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 Post subject: RC Circuit
PostPosted: Thu, 31 Jul 2003 12:59:40 UTC 
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I'll just throw this out here..see if i can handle the latex too:

Governing ODE of a RC circuit is:

R\frac{dI}{dT} + \frac{I}{C} = \frac{dV(t)}{dt}

we have conditions V(t) = V_{0}t^2 for 0 <= t <= 1
and v(t) = V_{0} for t > 1

Suppose I should check my general solution first of all...pretty sure it is

I(t) = 2V_{0}C(t-RC) + C_{1}\exp(\frac{-t}{RC}}) for 0 <= t <= 1

and

I(t) = C + C_{0}exp(\frac{-t}{RC}) for the other part

but im confident on those.

Questions start when we have inital conditions
I(t=0) = 0

so solve for the constant..not too hard.

Then i presume I(1) has to be the same for both eqs...so i can get the other constant..its just really ugly.

Then it asks for the maximum current and the time it occurs.

My first thoughts were that it would occur in the region 0 <= t <= 1, since the t > 1 decreases as t increases. but then i got thinking, if we chose appropriate constants couldn't we get the constant negative..and therefore the graph would increase exponetially after t = 1?

Assuming it doesnt, then we are left to decipher the first equation. Calculus is no help, since dI/dT = 0, when t = 0, and is a minimum.

Thus I am left to conclude that the max current occurs at t = 1, and is I(1)...whatever ugly equation that turns out to be. How am I going?


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PostPosted: Fri, 1 Aug 2003 01:41:09 UTC 
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I don't think you can assume that the currents given by each equation have the same value at t = 1. (For example, there could be a jump at t = 1 - the graphs would be discontinuous). At a glance, it seems to me that you need more info to be given in the question ....


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 Post subject:
PostPosted: Fri, 1 Aug 2003 02:35:25 UTC 
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yeah i was kinda figuring that would be the only way to extract two constants from 1 piece of info. maybe it just wants the solution for 0 <= t <= 1

However what no-one did notice was that I am stupid

Since when is the derivative of a constant (v[0]) 1 instead of 0..so the 2nd equation is just seperable.

ahh well..thoughts still hold for the 2nd question :oops:


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PostPosted: Fri, 1 Aug 2003 06:58:46 UTC 
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I checked your first solution (the harder DE) and it was fine. So I didn't even bother to check your second solution (the [much] easier DE)!! Just goes to show ...


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