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The capacitor in the circuit is initially uncharged and the switch S is suddenly closed at t=0. Find the current on resistor R₂ at t-co. Take R₁-4R and R₂=3R. A) = WIR w B) R w D) R 48 E) R S R₁ m www R₂ C

The capacitor in the circuit is initially uncharged and the switch S is suddenly closed at t=0. Find the current on resistor R₂ at t-co. Take R₁-4R and R₂=3R. A) = WIR w B) R w D) R 48 E) R S R₁ m www R₂ C

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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FIND THE CURRENT ON RESISTOR R2 (NEED NEAT HANDWRITTEN SOLUTION ONLY OTHERWISE DOWNVOTE).

**Problem Statement:** The capacitor in the circuit is initially uncharged, and the switch \( S \) is suddenly closed at \( t = 0 \). Find the current through resistor \( R_2 \) at \( t = \infty \). Given \( R_1 = 4R \) and \( R_2 = 3R \). **Answer Choices:** A) \(\frac{\mathcal{E}}{R}\) B) \(\frac{\mathcal{E}}{2R}\) C) \(\frac{2\mathcal{E}}{3R}\) D) \(\frac{\mathcal{E}}{4R}\) E) \(\frac{4\mathcal{E}}{3R}\) **Circuit Diagram Description:** The circuit schematic provided includes the following components: - A switch \( S \) that controls the circuit. - A resistor \( R_1 \) in series with the switch \( S \). - A power source with voltage \( \mathcal{E} \). - A parallel combination of a resistor \( R_2 \) and a capacitor \( C \). The voltage source \( \mathcal{E} \) is connected to the parallel combination of \( R_2 \) and \( C \) through the series resistor \( R_1 \). **Explanation:** 1. **Initial Condition at \( t = 0 \):** - The capacitor \( C \) is initially uncharged, so it behaves like a short circuit (wire) at the moment the switch is closed. 2. **As \( t \to \infty \):** - The capacitor \( C \) will be fully charged and will behave like an open circuit. - Current will only flow through resistors \( R_1 \) and \( R_2 \) in series. 3. **Resistor Values:** - \( R_1 = 4R \) - \( R_2 = 3R \) 4. **Simplifying the Circuit for \( t = \infty \):** - The total resistance in the circuit, which is the sum of \( R_1 \) and \( R_2 \), will be: \[ R_{\text{total}} = R_1 + R_2 = 4R + 3R = 7R
Transcribed Image Text:**Problem Statement:** The capacitor in the circuit is initially uncharged, and the switch \( S \) is suddenly closed at \( t = 0 \). Find the current through resistor \( R_2 \) at \( t = \infty \). Given \( R_1 = 4R \) and \( R_2 = 3R \). **Answer Choices:** A) \(\frac{\mathcal{E}}{R}\) B) \(\frac{\mathcal{E}}{2R}\) C) \(\frac{2\mathcal{E}}{3R}\) D) \(\frac{\mathcal{E}}{4R}\) E) \(\frac{4\mathcal{E}}{3R}\) **Circuit Diagram Description:** The circuit schematic provided includes the following components: - A switch \( S \) that controls the circuit. - A resistor \( R_1 \) in series with the switch \( S \). - A power source with voltage \( \mathcal{E} \). - A parallel combination of a resistor \( R_2 \) and a capacitor \( C \). The voltage source \( \mathcal{E} \) is connected to the parallel combination of \( R_2 \) and \( C \) through the series resistor \( R_1 \). **Explanation:** 1. **Initial Condition at \( t = 0 \):** - The capacitor \( C \) is initially uncharged, so it behaves like a short circuit (wire) at the moment the switch is closed. 2. **As \( t \to \infty \):** - The capacitor \( C \) will be fully charged and will behave like an open circuit. - Current will only flow through resistors \( R_1 \) and \( R_2 \) in series. 3. **Resistor Values:** - \( R_1 = 4R \) - \( R_2 = 3R \) 4. **Simplifying the Circuit for \( t = \infty \):** - The total resistance in the circuit, which is the sum of \( R_1 \) and \( R_2 \), will be: \[ R_{\text{total}} = R_1 + R_2 = 4R + 3R = 7R
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