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Find the Norton Equivalent with respect to Terminals A and B for: +280 V R1 2 ΚΩ R2 2 ΚΩ 0.2 *i_1 R3 2 KQ ww ww R4 ´ 5.6 ΚΩ A B

Find the Norton Equivalent with respect to Terminals A and B for: +280 V R1 2 ΚΩ R2 2 ΚΩ 0.2 *i_1 R3 2 KQ ww ww R4 ´ 5.6 ΚΩ A 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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Please help with circuit analysis question. 

**Title: Finding the Norton Equivalent for a Circuit** **Objective:** To find the Norton Equivalent with respect to Terminals A and B for the given circuit. **Circuit Description:** - **Voltage Source:** 280 V - **Resistors:** - R1 = 2 kΩ - R2 = 2 kΩ - R3 = 2 kΩ - R4 = 5.6 kΩ - **Current Source:** - A dependent current source with a value of 0.2 * i₁, where i₁ is the current through R1. **Circuit Layout:** 1. The 280 V source is connected in series with resistor R1 (2 kΩ). 2. Current i₁ flows through R1. 3. A branch connects from the junction between R1 and R2: - R2 (2 kΩ) is connected in parallel to the dependent current source (0.2 * i₁). 4. R3 (2 kΩ) is in series with the dependent current source, with its other terminal connecting to Terminal A. 5. R4 (5.6 kΩ) is connected from the junction of the dependent current source and R3 to Terminal B. 6. The circuit completes with resistors R2 and R4 bringing the paths back together. **Analysis Method:** 1. **Terminal Identification:** Terminals A and B are the points between which Norton Equivalent is to be found. 2. **Norton Equivalent:** Involves calculating the Norton current (I_N) and Norton resistance (R_N) between terminals A and B. - **I_N:** Magnitude of the current through a short circuit between terminals A and B. - **R_N:** Calculated by turning off all independent sources (setting 280 V to 0 V, i.e., short-circuit) and computing the total resistance seen from A to B. This setup forms the basics of determining how voltage, current, and resistance interact in complex circuit networks, allowing for simplification using Norton’s Theorem.
Transcribed Image Text:**Title: Finding the Norton Equivalent for a Circuit** **Objective:** To find the Norton Equivalent with respect to Terminals A and B for the given circuit. **Circuit Description:** - **Voltage Source:** 280 V - **Resistors:** - R1 = 2 kΩ - R2 = 2 kΩ - R3 = 2 kΩ - R4 = 5.6 kΩ - **Current Source:** - A dependent current source with a value of 0.2 * i₁, where i₁ is the current through R1. **Circuit Layout:** 1. The 280 V source is connected in series with resistor R1 (2 kΩ). 2. Current i₁ flows through R1. 3. A branch connects from the junction between R1 and R2: - R2 (2 kΩ) is connected in parallel to the dependent current source (0.2 * i₁). 4. R3 (2 kΩ) is in series with the dependent current source, with its other terminal connecting to Terminal A. 5. R4 (5.6 kΩ) is connected from the junction of the dependent current source and R3 to Terminal B. 6. The circuit completes with resistors R2 and R4 bringing the paths back together. **Analysis Method:** 1. **Terminal Identification:** Terminals A and B are the points between which Norton Equivalent is to be found. 2. **Norton Equivalent:** Involves calculating the Norton current (I_N) and Norton resistance (R_N) between terminals A and B. - **I_N:** Magnitude of the current through a short circuit between terminals A and B. - **R_N:** Calculated by turning off all independent sources (setting 280 V to 0 V, i.e., short-circuit) and computing the total resistance seen from A to B. This setup forms the basics of determining how voltage, current, and resistance interact in complex circuit networks, allowing for simplification using Norton’s Theorem.
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