How do I get to the Thevenin Equivalent of this circuit? ### Circuit 4.1 Explanation---#### Description:The diagram presents an electrical circuit termed “Circuit 4.1.” The main components in this circuit include resistors, voltage sources, and a load resistor connected in a complex network.#### Components and Configuration:1. **Voltage Sources:** - Two voltage sources are present. - One is 6V positioned at the start of the circuit. - The other is 15V, located downstream after a series of resistors.2. **Resistors:** - The circuit contains a series of resistors: - A 2.2kΩ resistor directly following the 6V voltage source. - A 680Ω resistor in parallel with the 470Ω resistor, which follows the initial 2.2kΩ resistor. - A 330Ω resistor between the 15V voltage source and the next component in the circuit. - A 1kΩ resistor at the end before point A.3. **Load Resistor:** - There is a labeled load resistor \( R_L = 4.55 kΩ \) present between points A and B, with a current \( I_L \) flowing through it.#### Diagram Details:- The 2.2kΩ resistor is connected in series after the 6V battery. - Then, a branch occurs where the 680Ω resistor is connected parallel to the current path featuring a 330Ω resistor in series with the 15V power supply. - The 470Ω resistor follows and combines back in series.- Finally, a 1kΩ resistor is positioned before the connection points A and B, across which the load resistor \( R_L \) is placed.#### Points Noted:- Points A and B signify key measuring points for current \( I_L \).- Ground symbols indicate that certain points of the circuit are in contact with a common reference point.Overall, this circuit showcases an example of complex resistor and node configuration, where analyzing voltage and current distributions will involve understanding principles such as Kirchhoff’s Voltage Law (KVL) and Ohm’s Law.---This transcription illustrates how to navigate an electrical circuit with multiple components in preparation for analysis and educational purposes.

Answered: Image /qna-images/answer/21e89760-9f17-48e6-8326-a6481d078ff0.jpg

Answered: Circuit 4.1 2.2ka 4700 1kQ 3302 IL RL=…

Engineering

Electrical Engineering

Circuit 4.1 2.2ka 4700 1kQ 3302 IL RL= 4.55k2 =6 V 680Ω - 15 V В

EBK ELECTRICAL WIRING RESIDENTIAL

19th Edition

ISBN:9781337516549

Author:Simmons

Publisher:Simmons

Chapter27: Service-entrance Equipment

Section: Chapter Questions

Problem 26R

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Concept explainers

Thevenin Theorem

Any linear circuit, no matter how complex, maybe simplified to an equivalent circuit utilizing only a single voltage source and series resistance coupled to a load, according to Thevenin's Theorem. The superposition principle says that all foundational equations must be linear, and the notion of "linear" is still the same (i.e., there should not be any exponentials or roots). This is particularly the case when operating with passive elements like resistors, inductors, and capacitors. However, some components (particularly some gaseous and semiconductor devices like tunnel diode, Schottky diode, transistors like BJT, MOSFET) are nonlinear, meaning their current resistance fluctuates with voltage and current. As a result, circuits made up of these parts are known as nonlinear circuits.

Question

How do I get to the Thevenin Equivalent of this circuit?

### Circuit 4.1 Explanation

---

#### Description:

The diagram presents an electrical circuit termed “Circuit 4.1.” The main components in this circuit include resistors, voltage sources, and a load resistor connected in a complex network.

#### Components and Configuration:

1. **Voltage Sources:**
- Two voltage sources are present.
- One is 6V positioned at the start of the circuit.
- The other is 15V, located downstream after a series of resistors.

2. **Resistors:**
- The circuit contains a series of resistors:
- A 2.2kΩ resistor directly following the 6V voltage source.
- A 680Ω resistor in parallel with the 470Ω resistor, which follows the initial 2.2kΩ resistor.
- A 330Ω resistor between the 15V voltage source and the next component in the circuit.
- A 1kΩ resistor at the end before point A.

3. **Load Resistor:**
- There is a labeled load resistor \( R_L = 4.55 kΩ \) present between points A and B, with a current \( I_L \) flowing through it.

#### Diagram Details:

- The 2.2kΩ resistor is connected in series after the 6V battery.
- Then, a branch occurs where the 680Ω resistor is connected parallel to the current path featuring a 330Ω resistor in series with the 15V power supply.
- The 470Ω resistor follows and combines back in series.
- Finally, a 1kΩ resistor is positioned before the connection points A and B, across which the load resistor \( R_L \) is placed.

#### Points Noted:

- Points A and B signify key measuring points for current \( I_L \).
- Ground symbols indicate that certain points of the circuit are in contact with a common reference point.

Overall, this circuit showcases an example of complex resistor and node configuration, where analyzing voltage and current distributions will involve understanding principles such as Kirchhoff’s Voltage Law (KVL) and Ohm’s Law.

---

This transcription illustrates how to navigate an electrical circuit with multiple components in preparation for analysis and educational purposes.

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