RC circuit, expression for voltage

[nossren]

Homework Statement

Assume you have a fully charged capacitor with initial condition V(0) = V_0 connected in series to two resistors R_1 and R_2. Derive an expression for the voltage over the capacitor with respect to time.

Homework Equations

1. Kirchhoff's voltage law \sum_n V_n = 0
2. Ohm's law: V=RI
3. Capacitance of capacitor: C = \frac{Q}{V}
4. Current I = \frac{dQ}{dt}

The Attempt at a Solution

If I "go" in the direction of the current, then the potential difference over both resistors are negative whereas it is positive over the capacitor, hence
V(t) - V_{R_{1}} - V_{R_{2}} = 0 = V(t) - R_1 I - R_2 I.
Which I can rewrite using the time derivative of Q (3 and 4 from above)
= V(t) - (R_1 + R_2)\frac{dQ}{dt} = V(t) - (R_1 + R_2)C\frac{dV(t)}{dt} = 0 \implies \frac{1}{(R_1 + R_2)C}V(t) - \frac{dV(t)}{dt} = 0.
By solving this differential equation, I get
V(t) = V_0e^\frac{t}{(R_1+R_2)C}
which is wrong, because the voltage should decrease over time, so a more logical solution would be
V(t) = V_0e^{-\frac{t}{(R_1+R_2)C}}.
Is there something I have overlooked? Is Kirchhoff's law setup incorrectly (sign conventions)?

Any help is appreciated.

 

[zoki85]

V_C(t) + V_R1(t) + V_R2(t) = 0 is the equation

 

[nossren]

Yes, I realize that has to be true for it to be correct, but I am just confused about the signs. My book (Sears and Zemansky's) mentions some sign conventions in relation to Kirchhoff's voltage law (see picture).

So why is it + rather than - ?

 

[zoki85]

You have 3 passive elements in the circuit and no driving emf. Don't know what confuses you.

 

[gneill]

nossren, have you considered what sign should be associated with dV/dt on the capacitor? Will V be increasing or decreasing given the direction of the current flow?