# Replace the emf+series resistance(s) by the Norton equivalent

• andrewmac
In summary, the conversation discusses replacing the emf+series resistance(s) with Norton equivalents in a circuit and finding the total power dissipated. The individual is struggling with understanding this concept and is seeking assistance. The suggestion is to replace the voltage supplies and their series resistances with Thevenin models and continue with writing the node-voltage equations.
andrewmac

## Homework Statement

Write down the node-voltage matrix equation for the following circuit. Hence find the total power dissipated. (Hint: replace the emf+series resistance(s) by the Norton equivalents.)

n/a

## The Attempt at a Solution

The problem I'm having is that I don't fully understand what is meant by replace the emf+series resistance with the norton equivalent. I have a very weak grasp on norton equivalents, and I'm more comfortable with them when I'm told "with respect to.." and a load resistor. The rest of the problem I'm sure I can do fine but this first part has stumped me. Thanks in advance for any assistance, I really appreciate it.

Presumably the idea is to replace the 8V supply and 10Ω resistor, and the 5V supply and its 10Ω resistor, and the 20V supply with its 25Ω and 30Ω resistors with their Norton equivalents, then proceed to write the node-voltage equations, etc.

Think of the voltage supplies with their series resistances as Thevenin models.

## 1. What is the purpose of replacing the emf+series resistance(s) by the Norton equivalent?

The purpose of replacing the emf+series resistance(s) by the Norton equivalent is to simplify a circuit and make it easier to analyze. The Norton equivalent replaces the emf and series resistance with a single current source and a single parallel resistance, which can make calculations and circuit analysis more efficient.

## 2. How can I determine the Norton equivalent of a circuit?

To determine the Norton equivalent of a circuit, you will need to calculate the short-circuit current and the equivalent resistance. The short-circuit current is the current that flows through the circuit when the load resistance is replaced with a short circuit. The equivalent resistance is the resistance seen by the short-circuit current when looking into the circuit from the load terminals. Once you have calculated these values, you can represent the circuit using the Norton equivalent.

## 3. What are the benefits of using the Norton equivalent in circuit analysis?

Using the Norton equivalent in circuit analysis can provide several benefits, including simplifying the circuit and reducing the number of calculations needed. The Norton equivalent also allows you to easily determine the behavior of a circuit under different load conditions, making it a useful tool for circuit design and troubleshooting.

## 4. Can the Norton equivalent be used for any type of circuit?

The Norton equivalent can be used for linear circuits that consist of only independent sources and resistors. Nonlinear elements, such as diodes or transistors, cannot be represented using the Norton equivalent. Additionally, the Norton equivalent is most useful for circuits with a single load, as multiple loads can complicate the calculation of the equivalent values.

## 5. How is the Norton equivalent related to the Thevenin equivalent?

The Norton equivalent and Thevenin equivalent are two different ways of representing a circuit, but they are mathematically equivalent. The Norton equivalent uses a current source and parallel resistance, while the Thevenin equivalent uses a voltage source and series resistance. Both can be converted to the other using simple equations. The choice between using the Norton or Thevenin equivalent depends on the type of circuit and the type of analysis being performed.

• Engineering and Comp Sci Homework Help
Replies
2
Views
2K
• Engineering and Comp Sci Homework Help
Replies
9
Views
4K
• Engineering and Comp Sci Homework Help
Replies
7
Views
2K
• Engineering and Comp Sci Homework Help
Replies
11
Views
5K
• Engineering and Comp Sci Homework Help
Replies
2
Views
1K
• Engineering and Comp Sci Homework Help
Replies
2
Views
913
• Engineering and Comp Sci Homework Help
Replies
5
Views
2K
• Engineering and Comp Sci Homework Help
Replies
7
Views
2K
• Engineering and Comp Sci Homework Help
Replies
35
Views
4K
• Engineering and Comp Sci Homework Help
Replies
17
Views
2K