 45°}{2\angle -30°}\)

(d) The magnitude and angle of \(Z = 5 + j12\)

Solution

(a) Addition (add real and imaginary parts separately):

\[(3+j4) + (1-j2) = 4 + j2 = \sqrt{4^2+2^2}\angle\arctan(2/4) = \sqrt{20}\angle 26.6° = 4.47\angle 26.6°\]

(b) Multiplication (expand):

[(3+j4)(1-j2) = 3 - 6j + 4j - 8j^2 = 3 - 2j + 8 = 11 - 2j] Polar: \(\sqrt{121+4}\angle\arctan(-2/11) = 11.18\angle -10.3°\)

(c) Division in polar form (divide magnitudes, subtract angles):

[\frac{10\angle 45°}{2\angle -30°} = 5\angle(45° - (-30°)) = 5\angle 75°] Rectangular: \(5\cos 75° + j5\sin 75° = 1.294 + j4.830\)

(d) Magnitude and angle:

[|Z| = \sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13] [\angle Z = \arctan(12/5) = \arctan(2.4) = 67.4°]

Problem 3 — RMS and Power Calculations

Three voltage signals are:

  • \(v_1(t) = 10\sin(2\pi \times 60t)\) V
  • \(v_2(t) = 15\cos(2\pi \times 60t + 45°)\) V
  • \(v_3(t) = 5\text{ V}\) (DC)

(a) Calculate the RMS value of each signal.

(b) Signal \(v_1\) is applied across a 50 Ω resistor. Calculate the average power.

(c) A signal has a DC component of 5 V and an AC component with amplitude 10 V. What is the total RMS value?

Solution

(a) RMS values:

  • \(V_{1,rms} = 10/\sqrt{2} = 7.07\text{ V}\)
  • \(V_{2,rms} = 15/\sqrt{2} = 10.61\text{ V}\)
  • \(V_{3,rms} = 5\text{ V}\) (DC RMS equals the DC value)

(b) Average power from \(v_1\):

\[P = \frac{V_{1,rms}^2}{R} = \frac{(7.07)^2}{50} = \frac{50}{50} = 1\text{ W}\]

(c) For a signal with both DC and AC components, the total RMS is:

\[V_{rms,total} = \sqrt{V_{DC}^2 + V_{AC,rms}^2} = \sqrt{5^2 + (10/\sqrt{2})^2} = \sqrt{25 + 50} = \sqrt{75} = 8.66\text{ V}\]

Problem 4 — Phase Relationships

Two sinusoidal signals are \(v(t) = 10\sin(1000t + 30°)\) V and \(i(t) = 2\sin(1000t - 45°)\) A.

(a) What is the phase difference between v and i?

(b) Does the voltage lead or lag the current?

(c) Express both signals as phasors (peak value form): \(\mathbf{V} = V_m\angle\phi_v\) and \(\mathbf{I} = I_m\angle\phi_i\).

(d) What is the impedance \(Z = \mathbf{V}/\mathbf{I}\)?

Solution

(a) Phase difference:

\[\Delta\phi = \phi_v - \phi_i = 30° - (-45°) = 75°\]

(b) Since \(\phi_v > \phi_i\), the voltage leads the current by 75°.

(c) Phasor representations:

\[\mathbf{V} = 10\angle 30°\text{ V}, \qquad \mathbf{I} = 2\angle -45°\text{ A}\]

(d) Impedance:

\[Z = \frac{\mathbf{V}}{\mathbf{I}} = \frac{10\angle 30°}{2\angle -45°} = 5\angle 75°\ \Omega = 1.294 + j4.83\ \Omega\]

The positive imaginary part indicates the circuit is inductive.

Problem 5 — Frequency Calculations

A sinusoidal signal completes 3.5 cycles in 70 ms.

(a) Calculate the period and frequency.

(b) Calculate the angular frequency in rad/s.

(c) If the peak amplitude is 8 V, write the mathematical expression for this signal (assuming zero phase, sine form).

(d) At t = 15 ms, what is the instantaneous voltage?

Solution

(a) Period and frequency:

[T = \frac{70\text{ ms}}{3.5} = 20\text{ ms} = 0.020\text{ s}] [f = \frac{1}{T} = \frac{1}{0.020} = 50\text{ Hz}]

(b) Angular frequency:

\[\omega = 2\pi f = 2\pi \times 50 = 314.2\text{ rad/s}\]

(c) Mathematical expression:

\[v(t) = 8\sin(314.2t)\text{ V}\]

(d) Instantaneous voltage at t = 15 ms:

\[v(0.015) = 8\sin(314.2 \times 0.015) = 8\sin(4.713) = 8\sin(270°) = 8 \times (-1) = -8\text{ V}\]

At 15 ms = 0.75T, the signal is at its negative peak. ✓