Solving Norton Equivalent Circuits - Help Needed

In summary, the student is having difficulty understanding and working with the circuit due to a lack of tutor support. They are having difficulty resolving the equations necessary to solve the circuit and are in need of help.
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Homework Statement



Having some Difficulty in starting the second question on this sheet. i have managed to find examples and worked my way through the first question, but unfortunately my tutor has decided to move onto pastures new, and we have an assignment to complete but no tutor which has left me in a slight pickle!

The main problem at the moment is reducing the circuit into a format where i can understand and work with it.
im assuming the switch is redundant as it states it is closed, but from there onwards, I am not sure which are in parallel and which are in series (as I am mechanically biased = clueless)

Homework Equations



i have all the equations necessary and i have worked out equivalent resistances etc before, but it is literally just knowing the order of things to do with this and what each are doing.

The Attempt at a Solution



My first attempt was to close the switch, assume that the 180 and 270 ohm resistors are in series, giving me 450ohm in parallel with the 90. and then having the equivalent (75) in series with the 135. giving me a 200ohm resistor, 6A and the load.

any help would be much appreciated as I am a bit aimless at the minute without a tutor to point me in the right direction.

thanks very much.
 

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  • #2
If the switch is closed, the 135 ohm resistor is shorted and no longer in the circuit.
 
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  • #3
Manners said:

Homework Statement



Having some Difficulty in starting the second question on this sheet. i have managed to find examples and worked my way through the first question, but unfortunately my tutor has decided to move onto pastures new, and we have an assignment to complete but no tutor which has left me in a slight pickle!

The main problem at the moment is reducing the circuit into a format where i can understand and work with it.
im assuming the switch is redundant as it states it is closed, but from there onwards, I am not sure which are in parallel and which are in series (as I am mechanically biased = clueless)

Homework Equations



i have all the equations necessary and i have worked out equivalent resistances etc before, but it is literally just knowing the order of things to do with this and what each are doing.

The Attempt at a Solution



My first attempt was to close the switch, assume that the 180 and 270 ohm resistors are in series, giving me 450ohm in parallel with the 90. and then having the equivalent (75) in series with the 135. giving me a 200ohm resistor, 6A and the load.

any help would be much appreciated as I am a bit aimless at the minute without a tutor to point me in the right direction.

thanks very much.

First determine what closing the the switch S1 accomplishes. For example, how is the 135 Ω resistor affected? Once you've sorted that out the network should look easier to solve for the Norton resistance.

When determining the Norton or Thevenin resistance you want to suppress the sources and then find the resistance of the resulting network as "seen" from the perspective of the load terminals (load removed, of course).

One suggestion: sometimes it is convenient to solve circuits in stages, converting Norton to Thevenin and back along the way, gobbling up components as you proceed across the circuit. Once you've figured out S1's effect you might find this hint helpful... :wink:
 
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  • #4
Ah, so the switch shorts out the 135 resistor, and because its an open circuit due to removing the current source I am just dealing with the top three resistors and the load.
Right ill have a bash at that tonight and see if i can find an old example somewhere to walk through with it. thank you both for the advice.
 
  • #5


I understand the importance of having a tutor to guide you through difficult concepts and assignments. However, it is also important to develop problem-solving skills and be able to work independently. I would suggest breaking down the circuit into smaller sections and using the basic principles of parallel and series circuits to simplify it. You can also use Kirchhoff's laws to analyze the circuit and determine the equivalent resistance. Additionally, it may be helpful to draw a schematic diagram of the circuit and label the components to better understand the connections. If you are still struggling, I recommend reaching out to your classmates or seeking help from a peer tutor or online resources. Keep practicing and don't give up, as solving circuits can be challenging but also very rewarding. Good luck with your assignment!
 

Related to Solving Norton Equivalent Circuits - Help Needed

1. What is a Norton equivalent circuit?

A Norton equivalent circuit is a simplified representation of a complex circuit that consists of a current source in parallel with a resistor. It is used to simplify calculations and analysis of a circuit while maintaining the same output characteristics as the original circuit.

2. How do you solve a Norton equivalent circuit?

To solve a Norton equivalent circuit, you need to follow these steps:
1. Calculate the equivalent resistance by combining all the resistors in parallel.
2. Calculate the equivalent current by dividing the source voltage by the equivalent resistance.
3. Draw the Norton equivalent circuit with the equivalent current source and equivalent resistance.
4. Analyze the circuit using standard circuit analysis techniques.
5. Convert back to the original circuit by replacing the equivalent current source with a short circuit and the equivalent resistance with an open circuit.

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

The main difference between a Norton equivalent circuit and a Thevenin equivalent circuit is the type of equivalent source used. A Norton equivalent circuit uses a current source, while a Thevenin equivalent circuit uses a voltage source. Additionally, in a Norton equivalent circuit, the resistor is in parallel with the source, whereas in a Thevenin equivalent circuit, the resistor is in series with the source.

4. When is it useful to use a Norton equivalent circuit?

A Norton equivalent circuit is useful in circuit analysis when you want to simplify a complex circuit and focus on a specific part of the circuit, such as the load. It can also be used to determine the maximum power transfer from a circuit to the load.

5. What are the limitations of using a Norton equivalent circuit?

There are a few limitations of using a Norton equivalent circuit, including the assumption that the circuit is linear and passive, meaning it does not contain active components such as transistors. Additionally, it only works for DC circuits and not AC circuits. Also, it does not take into account the effects of mutual inductance or capacitance in the circuit.

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