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Ry" 4.00 ΚΩ ww &₂-20.0 V R₁- 1.00 k ww &₁-12.0 V R₂-5.00 k ww &-60.0 V Using Kirchhoff's rules, U R 6.00 k www (a)find the current in each resistor (b)find the potential difference between points a and b.

Ry" 4.00 ΚΩ ww &₂-20.0 V R₁- 1.00 k ww &₁-12.0 V R₂-5.00 k ww &-60.0 V Using Kirchhoff's rules, U R 6.00 k www (a)find the current in each resistor (b)find the potential difference between points a and b.

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
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### Educational Content: Applying Kirchhoff’s Rules to Circuit Analysis #### Problem Statement Using Kirchhoff’s rules: - (a) Find the current in each resistor. - (b) Find the potential difference between points a and b. --- #### Circuit Diagram Description In the provided diagram, we have an electrical circuit with the following components and connections: 1. **Voltage Sources and Resistors**: - \( \mathbf{\varepsilon_1 = 12.0 \, V} \) with \( \mathbf{R_1 = 4.00 \, k\Omega} \) - \( \mathbf{\varepsilon_2 = 20.0 \, V} \) with \( \mathbf{R_2 = 5.00 \, k\Omega} \) - \( \mathbf{\varepsilon_3 = 60.0 \, V} \) with \( \mathbf{R_3 = 6.00 \, k\Omega} \) - \( \mathbf{R_4 = 1.00 \, k\Omega} \) 2. **Node Connections**: - Node **a** connects the positive terminals of \( \varepsilon_2 \) and \( R_4 \). - Node **b** connects the positive terminal of \( R_2 \) with the positive terminal of \( R_3 \). --- #### Applying Kirchhoff’s Rules 1. **Kirchhoff’s Current Law (KCL)**: - The sum of currents entering a junction must equal the sum of currents leaving the junction. 2. **Kirchhoff’s Voltage Law (KVL)**: - The sum of the electrical potential differences (voltage) around any closed loop is zero. By following these principles, we can set up equations to solve for the current through each resistor and determine the potential difference between points a and b. --- **Example Equations** (using hypothetical current designations \( I_1, I_2, I_3 \)): 1. For the loop involving \( \varepsilon_1 \), \( R_1 \), and \( R_2 \): \[ \varepsilon_1 - I_1 R_1 - I_2 R_2 = 0 \] 2. For the loop involving \( \varepsilon_2 \), \( R_2 \), and \(
Transcribed Image Text:### Educational Content: Applying Kirchhoff’s Rules to Circuit Analysis #### Problem Statement Using Kirchhoff’s rules: - (a) Find the current in each resistor. - (b) Find the potential difference between points a and b. --- #### Circuit Diagram Description In the provided diagram, we have an electrical circuit with the following components and connections: 1. **Voltage Sources and Resistors**: - \( \mathbf{\varepsilon_1 = 12.0 \, V} \) with \( \mathbf{R_1 = 4.00 \, k\Omega} \) - \( \mathbf{\varepsilon_2 = 20.0 \, V} \) with \( \mathbf{R_2 = 5.00 \, k\Omega} \) - \( \mathbf{\varepsilon_3 = 60.0 \, V} \) with \( \mathbf{R_3 = 6.00 \, k\Omega} \) - \( \mathbf{R_4 = 1.00 \, k\Omega} \) 2. **Node Connections**: - Node **a** connects the positive terminals of \( \varepsilon_2 \) and \( R_4 \). - Node **b** connects the positive terminal of \( R_2 \) with the positive terminal of \( R_3 \). --- #### Applying Kirchhoff’s Rules 1. **Kirchhoff’s Current Law (KCL)**: - The sum of currents entering a junction must equal the sum of currents leaving the junction. 2. **Kirchhoff’s Voltage Law (KVL)**: - The sum of the electrical potential differences (voltage) around any closed loop is zero. By following these principles, we can set up equations to solve for the current through each resistor and determine the potential difference between points a and b. --- **Example Equations** (using hypothetical current designations \( I_1, I_2, I_3 \)): 1. For the loop involving \( \varepsilon_1 \), \( R_1 \), and \( R_2 \): \[ \varepsilon_1 - I_1 R_1 - I_2 R_2 = 0 \] 2. For the loop involving \( \varepsilon_2 \), \( R_2 \), and \(
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