Need help solving this parallel RLC circuit:

Discussion in 'Homework Help' started by nishu_r, Jun 2, 2012.

  1. nishu_r

    Thread Starter Member

    Jun 2, 2012
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    I request help in solving this parallel RLC circuit which is a practice question in Engineering Circuit Analysis by Hayt,Kemmerly SIE . Practice question 9.6 , pg no 292
    Question and Circuit : [​IMG]

    Thank you
     
  2. WBahn

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    Mar 31, 2012
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    What have you done thus far?

    Have you gotten answers, even if wrong, for any of the three questions?

    Are you working the problem using differential equations or using Laplace transforms?

    Consider the following questions:

    Q1) What is the voltage on the capacitor (v_C) just before the switch is closed?

    Q2) What is the current in the inductor (i_L) (positive if flowing downward) just before the switch is closed?

    Q3) What is the voltage on the capacitor (v_C) just after the switch is closed? (If you don't have a numerical answer, express it in words in terms of the answer to Q1)

    Q4) What is the current in the inductor (i_L) (positive if flowing downward) just after the switch is closed? (If you don't have a numerical answer, express it in words in terms of the answer to Q2)

    Q5) What is the voltage across the resistor (v_R) (postive side is the top of the resistor) just after the switch has closed?

    Q6) What is the current in the resistor (i_R) (positive if going downward) just after the switch has closed?

    Q7) What is the current in the capacitor (i_c) (positive if going downward) just after the switch has closed? (If you don't have a numerical answer, express it in words in terms of the answers to Q4 and Q6)

    Q8) What is the rate of at which the voltage across the capacitor is changing? (Note that this is the answer to the first question)


    Let's get you this far and then we can tackle the other two questions from there.

    Q3)
     
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  3. nishu_r

    Thread Starter Member

    Jun 2, 2012
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    First of all , thanks a lot for your response.

    Yes i did try solving the question but in vain, i must have gone wrong somewhere.

    Using differential equations to solve the question.

    Q1. v_C (0-) = v_C (0+) = v_C (0) = 2V (using voltage division for the voltage across the 100 kiloOhm resistor)

    Q2. i_L (0-) = i_L (0+) = i_L (0) = -10 mA ( i have a doubt here , am i wrong? )

    Q3. & Q4. Doesn't voltage of the capacitor and the current through the inductor remain the same for abrupt changes in time? so, v_C and i_L should remain the same.

    Q5. voltage across v_R prior to t=0 is -5V. (across the 500 ohm resistor)

    Q6. i_R = 0 A? just after the switch turns right.

    Q7. is capacitor current , i_C = - i_L - i_R ? (nodal analysis), then it should be 10mA

    Q8. Since, it is underdamped , dv/dt= - (alpha)*B_1 + B_2 * (omega_d)

    Alpha, damping coeffecient = 1/ 2RC ; Omega_d, natural freq

    pls let me know where i am wrong and i will correct myself.

    thank you again
     
  4. WBahn

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    Mar 31, 2012
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    You are largely going down the right path, just making a few oversights along the way. Answering my questions has helped me see some of your difficulties very nicely -- thanks.

    Spot on.

    The right magnitude, but the wrong sign. Prior to t=0, the current comes out of the positive side of the 5V source (the left side), down through the inductor (the same direction that I defined for positive current in the inductor) and then up through the resistor.

    Yes, but the phrase "abrupt changes in time" is not correct. Time does not change "abruptly". It "changes" at the same rate all the time (no pun intended). The way to describe this is that capacitor voltage and inductor current are "continuous" with respect to time and therefore cannot change abruptly. Note that "abruptly" applies to the voltage and current, not time.

    Yes.

    Here is your first major conceptual error. So consider the following next questions:

    Q9) If the current in the resistor is 0A, then what is the voltage across it?

    Q10) In light of the answers to Q1 and Q3, what is the voltage on the capacitor at this same time?

    Q11) Given that, once the switch is closed and the 5V source is turned off (has 0V across it but has no restrictions on the current through it), the capacitor, inductor, and resistor are in parallel, what do you know about the voltages across all three from t=0+ on?

    Now go back and reanswer Q6 and, once we get a good answer there, take another shot at the rest.
     
