# Max values of current and charge using differential equations

1. May 21, 2017

### Arman777

1. The problem statement, all variables and given/known data

2. Relevant equations
Circuit Equations.
$U_C=Q^2\2C$
$U_L=Li^2\2$
3. 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

2. May 21, 2017

### kuruman

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.

3. May 21, 2017

### Arman777

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 , $q=\sqrt {2}. 10^{-1}$)

4. May 21, 2017

### kuruman

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?

5. May 21, 2017

### Arman777

Yeah now here comes that part.

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 couldnt get the previous results (I used desmos to graph the functions and it didnt give me the max values )

6. May 21, 2017

### kuruman

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.

7. May 21, 2017

### Arman777

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.

8. May 21, 2017

### Arman777

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

9. May 21, 2017

### kuruman

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?

10. May 21, 2017

### Arman777

wait..I thought we can also say 2A...

11. May 21, 2017

### kuruman

If the other question is not related to this particular problem, you need to post it in a separate thread.

12. May 21, 2017

### Arman777

13. May 21, 2017

### kuruman

Sorry, I have to logout now. Perhaps someone else will pick it up.

14. May 21, 2017

Ok,thanks :)