DE's in circuit analysis question

[zhongyan]

The question that I'm having trouble with involves an RC circuit with a wire down the middle and two switches, I've attached a picture of it.

There are two parts to the question and it's the second part of it I'm having trouble understanding but I'll put in the first one for context anyway:
Denote the signal vc by 'y' and denote y(0) = vc(0)=>v0 and tao=>RC. Write the differential equation governing the system in terms of y, y0 and tao when SW1 is open and SW2 is closed.

I assumed this was standard KVL in a loop with just a few variables replaced for naming conventions. So my own result for this question is:

V0 = iR + vc => V0 = RC(dvcdt) + vc => y0 = tao(dydt) + y

I'm hoping I'm not wrong yet. The second question is what has me stumped due to this being a "Signals and Systems" subject my classes do not go in depth into circuit analysis as it was recommended background knowledge (which I don't seem to have).

Now assume that at t = 0, SW1 closes and SW2 opens. Write down and solve the DE for this situation to obtain an analytic expression for y(t), assuming y(0) = v0 >0. This expression is known as the natural response.

If you can see the circuit that's attached I'm really not sure where to start when the voltage source isn't connected on both ends. Do I use KVL? KCL? Is there current actually running in this situation? If not how exactly do I construct the necessary DE's. I'm fine with solving them but if someone could help me understand how to construct the DE's I would be very grateful.

If I've left out anything that would help please let me know, it's my first time posting on these forums.
EDIT: forgot to upload picture so sorry

Homework Equations

i = C dvdt

 

[mfb]

There is no attached picture.

What does "=>" represent?

And do you mean the letter $\tau$ (tau)?
"dvcdt" is dV/dt?

V0 = RC(dvcdt) + vc => y0 = tao(dydt) + y

I don't understand what you did there, but maybe the attachment makes it clear.

Now assume that at t = 0, SW1 closes and SW2 opens.

That needs a sketch of the setup.

 

[zhongyan]

Really sorry about that, the picture didn't get uploaded but it's there now, hope it gives you gives you better context and I did mean tau (didn't know it was spelt like that). dvcdt is just dV/dt as you said, but the voltage in the capacitor's voltage specifically not the source's.

 

[mfb]

Okay. tau(dydt) + y should be V0 then, not y0.

If you can see the circuit that's attached I'm really not sure where to start when the voltage source isn't connected on both ends.

There is a wire connected to both ends. A wire is like a voltage source with 0V, if you like.

Do I use KVL? KCL?

Both.

Is there current actually running in this situation?

Sure.

 

[rezeew]

The Attempt at a Solution

Hello,

Thank you for providing the context and the picture of the circuit. Based on the information provided, it seems like this is a problem involving an RC circuit with two switches, SW1 and SW2. The first part of the question asks for the differential equation governing the system when SW1 is open and SW2 is closed. You have correctly applied KVL to the circuit and obtained the equation V0 = RC(dvcdt) + vc, where V0 = y0 and tao = RC. This is a valid differential equation for this circuit.

Now, for the second part of the question, we need to consider the situation when SW1 closes and SW2 opens, at t = 0. In this case, the circuit will have a different configuration, with the voltage source no longer connected to the circuit. To construct the necessary differential equation, we can use KCL at the node where the capacitor and the resistor are connected. This will give us the equation i = C(dvcdt) = 0, since there is no current flowing through the circuit in this situation. We can also use KVL in the loop to obtain the equation V0 = RC(dvcdt) + vc. Combining these two equations, we get the second differential equation as follows:

C(dvcdt) = 0
V0 = RC(dvcdt) + vc

Substituting V0 = y0 and tao = RC, we get:

C(dvcdt) = 0
y0 = tao(dvcdt) + y

This is the differential equation for the natural response of the circuit when SW1 closes and SW2 opens. To solve this equation and obtain an analytic expression for y(t), we can use standard methods for solving first-order differential equations, such as separation of variables or integrating factors. Once we have the solution, we can substitute the initial condition y(0) = v0 and obtain the final expression for y(t).

I hope this helps you understand how to construct the necessary differential equations for this circuit. If you have any further questions, please don't hesitate to ask. Best of luck with your studies!