Hard Circuit simplification problem

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Homework Help Overview

The discussion revolves around a circuit simplification problem, specifically related to Thevenin's theorem and the calculation of equivalent resistance and voltage in a given circuit configuration. Participants are preparing for a contest and are exploring the relationships between various circuit elements.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating total resistance and evaluating voltage across specific points in the circuit. There are attempts to apply Kirchhoff's laws and Thevenin's theorem, with some questioning the uniqueness of the Thevenin equivalent values. Others express confusion about eliminating variables and the relationships between different components in the circuit.

Discussion Status

Some participants have offered guidance by suggesting the use of Thevenin's theorem and encouraging others to share their work to identify where they are getting stuck. There is recognition of the complexity involved in manipulating the equations to eliminate certain variables, indicating a productive exploration of the problem.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the underlying principles rather than simply applying formulas.

rohanlol7
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Homework Statement


Everything is in the picture. Question 3(a)
This is for preparation for a contest.

FtiKaYw.png

http://i.imgur.com/FtiKaYw.jpg

Homework Equations


V=IR

The Attempt at a Solution


I calculated the total resistance of the first circuit and used that to evaluate V(AB). But this is in terms of the resistance of the load. I tried using kirchhoffs laws to eliminate it but i cannot seem to do so. Also i do not understand why there is a unique combination of R(th) and E(th) that works.
I also noticed that R(th) is in face the value of the resistance of the first circuit if the load has zero resistance but i don't understand why this is so
 
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Look at it from the point of view of the load. And post your work to see where you get stuck...
 
Look up Thevenin's theorem. Eth is the Thevenin equivalent voltage and and rth is the Thevenin resistance.
 
cnh1995 said:
Look up Thevenin's theorem. Eth is the Thevenin equivalent voltage and and rth is the Thevenin resistance.
Thanks ! i saw it. However i still don't understand how I'm supposed to figure that out in the problem
 
rohanlol7 said:
how I'm supposed to figure that out in the problem
You have two equations (already filling in ##\ \ V_{AB} = V_{PG} = V_L \ \ ## and ##\ \ I=I_0={V_L\over R_L}\ ##) : $$
V_L = V_{th}\; {r_{th} \over r_{th} + R_L}$$ in the right diagram; and in the left diagram you have $$
V_1 \; {R_3 \over R_1 + R_3} = V_L + R_2 {V_L \over R_L} $$and your task was to work around these two equations with four unknowns to two relationships by eliminating ##V_L## and ## R_L##. Not so easy, I grant you. But thereby you prove Thevenin's theorem (as opposed to making use of it)
 
BvU said:
You have two equations (already filling in ##\ \ V_{AB} = V_{PG} = V_L \ \ ## and ##\ \ I=I_0={V_L\over R_L}\ ##) : $$
V_L = V_{th}\; {r_{th} \over r_{th} + R_L}$$ in the right diagram; and in the left diagram you have $$
V_1 \; {R_3 \over R_1 + R_3} = V_L + R_2 {V_L \over R_L} $$and your task was to work around these two equations with four unknowns to two relationships by eliminating ##V_L## and ## R_L##. Not so easy, I grant you. But thereby you prove Thevenin's theorem (as opposed to making use of it)
Thanks ! i tried something similar.
The second equation come from equating the pd across R3 right ?
 
Across R3 and across the series Rload and R2 as well
 
BvU said:
Across R3 and across the series Rload and R2 as well
The right hand side of the equation represents the pd across the load and R2( right ? ) shouldn't the left hand side be v1- v1*R1/(total resistance for the circuit) ?
 
My bad ! Sorry. Well observed !
 
  • #10
BvU said:
My bad ! Sorry. Well observed !
No problem!
God so i knew how to do it its just the maths part that's killing me. I guess i'll have to work that out slowly. Thanks!
 

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