capacitor connected across DC voltage source?

Discussion in 'Homework Help' started by vead, Sep 2, 2016.

  1. vead

    Thread Starter Active Member

    Nov 24, 2011
    621
    8
    Hello
    Is that true, if capacitor connect across constant DC voltage source, output voltage of capacitor does not change, also the current through capacitor does not change. If yes than how to proof mathematically
    Let's take example, 4F capacitor which has internal resistance is 2 ohm connected across 10 volt DC supply
    1. Find out voltage across capacitor
    2. Find out current through capacitor
    According to definition, voltage across capacitor will be 10 DC.
    We know formula i=c(dv/dt)
    I=4*?
    Another formula v=i*r
    We know voltage drop across capacitor v=10v DC, internal resistance r=2 ohms
    Thus i=10/2=5 amps

    How to proof mathematically, if capacitor connected across DC voltage source neither voltage across capacitor nor current through capacitor change ,
    _20160902_210950.JPG
    please guide me on right direction.
     
    Last edited: Sep 2, 2016
  2. AlbertHall

    Well-Known Member

    Jun 4, 2014
    1,929
    381
    The 2 Ohms is in series with the capacitor not parallel. 10V across the capacitor, 0V across the resistor, no current (after the initial time while the capacitor charges).
     
  3. vead

    Thread Starter Active Member

    Nov 24, 2011
    621
    8
    Yes I know resistor is in series with capacitor
     
  4. AlbertHall

    Well-Known Member

    Jun 4, 2014
    1,929
    381
  5. WBahn

    Moderator

    Mar 31, 2012
    17,736
    4,789
    Who says that either the voltage or the current through the capacitor don't change?

    Look at the equations you got. Is the current through the capacitor changing? Is the voltage across the capacitor changing?
     
  6. vead

    Thread Starter Active Member

    Nov 24, 2011
    621
    8
    I read somewhere while searching on Internet , but where I don't know
    if I just put the value of t In equations, the value of current is decreasing. From the equations, it can be say that current change over time
     
  7. MrAl

    Well-Known Member

    Jun 17, 2014
    2,425
    490
    Hi,

    Yes but if you want to get exact then you have to figure what happen at all times before t=infinity and for what happens at t=infinity. They are two different animals.
    We often call t=1e7 for example infinity, but to be exact that's not infinity so look at what is happening at t=1e7 for example and then at t=infinity. You'll find a couple interesting results.
    If you look at values for t=1, t=10, t=100, t=1000, etc., up to maybe 1e50 you'll see how it changes, but when you go very high in value the usual 16 digits of precision used in most computer calculators wont be good enough. That is a hint about what is happening.
     
  8. WBahn

    Moderator

    Mar 31, 2012
    17,736
    4,789
    Instead of just trying memorize things that you read on the Internet, try understanding the underlying concepts.

    I think the thing you read somewhere is related to the concept that the voltage across a capacitor cannot change INSTANTANEOUSLY. If you plot the voltage versus time it must be a continuous plot. The current through it does NOT have to be continuous. The reverse is true for an inductor. For a resistor, neither the voltage nor the current need be continuous.

    When you connect a series RC branch to a DC voltage source, all of the voltage appears across the resistor since the voltage across the capacitor is initially zero (assuming it had no initial charge) and so the current through the resistor (and hence the capacitor) is initially the same as if just the resistor were placed across the voltage source. The capacitor initially behaves like a short circuit. As time goes by, the current flowing through the capacitor causes the voltage to increase. The reduces the voltage across the resistor which, in turn, results in less current in it (and thus in the capacitor as well). The result is that the rate at which the voltage builds across the capacitor slows, but it is still building. Eventually the voltage across the capacitor will be close enough to the source voltage that the current flowing in the resistor is negligible and, at that point, the voltage across the capacitor stops growing and now the capacitor is behaving like an open circuit.

    Mathematically, this takes an infinite amount of time. For most practical purposes it takes about five time constants since, at that point, the current is less than 1% of its initial value and we seldom know component values to better than 1%. But even if we said that we wanted the current to be down to one-part-per-trillion (so if the initial current was 1 A we want the current to fall to 1 pA), it would only take 28 time constants. In most cases, it's fallen below the noise floor well before this.
     
    vead likes this.
  9. vead

    Thread Starter Active Member

    Nov 24, 2011
    621
    8
    Please look at my first post. I have taken simple example. And trying to find out voltage across capacitor and current through capacitor over time. In image you will see, I tried to solve equations. But I am stuck here. Now I don't understand what I do, I need formula, but now I don't understand which formulas I have to apply so that I can find out voltage across capacitor and current through capacitor over time?
     
    Last edited: Sep 2, 2016
  10. WBahn

    Moderator

    Mar 31, 2012
    17,736
    4,789
    You HAVE the voltage across the capacitor AND the current through the capacitor over time! Look at the results of your work in the image you posted! Aside from your usual refusal to use units properly, you obtained

    <br />
i(t) \; = \; 5\.A \; e^{\frac{t}{8\,s}<br />

    If that isn't the current through the capacitor over time, then what do you think it is?

