Max values of current and charge using differential equations

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
Sign up Log in
13 replies · 2K views
Arman777
Insights Author
Gold Member
Messages
2,163
Reaction score
191

Homework Statement


upload_2017-5-21_14-32-29.png


Homework Equations


Circuit Equations.
##U_C=Q^2\2C##
##U_L=Li^2\2##

The Attempt at a Solution


For (a) I said ##100J## .But I think it might be ##200J## too.Here what I did;
##U_t=Q^2\2C## and I put ##Q=0.1C## and we know ##C##.Here I find ##100J##.But I think I should add also the energy of inductor (and when I do that total energy becomes ##200J##)
If I choose the later parts of the question when I try to find max values of current and charge using differantial equations.I get wrong results.I am really stucked.
Thanks
 
Physics news on Phys.org
kuruman said:
You are correct in saying that the total energy is 200 J. Both capacitor and inductor have energy stored in them at t = 0.
For the max values, you have to show us first what you did and then perhaps we can get you unstuck.
When capacitor has no charge currrent wlll have a max value so and all that value will gives the energy since its conserved, ##(i_{max})^2=200J.2/L## from there its ##i=2√2## and max charge is (from same logic , ##q=\sqrt {2}. 10^{-1}##)
 
kuruman said:
Your answers are what I got. So where did you get stuck? Is there more to this problem that you did not include in the original statement?
Yeah now here comes that part.
upload_2017-5-21_15-6-48.png

now part d is ##i(t)=dq_{upper}\dt## which its obvious.
part e ##i(t)=-dq_{lower}\dt##
part f ##w=\frac {1} {\sqrt{LC}}=20rad\s## and from there I can find easily f
now part g
İts a second order dif.equation and I thought solution should look like,
##Q(t)=Acos(wt)+Bsin(wt)##
and ##I(t)=Bwcos(wt)-Awsin(wt)##
we know that at ##t=0##, ##Q=0,1C##.and ##I=2A##
so when I put values I get
##Q(t)=0.1cos(20t)+0.01sin(20t)##
##I(t)=0.2cos(20t)-2sin(20t)##

but If you look the max values I couldn't get the previous results (I used desmos to graph the functions and it didnt give me the max values )
 
Arman777 said:
... so when I put values I get ...
You put in the values incorrectly. Start with the expression that gives you the charge on the capacitor at any time t, ##Q(t)=A \cos(\omega t) + B \sin(\omega t)##. You know that at time t = 0, ##Q_{upper} = 0.1 C##. This gives you ##Q(0)=A = 0.1.## That much you got right. What about B? Redo the algebra.

On edit: Also the current at t = 0 is -2A, not 2A.
 
kuruman said:
You put in the values incorrectly. Start with the expression that gives you the charge on the capacitor at any time t, ##Q(t)=A \cos(\omega t) + B \sin(\omega t)##. You know that at time t = 0, ##Q_{upper} = 0.1 C##. This gives you ##Q(0)=A = 0.1.## That much you got right. What about B? Redo the algebra.

On edit: Also the current at t = 0 is -2A, not 2A.
I choose the direction as the opposite sign.So in the question -2A means in the opposite direction 2A for my choosen direction ?

Edit:B=0.1

Ok and make sense now.
 
I understand the all parts thanks a lot.I have another question too If you can help me also on that I ll be very happy
 
The arrow in figure shows that the capacitor is discharging because positive charge is moving away from the positively charged upper plate. You have defined the current as the time derivative of ##q_{upper}##. Therefore the time derivative must be negative, ##\frac{dq_{upper}}{dt} = -2A##.

Did you sort out the value of B?
 
kuruman said:
The arrow in figure shows that the capacitor is discharging because positive charge is moving away from the positively charged upper plate. You have defined the current as the time derivative of ##q_{upper}##. Therefore the time derivative must be negative, ##\frac{dq_{upper}}{dt} = -2A##.

Did you sort out the value of B?
wait..I thought we can also say 2A...
 
kuruman said:
Sorry, I have to logout now. Perhaps someone else will pick it up.
Ok,thanks :)