Strange circuit

Discussion in 'Homework Help' started by victorhugo289, Aug 26, 2010.

  1. victorhugo289

    Thread Starter Member

    Aug 24, 2010
    49
    3
    <snip>, I find this circuit extremely difficult to solve, please have a look,

    I want to know the exact voltage across the capacitors at 40 seconds with both capacitors starting at 0V.

    I used Spice and it solves it perfectly, but I can't do it on my calculator, I need the formula, please.

    Thanks

    --------OMG, I almost forgot! the voltage of battery is 20V. : )
     
    Last edited by a moderator: Aug 26, 2010
  2. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    The math on this one is pretty involved, at least the way I solved it. Here is what I did:

    1. Transform the circuit into the frequency (s) domain.
    2. Perform nodal analysis to obtain three equations.
    3. Solve for the three unknown voltages (I used the simultaneous equation solver on my calculator).
    4. Subtract the voltages to obtain the voltages across each capacitor.
    5. Perform partial fraction decomposition on each.
    6. Take the inverse Laplace of each to return to the time domain.
    7. Substitute 40 for t and solve.
     
  3. Flow

    Member

    May 30, 2010
    37
    1
    Well, have you learnt about the laplace transform yet? That's the most ... hm, comfortable way. If not, you could go the route via differential equations and stay within the time domain.
     
  4. victorhugo289

    Thread Starter Member

    Aug 24, 2010
    49
    3
    Dude, I don't know what you're talking about when you say "Frequency domain" and "Laplace", I'm a newbie, I'm just reading through Tony Kuphaldts book and got to the part of capacitors, I made this circuit to practice but I couldn't figure it out.
    I knew it involved calculous or something like that.
    Hmmm, could you please just give me a formula... like the time constant formula that I do understand.

    Just show me how you did it, please!

    thanks.
     
  5. victorhugo289

    Thread Starter Member

    Aug 24, 2010
    49
    3
    I remember a couple of months ago that I thought learning how to solve simultaneous equations was difficult, but I still managed to solve them and it was pretty easy afterwards. So I don't believe this is so involved, there has got to be some kind of formula, like the Time Constant formula that I thought it was pretty involved at first. This is the only thing stopping me in DC, since I gotta move on to AC.
     
  6. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11


    In the time domain, we look at voltage or current or whatever over time (picture the typical x y graph where the x axis is time and the y axis is the whatever). In the frequency domain the x axis is frequency instead of time.

    The Laplace Transform converts functions from the time domain to the frequency domain using a method involving integration (calculus). Fortunately, lookup charts exist to convert typically encountered functions without actually having to do any integration. For example, the voltage source in your circuit of 20u(t) converts to 20/s. Notice the s variable instead of t, hence the terminology s domain. BTW, in case you are unfamiliar with u(t), that is the discontinuous function which equals 0 at t<0 and 1 at t>0.

    The benefit of transforming into the s domain is that, instead of solving problems with differential equations, we can use much more simple algebra.



    I have to be honest here, I am not certain how to solve this in the time domain, and perhaps the other respondent can be more helpful in that regard. Specifically, I am not sure how to combine the two capacitor legs to obtain V2, and then how to split it back apart to obtain the individual voltages. I can walk you through my solution, but it may take a considerable amount of time and effort if you've never worked with the s domain before. Also, I performed much of the more difficult aspects on a TI-89. I can make those leaps for you if you haven't got one or some advanced math software (MATHCAD, MAPLE, MATLAB, etc). To start, you need to draw out your circuit with s domain values and then lable the nodes. I labeled them just as they are labeled on your first post (V2, V3, and V4).

    To obtain s domain values, resistors remain the same, as stated earlier the voltage source becomes 20/s, and capacitors become 1/sC WITH THE INITIAL CONDITION OF COMPLETELY DISCHARGED CAPACITORS (as you specified in the onset).

    The next question is, are you familiar with nodal analysis? If yes, you need to come up with three equations, one for each of the unknown nodes. I can walk you through this if needed, and to help, your first equation should look like this:
    [​IMG]

    If anyone else knows a simpler method, please chime in.
     
  7. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    Me? I wanna learn how to input math using TeX. Here's that same equation (I hope):

    \frac{V_2-\frac{20}{s}}{100,000}+\frac{V_2-V_3}{\frac{1}{.00033s}}+\frac{V_2-V_4}{\frac{1}{.00022s}}
     
  8. victorhugo289

    Thread Starter Member

    Aug 24, 2010
    49
    3
    @Heavydoody your post is incredible, thanks for that. The time domain and frequency domain thing make sense to me, I can see how that could be the answer.
    All I know is basic stuff so I won't continue asking more about this.

