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.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.
Spot on.Q1. v_C (0-) = v_C (0+) = v_C (0) = 2V (using voltage division for the voltage across the 100 kiloOhm resistor)
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.Q2. i_L (0-) = i_L (0+) = i_L (0) = -10 mA ( i have a doubt here , am i wrong? )
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.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.
Yes.Q5. voltage across v_R prior to t=0 is -5V. (across the 500 ohm resistor)
Here is your first major conceptual error. So consider the following next questions:Q6. i_R = 0 A? just after the switch turns right.
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.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.
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.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?
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).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 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.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.
And you missed it again. B_1 does NOT equal 2. It equals 2V.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.
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.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.
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.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.
Good. Trust me, it will pay for itself many times over.You are right, I should be careful with the units.
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.combining the circuits the result would be either 10ms or 100s(or is it ms?)
Not sure what you mean by the differences only being "mathematical". The differences are real and can be significant.And i checked various instances of natural freq and resonant frequency and yes the differences between them is only mathematical.
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.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
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.
Yes Sir, but if we were to attach a forcing function to the dead network with the open circuit across the inductor, the electron has to flow through both , so doesn't it make it series? I did come across an instance of two resistors neither parallel nor series with each other, but that was in a larger circuit and was solveable.
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.
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.Yes Sir, but if we were to attach a forcing function to the dead network with the open circuit across the inductor, the electron has to flow through both , so doesn't it make it series?
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).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.
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?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.
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?
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.
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.
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.
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