RC circuit problem

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EEEZAY

Joined Dec 17, 2019
1
Capture.PNG
I'm unable to do this question. I get confused when do we have to take capacitor as an open circuit and when not.


I got t<-2 first which was V=0

then at t=-2
I was uable to solve it I had two resistors and one capacitor and one voltage source in series. I couldn't come up with an equation.

After what time interval do we take t---> infinity?
 

ci139

Joined Jul 11, 2016
1,088
The voltage source climbs to infinity - so you noticed . . .

We can estimate the persons brainfault (infinite voltage would require infinite energy . . .) - who compiled such excercise ::

1000µF caps are usually 50V range , also 50Ω resistors would be expected to be around 2W (← it's not the only option - but a quite likely 1)

  • V would reach 50V @ \( t=\frac{50V}{5·10^{-6}·1V·s^{-1}}-2s=10^{1-\left({-6}\right)}s-2=\frac{10Ms}{3600·24}-2≈115.7days-2s \) starting from 10µV (← a likely initial voltage on capacitor before t=0)
  • 50mA won't draw the cap to negative charge when V is greater than 2.5V = 50Ω · 50mA
    ( so the cap is started to be discharged ! from negative charge ! at that time \( t=\frac{2.5V}{5·10^{-6}·1V·s^{-1}}-2s \) )
  • constant current would cause a linear ramp on cap . . . so while the voltage is lower the 2.5 the (V.cap against time - ) lines would be quite stright from 10µV to 0V down to where it reaches before voltage catching up the current sink caused voltage drop at the junction of 3 resistors . . .
  • likely you need to integrate over currents and charges trough the resistors . . .

You likely need the 3 voltage source merging in (3 resistor junction) formula :: \[ U_\text{Junction}=\frac{R_\text{High}·U_\text{Med}·R_\text{Low}+U_\text{High}·R_\text{Med}·R_\text{Low}+R_\text{High}·R_\text{Med}·U_\text{Low}}{R_\text{High}·R_\text{Low}+R_\text{Med}·R_\text{Low}+R_\text{High}·R_\text{Med}} \] ← the current signs must be kept dependent on initial condition (R.Low sinks (U.Junction > U.Low it's current has "-" sign -- respective to U.Junction) the R.High and R.Med are sourcing (U.High > U.Junction , U.Med > U.Junction , -- their currents have "+" sign) . . . so the currents "turnover"/SUM in U.Junction is ZERO) where when assumed the current source being U.Low it's should (e.g. 50mA on your schematic) be changed with current induced values (i donno if such easily done) . . .

. . . i must be sleeping coz it's trivial \( I=Const → \frac UR=Const=I \ → U_\text{Junction}-U_\text{Low}=\Delta U=Const\ ,\ \frac{U_\text{Junction}-U_\text{Low}}R=\frac{\Delta U}R=50mA \) ↑Replace \( U_\text{Low}=U_\text{Junction}-2.5V \) and \( R_\text{Low}=\frac{2.5V}{50mA}=50Ω \) in the prev formula↑

. . . or define your own differential equation for the \( U_\text{Med} = U_\text{Capacitor} \)
 
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WBahn

Joined Mar 31, 2012
25,115
View attachment 194705
I'm unable to do this question. I get confused when do we have to take capacitor as an open circuit and when not.


I got t<-2 first which was V=0

then at t=-2
I was uable to solve it I had two resistors and one capacitor and one voltage source in series. I couldn't come up with an equation.

After what time interval do we take t---> infinity?
You treat capacitors as open circuits when the circuit operation is such that there is no current flow in them -- or at least that the current flow is negligible. Since current flows in a capacitor whenever the voltage across it changes, this is the same as saying that you can treat it as an open whenever it is a reasonable approximation that the voltage across the capacitor is not changing -- it doesn't have to be zero, just a constant.

Given that your sources are described in terms of step functions, I'm guessing that you are either expected to work the problem using differential equations or using Laplace/Fourier transforms. But I can't tell which since I don't know where you are in your curriculum. Please enlighten us.

If you are using transform methods, then just the voltage source as some Vo(t) and the current source (and the connected switch) as some Io(t) and solve for the output voltage in terms of Vo(s) and Io(s). Once you have that done, which is pretty straightforward, simply replace the symbolic transformed sources with their actual transformed expressions for these particular sources. The tricky part is then taking the inverse transform to get v(t).

One thing you should get in the habit of doing is asking if your circuit makes physical sense. Sketch the voltage output of the voltage source for all time -- does it make physical sense? What is the output of the current source before the switch closes? Does that make sense? On the one hand, exercises like this don't have to make physical as their primary purpose is to give you practice working with the solution methods, so they are contrived circuits to begin with. Still, realizable circuits should ideally be the goal. But even if it were, we wouldn't know the context of this circuit model -- it could be that it is modelling a circuit not over all time, but only over a fairly short period of time and all that is important is that it be a reasonable model during that time interval; we don't care if it is realizable or makes physical sense outside the time window of interest. That is an extremely common approach when working with highly nonlinear circuits or with transient signals.

