Another practice question: nortons theorem

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Discussion Overview

The discussion revolves around Norton's theorem in electrical engineering, specifically focusing on determining the current through a load resistor (RL), the voltage across RL, and the power developed in RL using the theorem. Participants seek clarification on the application of the theorem and the calculations involved.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant requests guidance on applying Norton's theorem to find current, voltage, and power in a circuit.
  • Another participant provides a formula for calculating Norton current and resistance, suggesting that R(Norton) equals R(Thevenin).
  • A participant expresses confusion about the notation used for parallel resistors, indicating a need for clarification on the double backslashes.
  • There is a discussion about the equivalence of Norton and Thevenin equivalents, with one participant explaining that Norton's equivalent consists of a current source in parallel with a resistor.
  • A participant questions why a specific resistor (R4) is not considered in the current calculation and seeks confirmation on applying the provided formula to the circuit diagram.
  • Another participant explains that all resistors except RL and R4 must be considered when calculating the short-circuit current, emphasizing the importance of the current's path in the circuit.
  • Instructions are provided for calculating the short-circuit current and output resistance as part of Norton's procedure.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the application of Norton's theorem, with some clarifying concepts while others seek further explanation. There is no consensus on the specific calculations or the treatment of certain resistors in the circuit.

Contextual Notes

Some participants express uncertainty about the notation and the role of specific resistors in the calculations, indicating potential gaps in understanding the theorem's application.

amy_a
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hi guys, sorry to be a bothersome newbie but i have another electrical query. this time its nortons theorem. i need to determine the current through RL, the voltage across RL and the power developed in RL using the theorem.

any pointers on where to start would be really appreciated :!)
 

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R(Norton) = R (Thevenin).
I(Norton) = [E/(R1+(R2//R3))]*[(R2//R3)/R3]

Regards,

Nacer.
 
thanks nacer, i'd best get to work :)
 
ok forgive my naivety but what do the double backslashes indicate? // like that? i know the single one is divide. i feel really retarded now.
 
amy_a said:
ok forgive my naivety but what do the double backslashes indicate? // like that? i know the single one is divide. i feel really retarded now.

I think he is referring to a parallel combination, i.e R1//R2 = (R1*R2)/(R1+R2)

Norton equivalent is no different from Thevenin. Norton equi. = Current Source with a resistor in parallel and Thevenin has voltage source and resistor in series. So, basically you can apply your knowledge of Thevenin equivalents from antoher thread :biggrin: and then just replace a thevenin equ that you'll get with Nortons
 
thanks guys :) i appreciate it! can anyone tell me why r4 is not considered when calculating the current (using the equation given by nacer) would i just have to apply that formula to the diagram above to find the norton current?
 
Every resistor excluding Rl&R4 has to be considered, because you're looking for so-called short-circuit current, so when you have a direct connection between the RL terminals, where does the current go? Since the current is looking for the easiest path, the path with least resistance, it will go just via a direct connection, rather then going via R4.

Check out this link to get more info.
http://en.wikipedia.org/wiki/Norton's_theorem

I suggest that you read up on the Nortons procedure.

1. Calculate the short-circuit current by applying the direct connection between output terminals
2. Calculate output resistance by replacing the voltage source with a short-circuit
3. Norton-source = short-circuit current & Output resistance = resistance from step 2.
 

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