I realize these problems may be difficult. If you get stuck at any point, please ask questions with a comment here. Probably some other people are stuck on the same thing? For 3a), do you know what T refers to and how to calculate it? What equation do you use for w? You can ask very simple questions. For example: what is T?, how do we calculate w and tau??, how do you calculate the current, I(t)?, etc.
(Instead of an online homework assignment this week, the class chose to instead work on some written practice/homework problems. Here they are. Feel free and encouraged to post any questions or comments here.)
PS. Even when the questions do not ask for graphs, i would strongly encourage you to graph everything you can think of, e.g., I v t, Q v t, tau v R, T v C, etc. Familiarity with graphs and their meanings is highly valued here. (I'll add that in now.)
1. In a series LRC circuit --which starts out with a charged capacitor -- which of the following statements is true:
a) the current starts out from the capacitor, then flows through the inductor and then through the resistor.
b) Current can flow either clockwise or counterclockwise, and it can change as a function of time, but at any given moment the current is the same everywhere in the circuit so it is not useful or helpful to talk about where it goes first.
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2.
a) What is an Ohm in terms of Coulombs, Volts and seconds?
b) What is a Farad in terms of Coulombs, Volts and seconds?
c) What is a Henry in terms of Coulombs, Volts and seconds?
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3. Consider a series LRC circuit with L= 1 H, R= 1 Ohm and C= 10^-2/(2*pi)^2 F*. Suppose that at t=0 the current is zero and the charge on the capacitor is 0.002 Coulombs.
(* This was 10^-6/(2*pi)^2 F earlier this week...)
a) What is tau? What is T? What is tau/T?
b) Write an equation for the charge on the capacitor as a function of time. (graph it)
c) At what time is the charge on the capacitor zero? What is the current in the circuit at that time? (graph it (current))
d) xc. why do you think i made C much smaller than R and L?
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4. Referring to the same circuit and initial conditions as in problem #3:
a) at the time when the charge on the capacitor has reached zero, where is the energy in the circuit and how much (energy) is there?
b) has any energy been lost? If so where did it go?
c) xc. draw a picture, a graph, which illustrates the lost energy as the area under a "curve". Describe in words what your picture represents and means.
5. Referring to the same circuit and initial conditions as in problem #3:
a) at approximately what time, after t=0, does the current return to zero? (graph it)
b) at that time, what is the energy stored in the inductor and in the capacitor respectively. (graph the energies as a function of time.)
c) has much energy been lost? how much? How does that compare to your result for energy lost from 4b?
6. Referring to the same circuit and initial conditions as in problem #3:
a) what is the voltage across the capacitor at t =0?
b) what is the voltage across the inductor at t = 0?
c) graph both voltages as a function of time.
7. a) Graph tau vs R.
b) Graph tau vs L.
c) Graph w vs C
d) Graph T vs C
e) extreme extra credit: graph T vs R. what is the range and domain that makes sense to you and why?
When is this due?
ReplyDeletei would suggest starting to work on them very soon. Maybe try to finish most of it before thanksgiving.
ReplyDeleteSo are these problems like a "do it an turn them all in at your leaser" kind of thing? I noticed you said to submit only certain problems, should we submit them all for full credit? Also how are we going to be evaluated on these problems?
ReplyDeletethanks!!
Looks like I got way carried away with making C small. I think we would still be in the "small R" regime with C = 1/(2*pi)^2. The problems might be "more fun" with C = 1/[100*(2*pi)^2] or even C= 1/(2*pi)^2 (since then there is more energy lost).
ReplyDeleteI need help on 4b.
ReplyDeleteIn part 4a I answered that at the time when the charge on the capacitor has reached zero the energy in the circuit is is in the inductor and I figured out that there is 0.008J of energy there.
to get that for 4a you would use U=(1/2)L*I^2, yes.
ReplyDeleteand I = dQ/dt is approximately w*Qo*exp(-t/tau) at that time since sin(wt)=1, yes?
I think to get the energy lost you have to include the exp(-t/tau) term (otherwise nothing will be lost). To see how much is lost compare to how much was in the capacitor to start with.
I don't think i agree with your result for I. Does it include the w factor?
I mean your .008 J...
ReplyDeleteI haven't seen any questions here, so i am assuming no one is stuck on these problems....
ReplyDeleteWhen you calculate I(t), from dQ/dt, remember to just keep one term, not both. This is an approximation, but it is pretty accurate (you can test how accurate if you like), and it is what I will expect you to calculate I(t) on the final. Keeping both I(t) terms adds very little to the accuracy or conceptualization of the phenomena, however, it makes all calculations much more difficult and messy. Thus the cost of keeping the 2nd, smaller term outweighs the benefit, in my opinion.
ReplyDeleteApproximations like this one are common, indeed pervasive, in physics, which has been described as "The art of systematic oversimplification*."
*Karl Popper
One of the answers to 5b is:
ReplyDelete8 pi^2 10^-4 (1-1/20). If you can get to that, and understand how to get to that, i think you are doing pretty good.
The (1-1/20) comes from the exponential and uses:
e^{-x} = 1-x, approximately, for small x.
