Time dependent Resistance in an RC-circuit

[gonzalesdp]

Homework Statement

You have an RC-circuit with emf E, resistance R, and Capacitance C. However, R = R(t) = R_o(t/tao+1)^-1, where R_o is the intitial resistance. Assume that other than the time-dependent resistance, the circuit behaves normally.

Homework Equations

Show that during discharge the charfe at time t is given by

q(t) = q_oe^{-t(t/2tao+1)/R_oC} where q_o is the initial charge on the capacitor.

The Attempt at a Solution

I know to find q(t) of a simple RC-circuit you need to use basic differential equations. (a course which I have not taken yet). I've tried to simply substitue R(t) for R in the equation

q(t) = q_oe^(-S dt/RC)
(the "S" is the great s for integration, I'm not sure how it should be notated.)

This isn't working though.

 

[SammyS]

Kirchhoff's Laws give: ℇ ‒ IR ‒ q/C = 0.

I = dq/dt .

Not quite sure about your R.

Is it: R(t)=R_0\left(\frac{t}{\tau}\ +1\right)^{-1}\ ?

 

[ehild]

I know to find q(t) of a simple RC-circuit you need to use basic differential equations. (a course which I have not taken yet). I've tried to simply substitue R(t) for R in the equation

q(t) = q_oe^(-S dt/RC)
(the "S" is the great s for integration, I'm not sure how it should be notated.)

If you did not learn how to solve a differential equation, do the opposite thing: show that the given q(t) function yields the same voltage across the capacitor (Q/C) and the resistor (IR). I hope you have learned to differentiate and you know that I=-dQ/dt in case of discharge. (The magnitude of charge removed from the capacitor in unit time (dQ/dt) is equal to the charge flowing through the loop in unit time (that is the current I). ehild

 

[gonzalesdp]

Thanks Sammy. I'll work on it a little more with your suggestions.

Thanks ehild. I don't know why but when I worked it out on my paper, I didn't realize the t/tao. I keep using t/(tao + 1). Such a dumb move on my part. Thanks again.

 

[gonzalesdp]

Yup it worked out. It's amazing that I worked on that problem for hours without realizing I wrote it down wrong to begin with. Thanks SammyS and ehild.

 

[gonzalesdp]

I'm still going to try it your way SammyS. I'd like to see how it unfolds that way. It's always good to know another way to solve a problem.