Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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### Educational Resource: Frequency-Domain Analysis of Series RLC Circuit

#### Electric Circuit Definition
An electric circuit consisting of:
- A 50 Ω resistor (R)
- A 5 mH inductor (L)
- A 1.25 μF capacitor (C)

These components are connected in series and are energized by a sinusoidal voltage source. The voltage of the source is given by the function:

\[ V(t) = 600 \cos(8000t + 20^\circ) \, V \]

#### Problem Statement: Impedance Calculation
**Objective:** Determine the impedances of the elements in the frequency-domain equivalent circuit.

**Instructions:** 
1. Express your answers in ohms (Ω) to three significant figures separated by commas.
2. Enter your answers in rectangular form.

#### Frequency-Domain Analysis
The impedance of each component in a series RLC circuit can be calculated as follows:

- **Resistor (R):** The impedance is purely real.
  \[ Z_R = R = 50 \, \Omega \]

- **Inductor (L):** The impedance is purely imaginary and given by \( Z_L = j\omega L \), where \( \omega = 8000 \, \text{rad/s} \).
  - Inductive reactance:
    \[ X_L = \omega L = 8000 \times 5 \times 10^{-3} = 40 \, \Omega \]
  - Impedance of the inductor:
    \[ Z_L = j40 \, \Omega \]

- **Capacitor (C):** The impedance is purely imaginary and given by \( Z_C = \frac{1}{j\omega C} \).
  - Capacitive reactance:
    \[ X_C = \frac{1}{\omega C} = \frac{1}{8000 \times 1.25 \times 10^{-6}} = 100 \, \Omega \]
  - Impedance of the capacitor:
    \[ Z_C = -j100 \, \Omega \]

#### Given Answer Format:
\[ Z_L, Z_C, Z_R = 40, -100, 50 \]

#### Submission 
Students are asked to submit their answers in the input box provided. Once entered, students should click the "Submit" button.

**Note:** There is an option to review previous answers or request an
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Transcribed Image Text:### Educational Resource: Frequency-Domain Analysis of Series RLC Circuit #### Electric Circuit Definition An electric circuit consisting of: - A 50 Ω resistor (R) - A 5 mH inductor (L) - A 1.25 μF capacitor (C) These components are connected in series and are energized by a sinusoidal voltage source. The voltage of the source is given by the function: \[ V(t) = 600 \cos(8000t + 20^\circ) \, V \] #### Problem Statement: Impedance Calculation **Objective:** Determine the impedances of the elements in the frequency-domain equivalent circuit. **Instructions:** 1. Express your answers in ohms (Ω) to three significant figures separated by commas. 2. Enter your answers in rectangular form. #### Frequency-Domain Analysis The impedance of each component in a series RLC circuit can be calculated as follows: - **Resistor (R):** The impedance is purely real. \[ Z_R = R = 50 \, \Omega \] - **Inductor (L):** The impedance is purely imaginary and given by \( Z_L = j\omega L \), where \( \omega = 8000 \, \text{rad/s} \). - Inductive reactance: \[ X_L = \omega L = 8000 \times 5 \times 10^{-3} = 40 \, \Omega \] - Impedance of the inductor: \[ Z_L = j40 \, \Omega \] - **Capacitor (C):** The impedance is purely imaginary and given by \( Z_C = \frac{1}{j\omega C} \). - Capacitive reactance: \[ X_C = \frac{1}{\omega C} = \frac{1}{8000 \times 1.25 \times 10^{-6}} = 100 \, \Omega \] - Impedance of the capacitor: \[ Z_C = -j100 \, \Omega \] #### Given Answer Format: \[ Z_L, Z_C, Z_R = 40, -100, 50 \] #### Submission Students are asked to submit their answers in the input box provided. Once entered, students should click the "Submit" button. **Note:** There is an option to review previous answers or request an
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