Reading of Ammeter for this circuit with Voltage Controlled Resistors

[Bling Fizikst]

Homework Statement

refer to image

Relevant Equations

refer to image

Let's say the elements connected in parallel have voltage drop of $V_x'$ each , hence they should have the same resistance $R_x'$as per the conditions of the problem (since it mentions that if voltage drop across an element is less than $1$ or greater than $1$ , it has some constant $R_x'$, i.e, $1$ and $2$ respectively) . Let the other element NOT connected in parallel have a voltage drop of $V_x$and resistance of $R_x$. We can easily deduce from here that $$i=\frac{V}{R_x+\frac{R_x'}{2}}$$ where $V$ is the voltage across the leads of the circuit . And hence , $$V_x=\frac{V R_x}{R_x+\frac{R_x'}{2}}$$ and $$V_x'=\frac{V\left(\frac{R_x'}{2}\right)}{R_x+\frac{R_x'}{2}}$$ . From the given graph , we can say , $$V=t , 0<t<10$$ and $$V=-t+20 , 10<t<20$$ . 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 $V_x<1$ and so on , but then it doesn't lead anywhere imo .

 

[BvU]

Hi,

So initially the $R_X$ are all 1 $\Omega$, right ?

Instead of 'a ton of cases': at what $t$ does an $R_X$ jump to 2 $\Omega$, and which one/ones is/are that ?

$\$

 

[Bling Fizikst]

Hi,

View attachment 351797

So initially the $R_X$ are all 1 $\Omega$, right ?

Instead of 'a ton of cases': at what $t$ does an $R_X$ jump to 2 $\Omega$, and which one/ones is/are that ?

$\$

Ok , i made these cases . let $(R_x,R_x')=(2,2),(1,2),(2,1),(1,1)$
For $(2,2)$ , we can conclude that the corresponding $V_x>1$ and $V_x'>1$
Solving for $V$ gives : $V>\frac{3}{2}$ and $V>3$ .
Hence , $V>3$
Now , if $t<10$ then , $V=t\in(3,10)$ and if $10<t<20$ then $10<t<17$.
Similarly , i did the calculations for the other three cases (I got a contradiction for $(1,2)$, so excluded that case) :

For $t\in(0,\frac{3}{2}) : i=\frac{2t}{3}$
For $t\in(\frac{3}{2},5] : i=\frac{2t}{5}$
For $t\in(3,10) : i=\frac{t}{3}$
For $t\in(10,17): i=\frac{20-t}{3}$
For $t\in[15,18.5): i=\frac{20-t}{\frac{5}{2}}$
For $t\in[18.5,20) : i=\frac{20-t}{\frac{3}{2}}$

Not sure what went wrong

 

[BvU]

The question was so simple, why did you ignore it ?

At $t=0$ you have a series circuit of $R_x$ on the left and $R_x// R_x = R_x/2$ on the right. So with $V=t$ the current will behave as $I=t/1.5$ A.

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

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


$\$

 

[BvU]

Ok , i made these cases . let $(R_x,R_x')=(2,2),(1,2),(2,1),(1,1)$

What is $R_x'$ ?
[edit] Oh my, what am I a bad reader. Post #1 says it's the resistance of each of the two on the right.

For $t\in(0,\frac{3}{2}) : i=\frac{2t}{3}$
For $t\in(\frac{3}{2},5] : i=\frac{2t}{5}$

Ah, at $t = 3/2$ something happens: left $R_x$ jumps to 2 $\Omega$, 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 $t=5$ ?

For $t\in(3,10) : i=\frac{t}{3}$

Don't understand. You had something different around $t=3$

I see that for $t$ from 6 to 10 all $R_x$ are 2 $\Omega$, so there $i=\frac{t}{3}$ 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 ).

$\$

 

[BvU]

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 ?

$\$

 

[DaveE]

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.

