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Homework Help: Flash frequency in a circuit

  1. Jun 7, 2006 #1
    Ok so I'm really terrible at circuit questions, and am in need of some help.

    Its a long question so I will just type out he gist of it.

    Q.A simple type of blinking light circuit (attached diagram) is constructed using a neon lamp. When the voltage is low in RC portion, the lamp does not conduct electricity (not there from electrical point of view). The RC cicuit will charge from the 110V power suply, and when the capacitor reaches 75V the neon lamp will become a good conductor and will immediately discharge the capacitor, energy stored will be given off as a flash of light. After the flash the charging process will start again.
    a)Determine the flash frequency with the resistance value shown.
    b) Make a sketch of the voltage across the capacitor in such a circuit, showing several periods.

    So for part (a) the only equations I could find in my text that I thought might apply are:

    time constant (T) = RC, therefore T = (2.5 x 10^4)(4 x 10^-6) = 0.1 sec

    V = Vo(e^-t/RC)
    75V = 110V (e^-t/0.1s)
    ln0.68 = -t/0.1
    t = 0.04 sec

    I'm unsure if this is the answer I'm looking for, that the frequency would be a flash every 0.04 seconds??

    Attached Files:

  2. jcsd
  3. Jun 7, 2006 #2
    This would be correct if the capacitor plates were *discharging*, but the capacitor plates are gaining charge so you need to rethink the equation to use. Just plug in t=0 for the equation. The plates are fully charged at 0 seconds? I think not! I'll have to wait until the document verifies to give you any other guidance though.
  4. Jun 8, 2006 #3
    Ok so the equation for voltage over Charging plate is:

    V = Vo(1 - e^-t/RC)
    And therefore I get an answer of t = 0.11 sec.

    I think I must be missing something since this seems a little too easy!
  5. Jun 8, 2006 #4
    Where did you get this new equation from?
  6. Jun 8, 2006 #5


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    at time infinity, the capacitor "would have been" fully charged, V=Vo .
    It's the CURRENT that always decreases exponentially with time (RC).
  7. Jun 9, 2006 #6


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    jg95ae - No that's all there is to it, except the second decimal is a bit off. First solve the equation for t and then subs the values.
  8. Jun 9, 2006 #7
    I still got the same answer. Solving for t:

    ln(1-(V/Vo)) = -t/RC
    t = -ln(1-(V/Vo))(RC)
    Sub in and get t = -ln(1- (75/110))(0.1)
    t = 0.1145 sec

    I've also realized that I forgot about part (b) of the question and I'm not sure what I'm supposed to draw. Isn't that what the diagram they gave all about?
  9. Jun 10, 2006 #8
    I'm still not sure about part (b) if any one has suggestions it would be greatly appreciated.

    But I also have another question about part (a) when I'm figuring out the flash frequency (flashes/sec) do I need to take into consideration the amount of time it takes to discharge/flash, and therefore it would actually be about 1 flash every 0.15seconds (0.04sec + 0.11sec) ??
  10. Jun 11, 2006 #9


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    Sorry, I get the same answ as you now , I think I took [itex]V_o = 100V[/itex] (the calculation is at work and I can't see what I did).

    The question states that the cap discharges immediately. So it is not necc to include any additional time. You need to calc the frequency for the answ. - you have the time per flash.
    Once it is discharged by the lamp the circuit becomes an RC charging cicuit. This means that the voltage over the cap will build up exponentially from 0V until it reaches 75 V. At which stage it will discharge again (drop to 0V) and the whole process wil repeat. The plasma in the lamp becomes conductive (it is ionized - some electrons are ripped from the atoms in the gas - and conducts electicity very well) when the potential over it reaches 75V. This means it shorts the cap out and results in its immediate discharge as the buildup charge on its plates can cancel each other out by flowing through it. Once the potential over it drops to 0 the ions in the plasma recombines and it again becomes non-conductive. This means as far as the RC circuit is concerned it does not influence the circuit - no conduction takes place through it.
    Last edited: Jun 11, 2006
  11. Jun 12, 2006 #10
    Ok about the sketch, I was thinking it was asking for a cicuit diagram but that didn't make sense seeing as thats what was given. So I'm thinking that it is asking for a graph - that should have a square-wave pattern.

    Can anyone verify if this is correct?
  12. Jun 13, 2006 #11


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    Part b) requires you o draw the graph of the voltage over the capacitor as it charges and discharges. The charging part is described by the equation [tex]V = V_o - V_o e^{-t/RC}[/tex]. Which is [tex]V_o[/tex] minus the discharge graph. As soon as the voltage reaches 75V the cap discharges and the voltage drops to 0V. Then process starts all over again.
    Last edited: Jun 14, 2006
  13. Jun 13, 2006 #12
    Forgive me maybe I'm just not getting this whole question. But in looking at the graph the time between each flash is the time it takes to discharge and charge again. And since I am asked to figure out the flash frequency, ie. the number of times that the light will flash/second. I'm still thinking I need to consider both times together, as it does take time for it to discharge and return to 0V (although its extremely quick, I calculated it to be 0.04sec).
    Maybe there's something I'm not getting?
  14. Jun 14, 2006 #13


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    The discharge takes place throught the neon tube. When the gas in it becomes conductive its resistance is very low. This means it effectively shorts the capacitor, which will discharge almost immediately - in a flash literally. This discharge energizes the atoms in the gas when the charge rush through it. The energized atoms emits the absorbed energy as light when the electrons drop back. This means the discharge is not an exponential process (if it were it would take an infinite amount of time to drop to 0V).

    If it were exponentialy discharge the equation would be

    [tex]V = V_o e^{-t/RC}[/tex]


    [tex]V_o = 75 V[/tex]


    [tex]V = 0 V[/tex]

    but you do not know what R is (very small indeed).
    Last edited: Jun 14, 2006
  15. Sep 8, 2006 #14
    would the graph of this seem linear for a few seconds, then stop, then begin again at the same point but farther down the x-axis (time)?
    any idea how to graph something like this?
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