    All i know is that at 40s the voltage at point (1,2) is 8.132 and then goes down gradually to 0, the voltage at point (2,0) is 11.868v and goes up to 20v after 300s or so (19.98 not infinity...).

    As you say, if anyone else knows a simpler --I would add intuitive-- method please comment.
     
  9. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    That's cool, it is a lot to take in all at once.
    I am guessing solution in the time domain is more complex, but I could definetly be wrong. To give you an idea of what you are shooting for, here is one of the solutions:

    \Huge{V_2-V_4=(20-17.55928946e}^{\large{-.01189014489t}}\Huge{-2.44071054e}^{\large{-.03861490561t}}\Huge{)u(t)}
     
    victorhugo289 likes this.
  10. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    You can find the voltage V2 with a single equation, then use the simple voltage divider equation to get the voltage across the caps if you want them.

    0 = \frac{V_2 - \frac{V_1}{s}}{R_1} + \frac{V_2}{R_2 + \frac{1}{sC_1}} + \frac{V_2}{R_3 + \frac{1}{sC_2}}

    That's just one unknown so you can get V2, then you can do this:

    V_3 = V_2\frac{R_2}{R_2 + \frac{1}{sC_1}}
    V_4 = V_2\frac{R_3}{R_3 + \frac{1}{sC_2}}
     
    victorhugo289 likes this.
  11. victorhugo289

    Thread Starter Member

    Aug 24, 2010
    49
    3
     
  12. victorhugo289

    Thread Starter Member

    Aug 24, 2010
    49
    3
    What does the little 's' stand for? Is it desired time in seconds??
     
  13. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    No, that's the 's' from the Laplace transform.
    It's related to frequency.

    Those equations would give you transfer functions which you can then convert to a time-domain equation, as seen in Heavy's post #9

    It was more a post for Heavy I guess, there's really not much else to say that's simple.
    If you want to derive the equation you're stuck using the transform or solving a differential equation.

    If you just want to have the equation then I might solve it later.
     
  14. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    These equations would probably be quite a bit easier to solve without software. I have a bad habit of immediately applying nodal analysis to everything, and, while it is a very effective technique, I frequently wind up doing too much work. As you pointed out though, these are still in the s domain. An earlier poster mentioned solving in the time domain, but I am guessing that would be a nightmare. I may attempt it just for practice. Class starts back up Monday.
     
  15. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    It's still nodal analysis, I just did the shortcut of not worrying about the nodes between resistor/capacitor.
    You only need to consider nodes with more than 2 connections so effectively that's a 1 node circuit.

    It's easy to formulate the equations in the time domain the problem is going to be solving the differential equation. Personally I don't remember how to do that and I honestly don't feel any motivation to refresh my memory.
     
  16. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    OK, I give up. Trying to solve this in the time domain is just plain silliness. Movin' on :)
     
  17. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    BTW, in case anyone is interested, here is my solution. I hate to let it go to waste, and, since this wasn't exactly an "official" sort of homework problem, I don't reckon its inclusion violates the principles of this forum.
     
  18. Flow

    Member

    May 30, 2010
    37
    1
    Victorhugo - not to discourage you, but there's a reason people study electrical engineering for more than 3 years full time.

    There's a lot of "basics" you need to know first, then you build on them, and on whatever you built then... you build something else. So you advance up an hierarchy until you're at a high enough level to be able to design things like TVs, Computers, mp3 players and so on.

    If you don't know what differential equations, complex numbers, fourier series and the fourier transforms are, the laplace transform will not really make sense to you. The math is quite involved, there's a lot of little things in the equations that you need to know, otherwise you're lost - and if nobody then helps you professionally you'd have to reinvent the wheel or study the stuff just like somebody who's studying electrical engineering, but without all the support of the university and fellow students.

    If you're interested in circuitry - why not pay your local university a visit? You'll have fun, believe me ;).
     
  19. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    Your solution is correct but for one quibble.

    When you're deriving the frequency domain solution, you shouldn't have the 1,4 element of your matrix as 20/s; it should just be 20. Your frequency domain expressions have an "s" in the denominator that shouldn't be there.

    You multiply by the transform of a unit step, 1/s, when calculating the time domain solution.
     
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