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  5. nishu_r

    Thread Starter Member

    Jun 2, 2012
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    Yes, you are correct (my bad, i deserve the pun :) ) . Abrupt changes in voltage across a capacitor or current through an inductor for a very small period of time is not possible as it would require an enourmous amount of current or voltage respectively.

    Q2. I have a doubt here, if (prior to t=0) current through the inductor is from the positive so the current is 10mA then is current through the 500 Ohm resister -10mA? I have troubles with the passive sign convention. How to minimise errors in that?

    Q10. still v_C = 2V

    Q11. & Q9. At t=0+ , since the connection is parallel the voltage across the other elements should also have the same voltage as the v_C , as it remains unchanged. Oh! So is this the same voltage cross the resistor? then it means i_R=(v_C / R) ?

    well now B_2 = -6 (correction) ; so i got dv/dt = -1400 V/s (correct answer)

    Thank you , i have got the answer through the questions. I really appreciate you spendin your time to correct me, most would have simply given me the answer.
     
  6. WBahn

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    Yes. More to the point, energy in a capacitor is related to the voltage and energy in an inductor is related to the current. Since energy cannot be created or destroyed, the quantities can therefore change only as fast as the energy stored in the device is changing and they physically cannot change instantaneously at all. Period. They can change fast, but not instantly.

    This way is perfectly valid and will produce correct answers. For every device in the circuit, classify it as either a source or a load. Don't worry if you aren't sure. You can make the classification randomly if you want to.

    For each device (let's stick with two-terminal devices) assign a voltage polarity to it. Again, you can flip a coin in making this decision for each device.

    Now assign the current flow for each device in accordance with the passive sign convention. This simply means that if you classified a given device as a load, then current is positive when it is flowing from the positive terminal (of THAT device) through the device and out the negative terminal. For a source, it is just the opposite and you draw the arrow indicating positive current through the device going from negative terminal to positive terminal.

    Now solve the circuit being careful to observe the polarities of the devices when writing your equations. At this point, it does not matter whether a given device is a battery, a resistor, or what. You have a black box with a defined polarity for both the (symbolic) current and the (symbolic) voltage. Use those and don't worry about what you do or don't know about the directions of the actual voltages and currents.

    Once you have solved for a particular quantity, say the current in the resistor, then either the solution is positive or it is negative. If it is positive, then you guessed right and the actual current is flowing in the same direction as the symbolic current. If it is negative, then you guessed wrong and the current is actually flowing the other direction.

    Similarly, multiply the voltage and current together and you get power, which may be positive or negative. If it is positive and the device was classified as a source, then the device is supplying power to the rest of the circuit. If it is negative, then it is absorbing power. The reverse is true for a device classified as a load -- positive power means it is absorbing power from the rest of the circuit and negative power means it is supplying power to the rest of the circuit.

    While you are allowed to do this randomly, you are much less likely to make stupid mistakes if you try to classify sources and loads correctly. For the voltages (and keep in mind that, for any given device, you are free to arbitrarily assign the direction of the symbolic current and then assign the polarity of the symbolic voltage accordingly), you should either take your best guess at the polarities or choose the polarities that will result in the cleanest equations. In this case, I simply chose to assign ALL of the symbolic currents as going downward through each component.

    In solving for the initial current in the inductor and resistor, I noted that, in steady state, the inductor would have no voltage across it and therefore the magnitude of the current around the source-inductor-resistor loop would be 5V/500Ω=10mA. I already had a sheet of paper with the circuit drawn on it and the symbolic voltages and currents shown on it. I then literally took my finger and went out of the positive terminal of the source, down through the inductor (in the same direction as the symbolic current, so it was +10mA) and then on around the bottom and up through the resistor (in the opposite direction as its symbolic current, so that was -10mA) and back to the negative terminal of the battery.

    Yes! Congratulations! Notice that I didn't answer any of the problem for you (though I did offer a couple of clarifications or explanations on points that I thought you basically had a handle on but weren't quite up to par yet).

    I can tell that you spotted the flaw in your reasoning and now understand so much better how to tackle problems like this (or even much more general) than you did.

    Having said that, what is B_2? And what should the units be?

