# LCR Circuit with Capacitor as diminishing power source

1. Sep 18, 2006

### FunkyDwarf

Hey guys

First off id like to say this is an awesome forum from what ive seen, and has since been bookmarked at the top of my list as i am an avid physics enthusiast (but **** at spelling :) )

Ok, to the nitty gritty. I have the following circuit (attached)

Now, there are two questions. Firstly to find the differential equation satisfied by Q(t) (where the charge for t=0 in the cap is Qmax).

Now i managed to get the following which i think is right (sorry i dunno how to do your cool maths type stuff)

Q/C -(dQ/dt)(R) - (L)(2nd derivative of q wrt t) = 0 (Kirchoff's loop law)

Now unfortunately i was sick the week we did DEs in maths :P so your gonna have to go slow. Im currently reviewing the notes from that part but the guys handwriting is shocking so its probably easier to just listen to you guys.

My first question is obviously is it right and is that the neatest most simplified form i can get it into?

Next question is i need to show that Q(t)=Qmax(e^(-at))(Cos(wt)) satisfies this differential equation. Now im assuming im not meant to simply take the required derivatives and sub it into the first eqn and see if it works, because that would be an algerbraic nightmare and ive seen glimpses of some nifty DE tricks so please, regail me!

Cheers guys!
-G

EDIT: Sorry guys, didnt see the no homework thing. feel free to move n stuff

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Last edited: Sep 18, 2006
2. Sep 18, 2006

### dextercioby

The way you said it ("show that Q(t)=Qmax(e^(-at))(Cos(wt)) satisfies this differential equation") implies just that: "simply take the required derivatives and sub it into the first eqn and see if it works".

Daniel.

P.S. You might (learn to) use the Latex compiler to cast your formulae into nicer form.

3. Sep 18, 2006

### FunkyDwarf

You sure? im just saying this because usually we dont get questions requiring long but simple computations and number crunching which essentially that would be.

is there not some way to use the first equation and do some crazy DE stuff (hey cmon i missed the lectures :P) to get the exponential form? I mean Q does follow an exponential relationship i know that, just how to get it i dont get.

4. Sep 18, 2006

### dextercioby

Yes, you can also do that, but it's a lot easier if you just plugged the function and its 2 derivatives into the ODE and from requiring that LHS=0 get the value of a and w in terms of R,L,C.

Daniel.

5. Sep 18, 2006

### FunkyDwarf

Well in the interest of learning ODE's could you please show me how to do the harder bottom up version? Also i need to find a and w so im assuming R,L and C will bugger off somehow :S

6. Sep 18, 2006

### CRGreathouse

$$\frac{Q}{C} -\frac{dQ}{dt}(R) - \frac{d^2q}{dt^2}(L) = 0$$ (Kirchoff's loop law)

I got this by typing:
[ tex ]\frac{Q}{C} -\frac{dQ}{dt}(R) - \frac{d^2q}{dt^2}(L) = 0[ /tex ]
without the spaces between the brackets [].

7. Sep 19, 2006

### FunkyDwarf

k, but that doesnt rly solve the problem :P

EDIT: Ok i went and talked to the maths lecturer today about the stuff i missed and i more or less understand ODEs now. As for this one, im still not sure. It appears that it would be from an ODE which has complex roots (in order for the trig functions to be there) however that usually includes a sine value too, so he suggested perhaps it was dropped to ignore the phase shift, which to me didnt make any sense as phase most certainly matters in a circuit like this.

Also, i find it hard to conceptulaise a method of solving this given it contains so many arbitary constants. I mean to solve for a and omega you need two equations or two starting conditions, we only have 1, or am i missing one?

Last edited: Sep 19, 2006
8. Sep 19, 2006

### Mute

The omega in Q(t) =Qmax(e^(-at))(Cos(wt)) is not an arbitrary constant, and neither is the "a" in the exponent there. Those two constants come about when solving the equation.

You will, however, have two arbitrary constants that you need to pin down with the initial conditions. Here, Q(0) = Qmax, as initially the switch is open and all of the charge is on the capacitor, correct? Similarly, Q'(0) = 0, as there can be no current flow when the switch is open.

As for why no term involving a sine appears, the second part of your question simply asks you to show that Q(t) =Qmax(e^(-at))(Cos(wt)) is a solution to the given equation, and it is. The trick is that a linear sum of solutions to a differential equation is also a solution, and the general solution to the DE, the one which satisfies the initial conditions, is such a sum, but each of its terms individually will also satisfy the DE (just not the initial conditions).

Of course, in this case, the arbitrary constant in front of e^(-at)sin(wt) is just going to evaluate to zero due to the initial conditions I mentioned above.

9. Sep 19, 2006

### FunkyDwarf

Ok you pretty much lost me from The trick is onwards. are you saying that because of the intial condtions the sine drops out?

Im planning to have a crack at this later doing a full backwards working from DE up assuming it has complex roots (which i think is a fair assumption due to the nature of capactiance and resistance)

10. Sep 21, 2006

### Mute

The general solution to the DE will be of the form

Q(t) = e^(-at)*(Acos(wt) + Bsin(wt))

and you have initial conditions Q(0) = Qmax and Q'(0) = 0. If you differentiate Q(t) and plug t = 0 into Q(t) and Q'(t), you'll get a system of equations to solve for A and B. You'll find that with these initial conditions, B = 0. That's why there is no sine term.

The "trick" business was just me saying that you'll find that e^(-at)cos(wt) will satisfy the DE on its own, as will e^(-at)sin(wt), but only a linear sum of the two will, in general, be able to satisfy the initial conditions as well (which is why we solve for A and B).

11. Sep 21, 2006

### FunkyDwarf

ah ok. yeh i tried plugging in the intial conditions earlier and it didnt work, ill give it another go

thanks guys!