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Please help me in this IES question in circuit theory (subject)

  1. Jun 30, 2011 #1
    i m trying to solve this question by method given in most of the standard books i.e.

    Vc(t) = Vc([itex]\infty[/itex]) - [Vc(0[itex]\hat{}[/itex]-) - Vc([itex]\infty[/itex]) ] e [itex]\hat{}[/itex] -t/RC

    first i found Vc(0[itex]\hat{}[/itex]-) by considering switch open and inductor behaving like short and capacitor open at steady state....then i find voltage acroos 4 ohm resistor i.e.

    V4 = (4 * 18)/6 = 12 volt

    now = Vc(0[itex]\hat{}[/itex]-) = [C(total)/2c ] * V4 = 4 volt

    now i m finding Vc([itex]\infty[/itex]) by considering switch closed ....and at steady state inductor will be shorted and both capacitors will be opened....and now inductor path and capacitor c bot are in parallel and i have confused what to do here next .....please help me....
     

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    Last edited: Jun 30, 2011
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  3. Jun 30, 2011 #2

    NascentOxygen

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    I don't like this question, because it involves 2 capacitors connected in what resembles a voltage divider across a DC supply. But if you treat them as perfectly ideal, then what you have done looks okay. The larger capacitance charges to 4v and the lower one reaches 8v across its plates.

    The inductor does indeed have zero volts across it in the steady state. But what you may be overlooking is the fundamental property of inductance---you can't change the current in an inductor instantly. So, after the switch is closed, the value of current flowing in the inductor continues at the same value and in the same direction as it was before the switch was closed.

    So, before the switch is closed, you determined the current through the inductor branch to be 3A, therefore after the switch is closed the current through the inductor is still 3A. (Though the closing of the switch may cause the current to be drawn from a different or additional path, the value of the current stays at 3A for the brief moment in time that we are considering.)

    Every other element in your circuit is happy to have its current changed instantaneously, but not the inductor. You can instantaneously change the voltage across an inductor, but you can't instantaneously change the current flowing through it.

    Perhaps if you think of the capacitances as being REALLY BIG (one is BIG and the other is DOUBLY BIG) it might help in the way you form an image in your mind of the transient conditions.
     
    Last edited: Jun 30, 2011
  4. Jun 30, 2011 #3
    This equation is used for "first order" circuits with one capacitor and no other reactive elements. It is not correct to use this equation for the circuit shown in the picture.

    Two things you could do
    1. write and solve the differential equations for the circuit after the switch is closed
    2. use the http://en.wikipedia.org/wiki/Laplace_transform#s-Domain_equivalent_circuits_and_impedances" for after the switch is closed
     
    Last edited by a moderator: Apr 26, 2017
  5. Jun 30, 2011 #4

    gneill

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    You're looking for the current through the 2C capacitor the instant after the switch is closed. So first you want to find the steady-state conditions for the circuit before the switch closes. You've figured out the voltage across the 4Ω resistor, which will also be the voltage on the 2C capacitor. What's the current through the inductor and voltage on capacitor C?
     
  6. Jul 3, 2011 #5
    There are no exact values for the reactive elements so I don't believe that s-domain will help here. I agree with the direction that NascentOxygen is going, inductors cannot have an instantaneous change in current and capacitors cannot have an instantaneous change in voltage. I do disagree with the assumption about the capacitors acting like a voltage divider. The voltage across C (the small capacitor) is not 4 volts at t = 0.

    And with V = I · R this problem becomes "intuitively obvious" ;)
     
    Last edited by a moderator: Apr 26, 2017
  7. Jul 3, 2011 #6

    NascentOxygen

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    No one said it is 4 volts at t=0. :wink:

    Not at t=0- nor at t=0+
     
  8. Jul 3, 2011 #7
    OK - you did state that it charges to 8 volts. But at t = 0, the voltage across the lower capacitor C is zero volts.
     
  9. Jul 4, 2011 #8

    NascentOxygen

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    Before S is closed, the voltage across capacitor C has stabilized at 8 volts. So immediately after S is closed, the voltage across capacitor C is still 8 volts.

    I can't see the series capacitors as anything other than a ghastly "ideal capacitor" voltage divider.

    Do we both agree the answer is 2 amps?

    P.S. I have mused how it is a bad question on two counts. People whose first language is not English could very easily read the question as the circuit having stabilized with S closed, and at time t=0 the switch is opened.

    I give the examiner 2/10 for this question! :devil: :devil: :mad: :mad:

    Must try harder! :devil:
     
  10. Jul 4, 2011 #9
    You are right I was looking at it wrong. The voltage drop across the inductor is zero volts before the switch is thrown, but when the switch is closed voltage across the inductor jumps to 8 volts because of the charge on C. I agree that it is 2 amps after looking at it closer.
     
