Thevenin equivalent of lattice network

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Discussion Overview

The discussion revolves around finding the Thevenin equivalent of a lattice network, specifically viewed from terminals As and Br. Participants explore various methods and transformations relevant to circuit analysis, including potential dividers and Y-Δ transformations.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant suggests starting without the resistance f to simplify the problem and then checking how it modifies the impedance.
  • Another participant proposes redrawing the circuit diagram to gain further insight into the problem.
  • Some participants mention the use of the Y-Δ transformation, indicating that it may lead to two triangles in parallel.
  • A participant notes that resistor e connects directly between the output nodes, implying it is in parallel with the rest of the network and suggesting a method to analyze the circuit by temporarily removing it.

Areas of Agreement / Disagreement

Participants express various approaches to the problem, with no consensus on a single method or solution. Multiple competing views on how to proceed remain evident.

Contextual Notes

There are unresolved assumptions regarding the configuration of the lattice network and the specific roles of the resistances involved, particularly the impact of resistance f and the configuration of resistor e.

Mbert
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Homework Statement


Find the Thevenin equivalent network of the lattice network (see attached), viewed from points As and Br (terminals Bs and Ar are not connected to anything).


Homework Equations





The Attempt at a Solution


Without the resistance f, the problem is quite simple. I can find voltages at nodes Bs and Ar as a function of voltage between As and Br (potential dividers). Then insert f and check how it modifies the impedance? I think it only works to check the Thevenin impedance between Bs and Ar, not As and Br, but I'm not sure. Any help would be appreciated.
 

Attachments

  • 2011-11-10 08.55.01 am.png
    2011-11-10 08.55.01 am.png
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Sometimes redrawing the diagram can provide insight. Relocate the node Bs to the interior of triangle AsArBr and see what you get.
 
I'm guessing I have to use the Y-D transformation, so I get 2 triangles in parallel?
 
Mbert said:
I'm guessing I have to use the Y-D transformation, so I get 2 triangles in parallel?

There are several ways to continue, one of which is certainly to employ a Y-Δ transform.

You might want to note that resistor e runs directly between the "output" nodes, so it is in effect in parallel with all the rest. You could always remove it for now, figure out the resistance of the rest, then put e back in parallel to finish up.
 
Yes, many thanks.
 

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