# Norton equivalen circuit problem

• terryds
In summary, @terryds found that the current through the 10 ohm resistor is 5A, and that ammeters have negligible resistance. To find the Norton current, he shorts the terminals A and B together and calculates the current from A to B. The Norton Resistance is (Δvab) / (Short Ckt Current from A to B).
terryds

## Homework Statement

Determine the norton equivalent Circuit

Norton theorem

## The Attempt at a Solution

First, I transform 5A and 10 ohm (in parallel) into 50 V and 10 ohm in series

So, there are three meshes,
for Mesh 1 (I assume the current is clockwise)

##30 i_1 - 10 i_2 = -20##

for Mesh 2

##40 i_2 - 20 i_3 = 50##

for Mesh 3

##40 i_3 - 20 i_2 = 20##Solving those three equations, I get ##i_1 = 0 A##, ##i_2 = 2 A##, ##i_3 = 1.5 A##

So, I_norton is i1 + i2 - i3 = 0 + 2 - 1.5 = 0.5 A

But, this is wrong answer. Please tell me where I got this wrong.

I cannot figure out where you got the equations for the Mesh Analysis. Where are your loops? Which is i1 i2 and i3? Can you label your picture? Getting zero for one of the currents is possible in some circuits, but I don't think this is one of them (I haven't actually gone through the analysis, but just from looking at the voltages, it doesn't look likely). You may want to revisit the steps for Mesh Analysis. Here are a couple of sources that may help. https://en.wikipedia.org/wiki/Mesh_analysis
https://ocw.mit.edu/courses/electri...pring-2006/lecture-notes/nodal_mesh_methd.pdf

terryds said:
40i2−20i3=5040i2−20i3=5040 i_2 - 20 i_3 = 50
You didn't include the voltage drop due to i1 in this equation. Other two equations look fine.

So I went through your calculations, and yes as @cnh1995 pointed out, you did not include the -10*i1 in Mesh 2. But I worked it out, and I arrived at i1 = zero Amperes (imagine that!).

So you can get the Open circuit voltage at A, and Open Circuit voltage at B. Find the difference between these for Δvab. So to get the Norton current, you need to short terminals A and B together, then run the calculations to find the current from A to B. The Norton Resistance is (Δvab) / (Short Ckt Current from A to B).

cnh1995
terryds said:
transform 5A and 10 ohm (in parallel) into 50 V and 10 ohm in series
I am no expert in this area, but I always understood that ammeters have negligible resistance. The 5A is the current not flowing through the 10 Ohm resistor, no?

haruspex said:
I am no expert in this area, but I always understood that ammeters have negligible resistance. The 5A is the current not flowing through the 10 Ohm resistor, no?
It is a 5Amp ideal current source. What @terryds did was replace a Norton circuit with a Thevenin Equivalent, which is a valid step in analysis. They behave the same.

scottdave said:
It is a 5Amp ideal current source. What @terryds did was replace a Norton circuit with a Thevenin Equivalent, which is a valid step in analysis. They behave the same.
Ok, thanks for the explanation.

scottdave said:
But I worked it out, and I arrived at i1 = zero Amperes (imagine that!).
Yes, I got the same result.
scottdave said:
So you can get the Open circuit voltage at A, and Open Circuit voltage at B. Find the difference between these for Δvab. So to get the Norton current, you need to short terminals A and B together, then run the calculations to find the current from A to B. The Norton Resistance is (Δvab) / (Short Ckt Current from A to B).
Another way to solve the problem is to find the Thevenin equivalent first (where ΔVab would be the Thevenin voltage), and convert it into Norton equivalent using source transformation.
(The resistors are connected in a special way, such that it is possible to find the Norton/Thevenin equivalent resistance only by visual inspection.)

Thanks @cnh1995 . Source Transformation was the term, which I couldn't think of.

## 1. What is a Norton equivalent circuit?

A Norton equivalent circuit is a simplified representation of a complex electrical circuit that contains a current source, a resistor, and a load. It is used to simplify circuit analysis and allows for easier calculation of circuit parameters.

## 2. How do I find the Norton equivalent circuit?

To find the Norton equivalent circuit, you need to determine the Norton current and the equivalent resistance. The Norton current is the short-circuit current at the load terminals, and the equivalent resistance is the resistance seen by the load when all the independent sources are turned off.

## 3. What is the difference between a Norton and Thevenin equivalent circuit?

The main difference between a Norton and Thevenin equivalent circuit is the type of equivalent source. A Norton equivalent circuit uses a current source, while a Thevenin equivalent circuit uses a voltage source. Additionally, the equivalent resistance in a Norton circuit is calculated at the load terminals, while in a Thevenin circuit it is calculated at the source terminals.

## 4. Why is the Norton equivalent circuit useful?

The Norton equivalent circuit is useful because it simplifies a complex circuit into a single current source and resistor, making it easier to analyze and calculate circuit parameters. It also allows for easier comparison and testing of different circuits.

## 5. Can a Norton equivalent circuit be used for both AC and DC circuits?

Yes, a Norton equivalent circuit can be used for both AC and DC circuits. However, the calculations for determining the Norton current and equivalent resistance may differ depending on the type of circuit. For AC circuits, the calculations may involve complex numbers and impedance, while for DC circuits, they may only involve resistance and Ohm's law.

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