Similarly, for 4a) U= 8*pi^2 x 10^-4 (1-1/40)
and for 3c) the answer for I includes a (1/1/80) term...
that is 1-1/80
ReplyDeletefor 3C i''m stuck on trying to figure out Qo..
ReplyDeleteI am not sure what you mean by Qo? What are you trying to figure out?
ReplyDeleteps. did you graph Q as a function of time?
ReplyDeleteI'm getting a negative number for 3c I(t) calculated using the equation I(t)= -w*Qnaught*(e^(-t/tau))*sinwt. Based on my graph it should be a positive number. Is my graph upside down or am I miscalculating?
ReplyDeleteyour equation looks fine. so your graph should be of -sin(wt), right? (when) is that positive or negative??
ReplyDelete-sin(wt), not sin(wt)
ReplyDeleteFor 5A are we supposed to use I(t)=-Q0we^-(t/tau)sin(wt) using a different point when the sin graph for I reaches 0?
ReplyDeleteAlso, in the equation for energy in the capacitor U=(Q^2)/2C, is Q equal to Q(t) or Qnaught (the initial charge on the capacitor)?
ReplyDeleteKelly: yes, (-sin(wt))
ReplyDeleteDani: it depends. what do you think?
well at t=0 it would be Qnaught, but at any other time it should be Q(t)?
ReplyDeleteYou expanded the equation for energy in an inductor from U=(LI^2)/2 to U=(L/2)(w^2)(Qnaught^2)sin^2(wt). Is it possible to expand the energy in a capacitor equation (the one i mentioned above) to include the oscillatory function? Also, where does the e^t/tau term go?
ReplyDeleteNevermind, found the answer to those questions. But still not sure why the e^t/tau term is not included?
ReplyDelete-Q0we^-(t/tau)sin(wt) so then changing it to -sin(wt) instead the equation would just be Q0we^-(t/tau)sin(wt)? Then I set I(t)=0 and solve for t?
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ReplyDeleteFor 5b I got that UL=UC when Q=0 at 1/4 of the period of cosine? As in 8pi^2=8pi^2 when time is .001/4 ? Am I totally off base?
ReplyDeleteDani: the exp{-t/tau} term is included! It gets squared to become exp{-2t/tau}. Does that make more sense?
ReplyDeleteKelly: the current starts out at zero and going negative. it reaches a minimum value, which is negative, when -sin(wt)= -1, then it heads back up toward zero and becomes positive for a while after that.
ReplyDeleteStill don't see exp{-2t/tau} in the equation for U.
ReplyDeleteIs the voltage across the inductor at t=0 zero?
ReplyDeletevery good question. what do you think? why?
ReplyDeleteDani: the exp(-t/tau) is in the expression for current (and for charge). U is proportional to I^2. That is why it should be there.
ReplyDeletere: "Still don't see exp{-2t/tau} in the equation for U. "
PS. what do you mean by "see"
snon student: So I think I might be misunderstanding one of the questions. I was hoping you could clarify. On problem number 4 it asks in the LRC circuit if any energy has been lost and I calculated that and got that a small amount had been lost,
ReplyDelete** good. that makes sense to me.
but actually the overall energy –energy of inductor was 0 (but with the 2nd part of the chain rule being ignored), that there was probably some error.
**that's fine. ignoring it is a good choice.
Then for 5 you ask for how much energy has been lost again in the same circuit. I was just wondering if you are talking about the energy loss in the circuit as a whole?
**yes, the energy lost in the circuit as a whole is what i wss meaning to ask about. I see what you mean though. It is ambiguous.
I too am getting 0.008J for 4b. My answer doesn't make sense, however because i am getting a larger inductor energy than a capacitor energy. I know this is wrong because the energy needs to be decreased from the capacitor to the inductor b/c it is being released through the resistor.
ReplyDeleteI want to clarify how I am solving for I^2. Each term in the equation I(t) needs to be squared correct? so it will look like:
-w^2*Qnot^2*e^2t/tau
I am using this equation to find I^2 so that I can find the energy in the inductor. I am getting an answer of 0.00911 J and the initial energy in the capacitor that i calculated was 0.008 J.
Is there anyway you can point me in the right direction to what i am doing wrong?
I am concerned about your understanding of the exponential function. Try graphing exp{x} and e{-x} as a function of x from x=0 to 2 or so. What do they look like? How are they different?
ReplyDeletePS. There is a post called:
ReplyDelete"Functions and graphs that are critically important." I think your issue may come from not having a deep enough familiarity with that.
(it says, in part, "It will be essential for you to become familiar and comfortable with these graphs, and with the functions that create them. If you can achieve that familiarity, then you will probably do well in this course.")
ReplyDeleteHave the solutions for this hw been posted? More than checking if I got the right answer, I would like to verify that I have solved the problems using the correct methods.
ReplyDeleteMiss Sarah: i don't think you can get the right answers with wrong methods. Its inconceivable!
ReplyDeleteThe reason the solutions won't be posted is because then some (many?) people might just wait for them instead of doing the problems themselves.
Feel free to ask any questions here or tomorrow. Are you stuck on anything? How is it going with these?