 

[SammyS]

Ok , i made these cases . let $(R_x,R_x')=(2,2),(1,2),(2,1),(1,1)$
For $(2,2)$ , we can conclude that the corresponding $V_x>1$ and $V_x'>1$
Solving for $V$ gives : $V>\frac{3}{2}$ and $V>3$ .
Hence , $V>3$
Now , if $t<10$ then , $V=t\in(3,10)$ and if $10<t<20$ then $10<t<17$.
Similarly , i did the calculations for the other three cases (I got a contradiction for $(1,2)$, so excluded that case) :

For $t\in(0,\frac{3}{2}) : i=\frac{2t}{3}$
For $t\in(\frac{3}{2},5] : i=\frac{2t}{5}$
For $t\in(3,10) : i=\frac{t}{3}$
For $t\in(10,17): i=\frac{20-t}{3}$
For $t\in[15,18.5): i=\frac{20-t}{\frac{5}{2}}$
For $t\in[18.5,20) : i=\frac{20-t}{\frac{3}{2}}$

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, $R_2$ , with its own set of characteristics. I think I'm using $R_X'$ as you did, so that in general, $R_2=R_X'/2$ and $V_2=V_X'$ .

For $\displaystyle \ \ V_2 \le 1 \text{ volt , } R_2 = \frac 1 2 \text{ ohm , }$ and for $\displaystyle \ \ V_2 > 1 \text{ volt , } R_2 = 1 \text{ ohm . }$

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, $R_{Eq}$ , are:
$1.5\ \Omega \ ,$ for $R_X=1\,,\ R_2=0.5\, ,\$
$2.5\ \Omega \ ,$ for $R_X=2\,,\ R_2=0.5\, ,\$ and
$3\ \Omega \ ,$ for $R_X=2\,,\ R_2=1\, ,\$

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 $R_X=2R_2\,,,\$ specifically when $R_{Eq} =1.5\ \Omega \,$ and $R_{Eq} =3\ \Omega \, ,$ we have $V_X=2V_2\,.\$
Furthermore, the source voltage, $V_0\,,\$ is related to $V_X$ and $V_2$ by $V_0=1.5V_X=3V_2\,.\$

Some additional details for $R_{Eq} =1.5\ \Omega \,:,$
As long as $V_X \le 1 \rm{ \ volt \,, \ then \ }$ $R_X = 1 \ \rm{\Omega}\$ and we have $R_{Eq} =1.5\ \rm{\Omega} \,$ This requires that $V_0\le 1.5 \ \rm{volts} \,.$ Also notice that if $V_X \le 1 \,,\$ then $V_2 \le 0.5 \,,\$ so that $R_{2} =0.5\ \rm{\Omega} \, .$

Similarly, some details for $R_{Eq} =3\ \Omega \,:,$
As long as $V_2 > 1 \rm{ \ volt \,, \ then \ }$ $R_2 = 1 \ \rm{\Omega}\$ and we have $R_{Eq} =3\ \rm{\Omega} \,$ This requires that $V_0> 3 \ \rm{volts} \,.$ Also notice that if $V_2 > 1 \,,\$ then $V_X > 2 \,,\$ so that $R_{2} =0.5\ \rm{\Omega} \, .$

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

The situation for $R_{Eq} =2.5\ \Omega \,:,$ Recall that in this case we have $R_X=2\ \Omega\,,\ R_2=0.5\ \Omega\, .\$

Here we have $R_X=4R_2\,,,\$ so that we have $V_X=4V_2\,.\$
For the source voltage we have $V_0=1.25V_X\,,\$ and $V_0=5V_2\,.\$

Some additional details for $R_{Eq} =2.5\ \Omega \,:,$
$\ R_2=0.5\ \Omega\, .\$ only if $V_2 \le 1\,, \$ requiring that $V_0\le 5 \ \rm{volts} \,.$

$\ R_X=2\ \Omega\,,\$ only if $V_X > 1\,, \$ requiring that $V_0\le 1.25 \ \rm{volts} \,.$

This circuit element does exhibit hysteresis.

 

[BvU]

I don't see any instability

You are quite right and I made a mess by mixing up $I$ and $V$.
My apologies to Bling

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?)

$\$

 

[Gavran]

(Question: anyone know what EstOPhC stands for?)

Estonian Olympiad in Physics Circuits