# How Do You Approach Nortons and Thevenins Equivalent Circuits?

• Engineering
• dcrisci
In summary, you are struggling with a practice problem for Nortons and Thevenins equivalents. You are confused when it comes to choosing a load, and reducing it into the final simple circuit. You need to pay close attention to what the question says: what are you supposed to find the equivalent for? There is a standard procedure to finding the Norton or Thevenin equivalent to a circuit. You will have this procedure in your course notes. Start at step one and work your way through. Show me where you get stuck.
dcrisci
Hello, I am dario and have a problem that deals with electrical circuits
I have a practice problem for Nortons and Thevenins equivalents, and don't really know how to approach it. I'm confused when it comes to choosing a load, and reducing it into the final simple circuit.

## Homework Statement

The problem is i the attached picture provided, it is a drawing of a circuit.

## Homework Equations

V = IR

Current divider: IRx = ( Rtotal / Rx ) Itotal

Voltage Divider: VRx = (Rx / Rtotal ) * Vtotal

Thevenins Resistance = Nortons Resistance

Nortons Current = Thevenins Voltage / Thevenins or Nortons Resistance

## The Attempt at a Solution

I do not have any progress really because I do not really know how to approach this circuit. Would i neglect the output on the right because no current will flow there and remove the load R1 to determine the circuit? I don't see how else the circuit could be reduced to a Nortons circuit.

#### Attachments

• Thevenin.jpg
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• Thevenin.pdf
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I have a practice problem for Nortons and Thevenins equivalents, and don't really know how to approach it. I'm confused when it comes to choosing a load, and reducing it into the final simple circuit.
In neither approach would you have to choose a load.

do not have any progress really because I do not really know how to approach this circuit. Would i neglect the output on the right because no current will flow there and remove the load R1 to determine the circuit?
In such circuit diagrams, the load would normally be attached to the terminals A and B.
You need to pay close attention to what the question says: what are you supposed to find the equivalent for?
Put that part of the circuit inside a box.

I don't see how else the circuit could be reduced to a Nortons circuit.
There is a standard procedure to finding the Norton or Thevenin equivalent to a circuit.
You will have this procedure in your course notes.
Start at step one and work your way through.
Show me where you get stuck.

If you are having trouble with your notes, then you can look up the procedures online.
i.e. for thevenin:
http://en.wikipedia.org/wiki/Thévenin's_theorem#Calculating_the_Th.C3.A9venin_equivalent
... there are two methods. The one I use is the second one.

For norton:
http://en.wikipedia.org/wiki/Norton's_theorem
... again, two methods: I use the second one.

Here is my attempt at a solution for Nortons equivalent. I am confused for one thing though. For the nortons current, the 30 A current will divide through R1 and R2 and 30 A will then flow through R3. I want to say that the norton's current will end up being 30 A as well but feel as if that is wrong.

For the norton's resistance I found it to be 18.33 ohms. I open circuited the current generator and simplified the circuit to find the total resistance.

#### Attachments

• Untitled.jpg
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Um OK: You did -

step 1. find the Norton resistance.
open circuit all current generators: there's just one.
find the resistance between A and B ... which you did from:

$$R_N= \left(\frac{1}{R_1}+\frac{1}{R_2}\right)^{-1} + R_3$$
... notice that this is just the equivalent resistance in the circuit?

You have a linear network with an ideal current source of 30A in parallel with an equivalent resistance RN ... the Norton equivalent to this is an ideal current source of IN in parallel with resistance RN ... sketch the two networks next to each other.

So what value does IN have to be for the Norton network to be the same as yours?

Simon Bridge said:
Um OK: You did -

step 1. find the Norton resistance.
open circuit all current generators: there's just one.
find the resistance between A and B ... which you did from:

$$R_N= \left(\frac{1}{R_1}+\frac{1}{R_2}\right)^{-1} + R_3$$
... notice that this is just the equivalent resistance in the circuit?

You have a linear network with an ideal current source of 30A in parallel with an equivalent resistance RN ... the Norton equivalent to this is an ideal current source of IN in parallel with resistance RN ... sketch the two networks next to each other.

So what value does IN have to be for the Norton network to be the same as yours?

Sorry for the really late response, I totally forgot about my new account on here. I better start remembering so I can put it to use more. Thank you for the help Simon, I actually got my professor to help me with this the following day so this problem is finished.

That's great - can I twist your arm to post the answer you worked out.
That way, anyone else who got stuck the same way you did and googled to this thread can get the answer they need?

Cheers :)

## 1. What is a Norton Circuit?

A Norton Circuit is a type of electrical circuit that is equivalent to a Thevenin Circuit. It consists of a current source in parallel with a resistance, and can be used to simplify complex circuits for analysis.

## 2. How is a Norton Circuit different from a Thevenin Circuit?

While both Norton and Thevenin Circuits are equivalent, they differ in the type of equivalent source they use. A Norton Circuit uses a current source, while a Thevenin Circuit uses a voltage source.

## 3. How can I find the Norton equivalent circuit of a complex network?

To find the Norton equivalent circuit, you can follow these steps:

1. Remove the load resistor from the network.
2. Find the open-circuit voltage at the load terminals.
3. Find the short-circuit current at the load terminals.
4. Use Ohm's Law to calculate the equivalent resistance at the load terminals.
5. Draw the Norton Circuit, with the calculated equivalent resistance in parallel with the short-circuit current source.

## 4. Are Norton and Thevenin Circuits only used for DC circuits?

No, Norton and Thevenin Circuits can be used for both DC and AC circuits. However, the calculations for equivalent resistance and source values may be different for AC circuits due to impedance.

## 5. How are Norton and Thevenin Circuits used in practical applications?

Norton and Thevenin Circuits are commonly used in circuit analysis and design, as they provide a simplified representation of a complex circuit. They are also used in electronic devices such as amplifiers and sensors, where it is important to match the source impedance to the load impedance for maximum power transfer.

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