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Transient circuit

  1. Dec 5, 2014 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    I had a question about transient circuits. I was asked to find an expression for ##v_C(t)## at any time ##t## for the following circuit. My work is shown in the image below.

    Then the response is interrupted at ##t = 50 ms## by the switch being opened and I'm asked to derive a new equation for ##v_C(t)## for ##t > 50 ms##.

    2. Relevant equations


    3. The attempt at a solution

    IMG_0454.jpg

    It seems like the capacitor is in steady state operation for any time ##t## regardless of the switch ##(i = 0)##. When I consider ##t \rightarrow \infty## in step 2, the extra wire seems to be confusing me for some reason.

    The ##4k## and ##1k## are actually in parallel (for convenience, ##R_{eq} = \frac{4k}{5}##), but how am I to find the voltage across the open terminals in this case?

    Is it simply the voltage across all of the components? Then ##v_c(t \rightarrow \infty) = 60V = V_F## since the remaining ##1k## and ##R_{eq}## are in series.
     
  2. jcsd
  3. Dec 5, 2014 #2

    gneill

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    While the switch is closed that horizontal path is all one node. You are free to move any connections to that node to any other location along that node. So see how the circuit looks if you slide the point where the 4 k resistor connects along to where the 1 k resistors connect:

    Fig1.gif

    Does that help? You should be able to tell immediately what the Thevenin values will be for that.
     
  4. Dec 5, 2014 #3

    Zondrina

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    Thank you for the reply. I understand what you've done there with the circuit topology somewhat.

    It's easy to see the Thevenin resistance as ##R_{th} = \frac{4k}{5} + 1k = \frac{9k}{5}##. So the time constant would be ##\tau_c = R_{th}C = \frac{9000}{5} (80 \times 10^{-6}) = 0.144 s = 144 ms##.

    I still can't seem to see ##v_C(t \rightarrow \infty) = V_F## for some reason. I know that ##60V## needs to drop across the ##\frac{4k}{5}## plus the capacitor so that ##v_c = 60V - v_R##. There's no way I can see to get that voltage though.

    EDIT: Wait wait, there's no current in that right side there is there? If there was no current, that would mean ##v_C(t \rightarrow \infty) = 0 V##!

    Is that the case?
     
    Last edited: Dec 5, 2014
  5. Dec 5, 2014 #4

    gneill

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    Ah, no. When you want to find the Thevenin resistance you first suppress the sources. In this case that means replacing the voltage source with a short circuit, so the 1 k resistor gets shorted out too...
    Not quite. The capacitor voltage will head towards the open circuit (Thevenin) voltage. At steady state the capacitor current is zero so the current through the 4/5 k resistance is zero, too. Zero current through a resistor means no change in potential from one end to the other.
     
  6. Dec 5, 2014 #5

    Zondrina

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    Sorry I seem to be speeding without thinking for some reason today.

    Actually short circuiting the voltage source this time, the ##1k## is shorted out. So ##R_{th} = \frac{4k}{5}## as viewed from the capacitor terminals.

    So ##\tau_C = R_{th}C = \frac{4000}{5} (80 \times 10^{-6}) = 0.064 s = 64 ms##.

    Yes I was intending to say this when I said there was no current. At steady state ##i_C = 0## because ##\frac{dv}{dt} = 0##. This means there is also zero current through the ##\frac{4k}{5}## and so ##V = IR = 0##.

    So if zero current is flowing through that part of the circuit, the current would be entirely across the ##1k## resistor ( This can be observed by applying KCL just above the ##1k## resistor ). So the voltage across the ##1k## resistor, which is equal to the source voltage, is also equal to the capacitor voltage.

    ##v_C(t \rightarrow \infty) = 60V = V_F##
     
  7. Dec 5, 2014 #6

    gneill

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    Right.

    Of course, the switch is to be opened at t = 50 ms, which is a good deal short of infinity. And since the time constant you calculated is just 64 ms you'll want to find what the capacitor voltage is at t = 50 ms; it won't be the full 60 V.

    Whatever that voltage turns out to be is where the next stage begins.
     
  8. Dec 5, 2014 #7

    Zondrina

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    Indeed, so this is what I got for that first portion:

    IMG_0455.jpg

    Then I continued by doing this:

    IMG_0456.jpg

    I stopped there since I just wanted to ask, since I'm back to the circuit in step 1, the Thevenin resistance would be ##2k## because the ##4k## gets shorted right?

    The time constant would then be easily found.

    Then I would simply plug everything into:

    $$v_C(t > 50 ms) = V_F + (V_I - V_F)e^{\frac{-(t - t_1)}{\tau_C}}$$

    Where ##t_1 = 50 ms##.

    EDIT: Disregard the plot. I forgot to tack on a negative sign when calculating graph points.

    I don't know what's going on today. Otherwise I think the rest of the process is correct.
     
    Last edited: Dec 5, 2014
  9. Dec 5, 2014 #8

    Zondrina

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    For the sake of not wanting to mess everything up.

    ##v_C(t = 50^-) = 41.69 V = v_C(t = 50^+) = V_I##

    ##v_C(t \rightarrow \infty) = 20V = V_F##

    ##\tau_C = R_{th}\cdot C = (2000)(80 \times 10^{-6}) = 0.16s = 160 ms##.

    ##v_C(t > 50) = 20 + (41.69 - 20)e^{\frac{-(t - 50)}{160}} = 20 + 21.69e^{\frac{50-t}{160}}##
     
  10. Dec 5, 2014 #9

    gneill

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    I'm getting confused as to which circuits pertain the step number. I didn't realize it before, because your text of the problem description did not say, that the switch was considered to be open for t < 0. I just noticed this indicated in your attachment.

    Anyways, when the switch is closed you found the Thevenin resistance to be 4/5 k. When the switch is open you have a simple voltage divider setup with 4k and 2k of resistance. The Thevenin resistance of that is NOT 2 k. The 4 k is NOT shorted by the source when it is suppressed.

    Why not list your Thevenin values and time constants for both configurations so that we can agree on them before moving on.
     
  11. Dec 5, 2014 #10

    Zondrina

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    Okay, I think I understand now what it really means when something is shorted. I was not entirely sure about that concept.

    I understand everything in the first image with the three circuits is correct now up to step 4 where I found ##v_C(t > 0)##. Up to this point I have:

    ##v_C(t = 0^-) = 20 V = v_C(t = 0^+) = V_I##
    ##v_C(t \rightarrow \infty) = 60V = V_F##
    ##R_{th} = \frac{4k}{5}##
    ##\tau_C = 64 ms##

    ##v_C(t > 0) = 60 - 40e^{\frac{-t}{64}}##

    The plot i did in step 5 was silly because i forgot to tack on a negative sign to the exponential.

    Now to find ##v_C(t > 50)##:

    ##v_C(t = 50^-) = 41.69 V = v_C(t = 50^+) = V_I##
    ##v_C(t \rightarrow \infty) = 20V = V_F##
    ##R_{th} = (\frac{1}{2k} + \frac{1}{4k})^{-1} = \frac{4k}{3}##
    ##\tau_C = R_{th}\cdot C = \frac{4k}{3}(80 \times 10^{-6}) = 0.107s = 107 ms##.

    ##v_C(t > 50) = 20 + (41.69 - 20)e^{\frac{-(t - 50)}{107}} = 20 + 21.69e^{\frac{50-t}{107}}##
     
  12. Dec 5, 2014 #11

    gneill

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    Yup, that looks good. Be sure to state somewhere in the solution that you hand in that t in your formulas is in milliseconds.
     
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