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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...
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**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: \[ v = i \times R_{eq} = 5 \, \text{mA} \times 50 \, \text{k}\Omega = 250 \, \text{V} \]
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