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)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa9b91490-7014-4990-b9a4-62583a5b405c%2F54506032-0a16-485c-8200-f272482e976d%2Fzso52o9_processed.jpeg&w=3840&q=75)
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa9b91490-7014-4990-b9a4-62583a5b405c%2F54506032-0a16-485c-8200-f272482e976d%2F3pszzbl_processed.jpeg&w=3840&q=75)
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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