Reading of Ammeter for this circuit with Voltage Controlled Resistors

[Bling Fizikst]

Homework Statement

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Relevant Equations

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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 ?

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[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.


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[BvU]

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

What is $R_x'$ ?

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 ).

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