Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
Problem 5 and v. Consider the following circuit operating under DC conditions. Compute i 2 mH all + 5 mA 30 k2 6 μF 20 k2

Problem 5 and v. Consider the following circuit operating under DC conditions. Compute i 2 mH all + 5 mA 30 k2 6 μF 20 k2

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
See similar textbooks
icon
Related questions
Question
**Problem 5:** Consider the following circuit operating under DC conditions. Compute \( i \) and \( v \). **Circuit Description:** - A current source provides a current of 5 mA. - The circuit contains: - A 30 kΩ resistor in series. - A 2 mH inductor in series. - A 6 μF capacitor in parallel with the 30 kΩ resistor. - A 20 kΩ resistor in parallel with the capacitor. - The voltage across the capacitor (and therefore across the 30 kΩ resistor and the 20 kΩ resistor) is labeled as \( v \). - The current through the inductor is labeled as \( i \). **Explanation:** Under DC conditions, the inductor behaves as a short circuit and the capacitor as an open circuit. Thus, you only need to consider the resistors to compute \( i \) and \( v \). 1. **Inductor Behavior**: Acts as a short circuit (0 Ω) under DC, so it has no voltage drop across it. 2. **Capacitor Behavior**: Acts as an open circuit, meaning no current flows through it. Since the capacitor is open, it does not affect the current paths. Therefore, the circuit simplifies to two series resistors: 30 kΩ and 20 kΩ. **Calculations:** To compute \( i \) and \( v \): - Find the equivalent resistance \( R_{eq} = 30 \, \text{k}\Omega + 20 \, \text{k}\Omega = 50 \, \text{k}\Omega \). - Use Ohm's Law to find the voltage \( v \) across the resistors: \[ v = i \times R_{eq} = 5 \, \text{mA} \times 50 \, \text{k}\Omega = 250 \, \text{V} \]
Transcribed Image Text:**Problem 5:** Consider the following circuit operating under DC conditions. Compute \( i \) and \( v \). **Circuit Description:** - A current source provides a current of 5 mA. - The circuit contains: - A 30 kΩ resistor in series. - A 2 mH inductor in series. - A 6 μF capacitor in parallel with the 30 kΩ resistor. - A 20 kΩ resistor in parallel with the capacitor. - The voltage across the capacitor (and therefore across the 30 kΩ resistor and the 20 kΩ resistor) is labeled as \( v \). - The current through the inductor is labeled as \( i \). **Explanation:** Under DC conditions, the inductor behaves as a short circuit and the capacitor as an open circuit. Thus, you only need to consider the resistors to compute \( i \) and \( v \). 1. **Inductor Behavior**: Acts as a short circuit (0 Ω) under DC, so it has no voltage drop across it. 2. **Capacitor Behavior**: Acts as an open circuit, meaning no current flows through it. Since the capacitor is open, it does not affect the current paths. Therefore, the circuit simplifies to two series resistors: 30 kΩ and 20 kΩ. **Calculations:** To compute \( i \) and \( v \): - Find the equivalent resistance \( R_{eq} = 30 \, \text{k}\Omega + 20 \, \text{k}\Omega = 50 \, \text{k}\Omega \). - Use Ohm's Law to find the voltage \( v \) across the resistors: \[ v = i \times R_{eq} = 5 \, \text{mA} \times 50 \, \text{k}\Omega = 250 \, \text{V} \]
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Capacitor
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,
SEE MORE TEXTBOOKS