For question seven. How do u graph these since they are just constants? Graph them as points? ( that's not your style ) R tau T C and I are constant. Please explain
ReplyDeleteI'm a bit confused by voltage across inductor in number 6b. I know that induced voltages are crated by magnetic fields that are induced by a change in current, but at t=0, there is no current therefore nothing to induce a voltage across the inductor?
ReplyDeleteI'm probably wrong haha, but just wondering.
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ReplyDeletefor part 4a my answer is -.061J ...is this correct?
ReplyDeleteI have U=LI^2/2 which for I i plugged in wQe^-2t/tau and w=20pi and t-.025. Is this all correct so far?
if so, for part 4b does that mean only .063J of energy was lost due to the resistor?
For 4a, I start with w=20*pi.
ReplyDeleteThen, using that, i get:
I=-w*.002 Coul/sec * sin(wt) epx{-t/tau}
= -.1256 sin(") * exp{"}
Then U =(1/2)L*I^2
=(1/2)(.0158)*exp{-2t/tau}
when sin(wt)=1.
exp{-2t/tau} ~ (1-2t/tau) is a very good approximation and one you should consider using because if you use a calculator you may miss out on a key thing.
"For question seven. How do u graph these since they are just constants? Graph them as points? ( that's not your style ) R tau T C and I are constant. Please explain "
ReplyDelete"Graph tau vs R" is asking you to express your understanding of how tau depends on R in a graph.
For an LR circuit, tau=L/R right? So you can graph that and it shows that tau-->infinity as R --> zero, and that tau goes to zero as R becomes very large. make sense?
(then there is also a very different graph for an LC circuit, for which tau=RC (it is linear...
For 4a I got 0.0019J and for part 4b for the energy lost i got 0.0059 J.... are these numbers correct?
ReplyDeleteBut then from Anya's answer you have to do Uc- U(inductor) to get the final energy loss from t=0 to when current=0 ? Is that correct?
ReplyDeleteok so then is my energy in the inductor .008J and therefore barely any energy has been lost in the circuit?
ReplyDeleteReally to get the energy lost it would be preferable to keep the exponential separate:
ReplyDeleteat t=0: U = .0079 Joules
at t=T/4: U = .0079*(1-2t/tau) Joules
so if 2t/tau = 1/40 =.025, then about 2.5% is lost.
(and so U is about .0077 Joules, i think...
The method is the important thing...
If we dont subtract Uc-Ul how do we compare it to the energy loss in question 5?
ReplyDeleteyou do subtract
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ReplyDeleteIs there supposed to be difference in the energy loss between when the charge is zero to when the current is zero. I keep recalculating and getting the same energy loss?
ReplyDelete4) .0079-.0077
5) .0079-.0077
Thank you
when the charge is zero comes at an earlier time, T/4.
ReplyDeleteWhen the current is zero happens later, T/2. There is less energy at the later time because the time, t, that goes in the exponential term,exp{-2t/tau}, is a bigger number so the exponential, which is approximately (1-2t/tau) is a little smaller. (it is maybe something like .96 instead of .98)
But in problem 4, at T/4 (assuming i'm correct) we get I^2=(20pi)^2(.002)^2(e^-(2*.025/2). The 2's in the exp expression cancel out so we end up with e^-(.025). When we plug our value in for I^2 into Ul=1/2LI^2 and solve we get .0077J. Then we solve for Uc=.0079 at time t=0 and so our difference from t=0 to t=T/4 is 2E-4 or 2.5%.
ReplyDeletefor question 5 at t=T/2, we use the equation Uc-Uc(e^-(t/tau) correct? if so we get (.0079J)-(.0079)*e^-(.05/2). Again, our value in the exp expression turns to e^-(.025) and so our final answer for Uc-Uc(e^-(t/tau) is 1.95E-4 which is practically 2E-4 (answer from question 4).
So either our answers are the same for 4 and 5, are wrong, or is the difference between question 4 and 5 (time T/4 and T/2) just that small?
i hope this all made sense.
for question 5 at t=T/2, we use the equation Uc-Uc(e^-(t/tau) correct?
ReplyDeletenot correct.
for question 5 at t=T/2, we use the equation Uc= (1/2C)Q^2 so you should get an (e^-(2t/tau). Do you see why?
So you are saying that we have a Uc-Uc (e^-2t/tau) ? Why would it be to the 2t, I think we need some help with that part
ReplyDeleteSo does that mean our equation turns into (1/2C)Qo^2e^-(2t/tau)cos(wt)?
ReplyDeletewhen would we use Uc-Uc(e^-(t/tau))?
it comes from Q^2.
ReplyDeleterederive it from Q^2/2C, where Q is a function of time. It is better to understand where it comes from.
when would we use Uc-Uc(e^-(t/tau))? "
ReplyDeletenever
so is (1/2C)Qo^2e^-(2t/tau)cos(wt) incorrect?
ReplyDeletehmm. you didn't square the cos(wt)
ReplyDeletewell cos^2(wt) would just become 1 anyway right?
ReplyDeleteYeah, exactly. Now you are being more assertive and confident. That is good!
ReplyDelete