    Thank you for acknowledging it. This approach is called the Socratic Method and, in loose terms, means asking small questions that only require the person to apply a single principle or concept (at least when working through difficult portions of a problem). That not only makes it easier for the student to avoid getting confused, but it also makes it clear to the instructor which concepts they have down and which concepts are causing the problem.

    It's an extremely powerful technique, but it requires both a willing teacher and a willing learner to work. Thank you for being such a learner.
     
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  7. nishu_r

    Thread Starter Member

    Jun 2, 2012
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    thank you for explaining passive sign convention, much clearer than i first read it from the text. You have nack of clearly explainng minute details, but now i learnt that these minute details for the basis of our understanding of circuits.

    B_1 is one of the two arbitrary constants that define the equation of any current or voltage in any branch of a parallel RLC network, say for the underdamped parallel/series RLC , v(t) = {e^[-(alpha * t)] * ( B_1.cos(Omega_d*t) + B_2.sin(Omega_d*t)}

    I found B_2 at time t=0, v(0)=B_2=2 ;
    and i had to differentiate v(0) to obtain B_1. I think rest you must know. And this is where i needed the i_R(0) for my calculation of i_C(0)

    It must be a real credit to be your student.
    Thanks again. I hope you don't mind me coming back with my queries and next time i will question myself. :)
     
  8. WBahn

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    Mar 31, 2012
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    Ah, thank you.

    So, in the example you give, B_2 is NOT 2, it is 2V. One of the most valuable lessons I could hope to teach you is to always do two things when working any problem: (1) Track your units from beginning to end. Treat them as part of the quantity, which they are. A desk isn't 3 long, it is 3 feet long. It is also 36 inches long and 1 yard long. These are all equal quantities. But 3 is not equal to 36 is not equal to 1. But "3" multiplied by "feet" is equal to "36" multiplied by "inches". Tracking units consistently will let you check your work, let you spot the majority of errors you will make, and will prevent you from making many of those errors in the first place. (2) Ask if the answer makes sense. In a case like this, you have a number of sanity checks you can make. You know what the time constants would be for an RC and an LR circuit. So calculate those and compare them to alpha. If alpha is much larger or much smaller than either of them, suspect that something is wrong. Similarly, you know the resonant frequency of an LC circuit, so compare that to your damped natural frequency.

    In fact, just by doing sanity checks, I can tell that your B_2 is probably wrong. I know that the voltage at t=0 is 2V. At t=0, v(t) = B_1, so B_1 is 2V, not B_2.

    And thank you for your final comment. It is appreciated. I am fortunate enough to know that a significant number of my students felt I did well by them. That is not to say that there aren't others who would stridently disagree.

    I hope to see you again in the future. Working with someone that has your attitude toward learning is always a pleasure.
     
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  9. nishu_r

    Thread Starter Member

    Jun 2, 2012
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    Yes, B_1 = 2; i did it correctly on my paper but typed it wrong and went by it. Am sorry. And these arbitrary constants that conform to initial conditions and the circuit element values have units. I missed that part too.

    So for an RLC ckt, to check, i should have the values of time constants of RC and LR ckts, and if the dampening coefficient is largely varying from either; then there is a mistake in the solution? But can you please tell me why this is so(relation between the both) in the circuit point of view and not just for correcting a solution. And also the comparison of the resonant frequency and the natural frequency? Yes, i do have a text but the way you explain gives me a new prespective.

    thanks again for your patience
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    And you missed it again. B_1 does NOT equal 2. It equals 2V.

    As others can attest, I am not going to relent on this point. I am, without a doubt and without apology, a Units Nazi.

    Its from the basis of applying common sense regarding what seems like reasonable behavior. For instance, let's say that you put a gallon of water into a bucket that has a hole in the bottom and it takes two minutes for it to all drain out. You then put a gallon of motor oil in it and it takes three minutes for it to all drain out. Someone then tells you that if you put a gallon of vegetable oil in it that it should take 30 minutes to drain out. Do you buy that? Probably not. Someone else tells you that it should take 5 seconds. Do you buy that? Probably not. Common sense says that vegetable oil is not enough different from water or motor oil to justify a significantly different drain time.

