Hard Circuit simplification problem

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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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on Phys.org
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 ?
 
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) ?
 
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!