# RC circuit, expression for voltage

1. Oct 6, 2014

### nossren

1. The problem statement, all variables and given/known data
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.

2. Relevant 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 $I = \frac{dQ}{dt}$

3. 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.

2. Oct 6, 2014

### zoki85

VC(t) + VR1(t) + VR2(t) = 0 is the equation

3. Oct 6, 2014

### 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 - ?

4. Oct 6, 2014

### zoki85

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

5. Oct 6, 2014

### Staff: Mentor

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?