An 50 resistor, a 5 mH inductor, and a 1.25 μF capacitor are connected in series. The series-connected elements are energized by a sinusoidal voltage source whose voltage is 600 cos(8000t +20°) V. Part A Determine the impedances of the elements in the frequency-domain equivalent circuit. Express your answers in ohms to three significant figures separated by commas. Enter your answers in rectangular form. ΨΕΙ ΑΣΦ | 11 Submit vec ZL. Zc. ZR 40,100,50 You have already submitted this answer. Enter a new answer. No credit lost. Try again. Previous Answers Request Answer ? Ω
An 50 resistor, a 5 mH inductor, and a 1.25 μF capacitor are connected in series. The series-connected elements are energized by a sinusoidal voltage source whose voltage is 600 cos(8000t +20°) V. Part A Determine the impedances of the elements in the frequency-domain equivalent circuit. Express your answers in ohms to three significant figures separated by commas. Enter your answers in rectangular form. ΨΕΙ ΑΣΦ | 11 Submit vec ZL. Zc. ZR 40,100,50 You have already submitted this answer. Enter a new answer. No credit lost. Try again. Previous Answers Request Answer ? Ω
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07dc3289-c262-41f1-9fc5-a8d18500de1b%2Fc211648f-8002-4d27-8bf6-2be0fe58507e%2F0s2u3lo_processed.png&w=3840&q=75)
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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