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DE's in circuit analysis question

  1. Aug 3, 2013 #1
    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

    2. Relevant equations
    i = C dvdt

    Attached Files:

    Last edited: Aug 3, 2013
  2. jcsd
  3. Aug 3, 2013 #2


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    There is no attached picture.

    What does "=>" represent?

    And do you mean the letter ##\tau## (tau)?
    "dvcdt" is dV/dt?
    I don't understand what you did there, but maybe the attachment makes it clear.

    That needs a sketch of the setup.
  4. Aug 3, 2013 #3
    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.
  5. Aug 3, 2013 #4


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

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