  11. Jul 4, 2011 #10

    gneill

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    Before the switch is thrown the voltage on C is zero because the inductor, with a steady current, has zero volts across it. The inductor current is 18V/6Ω = 3A. The voltage across the 2C capacitor is 12V (that is, 3A * 4Ω = 12V). There is no current flowing in either capacitor, and the 3A inductor current thus flows via the 2Ω resistor and the 4Ω resistor. There's 12V across the 4Ω resistor (same as capacitor 2C).

    When the switch closes, the current through the inductor will not immediately change. Neither will the voltage across the capacitor C (nor the capacitor 2C for that matter).

    So in the first instant after the switch closes, capacitor C "clamps" the voltage across the inductor at zero for the first instant. But it was already zero. So there's no voltage change for the inductor, no voltage change for the capacitors.

    The final steady state that the circuit will head for has 0V on the inductor and capacitor C, 12V on capacitor 2C, 3A flowing in the inductor via the 2Ω and 4Ω resistors. Well, this is exactly the state that the circuit is in before and after the switch closes! So what current has to flow through the 2C capacitor so that no change is made?

    EDIT: I've made an error in the above, in that with the switch open the capacitor C is *not* connected across the inductor, so its initial voltage should be 8V, not zero. It will still have this 8V potential at the instant the switch closes. That puts 8V across the inductor (which still carries 3A), and makes the drop across the 4Ω resistor 4V (that is, 12V - 8V). Of course, with a drop of 4V across 4Ω that yields a current of only 1A. But the inductor demands 3A. The additional 2A required must come from somewhere...

    Note that the potential at the top of 4Ω resistor is still 12V with respect to the battery negative terminal. So the current through the 2Ω resistor is still 3A. If only 1A is flowing via the 4Ω resistor, the other 2A must detour via the 2C capacitor.
     
    Last edited: Jul 4, 2011
  12. Jul 4, 2011 #11
    There is 12 volts across 2C and C. Use the voltage divider idea from NascentOxygen to get voltage across C. There is nothing connected between 2C and C to bring it to zero volts before t = 0. That what was tripping me up earlier.
     
  13. Jul 5, 2011 #12

    NascentOxygen

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    Why not edit your message so the mistake is either removed [strike]or at least clearly marked as wrong?[/STRIKE] There may be people coming into this topic who take on board your explanation, only to then find it in the next paragraph overturned. Confusion reigns!

    That's about how I see it.

    Any idea why I can't use the extended ASCII characters, such as omega, alpha, etc. As soon as I enter them, or try to quote a post that has some, the text in my posted message gets truncated at that point. I'm using Firefox. It probably hinges on the setting of a default character set somewhere, but I haven't discovered where it's hiding.
     
  14. Jul 5, 2011 #13

    NascentOxygen

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    Has munna007 who posted the question stayed with this to the bitter end? It is a fairly searching question for a multiple choice test!
     
  15. Jul 5, 2011 #14

    gneill

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    You can use latex to enter math, ("itex" delimiters for inline), or you can cut and paste symbols. Some people keep the symbols in a .signature so that they're readily available. I keep a set in a Word document that I leave open on my desktop. Here's a sampler:

    α β γ δ ε θ λ μ ν π ρ σ τ η φ χ ψ ω Γ Δ Θ Λ Π Σ Φ Ψ Ω

    ∂ ∏ ∑ ← → ↓ ↑ ↔

    ± − ÷ √ ∫ ½ ∞∴ ~ ≈ ≠ ≡ ≤ ≥ °

    π²³ ∞ ° → ~ µ ρ σ τ ω ∑ … √ ∫ ≤ ≥ ± ∃ · θ φ ψ Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô
     
  16. Jul 5, 2011 #15

    NascentOxygen

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    Yes, I know you can cut and paste maths symbols, but I'm bemoaning that I can't. :frown: To illustrate, I'll paste a Beta between two dollar signs farther on, and you'll see that nothing more that I type after that Beta shows up--my message gets truncated at the first extended ASCII character. Sure, I can type it, but when processed it gets chopped. That's why I'm asking if anyone knows why. :cry: Here's that Beta character between two dollar signs ......$
     
  17. Jul 5, 2011 #16

    NascentOxygen

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    ^^ see, you don't get to see the beta character, nor the rest of that post that I wrote after it. I can use itex, but that's a separate issue and is dodging the issue.
     
  18. Jul 5, 2011 #17

    gneill

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    I'm using Firefox and have no such difficulties. Here's a snapshot of my Firefox font and character set encoding settings (Tools --> Options --> Content --> (Fonts & Colors)Advanced.

    attachment.php?attachmentid=36962&stc=1&d=1309883753.gif

    I suggest that if you continue to have difficulties that you bring the matter to the "Forum Feedback and Announcements" forum. These sorts of issues have been thrashed out there before.
     

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