Finding R(t) in discharging RC circuit

[ddobre]

Homework Statement

A charged capacitor with capacitance C is being discharged through a variable resistor that has its resistance dependent on time: R = R(t). Find function R(t) if the current through the resistor remains constant until the capacitor is completely discharged and the resistance at the initial moment of the discharge process (t = 0) is equal to R₀

Homework Equations

(1) I = (Q₀/RC)e^-t/RC
(2,3) Q₀=Cε, Q = Cεe^-t/RC
t = RC
IR = Q/C

The Attempt at a Solution

Since I know I is contant, and at t = 0, R=R₀, I tried to use equation (1) for R and at t = 0, when R = R₀, so that I could equate the two equations and try to solve for R. This is what I started with:
Q₀/R₀C₀ = (Q/RC)e^-t/RC
I ended up with something like:
R = (QR₀C₀/Q₀C)e^-t/RC
But I was a little confused because there is still an R in the e expression. So I tried taking the natural log of each side, but what I ended up with didn't seem feasible. Any advice on what I should try to do?

 

[haruspex]

Isn't your eqn (1) a solution to a differential equation which assumes R is constant (the solution, not the equation)?

 

[ddobre]

Isn't your eqn (1) a solution to a differential equation which assumes R is constant (the solution, not the equation)?

I think so. But I was just trying to define R in some way. I'm having trouble trying to find an equation for R(t)

 

[haruspex]

I think so. But I was just trying to define R in some way. I'm having trouble trying to find an equation for R(t)

See the first equation under https://en.m.wikipedia.org/wiki/Capacitor#DC_circuits
It is an integral equation, and it is obviously true. The equation just below it is obtained by differentiating it, but on the assumption that R is constant, so that second equation does not apply here.
Instead, you have that the current is constant.

 

[ehild]

When a capacitor C charged with Q is connected to a resistor R, current I will flow, and the capacitor voltage is Uc-RI=0. The capacitor voltage is Uc=Q/C and the current is defined as I=dQ/dt. The current is flowing off the capacitor now, so it decreases the charge on it.If it is constant, I =-I₀, how does the charge change with time during the discharge?