Picking a direction for currents when doing nodal analysis involving capacitors

Question

Let's assume that I was given the following circuit:

Where the capacitor is fully charged at 6V, and a switch (not shown) will be closed at time t=0. From this, let's say that I pick the following directions of my currents in order to do nodal analysis (assume ground is on the middle bottom node):

My simplified KCL equations would end up being:

$I_{capacitor}(t) = -\frac{1}{4k} V_c(t)$

Since the capacitor is discharging, the equation becomes:

$ -C \frac{dV_c(t)}{dt} = -\frac{1}{4k} V_c(t) $

Solving this differential equation yields:

$ V_c(t) = 6e^{2.5t} $

This is clearly wrong, as this solution diverges.

However, if I reverse the directions of the resistor currents to this:

Then my simplified DE equation becomes:

$ -C \frac{dV_c(t)}{dt} = \frac{1}{4k} V_c(t) $

Which results in the solution:

$ V_c(t) = 6e^{-2.5t} $

Which is the correct solution.

So how do I know which direction to make my currents before I even solve the differential equation?

Answer 1

You don’t change the sign because the capacitor is discharging. The fundamental equation presupposes the current flowing into the capacitor (as you have it in your drawing). The negative sign will take care of itself in the solution (discharging).

$$ C \frac{dV_c(t)}{dt} = -\frac{1}{4k} V_c(t) $$

The fundamental capacitor equation being, $$i(t)=C\frac{dv}{dt}$$

If you want to define your current as leaving the capacitor then you need to include a negative sign in your dv/dt term.