    In this case, you have a circuit that is a combination of an LR and an RC circuit. The time constant is a reflection of the fact that the L and C store energy and the R dictates the rate at which that energy can be dissipated. It would seem intuitively unreasonable that if the LR time constant was 10ms and the RC time constant was 100ms, that combining them would lead to a time constant of either 10ms or 100s.

    Now, having said that, intuition can be wrong. I have found the following to be a very sound approach and I use it as much as possible. I first try to estimate what I expect the answer to be, or at least bounds on what I expect it to be. If I have an estimate, I try to see if I can decide whether I expect the actual answer to be above or below the estimate. Where possible, I try to estimate how close I expect the estimate to be (i.e., within 10%, within a factor of 2, within an order of magnitude, etc). One way to do this is to make two estimates, the first one you make simplifying assumptions that should make the answer smaller and the second on you make simplifying assumptions that should make the answer bigger. You now have a range within which you answer should lie. I then work the problem. If it agrees with my estimates, then I have a greater degree of confidence that I didn't make any mistakes or, if I have, that even though the answer is wrong is may be close enough so as to be workable.

    This is one where I would expect to make better estimates. The damped natural frequency for a second order system is always lower than the undamped frequency. So that right there gives you a sanity check on the answer. Furthermore, the less the damping, the closer the damped frequency is to the undamped.
     
    Last edited: Jun 6, 2012
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  11. nishu_r

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    Jun 2, 2012
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    Yes sir, am sorry . You are right, I should be careful with the units.

    And your analogy is simply brilliant, exactly explains the process of dampening an energy storage element.

    combining the circuits the result would be either 10ms or 100s(or is it ms?)

    And reading your estimating procedures, i understand that you have quite an experience in this as it must have only come from a lot of practice. And i checked various instances of natural freq and resonant frequency and yes the differences between them is only mathematical.

    Thanks again.
     
  12. WBahn

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    Good. Trust me, it will pay for itself many times over.

    I had a bit of a typo which made my point hard to really see. I meant to say that the RC and LR time constants were 10ms and 100ms, so it would be unreasonable if the calculated LRC time constant turned out to be 10us (three orders of magnitude smaller) or 100s (three orders of magnitude larger). I would expect something that would be in the range from 10ms to 100ms or perhaps somewhat outside of it. Since the sum of the energy is being dumped into the same resistor, I wouldn't be too surprised if the time constant were a bit longer (perhaps something around the sum of the two individual time constants) and since energy is also bouncing back and forth between the L and the C it might take a bit longer, yet.

    I forgot to make a point in my last post. As I said, intuition and common sense are not infallable, so occassionally you will estimate an answer to come out to be one thing and calculate (or measure) the value to be something wildly different. When this happens, you either made a mistake in the calculation or measurement that your estimate has flagged, or your estimate was way off. Either result is worthwhile. In the first case, you have caught an error and can correct it. In the second case, you have caught a shortcoming in your understanding and can correct that, instead. So don't just check the math three times and then move on. Stay on the problem until you understand WHY your estimate was so far off.

    Not sure what you mean by the differences only being "mathematical". The differences are real and can be significant.
     
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  13. nishu_r

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    Jun 2, 2012
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    A good hint you gave there, as far as circuits is concerned reasoning is as important as applying formulae and the maths that comes with it.

    And i just meant that even mathematically we can see the difference since the natural freq takes in resonant freq as a part in it's equation(omega_d=(sqrt)(omega_0^2 - alpha^2) for an underdamped response), so the relation is mathematical to see. And yes it is real.

    And sir I don't quite understand where i am going wrong in this question i will post now, i did find the thevinin's equivalent and then used the formula to find the i_L(t) for an RL circuit (only the steady state) but my result isn't even close . I am posting it below.
    Thanks [​IMG]
     
  14. WBahn

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    Please post a summary of your work, as well. What were your results? What was your Thevinen circuit? What order did you get your results in? These are all very useful and important to figuring out where you've gone astray.
     
  15. nishu_r

    Thread Starter Member

    Jun 2, 2012
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    [​IMG]

    I have posted the image above which shows the thevinin's equivalent circuit that i obtained, and i got an answer of i_L= 7.35 mA . pls let me know my mistakes.
    Thanks again Sir, since i am studying on my own i have troubles understanding completely. You seem to be my only hope.
     
