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Problem 2: Thevenin's equivalent circuit 2502 3002 4002 5000 20V 3002 10V 2000 °B

Problem 2: Thevenin's equivalent circuit 2502 3002 4002 5000 20V 3002 10V 2000 °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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**Problem 2: Thevenin’s Equivalent Circuit** ### Original Circuit Diagram - A circuit is shown with various components: - A 250Ω resistor - A 500Ω resistor - A 400Ω resistor - A 300Ω resistor - Two voltage sources (20V and 10V) ### Thevenin’s Equivalent Resistance: **Step 1** – Short circuit all voltage sources, replacing them with a wire. **Step 2** – Open circuit all current sources, removing the connection. ### Simplified Circuit Diagrams **Diagram 1:** - Components shown: - Resistors of 250Ω, 300Ω, 400Ω, and 500Ω - Terminals marked as A and B ### Resistor Arrangements - Combine series resistors: - 400Ω + 250Ω = 650Ω - 300Ω + 200Ω = 500Ω (but note this is short-circuited) **Diagram 2:** - Simplified to 650Ω and 500Ω resistors - Short circuit segment highlighted between the 300Ω and 200Ω combination ### Analysis In the modified circuit: - The series combination of 300Ω and 200Ω is short-circuited by a wire. - A wire with zero resistance ensures no current passes through the 500Ω resistor. ### Final Resistor Arrangement **Diagram 3:** - Resistors arranged as: - 650Ω in series with 500Ω - Parallel with 300Ω ### Thevenin Equivalent Resistance - Calculated as \([650Ω \parallel 500Ω] + 300Ω\). This step-by-step analysis provides the Thevenin equivalent resistance for the given circuit.
Transcribed Image Text:**Problem 2: Thevenin’s Equivalent Circuit** ### Original Circuit Diagram - A circuit is shown with various components: - A 250Ω resistor - A 500Ω resistor - A 400Ω resistor - A 300Ω resistor - Two voltage sources (20V and 10V) ### Thevenin’s Equivalent Resistance: **Step 1** – Short circuit all voltage sources, replacing them with a wire. **Step 2** – Open circuit all current sources, removing the connection. ### Simplified Circuit Diagrams **Diagram 1:** - Components shown: - Resistors of 250Ω, 300Ω, 400Ω, and 500Ω - Terminals marked as A and B ### Resistor Arrangements - Combine series resistors: - 400Ω + 250Ω = 650Ω - 300Ω + 200Ω = 500Ω (but note this is short-circuited) **Diagram 2:** - Simplified to 650Ω and 500Ω resistors - Short circuit segment highlighted between the 300Ω and 200Ω combination ### Analysis In the modified circuit: - The series combination of 300Ω and 200Ω is short-circuited by a wire. - A wire with zero resistance ensures no current passes through the 500Ω resistor. ### Final Resistor Arrangement **Diagram 3:** - Resistors arranged as: - 650Ω in series with 500Ω - Parallel with 300Ω ### Thevenin Equivalent Resistance - Calculated as \([650Ω \parallel 500Ω] + 300Ω\). This step-by-step analysis provides the Thevenin equivalent resistance for the given circuit.
**Thevenin’s Equivalent Voltage \( V_{th} \):** **Circuit Diagram Overview:** The diagram presents a circuit with the following components and configuration: - **Resistors:** - 250Ω (between A1 and top node of the 400Ω resistor) - 300Ω (between node A and a floating node on the right side) - 400Ω (in parallel with the 20V source) - 500Ω (between node C and node A) - 300Ω (below the 400Ω resistor) - 200Ω (connected between node B1 and node B) - **Voltage Sources:** - 20V in series with the 400Ω resistor - 10V source connected directly across node C and B1 **Current Flow:** - The current \( i_1 \) flows through the 500Ω resistor. **Analysis**: - Since the 300Ω resistor at the right is open at one end, no current flows through it. Hence, the voltage at nodes A and B is the same as at nodes A1 and B1. - **Highlighted Equation:** - The voltage between B1 and A1 is given by the equation: \[ \text{Voltage between A1 and B1} = (\text{Voltage between A1 and C}) + (\text{Voltage between C and B1}) \] - The voltage between node C and B1 is 10V, as it is directly connected to a 10V source. - To find the voltage across the 500Ω resistor, apply the mesh law with the current \( i_1 \): \[ V_{500} = 500 \times i_1 \] - Substitute values in the highlighted equation to determine \( V_{th} \) across nodes A and B.
Transcribed Image Text:**Thevenin’s Equivalent Voltage \( V_{th} \):** **Circuit Diagram Overview:** The diagram presents a circuit with the following components and configuration: - **Resistors:** - 250Ω (between A1 and top node of the 400Ω resistor) - 300Ω (between node A and a floating node on the right side) - 400Ω (in parallel with the 20V source) - 500Ω (between node C and node A) - 300Ω (below the 400Ω resistor) - 200Ω (connected between node B1 and node B) - **Voltage Sources:** - 20V in series with the 400Ω resistor - 10V source connected directly across node C and B1 **Current Flow:** - The current \( i_1 \) flows through the 500Ω resistor. **Analysis**: - Since the 300Ω resistor at the right is open at one end, no current flows through it. Hence, the voltage at nodes A and B is the same as at nodes A1 and B1. - **Highlighted Equation:** - The voltage between B1 and A1 is given by the equation: \[ \text{Voltage between A1 and B1} = (\text{Voltage between A1 and C}) + (\text{Voltage between C and B1}) \] - The voltage between node C and B1 is 10V, as it is directly connected to a 10V source. - To find the voltage across the 500Ω resistor, apply the mesh law with the current \( i_1 \): \[ V_{500} = 500 \times i_1 \] - Substitute values in the highlighted equation to determine \( V_{th} \) across nodes A and B.
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