Why is cap voltage negative in KVL for discharge circuit? I was just trying to derive the equation for a capacitor discharging through a resistor, and I've run into in a problem. 
If I set up my KVL, then I would say $iR = V_c$ (where $i$ is instantaneous current). Then you just replace $i$ with $dQ/dt$ and it's a separable DE. Easy. Except you get this:
$$
\frac{dQ}{dt}R=\frac{Q}{C}\\
\int_{t_0}^t{\frac{dt}{RC}}=\int_{Q_0}^Q{\frac{dQ}{Q}}\\
Q=Q_0e^{\frac{\Delta t}{RC}}\\
V_c=V_0e^{\frac{\Delta t}{RC}}
$$
This is the wrong sign for the exponent on $e$.
I looked it up, and in all other derivations, the original KVL is written as $-V_c-iR=0$.
But why should this be the case (besides the fact that it works)?
 A: This is a common question.   The issue is that the "Q" in $i = dQ/dt$ is not the same as the $Q$ that represents the charge on the capacitor.   The variable $Q$ in use here is simply the charge on the capacitor.  No problem.   When the capacitor discharges the quantity of charge that is introduced into the circuit after a time $\delta t$ has elapsed  is $$\delta q = Q_0-Q(\delta t)$$ a positive quantity.
However, the charge on the capacitor has decreased:  $$\delta Q = Q_f-Q_i = Q(\delta t) - Q_0 = -\delta q$$
That is:  the current increases when the charge on the capacitor decreases.  One book I saw recently gets it right at first, and then muddles it later.  Watch out.
A: I recently answered a similar question here.
The ideal capacitor equation
$$i_C = C\frac{dv_C}{dt}$$
assumes the passive sign convention which means that the reference direction for $i_C$ is into the positive labelled terminal.
When you write
$$iR = v_C$$
it is necessarily the case that
$$i_C = - i$$

To see this, assume that both positive labelled terminals are connected together (so that both negative labelled terminals are connected together).
Now, by the passive sign convention, the capacitor current $i_C$ is into the positive labelled terminal of the capacitor while the resistor current $i_R$ is into the positive labelled terminal of the resistor.
Since only the resistor and capacitor are connected there, it follows from KCL that
$$i_C + i_R = 0 \rightarrow i_C = - i_R$$
Since you've chosen $i$ to be in the direction of $i_R$, it follows that your second equation should be
$$\frac{dQ}{dt} = -i$$
which leads to the correct differential equation
$$\frac{dQ}{dt} + \frac{1}{RC}Q = 0 $$

The quick summary is this:  


*

*when the current $i$ in your first equation is positive, the charge $Q$ is
decreasing, i.e., $\frac{dQ}{dt}$ is negative.