  16. WBahn

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    In calculating the equivalent resistance, keep in mind the definition on series and parallel elements. If two elements are in series, then any electron that flows through one of them has to flow through the other. If two elements are in parallel, then any voltage that appears across one of them must also appear across the other. If neither of these apply, then they are neither series nor parallel.

    Also, you didn't track units when you plugged values into your equation.

    It admittedly makes it tough when the author is sloppy with units -- and most of them are. For instance, when he has v(t) = 40cos(8000t) V, at least he has the units on the overall quantity (and many authors would leave it off), but he also needs units on the 8000. There are two reasons for this: (1) The value of t could be expressed in one of any of a huge number of units, such as seconds, microseconds, hours, fortnights; (2) the argument to any transcendental function (trig functions, logs, etc.) must be dimensionless. This means that the units used to express t must have recipricol units to whatever the 8000 has. But if 8000 truly has no units, then we have to use a quantity of time that has not units and while I have no idea what those units might be, it can't be seconds. The units of 8000 should be 1/s or, more usefully, rad/s (radians are dimensiohless). When ever you start a problem, examine the units and, if necessary, supply them where the author has been too lazy to do their own work properly.
     
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  17. nishu_r

    Thread Starter Member

    Jun 2, 2012
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    [​IMG]

    I have included the units and i end up achieving Volt/Ohm, which acc to Ohm's law is Amp. And 8000 is rad per sec, Sir i came across another problem in this book, the frequency is in radians but the angle between the voltage and current ,say phi is in degrees. So doesn't it mean i have to convert the degree to radians and then solve ? I was saved because i kept a track of the units , thanks again to you. But with this particular problem, the result i have obtained is largely varying with the answer given i.e 18.71 mA .
     
  18. WBahn

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    Would it. Imagine applying a battery across the terminals you are looking into (the terminals where the inductor use to be). Does an electron leaving one terminal of the battery have to flow through both resistors to get back to the battery? It seems to me that it has to flow through one or the other, but not both. Also, if you apply, say, 1V to those terminals, what is the votlage across each resistor? If they are in series, then you would have to have a voltage divider.

    Good. It only takes a few times of seeing that tracking units can save your bacon for it to become an engrained habit. Until then, it is hard to view it as little more than a needless burden (at least emotionally).

    Note that, in your equation, it is showing up as rads^-1. I'm sure you meant that to be rad*s^-1 and that this is purely a typesetting artifact. I just wanted to make sure you were aware of it so that you can play with it and figure out how to get it to display the way you want it to.

    Worse, you still aren't actually checking your units. Look at the argument of your inverse tangent function. Is that dimensionless? Your 'seconds' cancel out in the numerator, leaving you with 'ohms' in the denominator. Inductance has units of Henries. Using V = L di/dt, this means that 'Henries' has units of (V dt/di), or Vs/A or Ωs. Then look at the denominator. The units of the first term in the radical are 'ohms-squared' but the units of the second term are dimensionless because the 'seconds' cancel. So you are adding two terms that don't have the same units. Can't do that.
     
    Last edited: Jun 7, 2012
  19. nishu_r

    Thread Starter Member

    Jun 2, 2012
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    Sorry sir, but i do understand what unit i will have after tan inverse of ohm inverse, will the tan inverse produce degrees ? i think radians, but correct me. or values inside tan inverse function are just dimensionless?

    And sir where am i wrong? apart from this that denies me the answer?
     
  20. nishu_r

    Thread Starter Member

    Jun 2, 2012
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    But sir, looking into the dead network, how can we work our assumption, and i was meaning before we make the voltage source ,V_s=0(short circuit) and after we open circuit the inductor, then is it a series connection?


    Yes i did mean rad*s^-1, it is quite hard working with paint to bring the image of my work which is in a paper.

    Sorry sir, but i do understand what unit i will have after tan inverse of ohm inverse, will the tan inverse produce degrees ? i think radians, but correct me. or values inside tan inverse function are just dimensionless?

    And sir where am i wrong? apart from this that denies me the answer?
     
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