Reading of Ammeter for this circuit with Voltage Controlled Resistors

  • #1
Bling Fizikst
107
13
Homework Statement
refer to image
Relevant Equations
refer to image
Screenshot 2024-10-03 160739.png


Let's say the elements connected in parallel have voltage drop of each , hence they should have the same resistance as per the conditions of the problem (since it mentions that if voltage drop across an element is less than or greater than , it has some constant , i.e, and respectively) . Let the other element NOT connected in parallel have a voltage drop of and resistance of . We can easily deduce from here that where is the voltage across the leads of the circuit . And hence , and . From the given graph , we can say , and . So , we have current as a function of time . But to write the resistances , all i can think of is make a ton of cases with and so on , but then it doesn't lead anywhere imo .
 
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  • #2
Hi,

1727954275162.png


So initially the are all 1 , right ?

Instead of 'a ton of cases': at what does an jump to 2 , and which one/ones is/are that ?

 
  • #3
BvU said:
Hi,

View attachment 351797

So initially the are all 1 , right ?

Instead of 'a ton of cases': at what does an jump to 2 , and which one/ones is/are that ?

Ok , i made these cases . let
For , we can conclude that the corresponding and
Solving for gives : and .
Hence ,
Now , if then , and if then .
Similarly , i did the calculations for the other three cases (I got a contradiction for , so excluded that case) :

For
For
For
For
For
For

Not sure what went wrong
 
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  • #4
The question was so simple, why did you ignore it ?

At you have a series circuit of on the left and on the right. So with the current will behave as A.

When will the first resistance jump from 1 to 2 occur and where ?

[edit] oops, my bad. You have the first second and a half just fine.


 
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  • #5
Bling Fizikst said:
Ok , i made these cases . let
What is ?
[edit] Oh my, what am I a bad reader. Post #1 says it's the resistance of each of the two on the right.

Bling Fizikst said:
For
For
Ah, at something happens: left jumps to 2 , so the current drops to 0.6 A, and the resistor jumps back, current up, etc etc. How long does this unstable situation last ?

Similarly: what happens at ?


Bling Fizikst said:
For
Don't understand. You had something different around

I see that for from 6 to 10 all are 2 , so there A.

But what from 5 to 6 seconds ?

Then: is this a mathematics exercise or a physics exercise ? In a physics situation the two resistors on the right don't 'jump' simultaneously, so the situation is even more complicated. ('My bad' was that I completely overlooked this unstability issue o:) ).

 
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  • #6
Is this from an electric circuit olympics set of exercises ?

I don't see stable solutions in spite of tip 9
fact 9: If there is more than one stable solution then the question of which solution is actually realized is resolved based on the history (e.g. if the voltage applied to the circuit has been increased or decreased) because internally, nonlinear elements obey inertia (for instance, the density of charge carriers can change fast, but not instantaneously) and it will not jump from one equilibrium state to another without a good reason(such as a loss of stability or disappearance of the current solution branch).
and, directly preceding problem 23:
The phenomenon when the system state depends on its history is called hysteresis. Hysteresis will typically appear if the
system can have more than one internal states; a simple example is provided by the following problem.

@DaveE ?

 
  • #7
OK, first, my normal rant. I think HW problems that are poorly documented and ask people to do extra work with simple tasks are annoying at best, and make me question the qualifications of the people writing them (most likely mathematicians or physicists that never have to actually do this stuff IRL). In this case, two annoying things.

1) Real EEs name each component in the schematic. I guess that X is a value in this case, so we choose our own names, like R1, R2, R3, etc. The problem is everyone will choose different names, so every time you read someone else's work you have to constantly translate, like "your R2 is really my R1". Or worse, they don't name them and we are left wondering "which X are you referring to?" It's an unnecessary PITA that has nearly no educational value. If you try that out in the real EE world, people will hate you and/or your CAD program won't let you.

2) What is the educational value in having two identical resistors in parallel. Do they think people are too dumb to just combine them? I dislike embedding easy problems into difficult ones; teach one thing at a time. What did anyone, who can actually solve this, learn by doing that extra bit of clerical work?

OK, sorry
-- End of Rant --

Brute force, for me. You just have to identify all of the breakpoints and what happens at each. Of course, everything changes linearly in between the breakpoints. Start at 0V and solve the circuit, then increase the voltage until something changes and do it again, etc. There may be a more elegant way, which I'd be interested in seeing. I'm reminded of the Ben Franklin quote "I have already made this paper too long, for which I must crave pardon, not having now time to make it shorter. "

Time isn't a factor, since it's only resistors, so stability isn't really correct in any sense. But it is definitely possible to have an undefined solution at the break points if they are ambiguously defined. This needs to be verified at each change in the network. In this case there is no ambiguity at the boundaries.

My brute force solution is attached, in it's original messy form. Sorry if it doesn't make sense. Also, algebra mistakes are a definite possibility.

Scan2024-10-04_1246471 copy.jpg

Scan2024-10-04_1246472 copy.jpg
 
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  • #8
Bling Fizikst said:
Ok , i made these cases . let
For , we can conclude that the corresponding and
Solving for gives : and .
Hence ,
Now , if then , and if then .
Similarly , i did the calculations for the other three cases (I got a contradiction for , so excluded that case) :

For
For
For
For
For
For

Not sure what went wrong
( I don't see any instability. )

I agree only with your first interval and its current. The sixth interval also looks good.

It was helpful for me to treat the two resistors in parallel as a single individual resistor, , with its own set of characteristics. I think I'm using as you did, so that in general, and .

For and for

There are only three resistance values possible for the given configuration of the three resistors. These three values for the equivalent resistance of the network, , are:
for
for and
for

Well, I see that @DaveE has chimed in, so I'll halt this for now. I may continue it later - - - or not.

Continuing on:

What we have here is a voltage divider.
The two cases in which specifically when and we have
Furthermore, the source voltage, is related to and by

Some additional details for
As long as and we have This requires that Also notice that if then so that

Similarly, some details for
As long as and we have This requires that Also notice that if then so that

Now for the complexity and what may seem to be contradictory results.

The situation for Recall that in this case we have

Here we have so that we have
For the source voltage we have and

Some additional details for
only if requiring that

only if requiring that

This circuit element does exhibit hysteresis.
 
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  • #9
SammyS said:
I don't see any instability
You are quite right and I made a mess by mixing up and .
My apologies to Bling o:) o:)

So all that remains in this exercise is the asymmetry snag (*)
And indeed the high artificiality: I agree with Dave's rant

Thanks @DaveE and @SammyS for setting me straight

------

(*) Which is pointed out in the PDF -- which isn't all that bad. Indeed Physics Olympiad stuff, of which we've seen more exercises pop up in PF (Question: anyone know what EstOPhC stands for?)

 
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  • #10
BvU said:
(Question: anyone know what EstOPhC stands for?)
Estonian Olympiad in Physics Circuits
 
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