# ODE RC Circuit Problem

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1. Sep 15, 2015

### keyermoond

1. The problem statement, all variables and given/known data
Suppose an RC series circuit has a variable resistor. If the resistance at time t is given by by R = a + bt, where a and b are known positive constants then the charge q(t) on the capacitor satisfies

(a+bt) q' + (1/C)q = V

where V is some constant. Also q(0) = q_0
Find q(t) as an explicit function of t.

2. Relevant equations

Now I have obtained the answer, however my main question is: am I allowed to treat C (capacitance) as a constant in this equation. It doesn't specify in the question, but to my knowledge (unless I am wrong of course) capacitance is a constant value and only depends on material and physical parameters of the capacitor itself (how it is build).

If I can't treat C as a constant then I believe there is no way to evaluate the integral in integrating factor and I'd have to leave it as it is.

Process is simple from there, I rewrite the equation in standart form, find the integrating factor and obtain a formula for q(t), evaluate an integration constant with q(0) = q_0 and obtain the overall solution q(t).

The answer looks quite frightening btw

3. My solution:
attached pdf file

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2. Sep 15, 2015

### tommyxu3

The capacitance must be constant during the all process, which just depends on its material and geometric shape, as what you say above,

3. Sep 16, 2015

### vela

Staff Emeritus
Is there some reason you're not simplifying $\frac{(a+bt)^k}{a+bt}$ to $(a+bt)^{k-1}$ and integrating the righthand side? Also, the $t$ in the integral is a dummy variable, so it's not correct to set it to 0.

4. Sep 17, 2015

### keyermoond

That's what my prof does. By setting it from 0 to t, we ensure evaluated integral is equal to the true value with t as a variable, how would setting it to something arbitary as t_o (I'm assuming that's what you mean)be any better? My understanding of it is that we need to choose some "convenient" interval, what's wrong with 0 to t?

And thank you for pointing out I can actually simplify it further, I completely missed that.

5. Sep 17, 2015

### vela

Staff Emeritus
What I'm saying is you can't do something like this:
$$\int (a+bt)^2\,dt = \int a\,dt$$ by claiming you're setting $t=0$. The $t$ inside the integral isn't the same $t$ that appears elsewhere.

6. Sep 17, 2015

### keyermoond

I see what you mean, my mistake, thank you for pointing it out. I see holes in my knowledge about understanding of basic integration now, will have to fill them in :)