But even then, asking whether the circuit makes sense and identifying what issues it has outside the window of interest can really help you validate that the circuit is correct within the window. For instance, if you are modelling a photodiode sensor as a current source charging a capacitor, then your model will probably have the capacitor charging to an infinite voltage as time goes forward. This makes sense, even if it is unrealizable. Our window of interest simply has to be narrow enough that this behavior is reasonable within it. But if the circuit we end up with doesn't have this behavior outside the window of interest, then there is probably something wrong with it.
 

WBahn

Joined Mar 31, 2012
25,115
? u(t-2) = ´ḻ́ḻ(t-2) , https://en.wikipedia.org/wiki/Heaviside_step_function → "usually denoted by H or θ (but sometimes u, 1 or ´ḻ́ḻ ) " ?
(( suppose ))
I'm not following what your question is.

The voltage source is explicitly given in terms of a step function, while the current source is implicitly given in terms of one due to the behavior of the switch that connects it to the rest of the circuit.
 

RBR1317

Joined Nov 13, 2010
512
I get confused when do we have to take capacitor as an open circuit and when not.
A capacitor becomes an open circuit when it is fully charged. But don't waste your time with differential equations or Laplace transforms for a circuit with a single energy storage element. You should know by now that the voltage across the capacitor will follow an exponential curve involving the term exp(-t/RC) where R is the Thevenin equivalent resistance of the charging circuit. But here you have a time segmented circuit where the initial conditions, i.e. voltage across the capacitor, will be the final condition from the previous segment. So each time segment will have its own equivalent circuit.
 

MrAl

Joined Jun 17, 2014
6,977
View attachment 194705
I'm unable to do this question. I get confused when do we have to take capacitor as an open circuit and when not.


I got t<-2 first which was V=0

then at t=-2
I was uable to solve it I had two resistors and one capacitor and one voltage source in series. I couldn't come up with an equation.

After what time interval do we take t---> infinity?
You are correct that at t=-2 seconds the voltage source is zero, then at t=-2+ seconds (just to the right of t=-2 seconds) the voltage jumps up to 5v. So you are right also that the voltage can be viewed as 5v at t=-2 seconds and the cap begins to charge. So you now have a 5v source, two 50 ohm resistors in series, and a cap that is charging through those two resistors.
Now are you stating that you dont know how to solve that circuit for the cap voltage?
That is just Step 1 for solving the entire problem. If you dont know how to do that then you will never be able to solve the entire circuit.
So the question then is what have you learned in the past? Havent you done any single source single resistor single capacitor circuits yet? If not then i guess you will need help with that before proceeding with the rest of the problem (when the 50ma source kicks in).
What methods have you used to solve RC problems in the past?
 

jarkky

Joined Jan 10, 2020
20
It should be probably 5u(t-2), because laplace is not valid for t<0 (if you want to use the Laplace).
But if its correct you should "rewind" the t -axis so that
5u(t-2) becomes 5u(t) and 50mA at t=0 becomes 50mA*u(t-2).
And once calculation done then "forward" the time axis results back by 2sec.
 
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jarkky

Joined Jan 10, 2020
20
Found the error and
vo(t)=(5/2*e^(-10*t) - 5/2)*u(t) + (5-5*e^(-20-10*t))*u(t+2)
should be checking faster...

This would give plot as
Screenshot from 2020-01-10 20-49-32.png
 

RBR1317

Joined Nov 13, 2010
512
I could not find yet any real application for the Laplace...
While I would not personally choose to solve a problem like this with the Laplace transform, the benefit of doing so is that it will allow you to master the techniques necessary to solve more complex switched circuits that can only be solved with Laplace methods (or with differential equations - but let's not go there.)
 

MrAl

Joined Jun 17, 2014
6,977
It is actually not very hard to solve using Laplace Transforms.
If i can get to it i can illustrate.
I just wondered what he meant by "cant find application for Laplace".
 

MrAl

Joined Jun 17, 2014
6,977
Here is a a quick rundown of one way to approach this.

Circuit A is the original, circuit B is just the same circuit with the times shifted.

Circuit 1 is mode 1 that has to be solved for the first 2 seconds. The cap voltage has to be solved for at t=2. I am sure anyone can figure out how to use Laplace on this. The approximate but very close solution to this is 5 volts, which becomes the initial current generator for the next mode.

Circuit 2 is mode 2, which has three sources and note the 'cap' starts with 0 volts again.
This is simply a 3 source circuit and we have to solve for Vo which means the new cap voltage plus the initial cap voltage. Although this is a little more difficult, if anyone has solved 3 source circuits before using Laplace then this should present no problem.

Generally circuits with switches that have given turn on and turn off times are relatively easy to solve. The ones where we have to solve for the switch timing too get much more difficult unless they happen to be simple by nature.
A simpler example is if the switch turned on once the cap voltage reached 1v. This would not be too hard to do we'd just have to solve for the time it takes for the cap to reach 1v in Mode 1. But if the source voltage was not constant it could get more difficult, just for example if the 5ut source was a square wave with frequency maybe 20Hz.

Give it a shot if you have time.
 

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