    You still need to evaluate the integral in the second part, but if that expression isn't the voltage across the capacitor over time, then what do you think it is?
     
  11. vead

    Thread Starter Active Member

    Nov 24, 2011
    621
    8
    If initial no charge across capacitor, capacitor act as short circuit.
    Look at attachments, sorry for poor quality
    IMG_20160903_021901_1.jpg
     
  12. WBahn

    Moderator

    Mar 31, 2012
    17,736
    4,789
    Why are you integrating from 0 to 1 (i.e., from 0 seconds to 1 second)? What is magical about 1 second? Why not 1 minute, or 1 hour? Why not 1 millisecond?

    If you had bothered to use units properly, you might have been led to ask yourself that question.

    If you want v(t), then integrate from 0 to t.

    If you want v(t=∞), then integrate from 0 to t=∞.
     
  13. vead

    Thread Starter Active Member

    Nov 24, 2011
    621
    8
    I just wanted to find out voltage across capacitor and current through capacitor at 1 second, after that I will calculate for 2s,3s,...etc so I can get table than I can easily figure out what is value of voltage and current at specific time
    Have you seen the answer, value of voltage is negative but it should be positive because capacitor is charging, when it begain to charge. It build some voltage
     
    Last edited: Sep 2, 2016
  14. WBahn

    Moderator

    Mar 31, 2012
    17,736
    4,789
    That's because you don't know how to integrate. And because you refuse to track your units, you didn't catch that you made a mistake that messed up the units on the very first line of your attempt to do so. As a result, you wasted a bunch of time coming up with an answer that was guaranteed to be wrong, but just went ahead and tacked on the units that you WANTED it to have and called it a day.
     
  15. vead

    Thread Starter Active Member

    Nov 24, 2011
    621
    8
    Actually I wanted to know how the voltage across capacitor and current through capacitor change over time. Than I take example to understand. The example that I have taken is hand made, I am not sure does it make any sense
     
  16. WBahn

    Moderator

    Mar 31, 2012
    17,736
    4,789
    It makes perfectly fine sense and you can certainly use it to see how the voltage across and the current through a capacitor change over time. But only if you do the math right and since your math skills are weak you make a lot of mistakes. That's fine; you have the opportunity to learn from your mistakes if you are willing to. But your steadfast refusal to track units, particularly give your weak math skills, indicates that you don't really care about finding your mistakes, let along learning from them.
     
  17. MrAl

    Well-Known Member

    Jun 17, 2014
    2,425
    490
    Hello again,

    If you are having trouble integrating you can do it numerically first to get an idea what is happening.

    If you start with 10v and 10 ohms and 1 Farad, after the first second an approximation with sample time T=1 second is:
    i1=(Vcc-Vc[0])/R=(10-0)/10=1
    Vc[1]=Vc[0]+i1*T*C=1*1=1v

    Then:
    i2=(Vcc-Vc[1])/R=(10-1)/10=0.9
    Vc[2]=Vc[1]+i2*T*C=1+0.9*1*1=1+0.9=1.9v

    Then:
    i3=(Vcc-Vc[2])/R=(10-1.9)/10=0.81
    Vc[3]=Vc[2]+0.81=1.9+0.81=2.71v

    Keep doing that and see what you get. Then later you can try to decrease the time value from 1 second to 0.1 seconds.
    i1=(Vcc-Vc[0])/R=(10-0)/10=1
    Vc[1]=Vc[0]+i*T*C=1*0.1*1=0.1v

    i2=(Vcc-Vc[1])/R=(10-0.1)/10=0.99
    Vc[2]=Vc[1]+i2*T*C=0.1+0.99*0.1*1=0.1+0.099=0.199

    and you can keep doing that up to 1 second or whatever you want. Then if you want to get finer detail, decrease the sample time T to 0.01 seconds and start over again.

    Doing this 1000 times with increment T=0.01 seconds we get for the last four calculations (time, voltage shown):
    9.970, 6.311993
    9.980, 6.315681
    9.990, 6.319365
    10.00, 6.323046

    The more exact result for 10 seconds is:
    6.321206

    so see that got pretty close.
     
    Last edited: Sep 2, 2016
  18. vead

    Thread Starter Active Member

    Nov 24, 2011
    621
    8
    So I have to calculate voltage across capacitor from zero to infinity
    _20160903_123248.JPG
     
  19. WBahn

    Moderator

    Mar 31, 2012
    17,736
    4,789
    Once again, if you would bother to track your units, you would discover that your first line of actual work, line 3, is wrong. Everything beyond that is wasted time and effort, for you and for anyone looking at your work.

    Is there a specific reason why you don't care to catch your mistakes?
     
  20. vead

    Thread Starter Active Member

    Nov 24, 2011
    621
    8
    Hello
    My math skill is very weak, but I am trying to improve and definitely I will improve
    Does this make sense
    IMG_20160904_001632.jpg
    _20160904_002127.JPG
     
Loading...